Properties of moist air numerical

Given

$t = 32$ °C, $\phi = 0.55$, $p = 101.325$ kPa, $p_s = 4.759$ kPa.

(a) Partial pressure of vapour

$$p_v = \phi\,p_s = 0.55 \times 4.759 = 2.618\ \text{kPa}$$

(b) Specific humidity

$$\omega = 0.622\,\frac{p_v}{p - p_v} = 0.622 \times \frac{2.618}{101.325 - 2.618} = 0.01649\ \text{kg/kg dry air}$$

(c) Dew point

The dew point is the saturation temperature at $p_v = 2.618$ kPa. From steam tables (interpolating between 20 °C, 2.339 kPa and 22 °C, 2.645 kPa):

$$t_{dp} \approx 21.8\ ^\circ\text{C}$$

(d) Degree of saturation

$$\omega_s = 0.622\,\frac{4.759}{101.325 - 4.759} = 0.03066,\qquad \mu = \frac{\omega}{\omega_s} = \frac{0.01649}{0.03066} = 0.538$$

(e) Enthalpy

$$\begin{aligned} h &= 1.005\,t + \omega\,(2501 + 1.88\,t)\\ &= 1.005 \times 32 + 0.01649\,(2501 + 1.88 \times 32) = 74.4\ \text{kJ/kg dry air} \end{aligned}$$

(f) Specific volume

$$v = \frac{R_a T}{p - p_v} = \frac{0.287 \times 305.15}{101.325 - 2.618} = 0.887\ \text{m}^3/\text{kg dry air}$$

(The wet bulb temperature is about 24.6 °C from the adiabatic saturation equation.)

Answer: (a) 2.62 kPa; (b) 0.01649 kg/kg; (c) about 21.8 °C; (d) 0.538 (53.8 %); (e) 74.4 kJ/kg; (f) 0.887 m³/kg.