Composite House Wall with Air Space: 22-Agric-A3 Heat Engineering, December 2014, Q2

Problem 2: Composite House Wall with Air Space (25 points)

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.

Layer (outer → inner)	Thickness	Conductivity / conductance
Common brick	10.16 cm	k = 0.0069 W/cm·K
Celotex sheathing	1.27 cm	k = 0.00048 W/cm·K
Air space (wall-stud cavity)	9.53 cm	h = 6.25×10⁻⁴ W/cm²·K (given directly)
Sheetrock	1.27 cm	k = 0.0074 W/cm·K
Outer brick surface temperature	4.44°C
Inner wall surface temperature	21.1°C

Find. The rate of heat loss per unit wall area (W/cm²).

Approach. Model the wall as four thermal resistances in series per unit area: three conduction resistances $L/k$ for the solid layers plus the air-space resistance $1/h$ given directly as a conductance; sum them for the total resistance, then divide the overall temperature difference by that total to get the steady-state heat flux.

Four resistances in series per unit wall area: brick, Celotex sheathing, the stud-cavity air space, and sheetrock; heat is lost from the warm inner face (21.1°C) to the cold outer face (4.44°C).

Conduction resistance of each solid layer, $R=L/k$ (per unit area). $$R_{brick} = \frac{10.16}{0.0069} = 1472.5\ \text{cm}^2\cdot\text{K/W}\qquad R_{Celotex} = \frac{1.27}{0.00048} = 2645.8\ \text{cm}^2\cdot\text{K/W}\qquad R_{sheetrock} = \frac{1.27}{0.0074} = 171.6\ \text{cm}^2\cdot\text{K/W}$$
Air-space resistance. The air space's conductance is given directly (not a bulk k), so its resistance is simply its reciprocal, $$R_{air} = \frac{1}{h} = \frac{1}{6.25\times10^{-4}} = 1600\ \text{cm}^2\cdot\text{K/W}$$
Total resistance. At steady state the same heat flux passes through all four layers in series, so the resistances add, $$R_{tot} = R_{brick}+R_{Celotex}+R_{air}+R_{sheetrock} = 1472.5+2645.8+1600+171.6 = \boxed{5889.9\ \text{cm}^2\cdot\text{K/W}}$$
Heat loss per unit area. With the overall temperature difference $\Delta T = 21.1-4.44 = 16.66\ \text{K}$, $$q = \frac{\Delta T}{R_{tot}} = \frac{16.66}{5889.9} = \boxed{2.83\times10^{-3}\ \text{W/cm}^2}$$ Converting for scale, $q = 2.83\times10^{-3}\times10^4 = 28.3\ \text{W/m}^2$ — the air space and Celotex sheathing together account for over 70% of the total resistance despite being the thinnest layers, which is exactly why they are the effective insulators in this wall.

Quantity	Result
Brick resistance, R_brick	1472.5 cm²·K/W
Celotex resistance, R_Celotex	2645.8 cm²·K/W
Air-space resistance, R_air	1600.0 cm²·K/W
Sheetrock resistance, R_sheetrock	171.6 cm²·K/W
Total resistance, R_tot	5889.9 cm²·K/W
Heat loss rate per unit area, q	2.83×10⁻³ W/cm² (28.3 W/m²)

Concept & Theory

Topic: Steady conduction through a composite (multi-layer) plane wall with a convective air gap.

Key relations: Layer conduction resistance $R=L/k$ (per unit area); a resistance given directly as a conductance $h$ is $R=1/h$; series layers add, $R_{tot}=\sum R_i$; overall flux $q=\Delta T/R_{tot}$.

Why this works: At steady state the same heat flux must pass through every layer in turn (none is stored or generated inside the wall), so the total temperature drop apportions itself across the layers in direct proportion to each layer's resistance — exactly analogous to resistors carrying the same current in a series electrical circuit, with $\Delta T$ playing the role of voltage and $q$ the role of current.

Common pitfall: Treating the air space as a conduction layer and dividing its thickness by an assumed conductivity, when its conductance is given directly — its resistance is just $1/h$, not $L/k$. A second pitfall is mixing centimetre-based conductivities with metre-based thicknesses; keeping every length in centimetres (matching the given k and h units) all the way through avoids a unit-conversion error compounding across four layers, with the final flux converted to W/m² only at the very end.

Source: Çengel & Ghajar, Heat and Mass Transfer, Ch. 3 (composite plane walls, thermal resistance networks).

Question 2 of 4: Composite House Wall with Air Space

Problem 2: Composite House Wall with Air Space (25 points)