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468 ENGINEERING THERMODYNAMICS

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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.)

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