478 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th10-2.pm5
Example 10.15. A cooling tower used in power plant consists of 10 big fans, m&water = 1000
kg/min. It is cooled from 35ºC to 30ºC. Atmospheric conditions are 35ºC DBT, 25ºC WBT. Air
leaves the tower at 30ºC, 90% RH. Find out the quantity of air handled per fan hour and the
quantity of make-up water required per hour. (AMIE Winter, 1999)
Solution. Refer Fig. 10.19.
Heat absorbed from the cooling tower
= m&water × c × ∆T
= (1000 × 60) × 4.186 × (35 – 30)
= 1.256 × 10^6 kJ/h
From psychrometric chart, we have
At 35ºC DBT and 25ºC WBT :
h 1 = 76.5 kJ/kg ; W 1 = 0.016 kg/kg of air
At 30ºC and 90% RH :
h 2 = 92.5 kJ/kg ; W 2 = 0.0246 kg/kg of air
Heat gained by air = Heat lost by water
m&air (h 2 – h 1 ) = 1.256 × 10^6
∴ Mass of air, m&air =
1256 10^6
21
.
()
×
hh−
= ×
−
1 256 10
92 5 76 5
.^6
(. .)
= 78.5 × 10^3 kg/h
∴ Quantity of air handled per fan =
78 5 10
10
. ×^3
= 7850 kg/h. (Ans.)
Quantity of make-up water = m&air (W 2 – W 1 )
= 78.5 × 10^3 (0.0246 – 0.016) = 675.1 kg/h. (Ans.)
SOLUTIONS USING PSYCHROMETRIC CHARTS
Example 10.16. The following data pertain to an air-conditioning system :
Unconditioned space DBT = 30ºC
Unconditioned space WBT = 22ºC
Cold air duct supply surface temperature = 14ºC.
Determine : (i) Dew point temperature.
(ii) Whether or not condensation will form on the duct.
Solution. Refer Fig. 10.20.
(i) To determine the dew point temperature for the given conditions, find the intersection of
30ºC DBT and 22ºC WBT and move horizontally (as shown by the arrow) to the dew point tempera-
ture scale. The dew point (tdp) is 18.6ºC. (Ans.)
(ii) Since the duct temperature (14ºC) is less than tdp (18.6ºC) therefore moisture will con-
dense on the duct surface. (Ans.)
Fig. 10.19