HEAT TRANSFER, FLUID MECHANICS

AND THERMODYNAMICS

STEAM POWER PLANTS AND THE RANKINE CYCLE

P R E S E N T E D B Y M AT T H I A S M A N D Ø
A S S O C I AT E P R O F E S S O R , P H D
This lecture

• Steam power plants

• Ideal simple Rankine cycle
• Reheat cycle and methods to improve efficiency

Prof William John Macquorn

Rankine (1820-1872)
21.08.2020 2
Steam power and the industrial revolution

Newcomens steam engine (1712)

• The worlds first practical steam engine

• Invented to pump water out of mines

As the steam is cooled inside of the cylinder, the pressure

decreases and the atmospheric pressure performs a work
on the piston.

Operating pressure = 0.1 bar

Efficiency = 0.5% (mechanical power)

From wikipedia

21.08.2020 3
Modern power plant.
Pressure: ~ 250 bar
Efficiency: ~40% (electric power)

21.08.2020 4
A more simple sketch of a power plant

Qin

Wout
Win

Wout
Wout − Win ηel = ≈ 40%
ηth = ≈ 39% Qin
Qin

Qout Qout
ηdistrict heating = ≈ 50%
Qin

21.08.2020 5
Modeling of thermodynamic processes: ideal cycles

The ideal cycle:

• The ideal cycle does not involve any friction
• Expansion and compression takes place in
quasi-equilibrium manner
• The pipes connecting components are well
insulated and no heat transfer takes place
• Ideal cycles are said to be internally reversible

21.08.2020 6
Example: real and ideal cycles - The Otto cycle

21.08.2020 7
Work and process path

• The work of a process can be found by integration in a P,V diagram

• For process path B, which go from 1 to 2, the work is positive
• For process path A, which go from 2 to 1, the work is negative

P P P

1 1 1

+ =
2 2

V V

Area under process path A = work done by this process

Area under process path B = work done by this process

21.08.2020 8
Work from a cycle

• The area enclosed by a cycle in both a P,v

diagram or a T,s diagram represent the work
done by the cycle

wnet = qnet = ∫ Pdv = ∫ Tds

21.08.2020 9
The simple ideal Rankine cycle

Isobaric heating

Isentropic
compression

Isentropic expansion

Isobaric cooling

21.08.2020 10
Example 9-7 using steam tables
• A steam power plant operates on the simple Rankine
cycle. Steam enters the turbine at 30bar and 350°C
and exits at 75kPa.
State 1:
P1 = 75kPa  h1 = h f @ 75kPa
 table A-5 (saturated water)
x1 = 0  v1 = v f @ 75kPa
State 2:
w pump ≅ v1 ( P2 − P1 )
h2 = h1 + w pump
State 3:
P3 = 3MPa  h3
 table A-6 (superheated water)
T1 = 350 C  s3
State 4:
s f @ 75kPa , s fg @ 75kPa
P4 = 75kPa} table A-5
h f @ 75kPa , h fg @ 75kPa
s4 − s f
s4 = s3 x4 = h4 = h f + x4h fg
s fg

21.08.2020 11
Example 9-7 cont.

• Net work
wnet = wturb − w pump
= 713 − 3 = 710 kJ kg
qin = h3 − h2
• Thermal efficiency
wnet q
ηth = = 1 − out
qin qin wturb = h3 − h4
• Net power
w pump = h2 − h1
W = wnet ⋅ m qout = h4 − h1

21.08.2020 12
Example 9-7 using EES

21.08.2020 13
Non-ideal cycles

• Friction causes pressure drop in pipes

• There are irrevesibilities in pumps and turbines,

see next slide

• Water is subcooled in condensers

• Auxiliary equipment takes power (fans, mills, etc.)

• Heat loss from pipes, steam leaks etc.

21.08.2020 14
Isentropic efficiency of a pump

• We consider the losses in pumps and turbines by

utilizing isentropic efficiencies.

• Pump work for an ideal pump (s2=s1):

ws = h2 s − h1 = v1 ( P2 − P1 )
• Pump work for an actual pump (s2>s1):
wa = h2 a − h1
• Isentropic efficiency:
ws h2 s − h1
ηP = =
wa h2 a − h1

21.08.2020 15
Isentropic efficiency of a turbine

• Turbine work for an ideal turbine (s4=s3):

ws = h3 − h4 s 3

• Pump work for an actual pump (s4>s3):

3
wa = h3 − h4 a 3

4
• Isentropic efficiency:
h 4a
w h −h 4
ηT = a = 3 4 a h4s
4
ws h3 − h4 s

4 3

21.08.2020 16
Example 9-7 expanded
• Determined the net work and thermal efficiency if the isen-
tropic efficiency of the pump and turbine is ηP=0.7 and ηT=0.8
• State 1: (same as before)
P1 = 75kPa  h1 = h f @ 75kPa
 table A-5 (saturated water)
x1 = 0  v1 = v f @ 75kPa
• State 2:
w pump ≅ v1 ( P2 − P1 ) h2 s = h1 + w pump
h2 a = ( h2 s − h1 ) ηP + h1
• State 3: (same as before)
P3 = 3MPa  h3
 table A-6 (superheated water)
T1 = 350 C  s3
• State 4:
s f @ 75kPa , s fg @ 75kPa
P4 = 75kPa} table A-5
h f @ 75kPa , h fg @ 75kPa
s4 − s f
s4 = s3 x4 = h4 s = h f + x4h fg
s fg
h4 a = h3 − ηT ( h3 − h4 s )

17
Example 9-7 Thermal efficiency comparison

Actual Ideal Carnot

Net work 566kJ/kg 710kJ/kg 1131kJ/kg
Thermal efficiency 21% 26% 42%

Tmin
ηth ,carnot = 1 −
Tmax

21.08.2020 18
How to improve the simple Rankine cycle?

21.08.2020 19
The ideal Rankine cycle with re-heating

21.08.2020 20
The ideal Rankine cycle with supercritical pressure

• Most modern power plants operate at

supercritical pressure (with re-heat)

• Critical point for water:

P=220 Bar
T=374°C

21.08.2020 21
The Carnot cycle (1823)

• The most efficient engine which cannot be made!

• Real engines can be compared with the Carnot efficiency

The Carnot efficiency:

Tlow
ηcarnot = 1 −
THigh

21.08.2020 22
Example: Esbjergværket

Source: DONG energy

• Carnot efficiency:

T 25 + 273 Fact sheet:

ηcarnot = 1 − low = 1 − = 64% • Efficiency (electric): 45%
THigh 561 + 273
• Steam temperature: 561 °C
• Steam pressure: 251 bar
• Electric power: 378 MW
• Fuel (Coal): 120 ton/h

21.08.2020 23
Heat Transfer, Fluid Mechanics and
Thermodynamics
REFRIGERATION, HEAT PUMPS AND THE
VAPOR-COMPRESSION REFRIGERATION CYCLE
PRESENTED BY MATTHIAS MANDØ
ASSOCIATE PROFESSOR, PHD
This Lecture

• Refrigerators and heat pumps

• The vapor compression refrigeration cycle

2 of 19
Refrigerator

Compressor

Hot line (outlet) (compressor discharge line)

Cold Line (inlet) (Suction line)

3 of 19
Heat pump

• Air source heat pump (air-air, air-liquid)

• Ground source heat pump (liquid-liquid)

4 of 19
Refrigeration –overview of components

5 of 19
A more simple sketch

Cold environment Hot environment

6 of 19
Coefficient of performance, COP

• The efficiency of refrigeration systems are referred to as COP

Cooling effect QL
COPR = =
Work input Wnet,in
Heating effect QH
COPHP = =
Work input Wnet,in

• We also have

COPHP = COPR +1

7 of 19
Example - COP

• A house is to be heated using a heat pump.

Determine the required Win and Qin:

• QH = heat loss from the house:

QH = 80 MJ h = 22kW

• COP definition:

Win = QH COPHP = 8.9kW

• Energy balance:

QL = QH − Win = 13.3kW

8 of 19
Ideal vapor compression refrigeration cycle

Cold environment Hot environment

1 2
Isobaric cooling

Isenthalpic 4 3
expansion

Isentropic compression

Isobaric heating

9 of 19
Refrigerants
Refrigerant vapor pressure

Criteria
• Non toxic
• Non ozone depleting
• Non flamable
• Non greenhouse gas
• Non corrosive
• Cheap

10 of 19
Refrigerants - continued
• Ammonia, NH3, R717
- Original refrigerant, still used for industrial facilities
- Poisonous, corrosive, explosive

• CFC, for example R12

- Ozone depleting and now banned

• HFC, for example R134a

- Potent greenhouse gas, R134a 1300 times that of CO2
- Is being phased out

• Hydrocarbons, for example isobuthane, R600a

- Used for modern refrigerators
- Is explosive but the quantity is small (24g in a 280L refrigerator)
- Is a greenhouse gas but only 3 times that of CO2

• Carbon dioxide, CO2, R744

- Used for large refrigeration systems, for example in supermarkeds
- High pressure put high demands to the equipment used

11 of 19
Example – Ideal vapor compression refrigeration cycle

• A refrigerator using R134a with a mass flow of

0.05kg/s is depicted on the figure. Determine all
temperatures as well as QL,QH, Win and COP.
• State 1: h1 = hg = 239.16 kJ kg
P1 = 0.14 MPa  A−12
  → T1 = −19 C
x1 = 1  s1 = sg = 0.94456 kJ kg ⋅ K
• State 2:
P2 = 0.8 MPa  A−13 h2 = 275.39 kJ kg
  →
s2 = s1  T2 = 39 C
• State 3:
P3 = 0.8MPa  A−12 h3 = 95.47 kJ kg
  →
x3 = 0  T3 = 31C
• State4:
T4 = T1 h4 = h3

12 of 19
Example – Ideal vapor compression refrigeration cycle

• Heat removal

QL = m ( h1 − h4 ) = 7.18kW
• Power input to compressor

Win = m ( h2 − h1 ) = 1.81kW
• Heat rejection
Q H = m ( h2 − h3 ) = 8.99kW
• Coefficient of performance
QL
COPR = = 3.97
Win

13 of 19
Actual vapor compression refrigeration cycle

Departure from ideal process:

• Pressure loss in pipes due to friction

• Subcooling in condensator
• Superheating in evaporator
• Irreversibilities in compressor

• Note process 1-2’:

- cooling can decrease entropy and the
work required for the compression

14 of 19
Example - actual vapor compression refrigeration cycle

• A refrigerator using R134a with a mass flow of

0.05kg/s is depicted on the figure. Determine the
isentropic efficiency of the compressor as well as
QL, Win and COP.
• Find enthalpies using tables or EES:
A−13
P1 , T1  → h1 = 243.40 kJ kg
A−13
P2 , T2  → h2 = 284.39 kJ kg
A−12
P3 , T3  → h3 ≈ h f @T3 = 85.75 kJ kg = h4
P2 s = P2  A−13
  → h2 s = 281.05 kJ kg
s2 s = s1 
• Heat transfer, power, COP and isentropic efficiency
QL = m ( h1 − h4 ) = 7.88kW COPR = QL Win = 3.84
Win = m ( h2 − h1 ) = 2.05kW ηC = ( h2 s − h1 ) ( h2 − h1 ) = 0.919

15 of 19
About expansion valves

• Temperature measurements from the

outlet of the evaporator is used to
regulate the flow and ensure that the
refrigerant is superheated before it
enters the compressor

• A capilary tube is a simple method used

to lower the pressure

16 of 19
About multiple systems

• The same compressor can be used for both cooling and freezing

17 of 19
About heat pump use

• A heat pump can be used for both heating and cooling

18 of 19
Problems, see course homepage on moodle

19 of 19
Heat Transfer, Fluid Mechanics and
Thermodynamics
PSYCHROMETRICS
AIR-WATER VAPOR MIXTURES
PRESENTED BY MATTHIAS MANDØ
ASSOCIATE PROFESSOR, PHD
This Lecture

• Mass and mole fractions, partial pressure

• Dalton’s and Amagat’s law
• On ideal and real gas mixtures
• Dry and wet air
• Absolute and relative humidity
• Molliers chart
• Dew point temperature

2 of 25
About gas mixtures

• Many engineering applications involves gas mixtures

• We are interested in finding the mixture properties

3 of 25
Mass and mole fractions
• The mass of the mixture, mm, is the sum of
the masses of the individual components
k
mm = m1 + m2 + m3 + ... = ∑ mi
i =1

• The mole number of the mixture, Nm, is the

sum of the moles of the individual components
k
Nm = ∑ Ni
i =1

• The mass fraction, mfi, and the mole fraction, yi:

mi Ni Note: Often the mass fraction
mf i = yi = is denoted by xi instead of mfi
mm Nm
as used by cengel

• The sum of the mass and mole fractions are equal to 1:

k k

∑ mfi = 1
i=1
∑y
i =1
i =1

4 of 25
Conversion between mass and mole fractions

• mass fraction � mole fraction !

• We convert between mass and moles using the molar mass:

m = NM ( mass = moles × molar mass )

• The apparent/average molar mass, Mm, and gas constant, Rm:
k
1 Ru
M m = ∑ yi M i = Rm =
k
mf i Mm
i =1
∑
i =1 M i

• Conversion between mass and mole fractions

Mi
mf i = yi
Mm

5 of 25
Partial pressures and Dalton’s law (Ideal gas)

• The total pressure in a mixture is equal to the sum of

the partial pressures of the component gases
k
Ptotal = P1 + P2 + P3 ... = ∑ Pi
i =1

kPa
kPa kPa kPa kPa kPa

6 of 25
Amagat’s law (ideal gas)

• The total volume is equal to the sum of the

partial volumes of the component gasses

NRuT RT
Vtot = = ( N1 + N 2 + N 3 + ...) u
P P
k
= V1 + V2 + V3... = ∑Vi The volume a component would
occupy if they existed alone at the
i =1 mixture temperature and pressure is
called the partial volume
• Note that:

Volume fraction = Pressure fraction = mole fraction

Vi P N
= i = i = yi
Vm Pm N m

7 of 25
Example 9-1

• A gas mixture consist of 3kg O2, 5kg N2, and 12kg CH4 at 100kPa. Determine mass
and molar fractions, average molar mass and partial pressures.
• Solution: make a table (for example in Excel)

m mf [-] y [-] M [kg/kmol] P

O2 3kg 0.15 0.092 32 9.2kPa
N2 5kg 0.25 0.175 28 17.5kPa
CH4 12kg 0.60 0.733 16 73.3kPa
20kg 1 1 19.6 100kPa

Step 1 Step 2 Step 4 Step 3 Step 5

k mi Mm 1 Pi = yi Ptot
mm = ∑ mi mf i = yi = mf i Mm = k
mm Mi mf i
i =1
∑
i =1 M i

8 of 25
Mixture properties

• Mixture properties of a gas can be found using the mass fraction:

k
Internal energy: um = ∑ mf i ui kJ kg
i =1
k
Enthalpy: hm = ∑ mfi hi kJ kg
i =1
k
Entropy: sm = ∑ mf i si kJ kg ⋅ K
i =1
k The intensive properties of
Specific heat: c p , m = ∑ mf i c p ,i kJ kg ⋅ K a mixture are determined
i =1 by weighted averaging

• Note: for ideal gasses the properties are independent on pressure

9 of 25
Example cont

• For the previous example determine the mixture enthalpy at a

mixture temperature of 20°C

mf [-] h
Note:
O2 0.15 -4.566 kJ/kg
Enthalpies found
N2 0.25 -5.188 kJ/kg using EES

CH4 0.60 -4661 kJ/kg

1 -2799 kJ/kg

k
hm = ∑ mf i hi
i =1

10 of 25
Example 9-2

• Determine the mixture temperature and pressure after the

partition is removed.

m T P1 R [kJ/kg K] V Cv [kJ/kg K] P2
O2 7kg 40°C 100kPa 0.2598 5.7m3 0.658 69kPa
N2 4kg 20°C 150kPa 0.2969 2.3m3 0.743 45kPa
mix 11kg 32.2°C 8.0m3 114kPa
Step 1
Step 3
mRT
V= mRTm
P1 P2 =
Step 2
Vm
Solve : mC v ( Tm − T1 ) + mC v ( Tm − T1 ) = 0
     
Nitrogen Oxygen

11 of 25
Alternative energy balance to find T m using EES

12 of 25
About ideal gasses

• Ideal gasses are gasses which obey the ideal gas law:

PV = NRuT

• Gasses at low pressure and high temperature can be considered ideal

• High pressure steam in power plants and refrigerants are not ideal gasses
• Water vapor in the atmosphere can be considered as an ideal gas

• If your gas is not ideal you have to use more elaborate equations of state!

13 of 25
Water vapor is an ideal gas ?

• Figure shows the percentage error if

water vapor is considered an Ideal
gas

• The partial pressure of water vapor in

atmospheric air is usually below
10kPa

14 of 25
Atmosheric air
Dry air - air that contains no water
o 78,03 vol. % N2
o 20,99 vol. % O2
o 0,94 vol. % Ar
o 0,03 vol. % CO2
o 0,01 vol. % H2

• The specific heat of air can be assumed to

be constant at cp=1.005 kJ/kg°C.
• The enthalpy can be found from:

hdry ,air = c pT

Atmosheric/moist air – air which contain some

amount of water vapor
• The enthalpy for water vapor can be taken to
be equal to the enthalpy of saturated vapor
at the same temperature:

hvap ≅ hg ( T )

15 of 25
Absolute or specific humidity (aka humidity ratio)

• The absolute humidity is defined as the

mass of water vapor present per unit mass
of dry air:

mv
ω= (kg water vapor/kg dry air)
ma

• Using the ideal gas law this can also be

expressed as:
For saturated air, the vapor pressure is
0.622 Pv equal to the saturation pressure of water
ω= (kg water vapor/kg dry air)
P − Pv

16 of 25
Relative humidity
• The absolute humidity does not say how much water the
atmoshere can hold at a given temperature.

• The relative humidity is defined as moisture in the air, mv,

relative to the maximum amount the air can hold at a given
temperature, mg:
mv Pv
φ= = (-) were Pg = Psat @ T
mg Pg

For the same absolute humidity we can have

different relative humidity as the air can hold
more water vapor at higher temperatures

17 of 25
Absolute and relative humidity

• Relation between absolute and relative humidity:

ωP 0.622φ Pg
φ= ω=
( 0.622 + ω ) Pg P − φ Pg

• The enthalpy of atmosheric air can be expressed as:

h = ha + ωhg (kJ/kg dry air)

18 of 25
Example – amount of water vapor in auditorium

• An auditorium have dimensions 8x9x4meters. The room is kept at T=25 °C, P=100kPa
and relative humidity,ϕ, of 75%. Determine partial pressure of the dry air, the absolute
humidity, the specific enthalpy, the mass of dry air and water vapor in the auditorium.

• Partial pressure: • Specific enthalpy:

Pv = φ Pg = φ Psat @ 25C = 2.38kPa h = ha + ωhv
Pa = P − Pv = 97.62kPa ≅ C p ,airT + ωhg = 63.8 kJ kg dry air
• Absolute humidity: • Mass of air and water vapor:
0.622 Pv Vroom = Va = Vv = 8 x 9 x 4 = 288m 3
ω= = 0.0152 kg kg dry air
P − Pv PV
ma = a a
= 339kg
RaT
PV
mv = v v
= 5.0kg
RvT

19 of 25
Saturation mixing ratio

• The maximum amount of water vapor which

the atmoshere can hold at a given
temperature is known as the saturation Temperature [°C] Vapor [g/kg dry air]
mixing ratio:
50 88.12
40 49.81
30 27.69
20 14.85
10 7.76
0 3.84

20 of 25
Dew point temperature

• The atmosheric air is cooled close to cold surfaces

o The vapor pressure, Pv, is constant.

o The saturation pressure, Psat, decreases with
decreasing temperature

• When Pv=Psat the water vapor in the air will

condense on the cold surface.

• The temperature when this

happens is the dew point
temperature:

Tdp = Tsat @ Pv

21 of 25
Example - Fogging of windows

• Air in a house is 20°C with a relative humidity ϕ=75%. At

what inside temperature of the windows will moisture
condense on the windows?

• The vapor pressure:

Pv = φ Pg @ 20 C = 0.75 ⋅ 2.339kPa = 1.754kPa

• The dew point temperature:

Tdp = Tsat @1.754 kPa = 15.4 C

22 of 25
Adiabatic saturation process

• A ”practical” method to determine the

humidity of air.

• The absolute humidity can be

determined by measurement of
pressure and temperature of inlet
and exit of the saturator:

C p ( T2 − T1 ) + ω2 h fg 0.622 Pg 2
ω1 = were ω2 =
hg1 − h f 2 P2 − Pg 2

23 of 25
Dry and wet bulb temperature

• An ”even more practical” method to determine

the humidity of air!

• The equations from the previous slide can be

used to determine the absolute humidity using
the wet and dry bulb temperature.

• Humidity can be measured directly using

electronic devices based on capacity change
of a polymerfilm as it absorbs water

24 of 25
• Mollier’s diagram

• for P=101.3kPa
• ρ is the mixture density

• Your can find high res

chart on moodle with a
brief guide

25 of 25
Heat Transfer, Fluid Mechanics and
Thermodynamics
AIR CONDITIONING

PRESENTED BY MATTHIAS MANDØ

ASSOCIATE PROFESSOR, PHD

1 of 19
This Lecture

• Indoor climate
• Adiabatic mixing of air steams
• Heating and cooling
• Humidification and de-humidification
• Drying

2 of 23
Human comfort

3 of 23
Mollier’s chart
• Air at 20°C and relative humidity 60%

• The Absolute humidity, ω, is found by ϕ

following the vertical line down
• The vapor pressure,Pv, is found at the
cross-section of the vertical line with
the brown diagonal line
• The saturation pressure, Pg=Psat, is
found following the isotherm to the
TWB
saturation curve and then vertical h
down to the brown diagonal line Tdew
• The dew point,Tdew, is found by going
vertical down to the saturation curve
and then horizontally to the left Pg
• The enthalpy, h, is found following the
diagonal down to the saturation curve
• The wet bulb temperature is found by Pv
going diagonally down to the
saturation curve and then horizontally
to the left ω 4 of 23
Simple heating and cooling

• During heating the absolute humidity

remains constant (thus the relative humidity
decreases)

• The required heat can be found by:

Q = m a ( h2 − h1 )

1
5 of 23
 Heating with humidifiation (using steam)

• Moisture can be added to the

airstream using water or steam

• The temperature will increase slightly

(more if using superheated steam)

• The added water content can be

found from:

m w = m a (ω3 − ω2 ) [ kg water/kg dry air ]

3
2

ω2 ω3 6 of 23
Humidifiation using recirculating water/
evaporative cooling

• Moisture will be added at almost

constant enthalpy

• The efficiency of the humidifier can

be expressed by:
ω2 − ω1
η=
ω3 − ω1
1

7 of 23
Humidifiation using cooled/heated water

• Point 3 indicates the waters

temperature

• Point 2 is the end state of the

conditioned air

• The efficiency of the humidifier can

again be found from
1
ω2 − ω1
η=
ω3 − ω1

2
3

8 of 23
Cooling with dehumidification

• The air is cooled below the dew point

and water condenses on cooling coils

• The water content removed can be

found from:
m w = m a (ω1 − ω2 ) [ kg water/kg dry air ]
• The cooling power can be found from:
1
Q = m a ( h1 − h2 ) − m w h f @ T2 [ kW ]

9 of 23
Adiabatic mixing of air m 1 = 2 kg s

• We can find the mixture properties from

the following relation: m 2 = 1 kg s
m a1 ω2 − ω3 h2 − h3
= =
m a2 ω3 − ω1 h3 − h1

• On the mollier diagram the ratio of

length is equal the the mass flow ratio
1
m a1 length 2-3
=
m a2 length 1-3
3

10 of 23
 Adiabatic mixing of air

• If the line crosses the saturation curve

condensation may occur.

• Point 3’ is found following the line of

constant enthalpy to the saturation
curve

• The amount of water vapor condensed

can be found from:
1
m w = m a (ω3 − ω3′ ) [ kg water/kg dry air ]
3’

3
2
h=cst

ω3’ ω3
11 of 23
Example 9-7(2edt) using Mollier’s chart
• Determine the rate of heat supply in the
heating section and the required mass
flow rate of steam.
• Dry air density:
101kPa
ρ1,dry = Pa RaT =
0, 270 kJ kg K ⋅ 283K
= 1.244 kg m3
• Mass flow rate:
m a = ρ1,dry V1 = 1.244 kg m3 ⋅ 45 m3 min
= 56.0 kg min
• Rate of heat supply:
Q = m ( h2 − h1 ) = 56.0 min
kg
( 28 kJkg − 16 kJkg ) 3
2
= 672 kJ min
• Mass flow rate of steam:
m w = m a (ω3 − ω2 ) =56.0 min
kg
(12.0 kgg − 2.4 kgg )
= 0.537 kg min 1

ω2 ω3 12 of 23
19
Example 9-7(2edt) determining density of dry air
By reading from Mollier chart (only for Ptot=1 atm):
• The density scale displayed is for the wet air density. You can find the dry air density
by using the following relation:

ρ wet = ρ dry (1 + ω )
By walking it past the Ideal gas law:
• Density can be found from:
Pa
ρ dry = where Pa = Ptot − Pv and Pv = φ Psat @ T
RaT
By using EES:

13 of 23
Example 9-10(2edt) using Mollier’s diagram
• Determine the properties of the mixture (3)
• Mass flow rate:
m a1 = ρ a1 V1 = 60.5 kg min
m = ρ V = 22.5 kg min
a2 a2 2

• Enthalpy and absolute humidity:

m a1 ω2 − ω3 h2 − h3
= =
m a2 ω3 − ω1 h3 − h1
2
ω3 = 0.0122 kg kg dry air
h3 = 50.1 kJ kg dry air
• Read values on chart: 3
T3 = 19 C 
 → ρ a ,3 = 1,185kg / m
3

φ3 = 90%  1
• Volume flow rate:
V = m a 3 ρ a 3 = 70.1 m3 min

14 of 23
Example – Cooling tower
• Determine Volume flow rate of air in and
mass flow rate of makeup water.
• Mass balance:
m makeup = m a (ω2 − ω1 )

• Energy balance:
m 3h3 = m a ( h2 − h1 ) + ( m 3 − m makeup ) h4

• Combine:
m 3 ( h3 − h4 )
m a = 2
( h2 − h1 ) − (ω2 − ω1 ) h4
• Values can be found from Molliers chart
and steam tables (Table A-4) 1
• Volume flow of air:
T1  m
 → ρ a ,1 V = a
φ1  ρ a1
15 of 23
CFD of aircraft cabin ventilation

16 of 23
HVAC and the spread of infectious diseases

17 of 23
Air drying – Conveyer belt drying

From spirax sarco

18 of 23
Air drying - Rotary drum drying

From GEA Process Engineering

19 of 23
Air drying – Fluid bed drying

From GEA Process Engineering

20 of 23
Air drying - Spraydrying

• Milk powder
• Enzymes
• Medical products

Rotating Atomizer in a Spray Dryer

From Sakav and GEA Process Engineering

21 of 23
Example – Drying of grain

• Air at T1=15°C and ω=0.0063kg/kg dry air is

heated to T2=29°C in a heat exchanger and
use to dry grain. The air absorbs 50kg water
per hour from the moist grain and exits the
dryer with a relative moisture of φ3=90%.
• The process can be considered adiabatic with
a constant pressure of P=101.3kPa

a) Draw the process in a Mollier chart and

determine ω2 ω3, h1, h2 and h3
b) Determine the required mass flow rate of dry
air
c) Determine the power requirement of the heat
exchanger in kW
d) Determine the volume flow rate of air at the
inlet (state 1)

22 of 23
Example – Drying of grain
• Draw process and read values
ω1=ω2=0.0063kg/kg dry air
h1=31kJ/kg
h2=h3=45kJ/kg 2
m w = 0.0139 kg s
• Mass flow rate of air:
m w = m a (ω3 − ω2 ) 3
m a = 3.1 kg s 1

• Power required:
Q = m a ( h2 − h1 ) = 43kW

• Volume flow rate:

m
V = a = 2.5 m3 s
ρ a1

23 of 23
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 This lecture
• Hydrostatics

2
 Hydrostatic Pressure
z
p1

g
ρg

p2
∑F z = ma z
⇔
p2 ∆x∆y − p1∆x∆y − ρ g ∆x∆z∆y = 0
⇔

p2 − p1 = ρ g ∆z
 Gauge Pressure
• For practical calculations the atmospheric pressure often appears on both
sides of the equation

4
 Example
• The pressure at the bottom of a lake
A lake has a depth of 10 m and a temperature of 4°C.
Calculate the absolute pressure on the bottom of the lake.
P1≈1atm
P2 = ρ gh + P1
= 1000kg/m3 ⋅ 9.8m/s2 ⋅ 10m + 101325Pa
=2.0 ⋅ 105 Pa P2
≈ 2atm

The pressure increases by approx. 1 atm for every 10 m a diver descents into the ocean

5
Pressure distibution

Pa = Pb = Pc = Pd
PA = PB = PC ≠ PD

6
Pascal’s principle

F1 F2 F A
p1 = p2 ⇔ = ⇔ 2 = 2
A1 A2 F1 A1

7
 Example
• The area ratio in a jack use for raising a car is A2/A1=10. Calculate the force
required, at the piston, to lift half of a 1 ton car.
F2 A2
=
F1 A1
⇔
A1 1
F1 = F2 = 500kg ⋅ 9.8 m s 2 ⋅ = 0.49kN
A2 10

8
Intermezzo: Pascal’s barrel experiment

Pascal's barrel is the name of a hydrostatics

experiment allegedly performed by Blaise
Pascal in 1646. In the experiment, Pascal
supposedly inserted a long vertical tube into
a barrel filled with water. When water was
poured into the vertical tube, the increase in
hydrostatic pressure caused the barrel to
burst. - wikipedia

9
 Pressure measurements

p A = ρ1 gh1 p A = ρ 2 gh2 − ρ1 gh1

𝑝𝑝𝐴𝐴 ≈ 𝜌𝜌2 𝑔𝑔ℎ2 𝑓𝑓𝑓𝑓𝑓𝑓 𝜌𝜌1 ≪ 𝜌𝜌2

10
 Example
The oil(ρ=900kg/m3) in a tank is pressurized by air, and the
pressure is measured by a manometer containing
mercury(ρ=13600kg/m3). Dimensions are h1=0,9m, h2=15cm
and h3=23cm. Determine the reading on the pressure gauge.
Pressure at 1:
p1 = pair + ρoile g ( h1 + h2 )
Pressure at 2:
p2 = ρ Hg g ( h3 )
Pascal’s principle:
p2 = p1
⇔
pair + ρ oile g ( h1 + h2 ) = ρ Hg g ( h3 )
⇔
pair = ρ Hg g ( h3 ) − ρ oile g ( h1 + h2 ) = 0.21bar

11
 Archimedes principle
• Buoyancy = weight of displaced fluid.
FB = ρ medium g V body
FB

• Resulting force on a submerged body:

ΣF = FB − Fg
⇔ Fg
FR = ( ρ medium − ρ body ) gVbody
Fg = ρ body g V body

12
 Archimedes and Hiero’s crown
The most widely known anecdote about Archimedes tells of how he
invented a method for determining the volume of an object with an
irregular shape. According to Vitruvius, a votive crown for a temple
had been made for King Hiero II, who had supplied the pure gold to be
used, and Archimedes was asked to determine whether some silver
had been substituted by the dishonest goldsmith. Archimedes had to
solve the problem without damaging the crown, so he could not melt
it down into a regularly shaped body in order to calculate its density.
While taking a bath, he noticed that the level of the water in the tub
rose as he got in, and realized that this effect could be used to
determine the volume of the crown. For practical purposes water is
incompressible, so the submerged crown would displace an amount of
water equal to its own volume. By dividing the mass of the crown by
the volume of water displaced, the density of the crown could be
obtained. This density would be lower than that of gold if cheaper and
less dense metals had been added. Archimedes then took to the
streets naked, so excited by his discovery that he had forgotten to
dress, crying "Eureka!" (Greek: "εὕρηκα, heúrēka!", meaning "I have
found [it]!"). The test was conducted successfully, proving that silver
had indeed been mixed in.
- from Wikipedia

13
 Resultant force on a vertical, rectangular wall

• Integration of pressure distribution:

1 Fres = ∫ PdA
Fres = ∫ PdA = ρ gh ⋅ A = Pav A A
A
2 h
= ∫ ρ gh ⋅ bdh
0
h
1 
=  ρ gh 2 ⋅ b 
2 0
1
= ρ gh ⋅ A
2

14
Location of resultant force on a vertical, rectangular wall
1
A ∫A
yC = ydA
• The resultant force pass through the
centroid of the pressure prism 1 h 1
1 h
= 1
bh ∫0
y ⋅ 2 bdy , b = ρ gy
yC = ∫ ydA =
2

A A 3 1 h
=
2 ρ gh
1 2 ∫ 1
0 2
ρ gy 2dy

1 h
= 
2 2
1
ρ g 1 3
y 
0
2 ρ gh
1 3

h
=
3

15
 Submerged vertical plate
• The resultant force found by adding
the force contributions:
Fres = ∫ PdA = F1 + F2
A

1
= ρ gh1 A + ρ g ( h2 − h1 ) A
2

• The location of the resultant force is

found by equating the moment of
forces about the vertical axis
Fres y A = F1 y1 + F2 y2

16
 Intermezzo: moment

M = d ⋅F [ Nm]
The moment is equal to the applied The sum of moments have to be zero
force times the ‘arm’ for the seesaw to be in balance

17
Example

18
19
20
 Tilted plate
• The resultant force:

Fres = ∫ PdA = ρ ghC A

A

• Location of resultant force:

I xc
y A = yc +
yc A
I xyc
x A = xc +
yc A

• Area moments of inertia can be found on figure 11-6 p.451 or the next slide

21
Centroids and moment of inertia

22
Example

23
24
25
26
Forces on a curved surface
• The resultant force is found by the hydrostatic force acting
on a projection of the curved surface
• The location of resultant force is again found by taking a
moment around an appropriate point.
• For a circular arc or segment thereof, the line of action
passes through the center of the arc.

27
Example

28
29
 Problems
• See course homepage on moodle

• SG= specific gravity, is often used in american textbooks .

Hvis SGA for a medium A is given, you can find the density as:

ρ A = SGA ⋅ ρ water @ 4°C

30
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 This lecture
• Conservation of mass
• Conservation of energy
• Bernoullis principle

2
 Basic principles
• Conservation of mass
• Conservation of energy
• Conservation of momentum

P1,

3
 Conservation of mass
• The mass flow at location 1 is the same as at location 2

m 1 = m 2 1

⇔ 2
ρV1 A1 = ρV2 A2
⇔
A1
V2 = V1
A2

4
 Conservation of energy
• The mechanical energy is conserved
– There is no pipe wall friction
– The flow is steady
– The fluid does not exchange heat with surroundings
– The flow is incompressible

Emek ,1 = Emek ,2

W flow,1 + Ekin ,1 + E pot ,1 = W flow,2 + Ekin ,2 + E pot ,2
2
1
P1,

5
 Mechanical energy
• Potential energy:
E pot = mgh = ρ gh ⋅ V , V = volume

• Kinetic energy:
1 1
Ekin = mV 2 = ρV 2 ⋅ V
2 2

• Flow work:
W flow = F ⋅ ∆s
= ( P ⋅ A) ⋅ ∆s
= P⋅V

6
 General Energy Equation
• The mechanical energy is conserved

W flow,1 + Ekin ,1 + E pot ,1 = W flow,2 + Ekin ,2 + E pot ,2


 1   1 
 P1 + ρV1 + ρ gh1  ⋅ V =  P2 + ρV2 + ρ gh2  ⋅ V
2 2

 2   2 

1 1
P1 + ρV12 + ρ gh1 = P2 + ρV22 + ρ gh2
2 2
2
1

P1,

7
 Example
• Determine the unknown values on the
2
figure water flow at room temperature
– Conservation of mass: 1
A1
V2 = V1 = 0.4 m s P1,
A2
– General energy equation:
1 1 P1=5.0 kPa P2=? kPa
P1 + ρV12 + ρ gh1 = P2 + ρV22 + ρ gh2
2 2 V1=10 cm/s V2=? m/s
 D1=4.0 cm D2=2.0 cm
1 h1=0 cm h2=30 cm
P2 = P1 − ρ (V22 − V12 ) − ρ g ( h2 − h1 )
2

P2 = 5000 Pa − 75Pa − 2940 Pa = 2.0kPa

8
 Conservation of momentum - Bernoulli’s equation

• Conservation of momentum of a fluid

particle moving along a streamline
∑ F = ma
⇓
1
P + ρV 2 + ρ gz = constant along a streamline
2

• For two locations on the same streamline

1 1
P1 + ρV12 + ρ gh1 = P2 + ρV22 + ρ gh2
2 2

9
Different representations of Bernoulli’s equation

• Units of energy per unit mass [J/kg] :

P V2
+ + gz = cst
ρ 2

• Units of pressure [Pa]:

V2
P+ρ + ρ gz = cst
2

• Units of head [m]:

P V2
+ + z = cst Daniel Bernoulli published the first work on fluid
ρ g 2g mechanics in 1738. Bernoullis equation does not
appear in this work and some believe that it was
actually Leonhart Euler who discovered this.

10
Static, dynamic and hydrostatic Pressure

V2
P+ρ + ρ gz = cst
2
Hydrostatic pressure
Static pressure

Dynamic pressure

11
 Stagnation and total pressure

• Stagnation pressure:
V2
Pstag = Pstatic + Pdyn = P+ρ
2
• Total pressure:
V2
Ptot = Pstatic + Pdyn + Phydro = P+ρ + ρ gz
2

12
Pressure measurement

13
Pitot tube mounted on a plane

A pitot tube mounted on a small airplane

14
 Example - Pitot tube

Pstag = Pstat + Pdyn

⇔
P2 = P1 + 12 ρV 2 1
⇔
2 ( P2 − P1 )
V=
ρ
2

You can measure the air velocity by a

single differential pressure measurement

15
 Flow rate measurement
• Different flow measurement
devices using the same principles
– Conservation of mass
V = V1 A1 = V2 A2
– Conservation of energy
1 1
P1 + ρV12 = P2 + ρV22
2 2

– Combine
2 ( P1 − P2 )
V = A2
ρ 1 − ( A2 A1 ) 
2
 

A discharge coefficient is needed to correct

for friction and compressional effects

16
Eksempel: venturimeter

17
18
 Example - reservoir
• Water is flowing out of an opening which is small compared to the size of
the hole. For the dimension given in the figure express the flow rate as a
function of the water level in the tank and the diameter of the hole.
– Conservation of energy
ρV12 ρV22 Notice that P1 and P2 are both equal to Patm
P1 + + g ρ h1 = P2 + + g ρ h2 The water level will decrease slowly thus v1≈0
2 2
⇔
V = 2 g ( h1 − h2 ) = 2 gh

– Flow rate
π
V = VA = 2 gh ⋅ D 2
4

19
 Torricelli’s law

V = 2 gh

By Matt Cook, wikipediea.com

20
 Hydraulic grade line
• HGL: Hydraulic Grade Line
• EGL: Energy Grade Line

21
 Vapor pressure
• If the local pressure decreases to the saturation pressure of the liquid it
will form vapor bubbles. We say it cavitates.
• Cavitation can quickly cause significant damage to flow equipment.

22
 Example: Siphon
• Calculate the maximum high, H, at which water can be siphoned from the tank
Energy balance between 1 and 3:
ρV12 ρV32
P1 + + g ρ h1 = P3 + + g ρ h3
⇔ 2 2
V3 = 2 g ( h1 − h3 ) = 2 g (20 ft)
Conservation of mass:
V2 = V3
Energy balance between 1 and 2:
ρV12 ρV22
P1 + + g ρ h1 = P2 + + g ρ h2
2 2
⇔
ρV22
Patm + 0 + g ρ h1 = Pvap + + gρH
2
⇔
 Patm − Pvap  V22 1ft ≈ 30cm
H = − + h1 = 28 ft = 8, 6m
 ρ g  2 g

23
 Problems
• See course homepage on moodle

24
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 This lecture
• Laminar and turbulent flows
• Pressure loss in pipes

2
 What do we know now?
• Energy-balance:
ρV12 ρV22
∆PPump + P1 + + g ρ z1 = P2 + + g ρ z2+ ∆PLoss
2 2 2

∆h

3
Laminar and turbulent flows

The Reynolds number

4
 Osborne Reynolds
• Reynolds made fluid dynamic experiments in the 1870’s

5
 Laminar and turbulent flow
• Reynolds original experiment
Laminar at low Reynolds numbers:
dye stay straight

Rapid mixing: color-blur to the naked eye

Spark photography: still distinct set of

curls and eddies moving at high frequency

6
 Reynolds number
• Characterization of internal flow

Re ≤ 2300 laminar
ρVD
Re = 2300 ≤ Re ≤ 10000 transitional
µ
10000 ≤ Re turbulent

• ρ – density
• μ – viscosity
• D – diameter
• V – velocity

7
 Laminar and turbulent flow
Laminar Turbulent

Molecular diffusion Macroscopisk mixing

8
 Example
• Is the flow in a water pipe laminar or turbulent? If we assume that it takes about 30
seconds to fill a one liter cup. The internal pipe diameter is assumed to be around 1
cm.
0.001m 3
V = = 0.000033m 3 / s
30sek

V π 
V= = V  D 2  = 0.42m / s
A 4 

ρVD 1000kg / m3 ⋅ 0.42m / s ⋅ 0.01m

Re = = = 3750
µ 1.12 ⋅10−3 kg / s ⋅ m

• The flow is in the transition interval.

9
Pressure loss in pipes

10
 Pressure loss
• The general energy equation expressed an energy balance
without friction loss
• To consider frictional losses we introduce ΔPloss together with the
kinetic correction factor α

ρV12 ρV22
P1 + α1 + g ρ z1 = P2 + α 2 + g ρ z2 + ∆Ploss
2 2

α = 2 for laminar flow 1 2

α = 1.05 for turbulent flow

11
 Loss...?
• We talk about pressure loss in pipes
– Friction between fluid and wall, the fluid sticks to the wall
– The mechanical energy of the fluid is converted into heat
– A pressure difference which we cannot recover
• Pressure loss is calculated as:
P2
L ρV 2 laminar: f = 64 / Re P1
∆P = f
D 2 turbulent: f = f ( ε/D,Re )
ΔP=P1-P2
• L – pipe length
• D – pipe diameter
• V – velocity
• f – friction factor

12
 Example: laminar flow
• Water (4°C) flows through a 10m steel pipe which
has a diameter of 1 cm. Calculate the pressure
loss at an average velocity of 10 cm/s
– The properties of water can be found in tables
– Reynolds number is calculated as:
ρ DVavg the flow is laminar!
Re = = 658
µ
– The friction factor for laminar flow is found by:
64
f = = 0.097
Re
– The pressure loss is determined as:
L ρV 2
∆P = f = 486 Pa
D 2

13
 About velocity profiles

Ideal (frictionsless) Laminar Turbulent

umax = 2Vavg

u y =0 = Vavg u y =0 = 0 u y =0 = 0

Does not exist!

14
 Colebrook equation
• For turbulent flow the friction factor, f, depends on Reynolds number, Re,
and the surface roughness, ε:

1  2.51 ε D 
= −2.0 log  + 
f  Re f 3.7 

15
 Moody chart
Moody chart

16
Friction factor using EES

Value of function moodychart in EES for RR=0.0045

17
 Example: turbulent flow
• Water (4°C) flows through a 10m steel pipe which
has a diameter of 1 cm. Calculate the pressure loss
at an average velocity of 1 m/s
– The properties of water can be found in tables
– Reynolds number is calculated as:
ρ DVavg the flow is transitional. We calculate
Re = = 6580
µ the pressure loss as if it was turbulent
– Roughness of commercial steel is found in table
– Relative roughness is calculated as:
ε D = 0.0045
– The friction factor for turbulent flow is found either by
Colebrook’s equation or Moody’s chart:
f = 0.041
– The pressure loss is determined as:
L ρV 2
∆P = f = 20kPa
D 2
18
 Minor and major loss
• ”Major” loss: Pressure loss in pipes
L ρV 2
∆P = f
D 2

• ”Minor” loss: Pressure loss in pipe components

ρV 2
∆P = K L
2

• Total pressure loss:

ρV 2 L ρV 2
∆Ptot = ∑ K L +∑ f
2 D 2

P1 P2

ΔP=P1-P2

19
Flow components

20
 Inlet and outlet

KL=0.8 KL=0.5 KL=1.0 KL=1.0

KL=0.2 KL=0.04 KL=1.0 KL=1.0

21
 Example: pressure loss
• A pipe system consists of 4m straight commercial steel pipes with an internal
diameter of 25mm, one 90 degree bend (threaded), two 45 degree bends
(threaded), a T-section, inlet and outlet as shown on the drawing. The height
difference between the water surface is 1m.
• Determine the pressure loss in the system if the flow is 60 liter per minute.
• Determine the power requirement of the pump if it has an efficiency η=60%

22
 Example cont.: pressure loss
• Velocity:
V
V= = 2m s
A
• Component loss:
ΣK L = 0.04 + 1.5 + 2 + 2 ⋅ 0.4 + 1 = 5.34
Inlet+Elbow+Tee+2x45°Elbow+outlet

• Friction factor:
ρVD
Re = = 50,000
µ
f = 0.026
ε
= 0.0018
D
• The pressure loss:

 ρV
2
 L
∆Ploss =  ∑ f + ∑ KL  = 20kPa
 D  2

23
Example cont.: power requirement
• The general energy equation between location 1 and 2:
ρV12 ρV22
∆PPump + P1 + α1 + g ρ z1 = P2 + α 2 + g ρ z2 + ∆PLoss
2 2
• The pumps hydraulic power requirement is found by:
W = ∆P V
hyd Pump
2
= ( g ρ z2 + ∆PLoss ) ⋅ V
= 30W

• The pumps electric power requirement: ∆h

Whyd = ηWelec 1
⇔ Whyd

Welec = = 50W
η

24
Example: pipe system

25
26
27
28
Example cont.

29
Example cont.

30
 Problems
• See the course homepage on moodle

31
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 This lecture
• Pump selection, pipe systems and NPSH

2
 What have we learned so far?
• Energy-balance:
ρV12 ρV22
∆PPump + P1 + α + g ρ z1 = P2 + α + g ρ z2 + ∆PLoss
2 2
2
• Pressure loss:
ρV 2 L ρV 2
∆Ploss = ∑ K L +∑ f
2 D 2

∆h

3
Pump selection

4
 Pump selection
• For a given pipe system how do we determine the pump size?

5
 Pump and system curves
ρV12 ρV22
∆PPump + P1 + α1 + g ρ z1 = P2 + α 2 + g ρ z2 + ∆PLoss
2 2

∆PPump = g ρ ∆h + ∆PLoss
  
pump curve system curve

∆P ∆P

2
V V
- The characteristics of the pump -Characteristics of the pipe system
- Provided by the manufacturer -v2 dependent (parabola) ∆h

6
 Operating point
P [kPa] Pump curve System curve

Operation point

Link:
WebCAPS

V [m /s]
3

• Pump curve: The pressure increase a pump can deliver at a given flowrate
• System curve: The pressure loss of a pipe system at a given flowrate
• Operation point: The pumps pressure increase = the pipe systems pressure loss

7
 About pump curves
• Pump curves are most
often given in terms of
H, head [m] and
Q, flowrate [m3/h]

• Conversion:
∆P
H=
ρg

From Grundfos’ pump handbook

8
 The centrifugal pump


Mechanical power V ∆Ptot ρ Vgh
W = = = pump

Pump efficiency η pump η pump

– Electronic speed control of RPM
– Differential pressure measurement to indicate the flow rate

9
 The positive displacement pump
P [kPa] Positive displacement pump

Centrifugal pump

V [m /s]
3

• The flowrate is directly related to rotational speed independent of system

pressure
• If operated against a closed valve, either pump or pipe will be damaged!
• A lot of different designs exists

10
Pump efficiency

11
 Example: Pump selection
• For a given pipe system the flow rate is 20
m3/h and the pressure loss 30kPa for
water@4°C
1. Can the pump on the figure be used?
2. Indicate the operation point and sketch the
system curve
3. The electronic control is switched off. Indicate
the new operation point. How much liquid is
pumped now?
4. Now a valve is closed and the pressure loss in
the pipe system increases to 54kPa. What is
the new flow rate?

12
 Case study
• AAU Esbjerg Laboratory
pump impeller test facility

13
 Net Positive Suction Head, NPSH
• If the pressure on the suction side of the pump is too low the pump will
cavitate.
• The criteria for cavitation to occur is Plocal<Pvapor
• We define an expression for the available head:

Ptot − PVapor 2
Plocal Vlocal P
NPSH A = = + − Vapor
ρg ρg 2g ρg

14
 NPSH
• There are two different NPSH
NPSHA  The availiable NPSH at the suction side of the pump
NPSHR  The required NPSH for the pump to avoid cavitation

• You should always ensure that:

NPSHA ≥ NPSHR

15
 Example: NPSH
• A system, as seen on the drawing, is to pump water with an average
velocity of 50cm/s at 40°C . Determine the maximum elevation z1 the
pump can operate at without cavitation if the pump has an NPSHR= 4.5cm.
• Vapor pressure of water is found in the steam tables
• Available NPSHA is found by:
P Pvap
NPSH A = atm − z1 − ΣH L −
ρg ρg
• Elevation z1 is found by:
NPSH A = NPSH R
⇔
z1 = 9,5m

16
 Example cont.
• Higher temperature, velocity or elevation makes
the pump more likely to cavitate
12 10

10
8
HLoss
8 z1

z1, HLoss [m]

6
z1 [m]

6
4
4

2
2

0
0 0 2 4 6 8
0 20 40 60 80 100
V [m/s]
T [C]

Plots made in EES using parametric tables

17
 Pipe systems

For example district heating

18
D’oh!

19
 Serial and parallel pipes
Serial connection: Same flow rate in the pipes

Parallel connection: Same pressure loss in the pipes

20
 Solution procedure
• Set up an energy equation for each branch
• Set up a mass balance for each node
• Solve the system of equations using an iterative procedure

21
 Example: loop and node
Determine the flow rate of
water in the steel pipes (2)
and (3) if D1=5cm, D2=1cm,
D3=4cm, L1=L2=L3=10m and
∆hA-B=1m

• Mass balance of node N:

Q1 = Q2 + Q3
• Energy balances:
ρVA2 ρVB2
PA + α A + g ρ z A = PB + α B + g ρ z B + ∆PLoss ,1+ 2
2 2
ρVA2 ρVB2
PA + α A + g ρ z A = PB + α B + g ρ z B + ∆PLoss ,1+3
2 2

22
 Example cont.
• Mass balance
V1 A1 = V2 A2 + V3 A3 V1
• Energy balance:
ρV12 L1 ρV12 ρV2 2 L2 ρV2 2
g ρ∆hA− B = ( 0.5 ) + f1 + (1.0 + 1.0 ) + f2 V2
 2 D1 2 
 2 D2 2
inlet branch + outlet

ρV12 L1 ρV12 ρV32 L3 ρV32

g ρ∆hA− B = ( 0.5 ) + f1 + (1.0 + 1.0 ) + f3 V3
 2 D1 2   2 D3 2
inlet branch + outlet

• Support equations:
ρ D1V1
Re1 =
µ
(
f1 = moodychart Re1 , Dε1 ) Solve three equations
Re 2 =
ρ D2V2
µ
( )
f 2 = moodychart Re 2 , Dε2 with three unknowns

Re3 =
ρ D3V3 f3 = moodychart ( Re , )
ε
3 D3
µ

23
Example: pipe system

24
25
26
27
28
 Loop and node
P0 • Insert nodes
• Set up equations for
Qa Qb P2 each loop and node
• • • • • Solve the system of
P1 equations using an
• iterative proceedure
Qc P1=…, P2=…, Qb=…
• • •
• • • •
• • • •

29
Type of Problems

30
 Problems
See course home page on moodle

31
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 External flow
• Lift and Drag forces
• A cyllinder in cross flow
• Flat plate boundary layer

2
 Flow past an airfoil

• Observation: The velocity is greater on the top of the airfoil

• From Bernoulli’s equation:

V2
P+ρ = constant
2
• High velocity = Low pressure
• Low velocity = High pressure

3
Lift and Drag forces
• We can integrate the pressure distribution
over the surface area to get a force:

Fres , P = ∫ PdA
A

• Friction results in shear stresses on the

surface. We can integrate these as well:

Fres ,τ = ∫ τ dA
A

• The total resulting force due to pressure

and shear stress is divided into two
components:
Fdrag, which is parallel with the flow
FLift, which is normal to the flow

4
 Lift and Drag coefficients
• The flow induced forces depends on:
– The characteristic area
– Flow speed and fluid properties
– Flow regime (e.g.. Reynolds number)

1
FD = CD

⋅ ρV 2 ⋅ 
A
Drag coefficient
2
 frontal area
Dynamic pressure

1
FL = CL

⋅ ρV 2 ⋅ A
Lift coefficient
2 
 Planform area
Dynamic pressure

5
 Lift and Drag coefficients
• Drag coefficients for common
geometries can be found in tables.

• Often the area used in the

expression is given in the tables.

• The coefficients given are only

valid in a specific regime of the
Reynolds number

6
 Example
• Two cyclists are racing at 36 km/h were one is riding
in the wake of the front rider. Determine the drag
force on the front rider and how much power is
saved by riding in the wake.
1. Fluid properties is found for air at room temperature V=10m/s
2. Calculate Reynolds number to determine flow regime D ≈ 1m (characteristic length)
ρVD Medium: Air (15°C)
Re = ≈ 6.8 ⋅105
µ ρ=1.23kg/m3
3. Find A and CD in tables in the book μ=1.8∙10-5 kg/(m ∙ s)
1 Tabel 15-2: (previous slide)
FD , front = CD ρV 2 A = 20 N , FD ,back = 11N A=0.36
2
CD,front=0.9 , CD,back=0.5
– Power is found by:

W front = F ⋅ V = 199W , Wback = 111W

7
 Flow regimes over a smooth circular cylinder

no separation steady separation unsteady vortex shedding

laminar boundary layer

wide turbulent wake

turbulent boundary layer

narrow turbulent wake

8
 Stokes flow
• At very low speeds the boundary layer
and wake will be laminar
• Also known as ”Creeping flow”
• CD is a linear fuction of Re

Sir G.G. Stokes

9
 Newtons regime
• At high speeds the boundary layer is laminar
but the wake is turbulent
• This is known as Newtons regime
• CD constant for Re = 103 to 105

10
 Critical Reynolds number
• The Drag coefficient suddenly drops as the boundary layer changes from
laminar to turbulent
• The Reynolds number this occurs at is the critical Reynolds number

Drag “crisis”

11
Case study: Golf ball - effect of surface roughness
• Disturbing the boundary layer by dimples
on the ball ”trips” the boundary layer

Link: The physics of golf Link: Aerodynamics of soccer 12

 Character of the drag coefficient
• Bluff bodies do not have a drag crisis

• For streamlined bodies the drag

coefficient tend to decrease with
increasing Reynolds numbers.

• For streamlined bodies (e.g. airfoils)

the drag/lift coefficient is assumed to
be constant for Re>106

13
 Character of the drag coefficient
• A sphere is associated with both pressure
and friction drag. It experiences
boundary layer separation.

• A streamlined body is mainly associated

with friction drag. There is no boundary
layer separation.

• A bluff body is mainly associated with

pressure drag. The boundary layer always
separates at the edge.

14
 Flow induced vibrations
• Stack in Horsens A Von Karman Vortex street

• Tacoma narrows

15
Prevention

16
Streamlining

17
 Drag coefficient of Cars
1.2

0.8
Drag Coefficient

0.6

0.4

0.2

0
1900 1920 1940 1960 1980 2000 2020
Year

18
 Car aerodynamics
Typical present day car Concept car with optimized aerodynamics

 Flow separation – wide wake  No flow separation – narrow wake

 Function and aesthetics is valued  Roof incline acts to reduce friction drag
higher than aerodynamics  Cd=0.189
 Cd=0.27

19
 Example: a drop in the ocean
• Determine the terminal velocity of a grain of sand dropped in the ocean
• Assume Stokes flow (check later): CD=24/Re
• Newtons 2 law:
dV 1
∑ F = m dt = CD ⋅ A ⋅ ρV 2 + ( ρ w − ρ s ) V g
2 V

 + ( ρw − ρs ) V
πµ
0 = 3

DV g
 Stokes Drag
Particle:
V = 0.00632 m s D = 0.1 mm
ρ=2300kg/m3
• Check if assumption is correct: Medium: Water (4°C)
ρ=1000kg/m3
ρVD
Re = = 0.564 μ=1.13∙10-3 kg/(m ∙ s)
µ
• Re < 1 - the flow can be considered as Stokes flow

20
 Example: Parachute jump
• Determine the maximum velocity for a human being in free fall
• Newtons 2 lov: (Assume Newtons Regime) 2 2
CDA(ft ) CDA(m ) Vter
1
0 = CD A ⋅ ρV 2 − mg 9 0.84 41m/s
 2
2.5 0.23 79m/s
V = 2mg ρ CD A
1.2 0.11 114m/s
• Check if range of :
Observations: 45-55m/s
ρVD
Re = ≈ 106
µ

21
 Example: Power of a car
• How much power does a car use to overcome air resistance if it is traveling
at 80km/h?
• Table 15-2: CD=0.3, Google: A≈2m2
• Drag force: V [km/h] P [kW]
1
FD = CD ⋅ A ⋅ ρV 2 = 182 N 50 1.0
2
80 4.1
• Power to overcome air resistance:
110 10.5
PD = FDV = 4.05kW 130 17.4
200 63.2

22
Intermezzo - Development of aerodynamics
Otto Lilienthal, 1891 Wright Brothers, 1903

WWI-plane, 1916 Spitfire, 1936

23
 Development of the boundary layer
• Three different flow regimes – laminar, transition, turbulent
• xcr – length of the laminar/stabile part of the boundary layer
• δ – Thickness of the boundary layer

24
 Terminology
V, Free stream velocity

Leading edge L, Length of plate

xcr, Length of laminar b. layer

x, local coordinate

ρVL
Re L = Reynolds number based on the length of the plate
µ
ρVxcr
Re x = Critical Reynolds number Re xcr = 500 000
cr
µ
ρVx
Re x = Local Reynolds number
µ

25
 Boundary Layer Growth
Air with a freestream velocity of 1.0 m/s
15
 µ  45
δ = 0.370   x
0.08
 U ρ 
0.06
µx
δ =5
δ (m)

ρU
0.04

0.02

0
0 0.5 1 1.5 2 2.5 3
x (m)

0.00800 0.059
C f ,x =
0.00600 0.664 Re1x 5
C f ,x =
Cf,x (-)

0.00400
Re1x 2
0.00200

0.00000
0 0.5 1 1.5 2 2.5 3
x (m)

26
 Friction for a flat surface
• For a flat surface we use the friction coefficient instead of the drag coefficient
CD = CD , f = C f

• The drag force is determined by:

1
FD = Cf ⋅ ρV 2 ⋅ A
 2 
 Surface area
Friction coefficient
Dynamic pressure

• The friction coefficient is found by integration:

Cf =
1
L (∫ 0
xcr L
C f , x ,lam dx + ∫ C f , x ,turb dx
xcr )
Integral friction coefficient ”Local” friction coefficient

27
 Friction coefficient
• Fortunately, Cengel have solved the integral for you!
• Laminar boundary layer:
1.33
Cf = for Re L < 5 ⋅105
Re1L 2
• Laminar plus turbulent boundary layer:
0.074 1742
Cf = − for 5 ⋅105 ≤ Re L ≤ 5 ⋅107
Re1L 5 Re L

28
 Example: a flag
• Calculate the drag force for a flag on a pole for V=15m/s and how far the
laminar boundary layer extends before it becomes turbulent.
• We assume the flag is a flat plate (hmm…)
• Determination of xcr:
ρVxcr Re µ
Re cr = ⇔ xcr = cr = 0.5m
µ ρV
• Thus the b.layer is laminar until 0.5 m.
• Determination of cf:
ρVL Cf =
0.074 1742
− = 0.0032
Re L = = 3.1 ⋅ 106
µ Re1L 5 Re L 227 x 300 cm.
• Determination of the drag force:
1
FD = 2 ⋅ C f ρV 2 A = 6.0 N
2

• Remember, the flag has two sides! 29

 Friction for a rough flat plate
• Surface roughness is important for
turbulent flows.

• For a rough flat plate Cengel suggest:

−2.5
 ε
C f = 1.89 − 1.62 log 
 L

• This is a generalization of this figure 

30
 Example: a rough flag
• Calculate the drag force for a flag on a pole for V=15m/s
• We assume that it is a rough flat plate with ε=1mm and that the boundary
layer is fully turbulent!
• Calculation of cf:
−2.5
ε  ε
= 0.00033, c f =  1.89 − 1.62 log  = 0.0064
L  L
• Determination of the drag force:
1
FD = 2 ⋅ C f ρV 2 A = 12 N
2
227 x 300 cm.
• Which is x2 compard to a smoot flag, but
x10 too low compared to empirical evidence!

From munson et al.

31
 Problems
• See course home page on moodle

32
Heat Transfer, Fluid Mechanics and
Thermodynamics
HEAT TRANSFER AND THERMAL
RESISTANCE NETWORKS
PRESENTED BY MATTHIAS MANDØ
ASSOCIATE PROFESSOR, PHD,
This lecture

• Fouriers, Newtons and Stefan-Boltzmanns laws

• Thermal resistance networks
• Heat transfer through pipes
• Thermal contact resistance

2 of 45
How to calculate heat loss through a wall

BASIC HEAT TRANSFER

3 of 45
Heat transfer

• Heat transfer takes place when you have a temperature difference

• There are three basic modes of heat transfer: conduction, convection and radiation

4 of 45
Heat conduction

• Heat can be seen as the kinetic energy of molecules

• When a molecule with high energy collides with

molecules with low energy some of the energy is
transferred

5 of 45
Heat convection

• Alongside heat conduction heat can be transfered by

macroscopic movement of fluids

• Natural convection – a temperature difference causes a

difference in density, ‘Hot air rises’

• Forced convection – Fluid flow is forced by the wind or

a blower

6 of 45
Heat radiation

• Heat is transferred by photons

• Everything above 0 Kelvin emits radiation

7 of 45
Thermal conductivity

• Thermal conductivity:

k [W / m ⋅ C ]

• How much heat (energy) per second, per

meter, per degree Celsius which can be
transfered by a given material.

• Metals are good heat conductors

• Gasses are bad heat conductors

8 of 45
Fouriers law

• Fouriers law of heat conduction:

dT T −T
Q cond = −kA = kA 2 1 [W ]
dx ∆x

• Heat always flows from hot to cold

• A is the area normal to the flow direction

9 of 45
Newton’s law of cooling

• Newton's law of cooling:

Q conv = hAs (Ts − T∞ ) [W ]

• h is the convective heat transfer coefficient
• As is the surface area
• Ts is the surface temperature Type of convection h, W/m2 ·K
• T∞ is the temperature of the surroundings
Natural convection of gasses 2-25

Natural convection of liquids 10-1000

• h [W/m2·°C] depends on the flow and is
not a property of the fluid! Forced convection of gasses 25-250

Forced convection of liquids 50-20,000

Boiling and condensation 2500-100,000

10 of 45
Case: Cooling of CPUs
• Newtons law of cooling:
Q conv = hAs (Ts − T∞ ) [W ]
• Problem:
Electronic equipment develops heat

• Solution:

• We can increase the surface

area (e.g. heat sink)
• We can increase the air velocity
• We can change the ( e.g. blower)
medium (e.g. water
instead of air)
11 of 45
Case: Cooling of CPUs

• Or perhaps?

12 of 45
Stafan-Boltzmann’s law

• Stefan-Boltzmann’s law:

Q rad = εσ A (Ts 4 − T∞ 4 ) [W ]
• ε is the emissivity (0<ε<1)
• σ is Boltzmann’s constant (5.670·10-8W/m2·K4)

• T4 dependency!

13 of 45
Example: Heat loss through a brik wall

Inside outside
(living room)

Convection Conduction Convection

Q =
(T ∞ ,1 − T1 )
=
(T1 − T2 ) = (T2 − T∞,2 )
1 hin A ∆x kA 1 hout A

=
(T ∞ ,1 − T1 ) + ( T1 − T2 ) + ( T2 − T∞ ,2 )
1 hin A + ∆x kA + 1 hout A
= UA (T∞ ,1 − T∞ ,2 )

14 of 45
 Comparison of two walls
150mm insulation using granules
Bare brick wall: between two brick walls
• kbrick= 0.72 W/m C • kbrick= 0.72 W/m C
• hin= hout= 20 W/m2 C • kins=0.026 W/m C
• ∆xwall= 10 cm • hin= hout= 20 W/m2 C
• ∆xwall= 2 x 10 cm
• ∆xins=150 mm

 U = 4.19 W/m2 C  U = 0.084 W/m2 C

For a temperature difference of 20 ºC: For a temperature difference of 20 ºC:

 Q = 84 W/m2  Q = 1.7 W/m2

15 ud af 42
Total heat loss from a house

16 of 45
Recommened overall heat transfer coefficients

Wall Windows Roof Floor

Denmark3 0.30 1.8 0.20 0.20
Germany1 0.35 1.3 0.24 0.30
UK2 0.25 2.2 0.13 0.20
Poland2 0.30 2.6 0.30 0.60
Spain1 1.22 5.7 0.65 0.69

1Compiled 2014 by www.entranze.eu for new single family houses

2Compiled 2007 by www.eurima.org
3Compiled 2018 by www.byggeriogenergi.dk (BR18)

17 of 45
Rules and tools

• Rules according to ‘bygningsreglementet’:

http://www.mur-tag.dk/index.php?id=351

• Calculation of heat loss from buildings according to standards:

http://www.rockwool.dk/beregninger/energiberegning

18 of 45
CFD simulation of

HEAT LOSS THROUGH

WINDOWS

19 of 45
Problem

• What is the optimal distance between two glass

sheets in a double-pane window?

• How can I reduce the heat transfer through a

window?

20 of 45
What is CFD?

Q: What is Computational Fluid Dynamics?

A: Computational fluid dynamics (CFD) is the

science of predicting fluid flow, heat transfer,
mass transfer, chemical reactions, and related
phenomena by solving the mathematical
equations which govern these processes
using a numerical process.

21 of 45
Case setup #1
∆x

External
Internal
convection
convection

Argon

A window A cut section of a window

22 of 45
Case setup #2

Solve for:
• Energy balance
• Mass balance
• Momentum balance
In each cell

Magnification of the tip!

The geometry is split up As the mesh appears

into small control volumes in the CFD software

23 of 45
Results: ∆x=5cm, t=34sec

section of vector field

24 of 45
Analysis of heat transfer through a double-pane window

• The simulation is repeated for different distances ∆x:

60
DeltaX
Case# [cm] Surface heat flux [W/m^2] 55
1 5 36

Surface heat flux [W/m^2

2 3 42 50
3 2 42
4 1 35 45
5 0.5 57
6 1.5 44 40
7 1.7 42
35
8
9 1.3 43 30
10 1.2 31
25
0 1 2 3 4 5 6
Gap distance [cm]

Sweet spot!

25 of 45
Concept:

THERMAL RESISTANCE

26 of 45
Thermal resistance concept

• Ohm’s law:
V1 − V2
Flow of electrons = I =
Re

• Fourier’s law:
T −T T −T
Flow of heat = Q cond = kA 1 2 = 1 2
∆x RWall

• Newton’s law:
T −T
Q conv = 1 2
Rconv

27 of 45
Serial resistance

28 of 45
Serial resistance - Procedure

• Determine the total resistance:

1 L 1
Rtotal = Rconv ,1 + Rwall + Rconv ,2 = + +
h1 A kA h2 A

• Calculate the heat flow:

T −T
Q = ∞1 ∞ 2
Rtotal

• Determine all other temperatures:

29 of 45
Example

• Total resistance:
Rtotal = Rconv ,in + 2 ⋅ Rwall + Rgap + Rconv ,out
1 2 L Lgap 1
= + + +
h1 A kA kair A h2 A
= 0.0167 + 0.0068 + 0.1333 + 0.0333 = 0.1902 K W
8.8% 3.6% 70% 17.4% 20°C
• Heat loss:
T −T -20°C
Q = ∞1 ∞ 2 = 210W = 18 MJ day
Rtotal
Rconv,in Rgap Rconv,out
• Temperature difference across gap: Rwall Rwall
∆T
Q = gap ⇔ ∆Tgap = Q ⋅ Rgap = 28 C Sketch of thermal network
Rgap

30 of 45
Parallel resistance

• The total resistance:

1 1 1
= +
Rtotal R1 R2

• The heat flow:

T −T
Q = 1 2
Rtotal

• Notice that for exactly two parallel resistances:

1 1 1 R1 R2
= + ⇔ Rtotal =
Rtotal R1 R2 R1 + R2

31 of 45
 Thermal resistance network

• The total resistance:

R1 R2
Rtotal = + R3 + Rconv
R1 + R2

• Where:
L1 L2 L3 1
R1 = R2 = R3 = R3 =
k1 A1 k2 A2 k3 A3 hA3

• The total heat flow:

T −T
Q = 1 2
Rtotal

32 of 45
Overall heat transfer coefficient

• Overall heat transfer coefficient: A heat exchanger has a complicated

geometry. It is only possible to measure the
U ~ W / m 2 ⋅°C  total resistance. This is given in terms of the
overall heat transfer coefficient, U.
• The heat transfer is found by:
Q = UA∆T

• Notice that:
1
UA =
Rtotal

33 of 45
Heat transfer through

PIPES

34 of 45
Heat transfer through a hollow cylinder

• Fouriers law in cylindrical coordinates:

dT
Q cond ,cyl = −kA
dr

• This can be integrated to give:

T −T T −T
Q cond ,cyl = 2π Lk 1 2 = 1 2
ln ( r2 / r1 ) Rcyl

35 of 45
Thermal resistance network for a pipe

• The total resistance:

Rtotal = Rconv ,1 + Rcyl + Rconv ,2 =
1 ln ( r2 / r1 ) 1
= + +
( 2π r1L ) h1 2π Lk ( 2π r2 L ) h2

• The heat flow:

T −T
Q = ∞1 ∞ 2
Rtotal

36 of 45
Heat loss from an insulated pipe

• The total resistance is calculated by:

Rtotal = Rconv ,1 + Rcyl ,1 + Rcyl ,3 + Rcyl ,3 + Rconv ,2 =
1 ln ( r2 / r1 ) ln ( r3 / r2 ) ln ( r4 / r3 ) 1
= + + + +
( 2π r1L ) h1 2π Lk1 2π Lk 2 2π Lk3 ( 2π r4 L ) h2
• The heat flow:
T −T
Q = ∞1 ∞ 2
Rtotal

37 of 45
Critical Radius of Insulation

• Adding more insulation to a wall always decreases heat transfer.

• Adding insulation to a cylindrical pipe or a spherical shell, however, is a

different matter.

• Adding insulation increases the conduction resistance of the insulation

layer but decreases the convection resistance of the surface because
of the increase in the outer surface area for convection.

• The heat transfer from the pipe may increase or decrease,

depending on which effect dominates!

38 of 45
Example: Heat loss from an insulated pipe

• The heat transfer can be expressed by:

T1 − T∞ T1 − T∞
Q = =
Rins + Rconv ln ( r2 / r1 ) 1
+
 2πLk h ( 2π r2 L )
increases with r

2 decreases with r2

39 of 45
The critical radius

• The critiske radius occurs when:

dQ
=0
dr

• This can be integrated to give:

k
rcr ,cylinder =
h

40 of 45
41 of 45
Thermal

CONTACT RESISTANCE

42 of 45
Thermal contact resistance

• When two surfaces are pressed together air filled

cavities wil cause a large thermal resistance

• Unfortunately there are no unambiguous empirical

relations to predict the contact resistance (material,
pressure, roughness, etc.)

• The use of thermal grease can noticeably decrease

the thermal resistance

43 of 45
44 of 45
Problems

• See course homepage on moodle

45 of 45
Heat Transfer, Fluid Mechanics and
Thermodynamics

5 ECTS

1
 This lecture
• Forced convection
– Thermal boundary layers
– Flow over cylinders
– Flow in pipes
– General thermal analysis

2
 Recap: Newton’s law of cooling
• Newton's law of cooling:

Q conv = hAs (Ts − T∞ ) [W ]

– h is the convective heat transfer
coefficient
– As is the surface area
Type of convection h, W/m2 ·K
– Ts is the surface temperature
– T∞ is the temperature of the Natural convection of gasses 2-25
surroundings
Natural convection of liquids 10-1000

• h [W/m2·°C] depends on the flow and is Forced convection of gasses 25-250

not a property of the fluid!
Forced convection of liquids 50-20.000

Boiling and condensation 2500-100.000

3 of 45
Recap: Natural vs forced convection

ρhot<ρcold

Heat rises due to density difference

4
 Thermal boundary layers
• Thermal boundary layer thickness
– Defined as 99% of freestream temperature:

T − Ts = 0.99 (T∞ − Ts )

5
 Prandtl’s number
• The thermal b.layer does not necessarily develop at
the same rate as the momentum b.layer
• The Prandtl number is the ratio between the
momentum and thermal b.layer

Molecular diffusivity of momentum ν µ c p

Pr = = =
Molecular diffusivity of heat α k

• Pr is a fluid property which can be found in tables:

6
 Thermal boundary layer
• Turbulence promotes both momentum, heat and mass transfer
• The empirical relations for all are similar in nature
• (See also Reynold’s and Chilton-Colburn’s analogy)
For example, for a laminar boundary layer:
Momentum transfer:
U∞ 1 2
τ w = 0.332 Re x
x
Heat transfer:
k
hx = 0.332 Re1x 2 Pr1 3
x
Mass transfer:
DAB 1 2 1 3
hmass , x = 0.332 Re x Sc
x

7
 About empirical correlations
• The convective heat transfer coefficient, h, is found from empirical correlations of
the Nusselt number:

hLc
Nu = = C Re mL Pr n
k

• The Nusselt number is usually a function of Reynold’s number and Prandtl’s

number. Fluid properties are to be evaluated at the average temperature.

• Lc is the characteristic length. For a pipe or cylinder this is the diameter, D. For a
flat plate the characteristic length is the plate length, L.

8
 Some correlations
• Constants C, m og n is tabulated for
mange situations.

• Note: I cannot show all correlations

on the slides. You have use the book!

Pitfall prevention:
This table is for flow over
cylinders in cross flow.
I have seen students confuse
it for the flow inside tubes!

9
 The Nusselt number
• Consider a fluid layer, see drawing
• Heat is conducted when the fluid is still
• Heat is convected when there is macroscopic movement in the fluid
∆T
qcond = k qconv = h∆T
L
• The Nusselt number is the ratio between
convection and conduction:

qconv h∆T hL
Nu = = =
qcond k ∆T / L k

• Thus, the Nusselt number indicates how many times more heat transfer
we get by convection compard to conduction.

10
 Nu for flow over a cylinder
• Nusselt number increases with increasing
Reynolds number

• Turbulent flow is associated with higher

Nusselt numbers compared to laminar flow

11
 Local vs integral Nusselt numbers
• Local Nusselt numbers indicates the
heat transfer at a specific point

• Integral Nusselt numbers provides an

average value for the entire surface
which you need for your calculations.

integral local

12
Empirical correlations – which one should I use?

• You need to carefully search the book for an appropiate correlation.

Guide:
– Is the flow pipe flow, b.layer flow or flow over an object?
Look at headings first
– Is the flow laminar or turbulent?
– Is the relation for an integral Nu number?
– Is the relation for uniform temperature or uniform heat flux? Look at individual
– Is the flow in the entrance region or fully developed? correlations in the
– Is the relation valid in the range of Re and Pr? section
– Is the relation accurate enough for my purpose?

13
 Example
• Hot air at 50°C is flowing in a square, thin metal
duct (100X150mm) at 120m3/h. 10 meters of the
duct goes through a room maintained at 20°C.
Take the temperature of the air to be constant.
– Determine the convective heat transfer coefficient
associated with the flow in the duct.
– Estimate the heat transfer to the room if the convective
heat transfer coefficient associated with natural
convection on the outside of the duct is taken at
h=20W/m2·°C and make a sketch of the thermal network
associated with this. Thin metal duct

1. This is internal flow! – need to determine if it is

laminar (19.7) or turbulent (19.8)
– Fluid properties is found in tables, here A-22:
ρ=1.092kg/m3, µ=1.963·10-5kg/m·s, Pr=0.7228,
k=0.02735W/m·°C

14
 Example cont.
– Flow speed is found by:
V 120 mh ⋅ 3600
3
1 h
V= = s
= 2.2 m s
Acs 0.1m ⋅ 0.15m

– Hydraulic diameter is found by:

4 Acs 4 ⋅ 0.1m ⋅ 0.15m
Dh = = = 0.12m
p 2 ⋅ 0.1m + 2 ⋅ 0.15m

– Reynolds number:
Thin metal duct
ρVDh
Re Dh = = 14 700
µ

Since Re is above 10 000 the flow is turbulent. We go to

section 19-8 in the book.

15
 Example cont.
– There are 9 different expressions given for turbulent flow
in tubes. You have to read the text to find the appropiate
expression.
– No really, you have to read the text!
– We note that:
Thin metal duct
Nu = 0.125 f Re Pr1 3  for fully developed turbulent flow
Nu = 0.023Re0.8 Pr1 3  ditto, but only smooth tubes
Nu = 0.023Re0.8 Pr n  only smooth tubes + considers heating or cooling
0.14
µ
Nu = 0.027 Re Pr    smooth tubes + considers large temperature difference
0.8 13

 µs 

Nu =
( f 8) Re Pr  More accurate expression
1.07 + 12.7 ( f 8) ( Pr 2 3 − 1)
0.5

Nu =
( f 8)( Re− 1000) Pr
 Even more accurate expression
1 + 12.7 ( f 8) ( Pr 2 3 − 1)
0.5

16
 Example cont.
– There are also two expressions for liquid metals which we
can dismiss without further consideration.
– We also note that:
• For developing turbulent flow we can use the same expressions
as for fully developed flow as an approximation
• For turbulent flow in non-circular tubes we can use the Thin metal duct
same expressions by using the hydraulic diameter.
– We can calculate the friction factor using the moody chart
or Colebrook’s equation. Looking at the moody chart, it
can be realized that the friction factor corresponds to that
of a smooth pipe.
– Thus all expressions given previously slide can be used.
– We might want to use the most simple expression for the
exam considering the time limit.
Nu = 0.023Re0.8 Pr 0.3 = 45
– The convective heat transfer coefficient:
hDh Nu ⋅ k
Nu = ⇔ h= = 10W m 2 ⋅ C
k Dh

17
 Example cont.
2. The heat transfer is calculated by:
T −T
Q = in out
Rtot
– The resistance from the heat conduction through the Thin metal duct
duct wall will be neglible.
Rtot = Rconv ,in + Rconv ,out 50°C

1 1 C 20°C
= + = 0.03
hin A hout A W
where
A = L ⋅ p = 10 ⋅ ( 2 ⋅ 0.1 + 2 ⋅ 0.15) = 5m 2 Rconv,in Rcond Rconv,out

– The heat transfer becomes:

T −T 50 − 20 Sketch of thermal network
Q = in out = = 1.0kW
Rtot 0.03

18
 Entrance region
• Higher pressure loss, higher heat transfer

• Entrance length:
– Laminar flow
Lvelocity = 0.05 Re D
Lthermal = 0.05 Re Pr D = Lvelocity Pr

– Turbulent flow
Lvelocity ≈ Lthermal ≈ 10 D

19
 Heat transfer in the entrance region
• Turbulent flow
– Use correlations for fully developed flow

• Laminar flow, Ts=constant

0.065 ( D L ) Re Pr
Nu = 3.66 +
1 + 0.04 ( D L ) Re Pr 
23

.
• Laminar flow, qs=constant
– No expression given in cengel!

20
 Parallel flow over isothermal, flat plates
• Calculate ReL based on the length, L, of the plate
• If ReL < 5·105 use: (laminar b.layer)
Nu = 0.664 Re0.5
L Pr
13

• If ReL > 5·105 use: (laminar+turbulent b.layer)

Nu = ( 0.037 Re0.8
L − 871) Pr
13

Pitfall prevention:
Nux denotes local Nusselt numbers
which you will need to integrate
over the length of the plate.
Nu denotes integral Nusselt
numbers which you can use right
away.

21
 Example
• Hot air flows over a 1,0m wide, flat plate
which is kept at 20°C. Determine the rate of
heat transferred to the plate.
– Fluid properties are evaluated at film temperature:
(60+20)/2=40°C

– Calculation of ReL:
ρVL
Re L = = 5.87 ⋅105
µ air @ 40 C :
The flow is laminar at the start of the plate and then
ρ = 1.127 kg m 3
transitions into turbulent flow
Pr = 0.7255
– Nusselt’s number: k = 0.02662 W m ⋅ K
Nu = ( 0.037 Re − 871) Pr
0.8
L
13
= 588 µ = 1.918 ⋅ 10−5 kg m ⋅ s

22
 Example cont.
– The convective heattransfer cofficient , h:
k
h= Nu = 3.13W m 2 ⋅ K
L

– The heat transferred to the plate:

Q = hA(T∞ − Ts ) = 626W

air @ 40 C :
ρ = 1.127 kg m 3
Pr = 0.7255
k = 0.02662 W m ⋅ K
µ = 1.918 ⋅ 10−5 kg m ⋅ s

23
 General thermal analysis
• Consider an energy balance for a fluid element

 p (Te − Ti )
Q = mC (W)
• The fluids average temperature has to change
between inlet and exit

• Now consider the heat-flux into the element:

qs = hx ( Ts − Tm ) (W/m 2 )

– if ts is constant, qs must change

– if qs is constant, ts must change

24
 Bulk mean temperature, Tm
• The temperature will not be uniform over the cross-section
• We define an average temperature for the fluid, Tm

25
 Constant surface heat flux
• if qs= constant :
 p (Te − Ti )
Q = qs As = mC (W)
then:
qs As
Te = Ti +
 p
mC
thus:

dT q s × perimeter
= = constant
dx  p
mC

constant heat flux = constant temperature gradient

26
 Constant surface temperature
• If Ts=constant :
Q = hAs (Ts − Tm )ave = hAs ∆Tln (W)

• The logarithmic mean temperature difference:

Ti − Te
∆Tln =
ln (Ts − Te ) (Ts − Ti ) 

• The fluidens average temperature at exit:

Te = Ts − (Ts − Ti ) exp ( −hAs mC
 p)

27
 Example
• Calculate the heat loss from a oil pipe running through a icy lake.
Take the surface temperature of the pipe to be constant 0°C.
1. Fluid properties evaluated at 20 °C initially
2. Calculate the Reynolds number:
ρUD 888 ⋅ 2 ⋅ 0.3
Re = = = 666 → Laminar
µ 0.8
3. Calculate the entrance length:
Lthermal = 0.05 Re Pr D = 0.05 ⋅ 666 ⋅10400 ⋅ 0.3 = 104 km

4. Find suitable Nu correlation, calculate the Nusselt number:

Oil @ 20C:
laminar flow, thermally developing flow, small temperature difference -
use correlation by Edwards et al 1979: ρ = 888 kg / m3
0.065 ( D L ) Re Pr k = 0.145 W / m ⋅ K
hD
Nu = = 3.66 + = 37.3 µ = 0.8 kg / m ⋅ s
1 + 0.04 ( D L ) Re Pr 
23
k
C p = 1880 J / kg ⋅ K
Note, this is the average Nu for the pipe flow Pr = 10400

28
 Example cont.
5. Calculate the convective heat transfer coefficient:
k 0.145
h= Nu = 37.3 = 18.0 W / m 2 ⋅ K
D 0.3
6. Calculate the exit temperature
As = π DL = 188.5 m 2
m = ρ AcU mean = 888 ⋅ π4 0.32 ⋅ 2 = 125.5 kg / s
Oil @ 20C:
Te = Ts − (Ts − Ti ) exp ( −hAs mC
 p ) = 19.71 C
ρ = 888 kg / m3
Note, this makes the bulk mean temperature k = 0.145 W / m ⋅ K
Tm=(20+19.71)/2=19.85°C. This low temperature
µ = 0.8 kg / m ⋅ s
difference makes it acceptable to evaluate fluid
C p = 1880 J / kg ⋅ K
properties at 20°C
Pr = 10400
7. Calculate the logarithmic temperature difference:
Ti − Te 20 − 19.71
∆Tln = = = −19.86 C
ln (Ts − Te ) (Ts − Ti )  ln ( 0 − 19.71) ( 0 − 20 ) 

29
 Example cont.
8. Calculate the heat loss:

Q = hAs ∆Tln = 18.0 ⋅188.5 ⋅ ( −19.85 ) = −67.4 kW

Oil @ 20C:
ρ = 888 kg / m3
k = 0.145 W / m ⋅ K
µ = 0.8 kg / m ⋅ s
C p = 1880 J / kg ⋅ K
Pr = 10400

30
Heat Transfer, Fluid Mechanics and
Thermodynamics
HEAT EXCHANGERS –
LMTD AND EFFECTIVENESS-NTU METHODS
PRESENTED BY MATTHIAS MANDØ
ASSOCIATE PROFESSOR, PHD
This Lecture
Notice that:
HX= Heat eXchanger
• Classification of heat exchangers
• Log Mean Temperature Difference method (LMTD)
• Fouling of heat exchangers
• Effectiveness - Number of Transfer Units method (Effectiveness-NTU)
Recommended literature

• Kays and London, Compact Heat Exchangers, McGraw-Hill, New York, 3rd Edition,
1984
• R.K. Shah and D.P. Sekulic, Fundamentals of Heat Exchanger Design, John Wiley,
New York, 2003

LINK:
http://www.vestas-aircoil.com

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 Know your heat exchanger

CLASSIFICATION OF HEAT
EXCHANGERS

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Plate-heat exchangers

• Compact design for water-water

heat exchange
• Easy to scale

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Shell and tube

• Designed for high-pressure applications

• Basically a tank with a lot of tubes inside

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Car radiator

• Compact design for liquid-air cooling

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Tube bundles in power plants
Højtryksoverheder Mellemtryksoverheder

• Gas-water/ Gas-vapor heat

transfer
Højtryksforoverheder
• Radiation contribution is

Fordampning i panelvægge
significant
• Fouling inhibits the heat Røggas Mellemtryksforoverheder
transfer

i panelvægge
Fordampning
Economizer 2

Economizer 1

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Design criteria for heat exchangers

• Perform the required heat transfer AND

• Minimize cost
• Minimize size and weight
• Minimize pressure drop Low
Low cost pumping
• Miscellaneous (lifetime, safety, power
quiet operation, corrosion
resistance, maintenance)

Compact

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Classification of HX - surface area density

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Classification of HX - Overall heat exchanger types

Co flow HX Counter flow HX Cross flow HX

For example For example

plate HX shell ‘n’ tube,
car radiator

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Co- and Counter flow HX

Co flow HX Counter flow HX

Counter flow configuration

is superior to Co flow

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 Analysis of heat exchangers

LOG MEAN TEMPERATURE DIFFERENCE

26.10.2020
Co-flow heat exchangers

• Hot – fluid which is to be cooled

• Cold – fluid which is to be heated
• ∆T1 – difference between inlet temperatures Tc ,out
• ∆T2 – difference between outlet temperatures ∆T2

• The following also applies:

Th ,in Th ,out
Q = m c c p ,c (Tc ,out − Tc ,in )

Q = m h c p ,h (Th ,in − Th ,out ) ∆T1

Tc ,in
• We wish to find the heat transfer as:
∆T1 − ∆T2
Q = UAs ∆Tlm ∆Tlm =
ln ( ∆T1 ∆T2 )
where

26.10.2020 14 ud af 42
Overall heat transfer coefficient, U

• We have seen that the total thermal resistance is:

Rtotal = ∑ Rj = Rconv ,i + Rcond , wall + Rconv ,o

1 ln ( Di / Do ) 1
= + +
hi Ai 2π kL ho Ao

• The total heat transferred can be written as:

∆T
Q = = UA∆T = U i Ai ∆T = U o Ao ∆T
Rtotal

• The overall heat transfer coefficient, U can be

defined using both the inner and outer area!

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26.10.2020 16 ud af 32
Fins

• For a gas-liquid HX fins are often used on

the gas side.

• If there are fins on one side the surface

area is found as:

 Aunfinned + Afin if the fins are short and has a high heat conductivity
A= 
 Aunfinned + η fin Afin if the fins are not as above

-Where ηfin is the fin-efficiency

26.10.2020 17 ud af 32
Energy balance for a
differential element

• Heat gain in cold flow:

δ Q = m c c p ,c dTc
• Heat loss in hot flow:
δ Q = −m h c p ,h dTh
• Heat transferred:
δ Q = U (Th − Tc ) dA
• Substitute, integrate, rearrange to give:

∆T1 − ∆T2
Q = UAs
ln ( ∆T1 ∆T2 )

26.10.2020 18 ud af 32
LMTD

• ΔT1 and ΔT2 is different for a co- or a counter flow HX

∆T1 − ∆T2
Q = UAs ∆Tlm , ∆Tlm =
ln ( ∆T1 ∆T2 )

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Example – counter flow flow HX, LMTD-method

1. Write down information given to give an overview

2. Find missing fluid properties from tables, use energy balances to find missing
temperature (or other value) and the heat transferred
3. Optional: make a sketch of the HX to identify ∆T1 and ∆T2
4. Calculate log mean temperature difference
5. Calculate either the surface area or overall heat transfer coefficient by: Q = UAs ∆Tlm

20
Example – counter flow HX, LMTD-method
1. Write down information given to give an overview

𝒎𝒎̇ 𝒄𝒄𝑷𝑷 𝑻𝑻𝒊𝒊𝒊𝒊 𝑻𝑻𝒐𝒐𝒐𝒐𝒐𝒐

Hot 720kg/h 2.2kJ/kg·K 150°C 40°C

Cold 540kg/h 4.18kJ/kg·K 10°C missing

2. Find missing fluid properties from tables, use energy balances to find missing
temperature (or other value) and the heat transferred
Energy balance no.1:

Q = m h c p ,h ( Th ,in − Th ,out ) = 48kW

Energy balance no.2:
Q
Q = m c c p ,c ( Tc ,out − Tc ,in ) ⇔ Tc ,out = Tc ,in + = 87 C
m c c p ,c

21
Example – counter flow HX, LMTD-method
3. Optional: make a sketch of the HX to identify ∆T1
and ∆T2
4. Calculate log mean temperature difference

∆T1 = Th ,in − Tc,out = 63 C , ∆T2 = Th ,out − Tc,in = 30 C

∆T1 − ∆T2
∆Tlm = = 30.7 C
ln ( ∆T1 ∆T2 )

5. Calculate the asked for surface area or overall heat

transfer coefficient
Q = UAs ∆Tlm ⇔ U = Q As ∆Tlm = 3340W m 2 ⋅ K
were
As = π Dl = 0.47m 2

22
Cross flow – heat exchanger

• Terminology

26.10.2020 23 ud af 32
LMTD - Cross flow / Shell and tube HX

• Vi introduce a correction factor F ∆Tlm is calculated in the same

way as for a counter flow HX
Q = UAs F ∆Tlm ,CF
𝛥𝛥𝑇𝑇1 − 𝛥𝛥𝑇𝑇2
𝛥𝛥𝑇𝑇𝑙𝑙𝑙𝑙,𝐶𝐶𝐶𝐶 =
ln 𝛥𝛥𝑇𝑇1 ⁄𝛥𝛥𝑇𝑇2
• F is tabulated as function of two temperature ratios:

T1 − T2 ( mc
 p)
t −t
∆T1 = Th ,in − Tc ,out
P= 2 1 , R= = tube side
T1 − t1 t2 − t1 ( mc
 p) ∆T2 = Th ,out − Tc ,in
shell side

• F is found using diagrams:

- see next slide

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Shell and Tube Heat Exchanger Cross Flow Heat Exchanger

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Example – Shell ‘n’ tube HX, LMTD method

• Overview:
𝒎𝒎̇ 𝒄𝒄𝑷𝑷 𝑻𝑻𝒊𝒊𝒊𝒊 𝑻𝑻𝒐𝒐𝒐𝒐𝒐𝒐
Hot (shell) missing 2.68kJ/kg·K 110°C 60°C

Cold (tube) 0.8kg/s 4.18kJ/kg·K 22°C 70°C

26.10.2020 26 ud af 32
Example – Shell ‘n’ tube HX, LMTD method

• Energy balances:
Q = m c c p ,c ( Tc ,out − Tc ,in ) = 161kW

Q = m h c p ,h ( Th ,in − Th ,out ) ⇔ m h = Q c p ,h ( Th ,in − Th ,out ) = 1.2 kg s

• Log mean temperature difference:

∆T1 = Th ,in − Tc,out = 40 C , ∆T2 = Th ,out − Tc,in = 38 C
∆T1 − ∆T2
∆Tlm = = 39.0 C
ln ( ∆T1 ∆T2 )
• Correction factor F:
t2 − t1 70 − 22 
P= = = 0.55 
T1 − t1 110 − 22 

T1 − T2 110 − 60
R= = = 1.04  F = 0.93
t2 − t1 70 − 22 

26.10.2020 27 ud af 32
Example – Shell ‘n’ tube HX, LMTD method

• Surface area:

Q = UAs F ∆Tlm ⇔ As = Q UF ∆Tlm = 15.8m 2

• If you are given the tube diameter you can now determine the amount of piping
required

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 Reduction of heat transfer over time

FOULING OF HEAT EXCHANGERS

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Fouling of Heat Exchangers

30
 Fouling of pipes

TH Fouling on inside
Tilsmudsning påofinderside
pipe

rørvæg
Pipe wall
Tilsmudsning
Fouling på yderside
on outside of pipe

TC
ri,f
ri • Fouling act as insulation
 heat conduction resistance
ro
ro,f Rf
Rcond , fouling =   C W 
A

31
 Ex. Fouling (ex 22-2 expanded)

• Consider a double pipe water-water HX:

D i= 1 . 6 c m , D o= 1 . 9 c m
h i= 8 0 0 W / m 2∙ k , h o= 1 2 0 0 W / m 2∙ k
k steel= 1 5 . 1 W / m ∙ k , k limestone= 2 . 9 W / m ∙ k ,
• Fouling is due to a layer of limestone on both sides:

Rtotal = Rconv ,i + R foul ,i + Rcond , wall + R foul ,i + Rconv ,o

1 ln ( Di / [ Di − 2t ]) ln ( Do / Di ) ln ([ Do + 2t ] / Do ) 1
= + + + +
hi Ai 2π klime L 2π k steel L 2π klime L ho Ao

• The heat transfer can be found as:

∆T
Q = = UA∆T = U i Ai ∆T = U o Ao ∆T
Rtotal

32
Ex. Fouling (ex 22-2 expanded)

500

Overall Heat Transfer Coefficient Ui [W/m2 ⋅ K ]

450

400

350

300

250

200
0 0.5 1 1.5 2 2.5 3 3.5 4
Thickness of kalklag
Tykkelse på fouling [mm]

33
 Analysis of heat exchangers

EFFECTIVENESS – NTU

26.10.2020 34 ud af 32
Design tasks

Known desired process, An existing HX is to be

Select/design suitable HX reused for a new process
• Inlet temperatures, flow rates together • Inlet temperatures, flow rates together
with the desired outlet temperature are with UA are known
known
• We have to determine new outlet
• We have to select a heat exchanger temperatures
type (U) and determined its size (As)
• The Effectiveness-NTU method is
• The LMTD-method is suited for this suited for this assignment
assignment

35
Task 1: Select/design a HX

𝒎𝒎̇ 𝒄𝒄𝑷𝑷 𝑻𝑻𝒊𝒊𝒊𝒊 𝑻𝑻𝒐𝒐𝒐𝒐𝒐𝒐

Hot Known Known Known Known

Cold Known Known Known Unknown

𝑈𝑈 =Known /Obtain 𝐴𝐴 =Unknown 𝑄𝑄 =Unknown

from table or producer

Q = m h c p ,h (Th ,in − Th ,out ) → Q

Use of LMTD leads to
Q = m c c p ,c (Tc,out − Tc,in ) → Tc,out straightforward
calculations
∆T1 − ∆T2
Q = UAs → As
ln ( ∆T1 ∆T2 )

36
Task 2: Repurpose a HX

𝒎𝒎̇ 𝒄𝒄𝑷𝑷 𝑻𝑻𝒊𝒊𝒊𝒊 𝑻𝑻𝒐𝒐𝒐𝒐𝒐𝒐

Hot Known Known Known Unknown

Cold Known Known Known Unknown

𝑈𝑈 =Known 𝐴𝐴 =known 𝑄𝑄 =Unknown

Q = m h c p ,h (Th ,in − Th ,out )

Use of LMTD requires an
iterative procedure.
Q = m c c p ,c (Tc,out − Tc,in ) Using EES/calculator you
will have to put up bounds
∆T1 − ∆T2
Q = UAs to successfully get a
ln ( ∆T1 ∆T2 ) solution

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Effectiveness – NTU method

1. Determine heat transfer by:

Q = ε ⋅ ( mc
 p) (Th ,in − Tc ,in )
min

2. Solve energy balances to find outlet temperatures:

Q = m h c p ,h (Th ,in − Th ,out )

Q = m c c p ,c (Tc ,out − Tc ,in )

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ε - heat transfer effectiveness

• Definition of heat transfer effectiveness:

Q Actual heat transfer rate

ε=  =
Q max Maximum possible heat transfer rate

• The effectiveness of a heat exchanger is dependent of:

• Geometry
• Flow configuration

• Effectiveness is found from relations of two dimensionless numbers:

ε = function ( NTU , c )

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Dimensionles numbers NTU and c

• NTU – Number of Transfer Units

UA s UA s
NTU= =
Cmin ( mc
 p) Heat capacity rate:
min
𝐶𝐶ℎ = 𝑐𝑐𝑝𝑝,ℎ 𝑚𝑚̇ ℎ
𝐶𝐶𝑐𝑐 = 𝑐𝑐𝑝𝑝,𝑐𝑐 𝑚𝑚̇ 𝑐𝑐

• c – Heat capacity rate ratio

𝐶𝐶𝑙𝑙𝑚𝑚𝑚𝑚 = min(𝐶𝐶𝑐𝑐 , 𝐶𝐶ℎ )
Cmin 𝐶𝐶𝑙𝑙𝑚𝑚𝑚𝑚 = max(𝐶𝐶𝑐𝑐 , 𝐶𝐶ℎ )
c=
Cmax

40
41
Effectiveness from diagrams

42
Example – counter flow HX, NTU method

1. Find fluid properties from tables, calculate heat capacity ratios

2. Determine Cmin and Cmax, calculate NTU and c
3. Determine effectiveness, ε, from relations or diagrams
4. Calculate heat transfer rate from Q = ε Cmin (Th ,in − Tc ,in )
5. Calculate temperatures out of the HX from energy balances

43
Example – counter flow HX, NTU method

1. Find fluid properties from tables, calculate heat capacity ratios

𝒎𝒎̇ 𝒄𝒄𝑷𝑷 𝑻𝑻𝒊𝒊𝒊𝒊 𝑻𝑻𝒐𝒐𝒐𝒐𝒐𝒐 C
Hot 0.3kg/s 1.01kJ/kg·K 90°C missing 303W/K Cmin
Cold 0.1kg/s 4.18kJ/kg·K 22°C missing 418W/K Cmax

2. Determine Cmin and Cmax, calculate NTU and c

UA s
NTU= = 0.12 were As = π Dl = 0.45m 2
Cmin
C min
c= = 0.725
C max

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Example – counter flow HX, NTU method

3. Determine effectiveness, ε, from relations or diagrams

1 − exp ( − NTU (1 − c ) )
ε= = 0.11
1 − c exp ( − NTU (1 − c ) )

- Often it is useful to do both to cross check the result -

4. Calculate heat transfer rate

Q = ε C ( T − T ) = 2.2kW
min h ,in c ,in

5. Calculate temperatures out of the HX from energy balances

Q = m h c p ,h ( Th ,in − Th ,out ) ⇔ Th ,out = Th ,in − Q m h c p ,h = 83 C
Q
Q = m c c p ,c ( Tc ,out − Tc ,in ) ⇔ Tc ,out = Tc ,in + = 27 C
m c c p ,c

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Observations

• NTU is a measure of heat transfer area –

the larger the NTU, the larger is the heat
exchanger

• A HX with NTU larger than 3 cannot be

justified economically

• For a given NTU the counter flow HX has

the highest effectiveness

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Observations

• Effectiveness is a maximum for c=0 and

minimum for c=1.

• c=0 corresponds to a phase-change

process in a condenser or boiler.

• For c=0 all effectiveness correlations

collapse into one, since the temperature
of one fluid is constant.

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For temperatures three, use LMTD
If you only have two, use NTU

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 End of lecture

THANK YOU FOR LISTENING

26.10.2020 49 ud af 32