24-Bld-A5 Building Science, May 2017, Q2

Question 2 (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.

Part (A) — condensation resistance of windows. A window's condensation resistance is governed by how warm its COLDEST interior-facing surface stays relative to the room's dew point, since condensation forms the instant any surface drops to or below the dew-point temperature of the adjacent air. The Condensation Resistance Factor (CRF, ASHRAE/NFRC-rated) and the newer Condensation Resistance (CR) rating both quantify this by reporting the minimum interior surface temperature a fenestration product reaches under a standard outdoor design condition, expressed as a percentage of the indoor–outdoor temperature difference recovered at the surface. Three physical levers control it. GLAZING: multiple panes, low-conductivity gas fill (argon/krypton) and low-emissivity coatings all raise the center-of-glass surface temperature by cutting conductive and radiative heat loss through the IGU. FRAME/SPACER: the frame and, especially, the glazing spacer at the edge of the IGU are usually the coldest points on the whole window — a conventional aluminum spacer is a severe thermal bridge that can drop the edge-of-glass temperature 5–10 °C below center-of-glass, so "warm-edge" (structural foam, stainless steel, or non-metallic) spacers and thermally-broken frames are the highest-leverage upgrade for condensation resistance specifically (as opposed to whole-window U-value). INSTALLATION/INTERIOR CONDITIONS: interior surface temperature also depends on local air movement (a window in a still corner or behind heavy drapery runs colder because room air can't sweep its surface) and on the indoor RH itself, which the OCCUPANT controls but the window's rated CR assumes at a fixed reference. Because the frame/spacer and glazing edge are almost always the limiting location, condensation resistance is fundamentally a THERMAL BRIDGING problem solved at the edge of the assembly, not a bulk-glazing U-value problem.

Part (B).

Given.

Wall assembly (interior → exterior) and boundary conditions
Layer / condition	Value
Gypsum board (interior finish)	10 mm, k ≈ 0.16 W/(m·K)
Fibreglass blanket	100 mm, k = 0.04 W/(m·K) (given)
Plywood siding (exterior)	25 mm, k ≈ 0.12 W/(m·K)
Inside / outside air temperature	20 °C (293.15 K) / −15 °C (258.15 K)
Wall area	300 m²
Interior finish emissivity	0.75

(1) k for plywood and gypsum board (a materials-property table was very likely supplied with the exam) — standard ASHRAE Fundamentals Ch. 26 figures are used. (2) Layer ORDER: "Gypsum board" is taken as the interior finish and "Plywood siding" as the exterior sheathing (normal wall construction, and required by part (iii)'s own wording, "between the Gypsum board and the house interior").

Find. (i) The total heat loss through the 300 m² wall; (ii) the interface temperature between the plywood and the fibreglass blanket; (iii) the radiation heat transfer coefficient at the interior gypsum surface, given ε = 0.75.

Interior air → gypsum → fibreglass batt → plywood siding → exterior air, all in series.

Approach. Sum the four conduction/surface-film resistances in series to get the total resistance and heat flux, step the temperature down layer by layer from the interior to find the plywood/fibreglass interface, then use the resulting interior gypsum surface temperature with the actual emissivity to isolate the radiative part of the interior film coefficient.

Surface and layer resistances. Standard ASHRAE winter-design surface films (still air inside, hᵢ=8.29 W/m²K; 24 km/h wind outside, hᵢ=34 W/m²K) and layer conduction R=L/k: $$R_{si}=\frac{1}{8.29}=0.1206,\ \ R_{gyp}=\frac{0.010}{0.16}=0.0625,\ \ R_{fg}=\frac{0.100}{0.04}=2.500,\ \ R_{ply}=\frac{0.025}{0.12}=0.2083,\ \ R_{so}=\frac{1}{34}=0.0294\ \ (\text{m}^2\text{K/W})$$
Total resistance and heat flux. $$R_{total}=0.1206+0.0625+2.500+0.2083+0.0294=2.921\ \text{m}^2\text{K/W}$$ $$q''=\frac{\Delta T}{R_{total}}=\frac{20-(-15)}{2.921}=11.98\ \text{W/m}^2$$
(i) Total heat loss. $$\boxed{Q = q''\times A = 11.98\times300 = 3595\ \text{W}\approx 3.60\ \text{kW}}$$
(ii) Step the temperature from the interior. T = T₀−q''R progressively: $$T_{s,i}(\text{gypsum surf.})=20-11.98(0.1206)=18.56^\circ\text{C}$$ $$T_{g/f}(\text{gypsum/fibreglass})=18.56-11.98(0.0625)=17.81^\circ\text{C}$$ $$\boxed{T_{f/p}(\text{fibreglass/plywood})=17.81-11.98(2.500)=-12.15^\circ\text{C}}$$ (continuing through the plywood and the outside film recovers −15.0 °C exactly, confirming the resistance bookkeeping).
(iii) Radiation coefficient at the interior gypsum surface. The standard Rᵣᵢ=0.1206 used above already lumps convection and radiation for a "typical" (ε≈0.9) surface; with the actual ε=0.75 given, the radiative COMPONENT is recomputed directly from the linearized radiation law using the interior surface temperature just found (Tᵣ=18.56 °C = 291.70 K) and the room air/surroundings temperature (Tₔ=20 °C = 293.15 K): $$h_{rad}=\varepsilon\sigma(T_s+T_r)(T_s^2+T_r^2)=0.75\times5.67\times10^{-8}\times(291.70+293.15)\times(291.70^2+293.15^2)$$ $$\boxed{h_{rad}=4.25\ \text{W/(m}^2\text{K)}}$$

Problem 2B — final results
Quantity	Value
(i) Total heat loss	3595 W (3.60 kW)
(ii) Plywood/fibreglass interface temperature	−12.15 °C
(iii) Radiation coefficient, gypsum/interior (ε=0.75)	4.25 W/(m²K)

Concept & Theory

Topic: Series thermal resistance through a composite wall, and decomposing a combined surface film coefficient into its radiative and convective parts.

Key relations: Rₜₖₜₕ = Σ(L/k) + Rᵣᵢ + Rᵣᵢ; q'' = ΔT/Rₜₖₜₕ; interior surface temperature Tᵣ = Tₔ−q''Rᵣᵢ; linearized radiation coefficient hₜ₊₌ = εσ(Tᵣ+Tₔ)(Tᵣ²+Tₔ²), equivalent to the Stefan–Boltzmann law Φ=εσ(Tᵣ⁴−Tₔ⁴) written as hₜ₊₌(Tₔ−Tᵣ).

Why this works: the combined film coefficient tabulated by ASHRAE bundles convection and radiation together for a "typical" indoor surface (ε≈0.9); once the surface temperature is known from the overall energy balance, the ACTUAL radiative exchange for any other emissivity can be recovered directly from the linearized radiation law evaluated at that temperature, without needing to iterate, because the small (≈1.5 K) surface-to-air temperature difference here makes the linearization essentially exact.

Common pitfall: assuming hₜ₊₌ must be solved iteratively; iteration is only needed when the surface temperature itself is unknown until hₜ₊₌ is fixed (a circular problem), but here the standard tabulated Rᵣᵢ already pins down a good first-pass surface temperature, so a single evaluation suffices for a 20-mark exam question.

Source: ASHRAE Handbook — Fundamentals, Ch. 26 (surface conductances, Table 1) and Ch. 4 (radiative heat transfer fundamentals); NFRC 500 (condensation resistance rating).

Question 2 of 6

Question 2 (20 marks)