Question 1 of 8: Summer plant with cooling coil, reheat and a face-and-by-pass rebuild (30 marks)

Question 1: Summer plant with cooling coil, reheat and a face-and-by-pass rebuild (30 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 once-through-plus-recirculation summer plant is fully specified by its two end states, its mixing ratio and the coil performance.

Find. (a) the plant diagram; (b) the cycle on the psychrometric chart; (c) the dry- and wet-bulb temperature of every significant state point; (d) the cooling-coil duty in kW; (e) the heating-coil duty; and (f) the percentage saturation of the air supplied by the rebuilt face-and-by-pass plant.

Approach. Fix the two given end states from the ASHRAE property relations, mix them by mass to reach the coil face, cut the coil line at the by-pass factor to get the leaving state, then take the two coil duties as enthalpy differences at the constant 2 kg/s dry-air flow; the rebuild is the same arithmetic run backwards, with the unknown split between coil air and by-pass air chosen so that the blend lands on the temperature the fan needs.

1. Part (a) — describe the plant. Outdoor air O meets recirculated room air R in the mixing box; the mixture M crosses the cooling coil, leaving at C; a heating coil raises it sensibly to H; the fan and the supply ductwork add the last 2 K to give the supply state S, which then picks up the room load and returns as R. The diagram above is the answer to part (a) — every state point that the rest of the question refers to is labelled on it.
2. Fix the room state R. Relative humidity is the ratio of partial pressures, so the humidity ratio follows from the ASHRAE saturation pressure at 20 °C. $$W = 0.621945\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}$$ With $p_{ws}(20^\circ\text{C}) = 2.339\ \text{kPa}$ and $\phi = 0.50$, $$W_{R} = 0.621945 \times \frac{0.50 \times 2.339}{101.325 - 1.170} = 0.007262\ \text{kg/kg}$$ and the moist-air enthalpy on the ASHRAE datum is $$h = 1.006\,t + W\,(2501 + 1.86\,t) = 38.55\ \text{kJ/kg dry air}$$ Solving the adiabatic-saturation relation for this state gives a wet bulb of 13.78 °C.
3. Fix the outdoor state O — percentage saturation, not relative humidity. The paper specifies $\mu = 50\ \%$, and percentage saturation is defined on the humidity ratio directly: $$\mu = \frac{W}{W_{s}(t)} \quad\Longrightarrow\quad W_{O} = 0.50 \times W_{s}(26^\circ\text{C}) = 0.50 \times 0.021352 = 0.010676\ \text{kg/kg}$$ so $h_{O} = 53.37\ \text{kJ/kg}$ and the outdoor wet bulb is 18.85 °C. (The same state expressed as a relative humidity is 50.8 %, which is the small but real difference flagged in the note above.)
4. Mix three parts room air with one part fresh air. Adiabatic mixing conserves dry-air mass, moisture and enthalpy exactly, so both $W$ and $h$ are mass-weighted: $$W_{M} = 0.75\,W_{R} + 0.25\,W_{O} = 0.75(0.007262) + 0.25(0.010676) = 0.008115\ \text{kg/kg}$$ $$h_{M} = 0.75(38.55) + 0.25(53.37) = 42.26\ \text{kJ/kg}$$ The dry bulb is then derived from those two, not weighted separately, because the $W t$ cross term in the enthalpy relation makes direct temperature-weighting inconsistent: $$t_{M} = \frac{h_{M} - 2501\,W_{M}}{1.006 + 1.86\,W_{M}} = \frac{42.26 - 20.30}{1.0211} = \boxed{21.51\ ^\circ\text{C}}$$ Weighting the dry bulb instead would have given 21.50 °C — a hundredth of a degree, but it is the derived value that satisfies $h(t_{M}, W_{M}) = h_{M}$ identically, which is what makes the rest of the arithmetic close. The mixed-air wet bulb is 15.15 °C.
5. Cut the coil line at the by-pass factor to reach C. A cooling coil is modelled as a stream that fully saturates at the apparatus dew point mixed with a fraction BF that slips through untouched, so the leaving state lies on the straight line from M to the ADP, a fraction BF of the way back from the ADP: $$t_{C} = \text{ADP} + \text{BF}\,(t_{M} - \text{ADP}) = 4 + 0.10(21.51 - 4) = 5.75\ ^\circ\text{C}$$ $$W_{C} = W_{s}(\text{ADP}) + \text{BF}\,\bigl(W_{M} - W_{s}(\text{ADP})\bigr) = 0.005034 + 0.10(0.008115 - 0.005034) = 0.005342\ \text{kg/kg}$$ giving $h_{C} = 19.20\ \text{kJ/kg}$, a wet bulb of 5.31 °C and a relative humidity of 93.9 % — a realistic off-coil condition, which is the first sign the model is being applied sensibly.
6. Work back from the supply state to size the reheat. The fan and ductwork add 2 K after the heating coil, so the heating coil must deliver $$t_{H} = t_{S} - \Delta t_{\text{fan}} = 15 - 2 = 13\ ^\circ\text{C}$$ Reheat is a sensible process, so $W_{H} = W_{C} = 0.005342\ \text{kg/kg}$ and $h_{H} = 26.57\ \text{kJ/kg}$; the wet bulb rises to 8.83 °C. The supply state itself is 15 °C at the same humidity ratio, i.e. 50.6 % RH and a 9.74 °C wet bulb.
7. Part (d) — cooling-coil capacity. The coil sees the whole 2 kg/s of dry air, and its duty is the enthalpy drop across it: $$\dot{Q}_{c} = \dot{m}_{a}\,(h_{M} - h_{C}) = 2.0\,(42.26 - 19.20) = \boxed{46.1\ \text{kW}}$$ Of this, $\dot{m}_{a} c_{pa}(t_{M} - t_{C}) = 2.0 \times 1.006 \times 15.76 = 31.7\ \text{kW}$ is sensible and the remaining 14.4 kW is latent, a sensible heat ratio of 0.69 across the coil.
8. Part (e) — heating-coil load. Along $C \rightarrow H$ the humidity ratio is unchanged, so the duty is again an enthalpy difference: $$\dot{Q}_{h} = \dot{m}_{a}\,(h_{H} - h_{C}) = 2.0\,(26.57 - 19.20) = \boxed{14.7\ \text{kW}}$$ This is the price of dehumidifying to 4 °C ADP and then reheating: a third of the cooling energy is thrown straight back in, which is exactly why part (f) proposes removing the heating coil.
9. Part (c) — the state points. Collecting the six significant points, with the mixing and coil arithmetic above:

   Dry- and wet-bulb temperature at each significant point

   Point	Description	Dry bulb, °C	Wet bulb, °C	W, g/kg	h, kJ/kg
   O	outdoor air	26.00	18.85	10.68	53.37
   R	room / recirculated air	20.00	13.78	7.26	38.55
   M	after mixing, on to the coil	21.51	15.15	8.12	42.26
   C	off the cooling coil	5.75	5.31	5.34	19.20
   H	off the heating coil	13.00	8.83	5.34	26.57
   S	supply to the room	15.00	9.74	5.34	28.60

   The apparatus dew point itself is the auxiliary point at 4.00 °C saturated, $W = 5.03\ \text{g/kg}$.

Part (b) is the plot of those six points, with the mixing line $O\!-\!M\!-\!R$ straight, the coil line $M\!-\!C$ aimed at the apparatus dew point, and the two horizontal sensible processes $C\!-\!H\!-\!S$.

The remaining part removes the heating coil altogether. Because the room humidity no longer has to be held, the plant may deliver drier or wetter air so long as the temperature is right; blending untreated room air back into the cold coil discharge does the reheating for nothing.

10. Part (f), first mixing box. The ratio is now one part fresh to one part recirculated, so the state entering the coil is the mid-point of $O\!-\!R$: $$W_{1} = 0.50(0.010676) + 0.50(0.007262) = 0.008969\ \text{kg/kg},\qquad h_{1} = 0.50(53.37) + 0.50(38.55) = 45.96\ \text{kJ/kg}$$ $$t_{1} = \frac{45.96 - 22.43}{1.0227} = 23.01\ ^\circ\text{C}$$ which is warmer and wetter than before, as it must be with twice the outdoor air on the coil face.
11. Part (f), the cooling coil. The apparatus dew point and by-pass factor are unchanged, so the same construction applies to the new entering state: $$t_{2} = 4 + 0.10(23.01 - 4) = 5.90\ ^\circ\text{C},\qquad W_{2} = 0.005034 + 0.10(0.008969 - 0.005034) = 0.005427\ \text{kg/kg}$$ with $h_{2} = 19.57\ \text{kJ/kg}$. The coil leaves the air a little warmer and a little wetter than in the original plant, because it is fed a heavier outdoor load.
12. Part (f), second mixing box — find the split. Let $\dot{m}_{c}$ be the flow over the coil face and $\dot{m}_{b} = 2 - \dot{m}_{c}$ the room air by-passed into mixing box 2. The blend must leave at $15 - 2 = 13\ ^\circ\text{C}$ so that the fan gain lands it on the required supply temperature. Writing the moisture and enthalpy balances and solving for the fraction $f = \dot{m}_{c}/\dot{m}_{a}$ that satisfies $t\bigl(h_{3}, W_{3}\bigr) = 13\ ^\circ\text{C}$ gives $$f = 0.4973 \quad\Longrightarrow\quad \boxed{\dot{m}_{c} = 0.995\ \text{kg/s},\qquad \dot{m}_{b} = 1.005\ \text{kg/s}}$$ so almost exactly half the air is treated and half by-passed. That is not a coincidence: the coil discharge is about 7 K below the target and the room air about 7 K above it.
13. Part (f) — the supply humidity, and the answer. With the split known, the humidity ratio of the blend follows from the same mass balance: $$W_{3} = f\,W_{2} + (1 - f)\,W_{R} = 0.4973(0.005427) + 0.5027(0.007262) = 0.006349\ \text{kg/kg}$$ The fan raises the dry bulb to 15 °C at that unchanged humidity ratio, and the saturation humidity ratio at 15 °C is 0.010647 kg/kg, so $$\mu_{S} = \frac{W_{S}}{W_{s}(15^\circ\text{C})} = \frac{0.006349}{0.010647} = \boxed{0.596 \equiv 59.6\ \%\ \text{saturation}}$$ Expressed as a relative humidity the same state is 60.0 %, and its wet bulb is 10.82 °C.
14. Check the rebuild is worth having. The coil now handles only 0.995 kg/s, and its duty is $\dot{m}_{c}(h_{1} - h_{2}) = 0.995(45.96 - 19.57) = 26.3\ \text{kW}$ against 46.1 kW before, with the 14.7 kW reheat abolished. Total plant energy falls from 60.8 kW to 26.3 kW — a 57 % saving — and the outdoor-air fraction of the supply is essentially unchanged at 24.9 %, so ventilation is not sacrificed to get it. The price is that the room humidity now floats, which the question explicitly permits.