Cooling Towers

PROBLEMS

1. A cooling tower is located on a mountain where barometric pressure is 90 kPa. The tower is to

cool 113.4 kg/s of water from 41°C to 28°C. Air enters at 36°C DB and a vapor pressure of 4.2

kPaa and leaves saturated at 39°C. Without using psychrometric chart, determine:

a) the mass flow rate of dry air required; and

 b) the mass

mass flow rate of make-up
make-up water required.
required.

1)
AirDiagram:
3

1 2 4 %M4 = 60%
Reheater Dryer

5 Co ra
%M5 = 5%

Required:

 =   
If

  =1
  = 1
Solution:

From;

 % =  100%

 = %
 =0.051
 =0.05
Also;

 =  
 =   
 =10.05
 =0.95=
And;

 =   
% = 
 = %
Thus;

0.6 =0.95
 =1.425
Therefore;

 =   
 =1.4250.05
 =.
b)

From;

 =   
% = 
 = %
% =  
0.05 =1
 =0.052632
%  =  

0.6 =1
 =1.5

So;
 =   
 =1.50.052632
 =.
2. Condenser cooling water is supplied to the forced-draft cooling tower at 40°C and is cooled to

3°C of approach temperature while falling through the tower. The air enters the tower at 35°C DB

and 28°C WB and leaves at 38°C DB and 60% RH. For 5,000 kg/min of condenser cooling water,

find:

a) the quantity of air required by the tower in kg/min;

 b) the amount

amount of make-up
make-up water required to compensate
compensate the water
water loss due to evaporation.
evaporation.

Required:

a) ma

 b) m5

Solution:

From:

 = 
∆ = ℎ  ℎ
 =5000  ∙ 160 ∙4.187 ∙ ∙ 403
403
 = 12 709.
709.9292 
For h2; h1:

@ 1 Pwb=3.782 kPa

 =3.782 101.31527, 8233528

253.41.782
283528 
 = 3.732732 
 = 0.02378 / /
ℎ =1.0062∙35 5 25001.8∙355
ℎ =96.18 /
For h2;

 =6∙6.632
 =3.9792 
 = 0.02543 / /
ℎ =103.54 
Thus;

 = ℎ ℎ = 103.125496.
909.921/88/
 =1754.07 
For MW;

= 
=1754.07  0.023780.02543 2543 
=2.894 /
3. Water at 55°C is cooled in a cooling tower which has an efficiency of 65%. The temperature of the

surrounding air is 32°C DB and 70% RH. The heat dissipated from the condenser is 2,300,000 kJ/hr.

Find the capacity in lps of the pump used in the cooling tower.

Diagram:

Given:

tdb3 = 55°C

eCT = 65%

tdb1 = 32°C

RH1 =

QR  = 2 300 000 kJ/hr

Required: V = ? in lps

Solution:

Psat @ 32°C = 4.759

0.7 =
.  ; Pwv = 0.7(4.759) = 3.3313

And;

Pwv = Pwb -
.
−−.−

ln(Psat) = 14.43509 -
.
 
  ln(Pwb) = 14.43509 -
. 
.− .  
Pwb = 100   
So,

.− .   .   .  

3.3313 = 100     .−.−
 − 
Pwv = Pwb -

−
twb = 27.48

thus,

eCT =
−−  ; 0.65 =
−
−.
tdb4 = 37.112

also;

QR  = mw(tdb3  –  tdb4) ; mw = m3

2300000kJ/hr = (m3)(4.1868)(55-37.112)

M3 = 30710.28kg/hr x
  = 8.5306kg/s

And;

V3 = vf  @ 32°C = 1.0050x10-3 m3/kg = 1.0050L/kg

Therefore;

V = V3m3 = (8.5306kg/s)(1.0050L/s) = 8.573 L/s

 Cooling Towers

6. Water is cooled in a cooling tower from a temperature of 38°C to 24°C. Air enters the cooling

tower at a temperature of 28°C DB with 40% RH and leaves at a temperature of 35°C DB with RH of

98%. Determine:

a) the amount of water cooled per kg of dry air;

 b) the percentage of water lost by evaporation; and

38℃
c) the efficiency or percent effectiveness of the cooling tower.

28℃
RH= 40%
35℃
RH=98%

Mass balance (water vapor)

24℃
  = 
 =  
 = =
 = 
Energy balance

ℎ ℎ =ℎ ℎ

ℎ ℎ =ℎ ℎ
ℎ ℎ =ℎ ℎ
℃
h3= hf at 38 = 159.21 kj/kg

h4=hf = at 24℃ =100.70 Kj/kg

For Pts. 1 & 2

= 0.6 22

ℎ=2500.91.82
For Pt. 1
 
1 = 0.8526 3

 =ΔΤ
for,
=   ; =
=1000  9020   1   1000   160
  3.785  1  
=569.07 
 =569.074.1872920
 =21 444.4  =
hence,


 =505.08  3
(a)

from, =
 =ℎ2 ℎ1
@1,

1 =1.7573 

1 =0.0110 
ℎ1 =44.15 
@2,

2 =3.567 

2 =0.02283 
ℎ2 =85.35 
thus,

 = ℎ2ℎ 1
21 444.4 
 = 85.35  49.15 
 =592.4 
from,

=  ;=
=592.4  0.8526 3
 1 =22
 =2 1
 =592.4  0.022830.0110 
 =7  (b)
 Drying Processes

PROBLEMS

1. Copra enters a dryer containing 60% water and 40% of solids and leaves with 5% water and

95% solids. Find the amount of water removed based on a kg of final product and a kg of bone-dry-

material (ME board Problem, October 1992).

Required:

MR  = M4  –  M5= ?

If a) GM5 = 1 kg

 b) BDM = 1 kg

Solution:

a) from;

% = 
 =% ) = 0.05(1 kg) = 0.05kg

Also;

GM5 = BDM5 + M5

BDM5 = GM5  –  M5 = 1 kg  –  0.05 kg = 0.95 kg = BDM4

And;

GM4 = BDM4 + M4

% =   = %

 ;
 9. In problem #5, if water flows at the rate of 10 kg/sec, air entering tower has a heat

enthalpy of 80 kJ/kg and exits at 125 kJ/kg, what is the required air flow rate in kgs/hr.

a) 55,000 b) 62,500

c) 60,300 d) 63,580

 =10 

ℎ1 =80 
ℎ2 =125 
 = ?
ℎ20 = 
(∆34) =∆h
  ∆ 4.1868   5027332273
 =ℎ20  ∆ℎ  =10   125  80    3600ℎ 
34  ∙  
 =60 289.92 ≈60 300 ℎ
10. In problem #5, the change in the humidity ratio of the incoming and exiting air is

0.00165. What is the required make-up water in kg/sec?

a) 0.156 b) 0.028

c) 0.037 d) 0.310

∆12 =0.00165 

 = ?
1 3  =2 4
 =24  1 3
 (

=  ;  =

also,
  = 6 4   3
 =1  21ℎ 
 =60 289.92 ℎ  3600 0.00165 
 =0.0276≈0.028 
11. The approach and efficiency of a cooling tower are 10°C and 65%, respectively. If the

temperature of water leaving the tower is 27°C, what is the temperature of water entering the

tower?

a) 45.57°C b) 47.55°C

c) 55.47°C d) 54.75°C

 =  
= 
10=27
 =17℃
0.65=  27
17
  =45.57℃ 
12. The change of temperature of water entering the cooling tower and the WB

temperature of surrounding air is 23°C, and the efficiency of the tower is 65%. If the mass flow rate

of the water is 15 kg/s, determine the heat carried away by the air, in kW.

a) 983.93 kW b) 938.93 kW

c) 993.83 kW d) 939.83 Kw

Q=?

=ℎ ℎ
Heat Balance

Heat Absorbed by Air=Heat Rejected by Water

ℎ ℎ= 

= 
 =15   

 =4.1868 
  =?
From,

= 
And,

  
=0.65
So,

0.65= 23
  =0.6523
  =14.98
Therefore,
GIVEN:

Tdb3=55C (16.12KPa)

ect = 65%

tdb1 = 32C (4.73407KPa)

twb1 = 26.4C (3.6189KPa)

Q 3to4 = 2300000 KJ/ hr

REQUIRED:

Ṽ4= ? at lph

SOLUTION:

Q 3to4 = m4or3Cpw (∆t3-4)

 = −−
 = + +
 = +. +
 =35.45℃ 
 = ∆ = . −.
  
 =28099.57725  
Ṽ =  = . 
 Ṽ =28 099.58   ℎ
Ṽ =7.805  
20. An atmospheric cooling tower is to provide cooling for the jacket water of a 4-stroke,

800 kW diesel generator. The cooling efficiency is 60% at a temperature approach of 10°C. If the

ambient air has a RH of 70% and DB temperature of 32°C, determine the amount of cooling water

supplied to the engine, in liters per hour. Generator efficiency is 97%, used work is 30%, and

cooling loss is 25%.

a) 39,804 lph b) 38,904 lph

c) 34,908 lph d) 34,809 lph

Given: Req’d: V 4= ? in L/hr

ec = 60%
CA = 10˚C
RH1 = 70%
tdb = 32˚C
ηg = 97%
%Q 3 = 25%
%Q 1 = 30%
Ec = Q 1 + Q 2 + Q 3 + Q 4 + Q 5

 ==    = 
Solution:
;

   ; But,

 = ;     =   
 =    


From:
 =   −−   100%
; Where:
CA = tdb4 – twb1 = 10˚C ; tdb4 = 10˚C + twb 1
10˚C + 25.9˚C = 35.9˚C

 = 
@Pt. 1
; Pwv = (RH)(Pda)
Pwv1 = 0.7(4.7301)
So,
Pwv1 = 3.31105 kPa
twb1 = 25.9˚C
%Q 1 = 30%
% =    100% = %
;

η =  = 

Also,
;

= . =824.742 

= %824. 742 
Then,
; Where: Q 1 = BP

= 0.30 =2799.14 

So,
% =    100% ; Q 3 = (%Q 3)(Ec)
Q 3 = (0.25)(2799.14 kW) = 687.285 kW
But,
Q 3 = Q water cooling  = mwCpw(∆t3-4)
Q 3 =mwCpw(tdb3 – tdb4) ;    =  
@ mw = 0
mwv4 = mwv3 = mwater

 =  35.9 ˚100%
So,

0.6=   25.9˚

tdb3 = 50.9˚C

 =  = 687.= 285      

Then,

 = 50.935.9˚4.1868 − =10.9437 

 =   
Therefore,

 = 10.1000 9437  3600

  
      
  =39397.
≈39804  3 2
 or LPH
 (a)