ANALYSIS OF EVAPORATIVE COOLERS AND CONDENSERS

by

Andre Alexis Dreyer

Thesis presented in partial fulfilment of the

requirements for the degree of Master of Engineering
at the University of Stellenbosch.

Thesis Supervisor: Prof. P.J. Erens

Department of Mechanical Engineering

University of Stellenbosch

November 1988
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Declaration

I, the undersigned, hereby declare that the work contained in this thesis
is my own original work and has not previously, in its entirety or in part,
been submitted at any university for a degree.

(Signature of candidate)

............... day of 1988.

j
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ABSTRACT

In this report various mathematical models for the thermal evaluation of

evaporative coolers and condensers are presented. These models range from
the exact model based on the work by Poppe [84P01] to the simplified
logarithmic models based on the work of McAdams [54Mcl] and Mizushina
et al. [67MI1], [68MI1].

Various computer programs were written to perform rating and selection

calculations on cross-flow and counterflow evaporative coolers and
condensers.

Experimental tests were conducted on a cross-flow evaporative cooler to

determine the governing heat and mass transfer coefficients. The
experimentally determined coefficients were cqrrelated and these
correlations are compared to the existing correlations. The two-phase
pressure drop across the tube bundle was also measured and a correlation
for two-phase pressure drop across a tube bundle is presented.
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ACKNOWLEDGEMENTS

The work described in this thesis forms part of an on-going research

program on wet and dry cooling in the Department of Mechanical Engineering
at the University of Stellenbosch.

I would like to acknowledge the valued contributions of each of the

following persons:

Prof. P.J. Erens for his guidance, support and assistance

throughout the project;

Prof. D.G. Kroger for his suggestions, patience and interest in the
project;

Mr.· D.C. Uys for the construction of the test section and practical
assistance with the experimental work;

Miss. E. Lekas for the typing of the thesis.

I would also like to thank the National Energy Council for their finanfcial
support.
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CONTENTS
Page
ABSTRACT ( i)
ACKNOWLEDGEMENTS (ii)
CONTENTS (iii)
NOMENCLATURE (v)
1 - INTRODUCTION 1.1
2 - LITERATURE SURVEY 2.1
3 - MATHEMATICAL MODELLING OF EVAPORATIVE COOLERS AND CONDENSERS 3.1
3.1) Basic theory for evaporative coolers 3.2
3.1.1) Exact analysis (Poppe model) 3.2
3.1.2) Merkel analysis 3.10
3.1.3) Improved Merkel analysis 3.11
3.1.4) Simplified model 3.12
3.2) Basic theory for evaporative condensers 3.15
3.2.1) Exact analysis (Poppe model) 3.15
3.2.2) Merkel analysis 3.18
3.2.3) Improved Merkel analysis 3.19
3.2.4) Simplified model 3.19
4 - HEAT/MASS TRANSFER AND PRESSURE DROP CORRELATIONS 4.1
4.1) Film heat transfer coefficient 4.1
4.2) Mass transfer coefficient 4.9
4.3) Pressure drop correlations 4.21
5 - COMPUTER SIMULATION 5.1
5.1) Determination of coefficients 5.2
5.2) Successive calculation models 5.4
5.2.1) Cross-flow evaporative cooler simulation 5.7
5.2.2) Counterflow evaporative cooler simulation 5.16
5.2.3) Combination cooler 5.17
5.3) Simplified models 5.25
5.4) Natural draft cooling tower 5.26
6 - EXPERIMENTAL DETERMINATION OF THE HEAT AND MASS TRANSFER 6.1
COEFFICIENTS IN A CROSS-FLOW EVAPORATIVE COOLER
6.1) Description of wind-tunnel and apparatus 6.1
6.2) Data logging and energy balance calculations 6.12
6.3) Experimental procedure 6.13
6.4) Observations and results 6.16
6.5) Determination of coefficients and correlations 6.29
6.6) Discussion of results 6.43
7 - CONCLUSION 7.1
REFERENCES R.1
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- APPENDICES
A - THERMOPHYSICAL PROPERTIES OF FLUIDS A.l
B - DEFINITION OF LEWIS NUMBER AND THE LEWIS FACTOR B.l
C - DEFINITION OF MASS TRANSFER COEFFICIENTS AND MASS TRANSFER C.l
POTENTIALS
D - SINGLE PHASE PRESSURE DROP CORRELATIONS ACROSS PLAIN TUBE D.l
BUNDLES IN CROSS-FLOW
E - DERIVATION OF THE DRAFT EQUATION FOR A NATURAL-DRAFT E.l
CROSS-FLOW EVAPORATIVE COOLING TOWER
F - SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS USING THE F.l
4TH ORDER RUNGE-KUTTA METHOD
G - CORRELATIONS FOR CONVECTIVE AND CONDENSATION HEAT TRANSFER G.l
COEFFICIENTS ON THE INSIDE OF TUBES
H - DETERMINATION OF THE AIR/WATER INTERFACE TEMPERATURE H.l
I - CORRECTION OF HEAT TRANSFER COEFFICIENT AT HIGH MASS I.l
TRANSFER RATES
J - EVALUATION OF CONVENTIONAL COOLING TOWER PACKING IN A J.l
COMBINATION EVAPORATIVE COOLER
K - RESULTS OF COMPUTER SIMULATIONS K.l
L - FORTRAN CODE FOR CROSS-FLOW EVAPORATIVE COOLER SIMULATION L.l
PROGRAM
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NOMENCLATURE
A Area, [m2]
a Effective suface area of tubes per unit volume, [m2;m3]
B Constant defined in section 4.3, [-]
b Slope of the air saturation enthalpy curve, [J/kgK]
c Concentration, [kg/m3]
q 'C2' .. Coefficients, [-]
Cp Specific heat at constant pressure, [J/kgK]
D Diffus~ity, [m2/s]
d Diameter (characteristic length), [m]
E Coefficient defined by equation 2.5, [-]
EP Dimensionless enthalpy potential defined by equation 4.2.21, [-]
F LMTD correction factor or force, [-] or [N]
f Friction factor, [-]
fa Arrangement factor (Appendix D), [-]
fn Correction factor for small number of tube rows (Appendix D), [-]
fz Correction factor for non-isothermal flow (Appendix D), [-]
fzn Correction factor for non-isothermal flow and small number
of tube rows (Appendix D), f-]
G Mass velocity, [kg/m2s]
g Gravitational acceleration, [m/s2]
h Heat transfer coefficient, [W/m2K]
ho Mass transfer coefficient based on Tw, [kg/m2s]
hoi Mass transfer coefficient based on Ti, [kg/m2s]
hoo Overall mass transfer coefficient, [kg/m2s]
hop Mass transfer.coefficient based ·on partial pressure
driving potential and Tw, [s/m]
hopi Mass transfer coefficient based on partial pressure
driving potential and Ti, [s/m]
i .Enthalpy, [J/kg]
~fg Latent heat of evaporation, [J/kg]
~fg
Corrected latent heat of evaporation, [J/kg]
lvo Latent heat of evaporation at ooc, [J/kg]
Ai "Entalpy potential" defined by equation 4.2.21, [J/kg]
Kwb "Wet bulb K" defined by equation 2.1, [-]
K Loss coefficient, [-]
Kg Coefficient defined by equation C.4, [s/m]
k Thermal conductivity, [W/mK]
kg Coefficient defined by equation C.S, [s/m]
kl Coefficient defined by equation C.6, [m/s]
L Length, [m]
LMED Log mean enthalpy difference, [J/kg]
LMTD Log mean temperature difference, [ C]
LVF Liquid void fraction, [-]
m Massflow rate, [kg/s]
N Constant defined by equation 2.6, [-]
n Number [-]
NTU Number of transfer units, [-]
p Pitch, [m]
p Pressure, [N/m2]
Ap Pressure drop, [N/m2]
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l\p* Pressure drop based on massflow rate of both phases

and the properties of one of the phases, [N/m2]
q Heat transfer rate, [W]
qll Heat flux, [W/m2]
R Universal gas constant, [J/kgK]
Re* Reynolds number based on the massflow rate of both phases
and the properties of one of the phases, [-]
RR Ratio defined by equation 4.2.19, [-]
Ry Characteristic flow parameter, [m-1]
r Ratio defined by equation B.21, [-]
T Temperature, [OC]
t Thickness, ~m]
u Overall heat transfer coefficient, [W/m K]
v Velocity, [m/s]
w Width, [m]
w Humidity ratio, [kg water/kg dry air]
X Mole fraction, [-]
X Vapour quality, [-]
y Ratio defined by equation 4.3.10, [-]
z· Height, thickness, [m]
a Thermal diffusivity, k/pc~, · [m2/s]
r Recirculating water massf ow rate per side per unit
length of tube, [kg/ms]
fG Ratio defined by equation 4.·3.14, [-]
0 Film thickness, [m]
oc Concentration boundary layer thickness, [m]
om Momentum boundary layer thickness, [m]
6t Thermal boundary layer thickness, [m]
E: Heat exchanger effectiveness [-]
r Parameter defined by equations 4.3.2_and 4.3.12,
Parameter defined by equation 6.17,
[-]
7] [-]
8 Angle, [ ]
J.L Dynamic viscosity, [kg~ms]
II Kinematic viscosity, [m /s]
p Density, [kg/m3]
!/) Parameter defined by equations 4.3.18 and 4.3.22, [-]
1/J Parameter defined by equation 4.3.6, [-]
Xtt Martinelli parameter defined by equation 4.3.21, [-]

Dimensionless Groups
Le Lewis number, a/D, Sc/Pr
Nu Nusselt number, hd/k
Pr Prandtl number, CpJ.L/k
Re Reynolds number, pvd/J.L
Sc Schmidt number,v/D

Abbreviations
BTF Back-To-Front
BTT Bottom-To-Top
FTB Front-To-Back
TTB Top-To-Bottom
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Subscripts
a Air
atm Atmospheric
as Saturated air
asi Air saturated at air/water interface temperature
asp Air saturated at process fluid temperature
asw Air saturated at bulk recirculating water temperature
c Convective or convection or condensate
ct Cooling tower
crit Critical
d Diagonal or downstream
db Dry bulb
de Drift eliminator
e Equivalant or effective
eb Equivalent (tube-) bundle
ec Equivalent constriction
ff Film cooler
fr Frontal
g Gas
go Gas only
he · Heat exchanger
hor Horizontal
i Inlet or inside or interface
il In-line
1 Longitudinal or liquid
lo Liquid only
lsl Laminar sublayer
m Mean or moist
max Maximum
min Minimum
a Outlet or outside
obl Oblique
p Process fluid (water)
r Refrigerant
rows Rows
rest Restrictions
st Staggered
t Tube or transverse
tp Two phase
thea Theoretical
v Vapour
ver Vertical
w Recirculating (spray) water
wb Wet bulb
~ Free stream
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CHAPTER 1

INTRODUCTION

The phenomenon of cooling by evaporation is well-known and it has found

many applications. Th~ ancient Egyptians used porous clay containers to·
keep water cool thousands of year ago.

Today evaporative cooling is used extensively in industry, ranging from

the cooling of power generating plants to the cooling of condensers in air-
conditioning systems.

In evaporative cooling, the medium which is being cooled can theoretically

reach the air wet bulb temperature whereas the minimum temperature which
can be reached in dry cooling would be the air dry bulb temperature. The
use of evaporative cooling can lead to major cost ·savings and improvements
in thermal efficiency because of the lower temperatures which can be
reached. ·.

In a conventional direct contact cooling tower (see figure 1.1) the

water to be cooled flows through the cooling tower where it is cooled by
counterflow or cross-flow airstream. The cooled water is then passed
through a heat exchanger or a condenser to cool a process fluid or condense
a vapour. This requires two separate units, i.e. the cooling tower
and the heat exchanger or condenser.

An evaporative cooler or condenser combines the heat exchanger or

condenser and the cooling tower in one unit with the evaporative cooler
or condenser tubes replacing the packing of the cooling tower. Figure 1.2
shows a schematic layout of a counterflow ev~porative cooler.

The operation of an evaporative cooler or condenser can be described as

follows: Recirculating water is sprayed onto a bank of horizontal tubes
containing a hot process fluid or a vapour which is to be condensed while
air is drawn across the wet tube bank. The recirculating water is heated
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by the hot process fluid or the condensing vapour inside the tubes while it
is cooled from the airside by a combined heat and mass transfer
process.

The airflow through the evaporative cooler or condenser may be horizontal,

in which case the unit is referred to as a cross-flow evaporative cooler or
condenser or vertically upwards through the tube bundle where it is known
as a counterflow evaporative cooler or condenser. Various other
configurations of evaporative coolers or condensers have been
proposed in the literature, but these are not commonly used.

In this report analytical models for the evaluation of cross-flow and

counterflow evaporative coolers and condensers are presented. The models
range ·from a comprehensive model which requires numerical integration and
successive calculations to a simplified model which allows easy and qui~k
sizing and rating calculations. Computer programs have been written to
analyse cross-flow and counterflow evaporative coolers and condensers.

Si nee corre 1at ions or data for heat and mass transfer coeffi ci.ents for
cross-flow evaporative coolers are practically non-existent, a series of
tests were performed on such a unit in order to determine the required
coefficients experimentally. The two phase pressure drop across the wet
tube bundle was also measured and compared with existing correlations.
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~-------------------------Airflow
.-----t--t-------,
~------------------- Tower she 11
..------------------- Drift e1 imina tor
..----------------Sprayers
~-----------Recirculating water
~=~~!:..____,

Process fluid

Figure 1.1 Conventional direct contact counterflow cooling tower layout.

U-------------
C::x!:t::::Jif------ Fan
Airflow

~mrrrrmJm~------- Drift e1 i mi nator

r---------h--,.--r-~~-,..~------ Rec i rcu 1at i ng water sprayers

>-------Process fluid (condensing vapour)

~-----Recirculating water pond

-------------------------Pump

Figure 1.2 Counterflow evaporative cooler (or condenser).

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CHAPTER 2

LITERATURE SURVEY

The mathematical modelling of an evaporative cooler or condenser is

complicated by the fa~t that three fluids, sometimes flowing in different
directions, interact with heat and/or mass transfer processes taking place.

Numerous modelling procedures each with varying degrees of approximation,

can be found in the literature. The older models often assumed the
recirculating water temperature to be constant. throughout the cooler and
most of the models used a one-dimensional modelling procedure.

Several authors have studied other types of evaporative coolers or

condensers where the air and the recirculating water flow inside the tubes
or through narrow slots between closely spaced plates. These· studies are
also of interest since the same interface phenomena occur in these
coolers or condensers as those corisidered in the current study.

Scott [29SC1] conducted a series of simple tests on a single tube

evaporative condenser to determine the coefficients involved in the heat
and mass transfer process.

The apparatus used consisted of a vertical tube with steam condensing on

the outside of the tube, while the recirculating water flowed as ·a thin
film on the inside of the tube concurrent with the air stream. A sample
design procedure for a single tube evaporative condenser was also
presented.The recirculating water temperature was assumed to be·constant
·throughout the tube.

One of the first attempts to evaluate a conventional horizontal tube

evaporative condenser was made by James [37JA1]. He described an
evaporative condenser in which the water was splashed up into the airstream
from a sump by a revolving perforated drum. The water was then carried to
the horizontal tube bundle by the air which flowed through the tube bundle.
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Since the air reached the coil almost saturated, he assumed that the
driving force for the heat transfer was the difference between the
condensing temperature and the air wet bulb temperature. Mathematically
his method stated

q = Kwb ( Tr - Tawb ) A (2.1)

where Kwb was called the "wet bulb K" and was defined as:

(2.2)
James noted that the tube to water and water to air coefficients, hw and
he, would be the controlling coefficients, and that a significant
improvement in performance of the condenser could be achieved if these
coefficients could be improved. A simple graphical design procedure was
also presented.

Goodman [38G01] and [38G02] gave the first useful procedure to rate or to ~

design counterflow evaporative condensers. In his analysis of the process

he uses the enthalpy potential, as first derived by Merkel [26ME1], ~

as the driving force for heat transfer from the recirculating water to air.

He used the difference between the condensing temperature and the

recirculating water temperature as the driving force for the heat
transfer from the refrigerant.

The assumption of a constant recirculating water temperature throughout the

condenser was justified as follows:

" In as much as the spray water wets the outside surface of the coil, the
heat is transferred through the wall of the coil to the water on its
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outer surface. But, the water as fast as it receives this heat trans=
fers it in turn to the air flowing over the coil. As the water is
neither heated nor cooled·while it is circulated, it must attain an
equilibrium temperature, but remains constant as long as the operating
conditions remain unchanged".

Although this is not strictly correct, it is still in fairly good approxi=

mation in the case of a counterflow evaporative condenser.

The design method relied on a graph which was used to determine the
recirculating water temperature if the condensing temperature and the
entering air wet bulb temperature was known.

The capacity of the condenser could then be determined from

(2.3)

or from

q = rna ( i asw - i ai ) E (2.4)

where

E = 1 - e-N (2.5)
and

N =
(2. 6) .

Note that if the.Lewis relation holds we have that

(2. 7)
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By substituting this into the relation for N it follows that

N =
(2.8)
This is similar to an £ - NTUa approach with the one fluid at a constant
temperature (Cmax/Cmin ~ ~ ) and N = NTUa

No correlations for any of the coefficients were given.

Thomson [39TH!] studied the heat and mass transfer processes in an

evaporative condenser and conducted a series of tests to determine the heat
transfer coefficient and the rate of evaporation from the water film to the
air.

A single horizontal tube (at a slight angle) in a horizontal airstream was

studied. It was found that the amount of water evaporated from the tube
was dependant on the film thickness (more water evaporates off a thinner
film) and that the total amount of water evaporated was always less than
8%.

Thomsen [46TH!] proposed a graphical design procedure for simple eva= X

porative condenser design calculations. He assumed that the recir=
culating water temperature stays constant throughout the condenser and used
the method given by Goodman [38G01] to determine the temperature of
the recirculating water.

He formulated the concept of a single resistance for latent and sensible

heat transfer by assuming that the driving force is the difference between
the spray water (recirculating water) temperature and the average air wet
bulb temperature. This approach is similar to that of James [37JA1].

Wile [SOWil] studied the operation of evaporative condensers proving that ~

the recirculating water temperature is not constant throughout the
condenser. For calculation purposes he assumed that the recirculating
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temperature can be represented by a single equivalent temperature which

would yield the same result as a varying temperature profile. He proposed .
a method by which the performance of an evaporative condenser at any
operating condition could be determined by using the results of a few tests
over the normal operating range. The representative test data can be
converted into rating tables or curves· that would apply over a wide
temperature range.

Wile [58Wil] discussed the operation of an evaporative condenser, covering

subjects like bleeding, scale deposits, winter control methods, desuper=
heating coils and general system performance. No design method was given.

Parker and Treybal [61PA1] gave the first accurate design procedure for )(
the evaluation and design of vertical counterflow evaporative coolers. The
model was kept simple by employing the following assumptions:

i) The air-water heat and mass transfer can be described using the
"Merkel" type equation

dq = ho ( i asw - i a ) dA, (2.9)

ii) The Lewis factor is equal to unity

he
= 1'
ho cpm (2.10)

iii)The amount of water evaporated from the cooler was considered

negligible,

iv) Air saturation enthalpy is a linear function of temperature.

In· solving the model Parker and Treybal realized that the recirculating
water temperature could not be constant, but since the variation in
recirculating water temperature is small the error introduced by assuming a
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linear relation between the ·air saturation enthalpy and the recirculating
water temperature is negligible.

After manipulation of the governing equations the set of three differential

equations could be solved analytically. The resulting three equations
could now be employed in the rating or selection of evaporative coolers.
The model as given by Parker and Treybal is not explicit, since the
coefficients in the design equations have to be found by simultaneous
solution of these equations.

Parker and Treybal also conducted a series of tests on a vertical airflow

evaporative cooler to determine the required mass and heat transfer
coefficients. Correlations for these coefficients were determined from the
test data.

\{· Harris [62HA1] and [64HA1] described the operation of a new type of cooler
~· which he called an "air-evaporative ·cooler."

According to Harris the definition of such a cooler is given by the

following:

"Air-evaporative cooler units have all or part. of their heat transfer

surface as finned tubing so it can operate as straight air coolers when the
~~

air temperature is low enough. When the air temperature is not

sufficiently low to produce the desired process temperature, a water spray
can be turned on to provide evaporative cooling."

He gave no design method of such a cooler. Various configurations were

proposed and simple cost comparisons were made between the air-evaporative
cooler and conventional dry coolers.

Mizushina et al. [67Mll] described an experimental study performed on a ~­

counterflow evaporative cooler. The controlling transfer coefficients:

i) between the process fluid and the tube wall,
ii) between the tube wall and the recirculating film and
iii)between the recirculating water and the air were determined.
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The experiments were conducted using three different tube sizes:

i) d0 = 12,7 mm, di = 10,7 mm,

ii) d0 = 19,05 mm, di = 16,05 mm and

iii)d 0 = 40,0 mm, di = 38 mm

The tubes were spaced in a 2 x d0 pitch triangular array in eight or twelve

tube rows. Mizushina et al. [67MI1] measured the recirculating water
temperature at various places inside the cooler and they observed a
temperature variation in the region of zoe. They used a simplified model
originally proposed by McAdams [54MC1] to determine an approximate average
recirculating water temperature. Using this approximate average
recirculating temperature the controlling transfer coefficients were found
by employing logarithmic type equations describing the heat and mass
transfer through the whole cooler.

Correlations for the required transfer coefficients were derived for the
numerical evaluation of evaporative coolers.

Mizushina et al.[68MI1] described the thermal design of vertical airflow ~

evaporative coolers. The one-dimensional model used was derived in detail
by evaluating the energy and mass balances of a single small element.

The main assumptions made were:

i) No change in recirculating water massflow (evaporation neglected),

ii) Lewis factor= 1,

iii)The saturation enthalpy of air is a linear relation of temperature in

the applicable temperature range.
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Two design methods were proposed. The first method is based on the method
given by McAdams [54Mcl]. Assuming recirculating water temperature to be
constant, the governing equations can be integrated analytically into a
single equation which can be used iteratively for rating or sizing
calculations.

For the setond design method the cooler is divided into a number of
vertical elements and the three governing differential equations are
integrated (using a numerical method) for every element. By a method of
successive calculation the whole cooler is then evaluated.

The paper gives a numerical example of each of the two design

methods.

Finlay and McMillan [70FI1] derived an analytical model .to evaluate the
;-?performance of a mist cooler, the mist cooler consisted of a horizontal
r ,J ~ /
~ ,,. tube bank with horizontal airflow across the tubes. Small amounts of
. L/ water spray was added to the air flow in order to wet the tubes. The
analytical model which is based on the work of Berman [61BE1] represents
the heat and mass transfer process in terms of five differential equations.
Separate equations were derived to describe the transfer
process when the air has become saturated.

By numeri ca 1 integration of the contra 11 i ng · equations the 1oca 1 air

properties, the cooling water and process fluid temperature could be
determined for every position in the cooler. This method was used for the
evaluation of a typical cooler and the effects of varying spray water inlet
temperatures and varying air velocity wer·e determined.

The required coefficients for heat and mass transfer were calculated from
dry tube data, employing the Lewis relation and by using Elperins'
[61Ell] equation for two-phase heat transfer.

Two-phase pressure drop measurements were made and the data compared
favourably with the simple theoretical model cited.
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Anastasov [67AN1] tested a vertical tube evaporative condenser where the

vapour condensed on the outside of the vertical tubes while both air and
recirculating water flowed downwards through the inside of the tubes. The
test results were discussed and guide values for the size and capacity of
vertical tube condensers were given.

~- Kals [71KA1] described an evaporative cooler where the air enters from the
'\-'' top and flows downwards over a tube bundle concurrent with a gravity flow
of recirculating water. The airstream is then turned upwards again before
it is discharged. The concurrent flow of the air and the water prevents
the breakup of the water blanket which covers the tubes. Changing the
direction of the airstream after-it has passed through the tube bundle
forces all the entrained water droplets to leave the
airstream. A simple graphical design procedure was provided.

Tasnadi [72TAI] was the first author to describe the operation of a *'
cross-flow evaporative cooler. His model employed the following
assumptions:

i) Lewis factor = = 1,
ho cpm
ii) The air at the air/water boundary is saturated at the bulk
recirculating water temperature and

iii) the water film flow is so turbulent that the temperature of the
film can be taken as the bulk water temperature {Tw = Til

Using these assumptions he derived a model in which the heat transfer

from the process liquid to the water film was given by

dq = U0 ( Tp - Tw ) dA ( 2. ll)

..
.'·'·
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and the heat transfer from the film to the air could be described by

(2.12)
By transforming the heat transfer driving force between the process fluid
and the recirculating water to an enthalpy driving force he could then
write the complete heat/mass transfer process as

(2.13)
where

(2.14)
Tasnadi gave no indication of how the heat and mass transfer coefficients
were determined, and no numerical solution or example was given.

Tezuka et al.[72TE1] modelled the operation· of an evaporative cooler· in ~

I

terms of an overall mass transfer coefficient and a Merkel type enthalpy

difference.

The overall heat/mass transfer coefficient approach used is similar to that

used by Tasnadi [72TA1] in that the governing equation for the heat
transfer from the process fluid to the air is written in terms of
an enthalpy driving force.

Tezuka also determined simple dimensional correlations for the overall

transfer coefficient and for the pressure drop across the coil.

Tez'uka [72TE2] continued his previous work on evaporative coolers by ~

conducting series of experiments on a counterflow evaporative cooler to
determine the film heat transfer coefficient which governs the heat
transfer from the tube wall to the recirculating water film. A dimensional
empirical correlation is given for the film heat transfer coefficient.
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Finlay and Grant [72FI1] formulated a comprehensive thermal design model ~

for evaporative coolers. This model did not assume the Lewis factor to be
unity and the evaporation of the recirculating water was also taken into
consideration. The cooler evaluation was performed by numerically
integrating the controlling differential equations through the
whole cooler.

A simplified model was also obtained by assuming the Lewis factor to be

equal to unity and by ignoring the evaporation of recirculating water. This
simplified model gave three controlling differential equations which still
had to be solved numerically.

A fairly comprehensive literature study summarized most of the important

contributions for the determination of the controlling heat- and mass
transfer coefficients.

An example of a typical ratiog solution was also given, evaluating the

effe~ts of varying heat and mass transfer coefficients. It was noted that
although the correlations of Mizushina et al. and Parker and Treybal vary
by up to 30 %, the ~pposing effect (hw predicted lower by Parker and
Treybal, while they predicted a higher mass transfer coefficient) of these
differences cancel and the overall agreement in performance prediction
between these two methods are good.

~inlay and Grant [72FI1] found that in the presence of fins on the outside
of the tubes the heat transfer coefficient from the tube to the
recirculating water film was reduced, but the mass transfer coefficient
between the recirculating water and the air was considerably enhanced. The
lower film heat transfer coefficient was attributed to the water held up in
between the fins by surface tension. It was consequently proposed that
the airflow should be arranged downwards to flow concurrently with the
recirculating water to assist in the transport of the recirculating water
through the tube bank.
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Finlay and Grant [74Fil] compared the accuracy of various design procedures
for evaporative coolers. As a reference the accurate model which was
introduced in a previous paper by the same authors [72FII] was used. The
mass and heat transfer coefficients required were obtained from the
correlations of Mizushina et al. [67MI1].

It was found that the simplified method of Parker and Treybal [61PA1] w~s
"in good agreement for most engineering purposes" to the accurate method.

According to Finlay and Grant the "usual logarithmic temperature driving

force does not ap~ly" because of the recirculating water profile that
exists. The methods of James [37JA1], Goodman [38G01] [38G02], Thomsen
[38TH!] and Wile [SOWil] were consequently not used for comparison since
these methods employed a single mean recirculating water
temperature.

The rating method of Tezuka et al. [76TE1] was shown to differ quite
significantly from the accurate solution.

They concluded that for tube banks of· less than seven rows and i small
cooling range the assumption of constant recirculating water temperature
would be reasonably valid as long as a close approach to air
wet bulb temperature is not required.

Mizushina et al.[74MI1] presented a simple design procedure for the design ~

of evaporative coolers or condensers. This model is similar to the
simplified model given by Mizushina et-al. [68MI1]. Flow charts of the
calculation procedure were also provided.
I

Tezuka et al.[76TE1] experimentally evaluated five different evaporative -~

cooling cores to determine correlations for the pressure drop across wet
tube bundles and the overall transfer coeffici~nt as defined in
a previous paper [72TE1].
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Correlations for pressure drop and the overall transfer coefficients were
presented for each of the five evaporative cooler coils. These
correlations were subsequently written in terms of dimensionless groups and
a single relation for the overall transfer coefficient was then derived to
unite the existing five correlations.

Kreid, Johnson and Faletti [78KR1] used a similar approach to Tasnadi ~

[72TA1] and Tezuka et al.[72TE1] to give the governing equations for the
operation of a wet surface finned heat exchanger in terms of an enthalpy
difference and an overall transfer coefficient.

The governing equations for a wet surface heat exchanger was shown to have
the same form as the corresponding dry surface equations, which then gave
the governing equation for a wet surface cooler (finned or unfinned) as

q = Fhoo ·( i asp - ia ) A (2.15)

where F is the conventional correction factor used in the LMTD approach
of heat exchanger design.

The design method for wet surface heat exchangers was then also extended to
the heat exchanger effectiveness form [ £ - NTU form ].

The wet heat and mass transfer coefficients were obtained from the analogy
between heat and mass transfer.

Kreid, Hauser and Johnson [81KR1] continued the previous work of Kreid et )(
al. [78KR1] by experimentally evaluating the unknown wet fin heat transfer
coefficient. This coefficient could not be determined from either first
principles or from existing empirical correlations.

Threlkeld [70TH1] discussed the operation of wet surface finned tube heat Jl
exchangers. A similar approach to that of Kreid et al. [78KR1] and
[81KR1] was used in that a fictitious saturation enthalpy at the process
fluid temperature was defined. The heat transfer from the process fluid to
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the air was then described in terms of an overall mass transfer coefficient
and a mean logarithmic enthalpy difference.

Leidenfrost and Korenic [79LE1] analyzed the operation of a counterflow ~

evaporative condenser. The analytical model derived was based on earlier
graphical method by Bosjnakovic [60801]. The only significant simplifying
assumption made in the analytical model was the assumption that the Lewis
factor is equal to unity.

The solution of this model involves a rather complicated integration

procedure, involving so-called "pulling points" which could be graphically
illustrated on a Mollier ia - wa chart. The model takes partial dryness of
certain tubes into consideration by a rather crude dryness factor which has
to be specified. The condenser to be evaluated is divided into elemental
modules. By a successive numerical evaluation of each module in the
condenser the operating point of the condenser can be found.

All the required coefficients were discussed in detail except the mass
transfer coefficient. The mass transfer coefficient is calculated from the
analogy between mass and heat transfer.

The results of evaporative condenser simulations show that evaporative

condensers can still operate at ambient dry bulb temperatures higher than
the condensing temperature and that close fin spacing is not required. The
amount of water evaporated is said to be about 1% of the recirculating
water flow.

In a later paper Leidenfrost and Korenic [82LE1] used the same model as ~

derived previously [79LE1] to evaluate finned counterflow evaporative

condensers. This paper discussed experimental work which was done on
evaporative condensers in order to verify the computer model which had been
set up.

In the computer model the Lewis factor was not assumed to be unity, but it
was calculated from a relation given by Bosjnakovic [60801]. ~
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The findings of the tests were described in detail and an empirical

relation was given for the film heat transfer coefficient. This
correlation gives values which fall between the values predicted by similar
correlations given by Mizushina et al. [67MI1] and Parker and Treybal
[61PA1].

/(
~A graphical representation of the measured pressure drop across th-e wetted
coil was presented. This showed an increase of up to 40 % in pressure
drop across the wet coil compared to the dry operation of the same coil.

It was experimentally shown that the amount of recirculating water needed

for complete wetting of the coil was sufficient to ensure maximum
/;
,; performance of the evaporative condenser. ~Increasing the air flow rate
increased the capacity of the condenser until up to a point where the
airflow caused the water film to break up.

Fisher, Leidenfrost and Li [83FI1] described the modelling and operation of )t

a vertical tube evaporative condenser. In the cooler described the air
flows upwards inside the vertical tube while the recirculating water flows
downward as a thin film inside the tube. Vapour is condensed on the
outside of the tubes. The condenser is similar to units described by
Anastasov [67AM1] and Perez-Blanco [82PE1] and [84PE1].

An experimental study was conducted to determine the controlling

coefficients used by the computer simulation program. The program used was
a modified version of the original program compiled by Leidenfrost and
Korenic [82LE1] for the evaluation of finned counterflow evaporative
condensers.

Perez-Blanco and Bird [82PE1] and [84PE2] studied the heat and mass
transfer process that occurs in a vertical tube evaporative cooler where
the air and the cooling water film flow countercurrently inside the tube.
An analytical model based on existing heat and mass transfer correlations
was developed. These transfer coefficients were then experimentally
varified.
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Perez-Blanco and Linkous [83PE1] studied a similar vertical evaporative

cooler to Perez-Blanco and Bird [82PE1]. They noted that the common
drawback in existing procedures to evaluate evaporative coolers lies in
the fact that the driving forces for heat and mass transfer differ.

They defined a fictitious air saturation enthalpy (at the process water
temperature) to formulate a single overall coeffi~ient. According to their
model the capacity is given by

q = h00 A LMED (2.16)

where

(2.17)
and

( iaspi - iai ) - ( iaspo - iao)

. LMED = -"1n--[-ia_s_p_i___i_a_i_l-----~

i aspo - i ao (2.18)

The formulation of a single transfer coefficient allowed the identification

of the controlling resistance in the transfer process. They identified the
controlling resistance to heat and mass transfer as being concentrated at
· the air/water interface.

When this model was experimentally verified it was found that the LMED
.
formulation could only be used for evaporative condensers or when the
cooling/recirculating water temperature change was small, otherwise a
stepwise evaluation would be necessary.
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Perez-Blanco and Webb [84PE1] noted from the work of Perez-Blanco and
Linkous [83PE1] that the controlling resistance at the air/water interface
has to be lowered in order to enhance-the performance of a vertical tube
evaporative cooler. They studied the effect of coiled wire. turbulence
promoters inside the vertical tube as an alternative to extended surfaces.
The turbulence promoters were placed away from the tube wall in order to
mix the air boundary layer and not the water film. Experimental work
shewed a marked increase in cooler performance. The spacing between
the promoter and the tube wall was found to be of critical importance.

Peterson [84PE3] studied the operation of a counterflow evaporative ~

condenser and modelled the heat and mass transfer processes at the air-
water interface very thoroughly. The complete model given by Peterson
required a set of eight differential equations to be solved, which would
require a numerical integration procedure. This model . was then
significantly simplified to give a model ·very similar to the model of
Parker and Treybal [61PA1] for an evaporative cooler. The major
simplifications were

i) The Lewis factor was taken as unity,

ii) The evaporation of recirculating water was ignored anq

iii)Air saturation enthalpy was taken as a linear function of temperature

for the operating temperature range.

Peterson obtained values for the controlling mass transfer and film heat
transfer coefficients after a series of experiments on an industrial
evaporative condenser, condensing Freon-22.

The correlation for the mass transfer coefficient agrees very well with
that obtained by Parker and Treybal [61PA1] but she could not corre·late
the film coefficient because of the scatter of the experimental readings.
Criticism could however be raised against the assumption of Peterson that
the condensation heat transfer coefficient for the Freon-22 condensing on
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the inside of the tubes is constant at 8000 W/m2K.In the condensation of

Freon the heat transfer coefficients are normally found to be in the
region of 1500 W/m2K_ because of the low thermal conductivity of liquid
Freon. The low condensation heat transfer coefficient would be the
governing resistance to heat transfer from the condensing Freon to the
water film on the outside of the tubes.

The fact that Peterson could not find a correlation for the film
coefficient after measuring the overall heat transfer coefficient could be
ascribed to this incorrect assumption.

Webb and Villacres [84WE1] and [84WE2] made the following assumptions l<
to simplify the evaluation of cooling towers, evaporative coolers
and evaporative condensers:

i) The total heat flux can be written in terms of the enthalpy difference
of moist air, the so-called "Merkel equation",

ii) The loss of water through evaporation, entrainment and blowdown

could be ignored,

iii)The saturation enthalpy of the air at the air/water interface can be

calculated at the bulk recirculating water temperature rather than at
the recirculating water interface temperature,

iv) Uniform and complete wetting of the packing or tubes and

v) Heat and mass transfer coefficients are constant through the

whole process.

The controlling heat and mass transfer coefficients governing the operation
of the cooling tower, evaporative cooler and evaporative condenser units
are discussed. The modelling procedures for evaporative coolers and
evaporative condensers assume a constant recirculating water temperature.
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This approximation makes it possible to integrate the controlling equations

and to give a single equation governing the operation of an evaporative
cooler or condenser. Mizushina et al. [67Mil], [68Mil] used a similar
approach to evaluate evaporative coolers.

Rating and selection procedures for all three types of cooler units are
described for both the simplified approach and the successive calculation
methods.

In a later paper Webb and Villacres [84WE2] used the methods described in
their previous work [84WE1] to set up computer programs for the rating of
any of the three units as described on the first article [84WE1] at off-
design conditions.

The heat and mass transfer characteristic for the cooler or condenser to be
rated is determined from the rating data at the design point. These
programs allow the user to evaluate the effects of various off-design
conditions on the cooler. The programs were able to predict the rating·
data given by the manufacturers within 3% for the coolers evaluated.
Various other papers [84WE3],[84WE1] and [85WE1] described the same work
as given in this article.

The complete Fortran program codes for all three the rating programs were
included in the paper.

Wassel et al. [84WA1] and [87WA1] modelled a countercurrent falling film

evaporative condenser consisting of closely spaced vertical metal plates.
On ·the air side of each plate a water film flows downwards while the
airstream flows upwards while vapour condenses on the other side of the
plate. The model developed does not assume a constant recirculating water
temperature through the cooler and it takes into account the cooling of
the recirculating water after it leaves the bottom of the plates until it
reaches th·e water sump. They found that the cooling of the recirculating
water between the sprayers and the top of the plates is
negligible.
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Rana et al. [86RAI], [87RAI] tested various counterflow single and multi-
tube evaporative coolers to evaluate the mass transfer from the
recirculating water to the air.

They compared their data to the theoretical mass transfer prediction

obtained by employing the Chilton Colburn heat/mass transfer analogy with a
Lewis factor of 0,92. It was found that the mass transfer from a single
tube evaporative cooler was between 200 % and 500 % higher than for a
multi-tube evaporative cooler.

The correlations given by Rana et al. for design purposes gives the
correction factor which should be used together with the heat/mass transfer
analogy to determine the mass transfer coefficient. The correlations given
include a term ( Ai/ifg) where Ai is a function of the air inlet and outlet
conditions. The fact that the outlet air enthalpy is required to determine
the mass transfer coefficient presents a complication if the ~quations are
to be used for cooler rating, since the outlet conditions are not known in
the rating calculations.

J Erens [87ER1] used the principles of the design method of Mizushina )t

[68Mil] to build a computer model for rating and sizin9 of evaporative
cooler units.

Block diagrams were presented to show the calculating procedure for the
rating and sizing calculations. The counterflow cooler was divided into a
number of elementary units; for each unit the controlling differential
equations were solved to obtain the inlet/outlet conditions for the next
element. By a method of successive calculation the whole cooler could then
be evaluated.

Since it is known that the inlet and outlet recirculating water streams
must have the same temperature, the solution procedure assumes a value of
the outlet recirculating water temperature and by the successive
calculating procedure the inlet recirculating water temperature is found.
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The correct choice.of inlet recirculating water temperature will give an

outlet recirculating water temperature which is equal to the chosen inlet
temperature.

Examples of the temperature profiles along the flow path were given as well
as numerical examples of the rating and the selection programs.

JErens [88ER1] realized that conventiona1 Munters type cool ina tower fill
could be used together with the bare coil of the evaporative cooler to
enhance the performance of the unit.

The packing has the effect of enlarging the mass/heat transfer area and
consequently the average recirculating water temperature is lowered
resultin~ in an improved cooler capacity. Two different variations were
compared to the bare tub~ cooler by employing modified versions of the
bare cooler rating program. The so-called "integral cooler" combined the
coils and the packing while the second layout consisted of a conventional
bare coil section with the packing placed underneath the coil.

A typical comparative calculation gave the following results:

Cooler Ca~acity

Bare tube cooler 147,1 kW

Integral fill cooler 199,2 kW
Bare tube + fill cooler 206,6 kW

Erens noted that by using fill together with the tubes it was possible to
use a considerable number of tube rows less than would be required for a
bare tube cooler of the same capacity.

~ Erens and Dreyer [88ER2] used the more accurate modelling procedure of 4f
Poppe [84P01] and Bourillot [83B01] to evaluate a typical element of an
evaporative cooler. This model did not include the Merkel assumptions of a
Lewis factor equal to unity and negligible water loss as result of
evaporation.
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Five controlling differential equations were given for the evaluation of a

typical element when the air is not saturated. If the air entering an
element is saturated, the mass transfer driving potential changes and
separate controlling equations were derived for this case.

The modelling procadure is similar to that of Leidenfrost and Korenic in

the sense that each element (module) was considered to be an imaginary
block around a length of tube.

By using a fourth order Runge-Kutta integration process with successive

calculations the whole cooler could be evaluated. Both cross-flow and
counterflow evaporative coolers were evaluated. Typical temperature
profiles were presented for both types of cooler units and it was shown
that the temperature variation of the recirculating water at the outlet
side ~as negligible in the case of a coun~erflow cooler, thus a one
dimensional analysis model would be sufficient.

The non-existance of correlations for heat and mass transfer coefficients

for cross-flow evaporative coolers was stated as the reason for the
application of the counterflow correlations for these coefficients. The
Lewis factor was calculated using the relation given by Bosjnakovic
[60B01].

In conventional cooling tower theory the Merkel type model has become the
accepted model for the analysis of a direct contact cooling tower.

Since 1970 various investigators, including Yadigaroglu and Pastor [74YA1]

Bourillot [83B01], Majumdar et al .[83MA1], Sutherland [83SU1] and Poppe
[84P01], have proposed more accurate models for. the analysis of
conventional wet cooling towers.

Webb [88WE1] gave a critical evaluation of current cooling tower

practice. The assumptions made in the different models were clearly shown,
and the effect of the different assumptions were discussed in detail.
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Although the conventional cooling tower theory is not directly applicable

to evaporative coolers or condensers the fundamentals of the heat and mass
transfer from the water to the air at the interface are similar.

The various modelling procedures for evaluating evaporative coolers and

condensers, given in the literature vary significantly in accuracy and
complexity of use. In many of the earlier models, the basic equations were
not explicitly stated which resulted in some dubious design models. The
first accurate mathematical model was presented by Parker and Treybal
[61PA1]. Various accurate numerical integration models have since then
been published in the literature. In many of the articles the mass and heat
transfer coefficients are not adequately_ defined and sometimes certain
coefficients are not defined at all. None of the models presented in the
literature has yet been established as the accepted model for the analysis
of evaporative coolers and condensers.
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CHAPTER 3

MATHEMATICAL MODELLING OF EVAPORATIVE COOLERS AND CONDENSERS

In the theoretical analysis of cross-flow and counterflow evaporative

coolers and condensers the following assumptions are made to obtain the
analytical model:
~

i) the system is in a steady state,

ii) radiative heat transfer can be ignored, 1

iii) low mass transfer rates (At high mass transfer rates the heat
transfer coefficient would be influenced by the mass transfer
rate; refer to Apppendix I);

iv) even distribution of recirculating water along each tube and

complete wetting of the tube surface, vi

v) the· water film temperature at the air/water interface is

approximately equal to the bulk film temperature~ (see Appendix H
for a discussion of this assumption), J

vi) the temperature rise of the recirculating water because of pump

work is negligible,.-

vii) the air/water interface area is approximately the same as the

outer surface of the tube bundle, i.e. the water films on the
tubes are very thin, and J

viii)the heat transfer to the surroundings from the U-bcnds

outside the cooler or condenser can be assumed to be negligible.

By employing these assumptions the analytical models for both

evaporative coolers and condensers can now be derived from basic
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principles.

The exact analytical method presented uses the same basic approach as
Poppe [84P01] and Bourillot [83801] to describe the transfer processes
between the air and the recirculating water in a conventional cooling
tower.

The more commonly used Merkel model can easily be found from the
controlling equations of the exact analytical model.

3.1 Basic theorv for evaporative coolers

3.1.1 Exact analvsis (Poppe model)

Consider a typical element of an evaporative ·cooler. The inlet and

outlet conditions of the cross-flow and counterflow elements are shown
in figure 3.1)a) and figure 3.1)b) respectively. This choice of inlet
and outlet conditions results in the same sign convention for cross-
flow and counterflow units, and consequently the controlling equations
would have the same signs for both cross-flow and counterflow models.

ia + dia
wa + dwa
Tp + dTp

ia + dia
wa + dwa

Tw + dTw Tw + dTw
mw + dmw mw + dmw

Figure 3.l)a) Control volume for Figure 3.l)b) Control volume

cross-flow evaporative cooler for counterflow evaporative
cooler analysis. analysis.
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The mass balance of the control volume gives

:. dwa =
(3.1.1)

The energy balance of the control volume gives

maia + mwcpwTw + mpcppTp = ,

rna ( ia+ dia) + ( mw + dmw) cpw ( Tw + dTw) + mp cpp ( Tp + dTP)

After simplification and by ignoring the second order terms the energy
balance gives

(3.1.2)

The controlling equation governing the heat and mass transfer from the
water film to the air is dependant on whether or not the air is
over-saturated (mist).

CASE 1 - Non-saturated moist air

The massflow of recirculating water evaporating from a typical element
into non-saturated air is given as

(3.1.3)

At the water/air interface simultaneous heat and mass transfer takes

place as given by

(3.1.4)
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By using equation (3.1.3) and noting that dq = madia this becomes

madia= ho ( wasw - wa) ivdAo +he ( Tw - Ta) dAo (3.1.5)

The following supplementary equations can now be used to simplify the

equation above:

i) cpm = cpa + wacpv (3.1.6)

i i ) iv = iva + cpvTw (3.1.7)

iii) ;a = ( cpa + wacpv ) Ta +waive (3.1.8)

iv) iasw = ( cpa + waswcpv ) Tw + wasw iva ( 3 .1. 9")

'

Rewriting equation (3.1.5) and employing equation (3.1.6) gives

madia = h0dA 0 [( wasw - Wa) iv + [ h::pm ] ( cpa + wacpv ) ( Tw Ta) J

= h0dA 0 [ ( wasw - wa ) iv +
[ h0 \J [( cpa + wa cpv ) Tw

- ( cpa + wa cpv J Ta~

By rewriting equation (3.1.8) as (cpa+ wa cpv) Ta = i~ - wa iva

and substituting it into the relation above, it follows that

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3.5

By rewriting equation (3.1.9) as

(cpa+ wasw cpv) Tw = iasw - wasw iva

and substituting it in the relation above, gives

-( wasw- wa )cpvTw- ia+ waiv~J

~ hodAo [ ( wasw- wa ) ; v + [ h:c cpm ] [ ( ; asw - ; a )

By noting from equation (3.1.7.) that iva+ cpvTw = iv, this becomes
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= h0dA 0 [( ; asw - ; a ) + [ [ h
0
h~pm ] - 1 ] [ ( ; asw - i a )

- ( wasw - wa ) ;v J]

h:~PJ J [(;
hD dA 0
:. di a=
rna [[ i asw- i a ) + [[ 1 asw - i a )

- ( wasw - wa ) ;v J] (3.1.10) '

CASE 2 - Saturated air

The massflow rate of recirculating water evaporating from the tube
surface of a typical element into saturated air is controlled by the
following equation

(3.1.11)
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At the air/water interface the simultaneous heat and mass transfer

can be described by

dq = -ivdmw +he ( Tw - Ta) dA 0

By employing equation (3.1.11) and noting that dq = madia it foliows
that

(3.1.12)
The following supplementary equations can be used to simplify the
equation above:

. i) cpm = cpa+ was cpv + ( wa - was) cpw (3.1.13)

ii) iv = iva + cpv Tw (3.1.14)

(3.1.16)
The last term in each of equations (3.1.13) and (3.1.15) constitutes a
correction to take into account the amount of water in the form of mist
in the saturated air.

Rewriting equation (3.1.12) and substituting (3.1.13) into it gives

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By noting that from equation (3.1.15) that

[cpa+ wascpv + ( wa- was) cpw] Ta = ia- wasivo

and from equation (3.1.16) that

c~ Tw = i asw - wasw cpv Tw - wasw ivo

this can be rewritten as

• hodAo [ ( Wasw - was ) i V + [ hD h~pm ] [ ( i asw - i a )

-[ wasw- was ) ( iva+ cpv1w) + ( wa - was ) cpw1~ J

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3.9

. di =
a

- ( wasw - was ) i v J h:~pm

+ [ ] ( wa - was ) cpwl w J
(3.1.17)

, The heat transfer from the process fluid to the recirculating water is
expressed by

dq . = U ( TP - Tw) dA 0
0

where

~p ~fi l ~w]
1 do d0 1n ( d0 / di ) 1
= + +-1 +
uo [[ : d.1 2 kt hfo/
/

The change in process water temperature can now be expressed as follows

by noting that dq = -mp Cpp dTp

(3.1.18)
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3.1.3 Imoroved Merkel analysis

Singham [83SI1] described an extra equation to use with the Merkel

equations to describe the air conditions more closely.

According to Singham the change in absolute humidity of the air in each

element can be given as

dwa =
[ ~asw - ~a ]
1 1
[
1 \asw ] dia
asw - a
By substituting equation (3.1.20) into the equation above, it can
be reduced to

(3.1.22)

The Merkel model is considerably improved.by using equation (3.1.22) in

conjunction with equations (3.1.19), (3.1.20) and (3.1.21). If the
Singham equation is used with the three Merkel equations two air
properties, humidity and enthalpy, are known at every position in the
cooler. If two properties of air are known then any other property can
be determined uniquely. This is of considerable importance in natural
draft coolers and condensers.

If the Singham equation is not employed with the Merkel method the
outlet air density has to be calculated after assuming that the
air leaving the evaporative coil is saturated.
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3.1.4 Simolified model

Consider the following two different evaporative cooler layouts.

Tp;

00000
7 _..,.__ Tp;
1a i
.... 00000
00000
...
i ao (
)
00000 _ _....,._ Tpo

Tpo/
i ai t ~Two
Figure 3.2)a) Cross-flow Figure 3.2)b) Counterflow
evaporative cooler layout. evaporative cooler layout.

Assuming that the recirculating water temperature is constant

throughout the cooler the cooler performance can be full~ described
(using the Merkel assumptions) by

mpcppdTP = U0 ( Tp - Twm) dA 0 (3.1.23)

rna di a= ho ( i asw - i a ) dAo (3.1.24)

Rewriting equation (3.1.23) and integrating between Tp; and

Tpo gives

dA 0 =
mp cpp
u0
[ dTP
TP - Twm
l
mp cpp
. A0 -- Tpi
••
uo [ ln ( Tp - Twm TI
Tpo

:. Ao =
mp cpp
uo
ln
[ TPl. - Twm
Tpo Twm
l (3.1.25)
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Rewrtting equation (3.1.24) gives

dA 0 =
ho

Integration between iai and iao gives

(3.1.26)
From equations (3.1.25) and (3.1.26) it follows that

(3.1.27)
By solving equation (3.1.27) iteratively for Twm [note that iasw = ias
(Twm)l and then using this value of Twm that satisfies equation
(3.1.27) _in either equations (3.1.25) of (3.1.26), the required cooler
surface area can be found.

This iterative procedure could be used for the rating of evaporative

coolers as well, but the rating procedure is greatly simplified
by rewriting the controlling equations as follows:

From equation (3.1.25) we have

Tpo = Twm + ( Tpi - Twm ) e -NTUP

Ao Uo
with NTUP = mpcpp
(3.1.28)
and from equation (3.1.26) we have

.
1 .
1 ( 1. . ) e -NTUa
ao asw - asw - 1 a i
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(3.1.29)
For the whole cooler we have

q = rna ( i ao - i a i ) = mp cpp ( Tpi - Tpo )

By substitution of the NTU relations into this equation we have

mp cpp ( Tpi - Twm - ( Tpi - Twm) e -NTUP )

:. rna ( ( iasw- iai ) ( 1 - e-NTUa) ) =

mp cpp [ ( Tpi - Twm ) ( 1_ e- NTU P ) J

rna ( iasw- iai ) ( 1 - e-NTUa)
:. Twm = Tpi - mp cpp (. 1 _ e-NTUP) (3.1.30)
Rating of a coil can now easily be done using equation (3.1.30) as
follows; A value of Twm is chosen and by using equation (3.1.30) the
value of Twm is corrected until Twm converges to a fixed value; By now
employing equations (3.1.28) and (3.1.29) the outlet conditions of
the cooler can be found.
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3.2 Basic theory for evaporative condensers

3.2.1 Exact analysis (Poppe model)

Consider the typical elements from a typical cross-flow and counterflow

evaporative condenser in figv.res 3.3)a) and 3.3)b) respectively. As
in section 3.1, the sign convention used results in the same equations
for both the cross-flow and counterflow models.

Tw i a + di a
ir + dir mw wa + dwa

ia + dia
wa + dwa
Tw + dTw Tw + dTw
mw + dmw mw + dmw

Figure 3.3)a) Control volume for Figure 3.3)b) Control volume

a cross-flow evaporative condenser. for a counterflow evaporative
condenser.

The mass balance of the control volume gives

(3.2.1)
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The energy balance for the control volume gives

maia+ mwcpwTw + mr ir =

( mw + dmw ) cpw ( Tw + dTw ) + rna ( i a + di a ) + mr ( i r + di r )

After simplification

(3.2.2)
Depending on whether the air is saturated or not, the controlling
equation for the heat and mass transfer from the water film to the air
is given by case 1 and case 2 respectively.

CASE 1 - Non-saturated moist air

The mass transfer from the water into the air is given by

(3.2.3)
From the exact analysis given in section 3.1.1 the change of air
enthalpy is given by

dia =
hD dA 0
rna r( iasw - i a ) +
[ he
ho cpm - I l
[ ( i asw - ia) - ( wasw - wa) iv J] (3.2.4)
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The heat transfer from the condensing refrigerant to the recirculating

water is given by

(3.2.5)
where

1
+-
hfo
J (3.2.6)

The change of refrigerant enthalpy can now be written as

By substituting equation (3.2.5) into the relation above, it follows

that we have

(3.2.7)

The five equations (3.2.1), (3.2.2), (3.2.3), (3.2.4), and (3.2.7)

fully describe the processes that take place in a single element of an
evaporative condenser if the air is not saturated.

CASE 2 - Saturated air

The mass transfer from the water to the saturated air is given by

(3.2.8)
From section (3.1.1) the Poppe-type analysis for the case of saturated
air results in

h dA
0
--·-

j
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(3.2.9)

The complete system for the case of saturated inlet air is now given by
equations (3.2.1), (3.2.2), (3.2.7), (3.2.8) and (3.2.9).

3.2.2 Merkel analysis

The main assumptions that need to be made to reduce the exact analysis
to the Merkel analysis are
i) the evaporation of the recirculating water is negligible and
ii) the Lewis factor is equal to unity.

The Merkel-type analysis does not involve separate equations for

the case of saturated air.

The governing equations of the Merkel analysis are given as

hD dA 0
. dia = ( i asw - i a )
rna (3.2.10)
1
dTw = ( -madia - mrdir )
mw cpw (3.2.11)

(3.2.12)
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3.2.3 Improved Merkel analysis

As seen in section (3.1.3) a supplementary equation for describing the

air conditions in a control volume has been proposed by Singham
[83SI1]. This equation (3.1.22) holds without alteration for the
application in an evaporative condenser, together with the three MP.rkel
equations (3.2.10), (3.2.11) and (3.1.12).

3.2.4 Simplified Model

The simplified modelling approach considers the evaporative condensers

as a single unit, using only the inlet and outlet values of the unit
for the analysis. The layout of a typical evaporative condenser is
shown in Figures 3.4)a) and 3.4)b).

Tw; ~
/ _..,.,..__ iri
i ai
...
00000
00000
00000
..
i ao (

(
00000
----- iro

iro/
ka
Figure 3.4)a) Cross-flow Figure 3.4)b) Counterflow
evaporative condenser layout. evaporative condenser la~out.

The load on the evaporative condenser will typically be specified as a

given vapour massflow, an inlet vapour quality, x;, and an outl8t
vapour quality, x0 .

q = rna ( i ao - i ai ) = mr ( X; - xo ) i fg
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If saturated vapour enter-the condenser and the refrigerant-leaves the

condenser as a saturated liquid the condenser load is

q = mrifg
By making the Merkel assumptions, the operation of an evaporative
condenser can be described by the following two relations

madia= Uo ( Tr - Tw) dAo (3.2.13)

madia= ho ( iasw - i~) dAo (3.2.14)

Assuming that the recirculating water is constant through the
condenser, integration of equation (3.2.13) between the condenser inlet
and outlet sides gives

(3.2.15)

and integration of equation (3.2.14) between the inlet and outlet sides
result in

(3.2.16)
Substituting equation (3.2.15) into equation (3.2.16) gives

ln [ :::: :
(3.2.17)'
By solving equation (3.2.17) iteratively for Twm and then using the
value of Twm which satisfies equation (3.2.17) in equation
(3.2.15), the required condenser area for the given load can be
determined.
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Equation (3.2.17) could be used for condenser rating by employing a

complicated iterative procedure. A simpler approach for condenser
rating can be found by rewriting equation (3.2.16) as follows

.[ ~ asw
1asw
- i a1'
i ao
l =e
--NTU a
....

fJ \. \...:.
where

ho Ao
NTU a =
rna

:. i ao = iasw - ( i asw - i ai ) e -NTU a (3.2.18)

From the energy balance of the condenser we have

q = rna ( i ao - i ai ) = Uo ( Tr - Twm ) Ao

Uo Ao
~ iao= iai + rna (3.2.19)

By equating equations (3.2.18) and (3.2.19) we have

Uo Ao
iai + rna ( Tr - Twm)

( Tr - Twm )

rna ( iasw - iai ) ( 1 e-NTUa)

Uo Ao· (3.2.20)

Rating of a condenser coil tan now easily be done by choosing an

initial value for Twm and then utilizing equation (3.2.20) to correct
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the previous choice of Twm until Twm converges. The outlet enthalpy of
the air can now be found by employing equations (3.2.18) or (3.2.19).

From the energy balance of the condenser the massfl ow of condensate·

condensed would be given by

rna ( i ao - i ai )
= ' '
('•
ifg \.. (3.3.21)
It is interesting to note that equation (3.2.20) could easily be found
using the £.-NTU design approach used for the rating of conventional
heat exchangers. For heat exchanger with one fluid at a constant
temperature the efficiency £. is given by

£. = 1 _ e -NTU (3.2.22)
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CHAPTER 4

HEAT/MASS TRANSFER AND PRESSURE DROP CORRELATIONS

Various correlations for the governing heat and mass transfer

coefficients and for pressure drop across tube bundles were found in
the literature. The majority of these correlations were based on
experimental results, but a few analytical models were also proposed.
In this chapter the various correlations for the required coefficients
and pressure drops, which are relevant in the evaluation of evaporative
coolers and condensers are summarised and graphically compared.

4.1 Film' heat transfer coefficient

The heat transfer between the cooler or condenser tube and the
recirculating water film is governed by the film heat transfer
coefficient. Various investigators have determined this coefficient
experimentally and analytically for vertical and horizontal tubes in
evaporative coolers or condensers and in so called "film"-, "trickle"-
. or "trombone" coolers.
In a "film" cooler there is no airflow through the cooler to cool the
water film flowing over the tubes as is the case in an
evaporative cooler or condenser.

Parker and Treybal [61PA1] studied horizontal tube counterflow

evaporative coolers and horizontal tube falling film coolers.
According to Parker and Treybal the film heat transfer coefficient in
an evaporative cooler or condenser is approximately 20 % less than in a
film cooler, expressed mathematically as

~ hw = 0,8 hff (4.1.1)

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Parker and Treybal correlated the film heat transfer coefficient in an

evaporative cooler with 19 mm 0.0. tubes as

I I I •
( Q' .L,c., t ~- "! t

hw = 704 (1,3936 + 0,02214 Tw)

I

\ ...\c. ,.,_ . v.· -....

(4.1.2)
for

and

r
1,36 < < 3 [ kg/m2s ]
do

McAdams [54Mc1] determined the following correlation for the film

coefficient i~· a horizontal tube film cooler as,

hff = 3334,6 [ ~o ] 1I 3
(4.1.3)
if

4r
< 2100
J.Lw

By employing the conversion given by Parker and Treybal the

correlation for an evaporative cooler or condenser would be

(4.1.4)
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Mizushina et al. [67MI1] found the following correlati<_m for the film
heat transfer coefficient in a counterflow horizontal tube
evaporative cooler

hw • 2102,9 [ ~O ] 113

(4.1.5)
with

r
0,195 < < 5,556 [kg/m2s]
do
The correlation given by Mizushina et al. was obtained from test data
using tube diameters of 12,7 mm, 19,05 mm, and 40,00 mm.

Conti [78C01] and Owens [78W01] evaluated the heat transfer to an

evaporating ammonia film flow over horizontal tubes.
They correlated the data with the following empirical relation

= 2.2
(4.1.6)

The exponent of the term (f/d 0 ) in the relation above differs

considerably from the exponents found for this term in the other film
coefficient correlations. This discrepancy might be attributed to the
influence of surface tension which would play a much larger role in
an ammonia film than it would in an water film.
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Nakoryakov et al.[79NAI] conducted a series of experiments to determine

the heat transfer coefficients between horizontal tubes and a falling
film. Nakoryakov et al. correlated the data with the following
relation.

Nuff = 1,06 Re Pr ~]
[ w w pd
0 (4.1.7)

with

I' 5 < [ R•w::w 6 ]

By employing the definition of the film Reynolds number and the long
established Nusselt equation for film thickness flowing down a vertical
surface the correlation above can be rewritten as follows:

(4.1.8)

This relation was derived for a film cooler, but by using the factor
proposed by Parker and Treybal [61PA1] this relation can be rewritten
for use in an evaporative cooler or condenser as follows

hw = 0,735
(4.1.9)

Rogers [81R01] studied the flow and heat transfer characteristics of

laminar falling films flowing over horizontal tubes using an
analytical approach.
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According to him .the transition from laminar to turbulerit flow of

falling films occurs in the film Reynolds number range between 1000 and
2000. Rogers divided the flow regions over the tube in two distinct
regions e.a. the development region and the developed region. · By
employing an integral method he determined local heat transfer
coefficients with the followiPg form

hw = f [~do [c(] ,
[ g p2 d3
r ~f ]] (4.1.10)

for both the developed and the developing regions. The actual
determination of the mean film heat transfer coefficient is rather
complicated since it requires a numerical integration procedure to
determine the required coefficients for the two regions. The mean film
heat transfer coefficient consists of a combination of the film
coefficients of the two regions.

A graphical example given by Rogers shows that the results obtained

using the analytical approach compares well with the
simple empirical correlation given by McAdams [54Mc1].

Ganic and Mastanaiah. [82GA1] gives an extensive survey of the

literature on the hydrodynamics and heat transfer in falling films up
to 1981. The subjects discussed include descriptions of the flow
regimes, correlations for mean film thickness and conditions for the
onset of turbulence and wavy flow. Correlations for the heat transfer
to subcooled and saturated film flowing over horizontal and vertical
tubes are compared with experimental data.

Leidenfrost and Korenic [82LE1] conducted a series of tests on an in-

line horizontal tube evaporative condenser to determine the film heat
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transfer coefficient. They correlated their test data with the

following relation

r ]0,252
hw = 2064,3 [ do (4.1.11)

where

r
2 < < 5,6 [kg/m2s]
do

Leidenfrost and Korenic used a tube diameter of 15,9 mm for all their
tests.

Dorokhov et al.[83D01] proposed a similar correlation as Nakoryakov et

al. [79NA1] for the film heat transfer coefficient based on a series of
experiments with a water and Li-Br mixture flowing as a film over
horizontal tubes.

The correlation given by Dorokhov et al. states

0,46

(4.1.12)

with

I' 6 < . Rew Prw [ !o l < 32

After simplification and using equation (4.1.1) this gives the film
heat transfer coefficient in an evaporati~e cooler or condenser as

0,67 k~ o- 0, 54 [ Jl.
l
Pwrw o' 46 l
[ rdo o' 46
(4.1.13)
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where

=
(4.1.14)

Peterson [84PE3] used a similar approach to that of Parker and Treybal

[61PA1] to determine the controlling coefficients in the operation of a
counterflow evaporative condenser. It was found that the film heat
transfer coefficient correlation given by Parker and Treybal
fitted the new test data very well in the following extended range,

~
1,3 < [
0 ] < 3,4 [kgjm2s]

Chyn and Berqles [87CH1] proposed a model for calculating the film heat
transfer coefficient between a saturated water film and a horizontal
tube. The model is based on three definite heat transfer regions: the
jet. impingement region, the thermal developing region and the fully
., region. The exponent of the term ( f/d 0 ) in the
develoP.ed
.

proposed model is about -0,22 in the wavy-laminar flow region.

This model correlated experimental data well when the liquid flowed
from one tube to the next as a sheet but not if the liquid feeds in
columns and droplets, in which case the model underpredicts the heat
transfer coefficient.

Discussion of film coefficient correlations

The correlatibns of Parker and Treybal [61PA1], Mizushina et al .[67MI1]

and Leidenfrost and Korenic [82LE1] were all determined for evaporative
coolers or condensers. Parker and Treybal [61PA1] however determined
their correlation for the film coefficient by operating the evaporative
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10000

/
v
.........·· ,...--
/
v .. 7'

-.~_,.··
........ ,........
/

/
.....·
~,- --- 1.-

v v ~ ~
_.-/

........ ,,'"'
,"'

v _,...... , ·" ~

/
v /

...." -
..-;·:.;·:''.,.,,
....·,

/
v
/ .....···

v ......,..,.....·
...........··
..-.....

.....··•·
.......·'
......

1000
0.1 1 10
f/do [ kgjm2s ]
- Parker et a l ..·--·-·-··Mizushina et al -McAdams
----- leidenfrost - - Nakoryakov ·-···•····· Oorokhov

Figure 4.1 Correlations for the film heat transfer coefficient.

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4.9

cooler as a "film" cooler i.e. without airflow. All three of these

correlations express the film heat transfer as being a function of rn
where the exponent n has a positive value. The correlations by McAdams
[54Mcl], Nakoryakov et al. [79NA1], Rogers [81R01] and Dorokhov et al.
[83001] also use a positive value for the exponent n. Since the film
thickness increases with higher water massflow rates (higher f) the
film coefficient should decrease with increasing water massflow, which
implies that the exponent, n, must be negative.
This approach however neglects the effects of evaporation of the film
and the entrance effects as the water strikes the tube from above.
Chyn and Bergles [87CH1] found that their analytical model correlated
experimental data well when the water flowed from the one tube to the
next as a sheet but not when the water flows from one tube to the next
in columns and drops. In evaporative coolers and condensers the water
flow from one tube to the next is almost always as columns or drops
which together with the fair degree of splashing and entrainment which
occurs can explain the difference found between exponent n used by
the various correlations.
In the evaluation of evaporative coolers and condensers it is advisable
to use one o( the correlatirins which was determined specifically for an
evaporative cooler or condenser.

The film coefficient correlations are compared graphically in figure

4.1. The fluid properties needed for the graphical comparison
was determined at 40oc and a tube diameter of 38,1 mm was assumed.

-4.2 Mass transfer coefficient

Following the approach of Tasnadi [72TA1], Tezuka et al. [72TE1] and

Perez-Blanco et al.[82PE1] the heat transfer from the process fluid or
the condensing refrig~rant to the airstream can be expressed as the
product of an overall heat transfer coefficient and an enthalpy driving
potential.
By evaluating the different terms in the overall heat transfer
coefficient relation it is possible to determine the relative
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contributions of each of the resistances to the flow of heat from the

tubes to the airstream. The mass transfer coefficient term in the
overall heat transfer relation contributes the largest resistance
according to Perez-Blanco [82PE1].

Very little information on the mass transfer from tube bundles is

available in the literature. Several investigators have determined the
heat transfer coefficient governing the heat transfer from tube bundles
to the fluid flowing through the bundle. By employing the analogy
between heat- and mass transfer it is possible to estimate the mass
transfer coefficient for a given geometrical layout if the heat
transfer coefficient for the same geometrical layout is known.

The analogy usually gives good results for well defined layouts such
as the heat~ and mass transfer from a flat plate, but in the case of
the flow of air through a wet surface tube bank the interfacial area
between the air and the water film would not be the same as the outside
surface of the tubes, because of the falling films and drops. The
heat- and mass transfer analogy would thus fail in the case of the wet
surface tube bundle, because of the non-similarity of the wet- and dry
surface areas.

Parker and Trevbal [61PA1] conducted a series tests on a

counterflow evaporative cooler. The cooler tested consisted of a bank
of sixty 19,05 mm O.D. tubes on a 2 x d0 triangular pitch, 6 tubes wide
and 10 tube rows deep in the direction of the airflow. Based on the
two film theory of Treybal [55TR1] the mass transfer coefficient was
assumed to be of the following form

1 b l-1
= [ hoi + hL (4.2.1)
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where

b =
(4.2.2)

The coefficient hL is the heat transfer coefficient between the

recirculating water and the air/water interface. Parker and Treybal
approximated this coefficient by

hL = 11 360 [W/m2 K]

The mass transfer coefficient hoi was correlated as

hDi = 49,35 x 10-3 [ ( 1 + wa ) Gmax J 0,905 (4.2.3)

The value of the slope b can be determined through differentiation of

the curve for air saturation enthalpy as a function of temperature
given by Stoecker and Jones [84ST1] or graphically from a psychrometric
chart.
According to Stoecker and Jones the air saturation enthalpy at sea
level is given by

ias = ( 4,7926 + 2,568 T 0,029834 T2 + 0,0016657 T3 ) x 10 3 (4.2.4)

Through differentation of equation (4.2.4) it follows that

diasw
dTw
= b = ( 2,568 - 0,059668 Tw+ 0,0049971 T! J x 10 3 (4.'2.5)

The mass transfer coefficient correlated by Parker and Treybal can

now be given as
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4.12

in the ranges 0,68 < Gmax < 5,02 [kg/m2s]

1,36 < f/d 0 < 3 [kgjm2s]

Peterson [84PE3] followed the work of Parker and Treybal [61PA1]

closely in determining the mass transfer coefficient in a counterflow
evaporative condenser. The condenser coil studied by Peterson
consisted of a bundle of tubes, six rows deep and 33 rows wide. The
tubes were spaced in a triangular array with the following dimensions:

di = 27 [mm]
do = 24,5 [mm]
Pl = 64 [mm]
Pt = 81,6 [mm]

The mass transfer coefficient determined by Peterson was correlated as

b l-1
1
G
[ ( 1 + wa ) max
J 0,905 + 11360
(4.2.7)
Peterson assumed throughout that the air saturation enthalpy at a level
of 1 700 m above sea level is given by

ias = 4593Tw - 2956,3 [J/kg ]

:.b = 4 593 [J/kgK] (4.2.8)

The correlation given by Peterson holds for the following flow ranges

6,3 < Gmax < 9,6 [kgjm2s]

r.
1, 3 <- < 3,4 [kg/m2s]
do
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Peterson found that the correlation for the mass transfer coefficient
given by Parker and Treybal [61PA1] holds with good approximation in
the range,

0,68 < Gmax < 9,6 [kg/m2s]

Mizushina et al.[67MI1] conducted a series of tests on a counterflow

evaporative cooler using three different tube bundles with tube
diameters of 12,7 [mm], 19,05 [mm] and 40 [mm] respectively. The
tubes were spaced in a 2 x d0 triangular array with either eight or
twelve tube rows along the path of the airflow. The following
volumetric correlation for the mass transfer coefficient was determined
by Mizushina et al. to fit the test data. ,1

8 9
ho a := 5,0278 x 10- (Rea) 0, ( Rew) o, 15 ( d0 ) -
2, 6 ( . . )
4 29
where

Pa v•~ . ...-. L
Rea = Gmax ;,.
L
~'a l' ~ .,-
4r l\1\...,..._

Rew = o:= ·1
;;. ·,I
~'w t5 .. .,•

with

1 500 < Rea < 8 000

r
0,195 < d < 5,6 kg/m2s
0

The interfacial area per unit volume of a tube bundle in a 2 x do

array can be expressed as

0,9069
a =
do (4.2.10)
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The mass transfer coefficient correlation can be rewritten by employing

the relation above, as

h
0
= 5,544 x 10- 8 (Rea) o,g ( Rew) 0, 15 ( d0 ) -1,6 (4.2.11)

Tezuka et al.[76TE1] determined a correlation for the overall heat

transfer coefficient in a counterflow evaporative cooler.
The overall transfer coefficient approach states that

dq = hoo ( i asp - i a ) dA (4.2.12)

where

hoo = [ ~D + !'_r
uo .

and

dias
b =
dTP

Tezuka et al. evaluated five different test sections with different

diameters and tube configurations. By defining an effective diameter
as

4 x flow area
de = wetted perimeter

( 2 Pl ) ( Pt ) - do

= -1r -d- - do
0 (4.2.13)
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4. 15

the overall transfer coefficient data can be expressed by one single

correlation

(4.2.14)

For a tube bundle with tubes spaced in a 2 x d0 triangular array the

mass transfer coefficient can be expressed as

(4.2.15)
where

-1

+ ~]
hw (4.2.16)

and

0,25
d 0,7]
T -0,75
= 1,1828 [ d:0,3 p
(4.2.17)
in the ranges

1 < [A:~ l < 2,22 [kg/m2s]

 Stellenbosch University
4. 16 http://scholar.sun.ac.za

1 < [::r l < 4, 2 [ kg/m2 s]

1 <
[ ~r l < 4,2 [kg/m2s]

Rana et al.[81RA1], [86RA1] and [87RA1] experimentally investigated the

mass transfer coefficient which governs the heat transfer in a
counterflow evaporative cooler or condenser. By using the heat and
mass transfer analogy and a Lewis factor of 0,92 they determined
the theoretical mass transfer coefficient from

ho,theo = 0,92 cpm

(4.2.18)
The predicted theoretical mass transfer coefficients were found to vary
qu~te significantly from the experimentally determined values. In
order to obtain a useful correlation for mass transfer
coefficient Rana defined the following ratio,

=
ho, thea (4.2.19)
Various studies were conducted by Rana et al.[81RA1], [86RA1] and
[87RA1] to find correlations for the ratio RR. The experimental work
was carried out on various counterflow evaporative cooler layouts,
including a single tube unit. According to Rana [87RA1] single tube
correlations developed by Rana [86RA1] overpredicts the mass transfer
coefficients by between 200 and 500 %.
Rana et al .[81RA1] determined the following correlation for a full
coil test unit

(4.2.20)
 Stellenbosch University
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where

4r
Rew =
'
~-'w

Pa va,max do
Rea =
~-'a

For a row of tubes in a counterflow evaporative cooler Rana [87RA1]

correlated the ratio of experimental to theoretical mass
transfer, RR, as

RR = 1, 7838 ( EP }0,39.85 ( Rew )0,3765 (Rea )-0,4114

(4.2.21)
where

Ai
EP

and

( i as , wall , i - i ai ) ( i as,wall ,o - i ao )
.Ai =
ln
[ i as , wa 11 , i - >i
.
1
as,wall,o 1
ao
l
in the ranqes

0,0544 < EP < 0,1971

41,9 < Rew ~ 294,3
692 < Rea ~ 2764
 Stellenbosch University
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Rana [87RA1] used the following correlation to determine the

convective heat transfer coefficient from the dry tube bundle

Nu c = ( Rea ) 0,628
(4.2.22)
Various other investigators determined correlations for the convective
heat transfer coefficient in dry tube bundles e.g.
Zukauskas [74ZHI] or Grimison [37GRI].

Discussion of mass transfer coefficient correlations

The mass transfer coefficient correlations given by Parker and Treybal

[61PA1], Mizushina et al.[67MI1], Tezuka et a1.[76TE1], Peterson
[84PE3] and Rana et al. [87RA1] are compared graphically in figures 4.2
and 4.3 for tube djameters of 19,05 mm and 38,1 mm respectively. The
fluid properties were evaluated at 35oc and a water flow rate of r =
300/3600 kg/ms was assumed.

The correlations of Parker and Treybal [61PA1], Mizushina et al .[67Mil]

and Peterson [84PE3] compare well for a tube diameter of 19,05 mm. At
the larger tube size the correlation by Mizushina et al. [67Mil]
predicts values which are much lower than the predictions by Parker and
Treybal [61PA1] and Peterson [84PE3]. It was noted by Finlay and
McMillan [74Fil] that the mass transfer correlation given by Parker and
Treybal [61PA1] gives mass transfer coefficients which are higher than
those predicted by the correlation of Mizushina eta] .[67Mil] but that
the film heat transfer coefficients given by the Parker and Treybal
[61PA1] correlation are lower than those found by Mizushina et al.
[67Mil]. Finlay and McMillan [74Fil] found that the models of Parker
and Treybal [61PA1] and Mizushina et al .[67MI1] are in good agreement
if each model uses its own correlations for the film heat transfer
coefficient and the mass transfer coefficients.
 Stellenbosch University
4.19 http://scholar.sun.ac.za

........
V)
N
s
~-
..........
0'
~

...... A
c ~
/
..s:
A

~
c
.,...
Q)
.,. •
~
~
u
.,...
Q)
0.1
:-,.. F?.·
..
....... ..--:-r
,,
--•.. .. ......
~
~
.~

-....- # v-·· ..-· ·· -.--

0
u ,
~
/ . , ..
Q) ,, .....
~ /
V)
,,
/, t:(:•,
c ,..··
res
~
.·· , .. •
~

£,j ,·
~;·
.....·"'-·
"./
•.
V)
V)
res .....
:::E: ~.. .... _ -·
.. ,•·~. ...··
, .··''
, ., .........·
.........
~/ --
............

..
···...··

0.01
0.1 1 10
Free stream air velocity, v~ [ m/s ]
--Parker et a l -Peterson ------·· Mi zush ina et al
----- Tezuka et al ............. Analogy - Rana et al

Figure 4.2 Correlations for the mass transfer. coefficient, d0 = 19,05 mm.
 Stellenbosch University
4.20 http://scholar.sun.ac.za

~-
,......, ,
......::::

,:;:/ ~
VI
N ......,...·
5 .•'
.......... ..,....."''
C'l
..:.~: , ......·"'"
........ 0.1 ...·
,
"• ..·· ....
...- -
,
Cl
,,
/ ./ -~·
.c / /' ........
.•
,,' / /
-
.··•
A

~
c: ..
, ,
,
/ v ..·· I"""' _,......~

.. ' , . .
.•
.,...
u
Q)

.,... ,
,
~ v ......
.........
...··
.
.... •.. "
~---·· .•.

~~
'+- , .... ,.... ......
'+-
Q) ...........·
_,.........
0 r;.....-
~A v~;:::~
u
s..
Q)
'+- _............-:./·
···
VI
c:
tU
0.01 ......-..........
...
s.. ..
~ .,

VI
VI
tU
:::E:

0.001
0.1 1 10
Free stream velocity, v~ [ m/s ]
--Parker et a1 -Peterson ·-·-···-··· Mi zush ina et ·a 1
----- Tezuka et al ·······•·-·· Analogy -Rana et al

Figure 4.3 Correlations for the mass transfer coefficient, d0 38,1 mm.
 Stellenbosch University http://scholar.sun.ac.za
4.21

The use of the correlations by Mizushina et al. [67Mil] are advisable

since the mass transfer coefficient correlation by Mizushina et al.
[67Mil] covers a much wider range of conditions and the effect of
recirculating water and tube diameter are taken into account.

The correlation of Tezuka et al. [76TE1] is rather dubious since it

does not give results which compares well with the other correlations.
It is also illogical that the overall mass transfer should depend on
the inner diameter of the tube. Similar criticism can be raised
against the correlation for the mass transfer coefficient given by
Rana et al.[SIRAI] since this correlation gives the mass transfer
coefficient, amongst other parameters, as a function of the process
fluid Reynolds number.

The mass transfer coefficient correlation given by Rana et al. [87RA1]

is rather cumbersome to use for cooler (or condenser) rating since the
correlation requires the outlet conditions of the cooler to be known in
order to determine the so called "enthalpy potential". Even if this
correlation is to be used for design with known outlet conditions the
tube wall temperatures has to be determined in order to evaluate the
mass transfer coefficient.

4.3 Pressure drop across horizontal tube bundles in cross-flow and

counterflow

The pressure drop across the tube bundle of an evaporative

cooler or condenser of importance in the estimation of the required fan
size in the case of a mechanical draft cooler or condenser and it is of
major • importance in the determination of the air massflow
rate through the cooler or condenser in a natural draft application.

Various researchers have studied the single phase pressure drop across
tube bundles. Appendix D gives an overview of the available
correlations for single phase pressure drop across a tube bundle.
 Stellenbosch University http://scholar.sun.ac.za
4.22

In an evaporative cooler or condenser the pressure drop calculations

are complicated by the presence of the recirculating water flow. The
airflow in an evaporative cooler and condenser can either be horizontal
(cross-flow) or vertical upwards (counterflow) or vertically downwards
(concurrent). The recirculating water normally flows
downwards under the influence of gravity only.

The pressure drop characteristics of a wet surface evaporative

cooler or condenser depends on various factors including the air
massflow rate, recirculating water massflow rate, average temperature
of recirculating water, tube array configuration etc.

Diehl [57DI1] proposed a method to calculate the two phase pressure

drop across a tube bundle with the air and the recirculating water
flowing concurrent by downwards through the tube bundle (downflow)
and he proposed two graphical correlations for pressure drop across in-
line tube banks and across staggered tube banks.

Diehl and Unruh [58DI1] tested various tube bundles to determine two-
phase pressure drop correlations for different tube layouts. Graphical
correlations were presented for staggered tubes with a 45° layout and
60° layout· as well as for in-line tubes. They found that the
correlations for the in~line tube bundle and the staggered bundle with
the 60° triangular layout were the same. The pressure drop for the
tube bundle with the 45° staggered layout was found to higher than
that for the other two layouts.

Simple regression analysis of the graphical correlations yielded the

following simple-to-use equations to determine the two phase pressure
drop across a horizontal tube bundle.
For cross-flow across banks of tubes spaced in a 60° staggered layout
or an in-line configuration the correlations are

= 52,167 r 3 - 26,677 r 2 + 2,788 r + 1,oo985 (4.3.1)

 Stellenbosch University http://scholar.sun.ac.za
4.23

where

(4.3.2)

in the range 0 ~ r < 0,25 and

Aptp 1,33 035

---*-
Apa
=
1+ r + o,o2oo16 r -0,257908
(4.3.3)
for the range 0,25 ~ r ~ 10.

For cross-flow across banks of tubes spaced in a staggered 45° layout

the following regression curves were found

152,961 (3 - 67,9895 r 2 + 7,274 r + 1,o2375

(4.3.4)
where 0 -< r -< 0,25 and

Aptp
---*- = [ 1,327] 0,0795 + 0,002888 r
Apa 1+r (4.3.5)

in the range 0,25 -< r -< 10

The counterflow pressure drop across staggered tube bundles were

correlated by Diehl and Unruh using the following parameter,

LVF
=
( Pal Pw J ( Re: J 0,5
mw
= [ rna + mw ( P/Pw J] ( Re; )0,5
(4.3.6)
 Stellenbosch University http://scholar.sun.ac.za
4.24

If 0 < 1/J < 0,007 the correlation gives

Aptp
~
upa
= 1,370042 + 44591,59 1/J -
[ 0,0000369198
0,0001 + ,hY'
l
- 103378,776 log 10 ( 1 .+ 1/J) (4.3.7)
and if 0,007 < 1/J < 1,0 the data was correlated by

0,00261965]
0,00376946 + 0,0087965111 1/J + [ 0,001 + 1/J

- 0,0052407713 1/J
2 (4.3.8)

Wallis [69WA1] presented a simple theoretical equation based on the

homogeneous flow mode 1 to determine the two phase pressure dr_op for
·horizontal cross-flow through a tube bank with a staggered tube layout.

(4.3.9)

where

(4.3.10)
Collier [72C01] rewrote the model of Wallis [69WA1] and compared the
result with the data given by Diehl and Unruh [57011],
 Stellenbosch University http://scholar.sun.ac.za
4.25

[58DI1]. Collier adapted the Wallis model as follows

Aptp 1 1
-*- = 1+ r (1 - (Pa/Pw))
::::::
1 + r (4.3.11)
Apa

where

(4.3.12)

Grant and Chisholm [79GR1] conducted a study on the two phase pressure
drop through the shell side of a segmentally baffled shell and tube
heat exchanger. The correlation presented is of the following form

= 1+ ( r~ 1) [s Y(2-n) ( 1_Y ) (2-n)/2 + Y2-n J (4.3.13)

where

1; 2
[ Ap:
rG =
Apw
l
The coefficient n is the exponent in the Blasius type
(4.3.14)
single phase
fluid friction equation as given by

f =
(4.3.15)
Grant and Chisholm uses the value of n = 0,46 in the cross-flow
pressure drop correlation. The correlation can consequently be
simplified as

[ B y0,77 (l-y)0,77 + yl,54 l

(4.3.16)
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For the flow regimes usually encountered in evaporative coolers or

condensers the following values of B are proposed:

B =1 for vertical up-and-down flow and

B = 0,75 for horizontal side-to-side flow.

Grant and Chisholm reports that the correlation matches the data of
Diehl and Unruh [57DI1] and [58DI1] to within 2 percent for y
~ 0,6 and

B = 0,75 + 3,5 y
10 (4.3.17)

Schrage et al .[87SC1] and [88SC1] measured void fractions and pressure

drop in two phase vertical cross-flow in a horizontal tube bundle.
From the experimental data correlations for the void fraction and two
phase friction multiplier were developed. Ishihara et al .[77IS1] first
proposed the use of a Martinelli type multiplier to determine the two
a
phase pressure drop across horizontal tube bundle. For spray liquid_
flows as encountered in an evaporative cooler or condenser Schrage et
al. proposed the following correlation,

2
~Ptp = 0w ~Pw (4.3.18)

where

c 0,205
1/)2 = 1 + +
w Xtt 2
Xtt (4.3.19)

with

C = 1180 Gmax - 1, 5 1n Xtt + 3,87 Gmax 0, 207

(4.3.20)
 Stellenbosch University
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and

1 ,8
1 - y Pa
2
Xtt = [ Y ] Pw (4.3.21)

This correlation holds only for Gmax ~ 43 kgjm2s. Schrage suggested

that the correlation of Ishihara et al.[77IS1] be used if Gmax < 43
kgjm2s. The Ishihara correlation gives the two phase pressure drop as

(4.3.22)
where

8 1
= 1 + - + -2
Xtt Xtt (4.3.23)
Very little pressure drop data measured on an actual ·evaporative
cooler or condenser have been supplied in the literature. Two
investigators reported pressure drop data for counterflow evaporative
coolers or condensers while no data has been found on horizontal
cross-flow pressure drop across an evaporative cooler or condenser.

Tezuka et al.[76TE1] correlated the pressure drop across five different

counterflow evaporative cooler coils using a correlation of
the form

~ ]0,32 [ ~ ]1,6
= 66,034 X 10 6 C1 [
Afr Afr (4.3.24)

A different C1 value was proposed for each of the five coils tested.
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The following table gives the C1 value for each of the coils tested:

Coil c1 d0 [mm] Pl/do Pt/do

A 1,1 X 1o-1 27,2 1,65 2,30

B 1,97 X 1o-1 34 1,44 2,18
c 1,91 X 10-7 42,7 1,17 2,25
0 0,84 X 1o-1 42,7 1,17 2,93
E 1,15 X 1o-1 42,7 1,12 2,08

Leidenfrost and Korenic [82LE1] reported that for the in-line

evaporative condenser tested the pressure drop increase due to the
recirculating water at the lowest air massflow was between 24% and 62%
when compared to the dry operation of the coil. At the maximum air
massflow rate the pressure drop increase due to the recirculating water
flow was only between 12% and 18% more than the corresponding pressure
drop across the dry tube bundle.

Discussion of pressure drop correlations

Many of the two phase pressure drop correlations require calculation of

single phase pressure drop in order to use the two phase correlation.
Single phase pressure drop across bundle of horizontal tubes have been
extensively studied by many authors. Refer to Appendix 0 for a survey
of the available single phase pressure drop correlations.

The correlation by Gaddis and Gnielinski [85GA1] is very comprehensive

but its complexity does not allow fast calculations. The correlations
,.
of Gunter and Shaw [45GU1] and Jakob [38JA1] are easy to use and
they are normally accurate enough for design purposes.
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The available correlations for two phase pressure drop_ across a

horizontal tube bank in cross-flow and counterflow are shown
graphically in figures 4.4 and 4.5.
None of these correlations except that of Tezuka et al. [76TE1] was
developed from tests on evaporative coolers or condensers and it is
therefore advisable to use the more conservative correlations for
design purposes.
It can also be seen from figure 4.5 that the two phase pressure drop
prediction· by Diehl and Unruh [58DI1] drops below the dry (air only)
pressure drop. The correlation by Diehl and Unruh [58DI1] holds for
concurrent flows which explains the low pressure drop at high water
loading.
 Stellenbosch University http://scholar.sun.ac.za
4.30

100

.......
N
5
.........
:z:: 80
......
0.
~
~

0.
0
s.. 60
"'0
Q)
s..
::::s
II)
II)
Q)
s..
. Q.. 40

0~~~--L_~L_~L_~--~--~L_~

0 2 4 6 8 10 12 14 16
Air massflow rate, rna [ kg/s ]
-Dry tube bundle -Diehl (60°) .
--Diehl (45°)
-·~·~·~·-·Wallis/Collier -----Grant et al

Figure 4.4 Pressure drop across a tube bundle in cross-flow with a

recirculating water massflow rate of 5 kg/s.
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300r---~r---.----.----.----.----.----~--~

,......
'N
e
~ 200
.......
c..
<I
~

c..
0
s..
"'C
150 ..
..·······•·
Ql
s.. ...·····•
:::::J
VI ...·•···
VI
Ql
..··'
s.. ........•.-·....
0..
100

0 1 2 3 4 5 6 7 8
Air massflow rate, rna [ kg/s ]
--rezuka et al -Grant et al ·-··-·-·-Ishihara et al
-----Diehl et al - Dry tube bundle

Figure 4.5 Pressure drop across a tube bundle in a counterflow layout with
a recirculating water massflow rate of 4 kg/s.
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CHAPTER 5

COMPUTER SIMULATION

Various computer programmes have been written in Fortran 77 source code

on a Digital VAX 785, to simulate the operation of evaporative coolers
and condensers.

The numerical models employed in these programs include the simplified

approach and successive calculation models with varying degrees of
approximation.

The following table lists the various programs and the solution method
used in each program.

Name Description Model Rating or

. Selection

CROSS Cross-flow evaporative cooler E,IM,M R

COUNTER Counterflow evaporative cooler E,IM,M R
COMBINE Counterflow evaporative cooler with M R+ S
packing
SCROSS Cross-flow evaporative cooler M* R+ S
SCOUNT Counterflow evaporative cooler M* R+ S
CSCROSS Cross-flow evaporative condenser M* R+ S
CSCOUNT Counterflow evaporative condenser M* R+ S
TOWER Natural draft cooling tower employing E,IM,M R
cross-flow evaporative cooling units

E - Exact (Poppe) model

IM - Improved Merkel model
M - Merkel model
M* - Simplified (Merkel) model
R - Rating
S - Selection
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5.1 Determination of coefficients

All the programs assume a 2 x d0 triangular tube spacing as shown in

figures 5.1)a) and 5.1)b) for cross-flow and counterflow
respectively.

Recirculating water flow

Recirculating water flow I I

lolol
---
bib 0
000
00
Air flow 0 0
-o 0 o 000
t t Air flOW
Figure 5.1) a) Tube layout for Figure 5.1) b) Tube layout
cross-flow evaporative cooler for counterflow evaporative
or condenser. cooler or condenser.

The massflow rate of recirculating water is usually defined in the

'·
literature by the massflow rate of recirculating water flowing down one
side of a tube p·er unit 1ength. According to this definition
the recirculating water massflow rate in a cross-flow evaporative
------~
cooler or condenser with triangular spacing is defined as

mw = 2 r L nhor (5.1.1)

or

mw
r = --~, --
2\L nhor (5.1.2)

and for a counterflow evaporative cooler or condenser the recirculating

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water massflow rate is defined as

mw = 4 r L nhor (5.1.3)

or

r- - - - -
- 4 L nhor (5.1.4)
The mass transfer coefficient correlation given by Mizushina et al.
[67MI1] (see Chapter 4.2) was used to determine the mass transfer
coefficient for both cross-flow and counterflow evaporative
coolers or condensers. The mass transfer coefficient correlation
given by Mizushina et al. [67MI1] was determined for a counterflow
evaporative cooler, but because of the lack of more suitable data this
correlation was also used for the cross-flow coolers and condensers.

The film heat transfer coefficients used in all the programs are
determined with the correlation ~resented by Mizushina et al. [67MI1].
Refer to Chapter 4.1 for a description of this film heat transfer
coefficient correlation.

The heat transfer coefficients on the inside of the tubes in the case
of an evaporative cooler are calculated from the correlations by
Gnielinski [75GN1] and Kays et al. [55KA1] for turbulent and laminar
flows respectively.

In the case of an evaporative condenser the correlations by Shah

[79SH1] and Chato [62CH1] were used to determine the condensation
coefficient inside the tubes. The correlation given by Chato [62CH1]
was used when the vapour Reynolds number at the tube
inlet was below 35000, otherwise the Shah [79SH1] correlation was
used.
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The Shah-correlation seems to predict very conservative condensation

heat transfer coefficients if fluorocarbons (Freon's) are used as
refrigerant, because of the low thermal conductivity of Freon liquid.
Refer to Appendix G for the heat transfer coefficient correlations and
condensation coefficient correlations.

5.2 Successive calculation models

The successive calculation models for the evaluation of evaporative

coolers employ a numerical integration procedure for the evaluation of
the governing differential equations. The cooler which is to be
evaluated is subdivided into imaginary blocks (control volumes) with
each block surrounding a length of tube.

By employing a fourth order Runge-Kutta integration procedure the

outlet conditions (temperature, enthalpy etc.) of a given element can
be determined if the inlet conditions of the element are known. The
governing coe~ficients for heat a~d mass transfer are calculated for
every block. Various flow geometries are possible for the process
fluid flow, each requiring a different calculation algorithm. Four
different process fluid patterns were considered in the case of cross-
flow evaporative coolers, i.e.

i) Single pass (Straight through),

i i) Top-to-bottom (TTB),
i i .i) Front-to- back (FTB) and
iv) Back-to-front (BTF)

Only two process fluid flow patterns were considered as options

for counterflow evaporative coolers, i.e.

i) Top-to-bottom (TTB) and

ii) Bottom-to-top (BTT)

The different process fluid flow patterns for cross-flow and

 Stellenbosch University http://scholar.sun.ac.za
5.5

'' ''
'' ''
'' ''
'' ''
'' '''
'
-~---....._..,,- -~--- .....--.,,-
'' ''
' '' ''
' ''

a b

Figure ·s. 2 Serpentining arrangements for cross-flow evaporative coolers:

a) Straight through (~ingle pass), b) top-to-bottom,
c) front-to-back and d) back-to-front.
 Stellenbosch University
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/
/
/
/
/
/
/
/ /
/
I " /
I "
v" /
/

"I /
/
"" I /
/ /

/
/
/
- -~------
/
/
/
/
/
/
/
/

lI
Figure 5.3 Top-to-bottom serpentining arrangement as used in a counterflow
evaporative cooler.
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counterflow evaporative coolers are shown in figures 5.2 and 5.3

respectively.

If the water flowing over the tubes is recirculated the inlet and
outlet recirculating water temperature should be the same, as soon as
the cooler is operating in a steady state. If the water flowing over
the tubes is not recirculated, but fresh water is sprayed over the
tubes, then the inlet spray water temperature has to be specified in
order to evaluate the cooler performance.

5.2.1 Cross-flow evaporative cooler simulation

The solution procedure for the coolers with single-pass, top-to-bottom

and front-to-back process fluid flow patterns are similar. Execution
proceeds from the top front corner element where all the inlet
conditions are known.

If the cooling water flowing over the outer surface of the tubes is
recirculated, a viable inlet recirculating water temperature is chosen
{the recirculating temperature at the inlet will always be larger than
the air inlet wet bulb temperature and smaller than the process fluid
inlet temperature).

By using a fourth order Runge-Kutta integration procedure the outlet

conditions for the first element is computed. The outlet conditions of
the first block is then used as inlet conditions for the surrounding
blocks eg. the outlet process fluid temperature of the first block is
used as the inlet process fluid temperature for the next block in the
top row factng the airstream. By continuing the calculations, all the
blocks along the top tube in the first row are evaluated. The
evaluation of the next tube in the first row"proceeds in a similar
fashion until all the tubes in the first row have been evaluated. The
next row of tubes can now be evaluated using the outlet air conditions
of the previous row as the inlet conditions for the current row. Since
the tubes are packed in a staggered array the inlet air conditions of a
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g]ven block are taken as the average of the two blocks

immediately in front of the bloc~ under evaluation.

As soon as all the blocks have been evaluated the average outlet
cooling water temperature can be determined. If the cooling water is
recirculated and the chosen inlet temperature of the recirculating '
water differs from the outlet temperature, a new value of inlet
temperature of the recirculating water is chosen and the whole
calculation is repeated l!ntil_the inlet and outlet recirculating_wat_g_r
temperatures are the same, giving the operating point of the cooler.

Interval halving could be used to determine the correct recirculating

water inlet temperature, but it was found that the number of iterations
could .be cut dramatically by using a modified interval halving
technique.

The modified technique is compared to the standard interval

halving technique in figure 5.4.

TL1 Twil Twol

I TL'
2

Figure 5.4 Graphical representation of the modified interval halving

procedure.

TL signifies a left boundary and TR a right-hand boundary. By using

the conventional interval halving procedure the first chosen inlet
temperature is Twil which.results in an outlet recirculating water
temperature of Twol· The conventional interval halving method uses TL2
as the new left-hand boundary while the modified interval halving
procedure uses Twol,as the new left-hand boundary. The same holds for
the right-hand boundary. It has been found that the outlet temperature
of the recirculating water after the first choice of inlet temperature
 Stellenbosch University
5.9 http://scholar.sun.ac.za

is very close to the final value which can be obtained by more

iterations. Since a low choice of inlet temperature would result in a
low outlet temperature and vice versa the operating point will always
be between the left-hand and right-hand boundaries even if the modified
interval halving method is used.

The modified interval halving method typically requires less than half
the number of iterations that would be required by the
conventional interval halving technique.

A simple flowchart showing the calculation procedure for the single

pass, top-to-bottom and front-to-back process fluid flow patterns is
showh in figure 5.5.

The evaluation of a cross-flow evaporative cooler with a back-to-front

process. fluid flow pattern is slightly more complicated than that of
the other three patterns since there is no element of which all the
inlet conditions are known even aft~r an initial choice of
recirculating water inlet temperature.

The solution is obtained by choosing the process fluid outlet

temperature for each of the elements in the first row facing the
airstream. By following a similar solution procedure as described for
the front-to-back flow pattern the process fluid inlet temperature can
be calculated. If the calculated average process fluid inlet
temperature differs from the specified inlet temperature or if the
calculated inlet temperature values very significantly from the average
calculated value, the chosen outlet temperature of the process fluid of
each horizontal row is changed by half the difference between the
specified and the calculated process fluid inlet temperature in the
given horizontal row.

Once the calculated and specified process fluid inlet temperatures are
equal the average inlet and outlet recirculating water temperatures are
compared and the inlet recirculating water temperature is adjusted
 5. 10 http://scholar.sun.ac.za
Stellenbosch University

B~IN

FIEAD IN
CODLEFI
DIMENSIONS
AND OPERATING
PAFIAMETEFIS

CHOOSE TWI

EVALUATE CODLEFI
FFIOM TOP ELEMENT AT
THE AIFI INLET SIDE.
CALCULATE TWD.TPD
ETC. USING EITHER
THE MEFIKEL. IMPROVED
MERKEL DR POPPE
MODEL

PFIINT FIESULTS

STOP

Figure 5.5 Program logic used in the determination of the operating point
of a cross-flow evaporative cooler with either a straight
through, top-to-bottom or a front-to-back process fluid flow
arrangement.
 Stellenbosch University
5.11http://scholar.sun.ac.za

accordingly as discussed above. The two iterative procedures for

recirculating water inlet temperature and the process fluid,outlet
temperature are repeated until the solution is found.

The modified interval halving temperature is again employed in the

determination of the recirculating water inlet temperature.

Figure 5.6 shows the calculation procedure for an evaporative

cooler with a back-to-front process fluid flow pattern.

If the cooling water flowing over the tubes is not recirculated, the
cooling water inlet temperature has to be specified and no iterative
solution method would be needed in determining the operating point
except in the case of back to front process fluid flow where the
iterative solution method for determining the process fluid outlet
temperature would still be needed.

Typical temperature profiles as determined with the program CROSS for

the different arrangements are shown in figures 5.7)a) to 5.7)c).

The performance of a given cross-flow evaporative cooler using a front-

to-back process fluid flow pattern would be very similar to the
performance of the same cooler with a back-to-front process fluid flow
pattern. The additional computer time needed for the evaluation of the
back-to-front flow case is often not justified by the improvement in
accuracy obtained by using the back-to-front algorithm instead of the
simpler front-to-back algorithm.

Apppendix K shows the results of the program CROSS for a few example
calculations to compare the different flow patterns and
the analytical models.
 Stellenbosch University
5.12http://scholar.sun.ac.za

ISI!lllN

READ IN
COOL.ER
DIMENSIONS
AND OPERATINIJ
PARAMETERS

CHOOSE Till

CNOOSE TPO

!VAL.UATE COOL.ER
FROM TOP EL.EMENT AT
THE AIR INL.ET SIDE. CORRECT TPO
CAL.CUL.ATE TWO.TPI (GUESS AN
ETC. USING EITHER OUTL.ET
THE MERKEL.. IMPROVED TEMPERATURE
MERKEL. OR POPPE PROFlL.EJ
MODEL.

NO

NO

STOP

Figure 5,6 Program logic used in the determination of the operating point
of a cross-flow evaporative cooler with a back-to-front process
fluid flow arrangement.
 Stellenbosch University
5.13 http://scholar.sun.ac.za

1 1\
Top I ·,
~
' \\
~ I
I
I
~
'
I \
~ '
I
\
I

6 -~

1-
I

'
I
I
\
I
l
I
I i
~
.c ~
I
l
....cur:n
.c
s..
cu
~

~ l!'
,.... 11 -~ j
0
- I

I
0
u
r:n
c:
- !I
0
,.... -
"'
c:
~ I
I

0
16 -~ I
~ J
1- !
V) i
0
c.. 1- i!
1- l
I
1- i
21 -- Ii
- ~
i .
i
-
- ii
~
i
j
I I I I I
26
30 35 40 45 50 55
Temperature [OC]
---- Tp (air inlet side) - - Tp (air outlet side)
·---·--- Tw (air in 1et side) ----- Tw (air outlet side)

Figure 5.7 a) Temperature profiles along the height of a cross-flow

evaporative cooler with a front-to-back process fluid flow
arrangement.
 Stellenbosch University
5014 http://scholar.sun.ac.za

1 Top ~ .
""-:--,

'\,'\I \
6
~
.c
....en J )
I /
Q)
.c
~

- 11
Q)

0
0
u i
i
:
•

I !
en
s::

-0
IU

s::
....
0 16
....
~

VI
0
Q.

i
;
i
I ,'
,
!
,
I '
21
I ,'
! :

26
I
f
I I!
,'

30 35 40 45 50 55
Temperature [°C]

--Tp (air inlet side) -Tp (air outlet side)

-··-··--· Tw (air in 1et side) ----- Tw (air outlet side)

Figure 5.7 b) Temperature profiles along the height of a cross-flow

evaporative cooler with a top-to-bottom process fluid flow
arrangement.
 Stellenbosch University http://scholar.sun.ac.za
5.15

1 Top

....cu
..c

....~
0
11
0
u
C')
c
....
0
IU
c
.:= 16
....
VI
0
0..

21

90 95 100 105 110 115 120

Air Enthalpy, [kJ/kg]

- FTB - BTF

·-·-··-··· TTB

Figure 5.7 c) Outlet air enthalpy profiles along the height of a cross-
flow evaporative cooler.
 s.
Stellenbosch University
16 http://scholar.sun.ac.za

5.2.2 Counterflow evaporative cooler simulation

In the solution of a counterflow evaporative cooler a two dimensional

model is used. The actual cooler consists of a number of similar
'
vertical elements alongside each other. It is only necessary to
analyse one of these vertical rows. The integration commences at the
'
bottom of the cooler since the air properties are fully defined at the
bottom of the cooler.

The solution of a counterflow evaporative cooler with a bottom-to-top

process fluid flow pattern starts by choosing an average outlet
~-

recirculating water temperature and a value for the outlet

recirculating water massflow rate (a given fraction of the
recirculating water evaporate into the airstream).
Through an iterative numerical integration procedure the recirculating
water outlet temperature and the recirculating water outlet massflow
rate are determined when the inlet conditions are satisfied.

If more than one block is chosen along the length of the cooler the
solution is further complicated by the fact that a different
recirculating water outlet temperature has to be selected for each
block to ensure that the calculated recirculating water inlet
temperatures are constant along the top tube.

The solution of a counterflow evaporative cooler is very sensitive to

the choice of outlet recirculating water temperature and double
precision variables are essential to obtain a solution.

If the counterflow evaporative cooler uses a top-to-bottom process

fluid flow pattern the solution would be even further complicated by
the fact that the outlet process fluid temperature has to be selected
and corfected after every integration through the cooler to ensure that
the calculated inlet temperature of the process fluid corresponds to
the specified value at the operating point of the cooler. The
iteration procedure for the evaluation of counterflow evaporative
 Stellenbosch University
5.17 http://scholar.sun.ac.za

coolers is shown in figures 5.8 and 5.9.

If the cooling water is not recirculated the solution of the cooler is

still fairly complicated since the variation in the outlet temperature
of the cooling water along the tube length must still be such that a
uniform cooling water inlet temperature corresponding to the specified
value is obtained at the top .of the cooler.

The results of a few sample calculations using the counterflow

evaporative cooler simulation program COUNTER are presented in
Appendix K.

Various temperature and enthalpy profiles as determined with COUNTER

are shown in figures S.lla) to S.ll)c). It can be noted that the
variation of recirculating water temperature along the outlet of the
cooler is so insignificant that a one-dimensional model could be used
without the loss of accuracy.

5.2.3 Combination cooler

A combination cooler is a counterflow evaporative cooler which employs

a section of conventional Munters-type cooling tower packing either
above or below the bare tube coil. The packing provides a large.
surface for mass transfer and this results in a lower average
recirculating water temperature.

The program COMBINE uses sections of the COUNTER program but several
simplific~tions hav~ been introduced to allow the practical use of the

program on a personal computer.

The simplifications include the following:
i) Only the Merkel solution method can be used,
ii) The model is one-dimensional and
iii) Only recirculating cooling water can be used.
 5. 18
Stellenbosch University http://scholar.sun.ac.za

RUO JH
COOL.~
OlME.NSIDNS
.t.NO OP!:R4TlN&
PARANI!iERS

CHOOSe: TPO

CHOOSE MWO

CHOOSE TWO

!VALUATE CDDI.~
~OM THE AIR INLeT
SIDE. CA~CIJU.TE
TWI. TPt. HWI ETC.
USING £ITHEA TH~
MERKEL. IMPROVED
MeRKEL DR POPPI!
MODEL

Figure 5.8 Program logic used in the determination of the operating point
of a counterflow evaporative cooler with a top-to-bottom (TTB)
process fluid flow arrangement.
 Stellenbosch University
5.19 http://scholar.sun.ac.za

III!IUN

AI!AD IN
CDDI..I!A
DIMI!NSIDNS
AND OPEAATINIII
PARANI!TI!RS

OIODSI! MWD

CHDDSI! TWD

EVAI..UATE CODI..EA
FADM THE AIR INI..ET
SIDE. CAI..CUUTE
TWI.TPD.MWI ETC.
USING EITHER THI!
MERKEl... IMPROVED
MERKEl. OF< POPPE
HODEl.

STOP

Figure 5.9 Program logic used in the determination of the operating point
of a counterflow evaporative cooler with a bottom-to-top (BTT)
process fluid flow arrangement.
 Stellenbosch University
5.20 http://scholar.sun.ac.za

The simplifications had to be made to lower program size,

storage needs and execution time.

The following features have been introduced in the program

to simplify counterflow evaporative cooler calculations:

i) Cooler selection or rating calculations can be performed,

.ii) The cooler can be evaluate as a bare tube unit,
iii) Conventional fill can be used above or below the bare tube coil,
iv) Bottom-to-top process fluid flOLO layout calculations
can be performed as a first approximation (Bottom-to-top
calculations are much faster than the top-to-bottom calculations.)

The evaluation of the cooler proceeds in a very similar fashion to that

of the COUNTER program discussed under section 3.2.2 except that the
recirculating water cooling through the packing has to be considered.
If the packing is placed above the tube section the temperature of the
water f~lling on the packing has to be equal to the temperature of the
water leaying the tubes. If the packing is placed below the coil the
recirculating water entering from above the coil must have the same
temperature as the water leaving the packing. Figure 5.10 shows the
evaluation algorithm used to evaluate a combination cooler. The
integration procedure through the packing is described in Appendix J.

It has been found that the addition of a section of fill material can
lead to a significant decrease in the number of tube rows required to
exchange a given amount of heat. Typical numerical examples are shown
in Appendix K.

Figures S.ll)a) to S.ll)c) show the temperature and enthalpy profiles

through a typical combination cooler.
 5.21 http://scholar.sun.ac.za
Stellenbosch University

I!II!IUN

READ IN
CDDL.I!R
DIMENSIONS
AND DPERATIMB
PIIAAMETERS

CHDDSI! TWO

TTB
CHDDSI! TPD

EVAL.UATI! CDDL.ER EVAL.UATE THE CDDL.ER

FROM THI! AIR INL.ET FROM THE AIR INL.ET
SIDE. CAL.CUL.ATE SIDE. CAL.CUL.ATE
Tlfi.TPD ETC. USING TMI. TPI ETC. USING
THE MERKEL. MODEL.. THE MERKEL. MODEL. •
INTEGRATE THROUGH INTEGRATE THROUGH
THE PACKING IF THE PACKING IF
REGUIRED. REGUIRED.

STOP

Figure 5.10 Program logic used in the determination of the operating point
of a counterflow evaporative cooler with conventional cooling
tower fill placed above or below the tubes.
 Stellenbosch University
5.22 http://scholar.sun.ac.za

1 Top
Packing section

....cu
..c
s..
....cu
0
0
u· 11 I
C'l I
c: I

....
0
IU
I
I
I Tube section
I
c: I

....
0
I
I

....
II)
I
I
I
I
0
c..
16

21
35 40 45 50 55
Temperature. [°C]

- - Tp (with packing)
. - Tw (with packing)
·-···-····-Tp (without packing) ----- Tw (without packing)

Figure 5.11 a) Temperature profiles along the height of a counterflow

evaporative cooler with a BTT process fluid flow
arrangement and conventional cooling tower packing placed
above the tubes.
 Stellenbosch University http://scholar.sun.ac.za
5.23

1 Top I Packing section

I
I
I

i
I
I
I
l
~
6 I
.=.
en
.,....
cv
.=.
I
I
I
s.. !
cv
,....
0
0
I
1
u !
I
en !
c: 11
0
,.... ··...
···...
cu ~
Tube section
\ .. I
c: ·.\. I
0 I
.,....
~ \\ ' '
.,....
VI
\ ''
0 '
Q. I I '

16
\ l I'
'
I '

li I
I

'
I
I ''
I I '
I
I
I

21
/
35 40 45 50 55
Temperature, [°C]
-Tp (with packing) ··~··~~~·· Tw (with packing)
-----Tp (without packing) -Tw (without packing)

Figure 5.11 b) Temperature profiles along the height of a counterflow

evaporative cooler with a TTB process fluid flow
arrangement and conventional cooling tower packing placed
above the tubes.
 Stellenbosch University
5.24 http://scholar.sun.ac.za

1
Top Packing section

~
.s:
C)
.,_ 11
C1)
.s:
Tube section
~
C1) I
..- I
I
0 I
0 I
u I

C) 16 , I

,,
s:::: I
0
..- ,
n:l
,,
s::::
0
, ,·
.,_
,,
,'
~ ,
.,_
V'l
21
0
0..

26

31
60 80 100 120 140 160 180
Air Enthalpy, [kJ/kg]
--ia {packing above) ······-··- ia {no packing)
----- ia {packing below) ·

Figure 5.11 c) Enthalpy profiles along the height of a counterflow

evaporative cooler with a TTB process fluid flow
arrangement.
 -----------------------.

Stellenbosch University
5.25 http://scholar.sun.ac.za

5.3 Simplified models

Four programs, SCOUNT, SCROSS, CSCOUNT and CSROSS using the simplified
analytical modelling procedure for evaluating evaporative coolers or
condensers have been written.

All four of these programs can be used for cooler or condenser rating
and sizing calculations, and the relative fast execution time allows
for the easy adaptation of these programs for execution on a personal
computer.

The iterative procedures for the rating and selection of evaporative

coolers are discussed in detail in section 3.1.4 and the rating and
selection procedures for evaporative condensers are discussed in detail
in section 3.2.4.

In the case of counterflow evaporative coolers the simplified model

yields results which are within 1% of the results obtained with the two
dimensional successive calculation numerical integration procedure.

The simplified model gives results which agree fairly well with the
results obtained with numerical integration model in cross-flow
evaporative coolers with relative sh~rt tube lengths. The discrepancy
in results at longer tube lengths is due to the three dimensional
nature of the recirculating water temperature profile, which cannot be
represented well enough by a single representative temperature.

The four ~ifferent process fluid flow patterns considered in the

numerical integration analysis of cross-flow evaporative coolers can be
compared to the results of the simplified method if the correct process
fluid velocity for the chosen process fluid flow pattern is used in the
calculation of the heat transfer coefficient on the inside of the
tubes.
 Stellenbosch University
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The simplified methods allow the easy evaluation of the effect of

serpentining on both cross-flow and counterflow evaporative coolers
since no complicated new integration procedure has to be used for
higher order serpentining. In order to evaluate a given cooler layout
with second order serpentining only the flow velocity of the process
fluid has to be doubled when calculating the heat
transfer coefficient on the inside of the tubes.

Refer to Appendix K for a comparison between the results obtained

with the simplified and accurate models.

The iterative selection and rating procedures for evaporative

condensers are discussed in detail in section 3.2.4. The simplified
method is expected to yield very good results in the counterflow model
since the recirculating water temperature in an evaporative condenser
would be almost constant because of the constant condensing
temperature.

It is expected that the simplified cross-flow evaporative condenser

simulation would yield fairly accurate results because of· the relative
flat recirculating water temperature profile which would be
prevailing in an evaporative condenser with a constant condensing
temperature.

Refer to Appendix K for some typical results which have been

obtained with the evaporative condenser simulation programs.

5.4 Natural draft cooling tower

The accurate numerical integration routines used in the program CROSS

for the simulation of a cross-flow evaporative cooler have been linked
to the natural draft equation for a cross-flow tower for the evaluation
of a cross-flow evaporative cooling tower.
 Stellenbosch University
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The proposed tower (shown in figure E.l) consists of large cross-flow

evaporative cooler modules placed around the outer perimeter of a
cooling tower shell. The cross-flow evaporative cooler modules may be
arranged in an A-frame configuration to obtain a larger surface area
without enlarging the tower base diameter too much.

In order to keep the process fluid velocity within allowable limits the
front-to-back or the back-to-front process fluid flow patterns are
normally employed. As mentioned before the relatively long execution
time of the back-to-front process fluid flow pattern compared to that
of the front-to-back flow pattern does not justify its use since there
is very little difference in the cooler capacities obtained with these
two flow patterns.

The draft equation for a typical cross-flow cooling tower is derived in

detail in Appendix E. The pressure drop coefficients are also
discussed in Appendix E, except for the pressure drop across the wet
tube bundle which is calculated from the cor~elation given by Collier··
[79COl]; The correlation of Collier is discussed in Chapter 4.

In analyzing a natural draft cooling tower an air massflow rate is

chosen, the cooler units evaluated, the total pressure drop
through the tower is computed and the available pressure difference
determined.

If the initial air massflow rate was chosen correctly the pressure drop
through the tower would be exactly balanced by the available draft, but
if the pressure drop is not matched by the available draft a new air
massflow rate has to be selected and the whole calculation process must
be repeated.

The calculating procedure is shown in figure 5.12.

Typical results of the natural draft tower simulation program is

presented in Appendix J. The air leaving the cooler units around the
 Stellenbosch University
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BEGIN

RI!AO IN TOWER
OIMI!NSION5
AND OPERATING
PARAMETERS

CHOOSE IN~ET
AIR MASSF~OW
RATE

I!VA~UATE COOLER
UNIT FROM THE AIR
IN~ET SIDE.
CA~CULATE TAO.TPO
ETC. USING EITHER
THE MERKEL. IMPROVED
MERKa DR POPPE
MOOEL.

CAL.CIJL.A TE THE
AVAI~AB~E PRESSURE
DIFFERENCE ACROSS
THE TOWER AND THE
PRESSURE DROP
THROUGH THE TOWER.

STOP

Figure 5.12 Program logic used in the determination of the operating point
of a natural draft cooling tower with cross-flow evaporative
cooling units placed around the outer perimeter of the tower.
 Stellenbosch University
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base has been found to be almost saturated, if not fully saturated, and
consequently the assumption of saturated outlet air to.determine the
outlet air density when employing the Merkel type analytical model is
normally a good assumption. The exact method does not employ this
assumption since the air properties are fixed at every part of the
cooler, therefore it is generally expected that the exact model will
yield more accurate results.
 Stellenbosch University
6.1 http://scholar.sun.ac.za

CHAPTER 6

EXPERIMENTAL DETERMINATION OF THE HEAT AND MASS TRANSFER COEFFICIENTS

IN A CROSS-FLOW EVAPORATIVE COOLER

No accurate data or correlations for the determination of the heat and

mass transfer coefficients in a cross-flow evaporative cooler or
condenser could be found in the literature. These coefficients can
only be found experimentally since the heat/mass transfer analogy
cannot be applied to an evaporative cooler because of the uncertainty
about the actual air/water interface area. The analogy fails because
of the geometrical dissimilarity of a dry tube bundle and a wet tube
bundle.

Various factors influence the transfer coefficients in an evaporative

cooler including process fluid temperature, air massflow rate,
·recirculating water massflow rate, process water massflow rate, and
inlet air conditions, and tube geometry.

A test tunnel was erected at the Department of Mechanical Engineering

of the University of Stellenbosch in order to conduct a series of
tests on a cross-flow evaporative cooler.

6.1 Description of test tunnel and apparatus

A horizontal tunnel with a 2 x 2 m cross section was built in order to

test wet heat exchanger coils and evaporative coolers.

The tunnel shown in figure 6.1 consists of an inlet section, a test

section, a mixing/measurement section and an induced draft fan. The
tunnel walls downstream of the test section are insulated to minimize
any change in the air temperature between the test section and the air
sampling station. The inlet air temperatures are measured in the
inlet section of the tunnel and the outlet air temperatures are
 Stellenbosch University http://scholar.sun.ac.za

r - - - - - - - - - - - - - - - D r i f t eliminator
r - - - - - - - - Mixers
,.----....,:__ Mesh

.
Cl'

Inlet section Settling chamber Fan section

Test
section '---- Measuring nozzles

Figure 6.1 Layout of the experimental tunnel for the cross-flow

evaporative cooler tests.
 Stellenbosch University
6.3 http://scholar.sun.ac.za

measured downstream of two sets of mixers to ensure good mixed air

temperatures.

The air massflow rate through the tunnel is infinitely variable

through the use of a stepless electronic speed .control device which
governs the speed of the centrifugal fan motor. Air massflow rates of
up to 15 kg/s through the tunnel can be obtained depending on the flow
resistance of the particular test section installed.

Two cross-flow evaporative cooler test sections were built. The first
test section (see figure 6.2) consisted of 250 galvanized steel tubes,
38,1 mm OD and 34,9 mm 10, spaced in a 2 x d0 tri.angular array of ten
vertical rows. The sides of the test section ~a~made of a 13 mm
thick transparent Perspex plate to allow observation of the test
section. Incomplete wetting of the lower tubes facing the airstream
was observed when testing the upright test section at high air
velocities and low recirculating water flow rates.

The second test section was suspended in a frame which pivoted around
the middle of the test section as seen in figure 6.3, this allowed the
test section to be rotated by up to 18,75° from the vertical. Only 22
vertical rows of tubes could be fitted in the rotating test· section in
a 2 x d0 triangular array. The same 38,1 mm 00 and 34,9 mm ID
galvanized steel tubes were used for the inclined and the upright test
sections.

The tubes were connected with flexible rubber hoses in a top-to-bottom

serpentining arrangement. . Drift eliminators were installed
downstream of the test section to prevent entrained water droplets
from travelling down the tunnel in the airstream.

The water in a 40 m3 underground tank was heated to the required

temperature by means of a two-pass oil burning boiler. The hot water
was then pumped from the surface of the tank to the inlet header of
the test section. After flowing through the test section the cooled
 6.4 http://scholar.sun.ac.za
Stellenbosch University

-l~~~V/;.-=-=-=t-=-=~----- Tube array

r - - - - - Air flow

\'---------..J
Figure 6.2 Upright test section layout. ·

r " ' r - + - - - - - - - - Tube array

r-----Air flow

Figure 6.3 Inclined test section layout.

 Stellenbosch University
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process water flowed back into the bottom of the water storage tank,
ensuring a stable process water inlet temperature at the
test section.

The recirculating water was pumped from the sump underneath the test
section to the spray tubes located above the test section. Each of
the tubes in the top row of the test section has a spray tube directly
above it to ensure an even distribution of recirculating water. The
layout of a spray tube is shown in figure 6.4. A spray tube consists
of a horizontal copper tube which has small diameter holes drilled
into the top of the tube along its length. The copper tube is
enclosed in a larger diameter plastic tube. The plastic tube has a
narrow slot machined at the bottom of the tube along its length. The
recirculating water which· is pumped from the sump undefneath the test
section is fed into a header which distributes the water to the copper
tubes of each spray tube. Since the pressure inside the copper tubes
is high, an equal amount of recirculating water is sprayed out of each
hole at the top of the copper tubes. The spraj water strikes the
insi~e of the larger diameter plastic tube and .flows downwards and out

through the slit in the bottom of the plastic tube and onto the top
tubes in the test section.

Special care was.taken to prevent the air stream from short circuiting
the test section. Galvanized plates were suspended underneath the
bottom row of tubes in the test section. The ends of the plates hung
in the.water in the recirculating water sump effectively stopping the
air from short circuiting underneath the test section.
Flat galvanized steel plates were placed on top of the spray tubes to
prevent short circuiting of the air through the gaps between the spray
tubes.
In the evaporative cooler tests the following quantities had to be
measured:

i) Process water - massflow rate, inlet and outlet

temperature
 Stellenbosch University
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ii) Recirculating water - massflow rate, inlet and outlet

temperatures
iii) Air - massflow rate, inlet and outlet
temperatures (wet bulb and dry bulb)
iv) Other - Atmospheric pressure

a) Massflow measurements

The process water massflow rate was measured using an orifice plate
placed in the process water supply line between the hot water tank and
the test section. The orifice plate was made and installed according
to the BS-1042 standard with pressure tappings at a distance equal to
one tube diameter upstream of the orifice plate and half a tube
diameter downstream of the orifice plate.

The pressure difference ·across the orifice plate was recorded with two
Foxboro differential pressure transducers. These two transducers
covered different pressure ranges and this allowed a wide massflow
range to be measured without having to change the orific~ plate. The
4 - 20 rnA signal delivered by the Foxboro pressure transducers were
converted to a voltage signal (between 1 and 5 V) by passing the
current through a high precision 250 ohm resistor. The pressure
transducers were calibrated by using a zero differential pressure
signal as the low range calibration point and a known pressure
difference near the pressure transducer full scale position as the
high range calibration point. The calibration of the transducers were
checked using a weighing drum and a stopwatch.

Two instruments were installed to measure the recirculating water

massflow rate i.e. a rotameter for measuring the low massflow rates
and an orifice plate for measuring the higher recirculating water
massflow rates. The recirculating water orifice plate was made using
the same BS-1042 standard as for the process water orifice plates.
The differential pressure across the recirculating water orifice plate
was also obtained with a calibrated Foxboro pressure transducer.
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Since the recirculating water orifice plate was installed at a

distance of about 20 diameters from the pump .it was deemed necessary
to install a flow straightener immediately downstream of the pump.
The straightener was also made according to the BS-1042 standard.

The rotameter which was installed can measure a water massflow rate of
up to 3,33 kg/s on a linear scale from 0 to 25. The rotameter ·was
consequently calibrated by using a stopwatch and a weighing drum and
a simple second order polynomial curve was fitted to the data and this
curve was then used as the calibration curve for the rotameter.

The air massflow rate was determined from the differential pressure
measured across the air measuring nozzles in the test tunnel (see
figure 6.1). The five elliptical nozzles were made according to the
ASHRAE 51 - 75 standard. The differential pressure readings across
the nozzles were taken with a calibrated low pressure Foxboro
transducer. As in the use of the other pressure transducers the
current signal of the transducer was converted to a voltage reading
through the use of a precision resistor. At low air massflow rates
one or more of the nozzles were closed up to give higher differential
pressure readings to ensure more accurate massflow determination.

The difference in pressure between the atmosphere and the pressure

inside the tunnel upstream of the nozzles was recorded for every test
and this value was subtracted from the atmospheric pressure in
calculating the density of the air entering the nozzles.

b) Temperature readings

The temperature readings were all made with calibrated copper-

constantan thermocouples. The thermocouples were calibrated by
determining the thermocouple readings at ice melting point at water
boiling point at atmospheric pressure. The calibration values were
then used to correct every temperature reading taken with each
thermocouple.
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The inlet·and outlet temperatures of the process water was measured

with two calibrated thermocouples in both the inlet and
outlet process water manifolds.

The inlet recirculating water temperature was measured with two

calibrated thermocouples placed in the recirculating water inlet
header. Special thermocouple probes were made to measure the bulk
temperature of the water film flowing over a tube. By using these
probes the average temperature of the recirculating water leaving the
coil could be determined.
The probes were made from a short piece of cylindrical Perspex with
holes drilled axially and radially into it as shown by figure 6.5. A
3 mm thermocouple fitted snugly into the axial·hole with the tip of
the thermocouple just visible through the radial holes. The
required dimensions of the radial holes were determined by a trial and
error method. If a temperature probe is held under a tube with the
Perspex jus~ touching the water film flowing over the tube, the
surface tension draws the water into the larger of the radial .holes at
the top of the probe. The water drops out of the bottom hole
continuously wetting the tip of the thermocouple. Ten thermocouples
were fitted with these probes and installed under every second tube in
the bottom tube layer on either side of the test section.

For energy balance calculations it is necessary to determine the

average temperature of recirculating water leaving the bottom for of
the cooler. The average of the ten film temperature probe readings
could be used as the average outlet recirculating water temperature,
but since the possibility of unequal recirculating water distribution
among the ten vertical tube rows exists, a temperature measuring
trough was installed under the tube bank. The trough collects the
water leaving all ten vertical rows and the mixed temperature in this
trough could be taken taken as the average recirculating water outlet
temperature.
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--1¥'1 I I Ll I ILl I ILl I I Ll I I rn I ILl I I,~ ILl I I L W . f -

-----Plastic tube
I
~~--Small diameter holes
I I . . .. . ... .. .. .. . . . . . I
·····················
\ I I I

---------------
.... ::::::::.·::::::::::::::::::::::.·:::....

I
'----- Narrow s1ot

Figure 6.4 Spray tube layout.

I ____J I I
3

a-~--~-J_
j_~·-, _..(ihj

~12w
, - - - - - - - - Perspex cylinder
,----- Thermocoup 1e

Figure 6.5 Recirculating water bulk temperature measuring probes.

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The air wet bulb and dry bulb temperature were measured upstream and
downstream of the test section. The wet bulb temperature readings
were taken using a simple sampling tube as shown in figure 6.6. The
wet bulb thermocouples were kept wet by a small cotton sleeve which
was pulled over the tip of the thermocouple while the other end of the
sleeve was suspended in a small water reservoir to keep it wet. To
ensure that the correct wet bulb temperature would be read with the
wetted sleeve thermocouple, a small fan was installed to draw the air
through the air sampling tube at between 3 and 5 m/s. In order to
read the average air temperature, five wet bulb and five dry bulb
thermocouples were installed at each air sampling point.

c) Other measurements

·The barometric pressure was recorded with a mercury column barometer

before every test. The two phase pressure drop across the wet tube
bundle was measured with a Betz manometer. At the upstream side of
the test section the walls of the tunnel always remained dry and
conventional pressure tappings could be installed. Downstream of the
test section the walls of the tunnel were wet because of splashing and
drop entrainment.
Special pressure tappings were needed to measure the static pressure
inside the tunnel in the presence of water on the inside wall, since
any water trapped in the pressure lines would result in faulty
pressure measurements. The downstream pressure tappings were
constructed from a copper tube with a relatively large diameter
installed flush with the inside wall of the tunnel as shown in figure
6.7. The copper tube enters the wall at goo but it is then bent
.
upwards to prevent any water from flowing down the pressure lines .
The relative large diameter of the copper tube ensured that water
drops do not close off the whole cross section of the tube as would be
the case with a small diameter tube.
 6 11 http://scholar.sun.ac.za
Stellenbosch University 0

, - - - - - - - - - - - - - - - D r y bulb thermocouple
, - - - - - - - - - - - - W e t bulb thermocouple
~--Air inlet

Water container
······•· . . .
·::::::t·::: ::
............
.... .. ... ....
..........
::::::.r:::. ::.
······!······
...... ········
............
... .... ...... .

Figure 6.6 Air sampling probe.

'

Copper tube

Figure 6.7 Layout of pressure tappings for static pressure readings on a

wet wall.
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6.2 Data logging and energy balance calculations

The data logging was performed using a Kayes Digilink 4 data logger
linked to an Olivetti M21 personal computer. The data logging system
layout is shown in figure 6.8.

transducers

Figure 6.8 Block diagram showing the data logging system layout.

The thermocouples were all directly connected to the Digilink, the

internal electronic ice point of the Digilink eliminating the need for
an ice- bath. The Digilink was programmed to convert all the
temperature readings to degrees Centigrade before transferring them
to the computer.

The pressure transducer signals were all converted to voltage signals

which could be measured with the Digilink. The Digilink was
programmed to convert the voltage signal of each pressure_ transducer
·into a pressure reading in Pascal through the use of the transducer
calibration data. The personal computer connected to the Digilink
could then read all the required temperatures and pressures directly
in degrees Centrigrade and Pascal respectively.

A computer program was written in TurboBasic to read all the data from
the Digilink and to perform the necessary energy balance calculations
on the data. The flow chart for the data logging program
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is shown in figure 6.9.

The temperature readings were corrected using the thermocouple

calibration data. The massflow rates were then computed from the
measured pressure differentials across the orifice plates and from the
rotameter reading if the rotameter was used for the recirculating
water massflow measurement. The energy balances were computed with
the following equation:

Energy balance = X 100 %

(6.1)

where

(6.2)

Aqw = mwi cpwi Twi - ( mwi - rna ( wao - wai ) ) cpwo Two (6.3)

Aqa = rna ( i ai - i ao ) (6.4)

The immediate processing of the data made it possible to continue the
tests until a completely steady state was reached. After each data
set was taken graphical displays of temperature and massflow rate
versus time could be displayed to show any fluctuations and
variations~

6.3 Experimental procedure

The following variable parameters have an influence on the performance

of the experimental evaporative cooler:
i) air massflow rate
ii) process water massflow rate
iii) recirculating water massflow rate
iv) inlet temperature of process water
v) inlet air conditions and
vi) the swing angle, 8, of the inclined test section.
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BEGIN

LOG

INITIAL.I:ZE THe
REQUIRED
OPERATING READ OATAI"ILE
PARAMETERS AND
YARIABL.ES.

CORRECT THE
TEMPERATURE
A~D DATA MEASUREMENTS USING
FADM THE THERMOCOUPLE
DIGIL.INK- CALIBRATION DATA.
DATAL.OGGER CAL.CUL.ATE THE
HASSFLDW RATES ANO
THE ENERGY BALANCE.

CORRECT THE
TEMPERATURE
MEASUREMENTS USING
THE THEAMOCDUPL.E DISPLAY THE
CAL.IBAATIDN DATA. RESULTS ON
CAL.CUL.ATE THE THE SCREEN.
MASSFL.OW RATES AND
THE ENEABY BAL.ANCE.

·aiSPL.AY THE SAVE THE

AESUL.TS ON RESULTS ON
THE SCREEN. DISK

SAVE THE
DATA ON DISK.

NO
YES

STDP

Figure 6.9 Program logic used in the data logging program.

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The inlet process water temperatures used for all the tests lie
between 38°C and ssoc which would be the normal operating temperatures
for evaporative coolers. Since the tunnel draws in fresh atmospheric
air the tests were all conducted without any control over the inlet
air conditions.

The swing angle of the inclined test section was always set to ensure
complete wetting of all the tube rows. The water distribution on the
tubes is dependant on the air and the recirculating water massflow
rates which implies that the optimum swing angle is a function of the
air and recirculating water massflow rates.

The only ~arameters which were freely variable ·were the massflow
rates of the air, process water and the recirculating water. The
process· water massflow rate could be varied between 5 and 16 kg/s.
For the upright test section the following massflow ranges were
possible, 1 ~ mw ~ 4 kg/s and 1 ~rna~ 6 kg/s and for the inclined
test section the following massflow ranges were covered: 1 < mw < 7
kg/s and 1 < rna < 12 kg/s.

Upon starting a new test the hot process water was allowed to
circulate through the test section and back to the hot water reservoir
to warm the piping and the test section. A low air massflow rate
through the tunnel ensured that the tunnel walls were sufficiently
warm in order to shorten the time needed to reach a steady operating
condition when the cooler operates as an evaporative cooler. After
about five minutes the recirculating water was started and the mass
'/·_.,...-· /" /

flow set to the required flow rate. The air massflow rate was then
increased to the required flow rate and the make-up line to the
recirculating water·sump was closed off.

The tests were run until the following stabilization criteria were met
i) an energy balance of better than 5% and
ii) the difference between the inlet and outlet recirculating water
temperatures stabilized.
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6.4 Observations and results

As expected it was ·found that the recirculating water flow was dragged
downstream by the cross-flow air at high air velocities which meant
that the bottom tubes facing airstream started to form dry patches.

At free stream air speeds of up to 1,25 m/s (rna z_6 kg/s) the
distribution of recirculating water among the tube rows in the upright
test section was still good. As the air velocity increased the first
few tube tows received less and less recirculating water until they
ran completely dry. The obvious solution to this problem was to swing
the test section through a small angle in order to align each
horizontal tube below and slightly downstream of the previous tube
above it. It was observed that the recirculating water flowed from
one tube to the next in the form of evenly spaced columns or droplets.
It was also noted that a recirculating water column falling from a
tube would adhere to the tube below if it only touched the lower tube.
If the airspeed was just slightly higher the deflected column would
miss the lower tube completely and it would be swept away. This
phenomena is graphically illustrated in figure 6.10. ·

Figure 6.10 Column deflection by cross-flow airstream

a) v~ = 0 m/s b) v~ z 1 m/s c) v~ > 1,5 m/s
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If the lower tube was placed slightly downstream of the upper tube the
water column which would just be swept away in an upright test
section, would strike the lower tube and a good distribution of
recirculating water would still be obtained.

Yung et al. [80YU1] studied the problem of liquid entrainment by a

cross-flow airstream where the liquid falls from one tube to the next
in either a droplet or a column mode. They presented a criterion for
the onset of the column formation and they derived equations for the
calculation of droplet and column deflection due to the gas cross-
flow.

Based on the work of Yung et al. [80YU1] and the current test section
dimensions (Pt = 2 x d0 , Pd = 2 x d0 ), two graphs were plotted to
determine the deflection of the· recirculating water flowing in droplet
and· column modes respectively. According to Yung et al. the liquid
flow in the droplet mode consists of a primary drop and four or five
smaller secondary drops. The smallest drops are obviously swept away
first by the cross-flowing air stream.

From figure 6.11 it can be seen that the smallest drops are swept
away from the lower tube in the upright test section at a free stream
velocity of about 2 m/s, if the test section is inclined at 18° the
smallest drops would be swept away only at a free stream velocity of
3 m/s.

Figure 6.12 shows the deflection of the water flow in the column flow
mode. According to the criterion given by Yung et al. the column mode
starts at r = 95 kg/m/hr for water at 40°C. It can be seen from
figure 6.12 that the maximum allowable free stream velocity (before
the water column is swept away) increases dramatically by inclining
the test section through relatively small angles from the vertical.

As the maximum obtainable free stream air velocity in the tunnel is 3

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7
~
v/
'•

~,..

,....., 6 ,........
v·
v/ V'
VI
g
,__,
5 1--/
./ ~
v
9
>
.. ..../
..,..·
/
~
+o)
.,...
u
0
.....-
4 ...........·
.--
,.,.
1.-·'
/.,.,,.,.

L
v ~
_.,.,.. ~
~

v/ V'
Q,)
>
s..
.,...
"'e 3
--
......
......./·····/ v ./'
V'
"'s..
..··•••··
v v~ .
/: /
Q,)

+o)
VI
. ... ··
Q,)
..
Q,)
s.. 2

fj /
1..1..

0
~
0 1 2 3 4 5 6 7 8 9 10
Droplet size, d [ mm ]
----a= oo ( in-line )
············· a = 18°

Figure 6.11 Maximum allowable freestream air velocity versus droplet size
for droplet deflection for in-line and inclined tubes.
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0

5~--------~----------~--------~--------~

..... 4
VI

~
.......
9
>
>,
..
.....,
....u 3 ..... -·
..·-- ---- _.....
....cv
0

..··..·
..·······
,., -
...
,- -~
> ... ··· _, ...
....s-
tU .·.
.· .·
, ·'
,,
~

.. ,. ,
stU ,. ,
cv 2 , ;
s-
....., ;

VI
cv
"
cv
s-
1.1..

0~--------~~--------~----------~----------~

0 200 400 600 800

Water massflow rate, r [ kg/m/hr ]
--8 = oo ----- 8 = 10°
···-·--·- 8 = 18°

Figure 6.12 Maximum allowable freestream air velocity versus water

massflow rate for column deflection for in-line and inclined
tubes.
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m/s it can be deduced from figures 6.11 and 6.12 that a swing angle of
18,75° should be sufficient to ensure very little water entrainment by
the airstream.

Since there was some uncertainty as to what size and how many of the
small spray holes should be drilled along the top edge of copper tubes -1
in the spray tubes, it was decided to drill 40 holes of 1,5 mm
diameter along the top of each copper tube. These spray tubes were
then used in the tests of the upright test section and this limited
the recirculating water massflow rate to 4 kg/s.

The second test section used tubes spaced in a 2 x d0 triangular array

but the whole tube bundle could be swung through an angle of up to
18,75° from the vertical. Much higher ai~ velocities were possible
without influencing the recirculating water distribution when using
the inclined test section than was possible with the upright
test section.

The holes in the top of the copper tubes inside the spray tubes were
increased to 2,25 mm in diameter and this allowed recirculating water
massflow rates of up to 7 kg/s.

The recirculating water outlet temperatures were measured using the

film bulk temperature probes and the averaging temperature measuring
trough installed under the bottom row of tubes along the middle of the
test section. It was found that the recirculating water temperatures
measured with the trough was always higher than the reading taken with
the film temperature measuring probes. This was probably due to
insufficient mixing of the water in the trough. The temperature
which was measured in the trough was not used in further
calculations.

The results of the tests are presented in table 6.1. Tests 908.1 to
1808.3 were conducted on the upright test section and tests 2610.1 to
411.8 were conducted on the inclined test section.
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Table 6.1 Experimental results of the tests conducted on the inclined and upright test sections.

Test Patm Tpi Tpo Twi Two Taidb Taiwb mp rna mw

908.1 101.690 48.900 45.• 024 43.411 43.720 15.602 10.777 14.285 2.309 3.905
908.2 101.690 46.746 43.394 41.778 42.110 15.223 10.701 14.285 2.323 2.996
908.3 101.690 44.982 42.099 40.396 40.814 16.583 11.550 14.245 2.377 1.603
1508.1 101.098 47.054 40.160 38.562 38.581 19.064 14.760 10.298 4.997 3.897
1508.2 101.098 45.507 39.292 37.645 37.675 20.064 14.671 10.373 5.013 2.996
1508.3 101.098 44.361 38.873 36.904 37.167 20.131 14.728 10.384 5.021 1.603
1508.4 100.584 47.276 42.204 40.254 40.493 19.340 15.348 14.861 4.960 3.902
1508.5 100.584 46.193 41.441 38.959 39.746 18.631 15.040 14.836 4.968 2.996
1608.1 100.566 49.998 42.186 40.406 40.380
39.170
19.266
19.362
14.628
14.485
10.134
10.131
4.800
4.847
3.916
2.996
......
0'
N
1608.2 100.566 47.915 40.896 39.047
1608.3 100.566 45.980 39.888 37.442 38.060 18.910 14.273 10.089 4.904 1.603
1608.4 100.566 43.999 39.719 38.584 38.619 18.281 14.104 9.783 2.448 3.858
1608.5 100.566 43.160 39.166 37.894 38.013 .18.258 13.932 9.808 2.487 2.996
1608.6 100.566 42.054 38.480 37.022 37.315 17.963 13.745 9.774 2.512 1.603
1708.1 100.792 50.024 41.334 39.385 39.328 18.588 14.052 10.470 6.021 3.837
1708.2 100.792 47.709 39.940 37.876 38.002 . 17.964 13.936 10.483 6.063 2.996
1708.3 100.792 45.573 38.908 36.231 36.879 17.553 13.679 10.457 6.135 1.603
1808.1 100.524 52.734 44.467 42.757 42.747 14.959 11.519 9.366 3.482 3.824
1808.2 100.524 50.276 42.900 40.985 41.220 14.992 11.381 9.402 3.553 2.996
1808.3 100.524 47.700 41.391 . 39.120 39.630 14.701 11.226 9.469 3.691 1.603
2610.1 100.919 47.582 39.508 36.747 37.515 23.563 14.967 12.420 8.011 6.861
2610.2 100.919 41.847 35.680 33.213 34.021 22.971 14.724 12.358 8.160 5.824
2610.3 100.919 40.266 34.861 32.466 33.186 22.423 14.579 12.289 8.224 4.653
2610.4 100.919 37.563 33.157 30.612 31.524 21.791 14.345 12.100 8.334 2.996
2610.5 100.581 53.087 39.720 35.863 36.975 11.773 9.773 10.204 10.002 6.927
2610.6 100.581 43.936 35.806 32.424 33.494 12.032 9.724 12.169 10.211 5.807
2610.7 100.581 40.691 33.882 30.641 31.533 11.274 9.355 12.467 10.305 4.828
2610.8 100.583 37.444 32.043 28.557 29.495 10.549 8.911 13.027 10.443 2.996
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Table 6.1 (cont.) Experimental results of the tests conducted on the inclined and upright test sections.

Test Patm Tpi Tpo Twi Two Taidb Taiwb mp rna mw

2710.1 99.645 49.943 42.116 38.422 39.939 20.219 14.670 14.177 7.495 6.865
2710.2 99.645 47.739 40.660 37.083 38.559 19.265 14.232 14.150 7.580 6.078
2710 .. 3 99.645 C) 45.517 39.043 35.372 36.840 19.160 13.870 14.155 8.330 4.614
2710.4 99.645 42.530 37.159 33.472 34.795 17.465 12.755 13.911 8.344 2.996
2710.5 99.915 48.139 39.095 35.506 36.639 13.813 11.450 13.218 9.434 6.790
2710.6 99.915 46.476 38.211 34.591 35.851 14.160 11.544 13.233 9.505 5.916.
2710.7 99.915 45.115 37.680 33.880 35.159 13.783 11.426 13.239 9.585 4.336
2710.8 99.915 42.943 36.608 32.474 33.961 13.722 11.553 13.246 9.688 2.996
2810.1 100.179 47.897 39.173 35.701 36.838 17.470 11.992
11.945
13.308
13.330
9.146
8.938
6.806
5.784
.
0'

2810.2 100.179 45.588 37.962 34.456 35.772 17.432 N

N
2810.3 100.179 43.633 36.912 33.388 34.684 17.754 11.969 13.332 9.129 4.493
2810.4 100.179 42.355 36.488 32.624 33.893 17.985 11.590 13.319 8.945 2.996
2810.5 100.179 49.554 40.047 36.011 37.630 17.330 11.984 13.264 9.116 6.764
2910.1 101.020 52.764 43.028 38.679 40.270 15.283 10.074 15.062 9.039 6.778
2910.2 101.020 49.949 41.601 37.327 39.123 14.907 10.009 15.084 8.712 6.029
2910.3 101.020 47.744 40.552 36.249 38.109 15.087 10.314 15.032 8.600 4.614
2910.4 101.020 46.053 39.872 35.220 36.856 14.991 10.347 15.033 8.567 2.996
2910.5 101.020 43.083 36.993 33.079 34.686 13.659 9.674 13.820 8.591 3.609
3010.1 100.438 53.889 43.559 39.227 40.617 18.372 12.891 14.852 9.700 6.769
3010.2 100.438 50.774 41.966 37.780 39.344 18.234 13.022 14.806 9.330 5.908
3010.3 100.438 48.785 40.916 36.710 38.249 17.672 12.773 14.801 9.418 4.742
3010.4 100.438 45.843 39.650 35.300 36.775 17.523 12.836 14.816 9.012 2.996
3010.5 100.438 44.344 38.326 34.291 35.927 16.924 12.536 14.804 9.014 3.610
3110.1 100.290 55.714 44.766 41.385 42.467 18.002 15.140 13.177 8.098 6.737
3110.2 100.290 53.014 43.358 39.748 40.894 17.698 14.857 13.186 8.226 5.798
3110.3 100.290 50.257 41.663 38.149 39.284 17.414 14.404 13.164 8.329 5.143
3110.4 100.290 47.725 40.152 36.815 37.861 17.420 14.767 13.183 8.405 4 .• 458
3110.5 100.290 45.858 39.103 35.688 36.785 16.519 14.219 13.178 8.424 3.610
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Table 6.1 ·(cont.) Experimental results of the tests conducted on the inclined and upright test sections.

Test Patm Tpi Tpo Twi Two Taidb Taiwb mp rna mw

3110.6 100.290 44.720 38.688 34.895 36.036 17.643 14.697 13.174 8.482 2.485
111.1 100.262 52.173 44.284 42.061 42.330 25.168 15.669 13.633 5.898 6.782
111.2 100.262 50.291 43.173 40.914 41.316 25_.481 15.588 13.650 5.969 5.970
111.3 100.262 48.443 42.064 39.725 40.151 24.476 15.149 13.674 6.069 4.952
111.4 100.262 46.364 40.730 38.401 38.854 24.377 15.179 13.674 6.079 3.935
111.5 100.262 44.069 39.061 36.900 37.274 23.021 14.703 13.630 6.141 3.610
111.6 100.262 42.560 37.492 35.768 36.050 23.167 14.823 13.638 6.092 5.957
111.7 100.262 41.775 36.850
39.095
35.267
37.120
35.524
37.275
23.039
12.652
14.886
10.376
13.659
12.925
6.170
6.926
6.643
6.596
.
211.1 100.189 46.369 N
w
211.2 100.189 44.156 37.644 35.730 35.870 12.532 10.484 12.910 7.005 5.956
211.3 100.189 43.026 36.907 34.934 35.174 12.395 10.413 12.900 7.013 5.310
211.4 100.189 41.779 36.091 34.099 34.377 12.612 10.572 12.922 7.051 4.906
311.1 100.874 52.728 44.045 41.575 41.930 16.847 11.558 13.852 6.531 6.795
311.2 100.874 50.625 42.843 40.297 40.695 16.566 10.940 13.806 6.609 5.530
311.3 100.874 47.377 40.937 38.306 38.757 .15.870 11.186 13.801 6.747 4.005
311.4 100.874 45.163 39.361 36.810 37.223 16.918 11.537 13.784 6.801 3.610
411.1 100.710 52.419 43.927 41.375 41.614 16.437 11.777 14.970 7.144 6.774
411.2 100.710 50.419 42.711 40.236 40.439 16.495 11.461 14.957 7.282 5.795
411.3 100.710 48.072 41.141 38.587 38.959 17.169 11.870 14.926 7.365 5.087
411.4 100.710 47.141 40.754 38.121 38.425 "17.347 11.417 14.945 7.457 3.974
411.5 100.710 45.961 40.225 37.375 37.739 17.859 12.122 14.864 7.537 2.996
411.6 100.710 44.146 37.969 36.053 \ 36.082 17.757 11.913 14.853 7.567 6.793
411.7 100.710 42.613 37.095 35.122 35.250 18.317 12.273 14.737 7.633 5.574
411.8 100.710 40.559 35.829 33.748 33.964 19.093 12.803 14.472 7.757 4.192
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The critical cross-flow velocity of the air was determined

experimentally at various inclination angles of the test section. The
experimental values are compared to the theoretical values based on
the model by by Yung et al. [80YU1] in table 6.2 It can be seen that
the theoretical maximum allowable cross-flow velocity is up to two
times as as high as the experimental maximum. It should however be
noted that it was difficult to determine accurately when the water
falling from one tube to the next would just miss the lower tube since
the water columns oscillated back and forth quite significantly.

Table 6.2 Experimentally determined critical cross-flow air velocity

compared with theoretical values.

Angle mw Vcrit Vcrit Vcrit Vcrit,theo

[kg/s] based on based on observed
droplet column experimentally Vcrit,exp
deflection deflection
[m/s] [m/s]
oo 1'0 2,08 1,50 0,80 1'9
4,0 2,08 2,12 1,00 2,1
7,0 2,08 2,44 1,10 1'9
so 1'0 2,36 1,73 1,31 1'3
4,0 2,36 2,45 1,48 1'6
7,0 2,36 2,81 1,53 1,5
10° 1,0 2,63 1,97 1,46 1'4
4,0 2,63 2,79 1,56 1,7
7,0 2,63 3,21 1,70 1,5
15° 1,0 2,90 2,23 1,70 1,3
4,0 2,90 3' 16 1,80 1'6
7,0 2,90 3,63 1,95 1,5
18,75° 1'0 3,11 2,45 2,04 1,2
4,0 3,11 3,47 2,16 1'4
7,0 3,11 3,99 2,40 1,3

The single phase pressure drop was measured across the tube bank and
the results are tabulated in table 6.3 · The pressure drop measurements
were all taken across the movable test section. The upstream pressure
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tappings were placed in a section of the tunnel where the cross

section area was 4m2, but the downstream readings were taken behind
the test section where the cross section area of the tunnel was
3,429m2. The measured pressure drop values were corrected to take the
contraction of the flow into account.

Table 6.3 Measured pressure drop across the dry tube bundle.

rna dPa APa(corr)

9.55 41.00 39.82

9.43 41.50 40.35
9.51 41.50 40.33
8.86 37.00 35.98
8.94 37.00 35.96
8.96 37.50 36.46
8.45 33.50 32.58
8.40 33.50 32.59
7.77 28.50 27.72
7.76 28.80 28.02
7.12 24.80 24.14
7.18 25.00 24.33
6. 72 21.40 20.81
6.62 21.00 20.43
6.59 21.00 20.44
5.94 18.00 17.54
5.46 15.00 14.61
4.81 12.00 11.70
4.82 12.20 11.90
4.16 9.50 9.28
4.22 10.00 9.77
3.60 7.20 7.03
3.57 7.00 6.83
3.04 5.20 5.08
3.01 5.20 . 5.08
2.41 3.90 3.82
2.46 3.80 3.72

The two phase pressure drop across the test section was measured for
various combinations of air and recirculating water massflow rates.
In all the pressure drop readings it was ensured that the
recirculating water distribution through the tube bundle was uniform.
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Table 6.4 Measured pressure drop across the wet tube bundle.

mw rna ~Ptp(corr)

5. 71 8.22 76.13
5.23 8.26 72.12
4.47 8.38 70.09
4.48 8.41 68.08
6.58 7.44 67.28
5.88 7.48 64.28
3.61 8.36 62.10
3.61 8.46 62.07
4.94 7.50 61.27
5.45 7.23 58.32
4.43 7.42 56.79
6.64 6.98 55.87
4.74 7.24 55.32
5.89 6.89 53.89
5.91 6.94 53.38
6.76 6.55 52.94
6.90 6.41 51.97
6.87 6.35 51.48
5.38 6.92 51.38
5.35 7.01 51.36
3.64 7.33 51.30
6.98 6.31 50.48
6.85 6.31 49.98
3.00 7.45 49.28
4.85 6.96 48.87
4.84 7.10 48.85
3.00 7.24 48.32
6.09 6.51 47.45
6.12 6.42 47.27
6.63 6.08 46.02
6.61 6.07 45.52
2.00 7.53 45.27
5.31 6.40 44.97
2.00 7.37 44.30
6.87 5.93 44.05
4.00 6.63 43.93
3.61 6.76 43.91
5.97 6.18 43.01
6.82 5.83 42.56
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Table 6.4 (cont.) Measured pressure drop across the wet tube bundle.

mw rna APtp(corr)

5.91 6.11 42.02

4.82 6.38 41.97
4.61 6.50 41.95
4.63 6.44 41.86
5.62 6.26 41.49
6.90 5.82 41.06
6.91 5.84 41.06
5.93 5.93 40.55
4.20 6.34 39.48
4.02 6.38 39.47
5.92 5.92 38.55
5.96 5.93 38.54
4.98 6.01 38.53
4.98 6.02 38.53
4.97 5.98 38.04
4.33 6.30 37.99
4.37 6.32 37.98
3.61 6.36 37.48
4.01 6.12 36.02
3.87 6.08 35.77
3.94 6.06 35.52
3.00 6.31 35.48
5.07 5.87 35.05
3.61 6.10 34.52
3.61 6.11 34.52
3.61 6.13 34.51
2.00 6.11 29.52
3.00 5.82 29.46
4.26 5.34 27.63
4.18 5.38 27.63
1.10 6.01 26.03
2.00 5.77 25.77
3.24 5.44 25.62
1.10 5.96 25.54
3.28 5.40 25.37
6.87 4.35 23.36
6.81 4.38 23.25
6.00 4.46 22.24
5.96 4.44 21.95
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Table 6.4 (cont.) Measured pressure drop across the wet tube bundle.

mw rna f1Ptp(corr)

1.10 5.40 21.62

5.00 4.44 21.25
1.10 4.44 21.25
5.32 4.46 21.24
5.06 4.41 20.95
1.10 5.32 20.63
6. 77 4.06 19.79
6.72 4.01 19.29
4.13 4.46 19.24
4.21 4.32 18.96
5.96 4.10 18.78
3.61 4.42 18.25
5.36 4.01 17.59
4.91 4.06 17.29
3.00 4.43 17.25
4.83 4.03 17.09
4.80 4.00 16.89
4.54 4.00 16.69
2.00 4.49 15.49
3.61 4.03 15.29
1.10 4.52 14.74
3.00 4.05 14.54
2.00 4.13 13.28
2.00 3.99 12.79
1.10 4.16 12.58
1.10 4.09 12.28
1.10 4.10 12.28
6.86 3.08 10.88
6.90 3.06 10.63
6.83 3.03 10.38
6.51 3.06 10.38
6.28 3.01 10.08
6.12 3.10 9.88
5.04 3.06 9.38
4.34 3.10 8.88
3.61 3.10 8.38
3.00 3.10 7.88
2.00 3.10 6.88
1.10 3.10 6.38
1.10 3.12 6.37
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Table 6.4 shows the measured two phase pressure drop across the tube
bundle at various combinations of air and recirculating water
massflow rates.

6.5 Determination of coefficients and correlations

A Fortran program, called COEFFS, was developed to calculate the

required heat and mass transfer coefficients from the experimental
data. This program uses the routines of the cross-flow evaporative
cooler rating program iteratively to determine the coefficients from
the known experimental temperatures and massflow rates. The program
logic for COEFFS is shown in figure 6.13.

All three analytical models availabl~ in the rating program have been
incorporated into COEFFS, i.e. the Merkel, the Improved Merkel and
the Poppe models.

As discussed in Chapter 3 an.d Appendi)(@s C and H, the basic relations

for heat and mass transfer between the water film and the tube and
between the water film and the air can be based on the bulk film
temperature

(6.5)

(6.6)
or it can be based on film/air interface temperature as follows

dq = hWl. ( Twall - Ti ) dA (6. 7)

dmw = hoi ( i as i - i a ) dA (6.8)

The program, COEFFS, calculates ho and hw values using the

Merkel model by default but the user can calculate ho and ho values
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B!:GlN

RI!AD IN TEST
DATA

CHDDS!: HD

CHDDS!: HW

EVALUATE THE CDDLER

USING THE SAME
ROUTINES AS FOR THE
RATING PROGRAM.
CALCULATE TPO AND
TWO.

STOP

Figure 6.13 Flow chart fa~ program COEFFS, showing program logic for the
iterative determination of the required coefficients from the
test data.
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using either the Poppe or the Improved Merkel models as well.

The program also allows the calculation of hwi and hoi by employing
the theory described in Appendix H to calculate the interface
temperature.
The program·was used to calculate the coefficients ho and hw based on
the bulk recirculating water temperature using both the Merkel and the
Poppe models. The coefficients hoi and hwi were also calculated using
the Merkel model. Table 6.5 summarizes the calculated
coefficients for each of the tests.

The following correlations were obtained through the use of Lotus 123

ho,Merkel = 6,72238 X 10- 5 Re a0,62 Re w0,20 (6.9)

ho,Poppe = 7,36673 X 10- 5 Re a0,61 Re w0,21 (6.10)

5 Re0,66 Re 0,20
hoi ,Merkel = 5,07155 X 10- a w (6.11)

where the Reynolds numbers are defined as

(6.12)

and

4f
Rew = ~w (6.13)

The correlations for the mass transfer coefficient holds in the

following ranges

2500 < Rea < 13500

230 < Rew < 1100
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Table 6.5 Calculated film heat transfer coefficients and mass transfer coefficients.

Test r Rea Rew ho(Merkel) hw(Merkel) ho(Poppe) hw(Poppe) hoi(Merkel) hwi(Merkel)

•

908.1 351.45 2546.7 638.97 0.02984 3701.57 0.03237 3738.40 0.03139 2358.26
908.2 269.64 2565.0 475.67 0.02781 3279.79 0.03005 3309.86 0.02922 2085.23
908.3 144.27 2616.5 247.98 0.02536 2958.66 0.02730 2990.27 0.02657 1882.92
1508.1 350.73 5485.6 582.05 0.05059 4144.73 0.05397 4211.76 0.05360 2673.18
1508.2 269.64 5484.8 439.57 0.04812 3718.14 0.05127 3776.73 0.05107 2389.72
1508.3 144.27 5492.9 231.79 0.04238 3202.19 0.04508 3256.89 0.04502 2048.53
1508.4 . 351.18 5446.2 602.00 0.04795 3843.58 0.05135 3891.41 0.05123 2465.37
1508.5 269.64 5464.5 450.92 0.04526 3660.53 0.04838 3705.04 0.04830 2340.62
1608.1 352.44 . 5265.2 605.91 0.04942 3856.67 0.05293 3917.75 0.05284
0.04985
2495.13
2474.64
.
0"

1608.2 269.64 5313.8 451.69 0.04690 3844.75 0.05011 3910.86

1608.3 144.27 5382.2 234.26 0.04167 3166.47 0.04439 3221.39 0.04441 2031.56
1608.4 347.22 2691.4 576.47 0.03362 3329.44 0.03595 3358.52 0.03527 2132.26
1608.5 269.64 2733.7 441.71 0.03156 3013.84 0.03372 3047.29 0.03316 1927.19
1608.6 144.27 2762.9 ' 232.33 0.02793 2909.93 0.02977 2950.46 0.02918 1860.75
1708.1 345.33 6612.2 582.25 0.05830 4045.10 0.06227 4116.66 0.06262 2619.84
1708.2 269.64 6670.4 441.56 0.05601 3888.85 0.05964 3958.03 0.05995 2507.91
1708.3 144.27 6755.7 228.72 0.04942 3204.03 0.05249 3260.43 0.05303 2056.25
1808.1 344.16 4367.2 618.25 0.04648 5428.06 0.05018 5505.13 0.04887 3530.10
1808.2 269.64 3930.7 468.65 0.03911 3890.13 0.04203 3958.96 0.04136 2505.29
1808.3 144.27 4086.3 242.01 0.03558 3307.30 0.03808 3365.16 0.03764 2124.41
2610.1 617.49 9828.9 989.02 0.08264 4494.71 0.08829 4524.31 0.08976 2942.95
2610.2 524.16 10026.3 781.74 0.07995 4265.03 0.08482 4284.49 0.08597 2774.54
2610.3 418.77 10120.0 614.97 0.07337 3738.80 0.07769 3752.91 0.07894 2415.25
2610.4 269.64 10271.9 380.82 0.06497 3179.74 0.06855 3188.51 0.06983 2040.57
2610.5 623.43 12642.5 981.15 0.09782 4093.38 0.10414 4133.51 0.10854 2715.03
2610.6 522.63 12894.9 766.82 0.08988 3709.77 0.09526 3730.41 0.09853 2420.50
2610.7 434.52 13039.0 614.07 0.08858 3150.60 0.09363 3165.21 0.09762 2047.77
2610.8 269.64 13236.8 364.33 0.08364 2466.87 0.08818 2473.79 0.09325 1588.54
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Table 6.5 (cont.) Calculated film heat transfer coefficients and mass transfer coefficients.

Test r Rea Rew ho(Merlcel) hw(Merlcel) ho(Poppe) hw(Poppe) hoi(Merlcel) hwi(Merlcel)

2710.1 617.85 9291.3 1022.58 0.07294 4294.84 0.07816 4319.65 0.07983 2806.67
2710.2 547.02 9418.4 881.98 0.07151 4122.43 0.07641 4143.90 0.07796 2685.04
2710.3 415.26 10347.4 647.14 0.07235 3757.88 0.07706 3776.44 0.07907 2434.31
2710.4 269.64 10402.3 404.29 0.06654 2882.88 0.07064 2893.07 0.07328 1855.04
2710.5 611.10 11876.2 954.90 0.09283 4181.99 0.09880 4209.57 0.10239 2739.34
2710.6 532.44 11953.8 816.77 0.08832 4099.57 0.09387 . 4124.41 0.09688 2677.03
2710.7 390.24 12067.0 590.03 0.08229 3540.67 0.08734 3560.42 0.09069 2296.25
2710.8 269.64 12201.5 396.04 0.07550 2872.04 0.07996 2883.95 0.08389 1853.50
2810.1 612.54 11387.3 960.89 0.09031 4325.63 0.09633 4353.78 0.09913 2831.03
2810.2 520.56 11128.8 796.36 0.08358 4143.25 0.08898 4166.89 0.09114 2700.23
2385.14
."'
2810.3 404.37 11355.3 605.26 0.07883 3677.37 0.08379 3694.75 0.08606
2810.4 269.64 11111.6 397.27 0.07217 2700.67 0.07662 2711.54 0.08025 1739.77
2810.5 608.76 11354.9 960.90 0.09087 4513.10 0.09704 4546.70 0.09985 2962.38
2910.1 610.02 11299.8 1014.65 0.08730 4682.09 0.09368 4722.33 0.09636 3062.32
2910.2 542.61 10903.3 879.07 0.07917 4553.10 0.08478 4587.53 0.08647 2964.65
2910.3 415.26 10761.6 658.58 0.07227 4064.34 0.07723 4093.91 0.07879 2632.89
2910.4 269.64 10724.2 418.92 0.06859 2719.29 0.07315 2731.76 0.07701 1749.46
2910.5 324.81 10790.3 483.09 0.06984 3364.13 0.07426 3379.53 0.07607 2168.78
3010.1 609.21 12058.8 1024.06 0.09335 4545.70 0.10010 4588.45 0.10418 2982.27
3010.2 531.72 11606.4 869.10 0.08551 4464.02 0.09149 4499.26 0.09429 2912.19
3010.3 426.78 11732.4 683.06 0.08139 3975.33 0.08690 4003.95 0.08991 2581.95
3010.4 269.64 11233.1 419.60 0.07130 2863.86 0.07592 2876.61 0.07990 1844.77
3010.5 324.90 11251.9 495.38 0.07224 3410.69 0.07685 3427.28 0.07928 2200.74
3110.1 606.33 10117.7 1061.79 0.08143 6394.40 0.09017 5576.01 0.09168 3618.45
3110.2 521.82 10282.6 885.95 0.07925 4854.96 0.08484 4902.00 0.08662 3171.81
3110.3 462.87 10412.8 762.03 0.07765 4505.11 0.08290 4541.70 0.08473 2931.35
3110.4 401.22 10514.1 643.49 0.07649 4187.04 0.08140 4216.47 0.08341 2713.28
3110.5 324.90 10558.7 509.54 0.07209 3682.75 0.07658 3705.29 0.07873 2381.59
3110.6 223.65 10601.5 345.20 0.06862 2773.41 0.07279 2784.53 0.07652 1782.76
•
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Table 6.5 (cont.) Calculated film heat transfer coefficients and mass transfer coefficients.

Test r Rea Rew ho(Merkel) hw(Merkel) ho(Poppe) hw(Poppe) hoi(Merkel) hwi(Merkel)

111.1 610.38 7209.5 1082.45 0.06858 5509.24 0.07406 5557.38 0.07365 3579.95
111.2 537.30 7288.1 932.62 0.06477 5345.74 0.06979 5388.56 0.06922 3465.03
111.3 445.68 7427.9 756.35 0.06152 4575.65 0.06610 4606.08 0.06599 2953.15
111.4 354.15 7442.9 585.90 0.05894 4202.60 0.06314 4228.18 0.06315 2702.66
111.5 324.90 7544.9 521.96 0.05784 3908.75 0.06176 3926.28 0.06190 2505.19
111.6 536.13 7482.8 842.16 0.06719 4868.84 0.07163 4888.91 0.07123 3139.77
111.7 597.87 7582.5 929.75 0.06850 5160.67 0.07293 5180.31 0.07234 3339.10
211.1 593.64 8737.5 957.84 0.07464 5264.20 0.07964 5301.91 0.07949 3413.92
211.2 536.04 8842.0 841.38 0.07312 4807.32 0.07780 4836.53 0.07788 3109.10
211.3 477.90 8855.2 738.21 0.07082 4710.36 0.07527 4736.94 0.07528 3039.09
211.4 441.54 8899.0 670.59 0.06948
0.07195
4394.84
5796.51
0.07374
0.07752
4416~ 11
5854.45
0.07394
0.07712
2830.78
3771.94
.
Cl'

311.1 611.55 8141.5 1074.74

311.2 497.70 8238.5 853.87 0.06756 5097.19 0.07262 5142.31 0.07250 3301.38
311.3 360.45 8432.6 595.22 0.06233 4100.14 0.06670 4126.33 0.06716 "2635. 77
311.4 324.90 8475.9 521.03 0.06273 ' 3730.59 0.06698 3750.53 0.06769 2398.35
411.1 609.66 8920.7 1067.42 0.07802' 5486.96 0.08395 5537.88 0.08439 3564.67
411.2 521.55 9086.9 893.75 0.07493 5172.01 0.08045 5216.12 0.08083 3347.98
411.3 457.83 9176.3 760.16 0.07370 4843.60 0.07891 4878.82 0.07941 3126.66
411.4 357.66 9279.1 588.50 0.06884 4144.68 0.07366 4172.00 0.07458 2664.01
411.5 269.64 9373.4 437.25 0.06432 3360.34 0.06870 3377.03 0.()7037 2154.23
411.6 611.37 9410.8 965.82 0.08382 4772.11 0.08939 4797.22 0.09017 3083.72
411.7 501.66 9481.1 777.86 0.07725 4370.83 0.08224 4391.06 0.08292 2820.29
411.8 377.28 9619.3 568.90 0.07044 3577.96 0.07478 3589.92 0.07596 2296.45
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The mass transfer coefficient data and correlations are shown in

figures 6.14, 6.15 and 6.16.
The film heat transfer coefficient was correlated through the use of
Lotus 123 as

0,32
hw,Merkel = 2946,494 [~al (6.14)

~0 l
0,33
hw,Poppe = 2937,132 [ (6.15)

r ·] o,3s
hwi,Merkel" 1843,035 do
[ (6.16)

The correlations for the film coefficient holds in the following range

140 < r < 650 [kg/m/hr]

Tests 1808.1, 2610.4 - 2610.8, 2710.5, 2710.8, 2810.4, 2910.4 3010.4,

3110.1, 3110.6 and 0411.5 were not considered in the film coefficient
correlation since they were either . conducted with non-uniform
recirculating water distribution or the tests did not.
stabilize due to limited hot water tank size.
At the higher air · ve 1oc it i es the coo 1i ng capacity provided by the
test section was so large that the hot water tank temperature cooled
down to fast to ensure a completely stable test. This did not seem to
influence the mass transfer coefficients significantly.

The film heat transfer coefficient data and correlations are shown
graphically in figures 6.17, 6.18 and 6.19.

The single phase pressure drop measurements are shown graphically in

figure 6.20 together with the single phase predictions by Jakob
[38JA1] and Gaddis and Gnielinski [85GA1].
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0.03 r - - - - - - - - - - - - - - - - - - - - ,

0.025 •

0.02

0
C\J

0.015

0.01

0.005

o~----~----~----~----~----~----~----~
0 2 4 6 8 10 12 14
Air Reynolds number, Rea (xlOOO)
• Experimental data --Carre 1at ion

Figure 6.14 Experimentally determined mass transfer coefficients based on

the Merkel model and the bulk recirculating-water temperature,
Tw.
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0.03r-----------------------------------~

•
0.025

0.02

-
N

0
A

3
~ 0.015
d
...c

0.01

0.005

o~--~~--~----~----~----~----~----~
0 2 4 6 8 10 12 14
Air Reynolds number, Rea (xlOOO)
• Experimental data --Carrel at ion

Figure 6.15 Experimentally determined mass transfer coefficients based on

the Poppe model and the bulk recirculating water temperature,
Tw.
 Stellenbosch University http://scholar.sun.ac.za
6.38

0.03 r - - : - - - - - - - - - - - - - - - - - - - - - ,
•

0.025

0.02

0
N
~

0
3:
c:::
ClJ
0.015
'
..c
0
•

0.01

0.005

o~--~~--~----~----~----~----~----~
0 2. 4 6 8 10 12 14
Air Reynolds number, Rea (xlOOO)
• Experimental data - - Carre 1at ion

Figure 6.16 Experimentally determined mass transfer coefficients based on

the Merkel model and the air/water interface temperature, T;.

I
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6.39

6r---------------------------~--------------~
•

-5
0

--
0
0
X
,.....
~
N
s
, ...
.......... 4 ,'
/, "
, ..
::::3:
.__.
3 ,,
~

... •
~

Q)
.,..
c:
3
, .........
u
.,.. , ... '·
,
,,
4-
4-
,
,,
Q)
0
u
s
,_..
.,.. 2
1.1..

o~--~~--~----~----~----~----~----~----~

0 100 200 300 400 500 600 700 800

Recirculating water massflow, r [kg/m/hr]
• Experimental data --Correlation
----- ± 15%

Figure 6.17 Experimentally determined film heat transfer coefficients

based on the Merkel model and the bulk recirculating water
temperature, Tw.
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6~------------------------------------------~
•

-5
0

--
0
0

.,
.. ........... ---..
X "' "'
........ , "'
~ , ,, • ,-·
,, .~-·

~4 , , ,' I
~ ,/
........... ..
•
~
--- ---
.
.....,
-~
·u
3
.,... , . -·
,.,"''
,,
.
...
4- ,.
4-
, ,;
,,
Q)
0
u

~2
.,...
1.1..

o~----~--~~--~----~----~----~----~----~
0 100 200 300 400 500 600 700 . 800
Recirculating water massflow, r [kg/m/hr]
• Experimental data --Correlation
----- ± 15%

Figure 6.18 Experimentally determined film heat transfer coefficients

based on the Poppe model and the bulk recirculating water
temperature, Tw.
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4r------------------------------------------
•

-
--
0
0
0
X 3
,....... ,
~ ,•'
N
s , ,'
........... , ,'
......
3
, ,'
..... ,
3: , ,,
..c
~
,./
.+-J ~~I
c
Q.l
2 ~
~

.,...
u
.,...
~
'+-
Q.l
0
u
s
.....
.,...
u...
1

o~--~----~--~----~--~----~--~--~
0 100 200 300 400 500 600 700 800
Recirculating water massflow, r [kg/m/hr]
• Experimental data --Correlation
± 15%

Figure 6.19 Experimentally determined film heat transfer coefficients

based on the Merkel model and the air/water interface
temperature, Ti·
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60~------------------------------------~

40
......
N
E
...........
z
.......

<I
c.30
~

Q.
0
~
"0
Q)
~
:::s
~ 20
Q)
~
Q..

10

o~--~--~--~--~--~--~--~--~--~--~~~

0 1 2 3 4 5 6 7 8 9 10 11
Air massflow rate, rna [kg/s]
• Experimental data ··--····-Jakob
--Gaddis et al

Figure 6.20 Single phase pressure drop values measured across the dry tube
bundle compared to existing correlations for single phase
pressure drop.
 Stellenbosch University http://scholar.sun.ac.za
6.43

The measured two phase pressure drop was correlated by defining the
following parameter

mw
TJ =
( rna + mw (P/ Pw) ) Re*a (6.17)

where Rea* =
[ ma +
Afr,min
~ l do
f.l.a
By simple regression analysis, using Lotus 123, the following
correlation for two phase pressure drop across a wet tube bundle could
be found

1,5482 10- 4 X
= -5 - 0,32773 (6.18)
T] + 9,25 X 10

in the ranges

2,15 X 10- 5 < TJ < 19 X 10- 5

rna
0,85 < A < 2,5 [kg/m2/s]
fr
100 < r < 630 [kg/m/hr]

The measured two phase pressure drop data and the pressure drop
correlation are shown graphically in figure 6.21.

6.6 Discussion of results

The correlations for the mass transfer coefficient presented in this

report has the same form as the correlations presented by Mizushina et
al. [67MI1]. The exponent of the air Reynolds number in the
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6.44

1.2 . . - - - - - - - - - - - - - - - - - - - - - - - - ,

\
\
\
\

0.8

ojC tU
c..
<I
'c.
+oJ
0.6
c..
<I

0.4

0.2

o~-~--~--~----._--~--~--~~--~--~--~

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

TJ (xlo-4)
• Experimental data - - Carre 1at ion
·----±7,5%

Figure 6.21 Two phase pressure drop values measured across the test
section.
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correlation by Mizushina et al. is 0,9. Parker and Treybal (61PA1]

and Peterson [84PE3] found the exponent of the air Reynolds number to
be 0,905 in their correlations for the mass transfer coefficient. In
the correlations for the mass transfer coefficient presented in this
report the. exponent of the air Reynolds number was found to be 0,62.
It should however be noted that th~ correlations by Parker and Treybal
[61PA1], Mizushina et al. [67Mil] and Peterson [84PE3] were
determined for counterflow evaporative coolers and condensers while
the current study was done on a cross-flow evaporative cooler.
The heat transfer correlations for heat transfer from a dry tube
bundle with staggered tubes spaced in a 2 x d0 triangular
array, typically gives heat transfer coefficient as

hc = C Re 0a ' 6 (6.19)

If the heat/mass transfer analogy is used to determine the mass

transfer coefficient governing the mass transfer from a wet tube
bundle the exponent of the air Reynolds number in this correlation
would be 0,6. The exponent of Rea in the mass transfer coefficient
correlation found in this study as 0,62 would then seem realistic.

Parker and Treybal [61PA1] and Peterson (84PE1] found the mass
transfer coefficient to be independent of the recirculating water flow
while Mizushina et al. [67MI1] found that the mass transfer
coefficient was dependant on the recirculating water Reynolds number
to the power of 0,15 which is in agreement with the findings of this
study where the exponent of the recirculating water Reynol~s number
was found to be 0,2.

The mass transfer coefficient correlations for the mass transfer

coefficients based on the Poppe theory and the Merkel theory, using
the interface temperature rather than the bulk recirculating water
temperature, shows similar dependencies on the air and water Reynolds
numbers as the correlation for the mass transfer coefficient based on
 Stellenbosch University http://scholar.sun.ac.za
6.46

the Merkel model and the bulk recirculating water temperature.

The three correlations determined in this study are graphically

compared to the correlation by Mizushina et al. [67MII] and the heat-
and mass transfer analogy in figure 6.22. The heat transfer
correlation for forced convection from a tube bundle by Grimison
[37GRI] was used to calculate the mass transfer coefficient in the
analogy approach. From figure 6.22 it can be seen that the current
correlations fall between the predictions by Mizushina et al. [67MII]
and the heat/mass transfer analogy. The correlation given- by
Mizushina was derived for a counterflow evaporative cooler where the
air/water interaction is expected to be more pronounced than in the
current study which was performed on a cross-flow evaporative cooler.
As expected the mass transfer coefficients in a counterflow
evaporative cooler are higher than for a cross-flow evaporative cooler
at the same air Reynolds number. The mass transfer coefficients
calculated from the analogy between heat transfer and mass transfer
does not take into account the air/water interaction and this
consequently results in lower mass transfer coefficients than the
coefficients determined experimentally for the cross-flow evaporative
cooler.

The experimentally determined film heat transfer coefficient shows a

fair amount of scatter. The scatter can be attributed to the fact
that the film heat transfer coefficient represents a relatively small
part of the overall thermal resistance and it is consequently very
sensitive to small variations in the recirculating water temperature.
A small variation of 0,2oC in the outlet recirculating water
temperature resulted in a variation of the film coefficient of up to
20%.
The correlations fitted on the experimental da~a represents the data
fairly well with only about 12,5% of the data points falling outside
the 15% variation zone -around the correlation as shown in figures
6.17, 6.18 and 6.19.
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0.14r-----------------------------------~
.•

..·····

o. 12

0.1
.,...
u
.,....
4-
~0.08
0
u
S-
Q)
4-
Vl
s:::
~ 0.06
~

Vl

. "'
Vl

::=:::
.

0;04

0.02

o~--~----~--~----~--~----~--~~
0 2 4 6 8 10 12 14
Air Reynolds number, Rea
- ho,Merkel - - ho, Poppe · ·-·---··-· Mi zushi na (counterflow)
-Analogy ----- ho; , Merke 1

Figure 6.22 Graphical comparison of the present mass transfer coefficient

correlations with existing correlations at Rew = 400.
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The correlations fitted to the data show the film heat transfer
coefficient to be dependant on the recirculating water massflow rate
per unit length (f) to the power of about 0,33, which is similar to
the findings of other investigators including McAdams [54Mc1], Parker
and Treybal [61PA1] and Mizushina et al. [67MI1].
The correlations for film coefficient determined in this study is
graphically compared to the correlations of McAdams [54Mc1] and
Mizushina et al. [67MI1] in figure 6.23. The correlation by McAdams
[54MI1] was determined for a film cooler (without airflow) while the
correlation given by Mizushina et al. [67MI1] was determined for a
counterflow evaporative cooler. The new correlation for hw based on
the Merkel model corresponds closely to the correlation for hw based
on the Poppe model as expected since they are both based on the same
driving force (Twall - Tw).
The correlation for hwi gives film coefficients which are lower than
those based on the bulk water temperature since the driving force for
the film coefficient based on the interface temperature is larger than
the driving force based on the bulk recirculating water temperature at
the same heat flux.

The single phase pressure drop measured across the tube bundle
corresponds very well to the correlations by Jakob [37JA1] and Gaddis
and Gnielinski [85GA1] as seen in figure 6.20. The two phase pressure
drop across the tube bundle was correlated by a parameter ~ given by

The correlation fitted through the data correlates the data-very well
with only 1 of 117 points differin~ from the correlation by more than
7,5% as seen in figure 6.21.

The correlation is compared to the other available cross-flow

correlations (see Chapter 4) in figure 6.24 for a bank of tubes spaced
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6r--------------------------------------------
..... -- -·
-· ---
5
--
-
0
0
0
...... ,,
-4 X
,......,
~ ,"
,. ,-'
... ...

N
5 , , ,'
~ , ,,
.........
,,
;

~ ,
s::
.....cu ,,
u
..... 3
1+-
1+-
cu
0
u
5
,.....
..... 2
u.

o~--~----~----~----~----~----L---~~
0 100 200 300 400 500 600 700
Recirculating water massflow rate, r [kg/m/hr]
-hw,Merkel - - hwi ,Merkel
-Mizushina ----- McAdams

Figure 6.23 Graphical comparison of the present film coefficient

correlations with existing correlations.
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60•.--------------------------------------~

50

c.
~

~40
C..
0
- ,,
s..
,,
,'
~

QJ ,,
s.. ,/
~ 30 ,/
VI
QJ
,/
,,
.'
s..
c.
QJ
VI

"'
-g_20
.0
3
1-

10

o~~--~--~--~~--~--~--~~~~--~~

2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8

Air massflow rate, ma [kg/s]
-New correlation --Dry tube bundle --Collier/Wallis
----- Di eh 1 & Unruh - Grant & Chi sho 1m + Experi menta1 data

Figure 6.24 Graphical comparison of the present pressure drop correlation

with existing correlations at a recirculating water massflow
rate of 5 kg/s.
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in a 2 x d0 triangular array with d0 = 0,0381 m and a frontal area

of 3,429m2. From figure 6.24 it can be seen that the new correlation
gives pressure drops which are significantly higher than those
predicted by the models of Collier [72C01], Wallis [69WA1] and Grant
and Chfisholm [79GR1] at high air massflow rates while at the new
correlation predicts lower values than the other correlations at low
air massflow rates.

The signifi~ant difference in the slope of the new two phase

correlation when compared to the existing correlations can be due to
the fact that the new correlation was derived for a set-up where the
water .and the air does not flow in the same direction {as was the case
for the other correlations) but the water flow across the airstream.
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CHAPTER 7

CONCLUSION

In this report complete theoretical models for the analysis of

evaporative coolers and condensers have been derived. These models
range from a very simple model, which allows for fast rating and sizing
calculations, to more accurate models requiring numerical integration
and successive calculating procedures.
Computer programs to analyse different cross-flow and counterflow
evaporative coolers and condensers have been compiled. It was found
that the simplified models usually gives results which are accurate
enough for simple engineering sizing and rating calculations.
The simplified model for the analysi~ of evaporative condensers is
expected to be very accurate since the assumption of constant
recirculating water temperature is a good approximation because of the
constant condensing temperature.

The simplified model can also be used in the analysis of cross-flow

evaporative coolers with a fair degree of accuracy. The simplified
model cannot be expected to yield accurate results in analysing cross-
flow evaporative coolers with relatively long tubes since the three-
dimensional recirculating water temperature profile becomes to ·complex
to describe with a single representative recirculating water
temperature.

A survey of the avaflable data for the heat- and mass transfer
coefficients was conducted and all the relevant correlations were
summarized and compared. The relevant correlations for two phase
pressure drop across a tube bundle, which could be found in the
literature are also presented.

An experimental study was conducted on a cross-flow evaporative cooler

to determine the governing heat- and mass transfer coefficients and the
two phase pressure drop across the tube bundle. The correlations which
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were fitted to the experimental data are compared to the other

available correlations.

The use of Mizushina's [67Mil] correlations for the heat- and mass
transfer coefficients, are recommended for the analysis of counterflow
evaporative coolers and condensers. The correlations presented here
should be used in the analysis of cross-flow evaporative coolers
and condensers.

The effect of tube diameter and tube spacing on the heat and mass
transfer coefficient could be the subject of further investigations.
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R.7 http://scholar.sun.ac.za

[74MI1] Mizushina, T., Design of Cooler Condensers and Evaporative

Condensers, in Heat Exchangers: Design and Theory Sourcebook, ed.
N.Afgan and A.E. SchlUnder, pp. 419- 440, McGraw-Hill, New York,
1974.

[74SK1] Skelland, A.H.P.~ Diffusional Mass Transfer, John Wiley and Sons,
New York, 1974.

[74ZU1] Zukauskas, A.A., Investigation of Heat Transfer in different

arrangements of Heat Exchanger Surfaces, Teploenergetika, Vol. 21,
pp. 40 - 46, 1974.

[75BE1] Berliner, P., KUhltUrme - Grundlagen der Berechnung und

Konstruktion, pp. 1 - 86, Springer - Verlag, Berlin, 1975.

[75GN1] Gnielinski, G., Forsch. Ing. Wesen, Vol. 41, No.1, 1975

[75SH1] Sherwood, T.K., Pigf6rd, R.L., and Wilke, C.R., Mass Transfer, pp.
178 - 297, McGraw-Hill, New York, 1975.

[76TE1] Tezuka, S., Kasai, S., and Nakamura, T., Performance of an

Evaporative Cooler, Journal of Refrigeration (Japan), Vol. 47, pp.
1 - 18, 1976.

[77CA1] Cale, S.A., Development of Evaporative Cooling Packing, EUR 7709

EN, 1977.

[77CH1] Chan, J., and Golay, M.W., Comparitive Performance Evaluation of

Current Design Evaporative Cooling Tower Drift Eliminators,
Atmospheric Environment, Vol. 11, pp. 775- 761, 1977.
I

[77IS1] Ishihara, K., Palen, J.W., and Taborek, J., Critical Review of
Correlations for predicting Two-Phase Flow Pressure Drop Across
Tube Banks, ASME-Paper No. 77-WA/HT-23, 1977.
 Stellenbosch University http://scholar.sun.ac.za
R.8

[77NA1] Nahavandi, A.N., and Dellinger, J.J., An Improved Model for the
Analysis of Evaporative Counterflow Cooling Towers, Nuclear
Engineering and Design, Vol. 40, pp. 327- 336, 1977.

[77VA1] Van Iwaarden, J.L., El~mentary Numerical Techniques for Ordinary

Differential Equations, pp. 123 - 131, Project COMPUTE, Hanover,
New Hampshire, 1977.

[78C01] Conti, R.J., Experimental Investigation of Horizontal-Tube Ammonia

Film Evaporation with Small Temperature Differentials, Proceedings
of the 5th Ocean Thermal Energy Conversion Conference, Miami Beach,
Florida, 1978.

[78KR1] Kreid, O.K., Johnson, B.M., and Faletti, D.W., Approximate

Analysis of Heat Transfer from the Surface of a Wet Finned Heat
Exchanger, ASME Paper No. 78-HT-26, Joint Thermodynamics and Heat
Transfer
.
Conference, Palo
.
Alto, California,1978.

[780W1] Owens, W.L., Correlation of Thin Film Evaporation Heat Transfer

Coefficients for Horizontal Tubes, Proceedings of the 5th Ocean
Thermal Energy Conversion Conference, Miami Beach, Florida, 1978.

[79GR1] Grant, I.D.R., and Chisholm, D., Two-Phase Flow on the Shell-Side
of a Segmentally Baffled Shell-and-Tube Heat Exchanger, Journal of
Heat Transfer, Vol. 101, pp. 38- 42,· February 1979.

[79LE1] Leidenfrost, W., and Korenic, B., Analysis of .Evaporative Cooling

and Enhancement of Condenser Efficiency and of Coefficient of
Performance, Warme-und Stoffubertragung, Vol. 12, pp. 5- 23,
1979.

[79NA1] Nakoryakov, V.Y., and Grigor'yeva, N.I., Combined Heat and Mass
Transfer in Film Absorpsion, in Heat and Mass Transfer and Hydrogas
dynamics in Boiling and Condensation, Siberian Division of the
 Stellenbosch University
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USSR Academy of Sciences Press, Novosibirsk, 1979.

[79SH1] Shah, M.M., A General Correlation for Heat Transfer During Film
Condensation inside Pipes, International Journal of Heat and Mass
Transfer, Vol. 22, pp. 547- 556, 1979.

[80F01] Foust, A.S., Wenzel, L.A., Clump, C.W., Maus, L., and Anderson,
L.B., Principles of Unit Operations, 2nd Ed, pp. 302 - 455, John
Wiley and Sons, New York, 1980.

[80YU1] Yung, D., Lorenz, J.J., and Ganic, E.N., Vapour/Liquid Interaction
and Entrainment in Falling Film Evaporators, Journal of Heat
Transfer, Vol. 102, pp. 20- 25, February 1980.

[81KE1] Kettleborough, C.F., The Thermal Performance of the Wet Surface

Plastic Plate Heat Exchanger Used as a Indirect Evaporative Cooler,
ASME Paper No. 81-WA/Sol-11, 1981.

[81KR1] Kreid, O.K., Hauser, S.G., and Johnson, B.M., Investigation of

Combined Heat and Mass Transfer from a Wet Heat Exchanger, Pacific
Northwest Laboratory Report SA-9501, Richland, Washington, 1981.

[81RA1] Rana, R.S., and Charan, V., Heat and Mass Transfer in an
Evaporative Tubular Heat Exchanger, VIth National Mass Transfer
Conference, I.I.T., Madras, India, 1981.

[81R01] Rogers, J.T., Laminar Falling Film Flow and Heat Transfer
Characteristics on Horizontal Tubes, The Canadian Journal of
Chemical Engineering, Vol. 59, pp. 213- 222, April 1981.

[82GA1] Ganic, E.N., and Mastanaiah, K., Hydrodynamics and Heat Transfer in
Falling Film Flow, in Low Reynolds Number flow Heat Exchangers, ed
S. Kakac et al, pp. 487- 527, Hemisphere Publishing Corporation,
Washington, 1982.
 Stellenbosch University
R.lO http://scholar.sun.ac.za

[82J01] Johannsen, A., Plotting Psychrometric Charts by Computer, The

South African Mechanical Engineer, Vol. 32, pp. 3- 11, July 1982.

[82LE1] Leidenfrost, W., and Korenic, B., Evaporative Cooling and Heat
Transfer Augmentation Related to Reduced Condenser Temperatures,
Heat Transfer Engineering, Vol. 3, pp. 38- 59, 1982.

[82PE1] Perez-Blanco, H., and Bird, W.A., Study of Heat and Mass Transfer
in Evaporative Coolers, ORNL/TM-8150, Oak Ridge, Tennessee, June
1982.

[83B01] Bourillot, C., TEFERI, Numerical Model for Calculating the

Performance of an Evaporative Cooling Tower, EPRI Report CS-3212-
SR, August 1983.

[83001] Dorokhov, A.R., and Bochagov, V.N., Heat Transfer to a Film Falling
over Horizontal Cylinders, Heat Transfer- Soviet Research, Vol.
15, pp. 96 ~ 101, March - April 1983.

[83FI1] Fisher, U;, Leidenfrost, W., and Li, J., Hybrid Evaporative -
Condenser Cooling Tower, Heat Transfer Engineering, Vol. 4, pp. 28
- 41, 1983.

[83KE1] Kern, D.Q., Process Heat Transfer, pp. 563- 623, McGraw-Hill, New
York, 1983.

[83MA1] Majumdar, A.K., Singhal, A.K., and Spalding, D.B., Numerical

Modelling of Wet Cooling Towers - Part 1: Mathematical and
Physical Models, Journal of Heat Transfer, Vol. 105, pp. 728- 735,
November 1983.

[83MA2] Majumdar, A.K., Singhal, A.K., Reilly, H.E., and Bartz, J.A.,
Numerical Modelling of Wet Cooling Towers - Part 2: Application to
Natural and Mechanical Draft Towers, Journal of Heat Transfer, Vol.
105, pp. 736 - 743, November 1983.
 Stellenbosch University
R.ll http://scholar.sun.ac.za

[83PE1] Perez-Blanco, H., and Linkous, R.L., Use of an Overall Heat

Transfer Coefficient to Calculate the Performance of an Evaporative
Cooler, ORNL/TM-8250, Oak Ridge, Tennessee, February 1983.

[83SI1] Singham, J.R., Cooling Towers, in Heat Exchanger Design Handbook,

ed. E. Gurney et al, Section 3.12, Hemisphere Publishing
Corporation, 1983.

[83SU1] Sutherland, J.W., Analysis of Mechanical-Draft Counterflow

Air/Water Cooling Towers, Journal of Heat Transfer, Vol. 105, pp.
576 - 583, August 1983.

[83ZU1] Zukauskas, A., and Ulinskas, R., Banks of Plain and Finned Tubes,
in Heat Exchanger Design Handbook, ed. E. Gurney et al, Hemisphere
Publishing Corporation, 1983.

[84CU1] Cussler, E.L., Diffusion -Mass Transfer in Fluid Systems, pp. 215
- 469, Cambridge University Press, Cambridge, 1984.

[84PE1] Perez-Blanco, H., and Webb, R.L., Enhancement of Combined Heat and
Mass Transfer in a Vertical Tube Evaporative Cooler, AIChE
Symposium Series- Heat Transfer, Niagara-Falls, Vol. 80, pp. 465-
469, 1984.

[84PE2] Perez-Blanco, H., and Bird, W.A., Study of Heat and Mass Transfer
in a Vertical Tube Evaporative Cooler, Journal of Heat Transfer,
Vol. 106, pp. 210- 215, February 1984.

[84PE3] Peterson, D., Develo~ment of a Method to Predict the Performance

of an Evaporative Condenser, M.S. Thesis, University of the
Witwatersrand, Johannesburg, 1984.

[84P01] Poppe, M., and Rogener, H., Evaporative Cooling Systems, VDI-
Warmeatlas, Section Mh, 1984.
 Stellenbosch University
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[84ST1] Stoecker, W.F., and Jones, J.W., Refrigeration and Air

Conditioning, pp. 40- 58, pp. 365- 378, McGraw-Hill, New York,
1984.

[84WA1] Wassel, A.T., and Raucher, J.W., Analysis of a Counter-Current

Falling Film Evaporative - Condenser, AIChE Symposium Series - Heat
Transfer, Niagara Falls, Vol. 80, pp. 354- 359, 1984.

[84WE1] Webb, R.L., A Unified Theoretical Treatment for Thermal Analysis of

Cooling Towers, Evaporative Condensers, and Fluid Coolers, ASHRAE
Transactions, KC-84-07, No. 3 (RP - 322), pp. 398 - 415, 1984.

[84WE2] Webb, .R.L., and Villacres, A., Algorithms for Performance

Simulation of Cooling Towers, Evaporative Condensers and Fluid
Coolers, ASHRAE Transactions, KC-84-07, No. 4 (RP - 322), pp. 416 -
458, 1984.

[84WE3] Webb, R.L., and Villacres, A., Cooling Tower Performance, ASHRAE
Journal, pp. 34- 40, November 1984.

[84WE4] Webb, R.L., and Villacres, A., Performance Simulation of

Evaporative Heat Exchangers, AIChE Symposium Series - Heat
Transfer, Niagara Falls, Vol. 80, pp. 458- 464, 1984.

[85AS1] ASHRAE, Handbook - 1981 Fundamentals, 5th Printing, Chapter 3,

1985.

[85GA1] Gaddis, E.S., and Gnielinski, V., Pressure Drop in Cross Flow
across Tube Bundles, International Chemical Engineering, Vol. 25,
pp. 1 - 15, January 1985.

[85WE1] Webb, R.L., and Villacres, A., Performance Simulation of

Evaporative Heat Exchangers (Cooling Towers, Fluid Coolers, and
Condensers), Heat Transfer Engineering, Vol. 6, No. 2, pp. 31 - 38,
1985.
 Stellenbosch University http://scholar.sun.ac.za
R.13

[86C01] Collatz, L., Differential Equations: An Introduction with

Applications, pp. 315 - 319, John Wiley and Sons, 1986.

[86K01] Kotze, J.C.B., Bellstedt, M.O., and Kroger, D.G., Pressure Drop and
Heat Transfer Characteristics of Inclined Finned Tube Heat
Exchanger Bundles, Proceedings of the 8th International Heat
Transfer Conference, San Francisco, 1986.

[86LE1] Leidenfrost, W., and Korenic, B., Principles of Evaporative Cooling

and Heat Transfer Augmentation , in Handbook of Heat and Mass
Transfer, pp. 1025 - 1063, Gulf Publishing Company, Houston, 1986.

· [86RA1] Rana, R.S., Charan, V., and Varma, H.K., Heat and Mass Transfer
from a Horizontal Tube of an Evaporative Heat Dissipater,
International Journal of Heat and Mass Transfer, Vol. 29, pp. 555-
561, 1986.

[87CH1] Chyn, M.-C., and Bergles, A.E., An Analytical and Experimental

Study of Falling Film Evaporation on a Horizontal Tube, Journal of
Heat Transfer, Vol. 109, pp. 783- 990, November 1987.

[87DA1] Datta, S., Sahgal, P.N., Subrahmaniyam, S., Dhingra, S.C., and
Kishorc, V.V.N., Design and Operating Characteristics of
Evaporative Cooling Systems, International Journal of
Refrigeration, Vol. 10, pp. 205- 208, July 1987.

[87ER1] Erens, P.J., A Procedure for the Design or Rating of Counterflow

Evaporative Cooler Cores, South African Institute of Mechanical
Engineering Research and Development Journal, Vol. 13, pp. 18- 25,
April 1987.

[87RA1] Rana, R.S., An Investigation of the Ratio of Experimental and

Theoretical Mass Transfer Coefficients from a Row of Tubes on an
Evaporative Heat Dissipater, XVIIth International Congress of
 Stellenbosch University
R.14 http://scholar.sun.ac.za

Refrigeration, Vienna, Austria, August 1987.

[87SC1] Schrage, D.S., Hsu, J.-T., and Jensen, M.K., Void Fractions and
·Two-phase Friction Multipliers in a Horizontal Tube Bundle, AIChE
Symposium Series - Heat Transfer, Pittsburgh, pp. 1 - 8, 1987.

[87WA1] Wassel, A.T., and Mills, A.F., Design Methodology for a Counter-
Current Falling Film Evaporative Condenser, Journal of Heat
Transfer, Vol. 109, pp. 784- 787, August 1987.

[88DU1] Du Preez, A.F., and Kroger, D.G., Proceedings of the 6th IAHR
Cooling Tower Workshop, Pisa, Italy, 1988.

[88ER1] Erens, P.J., Comparison of Some Design Choice for Evaporative

Coolers Cores, Heat Transfer Engineering, Vol. 9, 1988.

[88ER2] Erens, P.J., and Dreyer, A.A., A General Approach for the Rating of
Evaporative Closed Circuit Coolers, Internal Report, Department of
Mechanical Engineering, University of Stellenbosch, South Africa,
1988.

[88ER3] Erens, P.J., and Dreyer, A.A., An Improved Procedure for

Calculating the Performance of Evaporative Closed Circuit Coolers,
AIChE Symposium Series - Heat Transfer, Houston, Vol. 84, pp. 140 -
145, 1988.

[88KR1] Kroger, D.G., Dry Cooling Towers for Power Stations, Department of
Mechanical Engineering, University of Stellenbosch, South Africa,
1988.

[88SC1] Schrage, D.S., Hsu, S.-T., and Jensen, M.K., Two-phase Pressure
Drop in Vertical Crossflow across a Horizontal Tube Bundle, AIChE
Journal, Vol. 34, pp. 107- 115, January 1988.
 Stellenbosch University
R.15 http://scholar.sun.ac.za

[88WE1] Webb, R.L., A Critical Evaluation of Cooling Tower Design

Methodology, in Heat Exchange Equipment Design, ed. R.K. Shah et
al, pp. 547- 558, Hemisphere Publishing Corporation, Washington
D.C., 1988.
Stellenbosch University http://scholar.sun.ac.za

APPENDICES
 Stellenbosch University http://scholar.sun.ac.za
A. 1

APPENDIX A

pROpERTIES OF FLUIDS

A.l The thermophysica1 properties of dry air from 220 K to 380 K

Density

. {A.l. 1)
p p /RT, kg/m3
=
a a
where R = 287.08 J/kgK

Spec i f i c he.a t 1]"2AN a

c = a + bT + cT 2 + dT 3 , J/kgK (A. 1.2)
pa

a = 1.045356x10 3
1

b = -3.161783x10-l
4
c = 7.083814x10-
d = -2.705209x10-?

Dynamic viscosity ~2AN1]

a+ bT + cT 2 + dT3, kg/sm (A.1 .3)

)J
a =
6
a = 2.287973x10-
b = 6.259793x10-B
-11
c = -3. 131956xl0
15
d = 8.150380x10-

Thermal conductivity

2 3 W/mK (A. 1 .4)

k = a + bT + cT + dT ,
a

a = -4.937787x10- 4
b = 1 .018087x10- 4
c = -4.627937x10-B
d = 1 .250603x10-tl
 Stellenbosch University http://scholar.sun.ac.za
A.2

Table A.1: The thermophyslcal properties of dry air at standard

. 2
atmospheric pressure (101325 N/m )

T Pa c pa JJ• ka a Pr a
a a
K kg/m 3 J/kgK kg/ms W/mK
x10 5 x10 5

220 1. 60432 1007.20 1.46304 0.0197973 1.22518 0.744330

225 1. 56866 1006.99 1. 48797 0.0202127 1. 27957 0.741309
230 1. 53456 1006.81 1.51278 0.0206262 1.33500 0.738428
235 1 •50191 1006.66 1. 53746 0.0210378 1.39145 0.735680
240 1. 47062 1006.53 1. 56201 0.0214475 1. 44892 0.733056
245 1.44061 1006.43 1. 58643 0.0218553 1. 50738 0.730550
250 1• 41180 1006.35 1.61073 0.0222613 1.56684 o. 728156
255 1.38411 1006.30 1. 63490 0.0226655 1•62727 0.725867
260 1.35750 1006.28 1. 65894 0.0230678 1.68867 0. 723678
265 1.33188. 1006.28 1. 68286 0.0234683 1. 75103 0.721585
270 1.30722 1006.30 1. 70666 0.0238669 1.81433 0.719581
275 1~28345 1006.35 1. 73033 0.0242638 1. 87857 0.717663
280 1 .26053 1006.42 1. 75388 0.0246589 1.94373 0.715828
285 1. 23842 1006.52 1. 77731 0.0250521 2.00980 0.714070
290 1.21707 1006.64 1•80061 0.0254436 2.07677 0.712387
295 1 • 19644 1006.78 1•82380 0.0258334 2.14463 0.710776
300 1• 17650 1006.95 1. 84686 0.0262213 2.21336 0.709233
305 1• 15721 1007.14 1. 86980 0.0266075 2.28297 0.707755
310 1.13854 1007.35 1•89263 0.0269920 2.35342 0.706340
315 1.12047 1007.59 1.91533 0.0273747 2.42472 0.704985
320 1.10297 1007.85 1. 93792 0.0277558 2.49685 0.703688
325 1. 08600 1008. 13 1. 96039 0.0281351 2.56980 0.702446
330 1.06954 1008.43 1. 98274 0.0285127 2.64356 0.701258
335 1. 05358 1008.76 2.00498 0.0288886 2.71811 0.700122
340 1.03808 1009.11 2.02710 0.0292628 2.79345 0.699035
345 1.02304 1009.48 2. 04911 0.0296353 2.86957 0.697997
350 1.00842 1009.87 2.07100 0.0300062 2.94645 0.697004
355 0.99422 1010.28 2.09278 0.0303754 3.02408 0.696056
360 0.98041 1010.71 2. 11444 0.0307430 3. 10246 0.695151
365 0.96698 1011.17 2.13599 0.0311089 3.18156 0.694288
370 0.95392 1011 . 64 2.15743 0.0314732 3.26138 0.693465
375 0.94120 1012.13 2.17876 0.0318359 3.34191 0.692681
380 0 .·92881 1012.65 2.19998 0.0321970 3.42313 0.691935
 ·; ·")
0.032 1013 1.6

0.030 2.1 1012 1.5

~ 0.026 11'1 2.0 10 11

E E
........
~ 01
..X
;; ~
..X "'0 - 0\
-; ..X
:>. 0.026 :!. 1.9 ::::; 1010 1.3
> >. 0

1009
°·024 1.8 1.2
u ·-
.._ :::.::
1 ~ ! ~~ -~-lillllllll-1111111 :
a
E
(..
E
a ·-u 11'1
C1l c C1l c
f= 0.022 l; 1.7 ~ 1008 ~1.1
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0.020 1.6 1007 1.0

0.018 1.5 1006 0.9

-60 -40 -20 0 20 40 60 80 100
Temperature,~

2
Figure A.l: The thermophysical properties of dry air at standard atmospheric pressure (101325 N/m )
 Stellenbosch University http://scholar.sun.ac.za
A.4

A.2 The thermophysical properties of saturated water vapor from

273.15 K to 380 K

Vapor pressure ~6G01J

(A.2.1)
p
v =

z = · a ( 1- x) + b 1og
10
:J
( x) + c [i - 10d{ ( 1I x} - 1 + e ( 10f ( 1- x) - 1) + 9

X = 273.16/T

a = 1.079586x10
b = 5.028080
4
c = 1.504740x10-
d = -8.296920
e = 4.287300x10-l;
f = 4.769550
9 = 2.786118312

Specific heat

c = a +. bT + cT 5 + dT 6 , J/k9K (A.2.2}
pv

a = 1 .3605x103
b = 2.31334
c = -2.46784x10- 10
d = 5.91332xlo- 13

Dynamic viscosi~y

2 3 (A.2.3)
lJ v = a+ bT + cT + dT • kg/ms

a = 2.562435x10- 6
b = 1.816683x1o~ 8
11
c : 2.579066xlo-
-14
d = -1.067299x10 .
 Stellenbosch University http://scholar.sun.ac.za

A.5

Therma 1 co~duct i vi ty []2AN 1]

k ~ a + bT + cT 2 + dT 3 , W/mK (A.2.4)
v

a ~ 1.3046x10- 2
b = -3.756191x10-S
c ~ 2.217964x10- 7
d ~ -1.111562x10- 10

Vapor density IJOUKU

(A.2.5)

a = -4.062329056
b = 0.10277044
c = -9.76300388 X 10- 4
d = 4.475240795 X 10- 6
e = -1.004596894 X 10- 8
f = 8.9154895·x 10~ 12 ·
 Stellenbosch University http://scholar.sun.ac.za
A.6

Table A.2: The thermophysical properties of saturated water vapor

T Pv Pv cpv 'IJV kv a
v Pr v
K N/m2 kg/m 3 J/kgK kg/ms W/mK
x10 6 x10 5

275 697.820 0.00550 1864.29 9.28676 0.0171781 167.602 1 .00786

280 990.897 0.00767 1868.46 9.4368 0.0174774 121.992 1. 00887
285 1387.70 0.01056 1872.66 9.58775 0.0177831 90.0091 1. 00964
290 1918.11 0.01436 1876.92 9.73950 0.0180951 67.2667 1. 01023
295 2618.61 0.01928 1881 . 31 9.89208 0.0184134 50.8805 1. 01068
300 3533.19 0.02557 1885.89 10.0454 0.0187378 38.9260 1.01103
305 4_714.45 0.03355 1890.74 10.1996 0.0190684 30.1011 1.01135
310 6224.58 0.04360 1895.92 10.3546 0.0194049 23.5132 1.01168
315 8136.44 0.05611 1901.52 10.5104 0.0197474 18.5427 1.01207
320 10534.7 0.07155 1907.63 10.6670 0.0200957 14.7547 1~01259
325 13516.9 0.09045 1914.35 10.8244 0.0204498 11.8400 1.01329
330 17194.7 0.11341 1921.79 10.9825 0.0208095 9.57698 1.01425
335 21694.5 0. 14108 1930.04 11.1414 0.0211749 7.80452 1.01551
340 27158.9 0.17418 1939.25 11.3010 0.0215457 6.40488 1.01716
345 33747.7 0.21352 1949.63 11.4614 0.0219219 5.29095 1.01927
350 41638.4 0.26000 1961.03 11.6225 0.0223035 4. 39779 1.02191
355 51027.6 0.31455 ' 1973.90 11.7844 0.0226904 3.67653 1.02516
360 62131.3 0.37821 1988.29 11.9470 0.0230824 3.09016 1.02910
365 75186.3 0.45213 2004~37 12. 11 02 0.0234795 2.61037 1.03382
370 90450.0 0.53750 2022.33 12.2742 0.0238816 2.21538 1. 03940
375 108201 0.63568 2042.35 12.4389 6.0242886 1.888304 1.04595
380 128743 0.74799 2064.63 12.6043 0.0247005 1.615964 1.05355
 O.C!. 12.5 2140 14

o.::~; 12.0 2100 12

~ 0.022 ~ , 1.5 2060 10

e
01
~ .X
' N
. -# • p ~
> 0
.X 01 z
>C .X -# •
?: 0.02, :t , 1.0 :::, 2020 ~ 6
- >C
> >.. ; >
..... a. 0.
.....- ·- u
u Ill Cll
-
::I 0 ..... c..
"0 u 0 ::I
g 0.020 -~ 10.5 ~ 1960 6
u > )>
u .
u ·-- ........
0 .... ~ lllllllfU-HI~:IIIIIIIIIIIIIItumlllllll
e e c..
c.. 0 0
·-u
Cll c Cll 0.
(:= 0.019 . ti 10.0 a- 1940 g. 4

0.016 9.5 1900 2 lallllllffiil.

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0.017 9.0 1660 0

0 20 40 60 60 100
Temperature,"(
Figure A.2: The thermophysical properties of saturated water vapor
 Stellenbosch University http://scholar.sun.ac.za
A.8

A.3 The thermophysical properties of mixtures of dry air and water

vapor

Density (]2AS iJ
Pav = (1+w) [!-w/(w+0.62198>] (pab/RT), kg/m3 (A.3.1)

where R = 287.08 J/kgK

Specific heat U8FA 1]

c = (c + we )/(1+w), J/kgK (A.3.2)

pav pa pv

Dynamic viscosity ~4G00

(A.3.3)

.Therma 1 conductivity ~7LE 1]

0
kav = (X k M 0.33 + X k M 0.33)/(X M 0 • 33 + X .M • 33 ) ·W/mK (A 3 L)
a a a v v v a a v v ' • • ..

where

M
a
= 28.97 kg/mole
M
v
= 18.016 kg/mole
X = 1/(1+1.608w)
a
X
·v
= w/(w+0.622)

Humidity ratio

.
I
2501.6- 2.3263(Twb- 273.15) · ]
w = 2501.6 + 1.B577(Tdb- 273.15) - 4.184(Twb- 273.15)

X o.62509pvwb
[ pa b s. - 1.005p vw b
l
·~ l.00416(Tdb- Twb)
2501.6 + t.8577(Tdb- 273.15) - 4.t84(Twb- 273. 15) ' kg/kg
(A.3.5 )
l
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A.9

A.4 The thermophysica1 properties of saturated water liquid from

273. 15K to 380K

Density

p
w = (A.4.1)

a = 1.49343x10- 3
b = -3.7164x10- 6
c = 7.09782x10- 9
d = -1.90321x10- 20

Specific heat

c = a + bT + cT
2 6
+ dT , J/kgK (A.4.2)
pw

a = 8.15599x10 3
b = -2.80627x10
c = 5.11283x10-
2
d = -2.17582x10- 13

Dynamic viscosity ~2AN1]

. 10 b/(T-c) kg/ms (A.4.3)

"~w = a '

a = 2.414x10-S
b = 247.8
c = 140
 Stellenbosch University http://scholar.sun.ac.za
A. 10

Thermal conductivity

W/mK (A.~.4)
k = a + bT +
w

a = -6.1~255x10- 1
b = 6.9962x10-3
c = -1.01075x10-5
d = ~.7~737x10- 12

Latent heat of vaporization

(A.~.5)

6
a = 3.4831814 X 10
b = -5.8627703 X 10 3
c = 1.2139568 X 10.
2
d = -1.~.0290~31 ·x 10-

Critical pressure

6 (A.~.6)
p
we
= 22.09 X 10 ,

·surface tension []ouKQ

2
a = a+ ht + cT + dT 3 , N/m (A. 4. 7)

a = 10- 2
5.1~8103 X
4
b = 3.998714 X 10-
-6
c = -1.4721869 X 10
9
d = 1.21405335 X 10-
 Stellenbosch University http://scholar.sun.ac.za
A. 11

Table A.3: The thermophysical properties of saturated water 1iquid

T Pw c pw lJw kw ew Pr w
K kg/m3 J/kgK kg/ms W/mK 1/K
x10 4 x10 5

275 1000.03 4211 . 21 16.5307 0.572471 0.780333 12.1603

280 999.864 4202.04 14.2146 0.581432 6.184114 10.2730
285 999.422 4194.41 12.3510 0.590001 11 . 45765 8.78055
290 998.721 4188.27 10.8327 0.598179 16.59011 7.58474
295 997.768 4183.53 9.58179 0.605972 21 .57093 6.61511
3.00 996.572 4180. 10 8.54057 0.613383 26.38963 5.82026
305 995.141 4177.92 7.66576 0.620417 31.03593 5.16215
310 993.487 4176.87 6.92443 0.627079 35.49975 4.61225
315 991.618 4176.88 6.29125. 0.633372 39.771.22 4.14887
320 989.547 4177.83 5.74650 0.639300 43.84070 3.75534
325 987.284 4179.63 5.27468 0.644870 47.69877 3.41871
330 984.842 4182. 17 . 4.86348 0.650084 51.33626 3.12881
335 982.232 4185.32 4.50304 0.654948 54.74422 2. 87758
340 979.469 4188.98 4. 18540 0.659466 57.91392 2.65859
345 976.564 4193.01 3.90407 0.663644 60.83688 2.46665
350 973.532 4197.28 3.65373 0.667486 63.50480 2.29754
355 970.386 4201.67 3.43001 0.670997 65.90961 2.14781
360 967.141 4206.01 3.22924 0.674182 68.04338 2.01462
365 963.811 4210.17 3.04839 0.677046 69.89838 1. 89562
'370 960.409 4213.99 2.88488 0.679595 71.46697 1. 78884
375 956.952 4217.31 2.73656 0.681833 72.74164 1.69263
380 953.453 4219.96 2.60158 0.683767 73.71494 1. 60560
 0.70 16 4230

0.68 14 4220 1010

Ill
ve 0.66 e 12 4210 1000
. , ......
en
~ .X

~
.. u p..,
0 .
.X en Cpv
M .X
>..
.... 0.64 ~· 1.200 990
-:l 10
> )I
....>. u
n ...e
Ill
....u
:J 0 ..: 'en
'0 u 0 .X
c Ill Cl1
0 0.62 8 .c 4190 980
u > J
u
u )>
0 .... ....~
e ·-e Ill
c.. 0 u N
Cl1 c Cl1 c
.c
._ >- Cl1
0.60 0 6 a- 4180 0 970

0.58 4 4170 960

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0.56 2 4160 950

0 20 40 60 80 100
Temperature. "C

Figure A.]: The thermophysica1 properties of saturated water liquid

 Stellenbosch University http://scholar.sun.ac.za

A. 13

A.5 The thermophysical properties of saturated ammonia vapor.

Vapor pressure []6RA1], (230 K to 395 K)

2 3 ,. (A.5.1)
p = a + bT + cT + dT + eT ,
ammv
6
a= 1.992448 x 10.
b = -57.568140 X 103
C = 0.5640265 X 10
3
d = -2.337352
e = 3.541430 x 10-3

Density fz6RA 1] , (260 K to 390 K)

2 3 ,.
p
ammv
=a + bT + cT + dT + eT , (A.5.2)

2
a = -6.018936 x 10
b = 5.361048
2
C = -1.187296 X 10-
· d = -1 • 161 479 X .1 0- 5
8
e = 4.739058 x 10-

Specific heat J]2AS 1], (230 K to 325 K)

c =a + bT + cT
2 + dT3 J/kg K (A.5.3)
pammv '
4
a= -2.7761190256 x 10
2
b= 3.39116449 X 10
c = -1 .-3055687 . I

d r:: 1.728649' X 10~3-

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A. 14

Dynamic viscosity (]2AS1], (240 K to 370 K)

~ ammv =a+ bT + cT 2 + dT 3 + eT 4 , kg/sm (A.5.4)

a= ~2.748011 x 10-5
b = 2.82526 10-7
X
-10
C = -5.201831 X 10

d = -~.061761 X 10-
13
e = 2.126070 x 10-15

Thermal conductivity [72ASU, (245 K to 395 K)

k =a+ bT + cT 2 + dT 3 + eT ,
4 W/mK (A.5.6)
ammv

a = -0.1390216
b = 1.35238 X 10-3
C = -2.532035 X 10 -6
d = -4.~84341 X 10-9
e = 1.418657 x 1G-11
 Stellenbosch University http://scholar.sun.ac.za

A. 15

Table A.4: The thermophysical properties of saturated ammonia vapor

T Pammv Pammv c llammv k Pr

pammv ammv ammv
K N/m
2 kg/m3 J/kgK kg/sm W/mK
X 10- 3 X 106

230 60.58 2203.48

235 79.09 2265.33
240 102.08 2322.84 9.0376
245 130.32 2377.30 9.2605 0.019611 1• 12261
250 164.65 2430.01 9.4734 0.019920 1.15567
255 205.93 2482.27 9.6774 0.020185 1.19009
260 255. 10 1.78881 25.35.3 7 9.8737 0.020414 1. 22632
265 313.14 2.56766 2590.61 10.0635 0.020613 1. 26477
270 381.09 3.28963 2649.29 10.2480 0.020790 1.30588
275 460.03 3.98405 2712.69 10.4284 0.020954 1.35007
280 551.10 4.68094 2782.13 10.6060 0~021111 1 . 39774
285 655.51 5.41106 2858.89 10.7822 0.021270 1. 44924
290 774.50 6.20583 . 2944.27 10.9583 0.021439 1 .50491
295 909.36 7.09745 3039.57 11.1356 0.021627 1. 56503
300 1061 . 47 8.11876 3146.08 11.3156 0.021843 1. 62979
305 1232.21 9.30337 3265.10 11.4997 0.022096 1. 69932
310 1423.06 10.68557 3397.93 11.6894 0.022394 1. 77368
315 1635.53 12.30036 3545.87 11.8862 0.022748 1. 85280
320 1871.19 14.18346 3710.20 12.0917 0.023166 1.93654
325 2131 . 65 16.37130 .3892.24 12.3074 0.023660 2.02466
330 2418.59 18.90102 12.5349 0.024238
335 2733.73 21.81047 12.7758 o. 024911
340 3078.86 25.13821 13.0318 0.025690
345 3455.81 28.92352 13.3047 0.026586
350 3866.46 33.20637 13.5961 0.027608
355 4312.75 38.02745 13.9078 0.028769
360 4796.68 43.42818 14.24159 0.030080
365 5320.28 49.45067 . 14.5993 0.031551
370 5885.67 56.13774 14.9827 0.033196
375 6494.98 . 63.53294 0.035026
380 7150.43- 71.68050 0.037053
385 7854.27 80.62539 0.039289
390 8608.81 90.41328 0.041748
395 9416.42 o.o44441
 G.'JSO 15 4000 100 10
-

pommv 1 L
-
In -
0.044 E 13 3600 80 8
.........
01
~ .X -
N IJ OIMIY
E E
~· 00 z
0 ' - iJ
~
01
X .X ....
i 0.038 1t 3200 60 IO 6
.xo -,
i
0
' X -
::1 , j
.....>- ~ E i ( OIIIIIY
0 0
- 01 C1.
> .....>- uo.
.X
'
..... In C1l
u -
:::1 0.032 0 9 ..... 2800 40 c.. 4
u u 0 :::1
In C1l
i In )>
c a..o In .
0 .c
u > C1l ij_
u c.. -
u >- 0. II
0 .... ..... OIMIY
E c..
In
e0 u 0
c..
C1l c C1l c 0. POIIIIIY
>- 0. C1l 0
'.c
1- 0.026 CJ 7 tn 2400 0 20 > 2
~ ~
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0.020 5 2000 0 0
-40 -20 0 20 40 60 80 100 120
0
Temperature, (

Figure A.4: Thermophysical properties of saturated ammonia vapor

 Stellenbosch University http://scholar.sun.ac.za
A. 17

A.6 The thermophysical properties of saturated ammonia 1 iquid from

200 K to 405 K.

Density (]7YA 1]

r:( 1-T/T )0.285714-t

p = ab L.: c J, kg/m 3 (A.6.1)
amm

2
a = 2.312 x 10
b = 0.2471
Tc = 405.5 K

Specific heat [l?YA 1], (200 K to 375 K)

2
c =a + bT + cT + dT 3 J/kgK (A.6.2)
pamm '

3
a = -2.497276939 x 10
b ~ 7.7813907 X 10
1
C = -3.006252 X 10-
d = 4.06714 X 10~ 4

Dynamic viscosity []7YA1]

2
u = 0.001 x 10(a + b/T + cT + dT >, kg/sm (A.6.3)
amm
a = -8.591
b = 876.4
c = 0.02681
d = -3.612 x· 10-5
 Stellenbosch University http://scholar.sun.ac.za
A. 18

Thermal conductivity [)7YA1], (200 K to 375 K)

2 (A.6.4)
k = a + bT + cT , W/mK
amm

a = 1.068229
b = -1.576908 X 10- 3
C = -1.228884 X 10- 6

Latent heat of vaporization, 1]7YA1]

if =a [lb-T)/(b-c)]d, J/kg (A.6.5)

gamm .

a = 1.370758 X 10 6·
b = 405.55
c ·= 239.72

d = 6.38
 Stellenbosch University http://scholar.sun.ac.za
A. 19

Table A.5: The thermophysical properties of saturated ammonia liquid

T Pamm c J.lamm k Pr i
pamm amm anrn . fgamm
K kg/m 3 J/kgK kg/sm W/mk J/kg
x10 5 x1o- 3

200 731 . 094 4294.20 51.0740 0.703692 3.11673 1487.29

205 725.217 4324.69 45.9440 0.693319 2.86583 1473.44
210 719.282 4352.65 41.6429 0.682885 2.65428 1459.37
215 713.288 4378.38 37.9998 0.672389 2.47443 1445.08
220 707.232 4402.21 34.8841 0.661832 2.32034 1430.55
225 701 • 111 4424.42 32.1948 0.651213 2.18736 1. 415.78
230 694.923 4445.33 29.8529 0.640532 2.07181 1400.75
235 688.663 4465.24 27.7961 0.629791 1. 97076 1385.45
240 682.329 4484.46 25.9749 0.618988 1. 88184 1369.87
245 675.918 4503.28 24.3494 0.608123 1.80313 1354.00
250 669.424 4522.03 22.8875 0.597197 1. 73306 1337.82
255 662.844 4540.99 21.5630 0.586210 1.67035 1321 . 31
260 656.173 4560.48 20.3543 0.575161 1.61390 1304.46
265 649.406 4580.79 19.2438 0.564050 1.56284. 1287.25
270 642.537 4602.25 18.2170 . 0. 552878 1 . 51641 1269.65
275 635.560 4625.14 17.2617 0.541645 1.47399 .. 1251.65
280 628.469 4649.78 16.3678 0.530351 1.43503 1233.21
285 621.255 4676.47 15.5271 0.518994 1.39909 1214.31
290 613.912 4705.52 14.7325 0. 507577 1.36578 1194.92
295 606.428 4737.23 13.9783 0.496098 1.33478 1l75. 009
300 598.794 4771.90 13.2596 0.484557 1.30581 1154.52
305 590.999 4809.85 12.5727 0. 472955 1. 27861 1133.42 -

310 583.029 4851.37 11.9141 0.461292 1 .25300 1111~67

315 574.868 4896.77 11.2814 0.449567 ~.22879 1089. 19
320 566.500 4946.35 10.6723 ·o. 437781 1 .20584 1065.93
325 557.905 5000.43 10.0853 0.425933 1.18401 1041.82
330 549.060 5059.31 9.5189 0.414024 1.16320 1016.75
335 539.936 5123.28 8.9722 0.402054 1.14331 990.644
340 530.501 5192.66 8.4445 0.390022 1.12428 963.355
345 520.717 5267.75 7. 935.1 0.377928 1. 10604 934.743
350 510.534 5348.86 7.4438 0.365773 1. 08855 904.625
355 499.893 5436.29 6.9704 0.353557 1.07177 872.776
360 488.718 5530.35 6.5146 0.341279 1.05568 838.908
365 476.910 5631.34 6.0765 0.328940 1.04028 802.648
370 464.337 5739.56 5.6561 0.316539 1.02558 763.498
375 450.814 5855.32 5.2534 0.304077 1. 01160 . 720.764
380 I 436.074 4.8684 673.437
385 i 419.697 4.5011 I 619.951
390 I 4oo.967 4.1517 557.630
395 ; 378.461 3.8199 481.199
400 ! 348.445 i 3.5059 376.981
405 : 285.772
I I 3.2093 '
i 156.612
 :.J.75 so 6000 750

0.65 40 5600 650

Ill (J)OINII
~ E
E t:n
~ .Y.
'
.,. ~
POIIIQI
0 t:n
)( .Y.
.Y.a
I 0.55 -I 30 .........
-, 5200 550
a
::1. ,..
>- E
.... ~ 011'111
a t:n
> . >-- a. .Y.
.... u '
)>
Ill
....u .
::J 0.45 0 20 ....0 4800 llamm N
"0 u I 450 0
c VI GJ a..a
0 .c.
u > u
u ..__
-
....>-
0
E E
c... 0 u VI
GJ c GJ c
.c. >- o. GJ
1- 0.35 0 10 c.n 4400 0 350 1\
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0.25 0 4000 250

-60 -40 -20 0 20 40 60 80 100 120 140.
0
Temperature, (

Figur~ A.S: Thermophyslcal properties of saturated ammonia 1 iquid

 Stellenbosch University http://scholar.sun.ac.za

A.21

REFERENCES

46G01 Goff, J.A., Saturation Pressure of Water on the New Kelvin

Scale, Humidity and Moisture Measurement and Control in
Science and Industry, eds., A. Wexler and W.H. Wildhack,
Reinhold Publishing Co., New York, 1965.

54G01 Godridge, A.M., British Coal Utilisation Research Association

Monthly, Vol. 18, No. 1, 1954.

57LE1 Lehmann, H., Chemical Technology, Vol. 9, pp. 530, 1957.

70UK1 United Kingdom Committee on the Properties of Steam, U.K.

Steam Tables in Sl Units_ 1970,· Edward Ar-nold ltd., London,

1970.

72AS1 ASHRAE Handbook of Fundamenta 1s, Pub 1 i shed by the American

Society of Heating, Refrigeration ~nd Air Conditioning
Engineers, Inc., 1972.

76RA1 Raznjevic, K.t Handbook of Thermodynamic Tables and Charts,

McGraw-Hill Book Co., New York, 1976.

77YA1 Yaws, C.~., Physical Properties, Chemical Engineering Publ.,

McGraw-Hill Book Co.~ New York, 1977.

78FA1 Faires, V.M. and Simmang, C.M., Thermodynamics, 6th ed.,

Macmillan Publishing Co. Inc., 1978.

82AN1 Anon., Heat Transfer and Fluid Flow Data Book, General
Electric Co., Corporate Research Division, New York, 1982.
 Stellenbosch University http://scholar.sun.ac.za
B. 1

APPENDIX B

DEFINITION OF LEWIS NUMBER AND THE LEWIS FACTOR

In simultaneous heat and mass transfer the factor (hc/hDCpm) and the
Lewis number are often used as dimensionless parameters. In some of
the literature encountered there seems to be some confusion about ·the
definitions of these dimensionless numbers and the factor (hc/hDcpm)
is often incorrectly referred to as the Lewis number. The correct
definitions of both these parameters will now be presented to
clarify any misconceptions.

Definition of the Lewis number

The rate equations for the transfer of momentum, energy and mass are
given by

i) Newton's equation of viscosity,

[ ~) ~
__ [ a x]a:
or

(8.1)

ii) Fourier's equation of energy conduction,

or

(8.2)
 Stellenbosch University http://scholar.sun.ac.za
8.2

iii) and Fick's equation of diffusion,

(8.3)

The three coefficients v, a and D in these rate equations all have the
dimensions [ L2/T ]. Any ratio of these coefficients would result in
a dimensionless number.

In the study of systems undergoing simultaneous energy and momentum

transfer the ratio of v to a would be of importance. By definition the
Prandtl number is defined as

v
Pr = a =
(8.4)
In processes where simultaneous momentum and mass transfer occur the
Schmidt number is defined as the ratio of v to D, or

v
Sc = D (8.5)

The ratio of a to D would be important for simultaneous energy and mass

transfer processes. This ratio is called the Lewis number and it is
expressed as
a
Le = D (8.6)

These three dimensionless numbers can be seen as a measure of the

relative boundary layer thicknesses involved, e.g. the Lewis number can
be seen as the relative thickness of the thermal and concentration
boundary layers

Le (8.7)
 Stellenbosch University http://scholar.sun.ac.za
B.3

Similarly we have

om
Pr = (8.8)
ot

om
Sc = - (8.9)
oC

From the definitions above the Lewis number can be expressed in various
forms e.g.

Q k ot Sc
Le = D = = = Pr (8.10)
pep oC

Definition of the Lewis factor

W.K. Lewis [22LE1] tried to prove analytically that

= 1
ho cpm (8.11)
for gas/liquid systems.

In a later article Lewis [33LE1] showed that the previous relation

does not hold for all mixtures of liquid and gas, but that the
relation does, in fact, hold approximately for air/water mixtures.

Peterson [84PEI] concluded that the analytical proof of the Lewis

relation given by Lewis [22LEI] was mathematically incorrect.

Although the proof given by Lewis was incorrect the factor (hc/hocpm)
is today known as the Lewis factor and the relation hc/hocpm = 1 is
 Stellenbosch University http://scholar.sun.ac.za

known as the Lewis relation.

Many authors, including Arnold [33AR1], Threlkeld [70TH!], Berliner

[75BE1], Nahavandi and Dellinger [77NA1], Kettleboro~gh [81KE1], Kern
•
[83KEI] and Majumdar et al. [83MA1] erroneously refer to the Lewis
factor as the Lewis number.

The term (hc/hocpm) is called the "convective Lewis number" by Close

and Banks [74CL1] and Sutherland [83SU1], while Sherwood et al.[75SH1]
and Peterson [84PE1] refer to the Lewis factor as the
"psychrometric ratio".

Several investigators have studied the Lewis factor and various

empirical relations have been proposed.

Chilton and Colburn [34CHI] used experimental data to show that

(B.l2)
According to Cussler [84CU1] the exponent in the Chilton-Colburn
relation does not represent the best fit on the experimental data, but
it facilitated easier calculations with slide rules.

Bedingfield and Drew [SOBEl] obtained data on the heat and mass
transfer by studying solid cylinders of volatile solids such as
naphthalene in a normal gas flow. The data was correlated by the
following relation
he
= 1230,7 (Sc) 0 , 56
ho (B.l3)

1230,7 0 56
_ _ _(Pr)_ _,_ (Le)0,56
cpm
 Stellenbosch University http://scholar.sun.ac.za
8.5

If the non-condensable gas is air this simplifies to

(8.14)
in the temperature range normally encountered in evaporative coolers
and condensers.

Boelter et al.[65801] gave the following relation for the Lewis factor
for natural convection systems

c1
= ( Le)
(8.15)
where
2 3
3 < c1 < 4
n
For laminar and turbulent airflow Bos.inakovic [60801] proposed the
following correlation for the Lewis factor, i.e.

he
ho cpm
=
( r -
ln
1 )
r
_va
[ "rna
l[~2 ]
(Le)0,67
(8.16)
where

0,622 + wasw
r = 0,622 + wa

Assuming that ( va I "rna ) = 1 this becomes

0, 622 + wasw , - 1 l
l
[ 0,622, + wa
(Le)0,67
ln [ 0,622 + wasw
0,622 + wa (8.17)
 Stellenbosch University http://scholar.sun.ac.za
8.6

According to Berman [61BE1] the Lewis factor can be expressed as

Patm (B.l8)
for air/water systems.

Mizushina et al.[59Mll] assumed the following relation to hold in their

study on the operation of spray condensers.

he
= (Le)o,s
ho cpm (B.l9)
Threlkeld [70TH!] expressed the Lewis factor as

c
(Le) 1
(B.20)
where

o,6 < c1 < o,7

By using an analytical approach Arnold [33AR1] showed that the Lewis

factor can be expressed as

Le Pr +
Pr + ( 1
l (B.21)
where

The relation derived by Arnold shows some interesting points. If the

 Stellenbosch University http://scholar.sun.ac.za
8.7

free stream velocity nears zero then the ratio r would approach unity
and the Lewis factor would approach the Lewis number. If the free
stream velocity increases to infinity the ratio r becomes zero and the
Lewi~ factor would approach a value of unity, regardless of the Lewis
number.
The Arnold relation shows that the Lewis factor will have values
ranging from the Lewis number to unity depending on the free stream
velocity.

Various other investigators expressed the Lewis factor as a constant

value, eg.

(8.22)
Foust et al.[80F01] gave C1 as 0,98 < C1 < 1,13 for turbulent airflow,
while Sherwood [75SH1] reported values of C1 varying from 0,95 to 1,12.

In cooling tower theory it has been customary to assume a C1 value. of

unity since this simplified the theoretical model substantially.

According to the ASHRAE Handbook of Fundamentals [85AS1], the value of

CJ in equation (8.22) should be taken as unity for turbulent air flows
since the eddy diffusion in turbulent flow involves the same
macroscopic mixing action for heat exchange as for mass exchange, and
this completely overwhelms the contribution of molecular diffusion.
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c. 1

APPENDIX C

DEFINITION OF MASS TRANSFER COEFFICIENTS AND

MASS TRANSFER DRIVING POTENTIALS

Single phase mass transfer in a binary mixture takes place via a

phenomenon known as molecular diffusion. The basic relation describing
molecular diffusion is called Fick's law. This states that the mass
flux is proportional to the concentration gradient as
fallows

ac
-D ay (C.l)
The subscript rel in the massflux term indicates that the massflux
given by this relation is expressed in respect to moving coordinates.
This is the massflow observed by an observer travelling with the bulk
flow.

The absolute mass flux relative to a stationary observer would be given

by

( ~ ) abs • ( ~ ) re 1 + cv bu 1k (C.2)

Mass transfer between different phases is known as convective mass

transfer. Experiments on convective mass transfer have shown that the
transfer of mass across an interface can be expressed by a relation of
the form:

rate of transfer = transfer coefficient x area x driving potential

(C.3)
This form of rate equation corresponds to the form given to governing
mass transfer equations in the bulk of the literature. An equation of
this form expresses the mass transfer relative to stationary
 Stellenbosch University http://scholar.sun.ac.za
C.2

coordinates.

Bird, Stewart and Lightfoot [66BI1] stated that the mass transfer
coefficient, as defined by equation (C.3), is independent of the mass
transfer rate at only very low mass transfer rates. Thus mass transfer
coefficients defined with respect to stationary coordinates would be
dependant on the massflow rate at high massflow rates.
This effect arises from the distortion of the velocity and
concentration profiles by the high massflow rate across the interface.

Various driving potentials for mass transfer are employed in the

literature. The more popular driving potentials used include
concentration difference, mass fraction difference, mole fracture
difference, vapour· pressure difference and thermodynamic activity
difference.

In mass transfer processes across a phase interface three resistances

to the mass transfer are encountered; the liquid phase, the interface
itself and the gas (vapour) phase. Various authors including Treybal
[55TR1], Bird et al.[66BI1], Skelland [74SK1] and Foust et al. [80F01]
have studied interphase mass transfer by defining an overall mass
transfer coefficient and an overall driving potential.

Treybal [55TR1] used concentration differences as the driving potential

in the liquid phase and partial pressure differences as the driving
potential in the gas phase to derive a simple governing relation for
interphase mass transfer as

(C.4)
where
 Stellenbosch University http://scholar.sun.ac.za
C.3

The coefficients kg and kl are defined by the following single phase

mass transfer equations

DJ = kg ( Pg - p.
1 ) (C.5)

= kl ( cl - c.1 ) (C.6)
It is assumed that the vapour pressure at the interface is a linear
function of the liquid concentration at the interface as expressed by

pi = aci + b (C. 7)
ap.1
:. a =
ac.1 (C.8)
The composition p* does not physically exist, but it represents a gas
(vapour) phase composition which would be in equilibrium with the
average liquid composition at the point under consideration.

In cooling tower theory where the mass transfer involves the

evaporation of water into air, the driving potentials which are
normally used are humidity ratio differences or vapour pressure
differences. The governing mass transfer equation can thus be
expressed as

[~ ) = hoi ( wasi - wa ) (C.9)

or

[~ ) = hDpi ( Pasi - Pa ) (C.lO)

Since the interface temperatures are not always easy to determine in

 Stellenbosch University http://scholar.sun.ac.za
c.4

cooling tower applications it has been customary to use the average

water temperature instead of the interface temperature to define the
mass transfer coefficients. Equations (C.9) and (C.lO) could then be
written as

(C.ll)
and

(~) = hop ( Pasw - Pa ) . (C.12)

Berman [61BE1] showed ho~ hop values could be converted to ho values.

Following the method of Berman the relation between ho and hop can now
be determined.

From the definition of the absolute humidity ratio it follows that

Pa • [ wa :•0,622 ] Patm

Since the term ( wa/0,622 ) is much smaller than unity for air water
systems the term [ wa/(wa + 0,622) ] can be simplified as follows

wa
0,622
l
wa
:. Pa : : : 0,622 (C.13)
similarly

wasw wasw
Pasw - 0,622 [ 1 - 0,622
l Patm (C.l4)
 Stellenbosch University http://scholar.sun.ac.za
C.5

Setting equations (C.l3) and (C.l4) into equation (C.l2) leads to

[
m
A
l = hop
[ wasw
0,622 Patm

= hop
[ wasw - wa
0,622 -
( wasw + wa ) ( wasw - w.)
( 0,622)2
J Patm

= hop
[ wasw - wa
0,622 [I - [ wasw + w•]] ]
0,622 Patm

Comparing this result equation (C.ll) we note that

(C.l5)

If it is further more assumed that

equation (C.l5) can be further simplified to

Patm ]
[ 0,622 (C.l6)

Berman stressed that care should be taken when converting hop values
into ho values, since considerable errors may be introduced because of
the simplifications used. The reason for this lies in the fact that
the relatively small errors made in the simplifications may be
 Stellenbosch University http://scholar.sun.ac.za
C.6

significant when compared to the driving potential ( Pasw- Pa ).

Various analytical models for the determination of the mass transfer

coefficient exist in the literature, the most prominent models are
the "two-film" theory of Whitman [23WH1], the "penetration" model of
Higbie [35Hll], the "surface renewal" theory of Danckwerts [51DA1] and
the "film penetration" theory of Toor and Marchello .[58T01].

In cooling tower theory empirical relations are normally used to

determine the mass transfer coefficient. Chapter 3 gives a summary of
available mass transfer coefficient correlations which apply to the
operation of evaporative coolers and condensers.
 Stellenbosch University http://scholar.sun.ac.za
D. 1

APPENDIX D

SINGLE PHASE PRESSURE DROP ACROSS PLAIN TUBE BUNDLES IN CROSS-FLOW

The pressure drop, Ap, in cross-flow across a tube bundle is given by

pi
Ap = Kn - -
2 (D.l)
where

K = f ( Re, geometrical constants) and

pvd
Re = (D.2)
J.L

Here~ K is the pressure loss coefficient, n characterizes the number of

rows in the bundle, Re is the Reynolds number, d is the characteristic
length, v is the characteristic velocity, p is the density and J.L is the
dynamic viscosity of the fluid.

Various choices of d, n and v are used in the literature. E~uation

(D.l) is valid for an ideal tube bundle. An ideal tube bundle is

defined as a tube bundle which conforms to the following

i) the velocity in the free cross section is constant,

ii) the velocity is perpendicular to the tube bundle,
iii) the flow is isothermal,
iv) the number of tube rows ~ 10,
v) the number of tubes per row ~ 10,
vi) and the ratio of the tube length to diameter> 10.

Deviations from the ideal situation are allowed for by the use of
correction factors. The different tube configurations and the
geometrical parameters which have an influence on the pressure drop
coefficient are shown in Figure 0.1.
 Stellenbosch University http://scholar.sun.ac.za
0.2

b c

Figure 0.1 Tube array configurations: a) in-line, b) staggered with the

narrowest cross section perpendicular to the air flow and
c) staggered with the narrowest cross section along the
diagonals.

The following parameters are used in the calculation of the pressure

drop across a tube bundle:

(0.3)

For in-line tubes and for staggered tubes with the narrowest cross
section perpendicular to the flow the following parameters are used

(0.4)
1
= { 4a - 1r
4a (0.5)
a
vmax = ( a - 1 ) (0.6)
 Stellenbosch University http://scholar.sun.ac.za
0.3

For staggered tubes with the narrowest cross section along the diagonal
the following parameters are used

dec -i ~ I i 4c ~ wI do (0.7)
(a/(2c))
•
" { 4c - n } v~
vm
4c (0.8)

a
vmax = 2 ( c - 1 ) v~ (0.9)

According to Bell [63BEI] the flow through the tube bundle will be
laminar if Re < 100 and turbulent if Re > 4000. The flow is in the so-
called intermediate regime when 100 < Re < 4000.

Chilton and Generaux [33CHI] proposed different equations for pressure

loss coefficient in laminar and turbulent flow across tube bundles.

If the flow is laminar the proposed relations for this method are

L
n = deb , d = deb , v = vmax

and

106
K = Re

If the flow is turbulent then the following relations are to be used

n = "rows ' d = ( a -· 1 ) do ' v = vmax

and

1,32
K
 Stellenbosch University http://scholar.sun.ac.za
0.4

for a staggered configuration or,

3,0
K = Re0,2

for an in-line configuration.

Jakob [38JA1] proposed the following equation for determining the

pressure drop coefficient when the flow through the tube bundle is
turbulent.

For a staggered tube layout

1 0,47
K = Re 0,16 { I +
( a - 1 )1,06 }
and for an in-line tube 1ayout

1 0,32
K = Re0,15 { 0,176 +
( a - 1 )(0,43 + ( 1,13/b)) }
where

n = "rest' d = do ' v - v
max
Gunter and Shaw [45GUI] proposed the following equations to determine
the pressure drop coefficient of laminar flow across a tube bundle.

For a staggered layout

K
 Stellenbosch University http://scholar.sun.ac.za
D.S

and for an in-line layout

where

L
n = d = deb' v = vmax
deb '
Gunter and Shaw also proposed equations to determine the pressure drop
coefficient when the flow through the tube bank is turbulent.

For a staggered layout

K
- 1,92 J 4 •: - w 10,4 J =a l0,6
Re0,145 l J l J
and for an in-line layout

1,92
K = Re 0, 145·

where

L
n = deb ' d = deb' v = vmax

Bergelin et al.[50BE2] gave the following equation to determine the

pressure drop coefficient across a tube bundle if the flow is laminar.

If the layout is staggered with b < % ( 2a + 1 )0,5

= 280 J ~ l1,6
K
Re lcJ
If the layout is in-line or if .the layout is staggered with
 Stellenbosch University http://scholar.sun.ac.za
0.6

b ~ ~ ( 2a + 1 )0,5

where

280 J ~ ll,6
K
~
=
laJ
where

n = nrest' d = deb' v = vmax

Zukauskas [68ZU1] presented graphs to determine the pressure drop

coefficient for both laminar and turbulent flow across tube bundles.
According to Zukauskas the pressure drop coefficient can be written as

where k1 is a constant which is determined by the geometry of the tube

configuration. Zukauskas used the following characteristic values, n =
nrows' d = d0 and v = Vmax to determine the Reynolds number and
pressure drop coeff~cient. Figures 0.2 and 0.3 are reproductions of
the graphs given by Zukauskas.
 Stellenbosch University http://scholar.sun.ac.za
0.7

' -"~ 1 ! I I I I I I I I I I : I I I I I I I _
4

N f'-N_
I I Ill I I Ill I I II I I -
I 6
""'
-~
I
Re':'10'
I I

.,...
u 10 '
I"[' I~ I !!¢~!1! I I :~rill I I -
.,... ' ' - I

II )r:NI I i -
8
1+- I I I ! '-, I.: '-,I I I • =b I i ! I
6
1+- I f'{ "\-.. I I I II
I I ~ill
Q) 4 - !
10"
i i I ! I l i ' I 10', 10"
! liN"' ~~ ~/t 1 I II I
125
0
u
I
i 8
6
1I I I I I ~ I
I~ -
I
I , II I ! I II
Q.
0
s.. 10 . ' , I I I"' rt'' ~I I !' I. I I I I I I II I l'::--
8
I
:
I
'
I 4

-6·810'' 2 4
I
6 8 10" 2 4 6:
"C 6 ! i i i ' I N.. ......... I"'., I I I I io-11/lb-11
Q) ! I l! I II!~ ~f'i-.. !
'
I , 1.5? I I II I I I II -
s..
:::s
VI
4

1 I I,'I I I Ill l.~':r-'

2.00 t , I
I ill~
-4_ : I
.' I I
'
i I I
' ;

i
Ill I I II
VI

~
Q)
10' '
!
4 6 B10' 2. 4 6 8 10•
I I !
2 4
II
6 010'
I
:2 4
I II i 6 8 10"
i I I II . :2:So/ I Ill
2 4 6 810 2 4 6. 8 10

Reynolds number, Re

Figure 0.2 - Pressure drop coefficient for in-line tube banks.

6 ~~r=·~I:::=J::IJI::[I::[}IJI::r==c~~~~~~====~~~~
4 ~ "J. I J II I I I I I

i 2 r--'\,-:-"-~+-il\"-~-+~-:-~-+1-+1-i-1- ~ ¢ I :·: I <10'; o• 1 I Re = 1 O' J

~ 1 ~·
1+-
,· I' ~~~)...I "}
>- I i ! I 1
! ! !! ~
/ ' - - / Ji
/
II I' .:1 ..2 ~.~ I I
10
110
1

':v ! "'- I I I
t-~:'--f--+,"f:::l"'-:---"~,"--:---:,~-+~-+,-.:...,-,:--+1-+1-+1-+-1-+-+
/
. "-.J o• N /._,/ cr >- I
1
1 1

e I I! I ~I"~ N-·=, 1.2!5

t--il---:-++--+---=~~*-~~r--+-++;__+--+--+
I I 1.0~"' //
o/ I I I . . . _
__ __.--
I o•-...,
~
"U I
2 1 1 4

g- I I I"~ 6 8 ~ 4 10"
1
...
i; ~· ~~~~·~'~-=~~~=t'~~~·~E:~~~~,.._~±·~~·t£1j:±:::;t·;j':::j:;jl:tl=·~lbt:l=·~,:tt:t::3- .....
E 4 r-i'-rn'~-+~-~rt~~+-~~~~f=P•H-~-f~~~~~-~-~~~-,5v~~~~~~~~-r'~~-Ttllrl ~
1:.tjj~±=tWtJ=±'ftt~~t'ojjti~~~J~~-,~.~~~E. f~,
1 1 1 1 11 1 1 1 1

~
:::s
c..
I I I I 2 4 6810' 2
I 4
I I I 6810'
12.J,1 l
2 4
I
6810' 2 4 6810' 2 4 6810' 2 4 6810" 2

Reynolds number, Re

Figure 0.3 - Pressure drop coefficient for staggered tube banks.

Zukauskas and Ulinskas [83ZUI] presented previous data [68ZU1] in the

form of equations.
 Stellenbosch University http://scholar.sun.ac.za
0.8

Kast (74KA1] proposed an equation for the pressure drop coefficient for
staggered tube bundles with the narrowest cross section perpendicular
to the direction flow. Charts were given for in-line tube bundles and
staggered tube bundles where the narrowest cross section is along the
diagonal.

Kast used the following characteristics values for the determination of

the Reynolds number and pressure drop:

do
n = d "rows, v = vm
ec
The equation for K for staggered tube bundles with the narrowest cross
section perpendicular. to the direction of flow is given as

128 4 } '
K = Re + Re0,16
{

Note that this ~quation holds for all flow regimes from· laminar to
turbulent. Figure 0.4 and Figure b.S gives the charts for determining
the pressure drop coefficient for the in-line tube configuration and
the staggered layout (when the narrowest cross section is along the
diagonal) respectively.

,. I
.. '

. ..• ,. e.e-
I I I I I I I
~. -
-----
I I I
~
c: K~
.....QJu ['\r: ~
I
I I
I I
I
II - ee
.,... 111 ~ '
I
I .. o. i I I I i
' I I
4-
4- ' I "'-"'i-. x~ I I I I I
QJ
0
u ""'{.~ I I
·~zr
I
...
~~-~'
11-J.f)
Q. T
0
s.. .:: '
I d"(,-1, tJ.•U·ti-3./JLG.• •:J,JJ
-
II.•U~
'
"0
'·• I 1- ~.4¢ r--1'"4'=' t~.•IJII6b•~
QJ
s.. II "' ~~~
I I I

"'
1/11 b•IJ5 ·- a.•U6 b•t,JB I
::s z ~ ~ I
H~ ~.\%5
Ul
Ul
QJ 111
I I' ['...
I
II' I
I
I
' I I
'' 'I
'
I
'
.
~

s.. I I I I I! I I ! I I
'
I I
'' .
I I I I
Q.. '. ,,,, '. ',, '' ' ' ld
I • • '111 z • f I 111• l
• I I 1/J<I z • 4 ' 10• z • • 10• z • I 11/J

Reynolds number, Re

Figure 0.4- Pressure drop coefficient for in-line tube bundles.

 Stellenbosch University http://scholar.sun.ac.za
0.9

'. ' '

I I I I I ! I! I I Ill I I Ill I I i ,i I iif1'
I- I I I I I I II I' I Ill I I Ill I' I Ill I Ill
' I :!
I
I i'
' ' I:
I
! I I I ' I' ' I I
....
1
.
I
'.'
•
I I I
'
f 17D'
I

2
t
I
''I

•
I

f
I I

I 10 1
'
I
4
I I l

4 • 1oJ
I
'
2
I ' '
' • ro•
'
'
'
.
'' I I I
I''"
• ' 10'
i
z •
I

I HD

Reynolds number, Re

Figure 0.5 - Pressure drop coefficient for staggered tube bundles where
the narrowest cross section is along the diagonal.

Gaddis and Gnielinski [85GAI] developed comprehensive equations for the

pressure drop coefficient through in-line and staggered tube bundles.
These equations take into account the effect of number of rows of tubes
and the effect of heating/cooling of the fluid.

The following characteristic parameters were used

n = "rest ' d = do ' v = vmax

These equations are valid for the following ranges I~ Re ~ 3 x 10S and
0 rows ~ 5.

For an in-line configuration of tubes,

=
J Re + 1000 l}
K Ki,l fzn,l + ( Ki,t fz,t + fn,t) {I- exp l- 2000 J
 Stellenbosch University http://scholar.sun.ac.za
D. 10

and for a staggered configuration of tubes,

Re + 200 ~
K = K.1, 1 f zn, 1 + ( K.1, t f z, t + f n, t ) { 1 exp Jl - 1000 J
where

fa,1
i) Ki,1 = Re
For an in-line configuration or a staggered configuration with the
narrowest cross section perpendicular to the flow,

280 ~ ( ( b0 ' 5 - o,6 ) 2 + o,75)

f a, 1 =
( 4ab - ~ )a 1 , 6
while for a staggered arrangement with the narrowest cross section
along the diagonal

280 ~ ( ( b0' 5 - o,6) 2+ o,75)

fa,1 =
( 4 ab - ~ ) c 1, 6
ii) For an in-line configuration

Ki,t = Re0,1 (b/a)

and for a staggered configuration

f a,t,st

Ki,t = Re0,25
 Stellenbosch University http://scholar.sun.ac.za
D. 11

with

fa,t,il =
10,22 + 0,12
r- ~t 13
( a - 0,85 ) '
6

0,47 ( b/a _1,5 J)

+ 0,03 ( a - 1 ) ( b - 1 )

and

1,2
fa,t,st= 2, 5 + ( 1 06
a - 0,85 ) ' 3
- 0,01 { ~ - 1}

iii)If nrows < 10 then

0,25 .
0,57 "rows
{ 10 } .
where k • ll 4;b - 1 J Re J 0,25
and if nrows > 10 then

fzn,1. = { J.L: }k 0,57

~ where k • { { 4;b 0,25
- 1 } Re }

iv) fz,t- { :w }0,14

v) If 5 ~ nrows < 10 then

1
f
n,t
= K0 {
"rows - :o }
 Stellenbosch University http://scholar.sun.ac.za
D. 12

and if nrows ~ 10 then

fn,t = 0

For an in-line configuration and for a staggered configuration with the

narrowest cross section perpendicular to the direction of flow

and for a staggered configuration with the narrowest cross section

along the diagonal

c -
a - ::f
Comparison of the different correlations

The pressure drop across a typical bundle of tubes is evaluated with

the different correlations method in order to compare the methods.

Example: nrows = 10 staggered layout

a = 2
b = 3%
c =2
d 0 = 38,1 [ mm ]
J.Lw:=:::J.1.=1,8x 1o-5 [ kg/ms ]
p = 1,2 [ kg/m3 ]

In order to compare the different equations the pressure drop

coefficient is based on the so-called Ry-number, proposed by Kroger
[88KR1].
 Stellenbosch University http://scholar.sun.ac.za
0.13

The Ry-number is defined as

pv ro
Ry =

·since the different correlations are based on different characteristic

values of n,d and v the product of pressure loss coefficient and n was
calculated in order to make the results comparable.

The variation of the product of the pressure loss coefficient and the
characteristic number of tube rows vs Ry-number is shown in Figure 0.6.

*NOTE :For an in-line configuration and for a staggered configuration

where the narrowest cross section is perpendicular to the
direction of flow, nrest = nrows·

For ·a staggered configuration in which the narrowest ·cross

section is along the diagonals, nrest = nrows - 1.
 Stellenbosch University http://scholar.sun.ac.za
D. 14

4.5 """'r\
\
'
1\'I\
:\ ~ 1\ ' '
····... ~' i\,.
c. 4 "
.............. '
r.
r--. r-..\
:::...::
~... ' ~
~

l\
~

~
(ll. ··..... '
0
.--
r.\'·. ~-
~'
~
(ll
"0 3.5
~
"'
Q)

'
~
.u,...
~

0
.--
Q) 3 ~- .

'~
>
~'

2.5

2
10000 100000 100000
Ry - number [m-1]
-Jakob - - Gunter et a1 - Zukauskas et a1
----- Kast ~ Gaddis et al

Figure 0.6 Single phase pressure drop across a bundle of tubes.

 Stellenbosch University http://scholar.sun.ac.za
E. 1

APPENDIX E

DERIVATION OF THE DRAFT EQUATION FOR A NATURAL DRAFT

CROSS-FLOW EVAPORATIVE COOLING TOWER

Consider the cross-flow evaporative cooling tower shown in figure E.l,

with evaporative cooler units placed around the outer perimeter of the
tower base. The density of the heated air inside the tower is lower
than the density of the ambient air causing a lower pressure inside the
tower than the ambient pressure ai the same elevation.
An airflow is induced through the tower as a result of this pressure
differential. At the operating· point of the tower the air flowrate
through the tower would reach a value at which the pressure change due
to flow resistances encountered by the airstream and the changes in
elevation inside the tower would be in balance with the pressure
change, due to elevation change along the outside of the tower.

In the atmosphere outside the tower the following relation describes

the pressure change with changing elevation,

dp = - pg dz (E.l)

Assuming air to be a perfect gas the following holds

p
(E.2)

The dry adiabatic lapse rate in the atmosphere is

dT
dz = 0,00975 r·c;m]
resulting in the following temperature profile in the atmosphere ·

Ta = Tal - 0,00975 z (E.3)

 Stellenbosch University http://scholar.sun.ac.za
E.2

--------.- ®

Tower shell
-
c....

Evaporative
cool-er unit

Cooler unit

Figure E.l Layout of natural draft closed circuit cross-flow evaporative

cooling tower, showing the reference numbers used in the draft
equation.
 Stellenbosch University http://scholar.sun.ac.za
E.3

Substituting equations (E.2) and (E.3) into equation (E.l) and

integrating between positions 1 and 6 results in

(E.4)

with

Ra z 287,08 J/kgK and g = 9,8 m2/s

The air outside the tower accelerates from v1 = 0 m/s at point 1 to v2

at point 2. Application of the energy equation between points 1 and 2
gives

] = Pal 1-
(E. 5)
Between points 2 and 3 the air flows through the evaporative cooler
'
coils and the drop separators. If the coils are positioned in an A-
frame configuration there is an additional jetting or oblique flow
pressure drop. This can be expressed mathematically as

(E. 6)
 Stellenbosch University http://scholar.sun.ac.za
E.4

The flow between positions 3 and 4 changes direction and elevation,

expressed ·by the energy equation as

[ Pa3 +
P3 v~
2 ]-[ Pa4 +
~2]
2 + P34g ( H4 - H3 )
(E. 7)
Between positions 4 and S the airflow can be described by

2 2
4
P v4 ] [ Ps vs ]
[ Pa4 + 2 - Pas + 2 = P4s g ( Hs - H4 ) (E.S)

The pressure difference between positions 1 and S can be determined by

substituting equations (E.6), (E.7) and (E.8) into equation (E.s)· and
simplifying the result as follows,

2
P23 v23
Pal - Pas = Khe 2

- 0,0097S H2 ]0,0097S Ra
+ Pal Tal (E. 9)

At the operating point of the tower the pressures inside and outside
the tower must be in balance, i.e.

Pal - Pas= Pal - Pas (E.lO)

By substituting equations (E.4) and (E.9) into equation (E.lO) the
natural draft equation can now be determined as
 Stellenbosch University http://scholar.sun.ac.za
E.5

+ (E.ll)

where

P4 = P3

1
+-
p3
l- 1

Ps =R
. a
[ Ta3 + ~pa ( Hs - H2 ) l
P3s = ( P3 + Ps ) I 2

From continuity it follows that

v23 = P23
1 [ rna
Afr
l (E.12)

[ :;r l
1
v3 = p3 ( E.13)

v4 = p4
1
[ Afr
A4
][ ~
Afr
l (E.14)
 Stellenbosch University http://scholar.sun.ac.za
E.6

(E.15)
Equation (E.11) can be simplified by employing equations (E.12),(E.13),
(E.14) and (E.15) ..

2
P23] P23 [ Afr ]
= [ p3 + Kct p A + Ps
4 4

1
(E.16)
2 p 23

Equation (E.16) is the final form of the draft equat~on for natural
convection cooling towers.

The cooling tower loss coefficient for a tower with vertical heat
exchangers in the tower inlet was determined by Du Preez and Kroger
[88DU1] as
2

Kct = 2, 98 - 0, 44 [ :: ] + 0,11 [ :: ] (£.17)

Drift eliminator pressure loss coefficients range between 2,2 and 7,3
according to Chilton [52CH1] and Chan and Golay [77CH1]. A design
value of Kde = 5 was used throughout this investigation.
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E.7

The pressure drop and the associated pressure loss coefficient for
airflow across a wet tube bundle can be calculated with the
correlations presented in Chapter 3.

The oblique flow pressure loss coefficient for a heat exchanger with an
A-frame layout was correlated by Kotze el al. [86KOI] as

(E.l8)

where

and
_ 0,0019 [ ~ r+ 0,9133 [ ~ J _3,1558 (~.19)

• exp [ 5,488405 - 0,2131209 [ ~ J + 3. 533265 r~r

_ 0,2901016 [ ~ rl (E. 20)
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F. 1

APPENDIX F

SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS

USING THE 4TH ORDER RUNGE-KUTTA METHOD

For the single equation initial value differential equation problem

dy
dx = f ( x,y )

Y ( Xo ) = Yo

approximate values of Yn must be calculated at point Xn = x0 + nh

where n = 1,2,3 ... and h =step size.

The fourth order Runge-Kutta method allows the calculation of Yn+l at

the point Xn+l from the known function value Yn at xn~ According to
this method the new function value can be calculated by

where

al = h f( xn , Yn )

a2 = h f( xn + h; 2, Yn + a1; 2 )

a3 = h f( xn + h; 2, Yn+ a2; 2 )

a4 = h f( xn + h,yn + a3 )

According to Collatz [86C01] the step size, h, should be chosen such

that the values of k2 and k3 coincide to within at least two decimal
places.
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F.2

Van Iwaarden [77VA1] shows how the fourth order Runge-Kutta method can
be extended to a system of first order initial value problems.
Consider the following system of two differential equations and two
initial values

dy
dx = f{x,y,z)

dz
= g(x~y,z)
dx
y ( Xo ) = Yo

z ( Xo ) = zo

The fourth order Runge-Kutta method now becomes

Yn+l = Yn + ( a1 + 2a 2 + 2a 3 + a4 ) I 6

zn+l = zn + ( b1 + 2b 2 + 2b3 + b4 ) I 6

where

=h f( xn, y n, zn )

= h g ( xn, y n, zn )

a2 = h f( xn + h/2, Yn + a1;2, zn + b1/2 )

b2 = h 9( xn + h/2, Yn + a1/2, zn + b1/2 )

a3 = h f ( xn + h/2' yn + a2/2' zn + b/2 )

b3 = h 9( xn + h/2, Yn + a2/2, zn + b2/2 )

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F.3

a4 = h f( xn + ·h, Yn + a3 , zn + b3 )

b4 = h g( xn + h, Yn + a3 , zn + b3 )

This method i.s self starting (no initial estimates are needed) and the
new y and z values are calculated £fter calculating the required a's
and b's.

The fourth order Runge-Kutta method can easily be extended to solve any
number of simultaneous ordinary differential equations.
The following example shows how the fourth order Runge-Kutta method can
be used to solve the simultaneous differential equations governing the
heat and mass transfer processes of a single element.

The governing differential equations, according to the Merkel model,

are

dia = K2 ( i asw - ia )

dTw = -K3 ( i a·~w- i a ) + K4 ( Tp - Tw)

U0 dA 0
where Kl =
mp cpp

hD dA 0
K2 =
rna

hD dA 0
K3 =
mw cpw

U0 dA0
K4 = mw cpw
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F.4

Assume the following values for the governing variables:

dA 0 = 0,25m2
ho = 0,25 kg/m2s
rna = 0,25 kg/s
mw = 0,5 kg/s
mp = 0,6 kg/s
Cpw = 4190 JjkgK
Cpp = 4190 JjkgK
Tpi = 50°C
Twi = 35oc
iai = 55 kJ/kg
U0 = 1500 Wjm2K

The constants in the differential equation model can now be determined

as

KI = 0,1492
K2 = 0,25
K3 = 29,833 X Io-6
K4 = 0,1790

The Runge-Kutta method proceeds as follows

Step 1:

Ai 1 = ias ( Twi ) - iai = 74567 J/kg

AT 1 = Tpi - Twi = l5°C

a1 = -K1 AT1 = -0,1492 (15) = -2,238

b1 = K2 Ail = 0,25 (74567) = 18641,75

c1 = -K3 Ail + K4 AT= -29,833 x l0-6(74567) + 0,179 (15) = 0,4604

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F.S

Step 2:

Al 1 =las [ Twl + c~)- [ lal + b~) = 66860,125 Jfkg

ATw • [ Tpl + a~ ) - [ Twl + c! ) • 13,651 'C

~ a2 = -K1 ~T2 = -0,1492 (13,651) = -2,0367

b2 = K2 ~i2 = 0,25 (66860,125) = 16715,031
c2 = ~K3 ~i2 + K4 ~T2 = -29,833 x 10-6 (66860,125)
+ 0,179 (13,657) = 0,4489

Step 3:

Al .- \s [ Twl +
3
c~ ) - [ lal + b~ ) - 67785,095 J/kg

AT 3 = [ Tpl + a~ ) - [ Twl + c~ ) - 13 , 7573 'C

a3 = -K1 ~T3 = -0,1492 (13,651) = -2,0367

b3 = K2 ~i3 = 0,25 (67785,095) = 16946,294
c3 = -K 3 ~ij + K4 ~T 3 = -29,833 x 10-6 (67785,095)
+ 0,179 (13,7573) = 0,4403

Step 4:

~;4 = ias ( Twi + c3 ) - ( iai + b3 ) = 60645,526 J/kg

~T 4 = ( Tpi + a3 ) - ( Twi + c3 ) = 12,5071 oc

a4 = -K1 ~T4 = -0,1492 (12,5071) = -1,866
b4 = K2 ~i4 = 0,25 (60645,526) = 15161,382
c4 = -K 3 ~; 4 + K4 ~T 4 = -29,833 x 10-6 (60645,526)
+ 0,179 (12,5071) = 0,4295
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F.6

The other conditions of the element can now be determined as

Tpo = Tpi + ( a1 + 2 ( a2 + a3 ) + a4 ) I 6 = 47,953°C

iao = iai + ( b1 + 2 ( b2 + b3 ) + b4 ) I 6 = 71,854 kJ/kg

Two = Twi + ( c1 + 2 ( c2 + c3 ) + c4 ) I 6 = 35,4447°C

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G. 1

APPENDIX G

CORRELATIONS FOR CONVECTIVE AND CONDENSATION HEAT TRANSFER

COEFFICIENTS ON THE INSIDE OF TUBES.

Kroger [88KR1] presented a comprehensive summary of the available heat

transfer coefficient correlations for the flow of fluids inside ducts,
covering the laminar, transitional and turbulent flow regimes.

According to Kays [55KA1] the heat transfer coefficient during laminar

flow ( Rep < 2300 ) inside a duct with a constant wall temperature can
be expressed by

0, 104 ( Rep Pr P ( di I L ) )
= 3,66 + 08 (G.1)
1 + 0, 016 ( Rep Pr P ( di I L ) ) '

Gnielinski [75GN1] proposed the following equation for the heat

transfer coefficient on the inside of a tube in the turbulent flow
regime

( f of 8 ) ( Rep - 1000 ) Pr p ( 1 + ( d;f L ) 0' 67 )

Nu p =
1 + 12,7 ( fof8 )O,S ( Prp 0, 67 -1) (G.2)

where, the friction factor fD for smooth tubes is defined by Filonenko

[54FI1] as

-2
= ( 1,82 log 10 Rep - 1,64 ) (G.3)

Equation (G.2) is valid for the following ranges

2300 < Rep < 106

0,5 < Prp < 104

 --------------------------~--------------------~

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G.2

o < ( di I L ) <1

If the fluid properties vary significantly along the flow path the
following corrections must be made to the turbulent heat transfer
coefficient correlation and the fluid friction factor:

i) The right hand side of equation (G.2) must be multiplied by one

of the following correction factors

a = ( Pr I Prwall ) oP,ll (heating) (G.4)

a oP, 25 (G.5)
= ( Pr I Pr wall ) (cooling)
ii) The isothermal friction factor must be multiplied by the following
correction factor

a _ (
- ~wall
I ~
) o,
p
25 (G.6)

The following equatio·n proposed by Chato [62CH1] can be used to

determine the condensation heat transfer coefficient in essentially
horizontal tubes.

0,25

0,555
J (G. 7)
with

ifg = ifg + 0,68 cpc ( Tr - Twall )

This equation is only valid for relatively low vapour velocities
specified by the range

< 35000 at the tube inlet.

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G.3

For higher vapour velocities it is advisable to use the correlation

proposed by Shah [79SH1],

(G.S)

where hL is given as

04
hl = 0,023 Re~,a Prc ' ( kc 1 di ) (G.9)
with

Rec = [ Pc :: d; l (G.IO)

Pre = [ cpc ~c l (G.ll)

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H. 1

APPENDIX H

DETERMINATION OF THE AIR/WATER INTERFACE TEMPERATURE

The convective mass transfer coefficient between a water film and an

airstream (see Appendix C) is expressed as

J dmw = hD.1 (H. I)

but since the interface temperature Ti is difficult to determine, the
assumption Tw = Ti has often been made in cooling tower theory.
The mass transfer is then expressed as

(H.2)
A simple model is now proposed for the determination of the interface
temperature. The assumption of Ti = Tw does not have to be made when
this model is employed. Consider the typical temperature profile in
Figure H.l.

~--+-------Wall temperature, Twall

~-----Bulk temperature, Tw
~----- Air/Water interface temperature, Ti
Air temperature, Ta

Wall .[~ater fil~l- Air

Figure H.l Schematic representation of the water film flowing down a

vertical surface.
 Stellenbosch University http://scholar.sun.ac.za
H.2

If a linear temperature profile is assumed through the water film this

profile can be determined as

( Ti - Twall }
T(y) = o Y + Twa 11 (H.3)

From the Nusselt analysis of condensation on an inclined surface

the velocity profile in . the 1iquid film is given as

( Pw. - pa ) g sin 9
v =
x ~w (H.4)
According to the definition of bulk recirculating water temperature it
follows that

J ( cpw Pw vx T ) dy
0

(H.5)

By substitution of equations (H.3) .and (H.4) into equation (H.5) and

assuming Cpw to be constant it follows that

5 3
- T. + - T
8 1 8 wall (H.6)
By defining the film coefficient as

qn = hwi ( Twa 11 - Ti ) (H. 7)

it follows from equations (H.6) and (H.7) that

3 qll
T.1 = Tw 8 h
wi (H.8)
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H.3

If equation (H.S) is used together with the controlling differential

equations to evaluate a typical element of an evaporative
cooler or condenser, the assumption of Tw z Ti does not have to be
made.
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1.1 http://scholar.sun.ac.za

APPENDIX I

CORRECTION OF HEAT TRANSFER COEFFICIENT AT HIGH MASS TRANSFER RATES

If a water film is in contact with an air stream two heat transfer

mechanisms are involved in cooling of the water film, i.e. sensible
single phase heat transfer from the water surface to the air and the
latent heat transfer associated the mass transfer (evaporation) of
a part of the water into the air stream.

This can be mathematically expressed as

(I. I)

The first term in equation (I.l) accounts for the sensible heat
transfer and the second term accounts for the latent heat transfer. If
only sensible heat transfer took place equation (I.l) would become

qll = hc (T·1 -T)

a (I. 2)

Note that the 'F1 factor does not appear in equation I.2. The term F1
accounts for the effect of the mass transfer on the sensible heat
transfer when the heat and mass transfer processes take place
simultaneously.

Ackermann [37AC1] showed that

cl
Fl =
1 - e-el (I. 3)

. where

cpm ho; ( wasi - wa )

cl =
he
 - - - - - - - - - - - - - - - - - - - - - - -

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1.2

From the Chilton-Colburn analogy it follows that

2
= ( Le )/ 3

- 2I
:. cl = ( wasi - wa ) ( Le ) '3 (I. 4)

For air/water mixtures at ambient conditions the Lewis number is

typically Le = 0,866 ; it follows from equation (!.4) that

q z 1' 1 ( wasi - wa )

In the temperature range 20°C < Ti < sooc the correction factor
typically varies between 1,003 < F1 < 1,043

The correction F1 is always larger than unity, since the evaporation of

water from the surface gives a net mass flux from the surface in the
direction of the heat transfer which increases the heat transfer,
because of increased boundary layer activity.

The correction factor is negligible in the normal operating range of an

evaporative cooler ( 20° < Ti < sao ) since it influences the sensible
heat transfer, which represents only about 15% of the total heat
transfer, by less than 5%.

At higher temperatures this correction factor may become · more

significant, eg. at Ti z 60°C the correction factor is F1 z 1,08 and at
Ti z 70oc it is F1 z 1,155.
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J. 1

APPENDIX J

EVALUATION OF CONVENTIONAL COOLING TOWER PACKING

IN A COMBINATION EVAPORATIVE COOLER

Consider a horizontal slice of a conventional counterflow cooling tower

packing with airflow from below and recirculating water flowing through
the packing from above. The following Merkel-type equation describes
the heat and mass transfer in a section of cooling tower packing of
thickness dz,

dq = ho ( iasw - ia ) dA ( J .1)

From the energy balance of a section of fill it follows that

dq = rna dia

= mw Cpw dTw (J.2)

By rewriting equations (J.l) and (J.2) the following two controlling

differential equations can be found

(J .3)

and

( i asw - i a ) dA
(J.4)
For a typical 12mm Munters type extended film packing Cale [77CA1]
states that

a = 243 m2;m3
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J.2

and

~] [ mw ] -0,585
[ Afr rna (J.5)
The surface area of a typical element can be expressed as

dA = a Afr dz (J.6)

By employing equation (J.6) equations (J.3) and (J.4) can be expressed

as

ho a
dia = ( i asw - i a ) Afr dz ·
rna (J. 7)

and

ho a
dTw =
mw cpw ( i asw - i a ) Afr dz (J .8)

By using a numerical solution method such as the 4th order Runge-Kutta

method these two equations can be numerically integrated through the
fill, with the numerical integration starting from the air inlet side.
The outlet conditions of the water and air can be determined by a
simple iterative search procedure.
 Stellenbosch University http://scholar.sun.ac.za
K.l

APPENDIX K
RESULTS OF COMPUTER SIMULATIONS

A - CROSS-FLOW EVAPORATIVE COOLER SIMULATION

Example No. Flow pattern Analytical model No. of elements

AI TTB Merkel 1
A2 TTB Merkel 5
A3 TTB Improved Merkel 1
A4 TTB Poppe 1
AS FTB Merkel 1
A6 BTF Merkel 1
A7 Single pass Merkel 1
AS TTB Simplified model -
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K.2

Example AI
Simulation program CROSS
Process water flow layout TOP TO BOTTOM
Analytical model MERKEL
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ...•.... = 65.99 mm
Height of cooler unit .................. . = 0.80 m
Width of cooler unit ........•........... = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K·
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) ....... ·... = 116.017 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = 0. 0323806 kg/kg
Outlet air temperature (saturated) ..... = 32.844 oc
Outlet air density (saturated) ......... = 1.132 kgjm3
· Recirc·.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ........... . 1. 3333 kg/ s '
Recirc. water lost through evaporation . = 0. 0379 kg/s
Recirculating water temperature in ..... = 41.214 oc
Recirculating water temperature out ... . 41.214 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 1.587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.214 oc
Capacity of cooler unit ................ = 112.383 kW
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K.3 http://scholar.sun.ac.za

Example A2
Simulation program CROSS
Process water flow layout TOP TO BOTTOM
Analytical model MERKEL
Pipe outer diameter .........•.....•...•. = 38.10 mm
Pipe inner diam~ter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit .•................. = 0.80 m
Width of cooler unit ................... . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 5
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ·····!· = 19.500 oc
Inlet air density ...................... = 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air entha 1py in . . . . . . . .. . . . . . . . . . . . . . . .. . = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 116.015 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = 0. 0323806 kg/kg
Outlet air temperature (saturated) ..... = 32.844 "C
Outlet air density (saturated) ......... = 1.132 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
In 1et reci rc. water massfl ow ........... . 1.3333 kg/s
Recirc. water lost through evaporation . = 0.0379 kg/s
Recirculating water temperature in ..... = 41.217 oc
Recirculating water temperature out .... = 41.216 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes .. . 1. 587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.214 oc
Capacity of cooler unit ................ = 112.380 kW
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K.4

Example A3
Simulation program CROSS
Process water flow layout TOP TO BOTTOM
Analytical model IMPROVED MERKEL
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter •.......•..•....•..•. = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit ........•.......... = 0.80 m
Width of cooler unit ................... . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Fouling coefficient (inside) ........... . = 2oooo.oo w;ni2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................ ~. = 43.00 W/mK
Atmospheric pressure .................. . = 101.325 kPa
Inlet air temperature (dry bulb) ...... . = 25.000 oc
Inlet air temperature (wet bulb) ...... . = 19.500 oc
Inlet air density ..................... . = 1.175 kg/m3
Dry air massflow through cooler ....... . = 1.858 kg/s
Inlet air massflow (inc vapour) ..•..... = 1.880 kg/s
Air velocity through cooler ........... . = 2.499 m/s
Air enthalpy in ....................... . 55. 779 kJ/kg
Air enthalpy out (incl. mist) ......... . = 115.597 kJ/kg
Inlet air humidity ratio .............. . = 0.0120087 kg/kg
Outlet air humidity ratio (incl. mist) . 0. 0328293 kg/kg
Outlet air relative humidity .......... . = 1.0000000
Outlet air temperature (dry bulb) ..... . = 32.738 oc
Outlet air density ................... . 1.132 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Outlet recirc. water massflow .......... = 1. 2946 kg/s
Recirc. water lost through evaporation . = 0.0387 kg/s
Recirculating water temperature in ..... = 41.242 oc
Recirculating water temperature out ... . 41.241 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes .. . 1.587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.226 oc
Capacity of cooler unit ................ = 111.664 kW
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K.5

Example A4
Simulation program CROSS
Process water flow layout TOP TO BOTTOM
Analytical model POPPE
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter ......•..........•... = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit ................•.. = 0.80 m
Width of cooler unit ................... . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure .................... = 101.325 kPa
Inlet air temperature (dry bulb) ...... . 25.000 oc
Inlet air temperature (wet bulb) ...... . = 19.500 oc
Inlet air density ..................... . 1.175 kgjm3
Dry air massflow through cooler ....... . = 1.858 kg/s
Inlet air massflow (inc vapour) ....... . = 1.880 kg/s
Air velocity through cooler ........... . = 2.499 m/s
·Aif enthalpy in ....................... . = 55. 779 kJ/kg
Air enthalpy out (incl. mist) ......... . = 120.665. kJ/kg
Inlet air humidity ratio .............. . = 0.0120087 kg/kg
Outlet air humidity ratio (incl. mist) . = 0. 0405106 kg/kg
Outlet air relative humidity .......... . = 1.0000000
Outlet air temperature (dry bulb) ..... . = 33.429 oc
Outlet air density ................... . = 1.129 kgjm3
Recirc.water massflow I length ........ . 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Outlet recirc. water massflow .......... = 1. 2804 kg/s
Recirc. water lost through evaporation . = 0.0530 kg/s
Recirculating water temperature in ..... = 41.291 oc
Recirculating water temperature out ... . 41.290 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 1.587 m/s
Process water temperature in .......... . 50.000 oc
Process water temperature out .......... = 48.222 oc
Capacity of cooler unit .... ~ ........... = 111.884 kW
 Stellenbosch University http://scholar.sun.ac.za
K.6

Example AS
Simulation program CROSS
Process water flow layout FRONT TO BACK
Analytical model MERKEL
Pipe outer diameter .................•... = 38.10 mm
Pipe inner diameter .............•....... = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit .................. . = 0.80 m
Width of cooler unit ................... . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) ....... ~ .. . = 20000.00 Wjm2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kgjm3
Dry air massflow through ·cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in .............•........•. = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 115.784 kJ/kg
Inlet air humidity ratio ............... = 0 ._0120087 kg/kg
Outlet _air humidity ratio (saturated) .. = 0. 0322668 kg/kg
Outlet air temperature (saturated) ..... = 32.784 oc
Outlet air density (saturated) ......... = 1.132 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Recirc. water lost through evaporation . = 0.0376 kg/s
Recirculating water temperature in ..... = 41.524 oc
Recirculating water temperature out .... = 41.524 oc
Process water massflow through cooler .. 15.000 kg/s
Process water flow velocity in pipes ... = 1.587 m/s
Process water temperature in .......... . 50.000 oc
Process water temperature out ......... . 48.221 oc
Capacity of cooler unit ................ = 111.952 kW
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K.7

Example A6
Simulation program CROSS
Process water flow layout BACK TO FRONT
Analytical model MERKEL
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit .................. . = 0.80 m
Width of cooler unit ............. ~ ..... . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe •. = I
Fouling coefficient (inside) ..........•. = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.3'25 kPa
Inlet air tempe~ature (dry bulb) ····~·· = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ........... . 2.499 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 116.025 kJ/kg
Inlet air humidity ratio .............. . 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = 0.0323806 kg/kg
Outlet air temperature (saturated) ..... = 32.844 oc
Outlet air density (saturated) ......... = 1.132 kg/m3
Recirc.water massflow I length .......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Recirc. water lost through evaporation . 0.0379 kg/s
Recirculating water temperature in ..... = 41.481 oc
Recirculating water temperature out .... = 41.481 oc
Process water massflow through cooler .. 15.000 kg/s
Process water flow velocity in pipes .. . 1.587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.115 oc
Capacity of cooler unit ............... . 118.646 kW
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K.8

Example A7
Simulation program CROSS
Process water flow layout STRAIGHT THROUGH
Analytical model MERKEL
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm
Vertical spacing between pipes .......... = 76.20 mm
Horizontal spacing between pipes ........ = 65.99 mm
Height of cooler unit ................... = 0.80 m
Width of cooler unit .................... = 0.80 m
Number of rows of pipes across airstream ·= 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Fouling coefficient (inside) ............ = 20000.00 w;m2 K
Fouling coefficient (outside) ........... = 20000.00 w;m2 K
Pipe wall conductivity .................. = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density .................... .. 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through· cooler ............ = 2.499 m/s
Air enth~lpy in ·······~················ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 99.850 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = . 0.0272164 kg/kg
Outlet air temperature (saturated) ..... = 29.925 oc
Outlet air density (saturated) ......... = 1.146 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ........... . 1.3333 kg/s
Recirc. water lost through evaporation . 0. 0283 kg/s
Recirculating water temperature in ..... = 37.535 oc
Recirculating water temperature out .... = 37.535 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 0.159 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out ......... . 48.694 oc
Capacity of cooler unit ................ = 82.230 kW
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K.9 http://scholar.sun.ac.za

Example AS
Simulation program SCROSS
Process water flow layout TOP TO BOTTOM
Cooler type TUBES ONLY
Process fluid WATER
Pipe outer diameter ....................• = 38.10 mm
Pipe inner diameter ...•................. = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit .. .' ............... . = 0.80 m
Length of cooler unit .................. . = 0.80 m
Number of pipe rows along the airflow .. . = 10
Number of pipes facing the airstream ... . = 10
Order of tube serpentining ............ . = 1
Fouling coefficient (inside) ........... . = 20000.00 W/m2 K
Fouling coefficient (outside) .......... . = 20000.00 W/m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea level ) ............ . 0.000 m
Inlet air temperature (dry bulb) ....... 25.000
= oc
Inlet air temperature (wet bulb) ....... 19.500
= oc
Dry air massflow through cooler ........ =
1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through cool·er ............= 2.500 m/s
Air enthalpy in···········~············= 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 115.740 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Recirc. water lost through evaporation = 0. 0377 kg/s
Recirculating water temperature in ..... = 41.60 oc
Recirculating water temperature out .... = 41.60 oc
Process fluid massflow through cooler .. = 15.000 kg/s
Process fluid flow velocity in pipes ... = 1.587 m/s
Process fluid temperature in ........... = 50.000 oc
Process fluid temperature out ......... . 48.222 oc
Capacity of cooler unit ................ = 111.893 kW
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K. 10

B - COUNTERFLOW EVAPORATIVE COOLER SIMULATION

Example No. Flow pattern Analytical model No. of elements

B1 BTT Merkel 1
B2 BTT Merkel 5
B3 8TT Improved Merkel 1
B4 8TT Poppe 1
85 TTB Merkel 1
86 BTT+Packing Merkel 1
87 TTB+Packing Merkel 1
88 TTB Simplified model -
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K. 11

Example B1
Simulation program COUNTER
Process water flow layout BOTTOM TO TOP
Analytical model MERKEL
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm
Vertical spacing between pipes .......... = 66.00 mm
Horizontal spacing between pipes ........ = 76.20 mm
Width of cooler unit .................... = 0.80 m
Length of cooler unit ................... = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Atmospheric pressure . . . . . . . . . . . . . . . . . . . = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ....... ~ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in ........................ = 55.779 kJ/kg
Air enthalpy out (incl. mist) .......... = 118.231 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = 0. 0330753 kg/kg
Outlet air temperature (saturated) ..... = 33.203 oc
Outlet air density (saturated) ......... = 1.130 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.0391 kg/s
Recirculating water temperature in .~ ... = 41.39 oc
Recirculating water temperature out .... = 41.38 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes .. . 1.587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.237 oc
Capacity of cooler unit ................ = 110.955 kW
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K.12

Example 82
Simulation program COUNTER
Process water flow layout BOTTOM TO TOP
Analytical model MERKEL
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter .....•............... = 34.90 mm
Vertical spacing between pipes .......... = 66.00 mm
Horizontal spacing between pipes ........ = 76.20 mm
Width of cooler unit .................... = 0.80 m
Length of cooler unit ................... = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 5
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kgjm3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in ...........•............ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 118.222 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (saturated) .. = 0. 03307.53 kg/kg
Outlet air temperature (saturated) ..... = 33.203 oc
Outlet air density (saturated) ......... = 1.130 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.0391 kg/s
Recirculating water temperature in ..... = 41.38 oc
Recirculating water temperature out .... = 41.39 oc
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 1. 587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.237 oc
Capacity of cooler unit ................ = 110.972 kW
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K.13

Example 83
Simulation program COUNTER
Process water flow layout BOTTOM TO TOP
Analytical model IMPROVED MERKEL
Pipe outer diameter ...•............•.... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm
Vertical spacing between pipes .......... = 66.00 mm
Horizontal spacing between pipes ........ = 76.20 mm
Width of cooler unit .................... = 0.80 m
Length of cooler unit ..... o ••••••••••••• = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Inlet air density ...................... = 1.175 kg/m3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ........... . 2.499 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 117.697 kJ/kg
Inlet air humidity ratio ............... = 0. 0120087 kg/kg
Outlet air humidity ratio (incl. mist) = 0. 0335399 kg/kg
Outlet air relative hu~idity ........... = 1.0000000
Outlet air temperature (dry bulb) ...... = 33.092 oc
Outlet air.density .................... = 1.131 kgjm3
Recirc.water massflow I length ........ . 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Outlet recirc. water·massflow .......... = 2.6267 kg/s
Recirc. water lost through evaporation . 0.040Q kg/s
Recirculating water temperature in ..... = 41.43 c
Recirculating water temperature out .... = 41.43 oc
Process water massflow through cooler .. 15.000 kg/s
Process water flow velocity in pipes ... = I. 587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out .......... = 48.251 oc
Capacity of cooler unit ............... . 110.051 kW
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K.14

Example 84
Simulation program COUNTER
Process water flow layout BOTTOM TO TOP
Analytical model POPPE
Pipe outer diameter .......•............. = 38.10 mm
Pipe inner diameter ........•....•....... = 34.90 mm
Vertical spacing between pipes ......... . = 66.00 mm
Horizontal spacing between pipes ....... . = 76.20 mm
Width of cooler unit ......•............. = 0.80 m
Length of cooler unit .................. . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1
Atmospheric pressure ................... = 101.325 kPa
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 °C
Inlet air density ...................... = 1.175 kgjm3
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.880 kg/s
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in ........................ = 55.779 kJ/kg
Air enthalpy out (incl. mist) .......... = 119.121 kJ/kg
Inlet air humidity ratio ............... = 0.0120087 kg/kg
Outlet air humidity ratio (in<:l. ·mist) . = 0.0337335 kg/kg
Outlet air relative humidity .......... . 1.0000000
Outlet air temperature (dry bulb) : ..... = 32.897 oc
Outlet air density ·················~·· = 1.132 kg/m3
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Outlet recirc. water massflow .......... = 2.6263 kg/s
Recirc. water lost through evaporation . = 0.0404 kg/s
Recirculating water temperature in ..... = 41.75 °C
Recirculating water temperature out .... = 41.75 °C
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 1.587 m/s
Process water temperature in ........... = 50.000 oc
Process water temperature out ......... . 48.319 oc
Capacity of cooler unit ................ = 105.786 kW
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K. 15

Example 85
Simulation program COUNTER
Process water flow layout TOP TO BOTTOM
Analytical model MERKEL
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm
Vertical spacing between pipes .......... = 66.00 mm
Horizontal .spacing between pipes ........ = 76.20 mm
Width of cooler unit ..................... = 0.80 m
Length of coo 1er unit . . . . . . . . . . . . . . . . . . . = 0.80 m
Number of rows of pipes across airstream = 10
Number of pipes facing the airstream .... = 10
Number of elements along a single pipe .. = 1

Atmospheric pressure ................... = 101.325 kPa r-. I

Inlet air temperature (dry bulb) ....... = 25.000 °C .-u

Inlet air temperature (wet bulb) ....... = 19.500 oc -,~

Inle·t air density ...................... = 1.175 kg/m3 1
Dry air massflow through cooler ........ = 1.858 kg/s ···
Inlet air massflow (inc vapour) ........ = 1.880 kg/s ........
Air velocity through cooler ............ = 2.499 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 118,414 kJ/kg I ••

Inlet air humidity ratio ............... = 0.0120087 kg/kg •',.

Outlet air humidity ratio (saturated) .. = 0.0331230 kg/kg ...
Outlet air temperature (saturated) ..... = 33.228 oc ••
Outlet air density (saturated) ......... = 1.130 kgjm3 ~:.,

Recirc.water massflow I length ........ . 300.0000 kg/m/hr

Inlet recirc.water massflow ........... . 2.6667 kg/s
Recirc. water lost through evaporation . = 0.0392 kg/s
Recirculating water temperature in .... . 41.11 oc
Recirculating water temperature out ... . = '\:!1-11 :228 oc _,. u '·''. {X·
Process water massflow through cooler .. = 15.000 kg/s
Process water flow velocity in pipes ... = 1.587 m/s
Process water temperature in ........... = 50.010 oc
Process water temperature out .......... = 48.243 oc
Capacity of cooler unit ................ = ~111.225 kW

. '

,.,
/- '
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K.16

Example B6
Simulation program COMBINE
Process water flow layout BOTTOM TO TOP
Cooler type TUBES + 300 mm PACK ABOVE
Process fluid WATER
Pipe outer diameter ..•.......•.......... = 38.10 mm
Pipe inner diameter •.................... = 34.90 mm
Vertical spacing between pipes .......... = 65.99 mm
Horizontal spacing between pipes ........ = 76.20 mm
Width of cooler unit .................... = 0.80 m
Length of cooler unit ................... = 0.80 m
Number of pipes facing the airstream .... = 10
Number of pipe rows along the airflow ... = 10
Order of tube serpentining ............. = 1
Fouling coefficient {inside) ............ = 20000.00 W/m2 K
Fouling coefficient {outside) ........... = 20000.00 w;m2 K
Pipe wall conductivity .................. = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
A1t i tude {above sea 1eve 1) ............ · = 0 .· 00 0 'lie
Inlet air temperature {dry bulb) ....... = 25 000
Inlet air temperature .{wet bulb) ....... = 19.500 oc
Dry air massflow through cooler........ 1.858 kg/s
Inlet air massflow {inc vapour) ........ = 1.881 kg/s
Air velocity through cooler ............ = 2.5o·o m/s
Air enthalpy in ........................ = 55.779 kJ/kg
Air enthalpy out {incl. mist) .......... = 133.839 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ........... . 2.6667 kg/s
Recirc. water lost through evaporation . 0.0485 kg/s
Recirculating water temperature in ..... = 40.39 oc
Recirculating water temperature out ... . 40.39 oc
Process fluid massflow through cooler .. = 15.000 kg/s
Process fluid flow velocity in pipes ... = 1.587 m/s
Process fluid temperature in ..........• = 50.000 oc
Process fluid temperature out ......... . 47.751 oc
Capacity of cooler unit ............... . 141.559 kW
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K.17

Examole B7
Simulation program COMBINE
Process water flow layout TOP TO BOTTOM
Cooler type TUBES + 300 mm PACK ABOVE
Process fluid WATER
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spaciri'g between pipes ......... . = 65.99 mm
Horizontal spacing between pipes ....... . = 76.20 mm
Width of cooler unit .......•..•......... = 0.80 m
Length of cooler unit ......•............ = 0.80 m
Number of pipes facing the airstream ... . = 10
Number of pipe rows along the airflow .. . = 10
Order of tube serpentining ............ . = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea level) ............. = 0.000 m
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.~81 kg/s
Air velocity through cool.er ............ = 2.500 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 132.670 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.047~ kg/s
Recirculating water temperature in ..... = 39.99 c
Recirculating water temperature out .... = 40.00 oc
Process fluid massflow through cooler .. 15.000 kg/s
Process fluid flow velocity in pipes ... = 1.587 fli/S
Process fluid temperature in ........... = 50.012 c
Process fluid temperature out ......... . 47.796 oc
Capacity of cooler unit ................ = 139.469 kW
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K.18

Example B8
Simulation program SCOUNT
Process water flow layout TOP TO BOTTOM
Cooler type TUBES ONLY
·Process fluid WATER
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm I,
Vertical spacing between pipes .......... = 65.99 nim
Horizontal spacing between pipes ........ = 76.20 mm ..
Width of cooler unit .................... = 0.80 m
Length of cooler unit ................... = 0.80 m
Number of pipes facing the airstream .... = 10 ..
Number of pipe rows along the airflow ... = 10
Order of tube serpentining ............. = 1
Fouling coefficient (inside) ............ = 20000.00 w;m2 K
Fouling coefficient (outside) ........... = 20000.00 w;m2 K
Pipe wall conductivity .................. = 43.00 W/mK I •

Atmospheric pressure ................... = 101.325 kPa

A1t i tude (above sea 1eve 1) . . . . . . . . . . . . . = 0.000 Ill
Inlet air temperature (dry bulb) ....... = 25.000 c
Inlet air temperature (wet bulb) ....... = 19.500 oc
Dry air massflow through cooler ........ = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through cooler ······~····· = 2.500 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg.
Air enthalpy out (incl. mist) .......... = 115.740 kJ/kg
Recirc.water massflow I length ........ . 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.0377 kg/s
Recirculating water temperature in ..... = 41.60 °C .
Recirculating water temperature out .... = 41.60 oc
Process fluid massflow through cooler .. = 15.000 kg/s
Process fluid flow velocity in pipes .. . 1.587 fll/S ./ ,.
Process fluid temperature in .......... . 50.000 c
Process fluid temperature out ......... . 48.222 oc
Capacity of cooler unit ............... . = 111.893 kW

,.... 'I
t ... .. (..

..,
I

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. '-
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K.19

C - EVAPORATIVE CONDENSER SIMULATION

Example No. Airflow Refrigerant

Cl Cross Steam
C2 Cross R22
C3 Cross Ammonia
C4 Counter Steam
cs Counter R22
C6 Counter Ammonia
 Stellenbosch University http://scholar.sun.ac.za
K.20

Example C1
Simulation program CSCROSS
Condenser type TUBES ONLY
Refrigerant STEAM
Pipe outer diameter ....•................ = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of condenser unit ............... . = 0.80 m
Length of condenser unit ............... . = 0.80 m
Number of pipe rows along the airflow .. . = 10
Number of pipes facing the airstream ... . = 10
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea 1eve 1) . . . . . . . . . . . . . = 0.000 m
Inlet air temperature (dry bulb) ...... . 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Dry air massflow through condenser ..... = 1.858 kg/s
Inlet air massflow (inc Vapour) ........ = 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 119.749 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 1.3333 kg/s
Recirc. water lost through evaporation . = 0. 040Q kg/s
Recirculating water temperature (ave) .. = 42.46 c
Refrigerant massflow through condenser . 0.04989 kg/s
Condensing temperature ................. = 50.000 oc
Capacity of condenser unit ............. = 118.8800 kW
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K.21

Example C2
Simulation program CSCROSS
Condenser type TUBES ONLY
Refrigerant R22 (Freon 22)
Pipe outer diameter ..................... = 38.10 mm
Pipe inner diameter ..................... = 34.90 mm
Vertical spacing between pipes .......... = 76.20 mm
Horizontal spacing between pipes ........ = 65.99 mm
Height of condenser unit ................ = 0.80 m
Length of condenser unit ................ = 0.80 m
·Number of pipe rows along the airflow ... = 10
Number of pipes facing the airstream .... = 10
Fouling coefficient (inside) ............ = 20000.00 W/m2 K
Fouling coefficient (outside) ........... = 20000.00 W/m2 K
Pipe wall conductivity ...............•.. = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea level) ............. = 0.000 m
25. 0 ~ 0
500
Inlet air temperature (dry bulb) ....... = :cc
Inlet air temperature (wet bulb) ....... = 19
Dry air massflow through condenser..... 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/S
Air enthalpy in ........................ = 55.779 kJ/kg
Air enthalpy out (incl. mist) .......... = 87.996 kJ/kg
Recirc.water massflow I length ........ . = 300.0000 kg/m/hr
Inlet recirc.water massflow ........... . 1.3333 kg/s
Recirc. water lost through evaporation . = 0.0215 kg/s
Recirculating water temperature (ave) .. = 34.14 oc
Refrigerant massflow through condenser . = 0.38896 ~g/s
= 50.000 c
Condensing temperature . . . . . . . . . . . . . . . . .
Capacity of condenser unit ............. = 59.8717 kW
 Stellenbosch University http://scholar.sun.ac.za
K.22

Example C3
Simulation program CSCROSS
Condenser type TUBES ONLY
Refrigerant R717 (Ammonia)
Pipe outer diameter ·······~············· = 38.10 mm
Pipe inner diameter ...............•..... = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of condenser unit ............... . = 0.80 m
Length of condenser unit ............... . = 0.80 m
Number of pipe rows along the airflow .. . = 10
Number of pipes facing the airstream ... . = 10
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 Wjm2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea level) . . . . . . . . . . . . . = ·o.ooo m
= ·_25 ..000" c
0
Inlet air temperature (dry bulb) .......
Inlet air temperature (wet bulb) ....... = 19:'500 oc
Dry air massflow through condenser ..... = !.858 kg/s
Inlet air massflow (inc vapour) ....... . 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/s
Air enthalpy in ....................... . 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 117.180 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............. = 1.3333 kg/s
Recirc. water lost through evaporation . = 0. 038~ kg/s
Recirculating water temperature (ave) .. = 41.91 c
Refrigerant massflow through condenser . = 0.10860 ~g/s
Condensing temperature ................ . 50.000 c
Capacity of condenser unit ............. = 114.1063 kW
 Stellenbosch University http://scholar.sun.ac.za
K.23

Example C4
Simulation program CSCOUNT !A ...l. I
Condenser type TUBES ONLY "'· .J ~ I . )
Refrigerant STEAM
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 65.99 mm
Horizontal spacing between pipes ....... . = 76.20 mm
Width of condenser unit ............... .. = 0.80 m
Length of condenser unit ............... . = 0.80 m
Number of pipes facing the airstream ... . = 10
Number of pipe rows along the airflow .. . = 10
Order of tube serpentining ............ . = 1
Fouli-ng coefficient (inside) .......... .. = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
A1t itude (above sea 1eve 1) .............
= 0.000 m
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Dry air massflow through condenser ..... = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/s
Air enthalpy in ........................ = . 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 119.412 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.039~ kg/s
Recirculating water temperature (ave) .. = 42.39 c
Refrigerant massflow through condenser . 0.04963 ~g/s
Condensing temperature ................. = 50.000 c
Capacity of condenser unit ............ . 118.2548 kW
 Stellenbosch University http://scholar.sun.ac.za
K.24

Example CS
Simulation program CSCOUNT
Condenser type TUBES ONLY
Refrigerant R22. (Freon 22)
Pipe outer diameter ......•......•....... = 38.10 mm
Pipe inner diameter .................... . = 34.90 mm
Vertical spacing between pipes ......... . = 65.99 mm
Horizontal spacing between pipes .......• = 76.20 mm
Width of condenser unit ................ . = 0.80 m
Length of condenser unit ............... . = 0.80 m
Number of pipes facing the airstream ... . = 10
Number of pipe rows along the airflow .. . = 10
Order of tube serpentining ............ . = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wa 11 conductivity ................. . = 43.00 W/mK
Atmospheric pressure . . . . . . . . . . . . . . . . . . . = 101.325 kPa
Altitude (above sea level) ............. = 0.000 m
Inlet air temperature (dry bulb) ....... = 25.000 oc
Inlet air temperature (wet bulb) ....... = 19.500 oc
Dry air massflow through condenser ..... = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/s
Air enthalpy in ........................ = 55. 779 kJ/kg
Air enthalpy out (incl. mist) .......... = 87. 929 kJ/kg
Recirc.water massflow I length ......... = 300. 0000 kg/m/hr
Inlet recirc.water massflow ............ = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.021? kg/s
Recirculating water temperature (ave) .. = 34.12 c
Refrigerant massflow through condenser . = 0.38816 ~g/s
Condensing temperature . . . . . . . . . . . . . . . . . = 50.000 c
Capacity of condenser unit ............. = 59.7479 kW
 Stellenbosch University http://scholar.sun.ac.za
K.25

Example C6
Simulation program CSCOUNT
Condenser type TUBES ONLY
Refrigerant R717 (Ammonia)
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter •.................... = 34.90 mm
Vertical spacing between pipes ......... . = 65.99 mm
Horizontal spacing between pipes ....... . = 76.20 mm
Width of condenser unit ................ . = 0.80 m
Length of condenser unit ............... . = 0.80 m
Number of pipes facing the airstream ... . = 10
Number of pipe rows along the airflow .. . = 10
Order of tube serpentining ............ . = 1
Fouling coefficient (inside) ........... . = 20000.00 w;m2 K
Fouling coefficient (outside) .......... . = 20000.00 w;m2 K
Pipe wall conductivity ................. . = 43.00 W/mK
Atmospheric pressure ................... = 101.325 kPa
Altitude (above sea level) ............. = 0.000 m
Inlet air temperature (dry. bulb) ....... = 25.000 °C
Inlet air temperature (wet bulb) ....... = ·19.500 oc
Dry air massflow through condenser ..... = 1.858 kg/s
Inlet air massflow (inc vapour) ........ = 1.881 kg/s
Air velocity through condenser ......... = 2.500 m/s
Air enthalpy in ·········~·············· = 55. 779 kJ/kg
Air enthalpy out (incl. mist) ..... : .... = 117.181 kJ/kg
Recirc.water massflow I length ......... = 300.0000 kg/m/hr
Inlet recirc.water massflow ............ · = 2.6667 kg/s
Recirc. water lost through evaporation . = 0.038~ kg/s
Recirculating water temperature (ave) .. = 41.91 c
Refrigerant massflow through condenser . = 0.10860 ~g/s
Condensing temperature ................. = 50.000 c
Capacity of condenser unit ............ . 114.1073 kW
 Stellenbosch University http://scholar.sun.ac.za
K.26

D - CROSS-FLOW NATURAL DRAFT COOLING TOWER SIMULATION

Example D1
Simulation program TOWER
Process water flow layout FRONT TO BACK
Analytical model MERKEL
Total tower height ..................... . = ..147 .00 m
Inlet height at pond ................... . = 8.00 m
Diameter of tower at pond .............. . = 105.00 m
Diameter of tower at outlet ............ . = 60.85 m
Number of A-frames around tower; ....... . = 24
Included A-frame angle ................. . = 60.00
Face length of each A-frame side ....... . = 15.00 m
Pipe outer diameter .................... . = 38.10 mm
Pipe inner diameter ....... ~ ............ . = 34.90 mm
Vertical spacing between pipes ......... . = 76.20 mm
Horizontal spacing between pipes ....... . = 65.99 mm
Height of cooler unit .................. . = 8.00 m
Width of cooler unit ................... . = 15.00 m
Number of rows of pipes across airstream = 40
Number of pipes facing the airstream .... = 104
Number of elements along a single pipe = 1

Dry air massflow through A-frame side .. = 182.81 kg/s

Dry air massflow through tower ......... = 8775.00 kg/s ·
Air velocity through A-frame side ...... = 1. 53 m/s
Pressure drop across cooler ............ = 48.95 Pa
Pressure drop (oblique flow) ........... = 7.30 Pa
Pressure drop across drift eliminators . = 6.07 Pa
Tower pressure drop .................... = 8.68 Pa
Tower outlet pressure loss ............. = 4.79 Pa
Total pressure loss through tower ...... = 75.79 Pa
Total available buoyancy ............... = 76.28 Pa
Atmospheric pressure ................... = 84.000 kPa
Inlet air temperature (dry bulb) ...... . 15.450 oc
Inlet air temperature (wet bulb) ....... = 11.050 ° c
Inlet air density ...................... = 1.009 kg/m3
Inlet air massflow (inc vapour) ........ = 184.302 kg/s
Dry air massflow through cooler ........ = 182.813 kg/s
Dry air massflow through tower ......... = 8775.002 kg/s
Air enthalpy in ........................ = 36.165 kJ/kg
Air enthalpy out (incl. mist) .......... = 102.956 kJ/kg
Inlet air humidity ratio ............... = 0. 0081457 kg/kg
Outlet air humidity ratio (saturated) .. 0.02930~1 kg/kg
Outlet air temperature (saturated) .... . 27.907 c
Outlet air density (saturated) ........ . 0.955 kg/m3
 Stellenbosch University http://scholar.sun.ac.za
K.27

Recirc.water massflow I length ......... = 300.0000 kg/m/hr

Total inlet recirc.water massflow ...... = 4800.0000 kg/s
Recirc. water lost through evaporation . = 185.6741 kg/s
Recirculating water temperature in ..... = 29.45 C
Recirculating water temperature out .... = 29.46 oc
Process water massflow through tower ... = 12500.000 kg/s
Process water massflow through cooler .. = 260.417 kg/s
Process water flow velocity in pipes ... = 2.637 ~/s
Process water temperature in ........... = 39.440 c
Process water temperature out ......... . 28.216 oc
Capacity of cooler unit ................ = 12189.388 kW
Total capacity of tower ................ = 585.091 MW
 Stellenbosch University
L. 1 http://scholar.sun.ac.za

APPENDIX L

FORTRAN CODE FOR CROSS-FLOW EVAPORATIVE COOLER SIMULATION PROGRAM

 Stellenbosch University http://scholar.sun.ac.za
1

c =================================================
c = =
c = EVALUATION OF A =
c = CLOSED CIRCUIT CROSSFLOW
c = EVAPORATIVE COOLER
c = =
c = The evaluation of a evaporative cooler =
c = unit using a finite difference method. =
c = The evaporative cooler units are devided =
c = into small three dimensional blocks. =
c = The blocks are evaluated from a point where =
c = all the initial values for the block are =
c = known ; by continuing from this block =
c = the whole cooler unit may be evaluated.
c = =
c = ******************** =
c = =
c = Written by A.A.DREYER (8312818) =
c = =
c = DEPARTMENT OF MECHANICAL ENGINEERING =
c = UNIVERSITY OF STELLENBOSCH =
c = VAX 7·85 - FORTRAN Version 4.5 =
c = =
c = LAST REVISION 18 August 1988 =
c = =
c =================================================
c
C Tsp - Process water temperature (array) [ C]
C Tsw - Recirc. water temperature (array) [ C]
C Tsa - Dry bulb air temperature (array) [ C]
c sisa Enthalpy of air (array) [kJ/kg]
c swsa - Humidty of air (array) [kg water/kg air]
c smsw - Recirc. water massflow rate (array) [kg/s]
c Tspi Inlet process water temperature for element [ C]
c Tspo - Outlet process water temperature for element [ C]
c Tswi Inlet recirc. water temperature for element [ C]
c Tswo - Outlet recirc. water temperature for element [ C]
c Tsai Inlet air temperature for element [ C]
c Tsao - Outlet air temperature for element [ C]
c sisai Inlet air enthalpy for element [kJ/kg]
c sisao Outlet air enthalpy for element [kJ/kg]
c swsai Inlet air humidity for element [kg water/kg air]
c swsao - Outlet air humidity for element [kg water/kg air]
c smswi Inlet recirc. water massflow for element [kg/s]
c smswo - Outlet recirc. water massflow for element [kg/s]
c Tsa - Air temperature [ C]
c Tsai - Air temperature into element [ C]
c Tsao - Air temperature from element [ C]
c Tspi1 Inlet temperature of hot process water [ C]
c Tswil Inlet temperature of reciculating water [ C]
c TR - Right boundary of temperature interval [ C]
c TL - Left boundary of temperature interval [ C]
 Stellenbosch University http://scholar.sun.ac.za

C Tsadb - Dry bulb temperature of air [ C]

C Tsawb - Wet bulb temperature of air [ C]
C spsatm - Atmospheric pressure [Pa]
C L - Length of pipe (perpendicular to airstream) [m]
C H - Heigth of cooler (perpendicular to airstream) [m]
C sdsi - Inner diameter of pipe [m]
C sdso - Outer diameter of pipe [m]
C Kmax - Number of pipe rows along direction of airstream
C Lmax - Number of pipes per row
C Mmax - Number of elements along the length of a single pipe
C clrtype - 1-Recirc. cooling water, 2-Cooling water once through
C gradfile - Logical variable (1-print temp gradients,O-print nothing)
C rhosa - Density of air [kg/mA3]
C rhosw - Density of water [kg/mA3]
C rhosv - Density of water vapour [kg/mA3]
C rhosav - Density of air/vapour mixture [kg/mA3]
C musa - Dynamic viscosity of air [kg/ms]
C musw - Dynamic viscosity of water [kg/ms]
C musv - Dynamic viscosity of water vapour [kg/ms]
C musav - Dynamic viscosity of air/vapour mixture [kg/ms]
C smsa - Massflow of air [kg/s]
C smsp - Massflow of process water [kg/s]
C smsael - Massflow of air through single element [kg/s]
C smspel - Massflow of process water through element [kg/s]
C smswel - Massflow of recirculating water through element [kg/s]
C hspas- - Horizontal spacing of tubes [m]
C vspas - Vertical spacing of tubes [m]
C svsa Velocity of air [m/s]
C svsp - Velocity of process water [m/ s]
C shsf1 - Fouling heat transfer coefficient (inner) [W/mA2K]
C shsf2 - Fouling heat transfer coefficient (outer) [W/mA2K]
C skst - Thermal conductivity of tube wall [W/mK]
C gamma - Recirculating water massflowjunit length [kg/m/hr]
C sa - Contact area/ unit volume (mA2/mA3]
C dA - Air/water contact area per element [mA2)
C Aspi - Cross area of pipe (inner diameter) (mA2]
C Aspo - Cross area of pipe (outer diameter) (mA2]
C ReyC - Constant used in definition of Reynoldsnumber
C sdsperp - Gap between pipes perpendicular to airflow [m]
C sdsdiag - Gap on diagonal between pipes [m]
C Vstot - Total air volumeflow through cooler (mA3/s]
C Vseff - Effective air volumeflow through a row of pipes (mA3/s]
C Vseff2 - Effective air volumeflow through the cooler (mA3/s]
C Tsaodb Dry bulb temperature of outlet air [ C]
C Tsaowb - Wet bulb temperature of outlet air [ C]
C model - Solution model to be used (1-Merkel,2-Impr Merkel,3-Poppe)
C spssat - Saturation pressure of water [Pa]
C scspv - Specific heat of water vapour [J/kgK]
C scspa - Specific heat of dry air [J/kgK]
C scspw - Specific heat of saturated water [J/kgK]
C sissat Enthalpy of saturated air [kJ/kg]
C si svap Enthalpy of water vapour [kJ/kg]
 Stellenbosch University http://scholar.sun.ac.za
J

c sksw - Thermal conductivity of saturated water [W/mK]

c Reysp - Reynolds number for process water flow
c Reysw - Reynolds number for recirculating water flow
c Reysa - Reynolds number for air flow
c Pra - Prandtl number
c Lew - Lewis factor
c shsp - Heat transfer coefficient (process water - pipe) [W/m"2K]
c shsw Heat transfer coefficient (recirc. water - pipe [W/m"2K]
c koga - Overall capacity coefficient of mass transfer [kg/m"3s]
c kog - Mass transfer coefficient [kg/m"2s]
c Uo - Overall heat transfer coefficient [W/m"2K]
c K - Constant
c a - Runge-Kutta constant [kg]
c b - Runge-Kutta constant [kg/kg]
c c - Runge-Kutta constant [kJ/kg]
c d - Runge-Kutta constant [ C]
c e - Runge-Kutta constant [ C]
c ww - Temporary variable for Runge-Kutta approximation [kg/kg]
c ii Temporary variable for Runge-Kutta approximation [kJ/kg]
c TT Temporary variable for Runge-Kutta approximation [ C]
c www - Temporary variable for Runge-Kutta appro-ximation [kg/kg]
c swsasw - Saturation air humidity at Tsw [kg/kg]
c swsasa - Saturation air humidity at Tsa [kg/kg]
c sisasw Saturation enthalpy of air at Tsw [kJ/kg]
c sisasa Saturation enthalpy of air at Tsa [kJ/kg]
c Power - Capacity of cooler unit [kW]
c flowlayout- Process water flow layout through unit (1,2,3 or 4)
C Reserve storage space for five arrays i.e. Tsp,Tsw,sisa,swsa and smsw
DIMENSION Tsp(40,400,10)
DIMENSION Tsw(40,400,10)
DIMENSION Tsa(40,400,10)
DIMENSION sisa(40,400,10)
DIMENSION swsa(40,400,10)
DIMENSION smsw(40,400,10)
C Declare the changed data types
INTEGER gradfile,clrtype,flowlayout,gradplot
REAL L
CHARACTER*10 char
C Set the initial array values equal to zero
DO 30 i=l,Kmax+1
DO 20 j=1,Lmax+l
DO 10 k=1,Mmax+2
Tsp(i,j,k)=O.O
Tsw(i,j,k)=O.O
Tsa(i,j,k)=O.O
sisa(i,j,k)=O.O
swsa(i,j,k)=O.O
smsw(i,j,k)=O.O
10 CONTINUE
 Stellenbosch University http://scholar.sun.ac.za

20 CONTINUE
30 CONTINUE
C Call subroutine to set default values for a typical cooler
CALL INITIAL(spsatm,Tsadb,Tsawb,L,H,sdso,sdsi,Kmax,
+ Lmax,Mmax,vspas,hspas,smsp,PI,gamma,skst,Tspil,
+ clrtype,model,smsa,flowlayout,Tswil,shsfl,shsf2)
C Call subroutine to edit coolertype,flowlayout,model,size etc.
5 CALL MENUl{clrtype,model,H,L,spsatm,Tsadb,Tsawb,flowlayout,
+ Tswi1,shsf1,shsf2)
C Call subroutine to edit the cooler dimensions and operating parameters
CALL MENU2(sdso,sdsi,H,L,PI,svsa,vspas,hspas,Lmax,Kmax,Mmax,
+ spsatm,Tsadb,Tsawb,gamma,skst,Tspi1,smsp,svsp,sa,Aspi,
+ Aspo,clrtype,dA,model,smsa,flowlayout,Tswi1,shsfl,shsf2)
C Open result files for program results and cooler temperature gradients
gradfile=O ! 0 - print nothing , 1 - print gradients
gradplot=O ! 0 - print nothing , 1 - print gradients
OPEN (UNIT=1, FILE='CROSS.RES', STATUS='NEW')
IF (gradfile.EQ.l) THEN
OPEN (UNIT=4, FILE='CROSS.GRA', STATUS='NEW')
ELSE IF (gradplot.EQ.1) THEN
OPEN (UNIT=S, FILE='CROSS.PLO', STATUS='NEW')
END IF
C Determine the air flow parameters for cooler
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsai1)
CALL AirVapMixdensity{Tsadb,swsai1,spsatm,rhosail)
svsa=(smsa*(l.O+swsai1))/(rhosai1*L*(Lmax+0.5)*vspas)
ReyC=svsa*(vspas/sdso)*(vspas-sdso)
C Determine the massflow of each fluid for a typical element
CALL Waterdensity(Tspi1,rhosw)
smsael=smsa/(Mmax*(Lmax+O.S)) Air massflow I element
smspel=svsp*rhosw*Aspi Process water massflowjelement
smswel=2.0*gamma*L/Mmax Recirc. water massflow/element
smswi1=2.0*gamma*L*Kmax Total inlet recirc. water massflow
C Evaluate cooler with given flowlayout
C clrtype = 1 ==>Recirculating cooling water
C clrtype = 2 ==>Cooling water makes only single pass through cooler
CALL LIB$ERASE PAGE(1,1)
WRITE(*,*)' ITERATIVE CALCULATION IN PROGRESS'
WRITE(*,*)'***********************************'
IF (clrtype.EQ.l) THEN
TR=Tspi1 Set upper value for Tw(in)
TL=Tsawb ! Set lower value for Tw(in)
40 Tswi1=(TR+TL)/2~0 ! Halve the Tw(in) interval
IF (flowlayout.EQ.1) THEN
CALL FRONTTOBACK (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
 Stellenbosch University http://scholar.sun.ac.za

+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
ELSE IF (flowlayout.EQ.2) THEN
CALL BACKTOFRONT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
ELSE IF (flowlayout.EQ.3) THEN
CALL TOPTOBOTTOM {Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail, -
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2,
+ gradplot)
ELSE IF (flowlayout.EQ.4) THEN
CALL STRAIGHT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
END IF
IF (ABS(Tswil-Tswol).GT.O.OOl) THEN
IF (Tswil.LT.Tswol) THEN
TL=Tswol ! TL=Tswil
IF (TL.GT.Tspil) TL=Tspil
ELSE IF (Tswil.GT.Tswol) THEN
TR=Tswol ! TR=Tswil
END IF
GO TO 40
END IF
ELSE IF (clrtype.EQ.2) THEN
IF (flowlayout.EQ.l) THEN
CALL FRONTTOBACK (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
ELSE IF (flowlayout.EQ.2) THEN
CALL BACKTOFRONT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
ELSE IF (flowlayout.EQ.3) THEN
CALL TOPTOBOTTOM (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfil e, Kmax, Lmax ,Mmax, PI, mode 1, Tsaol, shsfl, shsf2,
 Stellenbosch University http://scholar.sun.ac.za

+ gradplot)
ELSE IF (flowlayout.EQ.4) THEN
CALL STRAIGHT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisai1,sisao1,Tspi1,Tspo1,Tswi1,Tswo1,swsai1,
+ swsao1,smswi1,smswo1,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsao1,shsf1,shsf2)
END IF
END IF
C Print final temperature,enthalpy etc. profiles
IF (flowlayout.EQ.3) THEN ! TTB flow pattern
DO j=1,Lmax+1
IF (j.EQ.1) THEN
tpi=Tspil
tpo=Tspi1
ELSE
iflag=j-2.0*INT(j/2.0)
IF (iflag.EQ.O) THEN
tpi=Tsp(1,j-1,3)
tpo=Tsp(Kmax,j-1,3)
ELSL
tpi=Tsp(1,j-1,1)
tpo=Tsp(Kmax,j-1,1)
END IF
END IF
twi=Tsw(l,j,2)
two=Tsw(Kmax,j,2)
siao=sisa(Kmax+1,j,2)
swao=swsa(Kmax+1,j,2)
WRITE(10,*)j,tpi,twi,tpo,two,siao,swao
END DO
WRITE(10,*)1
I

DO i=1,Kmax
WRITE(lO,*)i,Tsp(i,Lmax,3),Tsw(i,Lmax+1,2)
ENDDO
ELSE IF ((flowlayout.EQ.1).0R.(flowlayout.EQ.2)) THEN FTB&BTF
DO j=1,Lmax+1
tpi=Tsp(l,j,2)
tpo=Tsp(Kmax,j,1)
twi=Tsw(1,j,2)
two=Tsw(Kmax,j,2)
siao=sisa(Kmax+1,j,2)
swao=swsa(Kmax+1,j,2)
WRITE(11,*)j,tpi,twi,tpo,two,siao,swao
END DO
WRITE(ll,*)1
I

DO i=1,Kmax
jflag=i-2.0*INT(i/2.0)
IF (jflag.EQ.1) THEN
tp=Tsp(i,Lmax,3)
ELSE
 Stellenbosch University http://scholar.sun.ac.za

tp=Tsp(i,Lmax,l)
END IF
WRITE(ll,*)i,tp,Tsw(i,Lmax+l,2)
ENDDO
END IF
C Print solution model used, ambient conditions and results
IF (model.EQ.l) THEN
TR=Tspil
TL=Tsawb
50 Tsaol=(TR+TL)/2.0
CALL Satenthalpy(Tsaol,spsatm,sisasa)
IF (ABS(sisasa-sisaol).GT.O.l) THEN
IF (sisasa.GT.sisaol) THEN
TR=Tsaol
ELSE
TL=Tsaol
END IF
GO TO 50
END IF
CALL Airhumidity(Tsaol,Tsaol,spsatm,swsaol)
CALL AirVapMixdensity(Tsaol,swsaol,spsatm,rhosaol)
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsal)
CALL AirVapMixdensity(Tsadb,swsal,spsatm,rhosail)
smswol=smswil-(swsaol-swsal)*smsa
ELSE IF (model.EQ.2) THEN
CALL Satvappressure(Tsaol,spssat)
spsvap=spsatm*swsaol/(1.005*(0.62198+swsaol))
phio=spsvap/spssat
IF (phio.GT.l.O) THEN
phio=l.O
CALL Airhumidity(Tsaol,Tsaol,spsatm,swsao2)
CALL AirVapMixdensity(Tsaol,swsao2,spsatm,rhosaol)
ELSE
CALL AirVapMixdensity(Tsaol,swsaol,spsatm,rhosaol)
END IF
ELSE IF (model.EQ.3) THEN
CALL Satvappressure(Tsaol,spssasa)
spsvap=spsatm*swsaol/(1.005*(0.62198+swsaol))
phio=spsvap/spssasa
IF (phio.GT.l.O) THEN
phio=l.O
CALL Airhumidity(Tsaol,Tsaol,spsatm,swsao2)
CALL AirVapMixdensity(Tsaol,swsao2,spsatm,rhosaol)
ELSE
CALL AirVapMixdensity(Tsaol,swsaol,spsatm,rhosaol)
END IF
END IF
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsal)
CALL AirVapMixdensity(Tsadb,swsal,spsatm,rhosail)
CALL Cpw(Tspil,scsppi)
CALL Cpw(Tspol,scsppo)
 Stellenbosch University http://scholar.sun.ac.za

Power=smsp*((Tspi1)*scsppi-(Tspo1)*scsppo)/1000.0
CALL PRINT_RESULTS(Tspi1,Tspo1,smsp,sdsi,sdso,vspas,hspas,
+ Kmax,Lmax,Mmax,gamma,Vstot,rhosai1,Vseff2,sisai1,sisao1,
+ Tswi1,Tswo1,svsp,flowlayout,H,L,spsatm,PI,Tsawb,Tsadb,svsa,
+ swsai1,swsao1,smswi1,smswo1,model,Tsao1,rhosao1,phio,Power,
+ shsfl,shsf2,skst,smsa)
C Rerun program or return to DCL
WRITE(*,100)
100 FORMAT(' RERUN program or return to DCL (R/D) ?' ,$)
READ(*,'(A)')char
CLOSE (UNIT =1)
CLOSE (UNIT =4)
CLOSE (UNIT=5)
IF ((char.EQ.'R').OR.(char.EQ.'r')) GO TO 5
C End of main program
END

c *******************************************************************
c* *
c* COUNTER FLOW (FROM BACK TO FRONT OF COOLER) *
c* *
c *******************************************************************
C Subroutine to evaluate a cooler layout where the process fluid flows
C in a direction ~ounter to the direction of the airstream
SUBROUTINE BACKTOFRONT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisai1,sisao1,Tspi1,Tspo1,Tswi1,Tswo1,swsai1,
+ swsao1,smswi1,smswo1,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsao1,shsf1,shsf2)
DIMENSION Tsp(40,400,10)
DIMENSION Tsw(40,400,10)
DIMENSION Tsa(40,400,40)
DIMENSION sisa(40,400,10)
DIMENSION swsa(40,400,10)
DIMENSION smsw(40,400,10)
REAL L
INTEGER flag,flag2,gradfile
C Choose an average temperature for the outlet process water
Tspo1=(Tspi1+Tswil)/2.0
DO 50 j=1,Lmax
Tsp(1,j,2)=Tspo1
50 CONTINUE
C Initialize the arrays with the known temperature and enthalpy values
999 CALL Enthalpy(Tsadb,Tsawb,spsatm,sisail)
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsail)
 Stellenbosch University http://scholar.sun.ac.za

DO 20 j=1,Lmax
DO 10 k=2,Mmax+1
sisa(1,j,k)=sisai1
swsa(1,j,k)=swsai1
Tsa(1,j,k)=Tsadb
10 CONTINUE
20 CONTINUE
DO 40 i=1,Kmax
DO 30 k=2,Mmax+1
Tsw(i,1,k)=Tswi1
smsw(i,1,k)=smswel
30 CONTINUE
40 CONTINUE
C N.B. flag=1 for·backward process fluid flow
c L.W. flag=O for forward process fluid flow
· fl ag=O
C Start of the outer loop to evaluate each i-level of the model
DO 60 i=1,Kmax
flag2=i-2*INT(i/2.0)
C Flag2=1 in the, first row,O in the second row etc.
C Start of the middle loop to evaluate each j-level of the model
DO 70 j=1,Lmax
C Start of the inner loop to evaluate each each element of the model
IF (flag.EQ.O) THEN
C Process water flow is in a forward direction
DO so k=2,Mmax+1
C Determine the input values for each element
Tspo=Tsp(i,j,k)
IF((k.EQ.2).AND.(i.NE.1}) Tspo=Tsp(i-l,j,k-1)
Tswi=Tsw(i,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.1).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-1,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa{i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE
sisai=(sisa{i,j,k)+sisa{i,j-l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsai=(Tsa{i,j,k)+Tsa{i,j-l,k))/2.0
END IF
 Stellenbosch University http://scholar.sun.ac.za

END IF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) THEN
CALL MERKEL2 (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Kmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL2 (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,Kmax)
ELSE
CALL POPPE2 (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,sp~atm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,Kmax)
END IF
C Determine the exit values for each element
Tsp(i,j,k+l)=Tspi
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)~sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.l) THEN
. WRITE(4,*)i,j,k-1
WRITE(4,*)Tspo,Tspi
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.l) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
IF (model .EQ.3) THEN
WRITE(4,*)Tsai,Tsao
END IF
END IF
 Stellenbosch University http://scholar.sun.ac.za

END IF
80 CONTINUE
ELSE IF (Flag.EQ.l) THEN
C Start of the inner loop to evaluate each each element of the model
C Process water flow is backwards to the origin
DO 90 k=Mmax+1,2,-1
C Qetermine the input values for each element
Tspo=Tsp(i,j,k)
IF (k.EQ.(Mmax+l)) Tspo=Tsp(i-l,j,k+l)
Tswi=Tsw(i ,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.l).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-l,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j-l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsa i =(Tsa ( i , j, k) +Is a (.i , j -1, k)) /2.0
END IF
END IF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) THEN
CALL MERKEL2 (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Kmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL2 (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,Kmax)
 Stellenbosch University http://scholar.sun.ac.za It

ELSE
CALL POPPE2 (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,Kmax)
END IF
C Determine the exit values for each element
Tsp(i,j,k-l)=Tspi
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)=sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.l) THEN
WRITE(4,*)i,j,k-1
WRITE(4,*)Tspo,Tspi
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.l) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
IF (model.EQ.3) THEN
WRITE(4,*)Tsai,Tsao
END IF
END IF
END IF
90 CONTINUE
END IF
70 CONTINUE
IF (flag.EQ.O) THEN
flag=!
ELSE
flag=O
END IF
60 CONTINUE
C Determine the average inlet temperature of process water
suml=O.O
sum2=0.0
rem=Mmax+2
IF (flag.NE.l) rem=l
DO 120 j=l,Lmax
CALL Cpw(Tsp(Kmax,j,rem),scspp)
suml=suml+Tsp(Kmax,j,rem)*scspp
sum2=sum2+Tsp(Kmax,j,rem)
120 CONTINUE
CALL Cpw((sum2/Lmax),scspp)
Tspi2=suml/(Lmax*scspp)
 Stellenbosch University http://scholar.sun.ac.za f)

C Determine if the inlet conditions satisfies the known outlet condition

C If not choose an new inlet condition and repeat from 999
rem=Mmax+2
IF (flag.NE.1) rem=1
sum4=0.0
DO 121 j=1,Lmax
sum4=sum4+ABS(Tspi1-Tsp(Kmax,j,rem))
121 CONTINUE
WRITE(*,125)Tspi2,sum4/Lmax
125 FORMAT(' Tp(in) calculated =' ,F6.2,
+ ' Average Deviation =' ,F10.6)
IF ((sum4/Lmax).GT.0.1) THEN
DO 122 j=1,Lmax
dT=(Tspi1-Tsp(Kmax,j,rem))/2.0
Tsp(1,j,2)=Tsp(1,j,2)+dT
122 CONTINUE
GOTO 999
END IF
C Determine the average exit temperature of the process water
sum1=0.0
sum2=0.0
DO 51 j=1,Lmax
. CALL Cpw(Tsp(1,j,2),scspp)
sum1=sum1+Tsp(1,j,2)*scspp
sum2=sum2+Tsp(1,j,2)
51 CONTINUE
CALL Cpw({sum2/Lmax),scspp)
Tspo1=sum1/(Lmax*scspp)
C Determine the average exit temperature of recirculating water
sum1=0.0
sum2=0.0
DO 110 i =1, Kmax
DO 100 k=2,Mmax+1
CALL Cpw(Tsw(i,Lmax+1,k),scspw)
sum1=sum1+Tsw(i,Lmax+1,k)*scspw
sum2=sum2+smsw(i,Lmax+1,k)
100 CONTINUE
110 CONTINUE
CALL Cpw(Tswi1,scspw)
T.swo1=sum1/(Mmax*Kmax*scspw)
smswo1=sum2
C Determine the average exit enthalpy of the air
sum1=0.0
sum2=0.0
sum3=0.0
DO 140 j=1,Lmax
DO 130 k=2,Mmax+1
sum1=sum1+sisa(Kmax+1,j,k)
sum2=sum2+swsa(Kmax+1,j,k)
 Stellenbosch University http://scholar.sun.ac.za

sum3=sum3+Tsa(Kmax+I,j,k)
I30 CONTINUE
I40 CONTINUE
DO ISO k=2,Mmax+I
IF (flag2.EQ.O) THEN
sumi=sumi+sisa(Kmax,I,k)/2.0
sum2=sum2+swsa(Kmax,I,k)/2.0
sum3=sum3+Tsa(Kmax,I,k)/2.0
ELSE
sumi=sumi+sisa(Kmax,Lmax,k)/2.0
sum2=sum2+swsa(Kmax,Lmax,k)/2.0
sum3=sum3+Tsa(Kmax,Lmax,k)/2.0
END IF
ISO CONTINUE
sisaoi=sumi/(Mmax*(Lmax+.S))
swsaoi=sum2/(Mmax*(Lmax+.5))
Tsaoi=sum3/(Mmax*(Lmax+.S))
C Print the recirc.water inlet and outlet temperatures on the screen
WRITE(*,I60)Tswii,Tswoi
I60 FORMAT(' ',/' Tw(in) = ',F7.3,' Tw(out) = ',F7.3/)
RETURN
END
c *******************************************************************
c* *
C* PARALLEL FLOW (FROM FRONT TO BACK OF COOLER) *
c* *
c *******************************************************************
C Subroutine to evaluate a cooler layout where the process fluid flows
C in a direction parallel to the direction of the airstream
SUBROUTINE FRONTTOBACK (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,
+ smspel,smswel,sisaii,sisaoi,Tspii,Tspoi,Tswii,
+ Tswoi,swsaii,swsaoi,smswii,smswol,L,Hlsdsi,sdso,
+ dA,Tsadb,Tsawb,spsatm,gamma,Vstot,sa,skst,
+ svsp,Aspi,Aspo,ReyC,gradfile,Kmax,Lmax,Mmax,PI,
+ model,Tsaol,shsfl,shsf2)
DIMENSION Tsp(40,400,10)
DIMENSION Tsw(40,400,10)
DIMENSION Tsa(40,400,10)
DIMENSION sisa(40,400,IO)
DIMENSION swsa(40,400,IO)
DIMENSION smsw(40,400,10)
REAL L
INTEGER flag,flag2,gradfile
C Initialize the three arrays with the known temperature and enthalpy values
CALL Enthalpy(Tsadb,Tsawb,spsatm,sisail)
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsail)
DO 20 j=l,Lmax
 Stellenbosch University http://scholar.sun.ac.za

DO 10 k=2,Mmax+1
sisa(1,j,k)=sisai1
swsa(1,j,k)=swsai1
Tsa(1,j,k)=Tsadb
10 CONTINUE
20 CONTINUE
DO 40 i=1,Kmax
DO 30 k=2,Mmax+1
Tsw(i,1,k)=Tswi1
smsw(i,1,k)=smswel
30 CONTINUE
40 CONTINUE
DO 50 j=1,Lmax
Tsp(1,j,2)=Tspi1
50 CONTINUE
C N.B. flag=1 for backward process fluid flow
C L.W. flag=O for forward process fluid flow
flag=O
C Start of the outer loop to evaluate each i-level of the model
DO 60 i=1,Kmax
flag2=i-2*1NT(i/2.0)
C Flag2=1 in the first row,O in the second row etc.
C Start of the middle loop to evaluate each j-level of the model
DO 70 j=1,Lmax
C Start of the inner loop to evaluate each each element of the model
IF (flag.EQ.O) THEN
C Process water flow is in a forward direction
DO 80 k=2,Mmax+1
C Determine the input values for each element
Tspi=Tsp(i,j,k)
IF((k.EQ.2).AND.(i.NE.1)) Tspi=Tsp(i-1,j,k-1)
Tswi=Tsw(i ,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
.
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.1).AND.(i.NE.1)) THEN
IF (j.EQ.1) THEN
sisai=(sisa(i-1,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-1,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j-1,k))/2.0
 Stellenbosch University http://scholar.sun.ac.za

swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j-l,k))/2.0
END IF
END IF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) THEN
CALL MERKEL (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm~gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Kmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo~
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,Kmax)
ELSE
CALL POPPE (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,Kmax)
END IF
C Determine the exit values for each element
Tsp(i,j,k+l)=Tspo
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)=sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.l) THEN
WRITE(4,*)i,j,k-1
WRITE(4,*)Tspi,Tspo
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.l) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
IF (model.EQ.3) THEN
 Stellenbosch University http://scholar.sun.ac.za

WRITE(4,*)Tsai,Tsao
END IF
END IF
END IF
80 CONTINUE
ELSE IF (Flag.EQ.l) THEN
C Start of the inner loop to evaluate each each element of the model
C Process water flow is backwards to the origin
DO 90 k=Mmax+l,2,-l
C Determine the input values for each element
Tspi=Tsp(i,j,k)
IF (k.EQ.(Mmax+l)) Tspi=Tsp(i-l,j,k+l)
Tswi=Tsw(i,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.l).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-l,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE . .
sisai=(sisa(i,j,k)+sisa(i,j-l,k))/2.0
swsai=(swsa( i ,j, k·)+swsa( i ,j -1, k) )/2. 0
Tsai=(Tsa(i,j,k)+Tsa(i,j-l,k))/2.0
END IF
END iF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i;j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) T~EN
CALL MERKEL (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Kmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
 Stellenbosch University http://scholar.sun.ac.za

+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsf1,shsf2,Kmax)
ELSE
CALL.POPPE (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsf1,shsf2,Kmax)
END IF '
C Determine the exit values for each element
Tsp(i,j,k-1)=Tspo
Tsw(i,j+1,k)=Tswo
Tsa(i+1,j,k)=Tsao
sisa(i+1,j,k)=sisao
swsa(i+1,j,k)=swsao
smsw(i,j+1,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.1) THEN
WRITE(4,*)i,j,k-1
WRITE(4,*)Tspi,Tspo
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.1) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
IF (model.EQ.3) THEN
WRITE(4,*)Tsai,Tsao
END IF
END IF
END IF
90 CONTINUE
END IF
70 CONTINUE
IF (flag.EQ.O) THEN
flag=1
ELSE
flag=O
END IF
60 CONTINUE
C Determine the average exit temperature of recirculating water
sum1=0.0
sum2=0.0
DO 110 i=1,Kmax
DO 100 k=2,Mmax+1
CALL Cpw(Tsw(i,Lmax+1,k),scspw)
sum1=sum1+Tsw(i,Lmax+1,k)*scspw
sum2=sum2+smsw(i,Lmax+1,k)
100 CONTINUE
110 CONTINUE
 Stellenbosch University http://scholar.sun.ac.za

CALL Cpw(Tswi1,scspw)
Tswo1=sum1/(Mmax*Kmax*scspw)
smswo1=sum2
C Determine the average exit temperature of process water
sum1=0.0
sum2=0.0
rem=Mmax+2
IF (flag.NE.1) rem=1
DO 120 j=1,Lmax
CALL Cpw(Tsp(Kmax,j,rem),scspp)
sum1=sum1+Tsp(Kmax,j,rem)*scspp
sum2=sum2+Tsp(Kmax,j,rem)
120 CONTINUE
CALL Cpw((sum2/Lmax),scspp)
Tspo1=sum1/(Lmax*scspp)
C Determine the average exit enthalpy of the air
sum1=0.0
sum2=0.0
sum3=0.0
DO 140 J=1, Lmax
DO 130 k=2,Mmax+1
sum1=sum1+sisa(Kmax+1,j,k)
sum2=sum2+swsa(Kmax+l,j,k)
sum3=sum3+Tsa(Kmax+1,j,k)
130 CONTINUE
140 CONTINUE
DO 150 k=2,Mmax+1
IF (flag2.EQ.O) THEN
sum1=sum1+sisa(Kmax,1,k)/2.0
sum2=sum2+swsa(Kmax,1,k)/2.0
sum3=sum3+Tsa(Kmax,l,k)/2.0
ELSE
suml=suml+sisa(Kmax,Lmax,k)/2.0
sum2=sum2+swsa(Kmax,Lmax,k)/2.0
sum3=sum3+Tsa(Kmax,Lmax,k)/2.0
END IF
150 CONTINUE
sisaol=suml/(Mmax*(Lmax+.5))
swsao1=sum2/(Mmax*(Lmax+.5))
Tsao1=sum3/(Mmax*(Lmax+.5))
C Print the recirc.water inlet and outlet temperatures on the screen
WRITE(*,160)Tswil,Tswol
160 FORMAT(' ','Tw(in) = ',F7.3,' Tw(out) = ',F7.3)
RETURN
END
 Stellenbosch University http://scholar.sun.ac.za

c *******************************************************************
c* *
C* IMPROVED MERKEL METHOD TO EVAUALTE A SINGLE ELEMENT *
c* *
c *******************************************************************
C Subroutine to apply the Runge-Kutta method of solution to the three
C Merkel equations and one additional equation which controls the
C state of a single element
SUBROUTINE IMPMERKEL(Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,nrow)
REAL L,musav,musw,kog,koga,Kl,K2,K3,K4
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspi,musw)
CALL Waterdensity(Tspi,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) ! Reynoldsnumber of process water
CALL AirVapMixdensity(Tsai,swsai,spsatm,rhosav)
CALL AirVapMixviscosity(Tsai,swsai,spsatm,musav)
Reysa=ReyC*rhosav/musav ! Reynoldsnumber of airflow
CALL Waterviscosity(Tswi,musw) '
gammal=gamma*(smswi/smswel)
Reysw=4.0*gammal/musw Reynoldsnumber of recirc.water
C Determine the neccessary transfer-coefficients ·
CALL Waterconductivity(Tspi,sksp)
CALL Prandtl(Tspi,Prasp)
shsw=4.186*118.0*((gamma1*3600.0/sdso)**(l.0/3.0))/3.6
IF (Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/(L*nrow)
shsp=(3.66+0.104*(term)/(1.0+0.016*(term)**(0.8)))*sksp/sdsi
ELSE
sfsd=(1.82*LOGIO(Reysp)-1.64)**(-2.0)
terml=Prasp*(l.O+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*((sfsd/8.0)**(0.5))*(Prasp**(0.67)-1.0)
shsp=((sfsd/8.0)*(Reysp-1000.0)*terml/term2)*sksp/sdsi
END IF
koga=1.81E-4*((Reysa)**.9)*((Reysw)**.15)*((sdso)**(-2.6))/3600.
kog=kogajsa ! Mass-transfer coefficient
Uo=l.O/((sdso/sdsi)*((l.O/shsp)+(l.O/shsfl))+(l.O/shsw)
+ +(l.O/shsf2)+sdso*LOG(sdso/sdsi)/(2.0*skst))
C Determine the controlling constants Kl,K2,K3 and K4 .
CALL Cpw(Tspi,scspp)
CALL Cpw(Tswi,scspw)
Kl=kog*dA/(smsael)
K2=Uo*dA/(smswi*scspw)
K3=kog*dA*lOOO.O/(smswi*scspw)
K4=Uo*dA/(smspel*scspp)
 Stellenbosch University http://scholar.sun.ac.za
2(

C Determine the Runge-Kutta coefficients

CALL Satenthalpy(Tswi,spsatm,sisasw1)
a1=Kl*(sisasw1-sisai) _
b1=K2*(Tspi-Tswi)-K3*(sisasw1-sisai)
c1=-K4*(Tspi-Tswi)
CALL Satenthalpy((Tswi+b1/2.0),spsatm,sisasw2)
a2=K1*(sisasw2-(sisai+a1/2.0))
b2=KZ*((Tspi+c1/2.0}-(Tswi+b1/2.0))-K3*(sisasw2-(sisai+a1/2.0))
c2=-K4*((Tspi+c1/2.0}-(Tswi+b1/2.0))
CALL Satenthalpy((Tswi+b2/2.0},spsatm,sisasw3)
a3=K1*(sisasw3-(sisai+a2/2.0))
b3=K2*((Tspi+c2/2.0)-(Tswi+b2/2.0))-K3*(sisasw3-(sisai+a2/2.0})
c3=-K4*((Tspi+c2/2.0}-(Tswi+b2/2.0))
CALL Satenthalpy((Tswi+b3),spsatm,sisasw4)
a4=K1*(sisasw4-(sisai+a3))
b4=K2*((Tspi+c3}-(Tswi+b3))-K3*(sisasw4-(sisai+a3))
c4=-K4*((Jspi+c3}-(Tswi+b3))
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw1)
CALL Airhumidity((Tswi+b1/2.0},(Tswi+b1/2.0},spsatm,swsasw2)
CALL Airhumidity((Tswi+b2/2.0},(Tswi+b2/2.0),spsatm,swsasw3)
CALL Airhumidity((Tswi+b3},(Tswi+b3},spsatm,swsasw4)
d1=K1*(swsasw1-swsai}/(1.0-swsasw1)
d2=K1*(swsasw2-(swsai+d1/2.0))/(1.0-swsasw2)
d3=K1*(swsasw3-(swsai+d2/2.0))/(1.0-swsasw3)
d4=K1*(swsasw4-(swsai+d3))/(1.0-swsasw4)
C Determine the exit conditions of the element
sisao=sisai+(a1+2.0*(a2+a3)+a4)/6.0
Tswo=Tswi+(b1+2.0*(b2+b3}+b4)/6.0
Tspo=Tspi+(c1+2.0*(c2+c3)+c4)/6.0
swsao=swsai+(d1+2.0*(d2+d3)+d4)/6.0
smswo=smswi-smsael*(swsao-swsai)
C Determine the air outlet temperature-and saturation enthalpy
TR=Tspi
TL=O.O
10 Tsao=(TR+TL)/2.0
CALL Cpv(Tsao,scspv).
CALL Cpa(Tsao,scspa)
CALL Cpw(Tsao,scspw}
CALL Airhumidity(Tsao,Tsao,spsatm,swsasa)
IF (swsasa.GT.swsao} THEN
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0}
ELSE
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0}
+ +scspw*(swsao-swsasa)*Tsao/1000.0
END IF
IF ((ABS(sisao-sisasa)).GT.0.1) THEN
 Stellenbosch University http://scholar.sun.ac.za lL

IF (sisao.LT.sisasa) THEN
TR=Tsao
ELSE
TL=Tsao
END IF
GO TO 10
END IF
RETURN-
END

c *******************************************************************
c* *
C* IMPROVED MERKEL METHOD(2) TO EVAUALTE A SINGLE ELEMENT *
c*
c ********************************************************************
C Subroutine to apply the Runge-Kutta method of solution to the three
C Merkel equations and one additional equation which controls the
C state of a single element; BACKTOFRONT FLOW CASE
SUBROUTINE IMPMERKEL2{Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,nrow)
REAL L,musav,musw,kog,koga,Kl,K2,K3,K4
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspo,musw)
CALL Waterdensity(Tspo,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) ! Reynoldsnumber of process water
CALL AirVapMixdensity{Tsai,swsai,spsatm,rhosav)
CALL AirVapMixviscosity(Tsai,swsai,spsatm,musav)
Reysa=ReyC*rhosav/musav ! Reynoldsnumber of airflow
CALL Waterviscosity(Tswi,musw)
gammal=gamma*(smswi/smswel)
Reysw=4.0*gammal/musw Reynoldsnumber of recirc.water
C Determine the neccessary transfer-coefficients
CALL Waterconductivity(Tspo,sksp)
CALL Prandtl(Tspo,Prasp)
shsw=4.186*118.0*{(gammal*3600.0/sdso)**(l.0/3.0))/3.6
IF {Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/{L*nrow)
shsp={3.66+0.104*{term)/(1.0+0.016*(term)**{0.8)))*sksp/sdsi
ELSE
sfsd=(1.82*LOG1J(Reysp)-1.64)**(-2.0)
terml=Prasp*(l.O+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*({sfsd/8.0)**(0.5))*{Prasp**(0.67)-l.O)
shsp={(sfsd/8.0)*(Reysp-IOOO.O)*terml/term2)*sksp/sdsi
END IF
koga=l.81E-4*((Reysa)**.9)*((Reysw)**.l5)*((sdso)**(-2.6))/3600.
 Stellenbosch University http://scholar.sun.ac.za

kog=koga/sa ! Mass-transfer coefficient

Uo=l.O/((sdso/sdsi)*((l.O/shsp)+(l.O/shsfl))+(l.O/shsw)
+ +(l.O/shsf2)+sdso*LOG(sdso/sdsi)/(2.0*skst))
C Determine the controlling constants Kl,K2,K3 and K4
CALL Cpw(Tspo,scspp)
CALL Cpw(Tswi,scspw)
Kl=kog*dA/(smsael)
K2=Uo*dA/(smswi*scspw)
K3=kog*dA*lOOO.O/(smswi*scspw)
K4=Uo*dA/(smspel*scspp)
C Determine the Runge-Kutta coefficients
CALL Satenthalpy(Tswi,spsatm,sisaswl)
al=Kl*(sisaswl-sisai)
bl=K2*(Tspo-Tswi)-K3*(sisaswl-sisai)
cl=K4*(Tspo-Tswi)
CALL Satenthalpy((Tswi+bl/2.0),spsatm,sisasw2)
a2=Kl*(sisasw2-(sisai+al/2.0))
b2=K2*((Tspo+cl/2.0)-(Tswi+bl/2.0))-K3*(sisasw2-(sisai+al/2.0))
·c2=K4*((Tspo+cl/2.0)-(Tswi+bl/2.0))
CALL Satenthalpy((Tswi+b2/2.0),spsatm,sisasw3)
a3=Kl*(sisasw3-(sisai+a2/2.0))
b3=K2*((Tspo+c2/2.0)-(Tswi+b2/2.0))-K3*(sisasw3-(sisai+a2/2.0)}
c3=K4*((Tspo+c2/2.0)-{Tswi+b2/2.0))
CALL Satenthalpy((Tswi+b3),spsatm,sisasw4)
a4=Kl*(sisasw4-(sisai+a3))
b4=K2*({Tspo+c3)-(Tswi+b3)}-K3*(sisasw4-(sisai+a3))
c4=K4*({Tspo+c3)-{Tswi+b3)}
CALL Airhumidity(Tswi,Tswi,spsatm,swsaswl}
CALL Airhumidity((Tswi+bl/2.0),{Tswi+bl/2.0},spsatm,swsasw2}
CALL Airhumidity((Tswi+b2/2.0),{Tswi+b2/2.0),spsatm,swsasw3)
CALL Airhumidity((Tswi+b3),{Tswi+b3),spsatm,swsasw4)
dl=Kl*(swsaswl-swsai)/(1.0-swsaswl)
d2=Kl*(swsasw2-(swsai+dl/2.0})/(l.O-swsasw2}
d3=Kl*(swsasw3-(swsai+d2/2.0))/(l.O-swsasw3}
d4=Kl*(swsasw4-(swsai+d3}}/(l.O-swsasw4}
C Determine the exit conditions of the element
sisao=sisai+(al+2.0*(a2+a3}+a4)/6.0
Tswo=Tswi+(bl+2.0*(b2+b3)+b4)/6.0
Tspi=Tspo+(cl+2.0*(c2+c3)+c4)/6.0
swsao=swsai+(dl+2.0*(d2+d3)+d4)/6.0
smswo=smswi-smsael*(swsao-swsai)
C Determine the air outlet temperature and saturation enthalpy
TR=Tspi
TL=O.O
 Stellenbosch University http://scholar.sun.ac.za
4-

10 Tsao=(TR+TL)/2.0
CALL Cpv(Tsao,scspv)
CALL Cpa(Tsao,scspa)
CALL Cpw(Tsao,scspw)
CALL Airhumidity(Tsao,Tsao,spsatm,swsasa)
IF (swsasa.GT.swsao) THEN
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0)
ELSE
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0)
+ +scspw*(swsao-swsasa)*Tsao/1000.0
END IF
IF ((ABS(sisao-sisasa)).GT.0.1) THEN
IF (sisao:LT.sisasa) THEN
TR=Tsao
ELSE
TL=Tsao
END IF
GO TO 10
END IF
RETURN
END

c *************************************************************
c * *
C * INITIALIZE ALL THE NEEDED PARAMETERS *
c * *
·c *************************************************************
C Subroutine to set default values for a typical cooler
SUBROUTINE INITIAL(spsatm,Tsadb,Tsawb,L,H,sdso,sdsi,Kmax,
+ Lmax,Mmax,vspas,hspas,smsp,PI,gamma,skst,Tspi1,
+ clrtype,model,smsa,flowlayout,Tswi1,shsf1,shsf2)
REAL L
INTEGER clrtype,flowlayout
spsatm=101325.0 Atmospheric pressure [Pa]
Tsadb=25.0 Dry-bulb temperature of air [ C]
Tsawb=19.5 Dry-bulb temperature of air [ C]
sdso=38.1/1000.0 Pipe Outer Diameter [m]
sdsi=34.9/1000.0 Pipe Inner Diameter [m]
Kmax=10 Number of pipe rows
Mmax=1 Number of elements along each pipe
vspas=2.0*sdso Vertical spacing between pipes [m]
hspas=SQRT(3.0)*sdso Horizontal spacing between pipes [m]
PI=4.0*ATAN(1.0) Pi
gamma=300.0/3600.0 Recirc.~ater massflow/length [kg/m/s]
shsfl=20000.0 Fouling heat transfer coeff. [W/m"2 K]
shsf2=20000.0 Fouling heat transfer coeff. [W/m"2 K]
skst=43.0 Thermal conductivity of tube [W/m K]
Tspil=SO. 0 Process water inlet temperature [ C]
smsp=15.0 Total process water-massflow [kg/s]
smsa=11. 75388 Total air massflow [kg/s]
 Stellenbosch University http://scholar.sun.ac.za 2)

H=2.0 Inlet height of cooler [m]

L=2.0 Length of each pipe [m]
flowlayout=3 1 - front to back
2 - back to front
3 - top to bottom
4 - straight through
model=1 1 - Merkel model
2 - improved Merkel model
3 - Poppe model
clrtype=1 1 - Recirc. cooling water
2 - Single pass cooling water flow
Tswi1=35.0 Inlet cooling water temperature [ C]
RETURN
END

c *******************************************************************
c* . *
C* MENU (1) : EDIT CURRENT COOLER PARAMETERS *
c* *
c *******************************************************************
C Subroutine to edit the cooler dimensions
SUBROUTINE MENUI(clrtype,model,H,L,spsatm,Tsadb,Tsawb,
+ flowlayout,Tswil,shsfl,shsf2)
C Declare new variable types
REAL l
INTEGER clrtype,flowlayout
C Display the current cooler parameters on the screen
10 CALL LIB$ERASE PAGE(l,l)
WRITE{*,lS) -
lSFORMAT{'',
+ 'CROSSFLOW EVAPORATIVE COOLER - Menu 1'/
+I ---------------------------------------')
IF (clrtype.EQ.l) THEN
WRITE(*,*)'Cooling water flow RECIRCULATING'
ELSE IF (clrtype.EQ.2) THEN
WRITE(*,*)'Cooli~g water flow SINGLE PASS'
END IF
IF (flowlayout.EQ.l) THEN
WRITE(*,*)'Process water flow layout FRONT TO BACK'
ELSE IF (flowlayout.EQ.2) THEN
WRITE(*,*)'Process water flow layout BACK TO FRONT'
ELSE IF (flowlayout.EQ.3) THEN
WRITE(*,*)'Process water flow layout TOP TO BOTTOM'
ELSE IF (flowlayout.EQ.4) THEN
WRITE(*,*)'Process water flow layout STRAIGHT THROUGH'
END IF
IF (model .EQ.l) THEN
 Stellenbosch University http://scholar.sun.ac.za 2£

WRITE(*,*)'Analytical model MERKEL'

ELSE IF (model.EQ.2) THEN
WRITE{*,*)'Analytical model Improved MERKEL'
ELSE
WRITE(*,*)'Analytical model POPPE'
END IF
WRITE{*,*)' I

WRITE{*,20)H,L,spsatm/1000.0,Tsadb,Tsawb,shsfl,shsf2
20 FORMAT(
+' 0- Change cooling water flow (single pass/recirc.)'/
+ ' 1 - Change process water flow pattern'/
+ ' 2 - Change solution model (MERKEL/Improved MERKEL/POPPE)'/
+' 3- Cooler height .................... = ',F8.2,' m'/
+' 4- Cooler length (across airflow) ... = ',F8.2,' m'/
+' 5- Atmospheric pressure ............. = ',F8.2,' kPa'/
+' 6- Inlet air temperature (dry bulb) . = ',F8.2,' C'/
+' 7- Inlet air temperature (wet bulb) . = ',F8.2,' C'/
+ ' 8 Fouling coefficient inside tube .. = ',Fl2.2,' W/mA2 K'/
+' 9- Fouling coefficient outside tube = ',Fl2.2,' W/mA2 K')
C Display inlet cooling water temperature
IF (clrtype.EQ.2) THEN
WRITE(*,30)Tswil
30 FORMAT(' 10- Cooling water inlet temperature .. = ',F8.2,' C')
END IF
C Read keyboard to determine which dataset has to be changed
WRITE(*,35)
35 FORMAT(/' Which value has to be changed (15 - CONTINUE) ? ',$)
READ(*,*)number
C Change cooling/recirc. water option
. 999 IF (number.EQ.O) THEN
WRITE(*,*)' I

WRITE(*,*)'The following options are available'

WRITE(*,*)' 1 - Recirculating cooling water flow'
WRITE(*,*)' 2 - Single pass cooling water flow'
WRITE(*,*)' I

WRITE(*,*)'Enter choice (I or 2) ?'

READ(*,*)clrtype
IF ((clrtype.GT.2).0R.(clrtype.LT.l)) THEN
number=O
GOTO 999
END IF
GO TO 10
C Change process water flow pattern
ELSE IF (number.EQ.l) THEN
WRITE(*,*)' I

WRITE(*,*)'The following cooler layouts are available'

WRITE(*,*)' 1 - fronttoback'
 Stellenbosch University http://scholar.sun.ac.za

WRITE(*,*)' 2 - backtofront'
WRITE(*,*}' 3 - toptobottom'
WRITE(*,*}' 4 - straight through'
WRITE(*,*)' I

WRITE(*,*)'Enter choice (1,2,3 or 4) ?'

READ(*,*)flowlayout
IF ((flowlayout.GT.4).0R.(flowlayout.LT.l)) THEN
number= I
GOTO 999
END IF
GO TO 10
C Change model type
ELSE IF (number.EQ.2) THEN
WRITE(*,*}' I

WRITE(*,*)'The following models are available : '

WRITE(*,*}' 1 -MERKEL model ( 3 Equation )'
WRITE(*,*}' 2 - Improved MERKEL model ( 4 Equation )'
WRITE(*,*)' 3 - POPPE model'
WRITE(*,*}' I

WRITE(*,*}'Enter choice (1,2 or 3) ?'

READ(*,*)model
IF ((model.GT.3).0R.(model.LT.l)) THEN
number=2
GOTO 999
END IF
GO TO 10
C Change cooler height
ELSE IF (number.EQ.3) THEN
WRITE(*,*}' I

WRITE{*,*)'What is the new cooler height in m ?'

READ(*,*)H
IF (H.LE.O.O) THEN
number=3
GO TO 999
END IF
GO TO 10
C Change cooler width
ELSE IF (number.EQ.4) THEN
WRITE(*,*)' I

WRITE{*,*)'What is the new cooler width in m ?'

READ(*,*)L
IF (L.LE.O.O) THEN
number=4
GO TO 999
END IF
GO TO 10
C Change the value of atmospheric pressure
ELSE IF (number.EQ.S) THEN
 Stellenbosch University http://scholar.sun.ac.za

WRITE{*,*)I I

WRITE{*,*)'What is the new atmospheric pressure in kPa? '

READ{*,*)spsatm
C Ensure that input is not an absurd value
IF {spsatm.LT.60.0) THEN
WRITE{*,*)'The atmospheric pressure must be above 60 kPa'
number=S
GO TO 999
END IF
spsatm=spsatm*1000.0
GO TO 10
C Change the value of the inlet air temperature {dry bulb)
ELSE IF {number.EQ.6) THEN
WRITE{*,*)' I

WRITE{*,*)'Give the air inlet temperature {dry bulb C) ? '

READ{*,*)Tsadb
IF {Tsadb.LT.Tsawb) THEN
WRITE{*,*)'Wet bulb temperature > Dry bulb temperature'
WRITE{*,*)'T{dry bulb) larger {0) I T{wet bulb) smaller {1) ?'
READ{*,*)number3
IF {number3~EQ.O) THEN
number=Q Choose new dry bulb temperature
GO TO 999
ELSE
number=7 Choose new wet bulb temperature
GO TO 999
END IF
ELSE IF {Tsadb.GT.100) THEN
WRITE{*,*)'Air temperature {dry bulb) must be < lOO C'
number=6
GO TO 999
END IF
GO TO 10
C Change the value of the inlet air temperature {wet bulb)
ELSE IF {number.EQ.7) THEN
WRITE{*,*)' I

WRITE{*,*)'Give the air inlet temperature {wet bulb C) ? '

READ{*,*)Tsawb
IF {Tsadb.LT.Tsawb) THEN
WRITE{*,*)'Wet bulb temperature> Dry bulb temperature'
WRITE{*,*)'T{dry bulb) larger {0) I T{wet bulb) smaller {1) ?'
READ{*,*)number3
IF {number3.EQ.O) THEN
number=6 ! Choose new dry bulb temperature
GO TO 999
ELSE
number=7 Choose new wet bulb temperature
GO TO 999
END IF
END IF
 Stellenbosch University http://scholar.sun.ac.za

GO TO 10
C Change cooling water inlet temperature
ELSE IF (number.EQ.S) THEN
WRITE(*,*)' I

WRITE(*,*)'What is the new outer fouling coefficient ?'

READ(*,*)shsfl
IF (shsf1.LE.1000.0) THEN
WRITE(*,*)'Fouling coefficient> 1000 W/mA2 K'
number=8
GO TO 999
END IF
GO TO 10
C Change cooling water inlet temperature
ELSE IF (number.EQ.9) THEN
WRITE(*,*)' I

WRITE(*,*)'What is the new outer fouling coefficient ?'

READ(*,*)shsf2
IF (shsf2.LE.100Q.O) THEN
WRITE(*,*)'Fouling coefficient> 1000 W/mA2 K'
number=9
GO TO 999
END IF
GO TO 10
C Change cooling water inlet temperature
ELSE IF (number.EQ.10) THEN·
WRITE(*,*)' I

WRITE(*,*)'What is the new cooling water inlet temperature?'

READ(*, *)Tswil
IF ((Tswi1.LE.O.O).OR.(Tswi1.GT.100.0)) THEN
number=10
GO TO 999
END IF
GO TO 10
END IF
RETURN
END

c *******************************************************************
c* *
C* MENU (2) : EDIT CURRENT COOLER PARAMETERS *
c* *
c *******************************************************************
C Subroutine to edit the cooler dimensions
SUBROUTINE MENU2(sdso,sdsi,H,L,PI,svsa,vspas,hspas,Lmax,Kmax,
+ Mmax,spsatm,Tsadb,Tsawb,gamma,skst,Tspil,smsp,
+ svsp,sa,Aspi,Aspo,clrtype,dA,model,smsa,flowlayout,
+ Tswi1,shsf1,shsf2)
 Stellenbosch University http://scholar.sun.ac.za ]o

C Initialize data types

REAL L
INTEGER clrtype,flowlayout
C Display the current cooler parameters on the screen
10 CALL LIB$ERASE_PAGE(l,l)
WRITE(*, 15)
15 FORMAT(' I '
+ 'CROSSFLOW EVAPORATIVE COOLER --Menu 2'/
+ I ---------------------------------------')
IF (clrtype.EQ.1) THEN
WRITE(*,*)'Cooling water flow RECIRCULATING'
ELSE IF (clrtype.EQ.2) THEN
WRITE(*,*)'Cooling water flow SINGLE PASS'
END IF
IF (flowlayout.EQ.1) THEN
WRITE(~,*)'Process water flow layout FRONT TO BACK'
ELSE IF (flowlayout.EQ.2) THEN
WRITE(*,*)'Process water flow layout BACK TO FRONT'
ELSE IF (flowlayout.EQ.3) THEN
WRITE(*,*)'Process water flow layout TOP TO BOTTOM'
ELSE IF (flowlayout.EQ.4) THEN
WRITE(*,*)'Process water flow layout STRAIGHT THROUGH'
END IF
IF (model.EQ.1) THEN
WRITE(*,*)'Analytical model MERKEL'
·ELSE IF (model.EQ.2) THEN
WRITE(*,*)'Analytical model Improved MERKEL'
ELSE'
WRITE(*,*)'Analytical model POPPE'
END IF
WRITE(*,*)' I

WRITE(*,20)sdso*lOOO.O,sdsi*lOOO.O,vspas*lOOO.O,hspas*lOOO.O
20 FORMAT(
+ ' 0 -Go back to previous menu '/
+' 1 -Outer diameter of pipe ............... = ,F7.2,' mm'/ I

+·' 1 - Inner diameter of pipe ............... = 'F7. 2' mm'/ I I

+' 2 Vertical spacing between pipes ....... = 'F7. 2' mm'/ I I

+' 2- Horizontal spacing between pipes ..... = 'F7. 2' mm') I I

C Check whether the array-dimensions were large enough

Lmax=INT((H-O.S*vspas+O.OOl)/vspas)
IF (Lmax.GT.400) THEN
WRITE(*,*)'Max number of elements in vertical direction'
WRITE(*,*)'permitted = 400'
WRITE(*,*)'Choose a larger vertical spacing (0)'
WRITE(*,*)'or change the DIMENSION of the array (1)'
READ(*,*)iantw
IF (iantw.EQ.O) THEN
number=2
GO TO 999 Choose bigger vertical spacing
 Stellenbosch University http://scholar.sun.ac.za ](

ELSE
WRITE{*,*)'REMEMBER TO CHANGE THIS CONDITION AS WELL'
STOP ! Change the DIMENSION statement
END IF
C Check whether the new vertical spacing is allowable
ELSE IF (Lmax.LT.l) THEN
WRITE{*,*)'Vertical spacing too large to fit at least one'
WRITE{*,*)'pipe into the cooler'
WRITE{*,*)'Choose new vertical spacing'
number=2
GO TO 999
END IF
sa=PI*sdso/(vspas*hspas) ! Coolerarea/unit volume
dA=L*PI*sdso/(Mmax) ! Coolerarea/element
Aspi=PI*(sdsi/2.0)**2.0 ! Pipe inner area
Aspo=PI*(sdso/2.0)**2.0 ! Pipe outer area
C Determine the water veloctity inside tubes and massflow needed to give
C a water velocity of 1 m/s in tubes
CALL Waterdensity(Tspil,rhosw)
IF ((flowlayout.EQ.l).OR.(flowlayout.EQ.2)) THEN
svsp=smsp/(Aspi*rhosw*Lmax)
svspl=rhosw*Aspi*Lmax*l.O
ELSE IF (flowlayout.EQ.3) THEN
svsp=smsp/ (As pi *rhosw*Kmax)._
svspl=rhosw*Aspi*Kmax*l.O
ELSE IF (flowlayout.EQ.4) THEN
svsp=smsp/(Aspi*rhosw*Kmax*Lmax)
svspl=rhosw*Aspi*Kmax*Lmax*l.O
END IF
C Print the variable values on the screen in order to edit them if needed
WRITE(*;30)Kmax,Lmax,Mmax,Tspil,smsp,svsp,gamma*3600.0,
+ gamma*Kmax*L*2.0,smsa,skst
30 FORMAT(
+' 3- Number of pipe rows (passes) ......... - ',13/
+ Number of pipes facing the airstream . = ',13/
+ ' 4 - Number of elements along a single pipe= ',13/
+' 5 Process water inlet temperature ...... = ',F7.2,' t'/
+' 6- Process water massflow ............... = ',F7.2,' kg/~'/
+ Process water flow velocity in pipes . = ',F7.2,' m/s'/
+' 7 Recirc.water massflow I length ....... = ',F7.2,' kg/m.hr'/
+ Recirculating water massflow ......... = ',F7.2,' kg/s'/
+ ' 8 Dry air massflow rate ................ = ,F7 .2,' kg/s'/ 1

+ 9 -Thermal conductivity of tube wall

1
= ',F7.2, W/m K'/) 1

C Read keyboard to determine which dataset has to be changed

WRITE(*,35)
35 FORMAT(' Which value has to be changed (15 - CONTINUE) ? ',$)
READ(*,*)number

\
 Stellenbosch University http://scholar.sun.ac.za

C Change cooler unit layout

999 IF (number.EQ.O) THEN
CALL MENUI(clrtype,model,H~L,spsatm,Tsadb,Tsawb,flowlayout,
+ Tswil,shsfl,shsf2)
GO TO 10
C Change the pipe dimensions
ELSE IF (number.EQ.l) THEN
WRITE(*,*}' I

WRITE(*,*)'Enter the pipe outer diameter in mm ?'

READ(*,*}sdso
WRITE(*,*}'Enter the pipe inner diameter in mm ?'
READ(*,*}sdsi
sdso=sdso/1000.0
sdsi=sdsi/1000.0
C Check if the pipe size is physically allowable with given configuration
IF (sdsi.GE.sdso) THEN
WRITE(*,*}'inner diameter>= outer diameter'
number= I
GO TO 999
ELSE IF (sdso.GT.vspas) THEN
WRITE(*,*}'Element boandaries interfere !!!'
WRITE(*,*)'Choose smaller pipe diameter (0)
WRITE(*,*}'or choose a larger vertical spacing (1) ? '
READ(*,*)number2
IF (number2.EQ.l} THEN
number=2 Change spacing
ELSE
number=! Change pipe diameter
END IF
GO TO 999
ELSE IF (sdso.GT.hspas) THEN
WRITE(*,*)'Element boundaries interfere !!!'
WRITE(*,*)'Choose smaller pipe diameter (0)
WRITE(*,*)'or choose a larger horizontal spacing (1) ? 1

READ(*,*)number2
IF (number2.EQ.l) THEN
number=2 Change spacing
ELSE
number=l Change pipe diameter
END IF
GO TO 999
END IF
vspas=2.0*sdso
hspas=SQRT(3.0)*sdso
GO TO 10
C Change the spacing of the pipe array
ELSE IF (number.EQ.2) THEN
WRITE(*,*)' I

WRITE(*,*)'Give the vertical spacing between pipes in mm ? 1

 Stellenbosch University http://scholar.sun.ac.za

READ(*,*)vspas
WRITE(*,*)'Give the horizontal spacing between pipes in mm ?'
READ(*,*)hspas
vspas=vspas/1000.0
hspas=hspas/1000.0
C Check whether this configuration is physically possible
C with the chosen pipes
IF (sdso.GT.vspas) THEN
WRITE(*,*)'Element boundaries interfere !!!'
WRITE(*,*)'Choose smaller pipe diameter (0)
WRITE(*,*)'or choose a larger vertical spacing (1) ? '
READ(*,*)number2
IF (number2.EQ.O) THEN
number=! Change pipe diameter
ELSE
number=2 Change spacing
END IF
GO TO 999
ELSE IF (sdso.GT.hspas) THEN
WRITE(*,*)'Element boundaries interfere !!!'
WRITE(*,*)'Choose smaller pipe diameter (0)
WRITE(*,*)'or choose a larger horizontal spacing (lY? '
READ(*,*)number2
IF (number2.EQ.O) THEN
number=! Change pipe diameter
ELSE
number=2 Change spacing
END IF
GO TO 999
END IF
GO TO 10
C Change the number of pipe rows
ELSE IF (number.EQ.3) THEN
WRITE(*,*)' I

WRITE(*,*)'Number of pipe rows ? '

READ(*,*)Kmax
C Ensure that there is a positive number of pipe rows
IF (Kmax.LT.l) THEN
WRITE(*,*)'Minimum number of pipe= 1 - Choose again'
number=3
GO TO 999
C Check if array-DIMENSION size was sufficient
ELSE IF (Kmax.GT.40) THEN
WRITE(*,*)'Max number of pipe rows= 40'
WRITE(*,*)'Choose less pipe rows (0) OR '
WRITE(*,*)'change array DIMENSION (1) ?'
READ(*,*)iantw
C Determine if DIMENSION or number of rows must be changed
IF (iantw.EQ.O) THEN
number=3 ! Choose less then 40 pipe rows
GO TO 999
 Stellenbosch University http://scholar.sun.ac.za

ELSE
WRITE{*,*)'Change array DIMENSION in source code'
WRITE(*,*)'REMEMBER to change this condition too'
STOP
END IF
END IF J

GO TO 10
C Change the number of elements·across a single pipe
ELSE IF (number.EQ.4) THEN
WRITE(*,*)' I

WRITE(*,*)'Number of elements across a single pipe ? '

READ(*,*)Mmax
C Ensure a positive number of elements
IF (Mmax.LT.1) THEN
WRITE(*,*)'Minimum number of elements = 1 - Choose again'
number=4
GO TO 999
C Check if array-DIMENSION size was sufficient
ELSE IF (Mmax.GT.10) THEN
WRITE(*,*)'Max number of pipe rows = 10'
WRITE(*,*)'Choose less elements along pipe (0) OR '
WRITE(*,*)'change array DIMENSION (1) ?'
READ(*,*)iantw
C Determine if DIMENSION or number of elements must be changed
IF (iantw.EQ.O) THEN
· number=4 ~ Choose less than 10 elements
GOTO 999
ELSE
WRITE(*,*)'Change array DIMENSION in source code'
WRITE(*,*)'REMEMBER to change this condition too'
STOP
END IF
END IF
GO TO 10
C Change the process water inlettemperature
ELSE IF (number.EQ.S) THEN
WRITE(*,*)' I

WRITE(*,*)'Process water inlet temperature ( C) ? '

READ(*,*) Tspil
IF (Tspil.GT.100) THEN
WRITE(*,*)'Process water inlet temperature must be < 100 C'
number=S
GO TO 999
ELSE IF (Tspi1.LE.O) THEN
WRITE(*,*)'Process water inlet temperature must be > 0 C'
number=S
GO TO 999
END IF
GO TO 10
 Stellenbosch University http://scholar.sun.ac.za

C Change the process water massflow rate

ELSE IF {number.EQ.6) THEN
WRiTE{*,*)' I
WRITE{*,40)smsp,svsp1
40 FORMAT{
+ ' Process water massflow for cooler .. = ',F9.2,' kg/s'/
+ 'Massflow needed for velocity of 1 m/s = ',F9.2,' kg/s'/)
WRITE{*,*)'Give new total massflow {kg/s) ? '
READ{*,*)smsp
IF {smsp.LE.O) THEN
WRITE{*,*)'Total process water massflow must be> 0 kg/s'
number=6
GO TO 999
END IF
GO TO 10
C Change the recirculating water massflow rate
ELSE IF {number.EQ.7) THEN
. WRITE{*,*)'I

WRITE{*,45)gamma*3600
45 FORMAT{' Previous recirc. water massflow (kg/m/hr) = ',F8.2,/
+ ' New recirc. water massflow (kg/m/hr) ? ')
READ{*,*)gamma
gamma=gamma/3600.0
IF (gamma.LT.1.5*700.*sdso/3600.) THEN
WRITE{*,46)1.5*700.*sdso
number=7
GO TO 999
END IF
46 FORMAT(' Recirc. water massflow must be>' ,F8.2,' kg/m/hr')
GO TO 10
C Change the air massflow rate
ELSE IF (number.EQ.8) THEN
WRITE{*,*)' I

WRITE{*,*)'Dry air massflow {kg/s) ? '

READ{*,*)smsa
IF {smsa.LT.O) THEN
WRITE{*,*)'Total dry air massflow must be>= 0 kg/s'
number=8
GO TO 999
END IF
GO TO 10
C Change the tube wall thermal conductivity
ELSE IF {number.EQ.9) THEN
WRITE{*,*)' I

WRITE(*,*)'Thermal conductivity for different tube materials'

WRITE(*,*)'Kt {Aluminium) - 204 WymK'
WRITE{*,*)'Kt (Steel 0.5% C) - 54 W/mK'
WRITE(*,*)'Kt (Steel 1.0% C) - 43 W/mK'
WRITE{*,*)'Kt (Steel 1.5% C) - 36 W/mK'
 Stellenbosch University http://scholar.sun.ac.za

WRITE(*,*)'Kt (Copper) . - 376 W/mK'

WRITE(*,*)'Tube wall thermal conductivity (W/mK) ? '
READ(*,*)skst
IF (skst.LT.10.0) THEN
WRITE(*,*)'Conductivity must be > 10'
number=9
GO TO 999
END IF
GO TO 10
END IF
RETURN
END
c *******************************************************************
c* *
C* MERKEL METHOD TO EVAUALTE A SINGLE ELEMENT *
c* *
c *******************************************************************
C Subroutine to apply the Runge-Kutta method of solution to the three
C Merkel equations which controls the state of a single element
SUBROUTINE MERKEL(Tspi,Tswi,sisai,swsai1,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsf1,shsf2,nrow)
REAL L,musav,musw,kog,koga,K1,K2,K3,K4
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspi,musw)
CALL Waterdensity(Tspi,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) Reynoldsnumber of process water
Tsa=Tsadb
CALL AirVapMixdensity(Tsa,swsai1,spsatm,rhosav)
CALL AirVapMixviscosity(Tsa,swsai1,spsatm,musav)
Reysa=ReyC*rhosavjmusav ! Reynoldsnumber of airflow
CALL Waterviscosity(Tswi,musw)
Reysw=4.0*gammajmusw ! Reynoldsnumber of recirc.water

C Determine the neccessary transfer-coefficients

CALL Waterconductivity(Tspi,sksp)
CALL Prandtl(Tspi,Prasp)
shsw=4.186*118.0*((gamma*3600/sdso)**(1.0/3.0))/3.6
IF (Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/(L*nrow)
shsp=(3.66+0.104*(term)/(1.0+0.016*(term)**(0.8)))*sksp/sdsi
ELSE
sfsd=(1.82*LOG10(Reysp)-1.64)**(-2.0)
term1=Prasp*(1.0+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*((sfsd/8.0)**(0.5))*(Prasp**(0.67)-1.0)
shsp=((sfsd/8.0)*(Reysp-1000.0)*term1/term2)*sksp/sdsi
END IF
koga=1.81E-4*((Reysa)**.9)*((Reysw)**.15)*((sdso)**(-2.6))/3600.
 Stellenbosch University http://scholar.sun.ac.za

kog=koga/sa ! Mass-transfer coefficient

Uo=l.O/{{sdso/sdsi)*{{l.O/shsp)+{l.O/shsfl))+{l.O/shsw)
+ +{l.O/shsf2)+sdso*LOG{sdso/sdsi)/{2.0*skst))
C Determine the controlling constants Kl,K2,K3 and K4
CALL Cpw{Tspi,scspp)
CALL Cpw{Tswi,scspw)
Kl=kog*dA/{smsael)
K2=Uo*dA/{smswel*scspw)
K3=kog*dA*lOOO.O/{smswel*scspw)
K4=Uo*dA/{smspel*scspp)
C Determine the Runge-Kutta coefficients
CALL Satenthalpy{Tswi,spsatm,sisaswl)
al=Kl*{sisaswl-sisai)
bl=K2*{Tspi-Tswi)-K3*{sisaswl-sisai)
cl=-K4*{Tspi-Tswi)
CALL Satenthalpy((Tswi+bl/2.0),spsatm,sisasw2)
a2=Kl*(sisasw2-(sisai+al/2.0))
b2=K2*((Tspi+cl/2.0)-(Tswi+bl/2.0))-K3*(sisasw2-(sisai+al/2.0))
c2=-K4*((Tspi+cl/2.0)-(Tswi+bl/2.0))
CALL Satenthalpy((Tswi+b2/2.0),spsatm,sisasw3)
a3=Kl*{sisasw3-(sisai+a2/2.0))
b3=K2*((Tspi+c2/2.0)-(Tswi+b2/2.0))-K3*(sisasw3-(sisai+a2/2.0))
c3=-K4*({Tspi+c2/2.0)-(Tswi+b2/2.0))
CALL Satenthalpy((Tswi+b3),spsatm,sisasw4)
a4=Kl*(sisasw4-(sisai+a3))
b4=K2*((Tspi+c3)-(Tswi+b3))-K3*(sisasw4-(sisai+a3))
c4=-K4*((Tsp1+c3)-(Tswi+b3))
C Determine the exit conditions of the element
sisao=sisai+(al+2.0*(a2+a3)+a4)/6.0
Tswo=Tswi+(bl+2.0*(b2+b3)+b4)/6.0
Tspo=Tspi+(cl+2.0*(c2+c3)+c4)/6.0
RETURN
END

c *******************************************************************
c* *
C* • MERKEL(2) METHOD TO EVAUALTE A SINGLE ELEMENT *
c* *
c **************************************************~****************
C Subroutine to apply the Runge-Kutta method of solution to the three
C Merkel equations which controls the state of a single element
C BACKTOFRONT FLOW CASE
SUBROUTINE MERKEL2(Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
 Stellenbosch University http://scholar.sun.ac.za )g

+ ReyC,shsfl,shsf2,nrow)
REAL L,musav,musw,kog,koga,Kl,K2,K3,K4
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspo,musw)
CALL Waterdensity(Tspo,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) Reynoldsnumber of process water
Tsa=Tsadb
CALL AirVapMixdensity(Tsa,swsail,spsatm,rhosav)
CALL AirVapMixviscosity(Tsa,swsail,spsatm,musav)
Reysa=ReyC*rhosav/musav ! Reynoldsnumber of airflow
CALL Waterviscosity(Tswi,musw)
Reysw=4.0*gamma/musw ! Reynoldsnumber of recirc.water
C Determine the neccessary transfer-coefficients
CALL Waterconductivity(Tspo,sksp)
CALL Prandtl(Tspo,Prasp)
shsw=4.186*118*{(gamma*3600/sdso)**(l.0/3.0))/3.6
IF (Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/(L*nrow)
shsp=(3.66+0.104*(term)/(1.0+0.016*(term)**(0.8)))*sksp/sdsi
ELSE
sfsd=(l.82*LOG10(Reysp)-1.64)**(-2.0)
terml=Prasp*(l.O+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*((sfsd/8.0)**(0.5))*(Prasp**(0.67)-l.O)
shsp=((sfsd/8.0)*(Reysp-1000.0)*terml/term2)*sksp/sdsi
END IF
koga=l.81E-4*((Reysa)**.9)*((Reysw)**.l5)*((sdso)**(-2.6))/3600.
kog=koga/sa ! Mass-transfer coefficient
Uo=l.O/((sdso/sdsi)*((l.O/shsp)+(l.O/shsfl))+(l.O/shsw)
+ +(l.O/shsf2)+sdso*LOG(sdso/sdsi)/(2.0*skst))
C Determine the controlling constants Kl,K2,K3 and K4
CALL Cpw(Tspo,scspp)
CALL Cpw(Tswi,scspw)
Kl=kog*dA/(smsael)
K2=Uo*dA/(smswel*scspw)
K3=kog*dA*lOOO.O/(smswel*scspw)
K4=Uo*dA/(smspel*scspp)
C Determine the Runge-Kutta coefficients
CALL Satenthalpy(Tswi,spsatm,sisaswl)
al=Kl*(sisaswl-sisai)
bl=K2*(Tspo-Tswi)-K3*(sisaswl-sisai)
cl=K4*(Tspo-Tswi)
CALL Satenthalpy((Tswi+bl/2.0),spsatm,sisasw2)
a2=Kl*(sisasw2-(sisai+al/2.0))
b2=K2*((Tspo+cl/2.0)-(Tswi+bl/2.0))-K3*(sisasw2-(sisai+al/2.0))
c2=K4*((Tspo+cl/2.0)-(Tswi+bl/2.0))
 Stellenbosch University http://scholar.sun.ac.za

CALL Satenthalpy{{Tswi+b2/2.0),spsatm,sisasw3)
a3=Kl*{sisasw3-{sisai+a2/2.0)).
b3=K2*{{Tspo+c2/2.0)-{Tswi+b2/2.0))-K3*{sisasw3-{sisai+a2/2.0))
c3=K4*{(Tspo+c2/2.0)-{Tswi+b2/2.0))
CALL Satenthalpy{{Tswi+b3),spsatm,sisasw4)
a4=Kl*(sisasw4-{sisai+a3))
b4=K2*({Tspo+c3)-{Tswi+b3))-K3*{sisasw4-{sisai+a3))
c4=K4*{{Tspo+c3)-{Tswi+b3))
C Determine the exit conditions of the element
sisao=sisai+{al+2.0*{a2+a3)+a4)/6.0
Tswo=Tswi+{bl+2.0*{b2+b3)+b4)/6.0
Tspi=Tspo+{cl+2.0*{c2+c3)+c4)/6.0
RETURN
END

c *******************************************************************
c* *
C* POPPE METHOD TO EVAUALTE A SINGLE ELEMENT *
c* *
c *******************************************************************
C Subroutine to apply the Runge-Kutta method of solution to the five
C Poppe equations which controls the state of a single element
SUBROUTINE POPPE (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,nrow)
REAL L,Lew,musav,musw,kog,koga,Kl,K2,K3,K4,KS,K6,
+ iil,ti2,ii3,ii4,iv
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspi,musw)
CALL Waterdensity(Tspi,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) ! Reynoldsnumber of process water
CALL AirVapMixdensity(Tsai,swsai,spsatm,rhosav)
CALL AirVapMixviscosity(Tsai,swsai,spsatm,musav)
Reysa=ReyC*rhosav/musav · ! Reynoldsnumber of airflow
CALL Waterviscosity{Tswi,musw)
gammal={smswi/smswel)*gamma
Reysw=4.0*gammal/musw ! Reynoldsnumber of recirc.water
C Determine the neccessary transfer-coefficients
CALL Waterconductivity(Tspi,sksp)
CALL Prandtl(Tspi,Prasp)
shsw=4.186*118.0*((gamma1*3600.0/sdso)**(l.0/3.0))/3.6
IF (Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/(L*nrow)
shsp=(3.66+0.104*(term)/(1.0+0.016*(term)**(0.8)))*sksp/sdsi
ELSE
 Stellenbosch University http://scholar.sun.ac.za

sfsd=(l.82*LOGIO(Reysp)-1.64)**(-2.0)
terml=Prasp*(l.O+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*((sfsd/8.0)**(0.5))*(Prasp**(0.67)-l.O)
shsp=((sfsd/8.0)*(Reysp-1000.0)*terml/term2)*sksp/sdsi
END IF
koga=l.81E-4*((Reysa)**.9)*((Reysw)**.l5}*((sdso)**(-2.6))/3600.
kog=koga/sa ! Mass-transfer coefficient
Uo=l.O/((sdso/sdsi)*((l.O/shsp)+(l.O/shsfl))+(l.O/shsw)
+ +(1. O/shsf2)+sdso*LOG( sdso/sds i )/ (2. O*skst))
C Determine the controlling constants Kl,K2,K3,K4,K5 and K6
- CALL Cpw(Tswi,scspw)
CALL Cpw(Tspi,scspp)
Kl=kog*dA
K2=kog*dA/smsael
K3=kog*dA/smswi
K4=Uo*dA/(smswi*scspw)
KS=kog*dA/(smswi*scspw)
K6=Uo*dA/(smspel*scspp)
CALL Cpv(Tswi,scspv)
iv=2501.6+scspv*Tswi/1000.0
C Determine the humidity of saturated air
CALL Airhumidity(Tsai,Tsai,spsatm,swsasa)
C Determine the Lewis factor
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
term=(0.622+swsasw)/(0.622+swsai)
Lew=(0.90854253)*((term-l.O)/(LOG(term)))
C Determine the Runge-Kutta coefficients
IF (swsasa.GE.swsai) THEN ! Air not saturated
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
CALL Satenthalpy(Tswi,spsatm,sisasw)
wwl=swsasw-swsai
iil=(sisasw-sisai)
TTl= Tspi-Tswi
al=-Kl*wwl
bl=K2*wwl
cl=K2*(Lew*iil-(Lew-l.O)*wwl*iv)
dl=K3*Tswi*wwl+K4*TT1-KS*(cl/K2)*1000.0
el=-K6*TTI
CALL Airhumidity((Tswi+dl/2.0),(Tswi+dl/2.0),spsatm,swsasw)
CALL Satenthalpy((Tswi+dl/2.0),spsatm,sisasw)
ww2=swsasw-(swsaiTbl/2.0)
ii2=(sisasw-(sisai+cl/2.0))
TT2=(Tspi+el/2.0)-(Tswi+dl/2.0)
a2=-Kl*ww2
b2=K2*ww2
c2=K2*(Lew*ii2-(Lew-l.O)*ww2*iv)
d2=K3*(Tswi+dl/2.0)*ww2+K4*TT2-KS*(c2/K2)*1000.0
 Stellenbosch University http://scholar.sun.ac.za
4{

e2=-K6*TI2
CALL Airhumidity((Tswi+d2/2.0),(Tswi+d2/2.0),spsatm,swsasw)
CALL Satenthalpy((Tswi+d2/2.0),spsatm,sisasw)
ww3=swsasw-(swsai+b2/2.0)
ii3=(sisasw-(sisai+c2/2.0))
TT3=(Tspi+e2/2.0)-(Tswi+d2/2.0)
a3=-Kl*ww3
b3=K2*ww3
c3=K2*(Lew*ii3-(Lew-l.O)*ww3*iv)
d3=K3*(Tswi+d2/2.0)*ww3+K4*TT3-KS*(c3/K2)*1000.0
e3=-K6*TT3
CALL Airhumidity((Tswi+d3),(Tswi+d3),spsatm,swsasw)
CALL Satenthalpy((Tswi+d3),spsatm,sisasw)
ww4=swsasw-(swsai+b3)
ii4=(sisasw-(sisai+c3))
TT4=(Tspi+e3)-(Tswi+d3)
a4=-Kl*ww4
b4=K2*ww4
c4=K2*(Lew*ii4-(Lew-l.O)*ww4*iv)
d4=K3*(Tswi+d3)*ww4+K4*TT4-KS*(c4/K2)*1000.0
e4=-K6*TT4 ·
ELSE IF (swsasa.LT.swsai) THEN Air saturated
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
CALL Satenthalpy(Tswi,spsatm,sisasw)
wwl=swsasw-swsasa
iil~(sisasw-sisai)
TTl= Tspi-Tswi
wwwl=swsai-swsasa
al=-Kl*wWl
bl=K2*wwl
cl=K2*((Lew*iil-(Lew-l.O)*wwl*iv)
+ +wwwl*Lew*scspw*Tswi/1000.0)
dl=K3*Tswi*wwl+K4*TT1-KS*(cl/K2)*1000.0
el=-K6*TT1 .
CALL Airhumidity((Tswi+dl/2.0),(Tswi+dl/2),spsatm,swsasw)
CALL Satenthalpy((Tswi+dl/2.0),spsatm,sisasw)
ww2=swsasw-swsasa
ii2=(sisasw-(sisai+cl/2.0))
TT2=(Tspi+el/2.0)-(Tswi+d1/2.0)
www2=(swsai+bl/2.0)-swsasa
a2=-Kl*ww2
b2=K2*ww2
c2=K2* ( ( Lew*i i 2- ( Lew-1. 0) *ww2*i v)
+ +www2*Lew*scspw*(Tswi+dl/2.0)/1000.0)
d2=K3*(Tswi+dl/2.0)*ww2+K4*TT2-KS*(c2/K2)*1000.0
e2=-K6*TT2
CALL Airhumidity((Tswi+d2/2.0),(Tswi+d2/2),spsatm,swsasw)
CALL Satenthalpy((Tswi+d2/2.0),spsatm,sisasw)
 Stellenbosch University http://scholar.sun.ac.za
42_

ww3=swsasw-swsasa
ii3=(sisasw-(sisai+c2/2.0))
TT3=(Tspi+e2/2.0}-(Tswi+d2/2.0)
www3=(swsai+b2/2.0)-swsasa
a3=-K1*ww3
b3=K2*ww3
c3=K2*((Lew*ii3-(Lew-1.0)*ww3*iv)
+ +www3*Lew*scspw*(Tswi+d2/2.0)/1000.0)
d3=K3*{Tswi+d2/2.0)*ww3+K4*TT3-KS*(c3/K2)*1000.0
e3=-K6*TT3
CALL Airhumidity((Tswi+d3),(Tswi+d3},spsatm,swsasw}
CALL Satenthalpy((Tswi+d3},spsatm,sisasw)
ww4=swsasw-swsasa ·
ii4={sisasw-(sisai+c3))
TT4=(Tspi+e3)-{Tswi+d3)
www4=(swsai+b3)-swsasa
a4=-K1*ww4
b4=K2*ww4
c4=K2*{(Lew*ii4-(Lew-1.0)*ww4*iv)
+ +www4*Lew*scspw*(Tswi+d3)/1000.0)
d4=K3*(Tswi+d3)*ww4+K4*TT4-KS*(c4/K2}*1000.0
e4=-K6*TT4
END IF
smswo=smswi+(a1+2.0*{a2+a3}+a4)/6.0
swsao=swsai+(b1+2.0*(bZ+b3)+b4)/6.0
sisao=sisai+(c1+2.0*(c2+c3)+c4}/6.0
Tswo=Tswi+{d1+2.0*(d2+d3)+d4)/6.0
Tspo=Tspi+(e1+2.0*(e2+e3)+e4}/6.0
C Determine the air outlet temperature and saturation enthalpy
TR=Tspi
TL=O.O
10 Tsao=(TR+TL)/2.0
CALL Cpv(Tsao,scspv)
CALL Cpa(Tsao,scspa)
CALL Cpw{Tsao,scspw}
CALL Airhumidity(Tsao,Tsao,spsatm,swsasa}
IF (swsasa.GT.swsao} THEN
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0}
ELSE
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0)
+ +scspw*(swsao-swsasa)*Tsao/1000.0
END IF
IF ((ABS(sisao-sisasa)}.GT.0.1} THEN
IF (sisao.LT.sisasa} THEN
TR=Tsao
ELSE
TL=Tsao
END IF
GO TO 10
 Stellenbosch University http://scholar.sun.ac.za
4J

END IF
RETURN
END

c *******************************************************************
c* *
C* POPPE METHOD(2) TO EVAUALTE A SINGLE ELEMENT *
c* *
c *******************************************************************
C Subroutine to apply the Runge-Kutta method of solution to the five
C Poppe equations which controls the state of a single element
C BACKTOFRONT FLOW CASE
SUBROUTINE POPPE2 (Tspi,Tswi,Tsai,sisai~swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,nrow)
REAL L, Lew, musav, musw, kog, koga, Kl ,.K2, K3, K4, KS, K6,
+ iil,ii2,ii3,ii4,iv
C Determine the neccessary Reynoldsnumbers
CALL Waterviscosity(Tspo,musw)
CALL Waterdensity(Tspo,rhosw)
Reysp=rhosw*sdsi*svsp/(musw) ! Reynoldsnumber of process water
CALL AtrVapMixdensity(Tsai,swsai,spsatm,rhosav)
CALL AirVapMixviscosity(Tsai,swsai;spsatm,musav)
Reysa=ReyC*rhosav/musav ! Reynoldsnumber of. airflow
CALL Waterviscosity(Tswi,musw) •
gammal=(smswi/smswel)*gamma
Reysw=4.0*gammal/musw ! Reynoldsnumber of recirc.water
C Determine the neccessary transfer-coefficients
CALL Waterconductivity(Tspo,sksp)
CALL Prandtl(Tspo,Prasp)
shsw=4.186*118.0*((gammal*3600.0/sdso)**(l.0/3.0))/3.6
IF (Reysp.LT.2300.0) THEN
term=Reysp*Prasp*sdsi/(L*nrow)
shsp=(3.66+0.104*(term)/(1.0+0.016*(term)**(0.8)))*sksp/sdsi
ELSE
sfsd=(l.82*LOGIO(Reysp)-1.64)**(-2.0)
terml=Prasp*(l.O+(sdsi/(L*nrow))**(0.67))
term2=1.0+12.7*((sfsd/8.0)**(0.5))*(Prasp**(0.67)-l.O)
shsp=((sfsd/8.0)*(Reysp-IOOO.O)*terml/term2)*sksp/sdsi
END IF
koga=l.81E-4*((Reysa)**.9)*((Reysw)**.IS)*((sd~o)**(-2.6))/3600.
kog=koga/sa ! Mass-transfer coefficient
Uo=l.O/((sdso/sdsi)*((l.O/shsp)+(l.O/shsfl))+(l.O/shsw)
+ +(l.O/shsf2)+sdso*LOG(sdso/sdsi)/(2.0*skst))
C Determine the controlling constants Kl,K2,K3,K4,KS and K6
CALL Cpw(Tswi,scspw)
 Stellenbosch University http://scholar.sun.ac.za

CALL Cpw(Tspo,scspp)
Kl=kog*dA
K2=kog*dA/smsael
K3=kog*dA/smswi
K4=Uo*dA/(smswi*scspw)
KS=kog*dA/(smswi*scspw)
K6=Uo*dA/(smspel*scspp)
CALL Cpv(Tswi,scspv)
iv=2501.6+scspv*Tswi/1000.0
C Determine the humidity of saturated air
CALL Airhumidity(Tsai,Tsai,spsatm,swsasa)
C Determine the lewis factor
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
term=(0.622+swsasw)/(0.622+swsai)
lew=(0.90854253)*((term-l.O)/(LOG(term)))
C Determine the Runge-Kutta coefficients
IF (swsasa.GE.swsai) THEN ! Air not saturated
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
CALL Satenthalpy{Tswi,spsatm,sisasw)
wwl=swsasw-swsai
iil={sisasw-sisai)
TTl= Tspo-Tswi
al=-Kl*wwl
bl=K2*wwl
cl=K2*(lew*iil-(lew-l.O)*wwl*iv)
dl=K3*Tswi*wwl+K4*TT1-KS*(cl/K2)*1000.0
el=K6*TTI
CALL Airhumidity((Tswi+dl/2.0),(Tswi+dl/2.0),spsatm,swsasw)
CALL Satenthalpy((Tswi+dl/2.0),spsatm,sisasw)
ww2=swsasw-(swsai+bl/2.0)
ii2=(sisasw-(sisai+cl/2.0))
TT2=(Tspo+el/2.0)-(Tswi+dl/2.0)
a2=-Kl*ww2
b2=K2*ww2
c2=K2*(lew*ii2-(lew-l.O)*ww2*iv)
d2=K3*(Tswi+dl./2.0)*ww2+K4*TT2-KS*(c2/K2)*1000.0
e2=K6*TT2
CALL Airhumidity({Tswi+d2/2.0),{Tswi+d2/2.0),spsatm,swsasw)
CALL Satenthalpy({Tswi+d2/2.0),spsatm,sisasw)
ww3=swsasw-{swsai+b2/2.0)
ii3={sisasw-{sisai+c2/2.0))
TT3=(Tspo+e2/2.0)-(Tswi+d2/2.0)
a3=-Kl*ww3
b3=K2*ww3
c3=K2*(lew*ii3-(lew-l.O)*ww3*iv)
d3=K3*(Tswi+d2/2.0)*ww3+K4*TT3-KS*(c3/K2)*1000.0
e3=K6*TT3
 Stellenbosch University http://scholar.sun.ac.za

CALL Airhumidity((Tswi+d3),(Tswi+d3),spsatm,swsasw)
CALL Satenthalpy((Tswi+d3),spsatm,sisasw) ·
ww4=swsasw-(swsai+b3)
ii4=(sisasw-(sisai+c3))
TT4=(Tspo+e3)-(Tswi+d3)
a4=-Kl*ww4
b4=K2*ww4
c4=K2'k( Lew*i i 4- (Lew-1. O)*ww4*i v)
d4=K3*(Tswi+d3)*ww4+K4*TT4-KS*(c4/K2)*1000.0
e4=K6*TT4
ELSE IF (swsasa.LT.swsai) THEN Air saturated
CALL Airhumidity(Tswi,Tswi,spsatm,swsasw)
CALL Satenthalpy(Tswi,spsatm,sisasw)
wwl=swsasw-swsasa
iil=(sisasw-sisai)
TTl=Tspo-Tswi
wwwl=swsai-swsasa
al=-Kl*wwl
bl=K2*wwl
cl=K2*({Lew*iil-(Lew-l.O)*wwl*iv)
+ +wwwl*Lew*scspw*Tswi/1000.0)
dl=K3*Tswi*wwl+K4*TT1-KS*(cl/K2)*1000.0
el=K6*TT1
CALL Airhumidity((Tswi+dl/2.0),(Tswi+dl/2),spsatm,swsasw)
CALL Satenthalpy((Tswi+dl/2.0),spsatm,sisasw)
ww2=swsasw-swsasa
ii2=(sisasw-(sisai+cl/2.0))
TT2=(Tspo+el/2.0)-(Tswi+dl/2.0)
www2=(swsai+bl/2.0)-swsasa
a2=-Kl*ww2
b2=K2*ww2
c2=K2*((Lew*ii2-(Lew-l.O)*ww2*iv)
+ +www2*Lew*scspw*(Tswi+dl/2.0)/1000.0)
d2=K3*(Tswi+dl/2.0)*ww2+K4*TT2-KS*(c2/K2)*1000.0
e2=K6*TT2
CALL Airhumidity((Tswi+d2/2.0),(Tswi+d2/2),spsatm,swsasw)
CALL Satenthalpy((Tswi+d2/2.0),spsatm,sisasw)
ww3=swsasw-swsasa
ii3=(sisasw-(sisai+c2/2.0))
TT3=(Tspo+e2/2.0)-(Tswi+d2/2.0)
www3=(swsai+b2/2.0)-swsasa
a3=-Kl*ww3
b3=K2*ww3
c3=K2*((Lew*ii3-(Lew-l.O)*ww3*iv)
+ +www3*Lew*scspw*(Tswi+d2/2.0)/1000.0)
d3=K3*(Tswi+d2/2.0)*ww3+K4*TT3-KS*(c3/K2)*1000.0
e3=K6*TT3
CALL Airhumidity((Tswi+d3),(Tswi+d3),spsatm,swsasw)
 Stellenbosch University http://scholar.sun.ac.za

CALL Satenthalpy((Tswi+d3),spsatm,sisasw)
ww4=swsasw-swsasa
ii4=(sisasw-(sisai+c3})
TT4=(Tspo+e3)-(Tswi+d3)
www4=(swsai+b3)-swsasa
a4=-K1*ww4
b4=K2*ww4
c4=K2*((Lew*ii4-(Lew-1.0)*ww4*iv)
+ +www4*Lew*scspw*(Tswi+d3)/1000.0)
d4=K3*(Tswi+d3)*ww4+K4*TT4-KS*(c4/K2)*1000.0
e4=K6*TT4
END IF
·smswo=smswi+(a1+2.0*(a2+a3)+a4)/6.0
swsao=swsai+(b1+2.0*(b2+b3)+b4)/6.0
sisao=sisai+(c1+2.0*(c2+c3)+c4)/6.0
Tswo=Tswi+(d1+2.0*(d2+d3)+d4)/6.0
Tspi=Tspo+(e1+2.0*(e2+e3)+e4)/6.0
C Determine the air outlet temperature and saturation enthalpy
TR=Tspi
TL=O.O
10 Tsao=(TR+TL)/2.0
CALL Cpv(Tsao,scspv)
CALL Cpa(Tsao,scspa)
CALL Cpw(Tsao,scspw)
CALL Airhumidity(Tsao,Tsao,spsatm,swsasa)
IF (swsasa.GT.swsao) THEN
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0)
ELSE
sisasa=scspa*Tsao/1000.0+swsasa*(2501.6+scspv*Tsao/1000.0)
+ +scspw*(swsao-swsasa)*Tsao/1000.0
END IF
IF ((ABS(sisao-sisasa)).GT.0.1) THEN
IF (sisao.LT.sisasa) THEN
TR=Tsao
ELSE
TL=Tsao
END IF
GO TO 10
END IF
RETURN
END

c *******************************************************************
c* *
C* SUBROUTINE TO PRINT RESULTS OF CROSS.FOR *
c* *
c *******************************************************************
C Subroutine to print the results of the cooler calculations
SUBROUTINE PRINT RESULTS(Tspi1,Tspo1,smsp,sdsi,sdso,vspas,hspas,
+ Kmax,Imax,Mmax,gamma,Vstot,rhosail,Vseff2,sisail,
 Stellenbosch University http://scholar.sun.ac.za

CALL Satenthalpy((Jswi+d3),spsatm,sisasw)
ww4=swsasw-swsasa
ii4=(sisasw-(sisai+c3))
TT4=(Tspo+e3)-{Tswi+d3)
www4=(swsai+b3)-swsasa
a4=-Kl*ww4
b4=K2*ww4
c4=K2*((Lew*ii4-(Lew-l.O)*ww4*iv)
+ +www4*Lew*scspw*{Tswi+d3)/1000.0)
d4=K3*(Tswi+d3)*ww4+K4*TT4-KS*{c4/K2)*1000.0
e4=K6*TT4
END IF
smswo=smswi+(a1+2.0*{a2+a3)+a4)/6.0
swsao=swsai+(b1+2.0*(b2+b3)+b4)/6.0
sisao=sisai+(c1+2.0*(c2+c3)+c4)/6.0
Tswo=Tswi+(d1+2.0*{d2+d3)+d4)/6.0
Tspi=Tspo+(e1+2.0*(e2+e3)+e4)/6.0
C Determine the air outlet temperature and saturation enthalpy
TR=Tspi
TL=O.O
10 Tsao={TR+TL)/2.0
CALL Cpv(Tsao,scspv)
CALL Cpa{Tsao,scspa)
CALL Cpw(Tsao,scspw)
CALL Airhumidity{Tsao,Tsao,spsatm,swsasa)
IF (swsasa.GT.swsao) THEN
sisasa=scspa*Tsao/1000.0+swsasa*{2501.6+scspv*Tsao/1000.0)
ELSE
sisasa=scspa*Tsao/1000.0+swsasa*{2501~6+scspv*Tsao/1000.0)
+ +scspw*{swsao-swsasa)*Tsao/1000.0
END IF
IF ((ABS(sisao-sisasa)).GT.0.1) THEN
IF (sisao.LT.sisasa) THEN
TR=Tsao
ELSE
TL=Tsao
END IF
GO TO 10
END IF
RETURN
END

c *******************************************************************
c* *
C* SUBROUTINE TO PRINT RESULTS OF CROSS.FOR *
c* *
c *******************************************************************
C Subroutine to print the results of the cooler calculations
SUBROUTINE PRINT_RESULTS(Tspi1,Tspol,smsp,sdsi,sdso,vspas,hspas,
+ Kmax,Lmax,Mmax,gamma,Vstot,rhosail,Vseff2,sisail,
 Stellenbosch University http://scholar.sun.ac.za

+ sisao1,Tswi1,Tswo1,svsp,flowlayout,H,L,spsatm,PI,Tsawb,
+ Tsadb,svsa,swsai1,swsao1,smswi1,smswo1,model,Tsao),
+ rhosao1,phio,Power,shsfl,shsf2,skst,smsa)
C Initialize variable types
REAL L
INTEGER flowlayout
C Print the results on the screen or in file CROSS.RES
C Print cooler layout and dimensions
CALL LIB$ERASE PAGE{1,1)
WRITE(*, 10) -
WRITE(!, 10)
10 FORMAT(
+ ' CROSSFLOW EVAPORATIVE COOLER SIMULATION'/
+ '******************************************')
C Print. process water flow layout
IF (flowlayout.EQ~1) THEN
WRITE{*,*)'Process water flow layout FRONT TO BACK'
WRITE{1,*)'Process water flow layout FRONT TO BACK'
ELSE IF (flowlayout.EQ.2) THEN
WRITE{*,*)'Process water flow layout BACK TO FRONT'
WRITE{1,*)'Process water flow layout BACK TO FRONT'
ELSE IF (flowlayout.EQ.3) THEN
WRITE{*,*)'Process water flow layout TOP TO BOTTOM'
WRITE{1,*)'Process water flow layout TOP TO BOTTOM'
ELSE IF (flowlayout.EQ.4) THEN
WRITE{*,*)'Process water flow layout STRAIGHT THROUGH'
WRITE{1,*)'Process water flow layout STRAIGHT THROUGH'
END IF
IF (model.EQ.1) THEN
WRITE{l,*)'Analytical model .. MERKEL I

WRITE(*,*)'Analytical model MERKEL'

ELSE IF (model.EQ.2) THEN
WRITE{l,*)'Analytical model IMPROVED MERKEL'
WRITE{*,*)'Analytical model IMPROVED MERKEL'
ELSE
WRITE{l,*)'Analytical model POPPE'
WRITE{*,*)'Analytical model POPPE'
END IF
WRITE{l,ll)sdso*lOOO.O,sdsi*lOOO.O,vspas*lOOO.O,hspas*lOOO.O,
+ H,L,Kmax,Lmax,Mmax,shsfl,shsf2,skst
11 FORMAT{/
+ ' Pipe outer diameter·······!············· = ',F6.2,' mm'/
+ ' Pipe inner diameter ..................... = ',F6.2,' mm'/
+ ' Vertical spacing between pipes .......... = ,F6.2,' mm'/ I

+ ' Horizontal spacing between pipes ........ = ',F6.2,' mm'/

+ ' Height of cooler unit ................... = ',F6.2,' m'/
· + ' Width of cooler unit ................... . ',F6.2,' m'/
+ ' Number of rows of pipes across airstream = ,13/ I

+ ' Number of pipes facing the airstream .... ,13/ I

+ ' Number of elements along a single pipe .. = ,13/ I

 Stellenbosch University http://scholar.sun.ac.za

+1 Fouling coefficient (inside) ............ = 1 ,Fl2.2, 1 WlmA2 K1 1

· t (out s1'd e ) ........... = 1 ,Fl2.2, 1 WlmA2 K1 1
+1 Fou 1·1ng coe ff.1c1en
+ Pipe wall conductivity .................. = 1 ,F8.2, 1 Wlm K1 )
1

C Print ambient conditions and results in result file

IF (model.EQ.l) THEN .
WRITE(l,20)spsatmllOOO.O,Tsadb,Tsawb,rhosail,smsa,
+ smsa*(l.O+swsail),svsa,sisail,sisaol,swsail
WRITE(l,30)swsaol,Tsaol,rhosaol,gamma*3600.0,smswil,
+ smswil-smswol,Tswil,Tswol
ELSE IF (model.EQ.2) THEN
WRITE(l,20)spsatmllOOO.O,Tsadb,Tsawb,rhosail,smsa,
+ smsa*(l.O+swsail),svsa,sisail,sisaol,swsail
WRITE(l,40)swsaol,phio,Tsaol,rhosaol,gamma*3600.0,smswil,
+ smswol,smswil-smswol,Tswil,Tswol
ELSE
WRITE(l,20)spsatmllOOO.O,Tsadb,Tsawb,rhosail,smsa~
+ smsa*(l.O+swsail),svsa,sisail,sisaol,swsail
WRITE(l,40)swsaol,phio,Tsaol,rhosaol,gamma*3600.0,smswil,
+ smswol,smswil-smswol~Tswil,Tswol
END IF
WRITE(l,SO)smsp,svsp,Tspil,Tspol,Power
c Print the results on the screen
WRITE(*,l2)H,L,Lmax,Kmax,spsatmll000.0,svsa,
+ smsa,sisail,sisaol,smswil,smswil-smswol,
+ Tswil,Tswol,smsp,Tspil,Tspol,Power
12 FORMAT(
+ Height of cooler unit ................... = ,F8.3, mil
I
I I

+ Length of cooler tubes .................. = ,F8.3, mil

I I I

+ Number of pipe rows along cooler height . = ,I31

I I

+ Number of pipes along the airflow ....... = ,I31

I I

+ Atmospheric pressure ..................... = ,F8.3, kPa 1

I I I 1

+ I Air velocity through cooler.............. = ,F8.3, mls I I I 1

+ Dry air massflow through cooler ......... = ,F8.3, kgls I

I I I 1

+ Air enthalpy in ......................... = ,F8.3, kJikg I

1 I I 1

+ Air enthalpy out (incl. mist) ........... = ,F8.3, kJikg I

1 I I 1

+ Inlet recirculating water massflow ...... = ,F8.3, kgls I

I I I 1

+ Recirculating water evaporated ......... = ,F8.3, kgls I

I I I 1

+ Recirculating water temperature (in) ....

I
,F8.3, Cll I I

+ Recirculating water temperature (out) ... = ,F8.3, Cll

I I I

+ Process fluid massflow through cooler ...

I
,F8.3, kgls I I I 1

+ Process fluid temperature in ............ = ,F8.3, Cll

I I I

+ Process fluid temperature out ........... = ,F8.3, Cll

I I I

+ Total capacity of cooler unit ........... = ,F8.3, kW' I)

I I I

20 FORMAT( ,I I I

+ Atmospheric pressure ...................

1
= I
,F8.3, I
kPa 1
1

+ Inlet air temperature (dry bulb) .......

I
= I
,F8.3, I
Cll
+ Inlet air temperature (wet bulb) .......
I
= I
,F8.3, I
Cll
+ Inlet air density ......................
I
= I
,F8.3, I
kglmA3 I 1

+ Dry air massflow through cooler ........

I
= I
,F8.3, I
kgls I 1
 Stellenbosch University http://scholar.sun.ac.za

+ 1
Inlet air massflow (inc vapour) •....... = ,F8.3,
1
kgJ
1

+ 1
Air velocity through cooler .•.......... = ,F8.3,
1
m/5
1

+ 1
Air enthalpy in ••.••..•..•.•..........• = ,F8.3, kJ/
1 1

+ 1
Air enthalpy out (incl. mist) ......•... = ,F8.3, kJ; _ .
1 1

+ 1
I n1et a1·r hu m1'd't
1 y rat·10 ..•.......•.... = ,Fl2.7, kg/kg')
1 1

30 FORMAT(
+ Outlet air humidity ratio (saturated) ..
1
= ',Fl2.7,' kg/kg'/
, + ' Outlet air temperature (saturated) •.... = ',F8.3, C'/
1

+ 0utl et a1r
1
• dens1'ty ( sa t ura t ed) .•.•..... = ,F8.J, kg/m"J'//
1 1

+ Recirc.water massflow I length .....•...

1
= ,F9.4, kg/m hr'/
1 1

+ ' I n1et rec1rc.wa

. t er rna s s fl ow ............ = ',F9.4,' kg/s'/
+' Recirc. water lost through evaporation . = ',F9.4,' kg/s'/
+ ' Recirculating water temperature in ..... = ',F8.3,' C'/
+' Recirculating water temperature out .... = ',F8.3,' C')
40 FORMAT(
+' Outlet air humidity ratio (incl. mist) . = ',Fl2.7,' kg/kg'/
+' Outlet air relative humidity ........... = ',Fl2.7/
+' Outlet air temperature (dry bulb) ...... = ',F8.3,' C'/
+' Outlet air density .................... = ',F8.3,' kg/m"3'//
+' Recirc.water massflow /.length ......... = ',F9.4,' kg/m hr'/
+ ' Inlet recirc.water massflow ............ - ',F9.4,' kg/s'/
+ Outlet recirc. water massflow .......... = ',F9.4,' kg/s'/
+' Recirc. water 1ost through evaporation . = ',F9.4,' kg/s'/
+' Recirculating water temperature 1n ..... = ',F8.3,' C'/
+' Recirculating water temperature ~ut .... = ',F8.3,' C')
50 FORMAT(' ',/
+' Process water massflow through cooler ... = ',F8.3,' kg/s'/
+' Process water flow velocity in pipes ... = ',F8.3,' m/s'/
+' Process water temperature in ........... = ',F8.3,' C'/
+ ' Process water temperature out .......... = ',F8.3,' C'/
+ ' Capacity of cooler unit ................ = ',F8.3,' kW' /)
RETURN
END

c *******************************************************************
c* *
c* SINGLE STRAIGHT THROUGH PASS *
c* *
c *******************************************************************
C Subroutine to evaluate a cooler layout where the process fluid flows
C straight through the cooler in one pass
SUBROUTINE STRAIGHT (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,smspel,
+ smswel,sisail,sisaol,Tspil,Tspol,Tswil,Tswol,swsail,
+ swsaol,smswil,smswol,L,H,sdsi,sdso,dA,Tsadb,Tsawb,
+ spsatm,gamma,Vstot,sa,skst,svsp,Aspi,Aspo,ReyC,
+ gradfile,Kmax,Lmax,Mmax,PI,model,Tsaol,shsfl,shsf2)
 Stellenbosch University http://scholar.sun.ac.za

DIMENSION Tsp(40,400,10)
DIMENSION Tsw(40,400,10)
DIMENSION Tsa(40,400,10)
DIMENSION sisa(40,400,10)
DIMENSION swsa(40,400,10)
DIMENSION smsw(40,400,10)
REAL L
INTEGER flag,flag2,gradfile
C Initialize the three arrays with the known temperature and enthalpy values
CALL Enthalpy(Tsadb,Tsawb,spsatm,sisai1)
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsai1)
DO 10 j=1,Lmax
DO 20 k=1,Mmax
sisa(1,j,k+1)=sisai1
swsa(1,j,k+1)=swsai1
Tsa(1,j,k+1)=Tsadb
20 CONTINUE
10 CONTINUE
DO 30 i=1,Kmax
DO 40 k=1 ,Mmax
Tsw(i,1,k+l)=Tswil
smsw(i,l,k+l)=smswel
40 CONTINUE
30 CONTINUE
DO 50 i=l,Kmax
DO 60 j=1,Lmax
Tsp(i,j,l+l)=Tspil
60 CONTINUE
50 CONTINUE
C Start of the outer loop to evaluate each i-level of the model
DO 70 i=l,Kmax
flag2=i-2*INT(i/2.0)
C Flag2=1 in the first row,O in the second row etc.
C Start of the middle loop to evaluate each j-level of the model
DO 80 j=l,Lmax
C Start of the inner loop to evaluate each each element of the model
DO 90 k=2,Mmax+l
C Determine the input values for a given element
Tspi=Tsp(i,j,k)
Tswi=Tsw(i,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
 Stellenbosch University http://scholar.sun.ac.za

C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.l).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-l,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE .
sisai=(sisa(i,j,k)+sisa(i,j-l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j-l,k))/2.0
END IF
END IF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) THEN
CALL MERKEL (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfllshsf2,1)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,1)
ELSE
CALL POPPE (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,1)
END IF
C Determine the exit values for a given element
Tsp(i,j,k+l)=Tspo
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)=sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
 Stellenbosch University http://scholar.sun.ac.za

IF (gradfile.EQ.1) THEN
WRITE(4, *) i ,j, k
WRITE(4,*}Tspi,Tspo
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.1) THEN
WRITE(4,*}swsai,swsao
WRITE(4,*)smswi,smswo
IF (model.EQ.3) THEN
WRITE(4,*)Tswi,Tswo
END IF
END IF
END IF
90 CONTINUE
80 CONTINUE
70 CONTINUE
C Determine the average exit temperature of recirculating water
sum1=0.0
sum2=0.0
DO 100 i=1,Kmax
DO 110 k=1 ,Mmax
CALL Cpw(Tsw(i,Lmax+1,k+1},scspw)
sum1=sum1+Tsw(i,Lmax+1,k+1)*scspw
sum2=sum2+smsw(i,Lmax+1,k+1)
110 CONTINUE
100 CONTINUE
CALL Cpw(Tswi1,scspw)
Tswo1=sum1/(Mmax*Kmax*scspw)
smswo1=sum2
C Determine the average exit temperature of process water
sum1=0.0
sum2=0.0
DO 120 i=1,Kmax
DO 130 j=1,Lmax
CALL Cpw(Tsp(i,j,Mmax+1+1),scspp)
sum1=sum1+Tsp(i,j,Mmax+1+1)*scspp
sum2=sum2+Tsp(i,j,Mmax+1+1)
130 CONTINUE
120 CONTINUE
CALL Cpw((sum2/(Kmax*Lmax)),scspp)
Tspo1=sum1/(Kmax*Lmax*scspp)
C Determine the average exit enthalpy of the air
sum1=0.0
sum2=0.0
sum3=0.0
DO 140 j=1,Lmax
DO 150 k=1,Mmax
sum1=sum1+sisa(Kmax+1,j,k+1)
sum2=sum2+swsa(Kmax+1,j,k+1)
 Stellenbosch University http://scholar.sun.ac.za

sum3=sum3+Tsa(Kmax+l,j,k+l)
ISO CONTINUE
140 CONTINUE
DO 160 k=l,Mmax
IF (flag2.EQ.O) THEN
suml=suml+sisa(Kmax,l,k+l)/2.0
sum2=sum2+swsa(Kmax,l,k+l)/2.0
sum3=sum3+Tsa(Kmax,l,k+l)/2.0
ELSE
suml=suml+sisa(Kmax,Lmax,k+l)/2.0
sum2=sum2+swsa(Kmax,Lmax,k+l)/2.0
sum3=sum3+Tsa(Kmax,Lmax,k+l)/2.0
END IF
160 CONTINUE
sisaol=suml/(Mmax*(Lmax+.S))
swsaol=sum2/(Mmax*(Lmax+.5))
Tsaol=sum3/(Mmax*(Lmax+.5))
C Print the recirc.water inlet and outlet temperatures on the screen
WRITE(*,l70)Tswil,Tswol
· 170 FORMAT(' ','Tw(in) = ',F7.3,' Tw(out) = ',F7.3)
RETURN
END

c *******************************************************************
c* *
C* TOP TO BOTTOM PROCESS WATER FLOW *
c* *
c *******************************************************************
C Subroutine to evaluate a cooler layout where the process fluid flows
C downwards in a direction perpendicular to the direction of the airstream
SUBROUTINE TOPTOBOTTOM (Tsp,Tsw,Tsa,sisa,swsa,smsw,smsael,
+ smspel,smswel,sisail,sisaol,Tspil,Tspol,Tswil,
+ Tswol,swsail,swsaol,smswil,smswol,L,H,sdsi,sdso,
+ dA,Tsadb,Tsawb,spsatm,gamma,Vstot,sa,skst,
+ svsp,Aspi,Aspo,ReyC,gradfile,Kmax,Lmax,Mmax,PI,
+ model,Tsaol,shsfl,shsf2,gradplot)
DIMENSION Tsp(40,400,10)
DIMENSION Tsw(40,400,10)
DIMENSION Tsa(40,400,10)
DIMENSION sisa(40,400,10)
DIMENSION swsa(40,400,10)
DIMENSION smsw(40,400,10)
REAL L
INTEGER flag,flag2,gradfile,gradplot
C Initialize the arrays with the known values
CALL Enthalpy(Tsadb,Tsawb,spsatm,sisail)
CALL Airhumidity(Tsadb,Tsawb,spsatm,swsail)
 Stellenbosch University http://scholar.sun.ac.za

DO 20 j=1,Lmax
DO 10 k=2,Mmax+1
sisa(1,j,k)=sisai1
swsa(1,j,k)=swsai1
Tsa(1,j,k)=Tsadb
10 CONTINUE
20 CONTINUE
DO 40 i=1,Kmax
00 30 k=2,Mmax+1
Tsw(i,1,k)=Tswi1
smsw(i,1,k)=smswel
30 CONTINUE
40 CONTINUE
DO 50 i=1,Kmax
Tsp(i,1,2)=Tspi1
50 CONTINUE
C Start of the outer loop to evaluate each i-level of the model
DO 60 i=1,Kmax
flag=O
flag2=i-2*INT(i/2.0)
C Flag2=1 in the first row,O in the second row etc~
C Start of the middle loop to evaluate each j-level of the model
· DO 70 j=1,Lmax ·
C N.B. flag=1 for backward process fluid flow
C L.W. flag=O for forward process fluid flow
IF (flag.EQ.O) THEN
C Start of the inner loop to evaluate each each element of the model
C Process water flow is in a forward direction
DO 80 k=2,Mmax+1
C Determine the inlet values for a given element
Tspi=Tsp(i,j,k)
IF((k.EQ.2).AND.(j.NE.1)) Tspi=Tsp(i,j-l,k-1)
Tswi=Tsw(i ,j,k)
Tsai=Tsa(i,j,k)
sisai=sisa(i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.l).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-l,j,k)+sisa(i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE
 Stellenbosch University http://scholar.sun.ac.za

sisai=(sisa(i,j,k)+sisa(i,j-l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j-l,k))/2.0
END IF
END IF
IF (flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa(i,j,k)+sisa(i-l,j,k))/2.0
swsai=(swsa(i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa(i,j+l,k))/2.0
swsai=tswsa(i,j,k)+swsa(i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
IF (model.EQ.l) THEN
CALL MERKEL (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst;svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Lmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,stsao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,Lmax)
ELSE
CALL POPPE (Tspi,Tswi,Tsa,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ . smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,Lmax)
END IF
C Determine the exit values for a given element
Tsp(i,j,k+l)=Tspo
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)=sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.l) THEN
WRITE(4,*)i,j,k-1
WRITE(4,*)Tspi,Tspo
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.l) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
 Stellenbosch University http://scholar.sun.ac.za

IF (model.EQ.3) THEN
WRITE(4,*)Tswi,Tswo
END IF
END IF
ELSE IF (gradplot.EQ.l) THEN
IF (((i.EQ.l).OR.(i.EQ.lO)).AND.(k.EQ.2)) THEN
WRITE(5,3)i,j,k-l,Tspi,Tspo,Tswi,Tswo,sisai,sisao
3 FORMAT(3I4,6F9.3)
END IF
END IF
80 CONTINUE
flag=l
ELSE IF (flag.EQ.l) THEN
C Start of the inner loop to evaluate each each element of the model
C Process water flow is backwards to the origin
DO 90 k=Mmax+l,2,-1
C Determine the inlet values for a given element
Tspi=Tsp(i,j,k)
IF (k.EQ.{Mmax+l)) Tspi=Tsp{i,j-l,k+l}
Tswi=Tsw{i,j,k)
Tsai=Tsa{i,j,k)
sisai=sisa{i,j,k)
swsai=swsa(i,j,k)
smswi=smsw(i,j,k)
C Determine the enthalpy of air entering each element in the packed formation
IF ((flag2.EQ.l).AND.(i.NE.l)) THEN
IF (j.EQ.l) THEN
sisai=(sisa(i-l,j,k)+sisa{i,j,k))/2.0
swsai=(swsa(i-l,j,k)+swsa(i,j,k))/2.0
Tsai=(Tsa(i-l,j,k)+Tsa(i,j,k))/2.0
ELSE
sisai={sisa{i,j,k)+sisa{i,j-l,k))/2.0
swsai=(swsa(i,j,k)+swsa(i,j-l,k))/2.0
Tsai=(Tsa{i,j,k)+Tsa{i,j-l,k))/2.0
END IF
END IF
IF {flag2.EQ.O) THEN
IF (j.EQ.Lmax) THEN
sisai=(sisa{i;j,k)+sisa{i-l,j,k))/2.0
swsai=(swsa{i,j,k)+swsa(i-l,j,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i-l,j,k))/2.0
ELSE
sisai=(sisa(i,j,k)+sisa{i,j+l,k))/2.0
swsai=(swsa(i,j,k)+swsa{i,j+l,k))/2.0
Tsai=(Tsa(i,j,k)+Tsa(i,j+l,k))/2.0
END IF
END IF
C Call subroutine to determine outlet conditions of each element
 Stellenbosch University http://scholar.sun.ac.za

IF (model.EQ.l) THEN
CALL MERKEL (Tspi,Tswi,sisai,swsail,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,Aspi,Aspo,
+ ReyC,shsfl,shsf2,Lmax)
ELSE IF (model.EQ.2) THEN
CALL IMPMERKEL (Tspi,Tswi,sisai,swsai,L,H,sdsi,sdso,dA,
+ Tsadb,spsatm,gamma,Vstot,smsael,smspel,smswel,
+ sa,skst,svsp,Tspo,Tswo,sisao,swsao,Aspi,Aspo,
+ ReyC,smswi,smswo,Tsai,Tsao,shsfl,shsf2,Lmax)
ELSE
CALL POPPE (Tspi,Tswi,Tsai,sisai,swsai,smswi,L,H,sdsi,
+ sdso,dA,Tsadb,spsatm,gamma,Vstot,smsael,smspel,
+ smswel,sa,skst,svsp,Tspo,Tswo,Tsao,sisao,swsao,
+ smswo,Aspi,Aspo,ReyC,shsfl,shsf2,Lmax)
END IF
C Determine the exit values for a given element
Tsp(i,j,k-l)=Tspo
Tsw(i,j+l,k)=Tswo
Tsa(i+l,j,k)=Tsao
sisa(i+l,j,k)=sisao
swsa(i+l,j,k)=swsao
smsw(i,j+l,k)=smswo
C Write the temperature and enthalpy gradients to file CROSS.GRA
IF (gradfile.EQ.l) THEN
WRITE(4,*)i,j,k-1
WRITE(4,*)Tspi,Tspo
WRITE(4,*)Tswi,Tswo
WRITE(4,*)sisai,sisao
IF (model.NE.l) THEN
WRITE(4,*)swsai,swsao
WRITE(4,*)smswi,smswo
IF (model.EQ.3) THEN
WRITE(4,*)Tswi,Tswo
END IF
END IF
ELSE IF (gradplot.EQ.l) THEN
IF (((i.EQ.l).OR.(i.EQ.IO)).AND.(k.EQ.2)) THEN
WRITE(5,4)i,j,k-l,Tspi,Tspo,Tswi,Tswo,sisai,sisao
4 FORMAT(3I4,6F9.3)
END IF
END IF
90 CONTINUE
flag=O
END IF
70 CONTINUE
60 CONTINUE
C Determine the average exit temperature of recirculating water
suml=O.O
 Stellenbosch University http://scholar.sun.ac.za

sum2=0.0
DO 110 i =1, Kmax
DO 100 k=2,Mmax+l
CALL Cpw(Tsw(i,Lmax+l,k},scspw}
suml=suml+Tsw(i,Lmax+l,k}*scspw
sum2=sum2+smsw(i,Lmax+l,k}
100 CONTINUE
110 CONTINUE
CALL Cpw(Tswil,scspw}
Tswol=suml/(Mmax*Kmax*scspw}
smswol=sum2
C Determine the average exit temperature of process water
suml=O.O
sum2=0.0
rem=Mmax+2
IF (flag.EQ.O} rem=l
DO 120 i=l,Kmax
CALL Cpw(Tsp(i,Lmax,rem},scspp}
suml=suml+Tsp(i,Lmax,rem}*scspp
sum2=sum2+Tsp(i,Lmax,rem}
120 CONTINUE
CALL Cpw(( sum2/Kmax}, scspp J
Tspol=suml/(Kmax*scspp}
C Determine the average exit enthalpy of the air
suml=O.O
sum2=0.0
sum3=0.0
DO 130 j=l,Lmax
DO 130 k=2,Mmax+l
suml=suml+sisa(Kmax+l,j,k}
sum2=sum2+swsa(Kmax+l,j,k}
sum3=sum3+Tsa(Kmax+l,j,k}
130 CONTINUE
140 CONTINUE
DO ISO k=2,Mmax+l
IF (flag2.EQ.O} THEN
suml=suml+sisa(Kmax,l,k}/2.0
sum2=sum2+swsa(Kmax,l,k}/2.0
sum3=sum3+Tsa(Kmax,l,k}/2.0
ELSE
suml=suml+sisa(Kmax,Lmax,k}/2.0
sum2=sum2+swsa(Kmax,Lmax,k}/2.0
sum3=sum3+Tsa(Kmax,Lmax,k)/2.0
END IF
ISO CONTINUE
sisaol=suml/(Mmax*(Lmax+.S}}
swsaol=sum2/(Mmax*(Lmax+.5})
Tsaol=sum3/(Mmax*(Lmax+.5)}
C Print the recirc.water inlet and outlet temperatures on the screen
 Stellenbosch University http://scholar.sun.ac.za

WRITE(*,l60)Tswil,Tswol
160 FORMAT(' ','Tw(in) = ',F7.3,' Tw(out) = ',F7.3)
RETURN
END
c *********************************************************************
c * *
c * THERMOPHYSICAL PROPERTIES OF AIR, WATER, WATER-VAPOUR *
c * AND AIR WATER MIXTURES- *
c * *
c *********************************************************************
C Subroutine to calculate the saturation vapour~pressure of water
SUBROUTINE Satvappressure(tl,spssat)
T=tl+273 .16
a=1. 079586El
b=5.02808
c=1.50474E-4
d=-8.29692
e=4.2873E-4
f=4.76955
g=2. 786118312
X=273.16/T .
z=a*(1-x)+b*LOG10(x)+c*(1-10**{d*{(1/x)-1)))+e*(10**(f*(1-x))-l)+g
spssat=10**z ·
RETURN
END
c *******************************************************************
C Subroutine to calculate the specific heat of water-vapour
SUBROUTINE Cpv(tl,scspv)
T=tl+273.16
a=l.3605E3
b=2.31334
c=-2.46784E-10
d=5.91332E-13
scspv=a+b*T+c*T**5+d*T**6
RETURN
END
c *******************************************************************
C Subroutine to calculate the specific heat of air
SUBROUTINE Cpa{t1,scspa)
T=t1+273.16
a=l.045356E3
b=-3.161783E-l
c=7.083814E-4
d=-2.705209E-7
scspa=a+b*T+c*T**2+d*T**3
RETURN
END
c *******************************************************************
 Stellenbosch University http://scholar.sun.ac.za

C Subroutine to calculate the specific heat of water

SUBROUTINE Cpw(tl,scspw)
T=tl+273.16
a=8.15599E3
b=-2.80627El
c=5.11283E-2
d=-2.17582E-13
scspw=a+b*T+c*T**2+d*T**6
RETURN
END
c *******************************************************************
C Subroutine to calculate the saturation enthalpy of air
SUBROUTINE Satenthalpy(tt2,spsatm,sissat)
CALL Satvappressure(tt2,spssw2)
CALL Cpv(tt2,scspv2)
CALL Cpa(tt2,scspa2)
swsa=((0.62198)*1.005*spssw2)/(spsatm-(l.OOS*spssw2))
sisvap=swsa*(2501.6+(scspv2*tt2/1000))
sisdryair=scspa2*tt2/(1000)
sissat=sisdryair+sisvap
RETURN
END
c *******************************************************************
C Subroutine to calculate the enthalpy of air using the wb and db temps.
SUBROUTINE Enthalpy(ttl,tt2,spsatm,sisa)
CALL Cpv(ttl,scspv2)
CALL Cpa(ttl,scspa2)
CALL Airhumidity(ttl,tt2,spsatm,swsa2)
sisvap=swsa2*(2501.6+((scspv2*ttl)/(1000)))
sisdryair=(scspa2*ttl)/(1000)
sisa=sisdryair+sisvap
RETURN
END
c *******************************************************************
C Subroutine to calculate the humidity of air
SUBROUTINE Airhumidity(tttl,ttt2,spsatm,swsal)
CALL Satvappressure(ttt2,spssw2)
swsas=((0.62198*1.005*spssw2))/(spsatm-(l.OOS*spssw2))
q0=(2501.6-(2.3263*ttt2))*swsas
ql=l.00416*(tttl-ttt2)
q2=(2501.6+(1.8577*tttl)-(4.184*ttt2))
swsal=(q0-ql)/q2
RETURN
END
c *******************************************************************
 Stellenbosch University http://scholar.sun.ac.za
tz

C Subroutine to calculate the dynamic viscosity of air

SUBROUTiNE Airviscosity(tl,musa)
REAL musa
T=t1+273.16
a=2.287973E-6
b=6.259793E-8
c=-3 .131956E-11
d=8.15038E-15
musa=a+b*T+c*T**2+d*T**3
RETURN
END
c *******************************************************************
C Subroutine to calculate the dynamic viscosity of water
SUBROUTINE Waterviscosity(tl,musw)
REAL musw
T=tl+273 .16
a=2.414E-5
b=247.8
c=140
musw=a*IO**(b/(T-c))
RETURN
END
c *************~*****************************************************
G Subroutine to calculate the dynamic viscosity of water vapour
SUBROUTINE Vapourviscosity(tl,musv)
REAL musv
T=t1+273.16
a=2.562435E-6
b=1.816683E-8
c=2.579066E-ll
d=-1.067299E-14
musv=a+b*T+c*T**2+d*T**3
RETURN
END
c *******************************************************************
C Subroutine to calculate the dynamic viscosity of air/water vaopur mix
SUBROUTINE AirVapMixviscosity(t2,swsal,spsatm,musav)
REAL musav,musa,musv
.T=t2+273.16
xa=l.0*5.3824/(1.0+1.608*swsal)
xv=swsal*4.2445/(swsa1+0.622)
CALL Airviscosity(t2,musa)
CALL Vapourviscosity(t2,musv)
musav=(xa*musa+xv*musv)/(xa+xv)
RETURN
END
c *******************************************************************
 Stellenbosch University http://scholar.sun.ac.za

C Subroutine to calculate water-density

SUBROUTINE Waterdensity(tl,rhosw)
T=tl+273.16
a=l.49343E-3
b=-3.7164E-6
c=7.09782E-9
d=-1.90321E-20
rhosw=(a+b*T+c*T**2+d*T**6)**(-1)
RETURN
END
c *******************************************************************
C Subroutine to calculate air-density
SUBROUTINE Airdensity(tl,spsatm,rhosa)
T=t1+273 .16
rhosa=spsatm/(287.08*T)
RETURN
END
c *******************************************************************
C Subroutine to·calculate the density of an air/water vapour mix
SUBROUTINE AirVapMixdensity(t2,swsal,spsatm,rhosav)
T=t2+273.16
CALL Airdensity(t2,spsatm,rhosa)
rhosav=(l.O+swsal)*(l.O-swsal/(swsa1+0.62198))*rhosa
RETURN
END
c *******************************************************************
C Subroutine to calculate the conductivity of water
SUBROUTINE Waterconductivity(tl,sksw)
T=t1+273.16
a=-6.14255E-1
b=6.9962E-3
c=-1.01075E-5
d=4.74737E-12
sksw=a+b*T+c*T**2+d*T**4
RETURN
END
c *******************************************************************
C Subroutine to calculate the Prandtl-number of water
SUBROUTINE Prandtl(t3,Pra)
REAL musw2
CALL Waterconductivity(t3,sksw2)
CALL Waterviscosity(t3,musw2)
CALL Cpw(t3,scspp2)
 Stellenbosch University http://scholar.sun.ac.za

Pra=scspp2*musw2/sksw2
RETURN
END
c *******************************************************************