Determine specific enthalpy of moist air with DB 70F and WB

Determine specific enthalpy of moist air with DB = 70°F and WB = 60°F with sea level atmospheric pressure (do not use psychrometric chart to find the enthalpy, but you can use it to find W).

W=0.008 from psychrometric chart

Solution

Specific enthalpy of moist air can be expressed as:

h = h_a + x h_w         (1)

where

h = specific enthalpy of moist air (kJ/kg, Btu/lb)

h_a = specific enthalpy of dry air (kJ/kg, Btu/lb)

x = humidity ratio (kg/kg, lb/lb)

h_w = specific enthalpy of water vapor (kJ/kg, Btu/lb)

the specific enthalpy of dry air can be expressed as:

h_a = c_pa t         (2)

where

c_pa = specific heat of air at constant pressure (kJ/kg^oC, kWs/kgK, Btu/lb^oF)

t = air temperature (^oC, ^oF)

For air temperature between -100^oC (-150^oF) and 100^oC (212^oF) the specific heat can be set to

c_pa = 1.006 (kJ/kg^oC)

    = 0.240 (Btu/lb^oF)

the specific enthalpy of water vapor can be expressed as:

h_w = c_pw t + h_we         (3)

where

c_pw = specific heat of water vapor at constant pressure (kJ/kg^oC, kWs/kgK)

t = water vapor temperature (^oC)

h_we = evaporation heat of water at 0^oC (kJ/kg)

For water vapor the specific heat can be set to

c_pw = 1.84 (kJ/kg^oC)

    = 0.444 (Btu/lb^oF)

The evaporation heat (water at 0^oC) can be set to

h_we = 2501 (kJ/kg)

    = 1075 (Btu/lb)

h = (0.240 Btu/lb^oF) t + x [(0.444 Btu/lb^oF) t + (1061 Btu/lb)]         (1d)

where

h = enthalpy (Btu/lb)

x = mass of water vapor (lb/lb)

t = temperature (^oF)

h=(0.240 Btu/lb^oF) 70 + 0.008 [(0.444 Btu/lb^oF) 70 + (1061 Btu/lb)]

=16.8+0.248 +1061

answer = 1078.048