Window Air Conditioner — Psychrometric Dehumidification: 04-BS-10, Undated paper, Q9

Question 9: Window Air Conditioner — Psychrometric Dehumidification (Part B – 15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Inlet: $T_1=32\ ^\circ\text{C}$, $\phi_1=70\%$, $\dot V_1=3\ \text{m}^3/\text{min}=0.05\ \text{m}^3/\text{s}$, $P=101.325$ kPa. Exit: SATURATED air at $T_2=12\ ^\circ\text{C}$, i.e. $\phi_2=100\%$. Condensate removed as saturated liquid at $T_{cond}=12\ ^\circ\text{C}$ (the same temperature, since it leaves in equilibrium with the exit air).

Find. Rate of heat removal $\dot Q$; rate of moisture removal $\dot m_{w,removed}$.

Approach

Fix the humidity ratio at inlet and exit from the saturation-vapor-pressure/relative-humidity definition, then use dry-air mass conservation ($\dot m_a$ fixed at inlet from the given volumetric flow rate) and a moisture balance to get the condensate rate. An energy balance around the coil, including the condensate leaving at its own (lower) enthalpy, gives the heat-removal rate.

Humidity ratios. $P_{v,1}=\phi_1 P_{sat}(32\ ^\circ\text{C})=0.70\times4.760=3.332$ kPa; $\omega_1=0.622\,P_{v,1}/(P-P_{v,1})=\boxed{0.02115\ \text{kg/kg dry air}}$. The exit stream is SATURATED, so $P_{v,2}=P_{sat}(12\ ^\circ\text{C})=1.403$ kPa; $\omega_2=0.622\,P_{v,2}/(P-P_{v,2})=\boxed{0.008733\ \text{kg/kg dry air}}$.
Dry-air mass flow rate. Specific volume of the dry-air part of the inlet stream, $v_1=R_aT_1/(P-P_{v,1})=0.287\times305.15/(101.325-3.332)=0.8937\ \text{m}^3/\text{kg}$. $$\dot m_a=\frac{\dot V_1}{v_1}=\frac{0.05}{0.8937}=\boxed{0.05595\ \text{kg/s}}.$$
Moisture removal rate. $$\dot m_{w,removed}=\dot m_a(\omega_1-\omega_2)=0.05595\times(0.02115-0.008733) =\boxed{6.946\times10^{-4}\ \text{kg/s}}\ (0.0417\ \text{kg/min}=2.50\ \text{kg/h}).$$
Heat removal rate. Energy balance on the coil (moist air in = moist air out + condensate out + heat removed), with dry-air enthalpy from ideal-gas air tables (only the DIFFERENCE $h_{a,1}-h_{a,2}$ is physical) and the water vapour taken at its saturated-vapour enthalpy $h_g(T)$, which is the standard low-pressure moist-air approximation: $$\dot Q_{removed}=\dot m_a\big[(h_{a,1}-h_{a,2})+(\omega_1h_{v,1}-\omega_2h_{v,2})\big]-\dot m_{w,removed}\,h_{f,cond}$$ $$h_{a,1}-h_{a,2}=431.48-411.36=20.12\ \text{kJ/kg},\quad h_{v,1}=2559.15,\ h_{v,2}=2522.86,\ h_{f,cond}=50.41\ \text{kJ/kg}$$ $$\dot Q_{removed}=0.05595\times\big[20.12+(0.02115\times2559.15-0.008733\times2522.86)\big] -6.946\times10^{-4}\times50.41$$ $$\dot Q_{removed}=0.05595\times52.21-0.035=\boxed{2.886\ \text{kW}}\ \text{(heat leaves the air stream)}.$$ Sense check on the split: the sensible part is $\dot m_ac_p\Delta T\approx0.05595\times1.005\times20 =1.12$ kW and the latent part is $\dot m_{w}h_{fg}\approx6.946\times10^{-4}\times2472=1.72$ kW, which sum to 2.84 kW, within 1.5% of the full balance above (the residual is the temperature dependence of $h_g$, which the constant-$h_{fg}$ shortcut ignores) — confirming that most of this coil's duty is latent.

Quantity	Value
$\dot m_{w,removed}$ (moisture removal rate)	$6.946\times10^{-4}$ kg/s (0.0417 kg/min)
$\dot Q_{removed}$ (heat removal rate)	2.886 kW

Topic. Psychrometric dehumidification — combined sensible + latent cooling coil.

Key relations. $\omega=0.622\,P_v/(P-P_v)$; dry-air mass conservation $\dot m_a=\text{const}$; moisture balance $\dot m_{w,removed}=\dot m_a(\omega_1-\omega_2)$; energy balance including condensate enthalpy leaving the control volume.

Why this works. Dry air never leaves the control volume by any path other than the main duct, so its mass flow rate is the one invariant that lets every other stream (vapor in, vapor out, liquid condensate out) be referenced per kg of dry air — this is exactly why psychrometric properties are always defined "per kg dry air" rather than per kg of the mixture.

Common pitfall. Reading "leaves as saturated air at 12°C" as anything other than $\phi_2=100\%$ — the exit humidity ratio is fixed by the exit temperature alone, and guessing a relative humidity below 100% removes less moisture and understates the latent load. Second trap: omitting the condensate's own enthalpy leaving the control volume from the energy balance — it is a small term numerically but is required by the full steady-flow energy balance, and dropping it (or using the WRONG condensate temperature) shifts $\dot Q_{removed}$ by a percent or so.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., §14-5 (air conditioning processes).

Question 9 of 9: Window Air Conditioner — Psychrometric Dehumidification

Question 9: Window Air Conditioner — Psychrometric Dehumidification (Part B – 15 marks)

Approach