Equipment Design No.

12

DRYING

Michael Alex Sison Mabao

BS in Chemical Engineering – 5

October 15, 2018

I. Introduction

Drying a solid means the removal of relatively small amounts of water or other

liquid from the solid material to reduce the content of residual liquid to an

acceptably low value. Drying is usually the final step in a series of operations, and

the product from the dryer is often ready for final packaging [1].

In contrast to drying, evaporation refers to the removal of relatively large

amounts of water from material and is usually done by removing water vapor at its

boiling point. In drying, water is usually removed from the material by air [2].

Drying can be classified as batch or continuous. Drying can also be operated

in vacuum to reduce the drying temperature. The solid may be directly exposed to

hot gas, heated from an external medium such as condensing steam, or heated

by dielectric, radiant, or microwave energy. Dryers that expose the solids to a hot

gas are called adiabatic or direct dryers and those that transfer heat through an

external media is called non-adiabatic or indirect dryers [3].

To experimentally determine the rate of drying for a given material, a sample

is usually placed on a tray. If it is a solid material, it should fill the tray so that only

the top surface is exposed to the drying air stream. By suspending the tray from a
balance in a cabinet or duct through which the air is flowing, the loss of weight of

moisture during drying can be determined at different intervals without interruption

of operation. Data from the batch drying experiments is usually obtained as the

weight of the wet solid (W). This data would be converted to rate of drying data

using the equation below [2].

W −W s
Xt=
WS

For a given constant drying conditions, the equilibrium moisture content X*

kilogram equilibrium moisture per kilogram dry solid is determined. In order to

determine the free moisture content X in kg free water/kg dry solid can be

calculated for each value of Xt using the equation below [1].

¿
X t = X t− X

Using the data calculated above, a plot of free moisture content X versus time

(in hours) is made as shown in Figure 1.

Figure 1. Plot of free moisture content vs time. Source: Geankoplis,2003

 To obtain the rate-of-drying curve from this plot, the slopes of the tangents

drawn to the curve in Figure 1 can be measured, which gives values of dX/dt at a

given values of t. The rate R is calculated for each point by the equation below

where R is the drying rate in kg water/m2 hr, ms is the dry solid used, and A is the

exposed surface area for drying in m2 [1].

ms dX −ms ∆ X
R= =
A dt A ∆t

In Figure 2, the rate-of-drying curve for constant-drying conditions is shown.

At initial time, the initial free moisture content is represented as point A. Eventually at

point B, the temperature rises to its equilibrium value. If the solid is quite hot to start,

the rate may start at point A’. This is the unsteady state adjustment period.

Figure 2. Rate of Drying Curve as rate versus Free Moisture Content. Source:

Geankoplis, 2003

From point B to C, the line is straight, and this is the constant drying-rate period. At

point C, the drying rate the drying rate starts to decrease in the falling rate period

until it reaches point D. This is the first falling rate period which is represented as line
CD. At point D, rate of drying falls rapidly until it reaches E where the free moisture

content is zero [2].

II. Equipment Design Problems & Solutions

Problem:

A counter flow tunnel dryer is to be used to provide 227 kg/h product with 1

percent moisture content. The wet feed entering the dryer at 15 ℃ contains 1.5 kg of

water/kg of dry product. The dry bulk density is 560 kg/m 3. The specific heat of the

dry material is 1.25 kJ/kg∙K. Tests show that the critical moisture content of the

material is about 0.4 kg of water/kg dry product. The inlet air to the dryer has a

temperature of 149℃ , and the dried product leaves the dryer at 143℃ . Since fresh

air will be combined recirculated air, the entering air will enter with the moisture

content of 0.03 kg water/kg dry air. The air will leave the dryer at 60 ℃ . The

maximum mass velocity of air that can be used with the solids being dried is 2.71

kg/s∙m2. Spacing between drying trays (t ts) is 0.0381 m. Estimate the number of

transfer units that will be required for the constant rate section and the falling rate

sections of the dryer. Estimate the length of the dryer in meters using the empirical

relation:

Length of transfer unit = 72.77 ttsG0.2

where G = mass velocity of air in kg/s ∙m2 and tts is spacing between drying trays

in m.

3.5.3 Number of transfer units NTU

 Dryer performance may be evaluated considering the “number of transfer

units” associated with a dryer. Similar in concept to the NTU associated with heat

exchangers, rather than describing the heat transfer capacity, the number of transfer

units for a dryer describes the evaporative (mass transfer) capacity. The number of

transfer units for drying is given as (3)

outlet
dYa
NTU = ∫ Y s−Y a
(3.22a)
inlet

where Ya and Ys are the bulk air and surface humidity’s.

Required:

Number of transfer Units (NTU)

Length of Dryer

Illustration:

kg H 2 O
0.15
kg dry air
kg H 2 O 60 oC
0.03
kg dry air 149 oC
[Grab your Dryer
reader’s
Pre-Heater

kg H 2 O
1.5
kg dry air
15oC
kg
227
hr
1% moisture

143 oC
Solution:

We to calculate the heat-transfer requirement of the and the solid using heat balance.

For us to solve the heat transfer requirement we need to assume the following

conditions,

 Operation is at 1 atm.

 Shrinkage is negligible.

In calculating for the heat of water: (C p of water vapor = 1.884 kJ/kgK, Cp of liquid

water = 4.186 kJ/kgK) we will use the equation below,

q=m ∆ X C p ∆T Eq.1

Substituting all the given values into equation 1, we could get:

q H O= ( 227 kg/h )( 1.5−0.01 ) [ 100−15 ) ℃ ( 4.186 kJ /kgK ) +2256+1.884 (600−100) ¿

2

q H O=857 , 900 kJ /h
2

Calculation for the heat transfer of solid;

q=m ∆ X C p ∆T

q solid = ( 1.25 kJ / kgK )( 227 kg /h ) ( 143−15 ) ℃

q s=36,200 kJ /h

Since we have already calculated both heat transfer units of water and our solids, we

can now solve for the Total Heat transfer requirement:

 kJ kJ
q T =857 , 900 + 36,200
h h

q T =894 000 kJ /h

In order for us to calculate the mass of air required we need the following

assumptions,

 1% relative humidity of entering air with 0.03 kg H2O/kg dry air at 149°C.

And to calculate for the humid heat:

Cs ( kgK
kJ
)=1.005+1.88 H Eq. 2

From the equation above we get the value; Humid heat at 149°C and 1% RH is 1.06

kJ/kgK.

Mass of air is obtained through this formula:

q
mair = Eq. 3
C s ∆T

kJ
894 000
h
mair =
( 1.06 kJ /kgK ) (149−60 ) ℃

mair =9478 kg /h

In order for us to calculate the humidity of air leaving the dryer we need first to

calculate the rate of water removal:

m H O =( 227 kg dry solid /hr )( 1.5−0.01 ) kg water / kg dry solid

2
 mH O =338.2 kg/h
2

From the calculated value of water removal we can now calculate the total humidity

of air leaving the dryer:

338.2 kg /hr
H=0.03+
9478 kg / hr

mT =0.0656 kg water /kg dry air

In the calculation for the mass of air recirculated, the wet bulb temperature can be

obtained by getting its relative humidity first.

To calculate for the RH,

RH =100
( )
PA
P AS
eqn 4

The total pressure is assumed to be 101.325 kPa. RH is assumed to be 0.066 kg

water/kg dry air to get the partial pressure. After substituting the values to this

equation,

RH =
18.02
( PA
28.97 101.325−P A ) Eq. 5

We could get a partial pressure of 9.72 kPa.

RH =100 ( 19.940
9.72
)
RH =49 %
At RH = 49% and T = 60°C, wet bulb temperature is:

*Source; http://pathteacheroneword.blogspot.com.

9478(0.03−0.01)
m air =
(0.0656−0.01)

mair =3410 kg /h

*The heat required for the constant rate section is 642,000 kJ/h.

T =60+ ( 149−60 ) ( 642000

894100 )

T =123.9 °C

To obtain the number of units we use the equation below,

 T l −T wb
NTU =ln Eq.6
T c −T wb

123.9−46
ln
60−46

NTU =1.72

For the calculation of our falling rate section:

T C −T 2
NTU = Eq.7
∆Tm

Rearranging the equation above, we can now solve for ∆ T m:

( 123.9−46 ) −(149−143)
∆ T m=
(123.9−46.7)
ln
(149−143)

∆ T m=28.1° C

Substituting all given values,

149−123.9
NTU =
28.1

NTU =0.89

Total number of units:

NTU total=1.72+0.89

NTU total=2.61

To solve for the cross-sectional area:

9478
A=
( 2.71 ) ( 3600 )
 3
A=0.97 m

From the given empirical equation in the problem we can determine length of dryer.

L = 72.77ttsG0.20 Eq.8

L=( 72.77 )( 0.0381 ) ( 2.71 )0.2 ( 2.65 )

L=8.97 m

Problem:

The endothermic liquid phase elementary reaction A+ B →2 C proceeds,

substantially, to completion in a single steam-jacketed continuous reactor. If the

steady-state reactor temperature is 199℉ , calculate the reactor working volume in

gal if given the following:

Steam jacket area = 10 ft2

Jacket steam = 150 psig

U of jacket = 150 BTU/h∙ft2∙ ℉

Agitator shaft horsepower = 25 hp

Heat of reaction = 20,000 BTU/lbmol of A (independent of temperature)

A B C

Feed (lbmol/hr) 10 10

Feed temp, ℉ 80 80

Specific heat, 51 44 47.5

BTU/lbmol∙ ℉
 MW 128 94 111

Density lb/ft3 63 67.2 65

Required:

Illustration:
Fa = 10lbmol/hr

Fb = 10lbmol/hr

150psi

Solution:

To calculate for the reactor volume, we will utilize Equation 2 – 13 of the Fogler Book

(1st Edition) which is the general formula for CSTR,

F Ao X
V= Eq. 1
−r a

Set up energy balance using Equation 12 – 12 of the Fogler Book (1st Edition),

Q̇−Ẇ s−F A 0 ∑ θi C pi ( T −T i0 ) + ( r A V ) ( ∆ H Rx ) =0

˙ s is the shaft work done by the

Where Q̇ is the heat transferred by the steam and −W

stirrer expressed as:

 Q̇=UA (T s−T ) Eq. 2

Where U is the overall heat transfer coefficient, T s is the saturated temperature of

the steam, and A is the heat transfer area. Thus, the conversion is expressed as:

X=
UA ( T s −T ) −Ẇ s
−
∑ θi C pi ( T −T i 0 ) Eq. 3
F A 0 ∆ H Rx ∆ H Rx

For P = 150 psig,T s=359.6 ℉ .

Since it is equimolar feed, θ A =θ B=1because they have equimolar feed and θC

X=
[( 150
Btu (
hr ℉ ft 2) ][
10 ft 2 ) (365.9 ℉−199 ℉ ) + ( 25 hp )
2544.433
1 hp ( )]
(10 lbmol
hr )(
A
20,000
lbmol A )
Btu

[ (
− ( 1 ) 44
Btu
lbmol ℉ ) (
+ ( 1 ) 51
Btu
lbmol ℉ ) (
+ ( 0 ) 47.5
Btu
lbmol ℉
( 199 ℉−80℉ ) )]
Btu
20,000
lbmol A

X =0.999

−r a =k C a Eq. 4

Where,

F Ao
C a=C a 0 ( 1−X )= ( 1−X )
v0

FA 0
C A 0=
v0

To solve for the reactor volume, utilize equation

 V
τ= Eq. 5
v0

Where τ can be obtained from equation

τk
X= Eq. 6
1+ τk

v 0 is just the total volumetric feed expressed into:

F i 0 Mi F A 0 M A F B 0 M B
v 0=∑ = + Eq. 7
ρi ρA ρB

v=
( 10
hr )(
lbmol
128
lbmol ) (
lb
+
10
hr )( lbmol )
lbmol
94
lb

0
lb lb
63 67.2
ft 3 ft 3

ft 3
v 0=34.31
hr

The process is iterative. And for convenience, the following calculations were set up

in excel. The value for V was first assumed to get τ then obtain k then solve for −r A

. Several iterations were made until the assumed V matches the calculated V.

k can be taken from Fogler Book, Equation 5-8 . For the last iteration,

V assume =150 gal

τ =0.585 hr

k =1224.21 hr−1

lbmol
−r A =0.498 3
ft hr

V calculated =149.99 gal ≅ 150 gal

Concluding Remarks

Drying is an essential unit operation used in various process industries. The

mechanism of drying is well understood as a two-stage process and depends on the

drying medium and the moisture content of the product being dried.

I. References

[1] W. L. McCabe, J. C. Smith and P. Harriott, Unit Operations of Chemical Engineering,

New York: McGraw-Hill, 2006.

[2] C. J. Geankoplis, Principles of Transport Processes and Separation Processes, New

Jersey: Prentice Hall, 2003.

[3] R. K Sinnot, Chemical Engineering Design, 4 th Ed., Coulson & Richardson’s

CHEMICAL ENGINEERING, VOLUME 6.

[4] M. S. Peters, K. D. Timmerhaus and R. E. West, Plant Design Economics for

Chemical Engineers, 5th Edition ed., The McGraw-Hill Companies, Inc., 2003.

[5] D. W. Green and R. H. Perry, Perry's Chemical Engineers' Handbook, 8th Edition

ed., The McGraw-Hill Companies, Inc., 2008.

[6] R Turton, R.C Bailie, W.B Whiting, J.A Shaewitz, Bhattacharyya D. Analysis,

Synthesis, and Design of Chemical Processes. 4th ed. Upper Saddle River: Prentice-

Hall; 2012.