TITLE.PM5

472 ENGINEERING THERMODYNAMICS

dharm

\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