## Consider two walls, A and B, with the same surface areas and the same temperature drops across their thicknesses. The ratio of their thermal

Question

Consider two walls, A and B, with the same surface areas and the same temperature drops across their thicknesses. The ratio of their thermal conductivities is kA/kB=4 and the ratio of the wall thickness is LA/LB=2. The ratio of heat transfer rates through the walls qA/qB is:

(a) 0.5

(b) 1

(c) 2

(d) 4

(e) 8

(f) None of them

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1 year 2021-08-23T20:35:12+00:00 1 Answers 385 views 0

(c) 2

Explanation:

Heat transfer across the walls due to conduction is given by:

$$q = -KA\frac{\Delta T}{L}$$

where,

q = heat transfer rate

K = thermal conductivity

A = Area

ΔT = change in temperature

L = thickness

For wall A:

$$q_A = -K_AA\frac{\Delta T}{L_A}$$

For wall B:

$$q_B = -K_BA\frac{\Delta T}{L_B}$$

Because the change of temperature and area of walls are the same. Dividing both terms:

$$\frac{q_A}{q_B} = \frac{\frac{K_A}{L_A} }{\frac{K_B}{L_B}}\\\\ \frac{q_A}{q_B} =\frac{\frac{K_A}{K_B} }{\frac{L_A}{L_B}}$$

using values given in the question:

$$\frac{q_A}{q_B} = \frac{4}{2}\\\\\frac{q_A}{q_B} = 2$$