Question 13.3: Moist air at 310 K has a wet-bulb temperature of 300 K. If t......
Moist air at 310 K has a wet-bulb temperature of 300 K. If the latent heat of vaporisation of water at 300 K is 2440 kJ/kg, estimate the humidity of the air and the percentage relative humidity. The total pressure is 105 kN/m² and the vapour pressure of water vapour at 300 K is 3.60 kN/m² and 6.33 kN/m² at 310 K.
The humidity of air saturated at the wet-bulb temperature is given by:
\mathscr{H} _{w}=\frac{P_{\mathrm{w0}}}{P-P_{\mathrm{w0}}}\frac{M_{\mathrm{w}}}{M_{A}} (equation 13.2)
= \left(\frac{3.6}{105.0-3.6}\right)\left(\frac{18}{29}\right)=0.0220\mathrm{~kg/kg}
Therefore, taking (h/h_{D}\rho _{A}) as 1.0 kJ/kg K, in equation 13.8:
(\mathscr{H} -\mathscr{H} _{w})=-\frac{h}{h_{D}\rho_{A}\lambda}(\theta-\theta_{w}) (13.8)
(0.0220-\mathscr{H} )=\left(\frac{1.0}{2440}\right)(310-300)or: \mathscr{H} =\underline{\underline{0.018\ kg/kg}}
At 310 K, P_{w0} = 6.33 kN/m²
In equation 13.2: 0.0780={\frac{18P_{\mathrm{w}}}{(105.0-P_{\mathrm{w}})29}}
\mathscr{H} _{w}=\frac{P_{\mathrm{w0}}}{P-P_{\mathrm{w0}}}\frac{M_{\mathrm{w}}}{M_{A}} (13.2)
∴ P_{w} = 2.959 kN/m²
and the percentage relative humidity = \frac{(100\times2.959)}{6.33} = \underline{\underline{46.7\ per\ cent}}