07-Str-B6 · May 2018

Question 4 of 6: Temperature Gradient Through a Cavity Wall

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

Notes on this paper

Paper format: National Exams, May 2018 — 07-Str-B6 Building Engineering and Services. Three hours, open book, one Casio or Sharp approved calculator. Six questions of equal value (20 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here. This subject contains no structural analysis at all — the cover page names it Building Engineering and Services, and every question is HVAC, building physics, acoustics or electrical services.

Reference texts (the books an open-book candidate should have on the desk for this subject):

ASHRAE, Handbook — Fundamentals, 2021 — ch. 1 (psychrometrics, and the chart reproduced on page 5 of this paper), ch. 25–27 (thermal resistance of assemblies, surface films, air spaces), ch. 8 (sound and vibration).
F. C. McQuiston, J. D. Parker & J. D. Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design, 6th ed. — ch. 3 (moist-air processes, coil bypass factor, reheat), ch. 5 (wall U-values), ch. 15 (refrigeration cycles).
Y. A. Çengel & M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed. — ch. 11 (vapour-compression refrigeration, R134a tables), ch. 14 (gas–vapour mixtures).
D. Chadderton, Building Services Engineering, 6th ed. — room acoustics, site electrical distribution, ventilation systems for dwellings.
National Research Council Canada, National Building Code of Canada (NBCC) 2020, Part 6 and Subsection 9.32; National Energy Code of Canada for Buildings (NECB) 2020.
CSA C22.1, Canadian Electrical Code, Part I — Section 76 (temporary wiring on construction sites), Section 10 (grounding and bonding), Rule 26-700 (receptacle ground-fault protection).

Check — conventions adopted across this paper. (1) All psychrometry is worked at the 101.325 kPa sea-level barometric pressure printed on the supplied ASHRAE chart, using the standard moist-air relations rather than by scaling off the printed chart; the chart-read and calculated values agree to within the width of a pencil line, and calculating makes every number auditable. (2) Fan heat and duct gains are neglected, as the question intends: the supply-air state is taken as the state leaving the heating coil, and the return-air state as the room state. (3) In Question 3 the paper writes the second cycle as 4 → 1 → 2a → 3a → 4, reusing the label "4"; the state after throttling from 3a is not the same point as the state after throttling from 3, so it is called 4a here and the difference is exactly what changes the refrigerating effect. (4) Question 3 asks for "ideal COP" — taken as the Carnot COP between the stated evaporating and condensing temperatures, with the plotted vapour-compression cycle giving the "actual" COP and the ratio giving the COP efficiency. (5) Question 4 writes thermal conductivity in W/(m·°K); the degree sign on a kelvin is a typographic slip in the paper, and the units are read as W/(m·K). (6) Questions 2, 5 and 6 are answered in the Canadian frame — NBCC/NECB, CSA C22.1 and CSA/ANSI standards.

Question 4: Temperature Gradient Through a Cavity Wall (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 five-layer cavity wall separating a 20 °C interior from a −35 °C exterior, with the conductivity of each solid layer and the resistance of the cavity and of both surface films supplied in the question.

Given data — layers in order from inside to outside
Layer	Thickness L (m)	Conductivity k (W/m·K)	Resistance R = L/k (m²·K/W)
Interior surface film	—	—	0.1200 (given)
Lightweight plaster	0.013	0.16	0.0813
Lightweight concrete block	0.120	0.19	0.6316
Mineral fibre slab	0.040	0.035	1.1429
Air space (cavity)	0.010	—	0.1800 (given)
Brickwork	0.100	0.84	0.1190
Exterior surface film	—	—	0.0600 (given)
Indoor air 20 °C, outdoor air −35 °C, ΔT = 55 K			ΣR = 2.3347

Find. The heat flux through the wall and the temperature at every interface, that is, the full temperature gradient from the inside air to the outside air.

Approach. The layers are in series and the flow is one-dimensional and steady, so the same heat flux passes through every layer; sum the resistances, divide the overall temperature difference by the total to get the flux, then walk inwards to outwards subtracting the temperature drop of each layer in turn.

Convert each solid layer to a thermal resistance. For plane conduction the resistance of a layer of thickness $L$ and conductivity $k$ is $$R = \frac{L}{k}$$ so the plaster gives $0.013/0.16 = 0.0813$, the block $0.120/0.19 = 0.6316$, the mineral fibre $0.040/0.035 = 1.1429$ and the brickwork $0.100/0.84 = 0.1190\ \text{m}^2\!\cdot\!\text{K/W}$. The cavity and the two surface films are supplied as resistances directly, because neither a moving film of air nor a 10 mm ventilated gap transfers heat by conduction alone — radiation and convection dominate, and a single lumped resistance is the standard way to represent them.
Add the resistances in series and form the U-value. One-dimensional steady flow through layers in series simply adds: $$R_{total} = 0.1200 + 0.0813 + 0.6316 + 1.1429 + 0.1800 + 0.1190 + 0.0600 = 2.335\ \text{m}^2\!\cdot\!\text{K/W}$$ $$U = \frac{1}{R_{total}} = \boxed{0.428\ \text{W/m}^2\!\cdot\!\text{K}}$$
Compute the heat flux from the overall driving temperature difference. Because every layer carries the same flux, $$q = \frac{T_{in} - T_{out}}{R_{total}} = \frac{20 - (-35)}{2.335} = \frac{55}{2.335}$$ $$\boxed{q = 23.56\ \text{W/m}^2}$$
Walk the temperature down through the assembly. The drop across any layer is $\Delta T_i = q\,R_i$, so starting from the inside air at 20 °C and subtracting layer by layer: $$\Delta T_{film,in} = 23.56\times 0.1200 = 2.83\ \text{K} \;\Rightarrow\; T = 17.17\ ^\circ\text{C}\ \text{(inside surface)}$$ $$\Delta T_{plaster} = 23.56\times 0.0813 = 1.91\ \text{K} \;\Rightarrow\; T = 15.26\ ^\circ\text{C}$$ $$\Delta T_{block} = 23.56\times 0.6316 = 14.88\ \text{K} \;\Rightarrow\; T = 0.38\ ^\circ\text{C}$$ $$\Delta T_{fibre} = 23.56\times 1.1429 = 26.92\ \text{K} \;\Rightarrow\; T = -26.54\ ^\circ\text{C}$$ $$\Delta T_{cavity} = 23.56\times 0.1800 = 4.24\ \text{K} \;\Rightarrow\; T = -30.78\ ^\circ\text{C}$$ $$\Delta T_{brick} = 23.56\times 0.1190 = 2.80\ \text{K} \;\Rightarrow\; T = -33.59\ ^\circ\text{C}\ \text{(outside surface)}$$ $$\Delta T_{film,out} = 23.56\times 0.0600 = 1.41\ \text{K} \;\Rightarrow\; T = -35.00\ ^\circ\text{C}\ \checkmark$$ Landing exactly on the stated outdoor air temperature confirms both the resistance sum and the flux.

The profile below shows what the numbers mean. The gradient is not a straight line through the wall — it is a straight line through resistance, which is why the plot is drawn against cumulative R rather than against physical thickness. The 40 mm of mineral fibre, less than a sixth of the wall's physical thickness, carries 48.9% of the total temperature drop; the 100 mm of brickwork carries 2.8 K. That is the whole argument for insulated cavity construction in a Canadian climate, and it is also why the insulation position is a design decision and not an afterthought.

Figure 4.1 — the temperature profile plotted against cumulative thermal resistance, so each band's width is proportional to its resistance and the profile is a straight line. Note how the block and the mineral fibre dominate and the brickwork barely registers.

Check — condensation risk, which the arithmetic exposes but the question does not ask. The insulation is on the cold side of the concrete block, so the block runs at 0.4 °C on its outer face and the whole of the block, plaster and inner film sit above freezing. With indoor air at 20 °C and a winter indoor relative humidity of 40%, the dew point is 6.0 °C: the plane at which the wall temperature falls to 6.0 °C lies inside the concrete block, roughly two thirds of the way through it. Warm indoor vapour that reaches that plane will condense there. A vapour barrier on the warm side of the block — behind the plaster — is therefore required, and NBCC Subsection 9.25.4 places it accordingly. If the insulation were moved to the inside face, the block would run near −30 °C and the risk would be far worse; if the humidity were allowed to reach 50% the dew point rises to 9.3 °C and the condensation plane moves further into the block. The inside surface at 17.17 °C is 2.8 K below room temperature, which is comfortable (mean radiant asymmetry well within the ISO 7730 limits) and safely above the dew point, so surface condensation and mould are not a concern.

For context: an assembly at U = 0.428 W/m²·K would not comply with NECB 2020 or the BC Energy Step Code in any Canadian climate zone — Zone 6 (most of southern BC's interior) requires roughly U = 0.24 and Zone 7A something nearer 0.21 for above-grade walls. Reaching 0.24 from here means roughly doubling the insulation, to about 105 mm of the same mineral fibre, or moving to a continuous exterior insulation strategy that also addresses the thermal bridging this one-dimensional calculation ignores.

Final results — Question 4
Quantity	Value
Total thermal resistance ΣR	2.335 m²·K/W
Overall heat transfer coefficient U	0.428 W/m²·K
Heat flux q	23.56 W/m²
Inside air	20.00 °C
Inside surface of plaster	17.17 °C
Plaster / concrete block interface	15.26 °C
Block / mineral fibre interface	0.38 °C
Mineral fibre / cavity interface	−26.54 °C
Cavity / brickwork interface	−30.78 °C
Outside surface of brickwork	−33.59 °C
Outside air	−35.00 °C

Concept & Theory

Topic. One-dimensional steady conduction through a multi-layer building envelope: series thermal resistances, the U-value, and the interface temperature profile.

Key relations. Layer resistance $R = L/k$; series addition $R_{total} = \sum R_i$ including surface films and cavities; $U = 1/R_{total}$; heat flux $q = U\,\Delta T = \Delta T/R_{total}$; interface drop $\Delta T_i = q R_i$; dew point from the indoor humidity ratio via $p_w = P W/(0.621945 + W)$ and $t_{dp} = p_{ws}^{-1}(p_w)$.

Why this works. In steady state with no internal heat generation, conservation of energy forces the same flux through every layer of a plane wall — there is nowhere for energy to accumulate. That single constraint makes the layers behave exactly like resistors in series carrying a common current, with temperature playing the role of voltage. The consequence students find least intuitive is that a layer's share of the temperature drop is set by its resistance, not by its thickness or its temperature: 40 mm of mineral fibre outranks 100 mm of brick by an order of magnitude because their conductivities differ by a factor of 24. The surface films are handled the same way even though the physics inside them is convection and radiation, because the film coefficient $h$ is defined precisely so that $R_{film} = 1/h$ makes the analogy exact.

Common pitfall. Omitting the surface films is the most frequent error — together they contribute 0.18 m²·K/W here, 7.7% of the total, and they are what makes the inside surface temperature (and hence the comfort and condensation check) computable at all. The second is dividing the thickness by the resistance rather than the conductivity for the air space, which is given as a resistance already. The third is plotting the profile against physical thickness and drawing straight lines within each layer joined at the interfaces — that picture is correct, but it hides the point of the exercise, which is that the temperature drop follows resistance.

Source. ASHRAE Handbook — Fundamentals (2021), ch. 25 (heat, air and moisture control), ch. 26 tables 1–3 (material conductivities, surface film and air-space resistances), ch. 27 (condensation control); NBCC 2020 Subsection 9.25.4 (vapour barriers) and NECB 2020 Table 3.2.2.2 (maximum overall thermal transmittance by climate zone).

← Question 3 Next: Question 5 →