468 ENGINEERING THERMODYNAMICS
dharm
\M-therm\Th10-1.pm5
Enthalpy, h = cptdb + Whvapour
= 1.005 tdb + W [hg + 1.88 (tdb – tdp)]
= 1.005 × 35 + 0.0214 [2565.3 + 1.88 (35 – 26.1)]
= 90.43 kJ/kg of dry air.
For the air at 20°C DBT and 13°C dew point temperature :
pv is the vapour pressure corresponding to the saturation pressure of steam at 13ºC.
∴ pv = 0.0150 bar
W =
0 622 0 622 0 015
1 0132 0 015
...
..
p
pp
v
tv−
= ×
− = 0.00935 kg/kg of dry air
Enthalpy, h = cptdb + Whvapour
= 1.005 × 20 + 0.00935 [hg + 1.88 (tdb – tdp)]
= 20.1 + 0.00935 [2538.1 + 1.88 (20 – 13)]
= 43.95 kJ/kg of dry air
Now enthalpy per kg of moist air
=+L ×
NM
O
QP
1
3
90 43
1 0214
43 95 2
1 00935
.
.
.
.
= 58.54 kJ/kg of moist air
Mass of vapour/kg of moist air
=+
L ×
NM
O
QP
1
3
0 0214
1 0214
0 00935 2
1 00935
.
.
.
. = 0.01316 kg/kg of moist air
Specific humidity of mixture
−
0 01316
1 0 01316
.
. = 0.01333 kg/kg of dry air. (Ans.)
Example 10.6. Sensible heating : 90 m^3 of air per minute at 20°C and 75% R.H. is
heated until its temperature becomes 30°C. Calculate :
(i)R.H. of the heated air.(ii)Heat added to air per minute.
Solution. (i) For air at 20°C and 75% R.H. :
pvs = 0.0234 bar (from steam tables, at 20ºC)
pv = φ × pvs = 0.75 × 0.0234 = 0.01755 bar
tdp = 15 + (16 – 15) (.. )
(.. )
0 01755 0 017 ~
0 0182 0 017
−
−
−15.5ºC
W 1 =
0 622 0 622 0 01755
1 0132 0 01755
...
..
p
pp
v
tv−
= ×
− = 0.0109 kg/kg of dry air
Enthalpy, h 1 = cptdb + Whvapour
= 1.005 × 20 + 0.0109 [hg + 1.88 (tdb – tdp)]
= 1.005 × 20 + 0.0109 [2538.1 + 1.88(20 – 15.5)] = 47.85 kJ/kg of dry air
(i) Relative humidity of heated air :
For air at 30°C DBT :
Since the saturation pressure of water vapour at 30ºC is higher than the saturation pres-
sure of water vapour at 20ºC so it is sensible heating, where pv is same after heating.
∴ Relative humidity, φ =
p
p
v
vs
=
0 01755
0 0425
.
.
= 0.412 or 41.2%
(pvs = 0.0425 bar, corresponding to 30ºC)
i.e., Relative humidity of heated air = 41.2%. (Ans.)