Adiabatic Dehumidification of Moist Air: 04-BS-10, May 2017, Q9

Question 9: Adiabatic Dehumidification of Moist Air (15 marks)

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Given. Steady-flow dehumidifier, $P=1$ atm. Inlet: $T_1=35\ ^\circ$C, $RH_1=50\%$. Outlet moist-air stream: saturated ($RH_2=100\%$), $T_2=15\ ^\circ$C. Condensate: saturated liquid water, $T_2=15\ ^\circ$C.

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

Fig. Q9 — Steady-flow dehumidifier: moist air enters at 35°C/50% RH, exits saturated at 15°C alongside a separate liquid condensate stream at 15°C.

Approach

Evaluate humidity ratio and mixture enthalpy at the inlet and (saturated) outlet states directly via the ASHRAE moist-air formulation. The water balance gives the condensed mass directly from the drop in humidity ratio; the energy balance on the whole dehumidifier (moist air in, saturated moist air out, liquid condensate out) gives the heat rejected.

Inlet state. At $T_1=35\ ^\circ$C, $RH_1=50\%$, $P=101.325$ kPa: $$W_1=0.017852\ \text{kg/kg}_{da},\qquad H_1=81.00\ \text{kJ/kg}_{da}.$$
Outlet state. Saturated moist air at $T_2=15\ ^\circ$C: $$W_2=0.010694\ \text{kg/kg}_{da},\qquad H_2=42.12\ \text{kJ/kg}_{da}.$$
Water condensed (part b). Dry-air mass flow is conserved through the unit (only water crosses between the vapor and liquid streams): $$w_{cond}=W_1-W_2=0.017852-0.010694=\boxed{0.007159\ \text{kg}_{water}/\text{kg}_{da}}.$$
Heat transfer (part a). Condensate leaves as saturated liquid at 15°C, $h_{f}=62.98$ kJ/kg. Energy balance per kg dry air (moist air in $=$ saturated moist air out $+$ condensate out $+$ heat rejected): $$H_1=H_2+w_{cond}\,h_f+q_{out}$$ $$q_{out}=H_1-H_2-w_{cond}\,h_f=81.00-42.12-0.007159\times62.98=\boxed{38.44\ \text{kJ/kg}_{da}}.$$

Quantity	Result
(a) $q_{out}$	38.44 kJ/kg dry air
(b) $w_{cond}$	0.007159 kg/kg dry air

Concept & Theory

Topic. Steady-flow dehumidification: moist air is cooled below its dew point, condensing part of its water vapor, which exits as a separate liquid stream.

Key relations. Water (species) balance: $w_{cond}=W_1-W_2$. Overall energy balance per kg dry air: $H_1=H_2+w_{cond}h_f+q_{out}$.

Why this works. Expressing everything per unit mass of DRY air is what makes the balances linear and simple — the dry-air mass flow is the one quantity guaranteed constant through the device (water vapor condenses out, but no dry air is created or destroyed), so all humidity ratios and enthalpies are naturally normalized against it.

Common pitfall. Forgetting the condensate's own enthalpy term ($w_{cond}h_f$) in the energy balance and simply computing $q_{out}=H_1-H_2$ — this understates the heat rejected, since the condensing water also carries away (relatively small, but non-zero) enthalpy as a liquid.

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

Question 9 of 9: Adiabatic Dehumidification of Moist Air

Question 9: Adiabatic Dehumidification of Moist Air (15 marks)

Approach