Air Conditioner — Cooling and Dehumidification of Moist Air: 04-BS-10, December 2014, Q4

Question 4: Air Conditioner — Cooling and Dehumidification of Moist Air (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: $T=20\ ^\circ$C, $P=1$ atm, $\phi=65\%$. Exit moist-air stream: saturated, $T=12\ ^\circ$C. Condensate exit: saturated liquid water at $12\ ^\circ$C. Steady flow, per kg dry air.

Find. (a) $q_{out}$ [kJ/kg dry air]; (b) $w_{condensed}$ [kg/kg dry air].

Approach

Evaluate the humidity ratio and moist-air enthalpy per kg dry air at inlet and exit using the ASHRAE humid-air property formulation, then close a combined water and energy balance across the coil: the water lost from the air stream leaves as the condensate, and the energy it carries away (at the condensate's own enthalpy) must be subtracted from the inlet–outlet air-side enthalpy difference to get the coil's heat-transfer duty.

Inlet state. At $20\ ^\circ$C, 1 atm, $\phi=65\%$: humidity ratio $W_1=0.009515$ kg/kg dry air, moist-air enthalpy $H_1=44.258$ kJ/kg dry air.
Exit state. Saturated at $12\ ^\circ$C, 1 atm: $W_2=0.008768$ kg/kg dry air, $H_2=34.181$ kJ/kg dry air.
Water condensed (part b). Per kg dry air, the vapor lost from the air stream equals the drop in humidity ratio: $$w_{condensed}=W_1-W_2=0.009515-0.008768=\boxed{0.000748\ \text{kg/kg dry air}}.$$
Condensate enthalpy and coil energy balance (part a). The condensate leaves as saturated liquid water at $12\ ^\circ$C: $h_{cond}=50.41$ kJ/kg. The steady-flow energy balance per kg dry air ($H_1=H_2+w_{condensed}\,h_{cond}+q_{out}$) gives $$q_{out}=H_1-\left(H_2+w_{condensed}\,h_{cond}\right)=44.258-\left(34.181+0.000748\times50.41\right)=\boxed{10.039\ \text{kJ/kg dry air}}.$$

Quantity	Result
(a) $q_{out}$	10.039 kJ/kg dry air
(b) $w_{condensed}$	0.000748 kg/kg dry air

Concept & Theory

Topic. Steady-flow cooling-and-dehumidification (air-conditioning) psychrometric process with a two-stream exit (moist air + liquid condensate).

Key relations. Humidity ratio $W=0.622\,P_v/(P-P_v)$; combined water and energy balance across the cooling coil: $\dot m_{da}H_1=\dot m_{da}H_2+\dot m_{da}(W_1-W_2)h_{cond}+\dot Q_{out}$.

Why this works. Working per kg dry air (the one quantity conserved across the whole process, since dry air neither condenses nor is added) turns the two-stream exit into a single linear balance; the condensate must be assigned its own (liquid-water) enthalpy rather than folded into the exit moist-air enthalpy, since it leaves the control volume as a physically separate stream.

Common pitfall. Neglecting the condensate's enthalpy term entirely (i.e. computing $q_{out}=H_1-H_2$) — this is a small correction here (condensate carries away only about 0.04 kJ/kg dry air) but grows significant whenever the dehumidification duty is larger.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 14 (Gas–Vapor Mixtures and Air-Conditioning).

Question 4 of 9: Air Conditioner — Cooling and Dehumidification of Moist Air

Question 4: Air Conditioner — Cooling and Dehumidification of Moist Air (15 marks)

Approach