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22-Mec-B2 Environmental Control in Buildings · December 2019

Question 4 of 8: Composite wall with a plaster thermal bridge; moisture flow and vapour barriers

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

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Mec-B2 Environmental Control in Buildings. Three hours, open book: only textbooks and reference books are permitted (no notes and no solved problems), any non-communicating calculator is allowed, and candidates are expected to bring both an environmental-control text and steam tables because the tables and graphs in those books are needed. Eight problems are printed at 20 points each and only the first five in the exam book are graded, so the printed paper totals 160 points and a graded script totals 100. Psychrometric charts (IP and SI) and an R-134a pressure–enthalpy diagram are attached as the last three pages. All eight problems are worked below.

Reference texts for this subject.

Check — assumptions declared under cover-page instruction 1

Cover-page instruction 1 asks candidates to state any interpretive assumption with the answer. Four are needed on this paper and each is flagged again where it is used: the operating-room dry-bulb temperature in Problem 1 (not given — taken as 75 °F, the top of the ASHRAE 170 range, because it is the only part of that range that also satisfies the 60 % relative-humidity ceiling); the Winnipeg design conditions and degree-day base in Problem 6 (the paper says “select the design conditions”); the indoor design temperature and neutral pressure level in Problem 7(b); and the duct roughness and fitting allowance in Problem 8. Everything else in the paper is fully determined by the data given.

Question 4: Composite wall with a plaster thermal bridge; moisture flow and vapour barriers (20 marks: a — 15, b — 5)

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.

Part (a) — rate of heat transfer through the wall (15 marks)

Given.

QuantitySymbolValue
Wall height × widthH × W3 m × 5 m = 15 m²
Brick cross-section (thickness × height)—16 cm × 22 cm, k = 0.72 W/m·°C
Plaster between brick courses—3 cm total (1.5 cm above and below each brick), k = 0.22
Plaster each side of the brick—2 cm, k = 0.22 W/m·°C
Rigid foam, inner face—3 cm, k = 0.026 W/m·°C
Indoor / outdoor air temperatureT∞1 / T∞225 / −10 °C
Inside / outside film coefficienth1 / h210 / 25 W/m²·°C

Find. The total steady one-dimensional heat-transfer rate through the 15 m² wall, in watts.

Representative element — one brick course, 0.25 m high × 1 m deepBrick k = 0.7222 cm × 16 cmFoamk=0.026Plasterk=0.22Plasterplaster bridge3 cm2 cm16 cm2 cm22 cm1.5 cm1.5 cminside 25 °Chi = 10 W/m²·°Coutside -10 °Cho = 25 W/m²·°CHeat flows left to right; the thin plaster ribs above and below each brick are a parallel thermal bridge across the 16 cm core.
Figure 4.1 — the repeating element read off the paper's own section: 23 cm of wall thickness (3 cm foam, 2 cm plaster, 16 cm brick core, 2 cm plaster) by 25 cm of height (a 22 cm brick with a 1.5 cm plaster rib above and below — half of the 3 cm joint shared with each neighbouring course).

Approach. The wall repeats on a 25 cm pitch, so analyse one 25 cm × 1 m element as a series chain of resistances in which the 16 cm core is a parallel pair — brick over 22 cm of the height, plaster over the remaining 3 cm — then scale the element result by the number of elements in the wall.

  1. Identify the repeating element. Each brick course occupies 22 + 3 = 25 cm of height, sharing 1.5 cm of plaster with the course above and 1.5 cm with the one below. Taking 1 m of wall length, the element face area is $$A = 0.25 \times 1.0 = 0.25\ \text{m}^2$$ and the element is bounded by adiabatic planes through the middle of each plaster joint, which is what makes the one-dimensional assumption legitimate.
  2. Compute the two convection resistances. With R = 1/(hA), $$R_i = \frac{1}{10 \times 0.25} = 0.400\ {}^{\circ}\text{C/W}, \qquad R_o = \frac{1}{25 \times 0.25} = 0.160\ {}^{\circ}\text{C/W}$$
  3. Compute the full-width conduction layers. The foam and the two 2 cm plaster skins run the whole 0.25 m height, so R = L/(kA) with A = 0.25 m²: $$R_{foam} = \frac{0.03}{0.026 \times 0.25} = 4.615\ {}^{\circ}\text{C/W}, \qquad R_{plaster} = \frac{0.02}{0.22 \times 0.25} = 0.364\ {}^{\circ}\text{C/W}\ \text{(each)}$$ The foam alone is already two-thirds of the whole chain, which is the first thing worth noticing about this wall.
  4. Treat the 16 cm core as two parallel paths. Over that thickness, heat can cross either the brick (0.22 m of the element height) or the plaster rib (2 × 0.015 = 0.03 m): $$R_{brick} = \frac{0.16}{0.72 \times 0.22} = 1.010\ {}^{\circ}\text{C/W}, \qquad R_{pl,mid} = \frac{0.16}{0.22 \times 0.03} = 24.242\ {}^{\circ}\text{C/W}$$ $$\frac{1}{R_{mid}} = \frac{1}{1.010} + \frac{1}{24.242} = 0.9900 + 0.0413 = 1.0313 \quad\Longrightarrow\quad R_{mid} = 0.9697\ {}^{\circ}\text{C/W}$$ The rib carries only 4.0 % of the core's conductance because plaster is a third as conductive as brick over less than a seventh of the area — a real but minor bridge here, worth 0.6 % of the total heat flow.
  5. Sum the series chain. $$R_{total} = R_i + R_{foam} + R_{plaster} + R_{mid} + R_{plaster} + R_o$$ $$R_{total} = 0.400 + 4.615 + 0.364 + 0.9697 + 0.364 + 0.160 = \boxed{6.872\ {}^{\circ}\text{C/W}}$$
  6. Heat flow through one element. $$\dot Q_{element} = \frac{T_{\infty 1} - T_{\infty 2}}{R_{total}} = \frac{25 - (-10)}{6.872} = 5.093\ \text{W}$$
  7. Scale to the whole wall. The 15 m² wall contains $$n = \frac{15}{0.25} = 60\ \text{elements} \quad\Longrightarrow\quad \dot Q = 60 \times 5.093 = \boxed{306\ \text{W}}$$
  8. Report the wall in the form a designer uses, and check it. The overall coefficient is $$U = \frac{\dot Q}{A\,\Delta T} = \frac{305.6}{15 \times 35} = 0.582\ \text{W/m}^2\!\cdot\!\text{K} \quad (\text{RSI} = 1.72,\ \text{i.e. R-9.8 in IP units})$$ Marching the element temperature drop down the chain gives 25.0 → 22.96 → −0.54 → −2.39 → −7.33 → −9.19 → −10.0 °C, so the inside surface sits at 22.96 °C — comfortably above any realistic indoor dew point, which is the answer to the condensation question part (b) raises. Note also that the foam is on the warm side, so the whole masonry core runs below freezing; that placement is thermally efficient but it puts the vapour-control decision squarely in the middle of the assembly, and it is why the two parts of this question belong together.
Check — this wall would not comply with current Canadian code

U = 0.582 W/m²·K is roughly 2.8 times the maximum above-grade wall coefficient permitted for a Climate Zone 7A house by NBC 9.36 (≈ 0.21 W/m²·K, RSI 4.7). The question is a conduction exercise, not a compliance check, and the answer above is the answer to what was asked; the comparison is offered only because a 3 cm foam layer is what limits this wall, and tripling it to 9 cm would bring the assembly close to code.

Equivalent thermal-resistance network for the element1/(hi·A)0.400L/(kf·A)4.615L/(kp·A)0.364brick 1.0100.16/(0.72×0.22)plaster 24.2420.16/(0.22×0.03)in parallel → R mid0.9697L/(kp·A)0.3641/(ho·A)0.160T∞1T∞2R total = 6.872 °C/WQ̇ element = ΔT / R total = 5.093 W
Figure 4.2 — the equivalent resistance network for one element. Only the 16 cm core is a parallel pair; everything else is in series.

Part (b) — moisture flow through walls; vapour barriers (5 marks)

Moisture crosses a wall assembly by four mechanisms, and they differ enormously in magnitude. In rough order of importance they are bulk water (rain and snowmelt driven through the cladding, managed by flashings, a drainage plane and a drained-and-vented cavity), capillary suction (liquid wicking through porous masonry and concrete, broken by a capillary break or damp-proof course), air leakage (water vapour carried by air moving through holes in the assembly), and finally vapour diffusion (molecular transport driven by a difference in vapour pressure). The crucial quantitative point is that air leakage typically moves ten to a hundred times more moisture than diffusion does: a single square centimetre of continuous gap in a wall of a heated Canadian house can deposit litres of condensate over a winter, while diffusion through the same wall's intact area contributes a few tens of grams. This is why the air barrier is the more important of the two control layers, and why the two functions — air control and vapour control — should be thought about separately even when one membrane performs both.

Diffusion itself obeys a Fickian law directly analogous to the conduction chain solved in part (a): the vapour flow is the vapour-pressure difference divided by the sum of the layers' vapour resistances, where each layer's resistance is its thickness divided by its permeability. Condensation occurs wherever, marching that vapour-pressure profile and the temperature profile through the wall together, the local vapour pressure reaches the local saturation pressure — that is, wherever the wall's temperature falls below the dew point of the air within it. In the wall of part (a), with the foam on the inside face, the brick core sits between −2.4 and −9.2 °C in mid-winter, so any indoor moisture that reaches the core will condense and, at those temperatures, freeze.

A vapour barrier (properly, a vapour retarder) exists to keep interior moisture from reaching those cold layers by diffusion. Canadian practice classifies retarders by permeance: Class I below 1 perm (polyethylene sheet, sheet metal, foil facings), Class II from 1 to 10 perms (kraft paper, smart membranes, vapour-retarder paints), and Class III from 10 to 100 perms (latex paint on gypsum board). The National Building Code requires a vapour barrier with a permeance no greater than 60 ng/(Pa·s·m²) — about 1 perm — on the warm side of insulation in heated buildings, which is why 6-mil polyethylene became the Canadian default.

Installation is governed by one rule: the retarder goes on the warm side of the insulation, because that is where the vapour pressure is high and the temperature is above the dew point. In a Canadian heating climate that means the interior face, immediately behind the gypsum board. The quantitative version of the rule is the one-third / two-thirds guideline: at least enough of the assembly's thermal resistance must lie outboard of the vapour retarder to keep the retarder's own surface above the indoor dew point — roughly two-thirds of the RSI outside the plane of the retarder in Zone 7. Practically, the sheet is lapped at least 100 mm at joints over a solid backing, sealed with acoustical sealant or tape, carried continuously across framing intersections, floor rims and partition walls, and sealed at every penetration with gaskets or purpose-made boxes; electrical boxes and recessed lights are the classic failure points. Two errors are worse than omitting the retarder altogether. The first is double vapour barriers — polyethylene inside and an impermeable exterior sheathing or foil-faced board outside — which traps any water that does get in, with no direction in which the assembly can dry. The second is installing it on the cold side, which converts the retarder into a condensing surface. Where a wall must dry inwards in summer as well as outwards in winter, a smart (variable-permeance) membrane is the modern answer: it behaves as a Class II retarder in winter and opens up to 10 perms or more in humid summer conditions, so the assembly retains a drying path in both directions.