TITLE.PM5

466 ENGINEERING THERMODYNAMICS

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\M-therm\Th10-1.pm5

Solution. Corresponding to 30ºC, from steam tables, pvs = 0.0425 bar

∴ Relative humidity (R.H.), φ =

p

p

v

vs

i.e., 0.55 =

pv

0 0425.

∴ pv = 0.02337 bar.

Also the specific humidity,

W =

0 622 0 622 0 02337

1 0132 0 02337

...

..

p

pp

v

tv−

= ×
− = 0.01468 kg/kg of dry air.
The specific humidity after removing 0.004 kg of water vapour becomes,
0.01468 – 0.004 = 0.01068 kg/kg of dry air
and the temperature tdb is given as 20ºC.
The partial pressure of water vapour, pv, at this condition can be calculated as follows :

W

p

pp

v

tv

=

−

0 622.

0.01068 =

0 622

1 0132

.

.

p

p

v

− v

or, 0.01068 (1.0132 – pv) = 0.622 pv

or, 0.01082 – 0.01068 pv = 0.622 pv

0.6327 pv = 0.01082

∴ pv = 0.0171 bar

Corresponding to 20ºC, from steam tables, pvs = 0.0234 bar.

(i)Relative humidity, φ = p

p

v

vs

=0 0171

0 0234

.

.

= 0.73 or 73%. (Ans.)

(ii)Dew point temperature, tdp :

Corresponding to 0.0171 bar, from steam tables, tdp = 15 °C. (Ans.)

Example 10.4. The sling psychrometer in a laboratory test recorded the following readings :

Dry bulb temperature = 35°C

Wet bulb temperature = 25°C.

Calculate the following :

(i)Specific humidity (ii)Relative humidity

(iii)Vapour density in air (iv)Dew point temperature

(v)Enthalpy of mixture per kg of dry air

Take atmospheric pressure = 1.0132 bar.

Solution. For finding the partial pressure of vapour, using the equation :

pv = (pvs)wb – [()]( )

..

pp tt

t

tvswbdbwb

wb

−−

1527 4 1 3−

Corresponding to 25ºC (from steam tables),

(pvs)wb = 0.0317 bar

Substituting the values in the above equation, we get

pv = 0.0317 –

[.. ]( )

..

1 0132 0 0317 35 25

1527 4 1 3 25

−−

−×

= 0.0317 – 0.0065 = 0.0252 bar.