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22-Mec-B2 Environmental Control in Buildings · May 2013

Question 8 of 8: Parallel-path U-factor of a stud wall, and mould on an uninsulated concrete wall (20 marks)

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 2013 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for ammonia (R-717) are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to state any interpretation assumptions with the answer — that latitude is used explicitly below where the printed data are redundant.

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 are tighter than a graphical solution would be; chart-quality agreement (about ±0.2 K and ±0.0002 kg/kg) is all that an examiner expects.

Reference texts for this subject.

Psychrometric relations used throughout. At barometric pressure $p$, with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation,

$$W = 0.6220\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$

in SI (kJ per kg of dry air), and in the inch-pound system $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air. The thermodynamic wet-bulb temperature is obtained from the adiabatic-saturation equation, which is what a chart's constant-wet-bulb lines represent.

Question 8: Parallel-path U-factor of a stud wall, and mould on an uninsulated concrete 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 light wood-frame wall with insulating sheathing, and the fraction of the elevation occupied by each parallel heat-flow path.

Given data (part a)
Layer or elementDescription$R$ (h·ft²·°F/Btu)
Outside surfacewinter design, 15 mph wind0.17
Sidingwood bevel lapped, 0.5 in by 8 in0.81
Sheathingrigid foam insulating (given)4.00
Cavity3.5 in mineral fibre batt (given)13.00
Framing3.5 in softwood at 1.25 per inch4.38
Interior finish0.5 in gypsum wallboard0.45
Inside surfacestill air0.68
Area fractionscavity 75%, studs/plates/sills 21%, headers 4%—

Find. (a) the area-weighted overall coefficient of heat transmission $\bar{U}$ for the wall, and (b) a remediation strategy for mould on an uninsulated concrete solarium wall, with a discussion of moisture movement in an enclosed space.

bevel siding R 0.81foam sheathing R 4.00cavity / stud R 13.0 or 4.38gypsum R 0.45outsideair filminsideair filmPlan view of one framing bay -- the three parallel heat-flow paths75%R-13 batt21%studs, plates, sills4%Ū = Σ (area fraction) × U(path)
Section through the wall (outside at the left) and a plan view of one framing bay showing the three parallel heat-flow paths whose area fractions the question supplies.

Approach (part a). Use the parallel-path (isothermal-planes-free) method: sum the series resistances along each distinct path, invert each to a path $U$-factor, and weight the $U$-factors — never the resistances — by area fraction.

  1. Resistances common to every path. The surface films, siding, sheathing and gypsum are continuous: $$R_{common} = 0.17 + 0.81 + 4.00 + 0.45 + 0.68 = 6.11\ \text{h}\cdot\text{ft}^{2}\cdot{}^\circ\text{F/Btu}$$
  2. Path through the insulated cavity (75%). $$R_{cav} = 6.11 + 13.00 = 19.11 \;\Rightarrow\; U_{cav} = \frac{1}{19.11} = 0.0523\ \text{Btu/h}\cdot\text{ft}^{2}\cdot{}^\circ\text{F}$$
  3. Path through the framing (21% + 4%). The studs, plates and sills are 3.5 in of softwood, and the headers over openings are the same species and depth, so both share one path resistance: $$R_{fr} = 6.11 + 4.38 = 10.49 \;\Rightarrow\; U_{fr} = \frac{1}{10.49} = 0.0953\ \text{Btu/h}\cdot\text{ft}^{2}\cdot{}^\circ\text{F}$$ Wood conducts about three times as readily as the batt, so the quarter of the wall that is framing is the thermal weak link.
  4. Area-weighted average. Heat flows through the paths in parallel, so the conductances add in proportion to area: $$\bar{U} = \sum_i a_i U_i = 0.75(0.05233) + 0.21(0.09533) + 0.04(0.09533)$$ $$\bar{U} = 0.03925 + 0.02002 + 0.00381 = \boxed{0.0631\ \text{Btu/h}\cdot\text{ft}^{2}\cdot{}^\circ\text{F}}$$
  5. Express as a resistance and in SI. The effective overall resistance is $1/0.06308 = 15.9$ h·ft²·°F/Btu — not the 19.1 that the cavity alone would suggest, a 17% penalty from framing thermal bridging. In SI, $$\bar{U} = 0.06308 \times 5.678 = 0.358\ \text{W/m}^{2}\cdot\text{K}, \qquad \text{RSI} = 2.79\ \text{m}^{2}\cdot\text{K/W}$$
  6. Comment on the result. RSI 2.79 (nominal R-16 effective) is respectable for a 2 by 4 wall only because of the R-4 continuous sheathing, which is outboard of the framing and so benefits every path equally. Removing it would drop the wall to $\bar{U} = 0.0882$ Btu/h·ft²·°F (RSI 2.00), a 40% increase in loss. It still falls short of the RSI 3.6 to 4.3 that the National Energy Code of Canada for Buildings asks of an above-grade wall in southern Ontario, which is why current Canadian practice uses 2 by 6 framing, thicker exterior insulation, or both.
Final results (part a)
PathArea fraction$R$ (h·ft²·°F/Btu)$U$$a\,U$
Insulated cavity0.7519.110.05230.0392
Studs, plates, sills0.2110.490.09530.0200
Headers0.0410.490.09530.0038
Whole wall1.0015.9—0.0631

(b) Mould on the uninsulated concrete solarium wall.

The diagnosis follows directly from part (a). An uninsulated reinforced-concrete wall has an overall resistance of roughly RSI 0.4 (about R-2), so on a Toronto January design night at −18 °C its inside surface sits only two or three kelvin above outdoors — near 2 to 5 °C. Room air at 21 °C and a very ordinary 40% relative humidity has a dew point of 7 °C. The surface is therefore below the dew point for weeks at a time, condensation forms, and because the concrete stays damp and the enclosed solarium has little air movement and no exhaust, surface mould follows within a season. The room was designed as an exterior balcony, so nothing about it — no insulation, no vapour control, no ventilation, uninsulated slab edges and balcony-to-slab thermal bridges — was intended to face interior humidity.

Remediation must raise the surface temperature above the dew point and lower the dew point, in that order of priority, and must be done in a way that does not trap moisture inside the assembly. In practice, for a concrete wall in a cold climate: (i) remove the affected finishes, clean and dry the concrete and treat the mould per a recognised remediation protocol (IICRC S520 or the CCA 82 guidance), wearing appropriate protection and containing the work area; (ii) insulate on the inside face with a vapour-impermeable, non-absorbent insulation applied in direct contact with the concrete — closed-cell spray polyurethane foam or adhered extruded-polystyrene board — at RSI 1.8 to 2.6 (R-10 to R-15), which puts the first condensing plane at the warm face of the insulation, keeps the interior surface above 15 °C, and leaves no air space or gypsum in contact with cold concrete; avoid fibrous batt against bare concrete with a polyethylene sheet, which in a retrofit is very likely to trap moisture at the concrete face; (iii) insulate the slab edges, the underside of the balcony projection and any exposed columns for at least 600 mm back from the wall, since those bridges will otherwise become the new coldest surfaces; (iv) upgrade the glazing, which in an enclosed balcony is usually single or non-thermally-broken and will be the next condensation site; (v) air-seal the assembly, especially at the wall-to-slab and window perimeter joints, because air leakage carries orders of magnitude more moisture than diffusion; and (vi) give the room some conditioning and air change — a supply diffuser or a small fan-coil aimed at the glazing and wall, plus a path for return air — so that it is no longer a stagnant humidity trap. Where the humidity source is the dwelling itself, add or repair the bathroom and kitchen exhausts and consider a heat-recovery ventilator; a dehumidifier is a legitimate interim measure but not a substitute for insulation. In a condominium these are usually common-element repairs, so the work is the corporation's responsibility and should be designed by an engineer and documented, particularly since the enclosure was built by the developer as a “solarium” without the envelope that description implies.

Moisture flow in an enclosed environment. Moisture moves through and within a building by four mechanisms, in descending order of the quantity typically transported: bulk water (rain and plumbing leaks); capillary action through porous materials in contact with water, which is how a concrete slab edge wicks; air transport, in which humid air is carried through cracks by wind, stack effect or fan pressurisation; and vapour diffusion driven by the difference in vapour pressure across the assembly. The last two are the ones that matter here, and the important point is their disparity: an air leak through a 1 mm crack can move one hundred times as much water vapour as diffusion through the same area of intact wall, which is why air-tightness detailing outranks vapour-barrier permeance in modern practice. Inside an enclosed room the vapour generated by occupants, cooking, bathing and plants raises the vapour pressure until it is relieved by ventilation, by absorption into hygroscopic materials, or by condensation on the coldest surface — and it is always the coldest surface that fails first, whether that is a window, a thermal bridge or an uninsulated concrete wall. In winter the vapour drive is outwards, so vapour control belongs on the warm side; in a cooling climate or an air-conditioned space in summer it reverses, which is why an assembly with a low-permeance layer on both faces cannot dry in either direction and is the one detail to avoid. The design rule that follows is simple: keep liquid water out, keep air out with a continuous air barrier, keep every interior surface warmer than the room dew point by insulating outboard of thermal bridges, give the assembly one direction in which it can dry, and remove the moisture at source with ventilation. Mould needs a substrate, warmth and water; the only one of the three a designer controls reliably is the water.

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