Question

In an air conditioning space fresh air enters at ${45}^{\circ }$C and 30% relative humidity. Water has a saturation pressure of 0.09584 bar corresponding to ${45}^{\circ }$C. If superheated water vapour is assumed as ideal gas the enthalpy of moist air is equal to ______ (Take $P_{atm}=100kPa$)

• A. 92.8 kJ/kg of da
• B. 52.8 kJ/kg of da
• C. 45.2 kJ/kg of da
• D. 47 kJ/kg of da

Answer 1

The correct option is A 92.8 kJ/kg of da

Given: T = ${45}^{\circ }$C,
$p_{vs}$ = 0.09584 bar

Relative humidity, $\varphi$ = 30%

$P_{atm}$ = 100 kPa = 1 bar

$\varphi$ = $\frac{p_v}{p}$

$p_v=0.3\times 0.09584=0.028752$ bar

humidity ratio, $\omega$ = 0.622 $\times \frac{p_v}{p-p_n}$

= 0.622 $\times \frac{0.028752}{1-0.028752}$

= 0.0184 kg/kg da

Now,
enthalpy, h = 1.005t + $\omega (2500+1.88t)$
= 1.005 $\times$ 45 + 0.0184$(2500+1.88\times 45)$
= 92.816 kJ/kg da