‫‪Cooling tower‬‬
‫‪1‬‬
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‫‪TABLE OF CONTENT‬‬
‫‪.................................................................................................2‬‬
‫‪Summery‬‬
‫‪..................................................................................................3‬‬
‫‪Introduction‬‬
‫‪…….........................................................................................4‬‬
‫‪Theory‬‬
‫‪Experimental Procedure.........................................................................................6‬‬
‫‪….........................................................................................7‬‬
‫‪Schematic diagram‬‬
‫‪.................................................................................................9‬‬
‫‪Results‬‬
‫‪................................................................................................17‬‬
‫‪Discussion‬‬
‫‪...............................................................................18‬‬
‫‪Conclusion‬‬
‫‪............................................................................... 19‬‬
‫‪Appindex‬‬
‫\‬
Cooling tower
2
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1. Summary:
Cooling tower is the name of experiment that doing in the lab. mass and energy
balance was performed over a cooling tower also
the mean driving force was
determined the number of transfer unit and the over all mass transfer coefficient. In
this experiment warm water is contact with dry air and the heat transfer from water to
air and the out water be come cold. We use two heat source one at 1 kw and other at
1.5 kw
Experiment objective :
1) perform mass and energy balance
2) Determine the mean driving force.
3) Determine the number of transfer unit.
4) Determine the over all mass transfer coefficient.
Image (1) Cooling Tower
Cooling tower
3
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2. Introduction :
Open pond or by the spray pond technique in which it is dispersed in spray form and
then collected in a large, open pond. Cooling takes place both by the transference of
sensible Heat and by evaporative cooling as a result of which sensible heat in the
water provides the latent heat of vaporization.
On the large scale, air and water arc brought into countercurrent contact in a
cooling tower which may employ either natural draught or mechanical draught. The
water flows down over a series of wooden slats which give a large interfacial area
and promote turbulence in the liquid. The air is humidified and heated as it rises,
while the water is cooled mainly by evaporation.
The natural draught cooling tower depends. on the chimney effect produced by the
presence in the tower of air and vapor of higher temperature. and therefore of lower
density than
the surround. atmosphere. The
atmospheric conditions and the
temperature and quantity of the water will exert a very Important effect on the
operation of the tower. Not only will these factors influence the quantity of air drawn
through the tower, but they will also affect the velocities and flow patterns and hence
the transfer coefficients between and liquid.
Cooling tower
4
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3. Theory:
* Over all mass balance:
Input = output
L2  L1  G2  G1
Where L2 water inlet , L1 water outlet ,G1 air inlet G2 air outlet
* Water mass balance:
L2  L1  G2 * H 2  G1 * H1
G2  G1  G
L2  L1  G * ( H 2  H1 )
Where H2 humidity of outlet air , H1 humidity of inlet air
* Energy balance :
H Y  C S * (T T )  H * 
C S  1.005  1.88 * H
Q  G * (H Y 2  H Y1 )
Is latent heat
 Is enthalpy of air , cs is heat capacity of air H Y Where
Cooling tower
5
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* Determination of number of transfer unite :
N OG  
HY 2
HY 2
dH Y
H Y*  H Y
Where NOG is number of transfer unit (dimensionless)
* To calculate mean driving force :
H lom 
H Y 2  H Y1
ln( H Y 2 H Y 1 )
* To calculate over all mass transfer coefficient :
Z = NOG *HOG
H OG 
G
M B * P * KGa
Where is HOG is height of transfer unit (m) ,MB is molecular weight , P is the pressure
, and KGa is mass transfer coefficient .
Temperature and concentration profiles at interface
Figure 1: Temperature and concentration profile in upper part of cooling tower
Cooling tower
6
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4. Experimental procedure:
1- Introduce water and record its flow rate.
2- Put the heaters on so that water is heated to the required temperature.
3- Introduce air and record its flow rate.
4- Wait for steady state then record steady state dry and wet bulb
temperature of air at the entrance and exit.
5- Record the inlet and outlet temperature and flow rate of water also record
temperature at different stages.
6- Change the air flow rate and repeat step 3 on.
‫‪Cooling tower‬‬
‫‪7‬‬
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‫‪5. Schematic Apparatus:‬‬
‫‪Image 2 : cooling tower in the lab.‬‬
‫‪Cooling tower‬‬
‫‪8‬‬
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‫‪Dry bulb T3‬‬
‫‪Wet bulb T4‬‬
‫‪packing‬‬
‫‪T5‬‬
‫‪Dry bulb T1‬‬
‫‪Wet bulb T2‬‬
‫‪Fan‬‬
‫‪Flow meter‬‬
‫‪Heater‬‬
‫‪Pump‬‬
‫‪Manometer‬‬
‫‪Make up tank‬‬
‫‪Figure 2. setup of experiment‬‬
Cooling tower
9
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6. Results and calculation
A. Heater load =1 kw
Time min
0
5
10
15
20
T1 C
21
21
21
22
22
0
Table 1. air water temperatures at 1kw
T2 C 0
T3 C 0
T4 C 0
19
31
29
20
25
24
20
22
23
20
23
23
20
23
23
Initial pressure (P0)=31 mm h2o
Final pressure (P) =38 mm h2o
Water flow rate=40 g/sec
Total of water used= 1000ml
Water evaporated = 800 ml
From humidity chart :
H2 = 0.017 kg water/kg air ,
G1  0.0137(
H1 = 0.013 kg water/kg air
X
)
(1  H ) * v H
X = 38 – 31 = 7 mm H2O
H = 0.013 kg water/kg air & T= 273 + 22
v H  (2.38*10-3 + 4.56*10-3*H)(T)
v H  0.719 m3/kg air
G1 = 0.0424 kg/sec
L2 = 40 (g/sec)/1000(g/kg)  L2 = 0.04 kg/sec
T5 C 0
39
39
28
28
28
T6 C 0
26
22
22
22
22
‫‪Cooling tower‬‬
‫‪10‬‬
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‫‪* Water mass balance :‬‬
‫) ‪L2  L1  G * ( H 2  H1‬‬
‫)‪L1 = 0.04 - (0.0424)*(0.017 – 0.013‬‬
‫‪L2 = 0.03983 kg water /sec‬‬
‫‪* Energy balance :‬‬
‫) ‪Q  G * ( HY 2  HY 1‬‬
‫‪H Y1  (1.005  1.88 * H1 ) * (T1 )  2501.4 * H1‬‬
‫‪HY1 = (1.005 + 1.88*0.013)*(22) + 2501.4*0.013‬‬
‫‪HY1 = 55.165 KJ/kg air‬‬
‫‪HY 2  (1.005  1.88 * H 2 ) * (T2 )  2501.4 * H 2‬‬
‫)‪HY2 =(1.005+0.017*1.88)*(23) + (2501.4*0.017‬‬
‫‪HY2 = 66.3738 kJ/kg air‬‬
‫‪ Q = 1.71613 kJ/sec‬‬
‫)‪Q = 0.0424*(66.3738– 25.899‬‬
Cooling tower
11
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Calculate number of transfer unite (NTU) :
We neglible resistance to heat transfer in the liquid phase and we use correction factor
method :
TL2 = 28 oC  HY2 = 66.3738
TL1 = 22 oC  HY1 = 55.165
kJ/kg
kJ/kg
 TLm = 25oC
TLm = (28+22)/2
 HLm = 60.77 kJ/kg
HY1 = (HYi – HY)TL1 = ( 88.14 - 55.165 ) = 32.89 kJ/kg
HY2 = (HYi – HY)TL2 = (123 – 66.3738 ) = 56.63 kJ/kg
HYm =(HYi - HY)TLm = (104.89 – 60.77 ) = 44.12 kJ/kg
Hm/HY1 = 44.12/32.89
 Hm/HY1 = 1.34
Hm/HY2 = 44.12/56.63
 Hm/HY2 = 0.79
After using chart :
f =1
NTU = (HY2 – HY1)/(f*Hm)  NTU = (66.37 – 55.165)/(1*44.12)
NTU = 0.255
Cooling tower
12
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Calculate over all mass transfer coefficient (KGa) :
Z = NOG *HOG  HOG = Z / NOG
Z = 0.48 m , NOG = NTU = 0.255
HOG = 0.48/0.255
 HOG = 1.88 m
H OG 
G
M B * P * KGa
G = 0.0424 kg/sec , A = L2 = (150/1000)2
A = 0.0225 m2 MB = 29 kg/kgmol P = 1 atm
1.88 m = (0.0424/0.0225)(kg/s*m2)/(1 atm * 29 kg/kgmol*KGa)
KGa = 0.0344 kgmol/atm *m2*s
5) Calculate mean driving force
H lom 
H Y 2  H Y1
ln( H Y 2 H Y 1 )
Hlom = (66.37 – 55.165)/ln(66.37/55.165)

Hlom = 60.594 kJ/kg
Cooling tower
13
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B. Heater load =1.5 kw
Time min
0
5
10
15
20
Table 3. air water temperatures at 1kw
T1 C
T2 C 0
T3 C 0
T4 C 0
23
20
29
26
22
20
27
27
22
20
27
27
22
20
27
27
22
20
27
27
0
Initial pressure (P0)=31mm h2o
Final pressure (P) =38 mm h2o
Water flow rate=40 g/sec
Total of water used= 1000ml
Water evaporated = 850 ml
From humidity chart :
H2 = 0.022 kg water/kg air ,
G1  0.0137(
H1 = 0.013 kg water/kg air
X
)
(1  H ) * v H
X = 38 – 31 = 7 mm H2O
H = 0.013 kg water/kg air
v H  (2.83*10-3 + 4.56*10-3*H)(T)
v H  0.852 m3/kg air
G1 = 0.0390 kg/sec
L2 = 40 (g/sec)/1000(g/kg)  L2 = 0.04 kg/sec
T5 C 0
38
35
34
34
34
T6 C 0
25
24
24
24
24
‫‪Cooling tower‬‬
‫‪14‬‬
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‫‪* Water mass balance :‬‬
‫) ‪L2  L1  G * ( H 2  H1‬‬
‫)‪L1 = 0.04 - (0.039)*(0.022 – 0.013‬‬
‫‪L2 = 0.0396 kg water /sec‬‬
‫‪* Energy balance :‬‬
‫) ‪Q  G * ( HY 2  HY 1‬‬
‫‪H Y1  (1.005  1.88 * H1 ) * (T1 )  2501.4 * H1‬‬
‫‪HY1 = (1.005 + 1.88*0.013)*(22) + 2501.4*0.013‬‬
‫‪HY1 = 55.165 KJ/kg air‬‬
‫‪H Y 2  (1.005  1.88 * H 2 ) * (T2 )  2501.4 * H 2‬‬
‫)‪HY2 =(1.005+1.88*0.022)*(27) + (2501.4*0.022‬‬
‫‪HY2 = 83.282 kJ/kg air‬‬
‫‪ Q = 1.0965 kJ/sec‬‬
‫)‪Q = 0.039*(83.282– 55.165‬‬
Cooling tower
15
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Calculate number of transfer unite (NTU) :
We neglible resistance to heat transfer in the liquid phase and we use correction factor
method :
TL2 = 34 oC  HY2 =83.282
kJ/kg
TL1 = 24 oC  HY1 =55.165
kJ/kg
TLm = (34+24)/2
 TLm = 29oC
 HLm = 69.22 kJ/kg
HY1 = (HYi – HY)TL1 = ( 100.7 - 55.165 ) = 45.54 kJ/kg
HY2 = (HYi – HY)TL2 = (140.59 – 83.28 ) = 57.31 kJ/kg
HYm =(HYi - HY)TLm = (123.6 – 69.22 ) = 54.38 kJ/kg
Hm/HY1 = 54.38/45.54
 Hm/HY1 = 1.19
Hm/HY2 = 54.38/57.31
 Hm/HY2 = 0.95
After using chart :
 f = 0.98
NTU = (HY2 – HY1)/(f*Hm)
 NTU = (83.28– 55.165)/(0.97*54.38)
NTU = 0.51
Cooling tower
16
‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Calculate over all mass transfer coefficient (KGa) :
Z = NOG *HOG  HOG = Z / NOG
Z = 0.48 m , NOG = NTU = 0.51
HOG = 0.48/0.51
 HOG = 0.95 m
H OG 
G
M B * P * KGa
G = 0.0390 kg/sec , A = L2 = (150/1000)2
A = 0.0225 m2 MB = 29 kg/kgmol P = 1 atm
0.95 m = (0.0390/0.0225)(kg/s*m2)/(1 atm * 29 kg/kgmol*KGa)
KGa = 0.0598 kgmol/atm *m2*s
Calculate mean driving force
H lom 
H Y 2  H Y1
ln( H Y 2 H Y 1 )
Hlom =(83.28– 55.165) /ln(83.28/55.165)

Hlom = 68.260 kJ/kg
Cooling tower
17
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7. Discussion:
1)
In this experiment There are tow level of heat source first we use 1 kw then
we use 1.5 kw .
2)
The amount of water transfer to the air is very low.
3)
The air humidity inlet in second case is more than first case that mean the air
there is error coming from water evaporation from tower in side laboratory .
4)
the amount of heat transfer from liquid to the gas in second case less than
first case.
5)
The number of transfer unit in second case is larger than the first case.
6)
The mass transfer coefficient approximately constant.
7)
In this experiment no resistance to heat transfer in liquid phase was assumed.
Cooling tower
18
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8. Conclusions :
The mass transfer and heat transfer is decreased with increase inlet air humidity and,
the objective of use packing is to increase area of contact between water and air. The
cooling tower does not use in the place which the air have high humidity.
‫‪19‬‬
‫‪Cooling tower‬‬
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‫‪Appendix‬‬
‫‪20‬‬
‫‪Cooling tower‬‬
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‫‪21‬‬
‫‪Cooling tower‬‬
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Cooling tower
22
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B.Variable Listing:
NOG = number of transfer unit(dimensionless)
HOG = height of transfer unit (m)
L
= water flow (kg/s*m2)
G
= dry air flow (kg/s*m2)
H
= humidity of air (kg water/kg air)
Hy = enthalpy of air water mixture (J/kg air)
Cooling tower
23
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C. Reference:
1) Colunson , J.M and Rechardson , j.F “Chemical Engineering “ vol 1 , 3th edition ,
BUTTERWORTH HEINEMANN , 1999 .
2) Christie john geankoplis," transport processes and separation process principles",4th
edition, prentice hall (2003) .
3) "Chemical Engineering Laboratory 2",Dep. of Chemical engineering .