Enthalpy of Moist Air:

In accordance to Gibb's law, the enthalpy of a combination of perfect gases might be got by the total summation of the enthalpies of the respective constituents. Thus the enthalpy of the moist air h is equivalent to the summing up of the enthalpies of dry air and of the water vapour related with the air. Consequently,

                                                       h=h_a+ wh_v ...6.22

Per kg of dry air, here h_a is the enthalpy of the dry air part & wh_v is the enthalpy of the water vapour part. The modification in enthalpy of a perfect gas being referred only as a function of temperature, the enthalpy of the dry air part above a datum of 0⁰C is expressed as following:

h_a=C_pa t=1.005 t kJ/kg (=0.24 t Btu/lbm where t is in ⁰F)

here Cp= 1.005 kJ/kg.K refer for the specific heat of dry air, & t refer for the dry-bulb temperature of air in ⁰C.

Suppose the reference state enthalpy as zero for saturated liquid at 0⁰C, the enthalpy of water vapour on pt A in the above given Fig may be expressed as following:

h_v= h_A= C_pw t_d+ (h_fg)d + C_pv (t -  t_d) kJ/kg         ...6.24

here C_pw= specific heat of liquid water

         t_d = dew point temperature

   (h_fg)d = latent heat of vaporization at DPT

        C_pv=specific heat of superheated vapour

 By taking the specific heat of liquid water like 4.1868 kJ/kg K and that of water vapour is 1.88kJ/kg K, in the range 0 to 60⁰C, we have the following eq.

                                h_v=4.1868 t_d + (h_fg)d + 1.88 (t -t_d)

For an ideal gas at low pressure, the enthalpy is only a function of temperature. Therefore in given fig the enthalpies at point B & C are also the similar as the enthalpy at A. Hence, enthalpy of water vapour at point A, at DPT of t_d and DBT of t, might be find out more conveniently through the following two methods:

 (1)       h_A = h_C = (h_g)t                                      ...6.26

 (b)       h_A = h_B= (h_fg)0^o C+ C_pv ( t - 0)  ...6.27

           Figure: Evaluation of Enthalpy of Water Vapour Part

By using second expression and by taking the latent heat of vaporization of water at 0⁰C as 2501 k/kgK, we get the empirical expression for the enthalpy of the water vapour part

                       h_v=2501 + 1.88t kJ/kg  ...6.28

And joining Eq 6.23 & 6.24, we have the enthalpy of moist air

                        h =1.005t + ω (2500 + 1.88t) kJ/kg d.a.          ...6.29