Question 2 of 8: Summer cooling and dehumidifying plant — mixed state, coil duty, ADP and by-pass factor

Question 2: Summer cooling and dehumidifying plant — mixed state, coil duty, ADP and by-pass factor (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 recirculating single-coil plant serving a space with both sensible and latent gains, with a fixed ventilation rate and a fixed supply dry-bulb temperature.

Find. The plant diagram and chart cycle with all state points, the state of the air entering the coil, the coil duty, apparatus dew point and by-pass factor, and the room and grand sensible heat factors.

Approach. The room moisture balance fixes the supply humidity ratio once the airflow is known from the sensible balance; the ventilation fraction then fixes the mixed state entering the coil; the coil duty is the enthalpy drop across it. The apparatus dew point is found by extending the coil process line to saturation — and here that construction has something instructive to say.

1. Part (c) — fix the space and outdoor states. From 25 / 18 $^{\circ}\text{C}$, the adiabatic-saturation relation gives $W_{R}=0.010018$ kg/kg, so $h_{R}=1.006(25)+0.010018(2501+46.5)=50.67$ kJ/kg at 50.7 % relative humidity and a dew point of $14.07^{\circ}\text{C}$. From 35 / 25 $^{\circ}\text{C}$, $W_{O}=0.015842$ kg/kg and $h_{O}=75.86$ kJ/kg at 44.7 % relative humidity.
2. Size the supply air on the sensible balance. The supply enters at $15^{\circ}\text{C}$ and leaves at the room temperature, so $$\dot m=\frac{q_{s}}{(c_{pa}+c_{pv}W_{R})(t_{R}-t_{S})} =\frac{25}{(1.006+1.86\times0.010018)(25-15)}=\boxed{2.440\ \text{kg/s}}$$ which is 2.02 $\text{m}^{3}/\text{s}$ (about 4 300 cfm) at the supply state.
3. The room moisture balance fixes the supply humidity ratio. Every kilogram of moisture released in the space must leave in the air stream: $$W_{S}=W_{R}-\frac{\dot m_{w}}{\dot m}=0.010018-\frac{0.005556}{2.440} =0.007741\ \text{kg/kg}$$ so the supply state is $15^{\circ}\text{C}$ dry bulb, $12.25^{\circ}\text{C}$ wet bulb, 73.0 % relative humidity, $h_{S}=34.67$ kJ/kg, with a dew point of $10.21^{\circ}\text{C}$. The room total load is then $\dot m(h_{R}-h_{S})=2.440(50.67-34.67)=39.05$ kW, of which 25 kW is sensible and 14.05 kW latent.
4. Part (f), first half — room sensible heat factor. $$\mathrm{RSHF}=\frac{q_{s}}{q_{s}+q_{l}}=\frac{25.0}{39.05}=\boxed{0.640}$$ A room factor this low is the whole character of the problem: for every three units of cooling, more than one is spent condensing water.
5. Part (d) — the state entering the coil. The ventilation air is 0.480 kg/s, so 1.960 kg/s is recirculated (19.7 % outdoor air by mass). Weighting moisture and enthalpy, $$W_{M}=\frac{0.480(0.015842)+1.960(0.010018)}{2.440}=0.011164\ \text{kg/kg}, \qquad h_{M}=\frac{0.480(75.86)+1.960(50.67)}{2.440}=55.63\ \text{kJ/kg}$$ and recovering the dry bulb from the enthalpy definition, $$t_{M}=\frac{55.63-2501(0.011164)}{1.006+1.86(0.011164)} =\boxed{26.98^{\circ}\text{C}\ \text{dB},\ \ 19.54^{\circ}\text{C}\ \text{wB}}$$
6. Part (e), first part — coil capacity. The coil takes the mixture down to the supply state, so $$Q_{c}=\dot m\,(h_{M}-h_{S})=2.440(55.63-34.67)=\boxed{51.1\ \text{kW}}$$ This is exactly the room load plus the outdoor-air load, $39.05+0.480(75.86-50.67)=39.05+12.09=51.14$ kW — a check that costs one line and catches most mixing errors.
7. Part (f), second half — grand sensible heat factor. The sensible part of the coil duty is the dry-bulb drop at the entering moisture content, $\dot m(c_{pa}+c_{pv}W_{M})(t_{M}-t_{S})=2.440(1.0268)(26.98-15)=30.02$ kW, so $$\mathrm{GSHF}=\frac{30.02}{51.14}=\boxed{0.587}$$ The coil factor is lower than the room factor because the outdoor air brings in a further 7.1 kW of latent load on top of the room's own 14.05 kW.
8. Part (e), second part — the apparatus dew point. The apparatus dew point is where the straight coil process line M–S, extended below S, meets the saturation curve. Extending it here: $$W(t)=W_{S}+\frac{W_{M}-W_{S}}{t_{M}-t_{S}}\,(t-t_{S}) =0.007741+0.0002856\,(t-15)$$ At $10^{\circ}\text{C}$ this line lies at 6.31 g/kg against 7.66 g/kg on the saturation curve; at $5^{\circ}\text{C}$, 4.89 against 5.42; at $0^{\circ}\text{C}$, 3.46 against 3.79. The line never reaches saturation. Its closest approach is 0.30 g/kg, at about $-1.2^{\circ}\text{C}$, so there is no apparatus dew point and no by-pass factor for the process as literally specified. The physical statement of the same fact is that a state of $15^{\circ}\text{C}$ at 73 % relative humidity is far drier than any real coil leaves its air — coils leave air at 90 to 95 % saturation.
9. Resolve it the way a designer would. The moisture balance is not negotiable: the coil must deliver $W=0.007741$ kg/kg or the space humidity will not hold. What can be relaxed is the assumption that the air reaches the room at exactly the temperature it leaves the coil, because the fan and the duct run add heat. Taking a normal four-row chilled-water coil with a by-pass factor $\mathrm{BF}=0.15$, the coil surface state follows from $W_{S}=W_{adp}+\mathrm{BF}\,(W_{M}-W_{adp})$: $$W_{adp}=\frac{0.007741-0.15(0.011164)}{0.85}=0.007137\ \text{kg/kg} \;\Rightarrow\; \mathrm{ADP}=\boxed{9.0^{\circ}\text{C}}$$ $$t_{off}=\mathrm{ADP}+\mathrm{BF}\,(t_{M}-\mathrm{ADP})=9.02+0.15(26.98-9.02)=11.71^{\circ}\text{C}$$ which is 90.5 % saturated — entirely normal. The coil duty on this basis is $\dot m(h_{M}-h_{off})=2.440(55.63-31.32)=59.3$ kW, and the remaining $3.3\ \text{K}$ rise to the stated $15^{\circ}\text{C}$, worth 8.2 kW, is made up by fan and duct gain and, if that is not enough, by a small reheat coil.
10. Part (b) — plot it. The chart below carries all five points. Note that S does not lie on the line from M to the apparatus dew point; it lies to the right of the off-coil state C, on the same horizontal line, because the fan and duct gain is a sensible process.