Chapter 1, Problem 1.1P

On a cold winter day, the outer surface of a 0.2-m-thick concrete wall of a warehouse is exposed to temperature of –5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/m K. Determine the heat loss through the wall, which is 10-m long and 3-m high.

Problem 1.1

To determine

Heat loss through the wall

Answer to Problem 1.1P

Heat loss through the wall is 4.5 kW.

Explanation of Solution

Given Information :

Temperature of the inner surface of the wall T₁ = 20⁰C

Temperature of the outer surface of the wall T₂ = -5⁰C

Thickness of the wall (L) = 0.2m, Thermal conductivity of the wall (k) = 1.2 W/m.K, Length of the wall is 10m, and height of the wall is 3m.

The rate of heat loss from the inner surface to outer surface of the wall is obtained from equation 1

$\mathrm{q}=\mathrm{ }\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{\mathrm{R}}$ ——– Equation1

$\mathrm{q}\mathrm{ }=\mathrm{ }\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{s}$

$\mathrm{R}=\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{ }\mathrm{ }\mathrm{ }$

Thermal resistance to heat conduction for a plane wall can be obtained by the equation given below.

$\mathrm{R}=\frac{\mathrm{L}}{\mathrm{k}\mathrm{A}}$———–Equation2

“A” is an area of the wallperpendicular to the direction of heat flow.

Area of heat flow (A) =10 x 3 = 30m²

Substituting the values in Equation2, we get,

$\mathrm{R}=\frac{0.2}{1.2\mathrm{*}30}=5.55\mathrm{*}{10}^{-3}$

Substitute the value of R in Equation1

$\mathrm{q}=\mathrm{ }\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{\mathrm{R}}=\mathrm{ }\frac{20-\left(-5\right)}{5.55\mathrm{*}{10}^{-3}}=4504.5045\mathrm{ }\mathrm{W}\mathrm{a}\mathrm{t}\mathrm{t}=4.5\mathrm{ }\mathrm{k}\mathrm{W}\mathrm{a}\mathrm{t}\mathrm{t}$

Conslusion:

Heat loss through the wall is 4.5 kW.