Question 1 of 6: Auditorium Air-Conditioning — Recycle Humidity Balance

Question A1: Auditorium Air-Conditioning — Recycle Humidity Balance (25 points)

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. A mixing tee combines fresh air with recycled auditorium exhaust; the mixed air is cooled and dehumidified in the A/C plant and supplied to the auditorium, where occupants add moisture. Total pressure 1 atm throughout. The molar humidities (mol H₂O per mol dry air) are the load-bearing data:

Find. (a) the water condensed in the A/C plant; (b) the water picked up in the auditorium; (c) the recycle-to-fresh air ratio.

Approach. Dry air is conserved across every unit, so it is the natural tie: fix the dry-air molar flow from the supply stream, apply it to the humidity change across the plant and across the room, then close a water balance on the mixing tee to get the recycle ratio.

1. Dry-air molar flow through the plant and room. The 20,900 m³/hr supply is humid air at 17 °C (290.15 K), 1 atm. By the ideal-gas law the total molar flow is $$n_{tot}=\frac{P\dot V}{RT}=\frac{(101{,}325)(20{,}900)}{(8.314)(290.15)}=8.779\times10^{5}\ \text{mol/hr},$$ and removing the water gives the dry-air flow $$\dot n_{da}=\frac{n_{tot}}{1+H_{in}}=\frac{8.779\times10^{5}}{1.0163}=\boxed{8.638\times10^{5}\ \text{mol/hr}}.$$ This same dry air passes through the plant and then through the auditorium.
2. (a) Moisture removed in the A/C plant. Across the plant the dry air is unchanged while its humidity falls from $H_M$ to $H_{in}$, so the water condensed is $$\dot n_{H_2O}=\dot n_{da}(H_M-H_{in})=8.638\times10^{5}(0.0225-0.0163)=5{,}356\ \text{mol/hr},$$ i.e. $5{,}356\times18.015=$ 96.5 kg/hr of moisture removed.
3. (b) Moisture added in the auditorium. Between supply and exhaust the humidity rises from $H_{in}$ to $H_{out}$ on the same dry-air flow: $$\dot n_{H_2O}=\dot n_{da}(H_{out}-H_{in})=8.638\times10^{5}(0.0181-0.0163)=1{,}555\ \text{mol/hr}=\boxed{28.0\ \text{kg/hr added}}.$$
4. (c) Recycle ratio from the mixing-tee lever. On the tee, dry air and water both balance: with $F$, $R$, $M$ the dry-air flows of fresh, recycle and mixed streams, $F+R=M$ and $F\,H_F+R\,H_{out}=M\,H_M$. Eliminating $M$ gives the classic lever-arm result $$\frac{R}{F}=\frac{H_F-H_M}{H_M-H_{out}}=\frac{0.0405-0.0225}{0.0225-0.0181}=\frac{0.0180}{0.0044}=\boxed{4.09}.$$ About four moles of air are recycled per mole of fresh make-up. On a total (humid) mole basis the ratio is $4.09(1.0181)/(1.0405)=4.00$.
5. Consistency cross-check. The fresh feed of 4,500 m³/hr at 35 °C carries $\dot n_{da,F}=\dfrac{(101{,}325)(4{,}500)}{(8.314)(308.15)(1.0405)}=1.710\times10^{5}$ mol/hr dry air. Closing the dry-air balance on the tee ($R=\dot n_{da}-F=6.93\times10^{5}$) gives $R/F=4.05$. That agrees with the humidity lever to within 1%, so the volumetric flows and the humidity data are consistent. The problem is over-specified, and the lever value 4.09 is quoted because it uses the stated humidities directly.