Supply air quantity for a conditioned room

Room state

$$p_v = 0.5 \times 3.169 = 1.585\ \text{kPa},\qquad \omega_r = 0.622 \times \frac{1.585}{101.325 - 1.585} = 0.00988\ \text{kg/kg}$$ $$h_r = 1.005 \times 25 + 0.00988(2501 + 47) = 50.3\ \text{kJ/kg}$$

(a) Room sensible heat factor

$$\text{RSHF} = \frac{Q_s}{Q_s + Q_l} = \frac{18}{18 + 4.5} = 0.80$$

(b) Supply air mass flow (from the sensible load)

$$Q_s = \dot m_a\,c_{pm}(t_r - t_s),\qquad c_{pm} = 1.005 + 1.88\,\omega_s \approx 1.0215$$ $$\dot m_a = \frac{18}{1.0215 \times (25 - 14)} = 1.602\ \text{kg/s} = 5767\ \text{kg/h}$$

($\omega_s$ is found in step (c), and the two steps are iterated; the values converge after two or three trials.)

(c) Supply air humidity (from the latent load)

$$Q_l = \dot m_a(\omega_r - \omega_s)\,h_{fg}$$ $$\omega_r - \omega_s = \frac{4.5}{1.602 \times 2501} = 1.123 \times 10^{-3}$$ $$\omega_s = 0.00988 - 0.00112 = 0.00876\ \text{kg/kg}$$

Relative humidity at 14 °C: the vapour pressure of the supply air is $p_v = \omega_s p/(0.622 + \omega_s) = 1.407$ kPa, and the saturation pressure at 14 °C is about 1.598 kPa, so

$$\phi_s = \frac{p_v}{p_s} = \frac{1.407}{1.598} = 0.88\ \ (\approx 88\ \%)$$

The supply enthalpy is $h_s = 36.2$ kJ/kg.

(d) Volume flow of supply air

$$v_s = \frac{0.287 \times 287.15\,(1 + 1.608 \times 0.00876)}{101.325} = 0.8248\ \text{m}^3/\text{kg}$$ $$\dot V_s = 1.602 \times 0.8248 = 1.321\ \text{m}^3/\text{s}\ (\approx 79.3\ \text{m}^3/\text{min})$$

Answer: (a) RSHF = 0.80; (b) 1.60 kg/s; (c) $\omega_s$ = 0.00876 kg/kg, RH about 88 %; (d) about 1.32 m³/s (79 m³/min).