470 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th10-1.pm5
Example 10.8. Cooling and dehumidification : 120 m^3 of air per minute at 35°C DBT
and 50% relative humidity is cooled to 20°C DBT by passing through a cooling coil.
Determine the following :
(i)Relative humidity of out coming air and its wet bulb temperature.
(ii)Capacity of cooling coil in tonnes of refrigeration.
(iii)Amount of water vapour removed per hour.
Solution. For the air at 35°C DBT and 50% R.H. :
pvs = 0.0563 bar (At 35ºC, from steam tables)
pv = φ × pvs = 0.5 × 0.0563 = 0.02815 bar.
W 1 =
0 622 0 622 0 02815
1 0132 0 02815
...
..
p
pp
v
tv−
=
×
−
= 0.0177 kg/kg of dry air.
h 1 = cp tdb 1 + W 1 [hg 1 + 1.88 (tdb 1 – tdp 1 )]
tdp 1 ≈ 23ºC (Corresponding to 0.02815 bar).
∴ h 1 = 1.005 × 35 + 0.0177 [2565.3 + 1.88 (35 – 23)]
= 80.98 kJ/kg of dry dir.
For the air at 20°C. As the saturation vapour pressure at 20ºC is 0.0234 bar, less than the
vapour pressure 0.02815 bar at 35ºC, so that condensation takes place and air will be saturated at
20 °C.
(i)∴ Relative humidity of exit air is 100 per cent. (Ans.)
Since the air is saturated, wet bulb temperature is equal to dry bulb tempera-
ture = 20°C. (Ans.)
∴ pv = pvs = 0.0234 bar.
W 2 =
0 622 0 622 0 0234
1 0132 0 0234
...
(.. )
p
pp
v
tv−
= ×
− = 0.0147 kg/kg of dry air
h 2 = cp tdb 2 + W 2 [hg 2 + 1.88 (tdb 2 – tdp 2 )]
= 1.005 × 20 + 0.0147 [2538.1 + 1.88 (20 – 20)]
[Q When air is saturated tdb = tdp]
= 57.41 kJ/kg of dry air
The weight of water vapour removed per kg of dry air
= 0.0177 – 0.0147 = 0.003 kg/kg of dry air
Heat removed per kg of dry air
= h 1 – h 2 = 80.98 – 57.41 = 23.57 kJ/kg of dry air
Mass of dry air passing per minute
ma =
pV
RT
aa
aa
= −××
×+
()
()
- 0.0132 02815 10 120
287 35 273
5
= 133.7 kg/min
(ii)Capacity of the cooling coil in tonnes of refrigeration
=mh ha()^12 − = ××
14000
133.7 23.57 60
14000 = 13.5 TR. (Ans.)
(iii)Amount of water removed per hour
= ma (W 1 – W 2 ) × 60
= 133.7 (0.0177 – 0.0147) × 60 = 24.066 kg/h. (Ans.)