22-Mec-B2 Environmental Control in Buildings · May 2015

Question 5 of 8: Cavity wall with a window — the window's share of the heat loss

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

Notes on this paper

Paper format. Professional Engineers of Ontario / EGBC annual examination, 07-Mec-B2 (now 22-Mec-B2) Environmental Control in Buildings, May 2015 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate on the cover of the first workbook which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for the refrigerant are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to submit a clear statement of any interpretation assumptions with the answer — that latitude is used explicitly below wherever the printed data are redundant or incomplete.

All eight problems are worked here. Every psychrometric state has been recomputed from the ASHRAE formulation for saturation vapour pressure rather than scaled off a chart, so the numbers below are tighter than a graphical solution would be; chart-quality agreement (about $\pm 0.2$ K in temperature and $\pm 0.0002$ kg/kg in humidity ratio) is all that an examiner expects, and a candidate reading the appended charts should reproduce every answer to within that band.

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. 6 (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. 8 (energy estimating and degree-day methods), Ch. 12–13 (fluid flow, fans and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (fans and duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope), Ch. 30 (refrigerant properties).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality, and ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 (Part 6, and Appendix C for design temperatures and degree-days), National Energy Code of Canada for Buildings 2020, Health Canada Residential Indoor Air Quality Guidelines, and Environment and Climate Change Canada Canadian Climate Normals.

Psychrometric relations used throughout. At barometric pressure $p$ with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation, the humidity ratio, specific enthalpy and humid volume of moist air are

$$W = 0.621945\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t), \qquad v = \frac{0.287042\,(t + 273.15)\,(1 + 1.6078\,W)}{p}$$

with $h$ in kJ per kg of dry air, $t$ in $\,{}^{\circ}$C and $p$ in kPa. In inch-pound units the enthalpy becomes $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air with $t$ in $\,{}^{\circ}$F. Wet-bulb temperatures are obtained by solving the adiabatic-saturation equation, not by eye. Mixing two air streams is exact in moisture and in enthalpy, so $W$ and $h$ of the mixture are the mass-weighted averages and the mixed dry bulb follows from them; weighting the dry bulb directly is the usual shortcut and differs here by about $0.01$ K.

Question 5: Cavity wall with a window — the window's share of the heat 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.

Given. A cavity wall of two brick leaves, rendered outside and plastered inside, pierced by a single-glazed window. The two constructions span the same indoor-to-outdoor temperature difference, so they are thermal resistances in parallel.

Given data
Quantity	Symbol	Value
Inside surface coefficient	$h_i$	8.5 W/m$^2$K
Outside surface coefficient	$h_o$	31 W/m$^2$K
Brick leaves	$k$, $L$	0.43 W/mK, 125 mm each (two leaves)
Air cavity resistance	$R_{cav}$	0.15 m$^2$K/W
Internal plaster	$k$, $L$	0.14 W/mK, 10 mm
External cement render	$k$, $L$	0.86 W/mK, 5 mm
Glazing	$k$, $L$	0.76 W/mK, 1.5 mm
External wall size	—	4 m × 2.5 m
Window size	—	1.8 m × 1.2 m

Find. The proportion of the total heat transfer through the external wall that passes through the window.

Section through the cavity wall (layers to scale by resistance label, not by thickness), the elevation of the 4 m × 2.5 m external wall containing the 1.8 m × 1.2 m window, and the resulting split of the total UA between the two parallel paths.

Approach. Sum the series resistances of each construction to get its U-value, multiply each by its own area to get a UA, and take the window's share of the total UA — which is also its share of the heat flow, since both paths see the same temperature difference.

Build up the wall resistance. For one square metre of the opaque construction the series resistances are the inside film, the plaster, the inner brick leaf, the cavity, the outer brick leaf, the render and the outside film:$$R_{wall} = \frac{1}{h_i} + \frac{L_{pl}}{k_{pl}} + \frac{L_{br}}{k_{br}} + R_{cav} + \frac{L_{br}}{k_{br}} + \frac{L_{ce}}{k_{ce}} + \frac{1}{h_o}$$$$R_{wall} = \frac{1}{8.5} + \frac{0.010}{0.14} + \frac{0.125}{0.43} + 0.15 + \frac{0.125}{0.43} + \frac{0.005}{0.86} + \frac{1}{31}$$$$R_{wall} = 0.11765 + 0.07143 + 0.29070 + 0.15000 + 0.29070 + 0.00581 + 0.03226 = 0.95854 \text{ m}^2\text{K/W}$$ The two brick leaves and the cavity together contribute $0.7314$ m$^2$K/W, or $76$ % of the total — the render, at $0.00581$, is thermally negligible.
Wall U-value.$$U_{wall} = \frac{1}{R_{wall}} = \frac{1}{0.95854} = 1.0433 \text{ W/m}^2\text{K}$$
Window resistance and U-value. The glass itself is only 1.5 mm thick and highly conductive, so the two surface films dominate almost completely:$$R_{win} = \frac{1}{h_i} + \frac{L_{gl}}{k_{gl}} + \frac{1}{h_o} = 0.11765 + 0.001974 + 0.03226 = 0.15188 \text{ m}^2\text{K/W}$$$$U_{win} = \frac{1}{0.15188} = 6.5842 \text{ W/m}^2\text{K}$$ The glass contributes just $0.001974$ m$^2$K/W, about $1.3$ % of the window's resistance; a single sheet of glass is essentially a hole in the wall as far as conduction is concerned.
Areas of the two parallel paths.$$A_{total} = 4.0 \times 2.5 = 10.00 \text{ m}^2, \qquad A_{win} = 1.8 \times 1.2 = 2.16 \text{ m}^2$$$$A_{wall} = 10.00 - 2.16 = 7.84 \text{ m}^2$$ The window occupies $21.6$ % of the elevation by area.
Combine the paths by UA. Both constructions span the same inside-to-outside temperature difference $\Delta t$, so the heat flows add and the temperature difference cancels out of the ratio:$$(UA)_{wall} = 7.84 \times 1.0433 = 8.1791 \text{ W/K}$$$$(UA)_{win} = 2.16 \times 6.5842 = 14.2219 \text{ W/K}$$$$(UA)_{total} = 8.1791 + 14.2219 = 22.4009 \text{ W/K}$$
The window's share of the heat transfer.$$\frac{\dot{Q}_{win}}{\dot{Q}_{total}} = \frac{(UA)_{win}}{(UA)_{total}} = \frac{14.2219}{22.4009} = \boxed{0.6349 \;=\; 63.5\ \%}$$ The window covers only $21.6$ % of the elevation but carries $63.5$ % of the heat loss, because its U-value is $6.5842/1.0433 = 6.31$ times that of the wall. Put another way, each square metre of glazing loses as much heat as about $6.3$ m$^2$ of the cavity wall.

Check: the 1.5 mm glazing and the neglect of radiation. The paper prints 1.5 mm glass, which is thinner than any practical glazing (3–6 mm is normal), but the answer is insensitive to it: at 6 mm the window U-value would fall only from 6.5842 to 6.3371 W/m$^2$K and the window's share from 63.5% to 62.6%, because the glass contributes barely 1.3% of the resistance either way. The question also instructs that radiation be neglected; the surface coefficients given (8.5 and 31 W/m$^2$K) are the standard combined convective-plus-radiative film values, so the instruction is best read as excluding solar gain and long-wave exchange with the sky rather than as stripping radiation out of the films.

Final results
Quantity	Symbol	Result
Wall total resistance	$R_{wall}$	0.95854 m$^2$K/W
Wall U-value	$U_{wall}$	1.0433 W/m$^2$K
Window total resistance	$R_{win}$	0.15188 m$^2$K/W
Window U-value	$U_{win}$	6.5842 W/m$^2$K
Opaque wall area	$A_{wall}$	7.84 m$^2$
Window area	$A_{win}$	2.16 m$^2$
Wall UA	$(UA)_{wall}$	8.1791 W/K
Window UA	$(UA)_{win}$	14.2219 W/K
Total UA	$(UA)_{total}$	22.4009 W/K
Window's share of the heat transfer	—	63.5%

Topic. Series and parallel thermal resistance in a building envelope.

Key relations. Series (through one construction): $R = 1/h_i + \sum L_j/k_j + R_{cav} + 1/h_o$ and $U = 1/R$. Parallel (side-by-side constructions): $(UA)_{total} = \sum A_j U_j$, and each path's share of the heat flow is $A_j U_j / \sum A_k U_k$ because $\Delta t$ is common. Heat flow itself: $\dot{Q} = (UA)\,\Delta t$.

Why this works. Layers stacked through a wall carry the same heat flux and their resistances add; constructions sitting side by side in the same elevation see the same temperature difference and their conductances add. The question only asks for a ratio, so the temperature difference never has to be known — it cancels identically. What makes the answer striking is the arithmetic of the two U-values: adding brick and a cavity buys real resistance, while a single sheet of glass adds essentially none, leaving the window's U-value pinned near the value of the two surface films alone. That is why glazing dominates envelope heat loss out of all proportion to its area, and why double glazing with a low-emissivity coating and an argon fill is the single most effective envelope upgrade in a Canadian climate.

Common pitfall. Area-weighting the U-values instead of computing UA for each path separately, or averaging the two resistances — resistances in parallel do not add. Another frequent slip is counting only one brick leaf, which would raise the wall U-value to 1.4974 W/m$^2$K and drop the window's share to 54.8%.

Source. ASHRAE Handbook – Fundamentals (2021) Ch. 25–27 (heat, air and moisture transfer, and the parallel-path method); Jones, Air Conditioning Engineering, 5th ed., Ch. 5; NBC 2020 Appendix C and NECB 2020 for Canadian envelope requirements.

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