Question 1 of 8: Summer plant — mixing box, cooling coil, apparatus dew point

Problem 1: Summer plant — mixing box, cooling coil, apparatus dew point (30 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 once-mixed, single-coil constant-volume plant at sea level ($p = 14.696$ psia), with no fan or friction gain.

Find. The supply-air condition and quantity, the room and grand sensible heat factors, the refrigeration duty at the coil, and the coil apparatus dew point with its by-pass factor.

Approach. Fix the room and outdoor states from ASHRAE Ch. 1, convert the two stated volume flows to dry-air mass flows at their own states, mix them exactly, close the room sensible and latent balances to get the supply state, and then read the coil line $M \to S$ down to the saturation curve for the apparatus dew point.

1. Part (b) — fix the room state R. With a 78 °F dry bulb and 67 °F wet bulb, the ASHRAE wet-bulb relation $$W_R=\frac{(1093-0.556\,t^{*})W_s(t^{*})-0.240\,(t-t^{*})}{1093+0.444\,t-t^{*}}$$ with $t=78\,{}^{\circ}\text{F}$ and $t^{*}=67\,{}^{\circ}\text{F}$ gives $\boxed{W_R = 0.01163\ \text{lb}_w/\text{lb}_{da}}$, from which $h_R = 0.240t + W(1061+0.444t) = 31.46\ \text{Btu/lb}_{da}$, $v_R = 13.808\ \text{ft}^3/\text{lb}_{da}$, dew point 61.4 °F and relative humidity 56.8 %.
2. Fix the outdoor state O. At 90 °F the saturation pressure is 0.6989 psia, so $p_w = 0.70(0.6989) = 0.4892$ psia and $$W_O=\frac{0.621945\,p_w}{p-p_w}=0.02142\ \text{lb}_w/\text{lb}_{da},$$ giving $h_O = 45.18\ \text{Btu/lb}_{da}$, $v_O = 14.334\ \text{ft}^3/\text{lb}_{da}$, wet bulb 81.5 °F and dew point 78.9 °F. The outdoor air is both hotter and far wetter than the room, which is why the coil ends up latent-heavy.
3. Convert the two volume flows to dry-air mass flows. Each stream must be divided by the specific volume of that stream, not by a standard 13.5: $$\dot m_O=\frac{3000}{14.334}=209.3\ \frac{\text{lb}_{da}}{\text{min}},\qquad \dot m_R=\frac{11000}{13.808}=796.6\ \frac{\text{lb}_{da}}{\text{min}}$$ so $\dot m_S = 1{,}005.9\ \text{lb}_{da}/\text{min}$ and the outdoor-air mass fraction is $x = 209.3/1005.9 = 0.2081$ — slightly below the 3/14 = 0.214 volumetric fraction because the outdoor air is less dense.
4. Mix exactly to state M. Adiabatic mixing is exact in moisture and in enthalpy, so both are mass-weighted and the dry bulb is derived afterwards: $$W_M=xW_O+(1-x)W_R=0.01367,\qquad h_M=xh_O+(1-x)h_R=34.32\ \text{Btu/lb}_{da}$$ $$t_M=\frac{h_M-1061\,W_M}{0.240+0.444\,W_M}=80.53\,{}^{\circ}\text{F}$$ Weighting the dry bulb directly would be about 0.1 °F adrift because of the $W\!\cdot\!t$ cross term in the enthalpy relation. State M has a 70.5 °F wet bulb and a 66.0 °F dew point.
5. Part (d) — close the room latent balance for $W_S$. The room moisture gain must be absorbed by the supply air: $$Q_{RL}=\dot m_S\,(W_R-W_S)\,(1061+0.444\,t_R) \;\Rightarrow\; W_S=W_R-\frac{34{,}500}{60(1005.9)(1095.6)}=0.01111$$ $$\boxed{W_S = 0.01111\ \text{lb}_w/\text{lb}_{da}}$$
6. Close the room sensible balance for $t_S$. With the moist-air specific heat $c_p = 0.240+0.444W_S = 0.2449\ \text{Btu/lb}\cdot{}^{\circ}\text{F}$, $$t_S=t_R-\frac{Q_{RS}}{\dot m_S c_p}=78-\frac{195{,}500}{60(1005.9)(0.2449)} =\boxed{64.8\,{}^{\circ}\text{F db}}$$ The supply state is therefore 64.8 °F db, 61.8 °F wb, 85.0 % RH, $h_S = 27.65$ Btu/lb and $v_S = 13.457\ \text{ft}^3/\text{lb}_{da}$, i.e. $1005.9 \times 13.457 = 13{,}537$ cfm delivered at the supply state. As a check, $\dot m_S(h_R-h_S) = 230{,}000$ Btu/h exactly, which is the stated total room load.
7. Part (c) — the two sensible heat factors. The room factor is a pure ratio of the given loads, $$\text{RSHF}=\frac{Q_{RS}}{Q_{RS}+Q_{RL}}=\frac{195{,}500}{230{,}000}=\boxed{0.850}$$ The grand factor belongs to the coil, whose entering state is M and leaving state is S. Splitting the coil enthalpy drop at constant $W_S$, $$Q_{GS}=\dot m_S\big[h(t_M,W_S)-h(t_S,W_S)\big]=232{,}900\ \text{Btu/h},\qquad Q_{GT}=\dot m_S(h_M-h_S)=402{,}200\ \text{Btu/h}$$ $$\text{GSHF}=\frac{232{,}900}{402{,}200}=\boxed{0.579}$$ The grand factor is far below the room factor because the 3,000 cfm of 78.9 °F dew-point outdoor air is almost entirely a latent burden.
8. Part (e) — total energy input at the coil. $$Q_{GT}=\dot m_S\,(h_M-h_S)=60(1005.9)(34.32-27.65)=\boxed{402{,}200\ \text{Btu/h}=33.5\ \text{tons}}$$ This closes independently against the load-plus-ventilation route: the outdoor-air load is $\dot m_O(h_O-h_R)=60(209.3)(45.18-31.46)=172{,}200$ Btu/h, and $230{,}000+172{,}200 = 402{,}200$ Btu/h to within 40 Btu/h. Of the total, 232,900 Btu/h is sensible and 169,300 Btu/h latent.
9. Part (f) — apparatus dew point and by-pass factor. The apparatus dew point is the point where the straight coil line through M and S meets the saturation curve. Parameterising that line and solving $W(t)=W_s(t)$ gives $$\boxed{\text{ADP}=56.7\,{}^{\circ}\text{F}}$$ and the by-pass factor is the fraction of the entering air that leaves at the entering state, $$\text{BF}=\frac{t_S-\text{ADP}}{t_M-\text{ADP}}=\frac{64.8-56.7}{80.5-56.7}=\boxed{0.34}$$ A by-pass factor of 0.34 is high for a chilled-water coil — typical four- to six-row selections sit at 0.05 to 0.15 — which says the plant as specified needs a deeper coil or a lower face velocity than the stated flows imply. That observation is part of the answer, not a defect in it.