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PSYCHROMETRICS 471

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Example 10.9. Adiabatic humidification : 150 m^3 of air per minute is passed through
the adiabatic humidifier. The condition of air at inlet is 35°C DBT and 20 per cent relative
humidity and the outlet condition is 20°C DBT and 15°C WBT.
Determine the following :
(i)Dew point temperature (ii)Relative humidity of the exit air
(iii)Amount of water vapour added to the air per minute.
Solution. For air at 35°C DBT and 20% relative humidity.
pvs = 0.0563 bar (At 35ºC from steam tables)
pv = φ × pvs = 0.2 × 0.0563 = 0.01126 bar

W 1 =

0 622 0 622 0 01126

1 0132 0 01126

...

..

p

pp

v

tv−

= ×
− = 0.00699 kg/kg of dry air
(i) The dew point temperature of air which is the saturation temperature of steam corre-
sponding to the pressure 0.01126 bar is

8 + (9 – 8) ×

(.. )

(.. )

0 01126 0 01072

0 01150 0 01072

−

− = 8.7ºC

i.e., Dew point temperature = 8.7°C. (Ans.)

(ii)Relative humidity of the exit air :

For air at 20ºC DBT and 15ºC WBT.

pv = (pvs)wb – [()]( )

..

pp tt

t

tvswbdbwb

wb

−−

1527 4 1 3−

= 0.0170 – [.. ]( )

..

1 0132 0 0170 20 15

1527 4 1 3 15

−−

−×

= 0.0137 bar

W 2 = 0 622 0 622 0 0137

1 0132 0 0137

...

(.. )

p

pp

v

tv−

=

×

−

= 0.00852 kg/kg of dry air

Relative humidity = p

p

v

vs

=

0 0137

0 0234

.

.

= 0.585 or 58.5%. (Ans.)

(Q pvs = 0.0234 bar, corresponding to 20ºC, from steam tables)

The dew point temperature of air which is the saturation temperature of steam correspond-

ing to 0.0137 bar is 11 °C (from steam tables). (Ans.)

The amount of water vapour per kg of dry air

= W 2 – W 1 = 0.00852 – 0.00699 = 0.00153 kg

The mass of dry air in 150 m^3 of mixture

ma =

pV

RT

aa

aa

=

−××

×+

(.. )

()

1 0132 0 01126 10 150

287 35 273

5
= 170 kg
(iii)The amount of water vapour added to air per minute
= ma (W 2 – W 1 ) = 170 × 0.00153 = 0.26 kg/min. (Ans.)
Example 10.10. Adiabatic saturation process : An air-water vapour mixture enters an
adiabatic saturation chamber at 28°C and leaves at 18°C, which is the adiabatic saturation
temperature. The pressure remains constant at 1.0 bar.
Determine the relative humidity and humidity ratio of the inlet mixture.
Solution. The specific humidity at the exit

W

p

s pp

s

ts

2

0 622 0 622 0 0206

1 00 0 0206

=

−

=

×

−

...

(.. ) = 0.01308 kg/kg of dry air