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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Heat Transfer

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer

A hollow tube possesses a length l , inner radius R_1 , and outer...

A hollow tube possesses a length ll, inner radius R1R_1, and outer radius R2R_2. The material has a thermal conductivity KK. Determine the heat flowing through the walls of the tube if the flat ends are maintained at temperatures T1T_1 and T2T_2 (T2>T1T_2 > T_1).

Options

  1. AKπ(R22−R12)(T2−T1)l\dfrac{K \pi (R_2^2 - R_1^2)(T_2 - T_1)}{l}
  2. BKπ(R22+R12)(T2−T1)l\dfrac{K \pi (R_2^2 + R_1^2)(T_2 - T_1)}{l}
  3. CKπ(R2−R1)2(T2−T1)l\dfrac{K \pi (R_2 - R_1)^2(T_2 - T_1)}{l}
  4. D2πKl(T2−T1)ln⁡(R2/R1)\dfrac{2\pi K l (T_2 - T_1)}{\ln(R_2 / R_1)}

Correct answer

A. Kπ(R22−R12)(T2−T1)l\dfrac{K \pi (R_2^2 - R_1^2)(T_2 - T_1)}{l}

Step-by-step solution

The heat flows longitudinally along the length of the tube because the temperature difference is maintained across its flat ends.

The cross-sectional area of the hollow tube through which heat flows is the area of the annular region, given by A=πR22−πR12=π(R22−R12)A = \pi R_2^2 - \pi R_1^2 = \pi(R_2^2 - R_1^2).

The length of the tube is ll and the temperature difference is ΔT=T2−T1\Delta T = T_2 - T_1.

According to the law of thermal conduction, the rate of heat flow is H=KAΔTlH = \dfrac{KA\Delta T}{l}.

Substituting the values, we get H=Kπ(R22−R12)(T2−T1)lH = \dfrac{K \pi (R_2^2 - R_1^2)(T_2 - T_1)}{l}.
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