Question 1 of 8: Winter heating and humidifying plant — preheat, adiabatic spray, reheat

Question 1: Winter heating and humidifying plant — preheat, adiabatic spray, 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 once-through winter plant that mixes equal volumes of return and outdoor air, preheats the mixture, humidifies it adiabatically in a spray cabinet, reheats it and delivers it to a room whose entire load is sensible.

Find. The plant diagram and its cycle on the psychrometric chart with every state point characterised, then the total air mass flow, the preheater and reheater duties, the saturation efficiency of the spray cabinet and the make-up water rate.

Approach. Fix the room and outdoor states from the ASHRAE moist-air relations; mix them on a mass basis obtained from their specific volumes; size the airflow from the sensible balance (the room load is entirely sensible, so the supply humidity ratio must equal the room humidity ratio); then walk the plant backwards from the required supply state to find how much preheat the washer needs in order to deliver that moisture.

1. Part (c) — fix the room state R. The humidity ratio follows from the wet-bulb temperature through the adiabatic-saturation relation, with $W_{s}^{*} = 0.010648$ kg/kg at $15^{\circ}\text{C}$: $$W_{R}=\frac{(2501-2.326\,t^{*})W_{s}^{*}-1.006\,(t-t^{*})}{2501+1.86\,t-4.186\,t^{*}} =\frac{(2501-34.89)(0.010648)-1.006(20-15)}{2501+37.2-62.79}$$ which gives $W_{R}=0.008575$ kg/kg (8.58 g/kg), and hence $h_{R}=1.006(20)+0.008575(2501+37.2)=41.89$ kJ/kg, a relative humidity of 58.9 % and a dew point of $11.73^{\circ}\text{C}$. The humid volume is $v_{R}=0.8419\ \text{m}^{3}/\text{kg}$.
2. Fix the outdoor state O. A dew point of $4^{\circ}\text{C}$ fixes the moisture directly, because the humidity ratio is saturated at that temperature: $$W_{O}=\frac{0.621945\,p_{ws}(4^{\circ}\text{C})}{p-p_{ws}(4^{\circ}\text{C})} =\frac{0.621945(813.4)}{101325-813.4}=0.005034\ \text{kg/kg}$$ so $h_{O}=1.006(7)+0.005034(2501+13.0)=19.70$ kJ/kg, 81.2 % saturated, with a wet bulb of $5.56^{\circ}\text{C}$ and $v_{O}=0.8001\ \text{m}^{3}/\text{kg}$.
3. Mix on a mass basis, not a volume basis. "Equal volumes" is a statement about volumetric flow, and the two streams have different densities, so the mass fractions must come from the specific volumes. For a common volume flow $\dot V$ in each branch, $$x_{O}=\frac{\dot V/v_{O}}{\dot V/v_{O}+\dot V/v_{R}} =\frac{1/0.8001}{1/0.8001+1/0.8419}=0.5127$$ The colder outdoor air is denser, so it contributes slightly more than half the mass. Adiabatic mixing is exact in moisture and in enthalpy, so those are the two properties to weight: $$W_{M}=0.5127(0.005034)+0.4873(0.008575)=0.006759\ \text{kg/kg}$$ $$h_{M}=0.5127(19.70)+0.4873(41.89)=30.51\ \text{kJ/kg}$$ The dry bulb then follows from the enthalpy definition rather than from a weighted average of the two dry bulbs: $$t_{M}=\frac{h_{M}-2501\,W_{M}}{1.006+1.86\,W_{M}} =\frac{30.51-16.91}{1.0186}=\boxed{13.36^{\circ}\text{C}\ \text{dB},\ \ 10.54^{\circ}\text{C}\ \text{wB}}$$
4. Part (d) — total air mass flow. The room gains no moisture, so the supply air must carry the room humidity ratio, $W_{S}=W_{R}=0.008575$ kg/kg, and the whole 58 kW is absorbed by cooling the supply air from $40^{\circ}\text{C}$ to the room temperature: $$\dot m=\frac{Q_{s}}{(c_{pa}+c_{pv}W_{R})(t_{S}-t_{R})} =\frac{58}{(1.006+1.86\times0.008575)(40-20)}=\boxed{2.838\ \text{kg/s}}$$ As a check, the same figure comes from the enthalpy difference, $58/(62.33-41.89)=2.838$ kg/s. At the supply state this is 2.55 $\text{m}^{3}/\text{s}$, and 2.33 $\text{m}^{3}/\text{s}$ entering the plant at the mixed state — the two branches therefore handle about 1.16 $\text{m}^{3}/\text{s}$ each.
5. Test the washer before assuming a preheater duty. An adiabatic spray cabinet moves air along a line of constant thermodynamic wet bulb, so the wettest air it can possibly deliver is saturated air at the adiabatic saturation temperature of what enters it. For state M that temperature is $10.54^{\circ}\text{C}$, at which saturated air holds only $$W_{s}(10.54^{\circ}\text{C})=0.007916\ \text{kg/kg}\ \lt\ W_{R}=0.008575\ \text{kg/kg}$$ The mixture is 0.66 g/kg short of the room moisture even if the washer were perfect. Preheat is therefore genuinely required, and its purpose is to raise the wet bulb of the air offered to the cabinet, not to warm the supply air.
6. Fix state 3 leaving the cabinet. A single-bank spray cabinet does not saturate the air completely; taking the usual design figure of 95 % saturation at exit (see the callout below), the leaving state has $W_{3}=W_{R}=0.008575$ kg/kg at 95 % relative humidity, which is $t_{3}=12.51^{\circ}\text{C}$ and $h_{3}=1.006(12.51)+0.008575(2501+23.3)=34.23$ kJ/kg.
7. Work back through the cabinet to state 2. The cabinet is adiabatic, so the only energy crossing its boundary is the enthalpy of the make-up water, taken in at the leaving air temperature: $$h_{2}=h_{3}-(W_{3}-W_{2})\,c_{pw}t_{3}=34.23-(0.008575-0.006759)(4.186)(12.51)=34.14\ \text{kJ/kg}$$ Because the preheater is sensible, $W_{2}=W_{M}$ and $$t_{2}=\frac{34.14-2501(0.006759)}{1.006+1.86(0.006759)}=16.92^{\circ}\text{C}$$ whose adiabatic saturation temperature is $12.06^{\circ}\text{C}$ — and saturated air at that temperature holds 8.76 g/kg, comfortably more than the 8.58 g/kg required. The preheater has done its job.
8. Part (e) — preheater and reheater duties. Both are straightforward enthalpy rises at the plant mass flow: $$Q_{pre}=\dot m\,(h_{2}-h_{M})=2.838(34.14-30.51)=\boxed{10.3\ \text{kW}}$$ $$Q_{re}=\dot m\,(h_{S}-h_{3})=2.838(62.33-34.23)=\boxed{79.7\ \text{kW}}$$ The reheater carries almost the whole 90 kW of plant duty, which is the characteristic signature of this cycle: the preheater exists only to make the humidification possible, while the reheater does the heating.
9. Part (f) — saturation efficiency and make-up water. The adiabatic (saturation) efficiency of a washer compares the moisture it actually adds with the moisture full saturation at the same wet bulb would add: $$\eta=\frac{W_{3}-W_{2}}{W_{s}(t_{2}^{*})-W_{2}} =\frac{0.008575-0.006759}{0.008763-0.006759}=\boxed{90.7\ \%}$$ The temperature form of the same definition, $\eta=(t_{2}-t_{3})/(t_{2}-t_{2}^{*})=(16.92-12.51)/(16.92-12.06)=90.6\,\%$, agrees to within a tenth of a per cent, which confirms that states 2 and 3 lie on one wet-bulb line. The water evaporated — and therefore the make-up required — is $$\dot m_{w}=\dot m\,(W_{3}-W_{2})=2.838(0.008575-0.006759)=0.00515\ \text{kg/s}=\boxed{18.6\ \text{kg/h}}$$
10. Check the whole plant. Everything entering the plant must balance what leaves it: $\dot m(h_{S}-h_{M}) - \dot m_{w}c_{pw}t_{3} = 2.838(62.33-30.51)-0.00515(4.186)(12.51)=90.02$ kW, against $Q_{pre}+Q_{re}=10.3+79.7=90.0$ kW. Independently, the make-up water must equal the moisture that the outdoor-air branch is short of the room condition, $\dot m\,x_{O}(W_{R}-W_{O})=2.838(0.5127)(0.003541)=0.00515$ kg/s — the same figure, reached without using the washer analysis at all.