Question

The enthalpy of moist air with normal notations is given by
1. h = (1.005 + 1.88ω)t + 2500ω 
2. h = 1.88ωt + 2500ω 
3. h = 1.005ωt
4. h = (1.88 + 1.005ω)t + 2500ω 

Answer 1

Correct Answer - Option 1 : h = (1.005 + 1.88ω)t + 2500ω 

Concept:

Enthalpy of moist air is given by:

\(h = {h_{air}} + \omega {h_{vapour}}\)

where, h_air = C_p,air × t

Calculation:

∴ h_air = 1.005 × t

and  h_vapour at any temperature is:

h_vapour = latent heat of vapour + C_{p,vapour ­}× t

h_vapour = 2500 + 1.88t

h_air = 1.005t + ω(2500 + 1.88t)

∴ h_air = (1.005 + 1.88ω)t + 2500ω 

Therefore option (1) is correct.