Question

For heat conduction through a hollow cylinder, the heat transfer rate depends on:

• A. Difference in surface areas only
• B. Logarithmic mean radius
• C. Logarithmic mean temperature difference
• D. Logarithmic ratio of radii

Hint:

Logic Tip: Whenever you deal with radial flow (cylinders or spheres), the area is constantly changing, which means the math will involve calculus. For cylinders, the integral of $1/r$ always results in a natural logarithm ($\ln$) of the radii ratio.

Answer 1

The Correct Option is D

Concept:
Unlike a flat wall where the area available for heat transfer is constant, the area of a cylinder ($A = 2\pi r L$) increases as you move outward from the inner radius ($r_1$) to the outer radius ($r_2$). This changing area requires integration of Fourier's law.

Step 1: The equation is $Q = -kA \frac{dT}{dr}$. Substitute the area of a cylinder: $Q = -k(2\pi r L) \frac{dT}{dr}$.

Step 2: To solve for the total heat transfer ($Q$) across the cylinder wall thickness, we must integrate from the inner radius ($r_1$) to the outer radius ($r_2$): $$ \int_{r_1}^{r_2} \frac{Q}{2\pi k L} \cdot \frac{dr}{r} = -\int_{T_1}^{T_2} dT $$

Step 3: The integral of $1/r$ is the natural logarithm ($\ln(r)$). $$ \frac{Q}{2\pi k L} [\ln(r_2) - \ln(r_1)] = (T_1 - T_2) $$

Step 4: Using logarithm rules ($\ln(A) - \ln(B) = \ln(A/B)$), the final equation for radial heat conduction through a cylinder becomes: $$ Q = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)} $$

Step 5: The equation clearly shows that the heat transfer rate ($Q$) is inversely dependent on the term $\ln(r_2/r_1)$, which is the logarithmic ratio of the outer and inner radii.