Cooling and dehumidifying coil load

Entering air state 1 (30 °C, 60 %)

$p_s = 4.247$ kPa, $p_v = 0.6 \times 4.247 = 2.548$ kPa.

$$\omega_1 = 0.01605\ \text{kg/kg},\quad h_1 = 71.19\ \text{kJ/kg},\quad v_1 = 0.8808\ \text{m}^3/\text{kg}$$ $$\dot m_a = \frac{120}{0.8808 \times 60} = 2.271\ \text{kg/s}$$

ADP state A (8 °C, saturated)

$p_s(8\ ^\circ\text{C}) = 1.072$ kPa, so $\omega_A = 0.00665$ kg/kg and $h_A = 24.77$ kJ/kg.

(a) Leaving state 2

$$\text{BPF} = \frac{t_2 - t_A}{t_1 - t_A} = \frac{\omega_2 - \omega_A}{\omega_1 - \omega_A} = 0.25$$ $$\begin{aligned} t_2 &= 8 + 0.25(30 - 8) = 13.5\ ^\circ\text{C}\\ \omega_2 &= 0.00665 + 0.25(0.01605 - 0.00665) = 0.00900\ \text{kg/kg}\\ h_2 &= 1.005 \times 13.5 + 0.00900(2501 + 1.88 \times 13.5) = 36.30\ \text{kJ/kg} \end{aligned}$$

Relative humidity of the leaving air: $p_{v2} = 1.445$ kPa and $p_s(13.5\ ^\circ\text{C}) \approx 1.548$ kPa, so $\phi_2 \approx 0.93$ (about 93 %).

(b) Loads

Intermediate state $h'$ at ($t_2$, $\omega_1$): $h' = 1.005 \times 13.5 + 0.01605(2501 + 25.4) = 54.11$ kJ/kg.

$$\begin{aligned} Q_{total} &= \dot m_a(h_1 - h_2) = 2.271 \times (71.19 - 36.30) = 79.2\ \text{kW} = 22.5\ \text{TR}\\ Q_{sensible} &= \dot m_a(h_1 - h') = 2.271 \times (71.19 - 54.11) = 38.8\ \text{kW}\\ Q_{latent} &= \dot m_a(h' - h_2) = 2.271 \times (54.11 - 36.30) = 40.4\ \text{kW} \end{aligned}$$

(The enthalpy carried away by the condensate, under 1 kW, is small and neglected.)

(c) Sensible heat factor

$$\text{SHF} = \frac{Q_s}{Q_{total}} = \frac{38.8}{79.2} = 0.49$$

(d) Condensate

$$\dot m_w = \dot m_a(\omega_1 - \omega_2) = 2.271 \times (0.01605 - 0.00900) = 0.0160\ \text{kg/s} = 57.6\ \text{kg/h}$$

Answer: (a) 13.5 °C, $\omega$ = 0.00900 kg/kg, RH about 93 %; (b) total 79.2 kW (22.5 TR), sensible 38.8 kW, latent 40.4 kW; (c) SHF = 0.49; (d) 57.6 kg/h.