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Cooling tower and psychrometric chart questions and solutions, Study notes of Power Plant Engineering

University of Johannesburg Power Plant Engineering

Prof. Vuyo Hashe

Cooling tower and psychrometric chart questions and solutions

Typology: Study notes

2021/2022

Uploaded on 02/24/2023

nichoelas-nakedi 🇿🇦

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COOLING TOWERS

1. A mechanical-draft cooling tower receives 115 m3 per second of atmospheric air

at 103 kPa, 32 C dry bulb temperature, 55% RH and discharges the air saturated

at 36 C. If the tower receives 200 kg/s of water at 40 C, what will be the exit

temperature of the cooled water?

Solution:

skgm 200





Ct 40



saturated

smV

115



%55





at 1, for



kPap

799.4

kgkJh

9.2559



 

kPapp

639.2799.455.0





 

kgkg

0164.0

639.2103

639.2622.0622.0











     

kgkJhWtch

gdbp

2.749.25590164.0320062.1



Discover Study notes of Power Plant Engineering University of Johannesburg

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A mechanical-draft cooling tower receives 115 m^3 per second of atmospheric air at 103 kPa, 32 C dry bulb temperature, 55% RH and discharges the air saturated at 36 C. If the tower receives 200 kg/s of water at 40 C, what will be the exit temperature of the cooled water? Solution: m  (^) 3  200 kg s t (^) 3  40 C tdb (^) 2  36 C saturated V 1 (^)  115 m^3 s tdb (^) 1  32 C  1  55 % at 1, for tdb 1 (^) ^32 C pd  4. 799 kPa hg 1  2559. 9 kJ kg

p s   1 pd  0. 55  4. 799   2. 639 kPa

  kg kg p p p W t s s (^) 0. 0164 103 2. 639

h 1 (^)  cpt db 1  W 1 hg   1. 0062  32   0. 0164  2559. 9   74. 2 kJ kg

m kg p p RT v t s 3 1 0.^8722 103 2. 639

287 32 273       kg s v V ma 131. 85
8722 115 1  1     at 2, tdb (^) 2 ^36 C , saturated p (^) s  pd  5. 979 kPa hg 2  2567. 1 kJ kg   kg kg p p p W t s s (^) 0. 0383 103 5. 979

h (^) 2  cpt db 2  W 2 hg   1. 0062   36   0. 0383   2567. 1   134. 5 kJ kg At 3, t^ 3 ^40 C h (^) 3  hf at 40 C  167. 57 kJ kg To solve for m ^4 : By Mass Balance: m  3 (^)  m  a  W 1 m  a  m  4  m  a  W 2 m  a m  (^) 4  m  3   W 1  W 2  m  a  200  0. 0164  0. 0383  131. 85   197. 1 kg s To solve for h 4^ : m ^ (^) ah 1  m  3 h 3  m  4 h 4  m  ah 2  131. 85   74. 2    200  167. 57   197. 1  h 4   131. 85   134. 5  h (^) 4  129. 70 kJ kg  t^ 4 ^31.^9 C - exit water temperature.

In a cooling tower water enters at 52 C and leaves at 27 C. Air at 29 C and 47% RH also enters the cooling tower and leaves at 46 C fully saturated with moisture. It is desired to determine (a) the volume and mass of air necessary to cool 1 kg of water, and (b) the quantity of water that can be cooled with 142 cu m per minute of atmospheric air. Solution:

To solve for ma^ when m^ 3 ^1 kg By energy balance: Eq. (1) m^ a h 1  m 3 h 3  mah 2  m 4 h 4 By mass balance

m 3  m 4  ma  W 2  W 1 

Eq. (2) m 4^  m 3  ma ^ W 2  W 1 

Substitute in (1)

Eq. (3) m^ a h 1  m 3 h 3  mah 2 ^ m 3  ma ^ W 2  W 1 ^ h 4

 ma   59    1  217. 69   ma   225. 2   1  ma   0. 0692  0. 0116  113. 25 

ma  0. 654 kg

Volume of air = ^ ^ ^

3 V 1 (^)  ma v 1  0. 654 0. 873  0. 5709 m Mass of air required = ma ^0.^654 kg (b) Let m ^3 = quantity of water.

66 min
873 142 1 (^1) kg v V ma      Use Eq. (3), change ma^ ^ m  a , m 3^  m  3

m  ah 1  m  3 h 3  m  ah 2  m  3  m  a  W 2  W 1  h 4

 162. 66   59    m  3   217. 69   162. 66   225. 2   m  3  162. 66   0. 0692  0. 0116  113. 25 

9597  217. 69 m  3 (^)  36 , 631  113. 25 m  3  1061. 1 m  3 (^)  248. 7 kg min

 Quantity of water = m  3^ ^248.^7 kg min

A cooling tower receives 6 kg/s of water of 60 C. Air enters the tower at 32 C dry bulb and 27 C wet bulb temperatures and leaves at 50 C and 90% relative humidity. The cooling efficiency is 60.6%. Determine (a) the mass flow rate of air entering, and (b) the quantity of make-up water required. Solution:

m ^ (^) 3  6 kg s , t (^) 3  60 C tdb (^) 2  50 C  2  90 % tdb (^) 1  32 C twb (^) 1  27 C Cooling tower efficiency = 60.6% To solve for t 4^ : Efficiency (^31) 3 4 t t wb t t 

t t (^) 3  40 C at 1, tdb 1 (^) ^32 C , twb 1 (^) ^27 C h 1 (^)  85 kJ kg W 1 (^)  0. 0208 kg kg at 2, tdb (^) 2 ^50 C ,  2 ^90 % pd  12. 349 kPa

p s   2 pd   0. 90   12. 349   11. 114 kPa

hg 2  2592. 1 kJ kg   kg kg p p p W t s s (^) 0. 0766

325 11. 114

h (^) 2  cpt db 2  W 2 hg   1. 0062   50   0. 0766   2592. 1   248. 9 kJ kg