22-Mec-B2 Environmental Control in Buildings · December 2017

Question 1 of 8: Summer air-conditioning plant — cycle, load and energy input

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 1: Summer air-conditioning plant — cycle, load and energy input (20 points)

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.

Quantity	Symbol	Value
Room (space) state	$t_R,\ \phi_R$	$20\,{}^{\circ}\text{C}$ db, $50\%$ RH
Supply state	$t_S,\ \phi_S$	$14\,{}^{\circ}\text{C}$ db, $60\%$ RH
Supply mass flow (dry air)	$\dot m$	$1.8\ \text{kg/s}$
Outdoor design state	$t_O,\ \phi_O$	$27\,{}^{\circ}\text{C}$ db, $70\%$ RH
Recirculated : fresh air	$-$	$3:1$, so $25\%$ outdoor air by mass
Coil apparatus dew point	$t_{adp}$	$5\,{}^{\circ}\text{C}$ (saturated)
Refrigeration plant	$\text{COP}$	$2$
Losses, fan and pump work	$-$	neglected

Find. The complete plant diagram and psychrometric cycle with every state point characterised, the total room air-conditioning load, the total energy input to the plant, and the energy input when the heating coil is served by condenser heat recovery instead of a separate heat source.

Part (a) — the plant: outdoor air mixes with three parts of recirculated air at M, is cooled and dehumidified to C, sensibly reheated to S, and delivered to the room. The dashed lines show the chilled-water and condenser-water connections.

Approach. Fix the three given states from the chart, take the room load directly as the enthalpy the supply air absorbs, close the adiabatic mixing balance for M, place the off-coil state C on the line from M to the apparatus dew point at the supply humidity ratio, and then split the plant duty between the refrigeration machine and the heater.

Part (c) — fix the three given state points. With $p=101.325\ \text{kPa}$ and $p_{ws}$ from steam tables, $W=0.621945\,p_w/(p-p_w)$ and $h=1.006\,t+W(2501+1.86\,t)$:

Point	Description	$t$ (°C db)	$W$ (g/kg)	$h$ (kJ/kg)	$\phi$ (%)	$t_{wb}$ (°C)	$v$ (m³/kg)
O	outdoor air	27.00	15.72	67.26	70.0	22.78	0.8718
R	room / return	20.00	7.26	38.56	50.0	13.78	0.8402
S	supply to room	14.00	5.95	29.11	60.0	9.95	0.8212

Part (d) — the room load is the enthalpy the supply air picks up. The supply air enters at S and leaves at the room state R, so $$\dot Q_{\text{room}}=\dot m\,(h_R-h_S)=1.8\times(38.56-29.11) =\boxed{17.01\ \text{kW}}$$ Splitting it, the latent part is carried by the moisture the air picks up and the sensible part is the remainder: $$\dot Q_L=\dot m\,h_{fg}\,(W_R-W_S)=1.8\times2501\times(0.007263-0.005945) =5.93\ \text{kW}$$ $$\dot Q_S=\dot Q_{\text{total}}-\dot Q_L=17.01-5.93=11.07\ \text{kW}, \qquad \text{SHR}=\frac{11.07}{17.01}=0.651$$
Mix the outdoor and recirculated streams at M. A recirculated-to-fresh ratio of 3 means the outdoor-air fraction is $x=1/(1+3)=0.25$, i.e. $0.45\ \text{kg/s}$ of outdoor air and $1.35\ \text{kg/s}$ of return air. Adiabatic mixing is exact in moisture and in enthalpy, so both are mass-weighted: $$W_M=x\,W_O+(1-x)\,W_R=0.25(15.72)+0.75(7.26)=9.377\ \text{g/kg}$$ $$h_M=x\,h_O+(1-x)\,h_R=0.25(67.26)+0.75(38.56)=45.73\ \text{kJ/kg}$$ The dry bulb then follows from the enthalpy relation rather than from weighting $t$ directly: $$t_M=\frac{h_M-2501\,W_M}{1.006+1.86\,W_M} =\frac{45.73-2501(0.009377)}{1.006+1.86(0.009377)}=21.77\,{}^{\circ}\text{C}$$ so M is $21.77\,{}^{\circ}\text{C}$ db, $57.7\%$ RH.
Locate the off-coil state C. The cooling coil drives the air along the straight line from M toward its apparatus dew point — saturated air at $5\,{}^{\circ}\text{C}$, for which $W_{adp}=5.403\ \text{g/kg}$ and $h_{adp}=18.59\ \text{kJ/kg}$. The heating coil downstream is sensible only, so it cannot change the humidity ratio; the air must therefore leave the cooling coil already at the supply humidity ratio, $W_C=W_S=5.945\ \text{g/kg}$. The position along the coil line is the contact factor $$\lambda=\frac{W_S-W_M}{W_{adp}-W_M}=\frac{5.945-9.377}{5.403-9.377}=0.8635, \qquad \text{BF}=1-\lambda=0.137$$ $$t_C=t_M+\lambda\,(t_{adp}-t_M)=21.77+0.8635\,(5-21.77) =7.29\,{}^{\circ}\text{C}$$ which is $93.9\%$ RH — a realistic leaving condition for a four-row chilled-water coil, and the by-pass factor of $0.137$ confirms it. Then $h_C=22.28\ \text{kJ/kg}$.
Coil and heater duties. Both follow from the same mass flow across their own enthalpy differences: $$\dot Q_{cc}=\dot m\,(h_M-h_C)=1.8\,(45.73-22.28)=\boxed{42.21\ \text{kW}}$$ $$\dot Q_{hc}=\dot m\,(h_S-h_C)=1.8\,(29.11-22.28)=\boxed{12.28\ \text{kW}}$$ Check the whole plant: what the coil removes, less what the heater puts back, must equal the room load plus the load the outdoor air imposes, $\dot m x (h_O-h_R)=0.45(67.26-38.56)=12.92\ \text{kW}$. Indeed $42.21-12.28=29.93=17.01+12.92\ \text{kW}$ — the balance closes exactly.
Part (e) — total energy input. The refrigeration machine has an overall COP of 2, so it needs $$\dot W_{\text{ref}}=\frac{\dot Q_{cc}}{\text{COP}}=\frac{42.21}{2} =21.11\ \text{kW}$$ and the heating coil draws its $12.28\ \text{kW}$ from a separate source, giving $$\dot E_{\text{total}}=21.11+12.28=\boxed{33.39\ \text{kW}}$$
Part (f) — heating from condenser heat recovery. The condenser must reject everything the evaporator absorbed plus the work put in: $$\dot Q_{\text{cond}}=\dot Q_{cc}+\dot W_{\text{ref}}=42.21+21.11 =63.32\ \text{kW}$$ That is more than five times the $12.28\ \text{kW}$ the heating coil needs, and the condenser water leaves at roughly $35\,{}^{\circ}\text{C}$ — far above the $14\,{}^{\circ}\text{C}$ the reheat coil has to reach — so the reheat is available at no extra energy cost. The purchased input is then the compressor alone: $$\dot E_{\text{total}}=\dot W_{\text{ref}}=\boxed{21.11\ \text{kW}}$$ a saving of $12.28\ \text{kW}$, or $36.8\%$ of the energy input of part (e).

Part (b) — the cycle on ASHRAE Chart No. 1. M lies one quarter of the way along R→O, the coil line M→C aims at the 5 °C apparatus dew point (dashed extension), C→S is the sensible reheat, and S→R is the room load line at SHR = 0.651.

Final results.

Quantity	Result
(d) Total room air-conditioning load	$17.01\ \text{kW}$ (sensible $11.07$, latent $5.93$, SHR $0.651$)
Mixed state M	$21.77\,{}^{\circ}\text{C}$ db, $9.377\ \text{g/kg}$, $45.73\ \text{kJ/kg}$, $57.7\%$ RH
Off-coil state C	$7.29\,{}^{\circ}\text{C}$ db, $5.945\ \text{g/kg}$, $22.28\ \text{kJ/kg}$, $93.9\%$ RH
Coil by-pass factor	$0.137$ (contact factor $0.864$)
Cooling-coil duty	$42.21\ \text{kW}$
Reheat-coil duty	$12.28\ \text{kW}$
(e) Total energy input	$\mathbf{33.39\ \text{kW}}$ ($21.11$ refrigeration $+\ 12.28$ heating)
(f) Energy input with condenser heat recovery	$\mathbf{21.11\ \text{kW}}$ ($36.8\%$ saving)
Condenser heat available	$63.32\ \text{kW}$

Check: assumptions declared under cover-page instruction 1. (i) The plant order is mix → cooling coil → heating coil → fan, which is the order the question lists and the only order in which a coil with a $5\,{}^{\circ}\text{C}$ apparatus dew point can produce a $14\,{}^{\circ}\text{C}$ supply at $60\%$ RH. (ii) The heating coil is taken as purely sensible, which fixes $W_C=W_S$ and hence the off-coil state; without it the split between coil and heater is indeterminate. (iii) Fan and pump work and duct gains are neglected as instructed, so the state leaving the fan is the state entering the room. (iv) In part (f) the recovered heat is treated as free because the condenser must reject it anyway; in practice a desuperheater or a double-bundle condenser is needed, and the condenser water circuit then does slightly more pump work.

Topic. Adiabatic mixing, the coil process line, and the apparatus dew point.

Key relations. Mixing: $W_M=xW_O+(1-x)W_R$ and $h_M=xh_O+(1-x)h_R$, with $t_M=(h_M-2501W_M)/(1.006+1.86W_M)$. Coil: the process line runs from the entering state toward the apparatus dew point on the saturation curve, and the by-pass factor $\text{BF}=(W_C-W_{adp})/(W_M-W_{adp})=(t_C-t_{adp})/(t_M-t_{adp})$ measures how much air escapes contact with the fins. Loads: $\dot Q=\dot m\,\Delta h$ for the total, $\dot Q_L=\dot m\,h_{fg}\,\Delta W$ for the latent part. Plant: $\text{COP}=\dot Q_{\text{evap}}/\dot W$ and $\dot Q_{\text{cond}}=\dot Q_{\text{evap}}+\dot W$.

Why this works. Moisture and energy are both conserved when two air streams mix without heat transfer, which is why $W$ and $h$ — and only those two — combine linearly. A real cooling coil behaves as if a fraction $\lambda$ of the air were brought to saturation at the fin-surface temperature and the remaining fraction $\text{BF}$ passed through untouched; the mixture of those two streams lies on the straight line joining the entering state to the apparatus dew point, which is what makes the graphical construction exact rather than approximate. Because the downstream heater cannot add moisture, the humidity ratio required in the room fixes where on that line the coil must stop, and everything else — the off-coil temperature, the coil duty and the reheat — follows without further assumption. Finally, the condenser is not an incidental part of the machine: it is obliged to reject the evaporator load plus the compressor work, so any building that needs simultaneous cooling and reheat has that heat already paid for.

Common pitfall. Weighting the dry-bulb temperature directly to find the mixed state. It is the usual shortcut, but the $W\!\cdot\!t$ cross term in the enthalpy relation makes $h(t_M,W_M)$ disagree with the mass-weighted $h_M$, and the error propagates into the coil duty. The second trap is forgetting that reheat is sensible: candidates often place the off-coil state at the apparatus dew point itself, which gives a coil duty about $9\%$ too high and a physically impossible by-pass factor of zero.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 3 (mixing) and Ch. 7 (the cooling coil, apparatus dew point and contact factor); ASHRAE Handbook — Fundamentals, Ch. 1.

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