Heat conduction through a thick-walled cylinder — correct formula for Q. For a hollow cylinder of length l, inner radius r1, outer radius r2, thermal conductivity k, with inner surface at temperature T1 and outer surface at T2, select the correct steady heat-transfer rate Q expression.
-
AQ = 2 * pi * k * l * (T1 - T2) / ln(r2 / r1)
-
BQ = k * pi * (r2^2 - r1^2) * (T1 - T2) / l
-
CQ = 2 * pi * k * (T1 - T2) * ln(r2 / r1) / l
-
DQ = k * l * (T1 - T2) / [2 * pi * ln(r2 / r1)]
-
EQ = pi * k * l * (T1 - T2) * (r2 - r1)
Answer
Correct Answer: Q = 2 * pi * k * l * (T1 - T2) / ln(r2 / r1)
Explanation
Introduction / Context:Thermal resistance for radial conduction in cylindrical coordinates differs from that of a plane wall. Correctly identifying the logarithmic relationship between radii is crucial for pipes, insulation sizing, and heat-loss calculations in process plants and HVAC.
Given Data / Assumptions:
- Steady state, one-dimensional radial conduction.
- Constant thermal conductivity k.
- Inner surface temperature T1, outer surface temperature T2.
- No internal heat generation; length l is much larger than thickness.
Concept / Approach:From the differential form of Fourier’s law in cylindrical coordinates, the heat flow is not linear with radius. Integrating across r from r1 to r2 yields a logarithmic mean area effect, leading to a thermal resistance R_cond = ln(r2/r1) / (2 * pi * k * l).
Step-by-Step Solution:Start with Fourier’s law: Q = −k * A_r * dT/dr, where A_r = 2 * pi * r * l.Separate variables: (dT) = −(Q / (2 * pi * k * l)) * (dr / r).Integrate: ∫{T1}^{T2} dT = −(Q / (2 * pi * k * l)) ∫{r1}^{r2} (dr / r).Thus T2 − T1 = −(Q / (2 * pi * k * l)) * ln(r2/r1).Rearrange: Q = 2 * pi * k * l * (T1 − T2) / ln(r2 / r1).
Verification / Alternative check:Compare with the resistance analogy: Q = (T1 − T2) / R_cond with R_cond = ln(r2/r1)/(2 * pi * k * l), which yields the same expression.
Why Other Options Are Wrong:(b) is a plane-wall style form with area difference; not valid for radial conduction. (c) places the logarithm in the numerator incorrectly. (d) inverts the resistance structure. (e) uses linear thickness (r2 − r1) instead of the correct logarithmic dependence.
Common Pitfalls:Using average area or linear thickness for cylinders; this can significantly misestimate heat loss.
Final Answer:Q = 2 * pi * k * l * (T1 - T2) / ln(r2 / r1)