TITLE.PM5

474 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th10-2.pm5

At 18ºC :

hg 2 = 2534.4 kJ/kg, pvs = 0.0206 bar

∴ W 1 =

0 622 0 01657

1 0132 0 01657

..

..

×

− = 0.01037 kg/kg of dry air

Since enthalpy remains constant during the process

∴ ctpdb 1 + Wh 1 g 1 = ctpdb 2 + Wh 2 g 2

1.005 × 38 + 0.01034 × 2570.7 = 1.005 × 18 + W 2 × 2534.4

(Q At 18ºC, hg 2 = 2534.4 kJ/kg)

i.e., W 2 =

1 005 38 0 01034 2570 7 1 005 18

2534 4

.. ..

.

×+ × − ×

= 0.01842 kg/kg of dry air

∴ Amount of water added = W 2 – W 1 = 0.01842 – 0.01034

= 0.00808 kg/kg of dry air. (Ans.)

Also, 0.00808 =

0 622

1 0132

2

2

.

.

p

p

v

− v

or 0.00808 (1.0132 – pv 2 ) = 0.622 pv 2

0.00818 – 0.00808 pv 2 = 0.622 pv 2

∴ pv 2 = 0.01298 bar

∴ Final relative humidity =

0 01298

0 0206

.

. = 0.63 or 63%. (Ans.)
Example 10.13. Saturated air at 3ºC is required to be supplied to a room where the
temperature must be held at 22ºC with a relative humidity of 55%. The air is heated and then
water at 10ºC is sprayed to give the required humidity. Determine :
(i)The mass of spray water required per m^3 of air at room conditions.
(ii)The temperature to which the air must be heated.
Neglect the fan power. Assume that the total pressure is constant at 1.0132 bar.
Solution. (i) The flow diagram is shown in Fig. 10.17 (a) and the processes are shown in
Fig. 10.17 (b).

Room

22 C,

55 % R.H.

º

Saturated air

at 3 Cº

1 Heater 2 3

4

Spray

(a)