22-Mec-B2 Environmental Control in Buildings · May 2016

Question 8 of 8: Summer plant — supply air, coil capacity, apparatus dew point and bypass factor

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

Notes on this paper

Paper format. Professional Engineers of Ontario / Engineers Canada annual examination 07-Mec-B2 Environmental Control in Buildings, May 2016, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. Psychrometric charts and an R-134a p-h diagram are appended to the paper. All eight problems are solved here, because the set is intended as a study resource rather than an examination script.

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), Ch. 5–6 (heating and cooling loads), Ch. 10 (cooling towers), Ch. 15 (fans and duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 6 (heating loads and infiltration), Ch. 8 (energy estimating and the degree-day method), Ch. 12–13 (fluid flow, fans and duct design).
ASHRAE Handbook — Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 16 (ventilation and infiltration), Ch. 18 (residential and non-residential load calculations), Ch. 21 (duct design), Ch. 30 (thermophysical properties of refrigerants).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration, multistage systems and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality; ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 and its Appendix C design temperatures; National Energy Code of Canada for Buildings 2020; Environment and Climate Change Canada heating degree-day normals; CSA B52 Mechanical Refrigeration Code; Canadian federal halocarbon regulations (SOR/2003-289) for the refrigerant discussion in Problem 7.

Check: assumptions carried through this paper. Cover-page instruction 1 invites a clear statement of any assumption. Standard barometric pressure of 101.325 kPa is used throughout; moist-air properties follow the ASHRAE Handbook — Fundamentals Ch. 1 formulation (Hyland–Wexler saturation pressure, so results agree with the appended chart to chart-reading accuracy rather than being read off it); R-134a properties are on the IIR datum and agree with the appended p-h diagram. Problem-specific assumptions — climate data, fuel prices, emission factors, air-change rates, occupant density, duct roughness and the coil bypass factor — are stated where they are first used.

Question 8: Summer plant — supply air, coil capacity, apparatus dew point and bypass 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 space with a 12-ton cooling load of which 3 tons is latent, held at 78 °F dry bulb and 50 % relative humidity, requiring 1000 cfm of ventilation air on a day when the outdoor air is at 94 °F dry bulb and 55 % relative humidity; return air mixes with the ventilation air ahead of a filter, fan and cooling coil, at sea level and neglecting duct gain and fan temperature rise.

Given data
Quantity	Symbol	Value
Total room cooling load	—	12 tons = 144,000 Btu/h
Room latent load	$RLH$	3 tons = 36,000 Btu/h
Room sensible load	$RSH$	9 tons = 108,000 Btu/h
Room state	$t_R,\ \phi_R$	78 $^\circ$F, 50 %
Outdoor state	$t_O,\ \phi_O$	94 $^\circ$F, 55 %
Ventilation air	$\dot V_{oa}$	1000 cfm
Room sensible heat factor	$RSHF$	0.750

Find. (a) the plant arrangement, (b)–(c) the four significant states with dry- and wet-bulb temperatures, plotted on the chart, (d) the supply air rate, and (e) the coil capacity in kilowatts, the apparatus dew point and the coil bypass factor.

(a) Diagram of the system

Mixed-air plant. Return air from the space is split between relief and recirculation; the recirculated fraction mixes with 1000 cfm of outdoor air at M, and the mixture passes through the filter, fan and cooling coil to the supply state S.

(b), (c) State points and the operating cycle

Approach. The supply temperature is not given, so it cannot simply be assumed — instead use the Carrier effective-sensible- heat-factor construction, which fixes the apparatus dew point from the room state and the loads, and then derives the supply air quantity and state from an assumed coil bypass factor. The mixed state follows from a mass balance once the supply flow is known.

Check: assumed coil bypass factor. The question asks for the bypass factor as an output, but the supply air rate cannot be determined without either a supply temperature or a bypass factor, so one assumption is unavoidable under cover-page instruction 1. A bypass factor of 0.15 is assumed, typical of a four-row chilled-water coil at about 500 ft/min face velocity. The construction is then checked for self-consistency at the end: the bypass factor recovered geometrically from the finished state points is 0.151, reproducing the assumption, so the solution is internally closed rather than merely assumed.

Room and outdoor states. At 78 °F and 50 % relative humidity, $W_R=0.010219$ lb/lb and $h_R=29.92$ Btu/lb; at 94 °F and 55 %, $W_O=0.018986$ lb/lb and $h_O=43.50$ Btu/lb. The room sensible heat factor is $$RSHF=\frac{RSH}{RSH+RLH}=\frac{108,000}{144,000}=0.750$$
Load imposed by the ventilation air. Bringing 1000 cfm of outdoor air to the room state costs $$OASH=1.10\,\dot V_{oa}\,(t_O-t_R)=1.10(1000)(94-78) =17,600\ \text{Btu/h}$$ $$OALH=4840\,\dot V_{oa}\,(W_O-W_R)=4840(1000)(0.018986-0.010219) =42,433\ \text{Btu/h}$$ The latent term dominates, which is characteristic of a humid summer design day and is the reason the coil ends up so much larger than the room load.
Effective sensible heat factor. The bypassed fraction of the outdoor air reaches the room without being treated, so it behaves exactly like a room load; the treated fraction is handled at the coil. Adding only the bypassed part to the room loads, $$ERSH=RSH+BF\cdot OASH=108,000+0.15(17,600)=110,640\ \text{Btu/h}$$ $$ERLH=RLH+BF\cdot OALH=36,000+0.15(42,433)=42,365\ \text{Btu/h}$$ $$ESHF=\frac{ERSH}{ERSH+ERLH}=0.7231$$
Apparatus dew point. The $ESHF$ line drawn through the room state meets the saturation curve at the apparatus dew point. In $(t,W)$ coordinates that line has slope $$\frac{dW}{dt}=\frac{1-ESHF}{ESHF}\cdot\frac{1.10}{4840} =8.70\times 10^{-5}\ \text{lb/lb per }^\circ\text{F}$$ and solving $W_R+({dW}/{dt})(t-t_R)=W_{sat}(t)$ gives $$\boxed{\;t_{adp}=50.74\ ^\circ\text{F}\;}$$ This is a comfortable dew point for a chilled-water coil; had it come out below about 45 °F the design would have needed a lower bypass factor or dehumidification by other means.
Wet-bulb temperatures of the four states. Solving the adiabatic-saturation relation at each state gives room 65.02, outdoor 80.07, mixed 68.65 and supply 53.93 °F, collected in the table below.

(d) Air supply rate

Supply flow from the effective sensible load. Air leaving the coil is a blend of air that contacted the fins, at the apparatus dew point, and air that bypassed them, so the useful temperature difference is only $(1-BF)(t_R-t_{adp})$: $$\boxed{\;\dot V=\frac{ERSH}{1.10\,(1-BF)\,(t_R-t_{adp})} =\frac{110,640}{1.10(0.85)(78-50.74)} =4,341\ \text{cfm}\;}$$
Supply state. Back-substituting into the room sensible and latent balances, $$t_S=t_R-\frac{RSH}{1.10\,\dot V} =78-\frac{108,000}{1.10(4,341)}=55.38\ ^\circ\text{F}$$ $$W_S=W_R-\frac{RLH}{4840\,\dot V}=0.008505\ \text{lb/lb}$$ a supply condition of 55.38 °F at 91.3 % relative humidity, which is a normal leaving-coil state, and a 22.6 °F temperature difference into the room — comfortably within the range that ordinary ceiling diffusers can handle without dumping.
Mixed state. Converting to mass, the humid volumes are $v_S=13.162$ and $v_O=14.384$ ft³/lb, so $$\dot m_S=\frac{4,341}{13.162}=329.8\ \text{lb/min} \qquad \dot m_O=\frac{1000}{14.384}=69.52\ \text{lb/min}$$ an outdoor-air fraction of 0.2108, or 21.1 % by mass. Weighting moisture and enthalpy, $$W_M=0.012066\ \text{lb/lb} \qquad h_M=32.78\ \text{Btu/lb} \qquad t_M=81.41\ ^\circ\text{F}$$

State points
Point	Dry bulb ($^\circ$F)	Wet bulb ($^\circ$F)	$W$ (lb/lb)	$h$ (Btu/lb)
O — outdoor / ventilation air	94.00	80.07	0.018986	43.50
R — room and return air	78.00	65.02	0.010219	29.92
M — mixed air, entering the coil	81.41	68.65	0.012066	32.78
S — supply air, leaving the coil	55.38	53.93	0.008505	22.53
ADP — apparatus dew point	50.74	50.74	saturated	—

The summer cycle. O and R mix to M; the coil line runs from M through S and extends to the apparatus dew point on the saturation curve; the room line from S to R has slope RSHF, and the ESHF line from R fixes the ADP.

(e) Coil capacity, apparatus dew point and bypass factor

Coil capacity. The coil sees the mixed state and delivers the supply state, so $$\boxed{\;\dot Q_{coil}=\dot m_S\,(h_M-h_S)\times 60 =329.8\,(32.78-22.53)(60) =202,901\ \text{Btu/h}=59.46\ \text{kW}\;}$$ that is 16.91 tons. Checking against first principles, the coil must absorb the room load plus the outdoor-air load: 144,000 + 56,647 = 200,647 Btu/h, agreeing within 1 % — the residual is the density correction between the standard 1.10 and 4840 coefficients and the actual humid volume of 94 °F air.
Bypass factor, recovered from the geometry. The coil line M–S extended must strike saturation at the apparatus dew point, and the bypass factor is the fraction of the M-to-ADP interval that remains untreated: $$\boxed{\;BF=\frac{t_S-t_{adp}}{t_M-t_{adp}} =\frac{55.38-50.74}{81.41-50.74}=0.151\;}$$ which reproduces the assumed 0.15 to within 1 %. That agreement is the proof that the construction is closed: the assumption entered through the $ESHF$ and comes back out of an independent geometric relation.

The most instructive number in this problem is the ratio of the coil to the room load. The space needs 12 tons, but the coil must be selected for 16.91 — 27.9 % of its duty is spent on 1000 cfm of ventilation air, which is only 21.1 % of the air it handles. Two thirds of that penalty is latent, because the outdoor air at 94 °F and 55 % carries almost twice the moisture of the room air. It is exactly this term that an energy-recovery ventilator attacks: a total-enthalpy wheel at 70 % effectiveness on the 1000 cfm would recover about 41,000 Btu/h and cut the coil by more than three tons, for no change to the space conditions. In a Canadian climate the same device pays a second time in winter.

Final results — Question 8
Quantity	Value
Room sensible / latent load	108,000 / 36,000 Btu/h
Outdoor-air sensible / latent load	17,600 / 42,433 Btu/h
Room sensible heat factor	0.750
Effective sensible heat factor	0.7231
(e) Apparatus dew point	50.74 $^\circ$F
(d) Air supply rate	4,341 cfm
Supply state	55.38 $^\circ$F DB, 53.93 $^\circ$F WB (91.3 % RH)
Mixed state entering the coil	81.41 $^\circ$F DB, 68.65 $^\circ$F WB
Outdoor-air fraction by mass	21.1 %
(e) Coil capacity	202,901 Btu/h = 59.46 kW = 16.91 tons
(e) Coil bypass factor	0.151

Topic. The Carrier apparatus-dew-point and bypass-factor method for selecting a cooling coil.

Key relations. $RSHF=RSH/(RSH+RLH)$; $ERSH=RSH+BF\cdot OASH$ and $ERLH=RLH+BF\cdot OALH$ with $ESHF=ERSH/(ERSH+ERLH)$; the $ESHF$ line through the room state meets saturation at $t_{adp}$; $\dot V=ERSH/[1.10(1-BF)(t_R-t_{adp})]$; $BF=(t_S-t_{adp})/(t_M-t_{adp})=(h_S-h_{adp})/(h_M-h_{adp})$; $\dot Q_{coil}=\dot m_S(h_M-h_S)$.

Why this works. A finned coil cannot bring all of the air into contact with its wetted surface. The bypass-factor model idealises that imperfection in the simplest way that is still useful: it treats the leaving air as a mixture of a fraction $(1-BF)$ that reached equilibrium with the fin surface, and hence left saturated at the apparatus dew point, and a fraction $BF$ that passed through untouched at the entering state. Because it is a mixture of two states, the leaving point must lie on the straight line joining them, which is why the coil process line extended always strikes saturation at the ADP, and why the bypass factor can be read off as a simple length ratio in either temperature or enthalpy. The effective sensible heat factor is the companion idea applied to the outdoor air: the bypassed fraction of the ventilation air never sees the coil, so it arrives in the room untreated and must be carried by the supply air exactly as a room load. Adding $BF\cdot OASH$ and $BF\cdot OALH$ to the room loads accounts for that, and makes the ADP line pass through the room state, which is what allows the supply quantity to be determined without ever assuming a supply temperature.

Common pitfall. Drawing the RSHF line from the room state to saturation and calling the intersection the apparatus dew point. That is the room ADP; it ignores the bypassed ventilation air entirely and gives a supply quantity several per cent adrift. The second and more expensive pitfall is sizing the coil on the room load alone: here that would specify a 12-ton coil where 16.91 tons is required, and the plant would never hold the space on a design day. A third is computing the coil duty from $\dot V\times$ enthalpy difference using standard density — the mixed air at 81.41 °F is lighter than standard, so the mass flow must be found from the actual humid volume.

Source. Carrier Handbook of Air Conditioning System Design, Part 1 Ch. 8 (apparatus dew point, effective sensible heat factor and bypass factor); W. P. Jones, Air Conditioning Engineering, 5th ed., Ch. 6 and §3.9 (the cooling-coil process and contact factor); McQuiston, Parker & Spitler, 6th ed., Ch. 3.10 and Ch. 14; ASHRAE Handbook — Fundamentals (2021) Ch. 1.

← Question 7 Back to the paper →