Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-Bld-A5 Building Science — National Exam, December 2017. Six questions of 20 marks each were printed; per the paper's own NOTES only the first five in the answer book are graded, but all six are answered below as a complete study resource.
Reference texts: ASHRAE Handbook — Fundamentals (Chapters 1 Psychrometrics, 14 Climatic Design Information, 25 Thermal and Water Vapor Transmission Data, 26 Heat, Air, and Moisture Control in Building Assemblies); McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design; National Building Code of Canada (NBCC).
Part (A) — required physical properties of a thermal insulator. A thermal insulation material is chosen first for a low and STABLE thermal conductivity across the temperature range and moisture exposure it will see in service, since several common insulations lose a significant share of their rated resistance once even lightly wetted (fibrous battings especially), or as a foam's low-conductivity blowing gas diffuses out over the years and is replaced by air (long-term thermal resistance vs. the initial rating). It must be dimensionally and thermally stable — no meaningful shrinkage, sagging out of a cavity, or embrittlement over decades — and it needs mechanical properties matched to its application: real compressive strength where it carries load (under a slab, below grade, under a roof membrane), but little strength requirement in a stud cavity. Fire performance appropriate to the assembly and code (flame-spread/smoke-developed ratings, or a required thermal barrier such as gypsum board over foam plastic) is mandatory, and the material must be compatible with the layers on either side of it so the assembly does not trap moisture. Finally, a modern specification also weighs the environmental profile: low VOC off-gassing, and for foam insulations, a low-global-warming-potential blowing agent.
Part (B).
Given.
Wall assembly and boundary conditions
Layer / condition
Value
Plywood siding
25 mm, k ≈ 0.12 W/(m·K)
Fibreglass blanket
100 mm, k = 0.04 W/(m·K) (given)
Gypsum board
10 mm, k ≈ 0.16 W/(m·K)
Inside / outside air temperature
20 °C (293.15 K) / −20 °C
Wall area
300 m²
Check: two gaps in the printed data are adopted explicitly. (1) The question supplies k only for the fibreglass batt; the plywood and gypsum board conductivities used below are the standard ASHRAE Fundamentals Ch. 26 material-property figures for these products. (2) The printed outside temperature is −20 °C, but the parenthetical Kelvin value (258.15 K) converts to −15 °C, not −20 °C, so the Kelvin figure looks stale; the printed Celsius value (−20 °C = 253.15 K) is treated as governing throughout.
Find. (i) An expression/value for the total thermal resistance including surface films; (ii) the total heat loss through the 300 m² wall; (iii) the percentage increase in heat loss when the outside wind rises to 50 mph; (iv) which layer controls the heat flow.
Three-layer wall, interior air film to exterior air film, in series.
Approach. Model the wall as four conduction resistances plus two convective surface films, all in series (1-D steady-state conduction); sum the resistances, then apply Q = A·ΔT/Rₜₖₜₕ for the base case and repeat with a revised outside film coefficient for the wind-speed change.
Surface (convective) resistances. ASHRAE Fundamentals gives standard winter-design values for a vertical wall: inside, still air, hᵢ = 8.29 W/(m²·K) ⇒ Rᵣᵢ = 1/8.29 = 0.1206 m²·K/W; outside, "typical" 24 km/h (15 mph) winter design wind, hᵢ = 34 W/(m²·K) ⇒ Rᵣᵢ = 1/34 = 0.0294 m²·K/W.
(i) Total resistance, base case. Summing all six terms in series,
$$\boxed{R_{total} = R_{si}+R_{ply}+R_{fg}+R_{gyp}+R_{so} = 0.1206+0.2083+2.500+0.0625+0.0294 = 2.921\ \text{m}^2\text{K/W}}$$
(ii) Total heat loss, base case. With ΔT = 20−(−20) = 40 K and A = 300 m²,
$$\boxed{Q = \frac{A\,\Delta T}{R_{total}} = \frac{300\times 40}{2.921} = 4108\ \text{W}\approx 4.11\ \text{kW}}$$
(iii) Revise the outside film coefficient for 50 mph wind. Using the standard forced-convection wind correlation h₀ = 5.7 + 3.8V (V in m/s) to scale the ASHRAE base value proportionally: base wind 24 km/h = 6.67 m/s gives a correlation value of 5.7+3.8(6.67) = 31.0; 50 mph = 22.35 m/s gives 5.7+3.8(22.35) = 90.6. Scaling the tabulated h₀=34 W/(m²K) by this ratio,
$$h_{o,new} = 34\times\frac{90.6}{31.0} = 99.3\ \text{W/(m}^2\text{K)}\ \Rightarrow\ R_{so,new}=\frac{1}{99.3}=0.0101\ \text{m}^2\text{K/W}$$
Recompute R and Q at the higher wind speed.
$$R_{total,new} = 0.1206+0.2083+2.500+0.0625+0.0101 = 2.902\ \text{m}^2\text{K/W}$$
$$Q_{new} = \frac{300\times 40}{2.902} = 4136\ \text{W}$$
$$\boxed{\%\ \text{increase} = \frac{4136-4108}{4108}\times 100 = 0.67\%}$$
(iv) Controlling resistance. The fibreglass batt alone (R = 2.500) is 85.6% of Rₜₖₜₕ, so it is by a wide margin the CONTROLLING resistance — consistent with the near-negligible 0.67% change in heat loss found in (iii), since even nearly tripling the outside film coefficient barely dents a total dominated by the insulation layer.
