TITLE.PM5

(Ann) #1
472 ENGINEERING THERMODYNAMICS

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\M-therm\Th10-1.pm5

The specific humidity at the inlet (equation 10.18)

W 1 =

ct t W h h
hh

p db db s g f
g f

()() 21 22
12

−+ 2 −

= −+ −

1 005 18 28 0 01308 2534 4 75 6
2552 6 75 6

.( ). (. .)
(. .)

=42 211
2477

. = 0.01704 kg/kg of dry air


W 1 =

0 622 1
1

. p
pp


v
tv−

∴ 0.01704 =

0 622
100

1
1

.
.

p
p

v
− v
or 0.01704 (1.00 – pv 1 ) = 0.622 pv 1
or 0.01704 – 0.01704 pv 1 = 0.622 pv 1
or 0.0639 pv 1 = 0.01704
∴ pv 1 = 0.02666 bar

∴ Relative humidity =

p
p

v
s

1
1

0 02666
0 03782
=.

. = 0.7 or 70%. (Ans.)
Example 10.11. An air-water vapour mixture enters an air-conditioning unit at a pressure
of 1.0 bar. 38°C DBT, and a relative humidity of 75%. The mass of dry air entering is 1 kg/s. The
air-vapour mixture leaves the air-conditioning unit at 1.0 bar, 18°C, 85% relative humidity. The
moisture condensed leaves at 18°C. Determine the heat transfer rate for the process.


Solution. tdb 1 = 38ºC, R.H., φ 1 = 75%
ttb 2 = 18ºC, R.H., φ 2 = 85%
The flow diagram and the process are shown in Figs. 10.16 (a) and (b) respectively.
At 38°C
From steam tables : pvs = 0.0663 bar, hg 1 = 2570.7 kJ/kg
∴ pv = φ × pvs = 0.75 × 0.0663 = 0.0497 bar

W 1 =

0 622 0 0497
1 0 0 0497

..
..

×
− = 0.0325 kg/kg of dry air
At 18°C
From steam tables : pvs = 0.0206 bar, hg 2 = 2534.4 kJ/kg
hf 2 = 75.6 kJ/kg
pv = 0.85 × 0.0206 = 0.01751 bar

W 2 =

0 622 0 01751
1 0 01751

..
.

×
− = 0.01108 kg/kg of dry air
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