Walk-In Freezer — Composite Wall Heat Transfer: 22-Agric-B8 Food Process Engineering (Part 1), December 2013, Q10

Question 10: Walk-In Freezer — Composite Wall Heat Transfer (15 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.

Composite-wall construction and boundary conditions
Layer / boundary	Value
Room dimensions	4 m × 6 m × 3 m high
Stainless steel	1.7 mm, \(k=14.2\) W/(m·K)
Foam insulation	10 cm, \(k=0.034\) W/(m·K)
Cardboard ("wood side")	4.86 cm, \(k=0.043\) W/(m·K)
Outside film coefficient (cardboard side)	\(h_o=5\) W/(m²·K)
Inside film coefficient (steel side)	\(h_i=2\) W/(m²·K)
Outside / inside air temperature	32°C / -40°C

Find. The total rate of heat transfer through the walls and ceiling (no floor load specified).

Five resistances in series: two convective films plus three conduction layers (order shown outside-to-inside; layer order does not affect the series sum).

Approach. Sum all five resistances in series (per unit area), invert for the overall \(U\), find the wall + ceiling area (no floor specified), and apply \(Q=UA\Delta T\).

Overall resistance and \(U\). \(R_{tot} = \dfrac{1}{h_o}+\dfrac{1}{h_i}+\dfrac{x_{steel}}{k_{steel}}+\dfrac{x_{foam}}{k_{foam}}+\dfrac{x_{card}}{k_{card}}\) \(= \dfrac{1}{5}+\dfrac{1}{2}+\dfrac{0.0017}{14.2}+\dfrac{0.10}{0.034}+\dfrac{0.0486}{0.043}\) \(= 0.200+0.500+0.00012+2.941+1.130=\boxed{4.772\ \text{m}^2\text{K/W}}\). \(U=1/R_{tot}=\boxed{0.2096\ \text{W/(m}^2\text{K)}}\) — the foam layer alone contributes \(62\%\) of the total resistance.
Area (walls + ceiling, no floor). \(A_{walls}=2(4\times3)+2(6\times3)=24+36=60\ \text{m}^2\); \(A_{ceiling}=4\times6=24\ \text{m}^2\); \(A_{tot}=\boxed{84\ \text{m}^2}\).
Heat transfer rate. \(\Delta T = 32-(-40)=72\ \text{K}\). \(Q=UA\Delta T = 0.2096\times84\times72=\boxed{1267.5\ \text{W}}\;(\approx1.27\ \text{kW})\).

Final results
Quantity	Value
Overall resistance, \(R_{tot}\)	4.772 m²K/W
Overall coefficient, \(U\)	0.2096 W/(m²·K)
Total area (walls + ceiling)	84 m²
Rate of heat transfer, \(Q\)	≈ 1267.5 W (1.27 kW)

Concept & Theory

Topic: Steady-state conduction through a composite plane wall with surface convection (series thermal-resistance network).

Key relations: \(R_{tot}=\sum(1/h)+\sum(x/k)\); \(U=1/R_{tot}\); \(Q=UA\Delta T\).

Why this works: for one-dimensional steady conduction through layers in series (each carrying the same heat flux), thermal resistances add exactly like electrical resistances in a series circuit, regardless of the physical order the layers are installed in — only the individual \(1/h\) and \(x/k\) values matter, not which surface each film coefficient happens to be labelled against.

Common pitfall: forgetting either convective film resistance (treating the wall as conduction-only), which understates \(R_{tot}\) and overstates \(Q\); here the two air films together are actually \(0.70\ \text{m}^2\text{K/W}\), comparable to the cardboard layer.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 3 (composite walls, overall heat transfer coefficient).

Question 10 of 10: Walk-In Freezer — Composite Wall Heat Transfer

Question 10: Walk-In Freezer — Composite Wall Heat Transfer (15 marks)