Question 2 of 8: Winter preheat, humidify and reheat plant

Question 2: Winter preheat, humidify and reheat plant (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 winter preheat — adiabatic-humidify — reheat plant taking 25% outdoor air by mass, serving a space with an all-sensible heating load, as tabulated.

Find. The plant arrangement and its psychrometric cycle with every state identified, the total mass flow, the preheater and reheater ratings in Btu/h, and the adiabatic efficiency and make-up water demand of the spray cabinet.

Approach. Fix the room and outdoor states, obtain the mass flow from the all-sensible room load and the stated supply temperature, mix the two air streams, and then test whether the mixed air already has enough adiabatic-saturation capacity to reach the humidity the room needs. That test decides the preheater duty; the reheater then closes the sensible balance to 40 °C.

1. Room and outdoor states. At 21 °C the saturation pressure is 2.487 kPa, so at 30% RH $$W_R = \frac{0.622(0.30 \times 2.487)}{101.325 - 0.746} = 0.004615\ \text{kg/kg}, \qquad h_R = 32.85\ \text{kJ/kg}$$and saturated air at −12 °C (saturation pressure 0.2176 kPa over ice) carries $$W_O = 0.001337\ \text{kg/kg}, \qquad h_O = -8.76\ \text{kJ/kg}$$
2. (d) Total system mass flow. The load is entirely sensible, so the supply air need only carry a temperature difference into the room. With the moist-air specific heat $c_p = 1.006 + 1.86\,W_R = 1.015$ kJ/kg·K, $$\dot m = \frac{q}{c_p\,(t_6 - t_R)} = \frac{75}{1.015\,(40-21)} = \boxed{3.89\ \text{kg/s}}$$which is about 3.3 m³/s at room conditions. The outdoor-air stream is $0.25(3.89) = 0.973$ kg/s and the return stream 2.92 kg/s.
3. Supply humidity ratio. The room has no latent load, so in steady state the air leaves the room carrying exactly the moisture it brought in: $$W_6 = W_5 = W_R = 0.004615\ \text{kg/kg}$$The whole purpose of the spray cabinet is therefore to replace the moisture that the dry outdoor air dilutes out of the mixture, not to serve a room latent load.
4. (c) Mixed state 3. Mass-weighting the return and outdoor streams, $$t_3 = 0.75(21) + 0.25(-12) = 12.75\ ^\circ\text{C}, \qquad W_3 = 0.75(0.004615) + 0.25(0.001337) = 0.003795\ \text{kg/kg}$$$$h_3 = 1.006(12.75) + 0.003795\,[2501 + 1.86(12.75)] = 22.41\ \text{kJ/kg}$$The mixture is 3.8 g/kg short of the 4.6 g/kg the room needs.
5. (e) Preheater duty — test it, do not assume it. An adiabatic spray cabinet moves the air along a line of constant enthalpy, so the wettest state it could possibly deliver is saturation at the adiabatic saturation temperature of the entering air. For $h_3 = 22.41$ kJ/kg that temperature is $$t_3^{*} = 6.88\ ^\circ\text{C}, \qquad W^{*} = 0.006161\ \text{kg/kg}$$The mixed air can therefore reach 6.16 g/kg unaided, comfortably more than the 4.62 g/kg required. Preheating would raise the enthalpy and increase that capacity further, so it is not needed: $$\boxed{q_{preheat} = 0 \ \text{at the design point}}$$For completeness, an ideal washer discharging saturated air at the required humidity ratio would need entering air at only 4.8 °C — well below the 12.75 °C the mixing box already provides.
6. Rating the preheater anyway. The coil still has to be selected for its worst duty, which is the morning warm-up or purge condition with the outdoor damper fully open. Lifting the whole 3.89 kg/s from −12 °C to a +5 °C freeze-protection set point requires $$q_{preheat,rating} = \dot m\,c_p\,\Delta t = 3.89(1.008)(5-(-12)) = 66.7\ \text{kW} = \boxed{2.28\times10^{5}\ \text{Btu/h}}$$
7. State 5, leaving the spray cabinet. With no preheat, state 4 coincides with state 3 and the adiabatic process runs from $W_3$ to $W_5 = 0.004615$ kg/kg at constant enthalpy: $$h_5 = h_3 \;\Rightarrow\; t_5 = \frac{h_3 - 2501\,W_5}{1.006 + 1.86\,W_5} = \frac{22.41 - 11.54}{1.0146} = 10.7\ ^\circ\text{C}$$which is 58% relative humidity — the cabinet is deliberately operating well short of saturation.
8. (e) Reheater duty. The reheater raises state 5 to the 40 °C supply condition at constant humidity ratio: $$q_{reheat} = \dot m\,c_p\,(t_6 - t_5) = 3.89(1.015)(40 - 10.7) = 115.6\ \text{kW} = \boxed{3.94\times10^{5}\ \text{Btu/h}}$$This checks against an overall plant balance: the 75 kW room load, plus $0.973(1.008)(21-(-12)) = 32.4$ kW to warm the ventilation air from outdoor to room temperature, plus 8.0 kW of latent heat carried by the evaporating spray water, total 115.4 kW — the same figure to within rounding.
9. (f) Adiabatic efficiency and make-up water. The saturation efficiency of a spray cabinet compares the moisture actually picked up with the moisture an ideal, fully saturating washer would add on the same enthalpy line: $$\eta = \frac{W_5 - W_3}{W^{*} - W_3} = \frac{0.004615-0.003795}{0.006161-0.003795} = \boxed{0.35}$$The identical figure follows from the temperature form, $(t_3 - t_5)/(t_3 - t_3^{*}) = (12.75-10.7)/(12.75-6.88) = 0.35$. Make-up water is simply the moisture added: $$\dot m_w = \dot m\,(W_5 - W_3) = 3.89\,(0.000820) = 0.00319\ \text{kg/s} = \boxed{11.5\ \text{kg/h}}$$

A saturation efficiency of only 35% means the cabinet must be throttled — one bank of sprays in service, or a face-and-bypass damper — rather than run flat out. That is the practical control conclusion of the problem: the humidity is held by modulating the washer, and the preheater exists for frost protection rather than for humidification duty.

Taking the alternative reading of the incomplete printed line, with essentially dry outdoor air, the mixed humidity ratio falls to 3.46 g/kg, the required saturation efficiency rises to 46%, the reheater duty rises to 118.9 kW and the make-up water to 16.2 kg/h. The preheater duty remains zero and every qualitative conclusion is unchanged.