Question 13.6: Determination of Moist-Air Properties Calculate values of hu......
Determination of Moist-Air Properties
Calculate values of humidity ratio, specific volume, and enthalpy for air at dry-bulb temperature T_{db} = 60°F (15.6°C) and RH \phi = 80%.
Given: T_{db} = 60°F, \phi_{1} = 0.8, and gas constant R_{a} = 53.352 ft·lb_{f}/(lb_{m} · °R). Assumptions: The atmospheric pressure is the standard atmosphere, 14.696 psia.
Find: W, v, and h.
Lookup value from the steam tables (Table A3): saturated vapor pressure at 60°F is p_{sat} = 0.2563 psia.
The vapor pressure of the moist air is p_{vap} = p_{sat} . \phi = 0.2563 psia × 0.8 = 0.2050 psia.
From Equation 13.15, the humidity ratio is
W = 0.622 \times \left\lgroup \frac{\phi p_{sat}}{p_{tot} – \phi p_{sat}} \right\rgroup (13.15)
W = 0.622 \times \left\lgroup \frac{ p_{sat}}{p_{tot} – p_{sat}} \right\rgroup = 0.622 \times \left\lgroup \frac{0.205}{14.696 – 0.205} \right\rgroup= 0.0088 lb_{w}/lb_{a}
and the density
\frac{1}{v} = \rho = \frac{p_{a}}{R_{a} T_{db}} = \frac{(14.696 – 0.205) lb_{f}/in.^{2} \times 144 in.^{2}/ft^{2}}{53.352 ft \cdot lb_{f}/(lb_{a} \cdot °R) \times (60 + 459.67)°R}= 0.07526 lb_{a}/ft^{3}
Thus, specic volume is v = (1/0.07526) = 13.287 ft^{3}/lb_{a} (0.830 m^{3}/kg_{a}) .
Finally, enthalpy is determined from Equation 13.23:
= 0.240 Btu/(lb_{a} \cdot °F) \times 60°F + 0.0088 lb_{w}/lb_{a}
\times (1061.2 Btu/lb_{w} + 0.444 Btu/(lb_{w} \cdot °F) \times 60°F)
= 24.0 Btu/lb_{a} (55.8 kJ/kg_{a})
Often, we need to distinguish between the sensible and latent components of the enthalpy quantity. A simplified approach widely adopted by practicing HVAC engineers is to express enthalpy as
h = h_{sen} + h_{lat} – c_{p,m}(T_{db} _ T_{ref}) + W \cdot h_{liq-vap,ref} (13.24)
where the heat of vaporization is 2501.3 kJ/kg_{w} (corresponding to 0°C) in SI units and 1075 Btu/lb_{w} in IP units (corresponding to a reference temperature of 32°F).
The humid air specific heat at constant pressure c_{p,m} can be expressed as
= 1.0 kJ/(kg_{a} \cdot °C) + W kg_{w}/kg_{a} \times 1.86 kJ/(kg_{w} \cdot °C)
in SI units (kJ/kg_{a} \cdot °C)
= 0.24 Btu/(lb_{a} \cdot °F) + W lb_{w}/lb_{a} \times 0.444 Btu/(lb_{w} \cdot °F)
in IP units (Btu/lb_{a} \cdot °F) (13.25)
Note that since h_{liq-vap} depends on the temperature at which water evaporates into steam and not at a specic base temperature as assumed, these simplied expressions tend to introduce small errors into how the total energy is split into the sensible and latent components.