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=21−(−14)=35 K, $$q=\frac{35}{3.273}=10.69\ \text{W/m}^2$$ Stepping T = T₀ − q·R progressively from the interior: $$T_1(\text{int. surf.})=19.72^\circ\text{C},\ T_2(\text{conc/XPS})=18.71^\circ\text{C},\ T_3(\text{XPS/air})=-10.79^\circ\text{C},\ T_4(\text{air/brick})=-12.61^\circ\text{C},\ T_5(\text{ext. surf.})=-13.68^\circ\text{C}$$ (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): $$P_{sat}(21^\circ\text{C})=2482\ \text{Pa}\ \Rightarrow\ P_0 = 0.50\times2482 = 1241\ \text{Pa (interior)}$$ $$P_{sat}(-14^\circ\text{C})=181\ \text{Pa}\ \Rightarrow\ P_6 = 0.80\times181 = 145\ \text{Pa (exterior)}$$
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): $$w=\frac{P_0-P_6}{Z_{total}}=\frac{1241-145}{0.0980}=11\,180\ \text{ng/(s.m}^2\text{)}$$ $$\boxed{P_1(\text{conc/XPS})=843\ \text{Pa},\ \ P_2(\text{XPS/air})=247\ \text{Pa},\ \ P_3(\text{air/brick})=245\ \text{Pa}}$$ (the chain closes to 145 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	19.72 °C	1241 Pa	2293 Pa	54.1%
   Concrete / XPS	18.71 °C	843 Pa	2154 Pa	39.2%
   XPS / air space	−10.79 °C	247 Pa	242 Pa	102.1%
   Air space / brick	−12.61 °C	245 Pa	206 Pa	119.5%
   Exterior surface	−13.68 °C	145 Pa	186 Pa	77.7%

7. (iii) Condensation check. RH exceeds 100% at the XPS/air-space and air-space/brick interfaces — the theoretical (uncondensed) vapour-pressure profile crosses ABOVE the saturation curve there, so condensation forms within that zone, first appearing at the XPS/air-space interface (the cold face of the insulation).