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PSYCHROMETRICS 465

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0.009625 – 0.0095 pv = 0.622 pv

pv = 0.01524 bar. (Ans.)

(ii)Relative humidity φ :

Corresponding to 20ºC, from steam tables,

pvs = 0.0234 bar

∴ Relative humidity, φ =

p

p

v

vs

=0 01524

0 0234

.

. = 0.65 or 65%. (Ans.)
(iii)Dew point temperature, tdp :
The dew point temperature is the saturation temperature of water vapour at a pressure of
0.01524 bar,
tdp [from steam tables by interpolation]

= 13 +

()

(.. )

14 13

0 01598 0 0150

−

−

× [0.01524 – 0.0150]

= 13 +

0 00024

0 00098

.

. = 13.24°C. (Ans.)
Example 10.2. The air supplied to a room of a building in winter is to be at 17°C and have
a relative humidity of 60%. If the barometric pressure is 1.01325 bar, find : (i) The specific humid-
ity ; (ii) The dew point under these conditions.
Solution. Dry bulb temperature, tdb = 17ºC
Relative humidity, φ = 60%
Barometric or total pressure, pt = 1.01325 bar
Specific humidity, W :
Corresponding to 17ºC, from steam tables,
pvs = 0.0194 bar

Also, φ = p

p

v

vs

i.e., 0.6 =

pv

0 0194.

∴ pv = 0.6 × 0.0194 = 0.01164 bar.

Specific humidity, W =

0 622 0 622 0 01164

1 01325 0 01164

...

..

p

pp

v

tv−

= ×
−
= 0.00723 kg/kg of dry air. (Ans.)
Dew point temperature, tdp :
If the air is cooled at constant pressure the vapour will begin to condense at the saturation
temperature corresponding to 0.01164 bar. By interpolation from steam tables, the dew point
temperature, tdp is then

tdp = 9 + (10 – 9) ×

0 01164 0 01150

0 01230 0 01150

..

..

−
− = 9.18°C. (Ans.)
Example 10.3. 0.004 kg of water vapour per kg of atmospheric air is removed and tem-
perature of air after removing the water vapour becomes 20°C. Determine :
(i)Relative humidity (ii)Dew point temperature.
Assume that condition of atmospheric air is 30°C and 55% R.H. and pressure is 1.0132 bar.