One end of rod A is placed in a cold reservoir with a temperature of 5.00°C. The other end is held in a hot reservoir at 85.0°C. Rod A has a length L and a radius r. Rod B is made of the same material as rod A and the ends of rod B are placed in the same reservoirs as rod A. Rod B has a length 2L and a radius 2r. What is the ratio of heat flow through rod A to that through rod B?

1 / 2

Explanation:

This problem is a 1 - D steady state heat conduction with only one independent variable (x).

1 - D steady state:

Q = dT / Rc

Q = heat flow

dT = change in temperature between a pair of node

Rc = thermal resistance

Rc = L / k*A

Since in both cases Rod A and Rod B have identical boundary conditions:

dT_a = dT_b

So,

R_a = L / k * (pi*r^2)

R_b = 2L / k * (pi * (2r) ^2) = L / k * (2*pi*r^2)

Compute Q_a and Q_b:

Q_a = k * dT * (pi * r^2 * / L)

Q_a = k * dT * (2*pi * r^2 * / L)

Ratio of Q_a to Q_b

Q_a / Q_b = [k * dT * (pi * r^2 * / L) ] / [k * dT * (2*pi * r^2 * / L) ] = 1 / 2