Question

Consider that specific heat (0 to \(50~^\circ\mathrm{C}\)) of water, water vapour and air remains constant: \(4.48\), \(1.88\) and \(1.0~\mathrm{kJ/(kg\^\circ C)}\), respectively. Assuming the heat energy required to convert \(1~\mathrm{kg}\) of water to water vapour at \(0~^\circ\mathrm{C}\) is \(2000~\mathrm{kJ}\), the enthalpy (in kJ/kg dry air) of atmospheric air containing \(0.05~\mathrm{kg}\) water vapour per kg dry air at \(50~^\circ\mathrm{C}\) is ________. (rounded off to 1 decimal place)

Hint:

For quick psychrometrics with reference at \(0^\circ\mathrm{C}\): \(h \approx c_{pa}T + \omega(L_0 + c_{pv}T)\). Only add a liquid-water sensible term if liquid water is carried with the air.

Answer 1

Correct Answer: 152

Step 1: Enthalpy model (reference at \(0~^\circ\mathrm{C}\)).
Moist-air enthalpy per kg dry air: \[ h = c_{pa}T \;+\; \omega\big(L_0 + c_{pv}T\big), \] where \(c_{pa}=1.0\), \(c_{pv}=1.88~\mathrm{kJ/(kg\^\circ C)}\), \(L_0=2000~\mathrm{kJ/kg}\), \(\omega=0.05\), \(T=50^\circ\mathrm{C}\). (There is no liquid water term since only vapour is present.) 

Step 2: Substitute values.
\[ h = (1.0)(50) + 0.05\big(2000 + 1.88\times 50\big) = 50 + 0.05(2000+94) = 50 + 104.7 = 154.7~\text{kJ/kg dry air}. \] Final Answer:
\[ \boxed{154.7~\text{kJ/kg dry air}} \]