Question 2 of 6: Cooling and Dehumidification of Humid Air

Question A2: Cooling and Dehumidification of Humid Air (Part A — equal value)

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. Moist air enters at $T_1=311$ K, $P=1$ atm, 97% relative humidity, at $\dot V=8.5$ m³/s (taken as the inlet volumetric flow). It leaves saturated at $T_2=291$ K. Property data as listed; $M_{H_2O}=18.02$ g/mol.

Find. (a) the water condensation rate in kg/min; (b) the total cooling duty expressed in tons of refrigeration.

Approach. Fix the conserved dry-air flow from the ideal-gas inlet state, set the exit vapour to saturation at 291 K, close a water balance for the condensate (part a), then evaluate an overall enthalpy balance using the stated dry-air correlation and steam-table water enthalpies (part b).

1. Inlet molar flow (ideal gas). At the inlet state, $$\dot n=\frac{P\dot V}{RT_1}=\frac{(101{,}325)(8.5)}{(8.314)(311)}=333.1\ \text{mol/s}.$$
2. Split into water and dry air (Dalton). With $y_w=0.97\,p^{sat}_1/P=0.97(0.0654)=0.06344$, $$\dot n_{w,\text{in}}=y_w\dot n=21.13\ \text{mol/s},\qquad \dot n_{\text{dry}}=\dot n-\dot n_{w,\text{in}}=311.96\ \text{mol/s}.$$ The dry-air flow is conserved through the cooler.
3. Exit water (saturation at 291 K). The leaving air is saturated, so its molar humidity is $$Y_2=\frac{p^{sat}_2}{P-p^{sat}_2}=\frac{0.0204}{1-0.0204}=0.02082,\qquad \dot n_{w,\text{out}}=Y_2\,\dot n_{\text{dry}}=6.497\ \text{mol/s}.$$
4. Condensate rate (water balance, part a). $$\dot n_{\text{cond}}=\dot n_{w,\text{in}}-\dot n_{w,\text{out}}=21.13-6.497=14.63\ \text{mol/s},$$ $$\dot m_{\text{cond}}=14.63\times0.01802\times60=\boxed{15.8\ \text{kg/min}}.$$
5. Enthalpy inputs (part b). Using $H_{\text{dry}}=0.029(T_{{}^\circ\text{C}}-25)$ kJ/mol with $T_1=37.85\,{}^\circ\text{C}$, $T_2=17.85\,{}^\circ\text{C}$, and steam-table water enthalpies (kJ/kg × kg/s): $$\dot H_{\text{in}}=\dot n_{\text{dry}}(0.3727)+(\dot n_{w,\text{in}}M_w)(2570.8)=116.3+978.9=1095.2\ \text{kJ/s}.$$
6. Enthalpy outputs and duty. The exit carries cooled dry air, saturated vapour, and the liquid condensate: $$\dot H_{\text{out}}=\dot n_{\text{dry}}(-0.2073)+(\dot n_{w,\text{out}}M_w)(2534.5)+(\dot n_{\text{cond}}M_w)(75.5)=-64.7+296.8+19.9=251.9\ \text{kJ/s}.$$ The heat removed is $\dot Q=\dot H_{\text{in}}-\dot H_{\text{out}}=843.2$ kJ/s $=3.036\times10^{6}$ kJ/hr, so $$\text{Cooling}=\frac{3.036\times10^{6}}{12{,}660}=\boxed{239.8\ \text{tons}}.$$