Question

Q.2 A sling Psychrometer reads, 40°c dry bulb
Temprature and 28 degree Celsius Wet bulb temp. Calculate:
(1)Specific humidity
(2) Relative humidity
(3) Vapour density
(4)Dew point temperature
(5) specific Enthalpy.​

Answer 1

Answer:

Q.2 A sling Psychrometer reads, 40°c dry bulb

Temprature and 28 degree Celsius Wet bulb temp. Calculate:

(1)Specific humidity

(2) Relative humidity

(3) Vapour density

(4)Dew point temperature

(5) specific Enthalpy.

Answer 2

Given,

In a sling Psychrometer,

Dry bulb temprature = 40°C

Wet-bulb temperature = 28°C

To find,

1) Specific humidity

2) Relative humidity

3) Vapour density

4) Dew point temperature

5) Specific Enthalpy.

Solution,

1) Specific humidity: The mass of water vapour in a unit mass of wet air, commonly expressed in grammes of vapour per kilogramme of air, is known as specific humidity.

$W=\frac{0.622p_{v}}{p_{t}-p_{v} }$

$=\frac{0.622*0.0252}{1.0312-0.0252}\\= 0.01586 kg/kg$ of dry air.

2) Relative humidity: The amount of water vapour in the air as a percentage of the amount required for saturation at the same temperature.

Ф = $\frac{p_{v} }{p_{vs} } =\frac{0.0252}{0.0563}$

$=0.447$ or 44.7%

3) Vapour density: At the same pressure and temperature, the density of a gas or vapour in comparison to hydrogen.

Using the characteristic gas equation, we have

$p_{v} =V_{v} =p_{v} R_{v} T_{v} \\p_{v} =\frac{m_{v} }{V_{v}} R_{v} T_{v}=p_{v} R_{v} T_{v}$

where, $p_{v}$= Vapour density

$R_{v}$=$\frac{Universal gas constant}{Molecular weight of H2O} =\frac{8314.3}{18}$

$0.0252$ × $10$= $p_{v}$ × $=\frac{8314.3}{18}$ × $(273+35)$

$p_{v}$ $=\frac {0.0252*10^5*18}{8314.3*308}$

$=0.0177kg/m^3$

4) Dew point temperature $t_{dp}$:  Under constant air pressure and water content, the dew point is the temperature at which air must be chilled to become saturated with water vapour.

Corresponding to 0.0252 bar, from steam tables

$t_{dp}=21+(22-21)$× $\frac{0.0252-0.0249}{0.0264-0.0249}$

$= 21.2$°C

5) Specific Enthalpy, h: The sum of the specific enthalpies of each component and their mass fractions is the specific enthalpy of a mixture of gases.

$h= 1.005t_{db} + W (2500+ 1.88 t_{db})\\= 1.005 * 35 +0.01586 [2500+ 1.88 * 35]\\= 35.175 + 40.69$

$= 75.86kJ/kg$ of dry air