Composite wall — total resistance, heat loss and the controlling layer: 22-Mec-B2 Environmental Control in Buildings, December 2018, Q1

Question 1: Composite wall — total resistance, heat loss and the controlling layer (30 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.

Quantity	Symbol	Value
Plaster board (innermost layer)	$L_p$, $k_p$	10 mm, 0.17 W m$^{-1}$K$^{-1}$
Glass-fibre blanket, 28 kg m$^{-3}$	$L_b$, $k_b$	100 mm, 0.038 W m$^{-1}$K$^{-1}$
Plywood siding (outermost layer)	$L_s$, $k_s$	20 mm, 0.12 W m$^{-1}$K$^{-1}$
Inside film coefficient	$h_i$	30 W m$^{-2}$K$^{-1}$
Outside film coefficient, still air / violent wind	$h_o$	60 / 300 W m$^{-2}$K$^{-1}$
Air temperatures, inside and outside	$T_{\infty,i}$, $T_{\infty,o}$	$+20\,{}^{\circ}\text{C}$, $-15\,{}^{\circ}\text{C}$
Gross wall area	$A$	350 m$^2$

The three conductivities are not printed on the paper. The examination is open book and instruction 4 states that candidates are expected to have a text containing the property tables, so the values above are taken from Incropera Table A.3 at 300 K — the standard set for exactly these three materials.

Find. A symbolic expression for the total wall resistance, the heat loss it produces in still air, the percentage change when the outside film coefficient rises fivefold, and which single resistance governs the answer.

Figure 1.1 — The composite wall as drawn on the paper, with the equivalent series resistance network beneath it. The glass-fibre blanket alone carries 90.5% of the total resistance; the two film resistances together carry 1.7%.

Approach. The wall is a plane composite slab with one-dimensional steady conduction, so the two convective films and the three conductive layers are five resistances in series; sum them per unit area and divide the overall temperature difference by the total.

Part (a) — write the series resistance symbolically. Heat crosses the inside film, then each layer in turn, then the outside film. Resistances in series add, and each is referred to the same area $A$: $$R_{\text{tot}}=\frac{1}{h_i A}+\frac{L_p}{k_p A}+\frac{L_b}{k_b A}+\frac{L_s}{k_s A}+\frac{1}{h_o A}$$ It is more useful to factor the area out, giving the unit resistance $R''$ that is independent of how big the wall is: $$\boxed{\;R''_{\text{tot}}=A R_{\text{tot}}=\frac{1}{h_i}+\frac{L_p}{k_p}+\frac{L_b}{k_b}+\frac{L_s}{k_s}+\frac{1}{h_o}\;}$$ so that $q = A(T_{\infty,i}-T_{\infty,o})/R''_{\text{tot}}$ and the overall coefficient is $U = 1/R''_{\text{tot}}$.

Evaluate the five terms. Substituting the given thicknesses, conductivities and film coefficients term by term:

Resistance	Expression	Value (m$^2$K W$^{-1}$)	Share
Inside film	$1/h_i = 1/30$	0.03333	1.1%
Plaster board	$L_p/k_p = 0.010/0.17$	0.05882	2.0%
Glass-fibre blanket	$L_b/k_b = 0.100/0.038$	2.63158	90.5%
Plywood siding	$L_s/k_s = 0.020/0.12$	0.16667	5.7%
Outside film	$1/h_o = 1/60$	0.01667	0.6%
Total	$R''_{\text{tot}}$	2.9071	100%

Referred to the whole wall this is $R_{\text{tot}} = 2.9071/350 = 8.306\times10^{-3}$ K W$^{-1}$, and the overall coefficient is $U = 1/2.9071 = 0.344$ W m$^{-2}$K$^{-1}$.

Part (b) — heat loss in still air. The driving temperature difference is $20-(-15) = 35$ K across the whole assembly: $$q=\frac{A\,(T_{\infty,i}-T_{\infty,o})}{R''_{\text{tot}}}=\frac{350\times 35}{2.9071}$$ $$\boxed{\;q = 4214\ \text{W} \approx 4.21\ \text{kW}\;}$$ which is a flux of $q/A = 12.0$ W m$^{-2}$ — a sensible number for a 100 mm insulated timber-frame wall at a 35 K difference.
Part (c) — the effect of violent wind. Wind changes only the outside film. Replacing that one term, $1/300 = 0.00333$ in place of $0.01667$, gives $$R''_{\text{wind}} = 2.9071-0.01667+0.00333 = 2.8937\ \text{m}^2\text{K}\,\text{W}^{-1}$$ $$q_{\text{wind}} = \frac{350\times 35}{2.8937}=4233\ \text{W}$$ Substituting into the percentage change, $$\boxed{\;\frac{q_{\text{wind}}-q}{q}=\frac{4233-4214}{4214}=0.46\%\;}$$ A fivefold increase in the outside film coefficient buys less than half a percent of extra heat loss, because the film it improves was never the obstacle.
Part (d) — the controlling resistance. From the table in step 2 the glass-fibre blanket contributes 2.6316 of the 2.9071 m$^2$K W$^{-1}$ total, that is $$\boxed{\;\text{the 100 mm glass-fibre blanket controls the heat flow: }90.5\%\text{ of }R''_{\text{tot}}\;}$$ Everything else combined — both films and both sheathing boards — accounts for the remaining 9.5%. This is why part (c) came out so small, and it is the practical design message: to reduce this wall's heat loss you must add insulation thickness or lower $k_b$; nothing done to the surfaces, the plaster board or the siding will matter much. Doubling the blanket to 200 mm, for instance, raises $R''$ to 5.5387 m$^2$K W$^{-1}$ and drops the loss to 2212 W, a 47.5% reduction.

Result	Value
(a) Unit resistance, symbolic	$R'' = 1/h_i + L_p/k_p + L_b/k_b + L_s/k_s + 1/h_o$
(a) Unit resistance, evaluated	2.9071 m$^2$K W$^{-1}$
(a) Whole-wall resistance / overall coefficient	$8.306\times10^{-3}$ K W$^{-1}$ / $U = 0.344$ W m$^{-2}$K$^{-1}$
(b) Total heat loss, $h_o = 60$	4214 W (12.0 W m$^{-2}$)
(c) Total heat loss, $h_o = 300$	4233 W
(c) Percentage increase	0.46%
(d) Controlling resistance	Glass-fibre blanket, 90.5% of $R''$

Check: the three conductivities are table values, not given data. They are taken from Incropera Table A.3 at 300 K: plaster board 0.17, glass-fibre blanket at 28 kg m$^{-3}$ 0.038, plywood 0.12 W m$^{-1}$K$^{-1}$. Any equivalent handbook set gives the same answer to within a few watts, and the result is insensitive to the two minor layers: a 10% error in $k_s$ shifts $q$ by 0.5% and a 10% error in $k_p$ by 0.2%. Only $k_b$ matters — a 10% error there moves $q$ by 8.9% — so state which blanket conductivity you used, exactly as cover-page instruction 1 requests.

Topic. One-dimensional steady conduction through a plane composite wall, expressed as a series thermal-resistance network.

Key relations. Conduction resistance of a slab $R_{\text{cond}} = L/(kA)$; convection resistance of a surface film $R_{\text{conv}} = 1/(hA)$; resistances in series add, $R_{\text{tot}} = \sum R_i$; heat rate $q = \Delta T / R_{\text{tot}}$; overall coefficient $U = 1/R''_{\text{tot}}$ with $q = UA\,\Delta T$.

Why this works. With no internal generation and steady conditions the same heat rate $q$ crosses every layer, so the temperature drops add while the flow stays constant — exactly the behaviour of resistors carrying one current in series. That equivalence is what lets you write the answer symbolically before touching a calculator, and it is what makes the percentage shares in step 2 meaningful: because the drops are proportional to the resistances, the layer holding 90.5% of the resistance also holds 90.5% of the temperature difference. The network form also makes the answer to part (c) obvious in advance, since improving a resistance that is already 0.6% of the total cannot change $q$ by more than 0.6%.

Common pitfall. Two errors recur. The first is averaging or adding the film coefficients themselves rather than their reciprocals — $h$ is a conductance, so it is $1/h$ that enters a series sum. The second is dropping the area: a symbolic expression written in m$^2$K W$^{-1}$ and then used as if it were K W$^{-1}$ understates the loss by a factor of 350 here. A third, specific to part (d), is naming the plywood siding as controlling because it is the thickest of the two boards; thickness alone decides nothing, only the ratio $L/k$ does.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, 8th ed., §3.1 (the plane wall and the thermal circuit) with conductivities from Table A.3; equivalent treatment in Cengel, Heat and Mass Transfer, 6th ed., §3.1–3.2.

Question 1 of 8: Composite wall — total resistance, heat loss and the controlling layer

Question 1: Composite wall — total resistance, heat loss and the controlling layer (30 marks)