Air Dehumidifier: 04-BS-10, December 2019, Q6

Question 6: Air Dehumidifier (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: $T_1=25^\circ$C, $P=1$ atm, $\phi_1=65\%$. Exit moist-air stream: saturated at $T_2=12^\circ$C, same pressure. Condensate exits at $T_2=12^\circ$C as saturated liquid. KE/PE negligible.

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

Approach

Get the inlet and exit humidity ratios from the given/saturation relative humidities, take the difference as the condensed water per kg dry air, then close a steady-flow energy balance (moist-air enthalpy in vs. moist-air enthalpy out plus the condensate's enthalpy) to isolate the heat rejected.

Humidity ratios. Inlet at 65% RH, 25°C; exit saturated at 12°C: $$\omega_1=0.6222\,\frac{\phi_1 P_{sat}(25^\circ\text{C})}{P-\phi_1P_{sat}(25^\circ\text{C})}=0.012966\ \text{kg/kg-da}$$ $$\omega_2=0.6222\,\frac{P_{sat}(12^\circ\text{C})}{P-P_{sat}(12^\circ\text{C})}=0.008768\ \text{kg/kg-da}.$$
Water condensed (part b). $$w_{cond}=\omega_1-\omega_2=0.012966-0.008768=\boxed{0.004198\ \text{kg/kg-da}}.$$
Energy balance for heat rejected (part a). Per kg dry air, moist-air enthalpy in equals moist-air enthalpy out, plus the condensate's enthalpy, plus the heat removed: $$q_{out}=h_1-h_2-w_{cond}\,h_{f}(12^\circ\text{C})=(h_1-h_2)-0.004198(50.4)$$ Evaluating the moist-air enthalpies (dry-air + water-vapor contributions) at states 1 and 2 gives $$q_{out}=\boxed{23.77\ \text{kJ/kg-da}}.$$

Quantity	Result
(a) $q_{out}$	23.77 kJ/kg dry air
(b) $w_{cond}$	0.004198 kg water/kg dry air

Concept & Theory

Topic. Steady-flow dehumidification: cooling moist air below its dew point to condense out excess moisture.

Key relations. Humidity ratio $\omega=0.622\,P_v/(P-P_v)$; mass balance on water $w_{cond}=\omega_1-\omega_2$; energy balance $h_1=h_2+w_{cond}h_f+q_{out}$ (condensate leaves at the exit temperature as saturated liquid).

Why this works. Cooling moist air along a constant-pressure line eventually hits the dew point; further cooling forces excess vapor to condense, so the EXIT air stream is saturated (100% RH) at the lower exit temperature, while the condensate leaves as a separate liquid stream in thermal equilibrium with it.

Common pitfall. Forgetting the condensate's own enthalpy term in the energy balance — it is a comparatively small correction here, but omitting it entirely biases $q_{out}$ the wrong way.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 14 (Psychrometrics, Dehumidification).

Question 6 of 9: Air Dehumidifier

Question 6: Air Dehumidifier (15 marks)

Approach