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Difficulty: Very hardModes of Heat Transfer (Conduction, Convection, and Radiation)

A composite cylindrical bar consists of two uniform sections of equal length joined end-to-end. Section A has a radius of 2.0 cm2.0\text{ cm} and a thermal conductivity of 300 W m−1K−1300\text{ W m}^{-1}\text{K}^{-1}. Section B has a radius of 4.0 cm4.0\text{ cm} and a thermal conductivity of 150 W m−1K−1150\text{ W m}^{-1}\text{K}^{-1}. The outer end of Section A is maintained at a constant temperature of 120∘C120^\circ\text{C}, while the outer end of Section B is held at 0∘C0^\circ\text{C}. Assuming the curved surfaces of both sections are perfectly insulated and heat flow is steady, what is the temperature at the junction between the two sections in ∘C^\circ\text{C}?

Answer: 40 °C

Answer

The steady-state temperature at the junction between Section A and Section B is 40.0°C.
At steady state, the rate of heat conduction through Section A equals that through Section B. Because Section B has double the radius of Section A, its cross-sectional area is four times as large. Equating the heat flow rates gives 300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J, which simplifies directly to 120 - T_J = 2 T_J, yielding a junction temperature of 40.0°C.

Step-by-Step Solution

1
Determine the relationship between the cross-sectional areas of Section A and Section B.
The area ratio A_B / A_A = (r_B / r_A)² = (4.0 cm / 2.0 cm)² = 4.
The cross-sectional area of a cylinder is proportional to the square of its radius.
2
Write the steady-state heat flow equation for each section.
H_A = (k_A * A_A * (120 - T_J)) / L and H_B = (k_B * A_B * (T_J - 0)) / L.
According to Fourier's law of thermal conduction, the heat transfer rate through a uniform layer is proportional to thermal conductivity, cross-sectional area, and temperature difference, and inversely proportional to length.
3
Equate the heat transfer rates H_A and H_B and solve for the junction temperature T_J.
300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J => 300 * (120 - T_J) = 600 * T_J => 120 - T_J = 2 T_J => 3 T_J = 120 => T_J = 40.0°C.
At steady state with insulated sides, heat does not accumulate or escape, so the rate of heat conduction through both sections must be identical.

Key Concept

Steady-state thermal conduction through composite conductors with differing cross-sectional areas and thermal conductivities
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Topics

Modes of Heat Transfer (Conduction, Convection, and Radiation)

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