Question 3 of 8: Summer plant — mixing, chilled-water coil, reheat and plant energy

Question 3: Summer plant — mixing, chilled-water coil, reheat and plant energy (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.

Find. The plant diagram and cycle, every state point with dry- and wet-bulb temperature, the total room air-conditioning load, the total plant energy input, and the reduced input when the reheat is taken from condenser cooling water.

Approach. Fix O, R and S from their dry-bulb and relative-humidity pairs; obtain the mixed state from exact moisture and enthalpy balances at 25% outdoor air; use the given apparatus dew point with the supply humidity ratio to find the coil by-pass factor and hence the off-coil state; then take the coil and reheat duties as enthalpy differences on the 2.0 kg/s supply flow, and divide the refrigeration duty by the COP.

1. Fix the three stated air states. With $p = 101.325$ kPa, $W = 0.621945\,p_w/(p-p_w)$ and $h = 1.006t + W(2501+1.86t)$ kJ/kg$_{da}$:

   Point	$t$ (°C)	$\phi$	$W$ (g/kg)	$h$ (kJ/kg)	$t_{wb}$ (°C)
   O outdoor	27.00	70%	15.715	67.25	22.8
   R zone / return	24.00	50%	9.299	47.81	17.1
   S supply	13.00	60%	5.566	27.13	9.1

2. Mix the return and fresh streams. Adiabatic mixing conserves moisture and enthalpy exactly, so with an outdoor-air mass fraction $x = 0.25$ both are mass-weighted: $$W_M=xW_O+(1-x)W_R = 0.25(15.715)+0.75(9.299)=10.903\ \text{g/kg}$$ $$h_M=xh_O+(1-x)h_R = 0.25(67.25)+0.75(47.81)=52.68\ \text{kJ/kg}$$ The dry bulb is then derived from those two, not weighted: $$\boxed{\;t_M=\frac{h_M-2501\,W_M}{1.006+1.86\,W_M}=24.76\,{}^{\circ}\text{C},\quad t_{wb,M}=18.6\,{}^{\circ}\text{C}\;}$$ Weighting the dry bulb directly would give $24.75\,{}^{\circ}\text{C}$ — close, but it makes $h(t_M,W_M)$ disagree with the mass-weighted $h_M$, because of the $Wt$ cross-term in the enthalpy relation. Deriving $t_M$ instead makes $h(t_M,W_M) = h_M$ identically, which is a free check that the mixing arithmetic is right.
3. Part (d) — total room air-conditioning load. The zone loads are stated directly, and the total load is their sum: $$\boxed{\;Q_{\text{room}} = q_s + q_l = 22.5+9.8 = 32.3\ \text{kW}\quad(\text{SHR}=0.697)\;}$$
4. Find the coil by-pass factor and the off-coil state. The apparatus dew point is the saturated state the coil surface behaves as, here $5\,{}^{\circ}\text{C}$ with $W_{adp} = 5.402$ g/kg. The by-pass factor is the fraction of the air that leaves the coil untreated, so on humidity ratio $$\text{BF}=\frac{W_S-W_{adp}}{W_M-W_{adp}}=\frac{5.566-5.402}{10.903-5.402}=0.0298$$ and the same fraction applies to the temperature approach: $$t_C = t_{adp}+\text{BF}\,(t_M-t_{adp}) = 5+0.0298(24.76-5)=5.59\,{}^{\circ}\text{C}$$ giving $h_C = 19.60$ kJ/kg at 98.9% RH. A by-pass factor of only 3% means a deep, many-row coil that takes the air almost exactly to its apparatus dew point — which is what the given supply humidity ratio, barely above $W_{adp}$, demands.
5. Part (c) — the state points. Adding the mixed and off-coil states to the table of step 1 completes the set plotted in Figure 3.2:

   Point	Where	Dry bulb (°C)	Wet bulb (°C)	$W$ (g/kg)	$h$ (kJ/kg)
   O	outdoor air	27.00	22.8	15.715	67.25
   R	zone / return air	24.00	17.1	9.299	47.81
   M	mixing-box exit / coil face	24.76	18.6	10.903	52.68
   C	off cooling coil	5.59	5.5	5.566	19.60
   S	supply to zone	13.00	9.1	5.566	27.13
   ADP	apparatus dew point	5.00	5.0	5.402	18.59

6. Cooling-coil duty. The coil treats the whole 2.0 kg/s from M to C: $$Q_{cool}=\dot m (h_M-h_C)=2.0\,(52.68-19.60)=\boxed{\;66.15\ \text{kW}\;}$$ of which the moisture removed is $\dot m\,\Delta W = 2.0(10.903-5.566)/1000 = 0.01067$ kg/s, or 38.4 kg per hour of condensate.
7. Reheat-coil duty. From C to S at constant humidity ratio: $$Q_{heat}=\dot m (h_S-h_C)=2.0\,(27.13-19.60)=15.07\ \text{kW}$$ Because no moisture changes, this must also equal $\dot m c_p \Delta t = 2.0(1.0163)(13.00-5.59) = 15.07$ kW, which it does.
8. Part (e) — total energy input. The refrigeration plant delivers 66.15 kW of cooling at a COP of 2, so it absorbs 33.08 kW; the heating coil, supplied conventionally, absorbs its full 15.07 kW: $$\boxed{\;E_{\text{total}}=\frac{Q_{cool}}{\text{COP}}+Q_{heat}=\frac{66.15}{2}+15.07=48.14\ \text{kW}\;}$$
9. Part (f) — reheat from condenser cooling water. The condenser must reject everything the plant absorbs plus everything it removes from the air, $Q_{cond} = 66.15+33.08 = 99.23$ kW. That is 6.6 times the 15.07 kW the reheat coil needs, so the reheat is available as waste heat at no additional energy cost and the input collapses to the compressor alone: $$\boxed{\;E_{\text{recovered}}=\frac{Q_{cool}}{\text{COP}}=33.08\ \text{kW}\;}$$ a saving of 15.07 kW, or 31.3% of the total input. The caveat worth stating is one of temperature rather than quantity: condenser water leaving at around $35\,{}^{\circ}\text{C}$ is warm enough to raise air from $5.6$ to $13\,{}^{\circ}\text{C}$ with a comfortable approach, so the recovery is genuinely practical here.