Question 2 of 8: Winter plant with preheat, mixing, reheat and steam humidification

Question 2: Winter plant with preheat, mixing, reheat and steam humidification (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 air-handling plant serving a space held at 21 °C dry bulb and 35 % relative humidity against a total heating load of 215 kW; outdoor air enters a preheater at −14 °C and 0 % relative humidity and leaves it at 16 °C, is mixed with return air in the ratio 25 % outdoor by mass, and the mixture is then heated and steam-humidified to a supply state of 40 °C dry bulb and 30 % relative humidity.

Find. (a) the plant arrangement, (b) the six state points with their dry- and wet-bulb temperatures plotted on the psychrometric chart, (c) the supply air volume flow, (d) the duty of the preheater and of the heater separately, and (e) the steam flow rate to the humidifier.

(a) Diagram of the system

The preheater sits upstream of the mixing box, which is the point of the arrangement: it lifts the outdoor air from −14 °C to 16 °C before it can meet the return air, so no part of the mixing box, filter or downstream coil ever sees an air stream below freezing. Face-and-bypass or freeze-stat protection on the preheater is what keeps the plant safe on a design night.

(b) State points and the operating cycle

Approach. Fix each state from its two given properties using the ASHRAE moist-air relations, take the mixed state from exact moisture and enthalpy balances, obtain the mass flow from the room load and the supply-to-room enthalpy difference, and split the last process into a sensible coil duty plus the enthalpy carried in by the injected steam.

1. Humidity ratio and enthalpy of each fixed state. For any moist-air state, $$W=0.621945\,\frac{\phi\,p_{ws}(t)}{p-\phi\,p_{ws}(t)} \qquad h=1.006\,t+W\,(2501+1.86\,t)$$ At the room state, $p_{ws}(21\,{}^\circ\text{C})=2.488$ kPa, so $W_3=0.005391$ kg/kg and $h_3=34.82$ kJ/kg. At the supply state, $p_{ws}(40\,{}^\circ\text{C})=7.384$ kPa, giving $W_5=0.013900$ kg/kg and $h_5=76.04$ kJ/kg. The outdoor air is dry, so $W_1=W_2=0$ and $h_1=-14.08$ kJ/kg, $h_2=16.10$ kJ/kg.
2. Mixed state, from mass and energy balances. Adiabatic mixing conserves moisture and enthalpy exactly, so both are mass-weighted, $$W_4=x\,W_2+(1-x)\,W_3=0.25(0)+0.75(0.005391)=0.004043\ \text{kg/kg}$$ $$h_4=x\,h_2+(1-x)\,h_3=0.25(16.10)+0.75(34.82) =30.14\ \text{kJ/kg}$$ The dry bulb then follows by inverting the enthalpy relation, not by weighting the temperatures: $$t_4=\frac{h_4-2501\,W_4}{1.006+1.86\,W_4}=19.76\ ^\circ\text{C}$$
3. Wet-bulb temperature of each state. The thermodynamic wet bulb is the root of the adiabatic-saturation equation $$W=\frac{(2501-2.326\,t^{*})\,W_s(t^{*})-1.006\,(t-t^{*})} {2501+1.86\,t-4.186\,t^{*}}$$ solved for $t^{*}$ at each state. The results are collected below.

The two dry states carry no moisture at all, so their dew points are undefined; the table reports the solver's lower bound of −80 °C for them, which should be read as "bone dry". Their wet bulbs are nevertheless real: dry air at 16 °C passed through an adiabatic saturator would cool to 3.75 °C as water evaporated into it, and that is the number a sling psychrometer would read at state 2.

(c) Total air supply volume

1. Mass flow from the room energy balance. The supply air must carry the whole 215 kW into the space as it falls from state 5 to the room state 3, $$\dot m=\frac{q}{h_5-h_3}=\frac{215}{76.04-34.82} =5.216\ \text{kg/s of dry air}$$
2. Convert to volume at the supply state. The humid volume at 40 °C and $W_5=0.013900$ is $$v_5=\frac{R_a T}{p}\,(1+1.6078\,W_5) =\frac{0.287\times 313.15}{101.325}\,(1+1.6078\times 0.013900) =0.9069\ \text{m}^3/\text{kg}$$ so the volume the fan must handle is $$\boxed{\;\dot V=\dot m\,v_5=5.216\times 0.9069 =4.731\ \text{m}^3/\text{s}=4,731\ \text{L/s} \;(10,024\ \text{cfm})\;}$$ Note that the volume must be evaluated at the supply state, where the fan and ductwork actually see the air; using standard density here would understate the duct size by about 10 %.

(d) Energy input to each coil

1. Preheater. Only the outdoor-air stream passes through it, so $$\dot m_{oa}=x\,\dot m=0.25\times 5.216=1.3040\ \text{kg/s}$$ $$\boxed{\;q_{pre}=\dot m_{oa}\,(h_2-h_1) =1.3040\,[16.10-(-14.08)]=39.35\ \text{kW}\;}$$ This is a purely sensible duty, since the outdoor air carries no moisture.
2. Total duty of the heating and humidifying section. Between the mixed state and the supply state the whole air stream gains $$q_{4\to 5}=\dot m\,(h_5-h_4)=5.216\,(76.04-30.14) =239.41\ \text{kW}$$ but part of that enthalpy walks in with the steam rather than through the coil, so this figure must be split before the coil can be selected.
3. Subtract the enthalpy carried by the steam. Saturated vapour at 2 psig is at $14.696+2=16.70$ psia $=115.1$ kPa, for which the steam tables give $t_{sat}=103.59\ ^\circ$C and $h_g=2,681.2$ kJ/kg. With the steam flow found in part (e), $$q_{steam}=\dot m_{st}\,h_g=0.05141\times 2,681.2 =137.85\ \text{kW}$$ $$\boxed{\;q_{heater}=q_{4\to 5}-q_{steam} =239.41-137.85=101.56\ \text{kW}\;}$$ The heater therefore lifts the air from 19.76 °C to 38.97 °C at constant moisture, and the steam supplies the last degree of dry bulb together with all of the moisture. As a check, the same duty follows from a pure sensible balance, $\dot m\,(1.006+1.86\,W_4)(t_{4'}-t_4)=101.56$ kW.

(e) Steam flow rate

1. Moisture balance across the humidifier. Every kilogram of moisture that leaves in the supply air above the mixed-air level must have been injected, so $$\boxed{\;\dot m_{st}=\dot m\,(W_5-W_4) =5.216\,(0.013900-0.004043) =0.05141\ \text{kg/s}=185.1\ \text{kg/h}\;}$$

It is worth reading the two coil duties against each other. The preheater handles only a quarter of the air but takes it through a 30 K rise, and still costs only 39.35 kW; the heater handles all of the air through a 19 K rise and costs 101.56 kW. The steam is the quiet third player: at 137.85 kW it delivers more than half of the plant's total input downstream of the mixing box, which is why a steam humidifier is never a trivial addition to a boiler's load.