Question 4 of 6

Question 4 (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.

Find. (i) The steady-state vapour pressure at each material interface; (ii) the relative humidity at each interface; (iii) whether — and where — condensation occurs within the wall.

Approach. Compute the steady-state temperature at each interface from the thermal-resistance chain (Q/A = ΔT/Rₜₖₜₕ), compute the steady-state vapour pressure at each interface from the analogous vapour-resistance chain (w = ΔP/Zₜₖₜₕ), then compare each interface's actual vapour pressure against the saturation pressure at that interface's temperature.

1. Thermal resistances (R=L/k, plus the airspace's tabulated R): $$R_{conc}=\frac{0.160}{1.7}=0.094,\ R_{xps}=\frac{0.080}{0.029}=2.759,\ R_{air}=0.170,\ R_{brick}=\frac{0.090}{0.9}=0.100\ \ (\text{m}^2\text{K/W})$$ $$R_{total}=R_{si}+R_{conc}+R_{xps}+R_{air}+R_{brick}+R_{so}=0.12+0.094+2.759+0.170+0.100+0.03=3.273\ \text{m}^2\text{K/W}$$
2. Heat flux and interface temperatures. With ΔT=22−(−14)=36 K, $$q=\frac{36}{3.273}=11.00\ \text{W/m}^2$$ Stepping T = T₀ − q·R progressively from the interior: $$T_1(\text{int. surf.})=20.68^\circ\text{C},\ T_2(\text{conc/XPS})=19.65^\circ\text{C},\ T_3(\text{XPS/air})=-10.70^\circ\text{C},\ T_4(\text{air/brick})=-12.57^\circ\text{C},\ T_5(\text{ext. surf.})=-13.67^\circ\text{C}$$ (the chain closes to −14.00°C at the outside air, confirming the resistance bookkeeping).
3. Vapour resistances, Z=L/μ (m²·s·Pa/ng): $$Z_{conc}=\frac{0.160}{4.5}=0.0356,\ Z_{xps}=\frac{0.080}{1.5}=0.0533,\ Z_{air}=\frac{0.030}{194}=0.00015,\ Z_{brick}=\frac{0.090}{10}=0.0090$$ $$Z_{total}=0.0356+0.0533+0.00015+0.0090=0.0980$$
4. Boundary vapour pressures (Magnus-Tetens saturation curve, over water for T≥0°C and over ice for T<0°C): $$P_{sat}(22^\circ\text{C})=2639\ \text{Pa}\ \Rightarrow\ P_0 = 0.55\times2639 = 1451\ \text{Pa (interior)}$$ $$P_{sat}(-14^\circ\text{C})=181\ \text{Pa}\ \Rightarrow\ P_6 = 0.20\times181 = 36\ \text{Pa (exterior)}$$
5. (i) Vapour flux and interface vapour pressures, stepping P = P₀−w·Z progressively (surface-film vapour resistance neglected, so P at the interior wall surface equals the room's vapour pressure): $$w=\frac{P_0-P_6}{Z_{total}}=\frac{1451-36}{0.0980}=14\,432\ \text{ng/(s.m}^2\text{)}$$ $$\boxed{P_1(\text{conc/XPS})=938\ \text{Pa},\ \ P_2(\text{XPS/air})=168\ \text{Pa},\ \ P_3(\text{air/brick})=166\ \text{Pa}}$$ (the chain closes to 36 Pa at the exterior, matching P₆ and confirming the vapour-resistance bookkeeping).
6. (ii) Relative humidity at each interface, RH = Pₕ₋ₔ₊₌ₓ/Pₛ₊ₜ(T):

   Interface temperature, actual vapour pressure and saturation vapour pressure

   Interface	T	Pₕ₋ₔ₊₌ₓ	Pₛ₊ₜ(T)	RH
   Interior surface	20.68 °C	1451 Pa	2434 Pa	59.6%
   Concrete / XPS	19.65 °C	938 Pa	2283 Pa	41.1%
   XPS / air space	−10.70 °C	168 Pa	244 Pa	69.0%
   Air space / brick	−12.57 °C	166 Pa	206 Pa	80.6%
   Exterior surface	−13.67 °C	36 Pa	187 Pa	19.4%

7. (iii) Condensation check. RH stays below 100% at every interface, peaking at 80.6% at the airspace/brick interface — the theoretical vapour-pressure profile never crosses the saturation curve, so no condensation occurs within this wall. The exterior air is very dry (20% RH), which pulls the boundary vapour pressure P₆ and the whole interior vapour-pressure profile down far enough that it stays under the saturation curve everywhere.