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

Question 6 of 8: Cavity wall with a window, and moisture flow through walls

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, December 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-22 p-h diagram are appended to the paper. All eight problems are solved here.

Reference texts for this subject.

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-22 properties are on the IIR datum and agree with the appended p-h diagram. Problem-specific assumptions — the coil bypass factor, climate and degree-day data, fuel prices and equipment efficiencies, duct roughness, and the CLTD / SCL / CLF table entries — are stated where they are first used.

Question 6: Cavity wall with a window, and moisture flow through walls (20 points: 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.

Given. A cavity brick wall of gross area 10.0 m² containing a single-glazed window of 2.16 m², with the layer thicknesses and conductivities below and the same surface coefficients on wall and glass.

Given data, Problem 6(a)
Layer or coefficientThicknessk, W/m·KR, m²·K/W
Inside surface film—hi = 8.5 W/m²K0.11765
Plaster10 mm0.140.07143
Brick (inner leaf)125 mm0.430.29070
Air cavity——0.16000
Brick (outer leaf)125 mm0.430.29070
Cement rendering5 mm0.860.005814
Outside surface film—ho = 31 W/m²K0.032258
Glass1.5 mm0.760.001974
Wall 4.0 m × 2.5 m; window 1.8 m × 1.2 mgross 10.0 m², glass 2.16 m², net wall 7.84 m²

Find. The proportion of the total heat transfer through the external wall that passes through the window, and then an explanation of moisture flow through walls and the role and installation of vapour barriers.

inside film R 0.1176 plaster 10 R 0.0714 brick 125 R 0.2907 cavity R 0.1600 brick 125 R 0.2907 cement 5 R 0.0058 outside film R 0.0323 section through the cavity wall in out the two paths in parallel, per kelvin of temperature difference inside wall UA 8.095 glass UA 14.222 outside total UA = 22.316 W/K the glass carries 63.73 % of the loss on only 21.6 % of the wall area U_wall 1.0325, U_glass 6.5842 W/m²K the glass is 6.4 times as conductive per unit area
Part (a) — the layer build-up of the cavity wall, and the two parallel conductance paths that carry the room heat loss. Resistances add in series through the wall and the two paths add in parallel.

Approach. Add the series resistances of each construction to get its overall coefficient, multiply by the respective areas to get two parallel conductances, and take the ratio — the temperature difference is common to both paths and cancels, so no design temperatures are needed.

  1. Build up the wall resistance. Conduction and surface resistances add in series:$$R_{\text{wall}}=\frac{1}{h_{i}}+\frac{L_{p}}{k_{p}}+\frac{L_{b}}{k_{b}}+R_{\text{cav}}+\frac{L_{b}}{k_{b}}+\frac{L_{c}}{k_{c}}+\frac{1}{h_{o}}$$Substituting: $0.11765+0.07143+0.29070+0.16000+0.29070+0.00581+0.03226$, so $R_{\text{wall}}=0.96854$ m²K/W and$$U_{\text{wall}}=\frac{1}{R_{\text{wall}}}=1.0325\ \text{W/m}^{2}\text{K}$$The two brick leaves and the cavity carry three-quarters of the resistance; the 5 mm of cement rendering contributes essentially nothing.
  2. Build up the glass resistance. A 1.5 mm pane of glass is almost no resistance at all — the two surface films are nearly the whole story:$$R_{\text{glass}}=\frac{1}{h_{i}}+\frac{L_{g}}{k_{g}}+\frac{1}{h_{o}}=0.11765+0.001974+0.03226=0.15188\ \text{m}^{2}\text{K/W}$$$$U_{\text{glass}}=6.5842\ \text{W/m}^{2}\text{K}$$The glass itself accounts for barely one per cent of that resistance, which is why single glazing of any thickness performs the same and why the only useful improvement is to add a second pane and an air space.
  3. Form the two parallel conductances. The net wall area is $10.0-2.16=7.84$ m², so$$(UA)_{\text{wall}}=1.0325\times7.84=8.0946\ \text{W/K},\qquad (UA)_{\text{glass}}=6.5842\times2.16=14.2219\ \text{W/K}$$$$(UA)_{\text{total}}=22.3165\ \text{W/K}$$
  4. Take the ratio. Both paths see the same inside-to-outside temperature difference, so it cancels and the answer is a pure ratio of conductances:$$\frac{q_{\text{glass}}}{q_{\text{total}}}=\frac{(UA)_{\text{glass}}}{(UA)_{\text{total}}}=\frac{14.2219}{22.3165}$$$$\boxed{\ \frac{q_{\text{glass}}}{q_{\text{total}}}=63.73\ \%\ }$$The window occupies only 21.6 % of the wall area but carries 63.73 % of the heat loss, because its overall coefficient is 6.38 times the wall's. Put concretely, at a 25 K design difference the room loses 557.9 W in total, of which 355.5 W is through the glass and only 202.4 W through the masonry.
Problem 6(a) — results
QuantityValue
Wall resistance / coefficient0.96854 m²K/W, U = 1.0325 W/m²K
Glass resistance / coefficient0.15188 m²K/W, U = 6.5842 W/m²K
Wall conductance (UA)wall8.0946 W/K over 7.84 m²
Glass conductance (UA)glass14.2219 W/K over 2.16 m²
Total conductance22.3165 W/K
Proportion of loss through the window63.73 % (on 21.6 % of the area)

(b) Moisture flow through walls, and vapour barriers. Moisture crosses a wall by four mechanisms, and it is worth ranking them because they differ by orders of magnitude. Bulk water — rain driven against the face and leaking through joints and penetrations — delivers by far the most water and is controlled by drainage, flashing, a drained cavity and a rain-screen, not by any vapour-control layer. Capillary rise draws liquid water through porous masonry from wet ground and is stopped by a damp-proof course. Air leakage carries water vapour dissolved in moving air through cracks, gaps and service penetrations, and in a Canadian winter it typically transports ten to a hundred times more moisture into a wall than diffusion does; it is controlled by a continuous air barrier. Vapour diffusion, the slow migration of water molecules through solid materials down a vapour-pressure gradient, is the smallest of the four but the one the vapour barrier addresses. It follows a Fick's-law form directly analogous to the conduction calculation in part (a): the flow rate is the vapour-pressure difference divided by the sum of the layer vapour resistances, so a vapour profile can be plotted through the wall in the same way as a temperature profile. Where the calculated vapour pressure inside the construction reaches the saturation pressure at the local temperature, interstitial condensation occurs — and that dew-point plane is what the design must keep out of the moisture-sensitive layers.

A vapour barrier, more properly a vapour retarder, is a low-permeance layer — polyethylene sheet, foil-backed board, kraft-faced batt or a vapour-retarding paint — installed to keep the vapour pressure inside the assembly below saturation at the temperature that assembly actually reaches. The governing rule in a heating-dominated Canadian climate is that the retarder goes on the warm side of the insulation, because in winter the vapour drive is outward from the warm humid interior, and a retarder placed on the cold side would trap that vapour against a cold surface and guarantee condensation. The National Building Code of Canada requires a vapour barrier on the warm side of insulation in insulated assemblies, with a permeance not exceeding 60 ng/(Pa·s·m²), and the practical version of the rule is that no more than about one third of the total thermal resistance may lie inboard of the retarder. Installation matters as much as selection: the sheet must be continuous, lapped at joints over framing, sealed at penetrations for outlets, plumbing and ducts, and carried across floor and ceiling junctions, because a small unsealed area passes far more moisture by air leakage than the intact remainder passes by diffusion. Two further points belong in a complete answer. First, the vapour barrier and the air barrier are different functions that a single polyethylene sheet may or may not perform together; the air barrier must additionally be continuous, structurally supported against wind load, and may sit anywhere in the assembly. Second, a wall must be able to dry, so a construction sandwiched between two impermeable layers is a defect even if neither layer leaks; the modern preference is a vapour-retarding but not vapour-tight interior layer, sometimes a smart retarder whose permeance rises with humidity, over exterior insulation that keeps the sheathing above the dew point — an arrangement that tolerates the imperfections real construction always contains.