Question 1 of 6: Air-Handling Unit with Coil Bypass and Reheat

Question 1: Air-Handling Unit with Coil Bypass and Reheat (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 conditioned space held at 26 °C and 50% relative humidity carries a total load of 40 kJ/s and a moisture load of 8 g/s; the outdoor air is hot and humid, one quarter of the supply air is drawn from outdoors, and inside the air-handling unit one quarter of the mixed air bypasses the cooling coil while the remaining three quarters leave the coil at 10 °C and 97% RH.

Find. The dry-bulb temperature and relative humidity of every labelled state (A, MA, B, C, SA) plotted on the psychrometric chart, the total supply-air mass flow rate, and the duties of the cooling and heating coils.

[Figure not reproduced: Figure 1.1 — the air-handling unit, redrawn from the paper. Air leaves the space at A, three quarters of it returns to the unit and mixes with 25% fresh air to give MA; MA then splits, 25% bypassing the coil and 75% being cooled to B; the two streams re-mix at C and are reheated to SA. See the official exam paper.]

Approach. Fix the humidity ratio and enthalpy of the three states the question states outright (A, FA, B), carry them through the two adiabatic mixing junctions to get MA and C, note that the reheat coil cannot change the humidity ratio, and then close the room moisture and energy balances to get the supply flow rate and the supply state; the two coil duties follow from steady-flow energy balances on the streams that actually pass through them.

1. Fix the three stated states from the psychrometric relations. The humidity ratio and enthalpy of moist air at pressure $P$ follow from $$W = 0.621945\,\frac{p_w}{P - p_w}, \qquad p_w = \mathrm{RH}\times p_{ws}(t), \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$ with $h$ in kJ per kilogram of dry air. At 26 °C the saturation pressure is 3.363 kPa, so $p_w = 1.682$ kPa and $W_A = 0.621945\times 1.682/(101.325-1.682) = 0.010496$ kg/kg. Repeating for the fresh air (35 °C, 70%) and for the coil outlet (10 °C, 97%) gives the first three rows of the state table below.
2. Mix return air with fresh air to find MA. Adiabatic mixing of two moist-air streams is a mass-weighted average on both the humidity ratio and the enthalpy, so with the fresh-air fraction $f_{FA} = 0.25$, $$W_{MA} = f_{FA}W_{FA} + (1-f_{FA})W_A, \qquad h_{MA} = f_{FA}h_{FA} + (1-f_{FA})h_A$$ Substituting, $W_{MA} = 0.25(0.025159) + 0.75(0.010496) = 0.014162$ kg/kg and $h_{MA} = 0.25(99.771) + 0.75(52.914) = 64.628$ kJ/kg. Inverting the enthalpy relation for temperature gives $T_{MA} = 28.29\ ^\circ\text{C}$ at 58.6% RH — the mixed state lies one quarter of the way from A to FA along the straight mixing line, which is the geometric check worth making on the chart.
3. Cool three quarters of the flow, then re-mix with the bypass to find C. The bypassed quarter is still at MA and the cooled three quarters are at B, so the same lever rule applies again with a bypass fraction of 0.25: $$W_C = 0.25\,W_{MA} + 0.75\,W_B = 0.25(0.014162) + 0.75(0.007398) = 0.009089\ \text{kg/kg}$$ $$h_C = 0.25(64.628) + 0.75(28.701) = 37.683\ \text{kJ/kg} \;\Rightarrow\; T_C = 14.62\ ^\circ\text{C},\ \mathrm{RH}_C = 87.7\%$$ On the chart, C sits on the line MA–B three quarters of the way towards B — the cooling process line and the bypass mixing line are the same straight line, which is why a bypass arrangement can never reach a state below that line.
4. Recognise that the reheat coil fixes the supply humidity ratio. A heating coil transfers sensible heat only; no moisture is added or removed between C and SA, so $$\boxed{W_{SA} = W_C = 0.009089\ \text{kg/kg} = 9.089\ \text{g/kg}}$$ This is the pivot of the whole question: the supply humidity ratio is known before the supply flow rate is, which is what makes the room moisture balance solvable in one step.
5. Close the room moisture balance for the supply mass flow rate. Steady state in the space requires the moisture picked up by the supply air to equal the moisture generated in it: $$\dot m_{SA}\,(W_A - W_{SA}) = \Delta W \;\Longrightarrow\; \dot m_{SA} = \frac{0.008}{0.010496 - 0.009089}$$ $$\boxed{\dot m_{SA} = 5.69\ \text{kg/s}}$$ Of this, 1.42 kg/s is fresh air and 4.27 kg/s passes over the cooling coil.
6. Close the room energy balance for the supply state. With the flow rate now known, the total load fixes the supply enthalpy: $$h_{SA} = h_A - \frac{\Delta H}{\dot m_{SA}} = 52.914 - \frac{40}{5.6875} = 45.881\ \text{kJ/kg}$$ and inverting $h = 1.006\,t + W(2501 + 1.86\,t)$ at $W = 0.009089$ gives $$\boxed{T_{SA} = 22.63\ ^\circ\text{C}, \qquad \mathrm{RH}_{SA} = 53.1\%}$$ The supply air is only 3.4 K cooler than the room because this space is latent-dominated: the sensible part of the load is 19.3 kW and the latent part 20.0 kW, a sensible heat ratio of 0.49.
7. Take an energy balance across the cooling coil. Only three quarters of the supply air passes over it, and it takes that stream from MA to B: $$Q_c = 0.75\,\dot m_{SA}\,(h_{MA} - h_B) = 0.75(5.6875)(64.628 - 28.701)$$ $$\boxed{Q_c = 153.3\ \text{kW}}$$ The coil also condenses $0.75\,\dot m_{SA}(W_{MA} - W_B) = 28.9$ g/s of water, which must be drained; the latent share is most of the coil duty.
8. Take an energy balance across the heating coil. The whole supply stream passes through it, from C to SA: $$Q_H = \dot m_{SA}\,(h_{SA} - h_C) = 5.6875\,(45.881 - 37.683)$$ $$\boxed{Q_H = 46.6\ \text{kW}}$$

The arithmetic is worth stepping back from. A space that needs only 40 kW of cooling is being served by a coil doing 153 kW — a factor of 3.8 — and then 47 kW of heat is deliberately put back in. That is the cost of using a fixed 25% bypass to control humidity: the coil must overcool the whole 75% stream far below the required supply condition so that the re-mixed stream is dry enough, and the reheat coil then undoes the overcooling in temperature terms only. The five processes are plotted below.