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/R_total), compute the steady-state vapour pressure at each interface from the analogous vapour-resistance chain (w = ΔP/Z_total), 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): $$R_{conc}=\frac{0.160}{1.7}=0.094,\ R_{xps}=\frac{0.080}{0.029}=2.759,\ R_{air}=0.170,\ R_{brick}=\frac{0.080}{0.9}=0.089\ \ (\text{m}^2\text{K/W})$$ $$R_{total}=R_{si}+R_{conc}+R_{xps}+R_{air}+R_{brick}+R_{so}=0.12+0.094+2.759+0.170+0.089+0.03=3.262\ \text{m}^2\text{K/W}$$
2. Heat flux and interface temperatures. With ΔT=21−(−14)=35 K, $$q=\frac{35}{3.262}=10.73\ \text{W/m}^2$$ Stepping T = T₀ − q·R progressively from the interior: $$T_1(\text{int. surf.})=19.71^\circ\text{C},\ T_2(\text{conc/XPS})=18.70^\circ\text{C},\ T_3(\text{XPS/air})=-10.90^\circ\text{C},\ T_4(\text{air/brick})=-12.72^\circ\text{C},\ T_5(\text{ext. surf.})=-13.68^\circ\text{C}$$ (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): $$Z_{conc}=\frac{0.160}{4.5}=0.0356,\ Z_{xps}=\frac{0.080}{1.5}=0.0533,\ Z_{air}=\frac{0.030}{194}=0.00015,\ Z_{brick}=\frac{0.080}{10}=0.0080$$ $$Z_{total}=0.0356+0.0533+0.00015+0.0080=0.0970$$
4. Boundary vapour pressures (Magnus-Tetens saturation curve, over water for T≥0°C and over ice for T<0°C): $$P_{sat}(21^\circ\text{C})=2482\ \text{Pa}\ \Rightarrow\ P_0 = 0.60\times2482 = 1489\ \text{Pa (interior)}$$ $$P_{sat}(-14^\circ\text{C})=181\ \text{Pa}\ \Rightarrow\ P_6 = 0.20\times181 = 36\ \text{Pa (exterior)}$$
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): $$w=\frac{P_0-P_6}{Z_{total}}=\frac{1489-36}{0.0970}=14\,972\ \text{ng/(s.m}^2\text{)}$$ $$\boxed{P_1(\text{conc/XPS})=957\ \text{Pa},\ \ P_2(\text{XPS/air})=158\ \text{Pa},\ \ P_3(\text{air/brick})=156\ \text{Pa}}$$ (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 = P_actual/P_sat(T):

   Interface temperature, actual vapour pressure and saturation vapour pressure

   Interface	T	Pactual	Psat(T)	RH
   Interior surface	19.71 °C	1489 Pa	2292 Pa	65.0%
   Concrete / XPS	18.70 °C	957 Pa	2152 Pa	44.5%
   XPS / air space	−10.90 °C	158 Pa	240 Pa	66.1%
   Air space / brick	−12.72 °C	156 Pa	203 Pa	76.7%
   Exterior surface	−13.68 °C	36 Pa	186 Pa	19.4%

7. (iii) Condensation check. RH stays below 100% at every interface, peaking at 76.7% at the airspace/brick interface — the theoretical vapour-pressure profile never crosses the saturation curve, so no condensation occurs within this wall under these conditions. The result is sensitive to the interior RH assumed: at this wall's moderate 60% interior RH the peak stays comfortably under saturation, but a higher winter humidification set-point would push the same profile over 100% at this same airspace/brick interface, since that is consistently the coldest point at which vapour pressure is still relatively high.