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

Question 3 of 8: Ventilation and infiltration heat loss, and the effect of attic ventilation on ceiling loss

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 3: Ventilation and infiltration heat loss, and the effect of attic ventilation on ceiling loss (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.

(a) Winter ventilation and infiltration load

Given. A single-storey fast-food cafeteria in Ottawa, 30 × 100 × 9 ft, with windows and doors on two of its four sides, held at 72 °F and 30 % relative humidity against a −7 °F outdoor design temperature, and fitted with a humidifier.

Given data
Quantity	Symbol	Value
Floor plan and height	—	30 × 100 × 9 ft
Gross volume	$\mathrm{Vol}$	27,000 ft³
Floor area	$A_f$	3,000 ft²
Exposed sides with openings	—	east and north (two)
Indoor design condition	$t_i,\ \phi_i$	72 $^\circ$F, 30 %
Outdoor design dry bulb	$t_o$	−7 $^\circ$F

Find. The winter design heating load attributable to outdoor air entering the building, both as uncontrolled infiltration and as the ventilation the occupancy requires, together with the humidifier duty that accompanies it.

Approach. Estimate infiltration by the air-change method for a building with two exposed sides; separately compute the outdoor air the occupancy demands under the ASHRAE Standard 62.1 ventilation-rate procedure; then load each air quantity with a sensible and a latent term and take the design value as the larger of the two, because a positively pressurised building does not do both at once.

Check: assumed air-change rate and occupant density. The paper gives no crack data, occupancy or ventilation rate, so two assumptions are declared under cover-page instruction 1. First, 1.0 air change per hour is taken for a building of average construction with openings on two sides (ASHRAE Handbook — Fundamentals, Ch. 16, air-change table). Second, the default occupant density for a cafeteria or fast-food dining room, 100 persons per 1000 ft², is taken from ASHRAE Standard 62.1 Table 6-1, with its rates of 7.5 cfm per person and 0.18 cfm per ft². A lower assumed occupancy scales the ventilation result proportionally but does not change which of the two mechanisms governs.

Moisture content of the two air streams. Indoors at 72 °F (22.22 °C) and 30 % relative humidity, $W_i=0.004976$ lb/lb. The outdoor design condition is quoted as a dry bulb only; at −7 °F (−21.7 °C) the air can hold almost nothing, and taking it as saturated over ice gives $W_o=0.000540$ lb/lb. Hence $$\Delta t=72-(-7)=79\ ^\circ\text{F} \qquad \Delta W=0.004976-0.000540=0.004436\ \text{lb/lb}$$ Assuming the outdoor air saturated is conservative for the humidifier and makes no measurable difference to the sensible term.
Infiltration by the air-change method. With a gross volume of 27,000 ft³ at one air change per hour, $$\dot V_{inf}=\frac{\mathrm{Vol}\times \mathrm{ACH}}{60} =\frac{27,000\times 1.0}{60}=450\ \text{cfm}$$ The standard sea-level coefficients then give $$q_s=1.10\,\dot V\,\Delta t=1.10(450)(79) =39,105\ \text{Btu/h}$$ $$q_l=4840\,\dot V\,\Delta W=4840(450)(0.004436) =9,662\ \text{Btu/h}$$ $$q_{inf}=48,767\ \text{Btu/h}=14.29\ \text{kW}$$
Required ventilation air. A cafeteria is a people-dominated occupancy, and the code minimum is far larger than the crack flow. At 300 persons on 3,000 ft², $$\dot V_{vent}=R_p N+R_a A_f=7.5(300)+0.18(3,000) =2,790\ \text{cfm}$$
Load carried by the ventilation air. The same coefficients applied to the larger flow give $$q_s=1.10(2,790)(79)=242,451\ \text{Btu/h}$$ $$q_l=4840(2,790)(0.004436)=59,902\ \text{Btu/h}$$ $$\boxed{\;q_{vent}=302,353\ \text{Btu/h} =88.6\ \text{kW}\;}$$
Decide which governs. The infiltration estimate is only 16.1 % of the ventilation requirement. A building supplied with 2,790 cfm of outdoor air and relieved through a controlled path runs at positive pressure, and the two mechanisms are therefore not additive: the supply air suppresses the crack flow rather than adding to it. The design figure is the ventilation load. On the east and north faces, however, the wind-driven pressure is highest at exactly the same hours as the design temperature, and those are the two faces with the doors — so the outdoor-air quantity should be confirmed to exceed the wind-driven crack flow on that side, and vestibules or air curtains should be fitted at the entrances to keep it doing so.
Humidifier duty. The latent term above is the energy; the water itself is the mass the humidifier must evaporate, $$\dot m_w=\frac{60\,\dot V_{vent}}{v_o}\,\Delta W =\frac{60(2,790)}{11.42}\,(0.004436) =65.0\ \text{lb/h}=29.5\ \text{kg/h}$$ which is the number the humidifier is selected on.

(b) Ceiling heat loss with and without attic ventilation

Given. A ventilated attic receiving 59 L/s of outside air at −13 °C, bounded below by a 203 m² ceiling of $U=0.30$ W/m²·K over a space at 22 °C, and above by a 244 m² roof of $U=2.7$ W/m²·K.

Find. The heat lost through the ceiling with the ventilation running, and the same loss if the attic were sealed.

Steady-state energy balance on the attic. Heat arriving through the ceiling leaves through the roof and with the ventilation air; the attic temperature is whatever makes the two sides equal.

Approach. The attic is an unheated buffer space, so its temperature is not given — it floats to whatever value balances the heat arriving through the ceiling against the heat leaving through the roof and with the ventilation air. Solve that balance for the attic temperature, then evaluate the ceiling loss.

Conductances and the ventilation capacity rate. At the outdoor condition the air density is $$\rho_o=\frac{p}{R_a T_o}=\frac{101.325}{0.287\times 260.15} =1.3569\ \text{kg/m}^3$$ so the ventilation stream is $\dot m=0.059\times 1.3569=0.08006$ kg/s, and $$\dot m c_p=0.08006\times 1006=80.54\ \text{W/K} \qquad U_cA_c=0.30\times 203=60.90\ \text{W/K} \qquad U_rA_r=2.7\times 244=658.80\ \text{W/K}$$ The flow is quoted at the outdoor state, so it must be converted to mass at the outdoor density, not at 1.2 kg/m³.
Energy balance on the attic. In steady state, $$U_cA_c\,(t_i-t_a)=\left(U_rA_r+\dot m c_p\right)(t_a-t_o)$$ $$60.90\,(22-t_a)=(658.80+80.54)\,(t_a+13)$$ Solving for the attic temperature, $$t_a=\frac{U_cA_c\,t_i+(U_rA_r+\dot m c_p)\,t_o} {U_cA_c+U_rA_r+\dot m c_p}=-10.34\ ^\circ\text{C}$$
Ceiling loss with ventilation. $$\boxed{\;q_c=U_cA_c\,(t_i-t_a)=60.90\,[22-(-10.34)] =1,969\ \text{W}\;}$$ As a check the attic balances: 1,755 W leaves through the roof and 215 W with the ventilation air, and the two sum to 1,969 W.
Ceiling loss without ventilation. Deleting the $\dot m c_p$ term, $$t_a=\frac{U_cA_c\,t_i+U_rA_r\,t_o}{U_cA_c+U_rA_r} =-10.04\ ^\circ\text{C} \qquad \boxed{\;q_c=60.90\,[22-(-10.04)]=1,951\ \text{W}\;}$$
Compare. Ventilating the attic raises the ceiling loss by only $$\Delta q=1,969-1,951=18.2\ \text{W} \qquad\text{that is, }0.93\ \%$$

The reason the penalty is so small is worth stating plainly, because it is the point of the question. The roof conductance, 658.80 W/K, is more than ten times the ceiling conductance, 60.90 W/K — an uninsulated deck over a well-insulated ceiling, which is the normal arrangement. The attic is therefore already tightly coupled to outdoors: even sealed it settles only 2.96 K above the outdoor air, and adding a ventilation conductance of 80.54 W/K to an existing 658.80 W/K pulls it a further 0.30 K down, to 2.66 K above outdoors. In effect, the ceiling already sees very nearly the full 35 K indoor-to-outdoor difference: the ratio of the actual loss to the loss it would suffer with the attic held at the outdoor temperature is 0.924. The engineering conclusion is that the attic ventilation required to control condensation costs almost nothing in heating energy and should not be traded away — the moisture risk it removes is real, and the 18.2 W it costs is not.

Final results — Question 3
Quantity	Value
(a) Infiltration at 1.0 ACH	450 cfm
(a) Infiltration load (sensible + latent)	39,105 + 9,662 = 48,767 Btu/h (14.29 kW)
(a) Required ventilation air	2,790 cfm
(a) Design ventilation load — governs	242,451 + 59,902 = 302,353 Btu/h (88.6 kW)
(a) Humidifier evaporation rate	65.0 lb/h (29.5 kg/h)
(b) Attic temperature, ventilated / sealed	-10.34 / -10.04 $^\circ$C
(b) Ceiling heat loss, ventilated	1,969 W
(b) Ceiling heat loss, sealed attic	1,951 W
(b) Penalty attributable to attic ventilation	18.2 W (0.93 %)

Topic. Outdoor-air heating loads, and the floating temperature of an unheated buffer space.

Key relations. Air-change infiltration $\dot V=\mathrm{Vol}\times\mathrm{ACH}/60$; sea-level sensible and latent coefficients $q_s=1.10\,\dot V\,\Delta t$ and $q_l=4840\,\dot V\,\Delta W$ (which are $60\rho c_p$ and $60\rho h_{fg}$ at standard density); ASHRAE 62.1 breathing-zone outdoor air $\dot V=R_pN+R_aA_f$; buffer-space balance $U_cA_c(t_i-t_a)=(U_rA_r+\dot m c_p)(t_a-t_o)$.

Why this works. The 1.10 and 4840 coefficients are not arbitrary: at standard air density 0.075 lb/ft³, $60\times 0.075\times 0.244=1.10$ Btu/h per cfm per °F, and $60\times 0.075\times 1076=4840$ Btu/h per cfm per unit of humidity ratio. They therefore embed a density, and strictly should be corrected for altitude or for very cold entering air — here the correction is small enough to leave in the noise. The buffer-space balance works because an unheated space has no source or sink of its own: whatever crosses its lower boundary must leave through its upper boundaries, and the temperature is simply the conductance-weighted mean of the two environments it connects. Written that way it is an exact analogue of a voltage divider, with $UA$ playing the part of conductance, and it shows at a glance that the attic temperature is pulled towards whichever side has the larger $UA$.

Common pitfall. Adding the infiltration load to the ventilation load. In a mechanically ventilated, positively pressurised building the outdoor air enters once; counting it twice inflates the plant by more than 15 % here. The second pitfall is computing the ceiling loss as $U_cA_c(t_i-t_o)$ — that is, assuming the attic is at outdoor temperature. It happens to be within 8 % of the truth in this problem because the roof is uninsulated, but on a building with an insulated deck the same shortcut can overstate the ceiling loss by half.

Source. ASHRAE Handbook — Fundamentals (2021) Ch. 16 (ventilation and infiltration) and Ch. 18 (load calculations); ANSI/ASHRAE Standard 62.1 Table 6-1; McQuiston, Parker & Spitler, 6th ed., Ch. 6 (heating loads, unheated spaces and infiltration).

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