Question 1 of 8: Summer cooling plant with a governing ventilation rate

Question 1: Summer cooling plant with a governing ventilation rate (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. A space held at 78 °F dry bulb and 50% relative humidity, carrying the cooling loads and the outdoor-air requirement listed below, on a design day of 95 °F dry bulb / 75 °F wet bulb at sea level. Duct heat gain and fan temperature rise are neglected.

Find. The system arrangement and its psychrometric cycle, the supply air rate, the coil capacity, apparatus dew point and by-pass factor, and the grand sensible heat factor.

Approach. Fix the room and outdoor states from the chart, draw the room sensible heat factor (RSHF) line through the room point, show that the ventilation requirement — not a chosen supply temperature difference — sets the air quantity, then close a sensible and a total energy balance across the coil and extend the coil process line to saturation for the apparatus dew point.

1. Split the room load and fix the room state. The latent part is the balance of the total: $$q_L = q_T - q_S = 250{,}000 - 220{,}000 = 30{,}000\ \text{Btu/h}$$so the room sensible heat factor is $$\mathrm{RSHF} = \frac{q_S}{q_T} = \frac{220{,}000}{250{,}000} = 0.88$$At 78 °F and 50% RH the saturation pressure is 0.4738 psia, so $p_w = 0.5\,(0.4738) = 0.2369$ psia and $$W_R = \frac{0.622\,p_w}{p - p_w} = \frac{0.622\,(0.2369)}{14.696-0.2369} = 0.01022\ \text{lb/lb}$$$$h_R = 0.240\,t + W(1061 + 0.444\,t) = 0.240(78) + 0.01022(1095.6) = 29.92\ \text{Btu/lb}$$
2. Fix the outdoor state. Entering the chart at 95 °F dry bulb and 75 °F wet bulb (or using the ASHRAE wet-bulb relation), $$W_O = \frac{(1093 - 0.556\,t^{*})W_s^{*} - 0.240\,(t - t^{*})}{1093 + 0.444\,t - t^{*}} = 0.01407\ \text{lb/lb}, \qquad h_O = 38.32\ \text{Btu/lb}$$
3. Locate the room apparatus dew point. Drawing the RSHF line of slope 0.88 through the room point and extending it to the saturation curve gives the lowest temperature at which supply air could be delivered, $\boxed{t_{ADP,R} \approx 56\ ^\circ\text{F}}$. Any supply state on that line between 56 °F and 78 °F satisfies the room simultaneously in sensible and latent terms.
4. (c) Establish the air supply rate. A conventional 20 °F supply temperature difference would call for only $220{,}000/(1.10 \times 20) = 10{,}000$ cfm — less than the 14,000 cfm of outdoor air the indoor-air-quality requirement demands. Supply air can never be less than the outdoor air it must carry, so the ventilation requirement governs and the economical choice is to supply exactly that quantity as 100% outdoor air: $$\boxed{\dot V_S = \dot V_{OA} = 14{,}000\ \text{cfm}}, \qquad \dot m_a = 4.5\,\dot V_S = 4.5(14{,}000) = 63{,}000\ \text{lb dry air/h}$$
5. Fix the supply state on the RSHF line. The sensible balance with $c_p = 0.240 + 0.444\,W = 0.2443$ Btu/lb·°F gives $$\Delta t = \frac{q_S}{\dot m_a c_p} = \frac{220{,}000}{63{,}000(0.2443)} = 14.29\ ^\circ\text{F} \;\Rightarrow\; t_S = 78 - 14.29 = \boxed{63.7\ ^\circ\text{F}}$$and the latent balance gives the supply humidity ratio $$W_S = W_R - \frac{q_L}{\dot m_a (1061 + 0.444\,t_R)} = 0.01022 - \frac{30{,}000}{63{,}000(1095.6)} = 0.009784\ \text{lb/lb}$$which on the chart is 63.7 °F dry bulb at 77.9% relative humidity, with $h_S = 25.95$ Btu/lb. This is a comfortable coil-leaving condition, well above the 56 °F room apparatus dew point.
6. (d) Coil capacity. With no return air the coil sees outdoor air on its face, so the grand total load is the enthalpy drop from O to S: $$q_{coil} = \dot m_a (h_O - h_S) = 63{,}000\,(38.32 - 25.95) = \boxed{7.79 \times 10^{5}\ \text{Btu/h}} \;(64.9\ \text{tons})$$The result checks against the load definition: the room load plus the ventilation load is $250{,}000 + 63{,}000(38.32-29.92) = 250{,}000 + 529{,}000 = 779{,}000$ Btu/h, the same figure. Almost 70% of the coil duty is spent conditioning ventilation air — the direct consequence of the 14,000 cfm requirement.
7. Sensible and latent split, and the GSHF. $$q_{coil,S} = \dot m_a\,c_p\,(t_O - t_S) = 63{,}000(0.2462)(95-63.7) = 4.85\times10^{5}\ \text{Btu/h}$$$$q_{coil,L} = q_{coil} - q_{coil,S} = 2.94\times10^{5}\ \text{Btu/h}, \qquad \mathrm{GSHF} = \frac{4.85}{7.79} = \boxed{0.62}$$The coil line is much steeper than the room line because the outdoor air arrives with 0.01407 lb/lb of moisture and must leave at 0.00978 lb/lb.
8. Apparatus dew point and by-pass factor. Extending the straight line O–S to the saturation curve locates the coil apparatus dew point at $\boxed{t_{ADP} = 51.7\ ^\circ\text{F}}$. The by-pass factor is the fraction of the entering air that leaves the coil unchanged, $$\mathrm{BF} = \frac{t_S - t_{ADP}}{t_O - t_{ADP}} = \frac{63.7-51.7}{95-51.7} = \boxed{0.28}$$A by-pass factor near 0.28 corresponds to a four-row coil at about 500 fpm face velocity — a standard selection, so the design is realisable.