Question 1 of 8: All-outdoor-air operating-room plant — state points, air quantity, coil duty

Question 1: All-outdoor-air operating-room plant — state points, air quantity, coil duty (20 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.

Find. The supply air rate in cfm, the total cooling-coil capacity in Btu/h, and the relative humidity of the air leaving the operating room, with every significant state point labelled by dry-bulb and wet-bulb temperature on both a system diagram and the psychrometric chart.

Approach. Fix the supply state from saturation at 55 °F and the outdoor state from its wet bulb, size the dry-air mass flow on the sensible balance between supply and room, close the moisture balance to get the room humidity ratio, then take the coil duty as the whole enthalpy drop from O to S because the system is once-through.

1. Part (a) — describe the plant and fix the notation. Outdoor air is filtered, cooled and dehumidified in one coil, moved by the supply fan and delivered to the room; all of it is exhausted. Three states matter: O outdoors, S leaving the coil (and, because duct gain and fan rise are to be neglected, also entering the room), and R in the room and in the exhaust. Figure 1.1 is the answer to part (a).
2. Fix the supply state S from saturation at 55 °F. With φ = 100 % the vapour pressure equals the saturation pressure, $$p_{ws}(55\,{}^{\circ}\text{F}) = 0.2141\ \text{psia}, \qquad W = 0.621945\,\frac{p_w}{p - p_w}$$ so that $$W_S = 0.621945 \times \frac{0.2141}{14.696 - 0.2141} = \boxed{0.009195\ \text{lb/lb da}}$$ The enthalpy and specific volume follow from the ASHRAE moist-air relations, h = 0.240t + W(1061 + 0.444t) and v = 0.370486 T(1 + 1.6079W)/p: h_S = 23.18 Btu/lb and v_S = 13.167 ft³/lb da. A saturated state is its own wet bulb, so t_wb,S = 55.0 °F.
3. Fix the outdoor state O from its wet bulb. The adiabatic-saturation relation inverted at t = 95 °F, t* = 75 °F gives $$W_O = \frac{h(t^{*}, W_{sat}(t^{*})) - 0.240\,t - W_{sat}(t^{*})\,h_f(t^{*})}{(1061 + 0.444\,t) - h_f(t^{*})} = \boxed{0.014065\ \text{lb/lb da}}$$ whence h_O = 38.32 Btu/lb and φ_O = 39.8 % — a hot, only moderately humid design day, which is what makes the outdoor air worth cooling in one pass.
4. Part (d) — size the dry-air flow on the sensible balance. The supply-to-room sensible gain is taken at constant humidity ratio, so the exact moist-air specific heat at W_S applies: $$q_s = \dot m_{da}\,(0.240 + 0.444\,W_S)\,(t_R - t_S)$$ $$\dot m_{da} = \frac{40\,000}{(0.240 + 0.444 \times 0.009195)(75 - 55)} = \frac{40\,000}{4.8817} = \boxed{8\,194\ \text{lb da/h}}$$ Converting to volume at the supply state, $$\dot V_S = \dot m_{da}\,\frac{v_S}{60} = 8\,194 \times \frac{13.167}{60} = \boxed{1\,798\ \text{cfm}}$$ The familiar shortcut cfm = q_s/(1.10 Δt) returns 1 818 cfm; the 1 % difference is the standard-density coefficient standing in for the actual 13.167 ft³/lb of the cold supply air.
5. Close the moisture balance to reach the room state R. All the latent load is picked up between S and R, so $$W_R = W_S + \frac{q_l}{\dot m_{da}\,(1061 + 0.444\,t_R)} = 0.009195 + \frac{10\,000}{8\,194 \times 1\,094.3} = \boxed{0.010311\ \text{lb/lb da}}$$ Two checks close on this: the total enthalpy rise ṁ_da(h_R − h_S) = 8 194 × (29.283 − 23.180) = 50 000 Btu/h, exactly the stated total load, and the room sensible-heat factor SHF = 40 000/50 000 = 0.80, which is the slope of the S–R line drawn on Figure 1.2.
6. Part (f) — the humidity of the air leaving the operating room. The air leaving the room is at the room state, so its relative humidity follows from W_R: $$p_w = \frac{p\,W_R}{0.621945 + W_R} = \frac{14.696 \times 0.010311}{0.632256} = 0.2397\ \text{psia}, \qquad \phi_R = \frac{p_w}{p_{ws}(75\,{}^{\circ}\text{F})} = \frac{0.2397}{0.4300}$$ $$\phi_R = \boxed{55.7\ \%}$$ with a wet-bulb temperature of 64.2 °F. Repeating steps 4–6 across the ASHRAE 170 band shows why the top of the band was chosen: at 68 °F the room would sit at 68.0 % RH, at 70 °F at 64.2 %, and the 60 % limit is not reached until 72.4 °F. A colder room needs more air, but the extra air removes proportionally less moisture relative to its own saturation limit, so the room gets damper, not drier.
7. Part (e) — the coil capacity. Because the system is once-through, the coil processes the whole supply flow all the way from the outdoor state: $$\dot Q_{coil} = \dot m_{da}\,(h_O - h_S) = 8\,194 \times (38.316 - 23.180) = \boxed{124\,000\ \text{Btu/h}} \;=\; 10.3\ \text{tons}$$ Splitting it the way a coil-selection sheet does, the sensible part is ṁ_da[h(t_O,W_O) − h(t_S,W_O)] = 80 700 Btu/h and the latent part 43 300 Btu/h, a coil sensible-heat factor of 0.651. The coil is therefore 2.5 times the room load, and every bit of that multiplier is the price of the infection-control decision to recirculate nothing.
8. Parts (b) and (c) — the cycle and its state points. On the chart the process is a straight coil line O → S followed by the room line S → R at SHF 0.80. The coil line terminates on the saturation curve, so the apparatus dew point coincides with S and the coil by-pass factor is $$\text{BF} = \frac{W_S - W_{adp}}{W_M - W_{adp}} = 0\ ,\qquad \text{ADP} = 55\,{}^{\circ}\text{F}$$ A real coil cannot achieve this — four- to six-row coils leave air at 90–95 % saturation — so in practice the coil would be selected for an apparatus dew point of about 52 °F with a by-pass factor near 0.10, giving the stated 55 °F saturated condition after a small amount of mixing. That refinement changes the duty by under 2 % and is noted rather than carried.