Question

Dew point temperature of air at one atmospheric pressure (1.013 bar) is ${18}^{\circ }C$. The air dry bulb temperature is ${30}^{\circ }C$. The saturation pressure of water at ${18}^{\circ }Cand{30}^{\circ }C$ are 0.02062 bar and 0.04241 bar respectively. The specific heat of air and water vapour respectively are 1.005 and 1.88 kJ/kgK and the latent heat of vaporization of water at $0^{\circ }C$ is 2500 kJ/kg. The specific humidity (kg/kg of dry air) and enthalpy (kJ/kg of dry air) of this moist air respectively, are

• A. 0.01051,52.64
• B. 0.01291,63.15
• C. 0.01481,78.60
• D. 0.01532,81.40

Answer 1

The correct option is B 0.01291,63.15

Given, p = 1.013 bar

$p_v=0.02062$ bar (at dew point)

We know that,

Specific humidity,
$\omega =\frac{0.622p_v}{p-p_v}=\frac{0.622\times 0.020620}{1.013-0.02062}$

= 0.012924 kg/kg of dry air

Enthalpy: $h=1.005t_d+\omega (h_{fg}+1.88t_{db})$

$=1.005\times 30+0.01291(2500+1.88\times 30)$

$=1.005\times 30+33.00$

$=63.15\text{kJ/kg}ofdryair$

POINT TO REMEMBER

• It is very important to keep in mind that the temperature, used in the formula $=c_{pa}t+\omega (2500+1.88t),$ is the Dry bulb temperature.
• Enthalpy of moist air is the summation of enthalpy of dry and enthalpy of vapour.