Current-carrying conductor with internal heat generation: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2019, Q6

Question 6 — Current-carrying conductor with internal heat generation (Part B, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Radius	$r_0$	0.0375 m
Conductivity	$k$	70 W/m·°C
Fluid temperature	$T_\infty$	27 °C
Surface coefficient	$h$	568 W/m²·°C
Max (centreline) temperature	$T_\text{max}$	540 °C

Find. (a) maximum $\dot g$ expressed per unit length; (b) surface temperature.

Approach. For uniform generation in a solid cylinder the hottest point is the centreline; write the total centreline rise as the sum of the surface-convection rise and the internal conduction rise, set it to the limit, solve for $\dot g$, then back out the surface temperature.

Centreline temperature relation. Surface balance gives $T_s-T_\infty=\dot g\,r_0/(2h)$; internal conduction gives $T_\text{max}-T_s=\dot g\,r_0^2/(4k)$. Adding: $$T_\text{max}-T_\infty=\dot g\!\left[\frac{r_0}{2h}+\frac{r_0^2}{4k}\right]$$
Solve for the volumetric generation. $$\dot g=\frac{540-27}{\dfrac{0.0375}{2(568)}+\dfrac{0.0375^2}{4(70)}}=\frac{513}{3.803\times10^{-5}}=1.349\times10^{7}\ \text{W/m}^3$$
Per-unit-length generation. $$q'=\dot g\,(\pi r_0^2)=1.349\times10^{7}\,\pi(0.0375)^2=\boxed{59.6\ \text{kW/m}}$$
Surface temperature. $$T_s=T_\infty+\frac{\dot g\,r_0}{2h}=27+\frac{1.349\times10^{7}(0.0375)}{2(568)}=27+445=\boxed{472\ ^\circ\text{C}}$$ Check: $T_s+\dot g r_0^2/(4k)=472+68=540\ ^\circ\text{C}$ ✓.

Quantity	Result
Volumetric generation $\dot g$	1.35 × 10⁷ W/m³
(a) Per-length generation $q'$	59.6 kW/m
(b) Surface temperature	472 °C

Concept & Theory — Q6

Topic: Steady conduction in a solid cylinder with uniform volumetric heat generation and convective cooling.

Key relations: Centre-to-surface rise $T_0-T_s=\dot g r_0^2/(4k)$; surface energy balance $\dot g\,\pi r_0^2=h(2\pi r_0)(T_s-T_\infty)$ ⇒ $T_s-T_\infty=\dot g r_0/(2h)$; per-length rate $q'=\dot g\,\pi r_0^2$.

Why this works: Solving the 1-D generation equation with a symmetric (zero-flux) centreline gives a parabolic temperature profile whose peak is at the axis; the two rises — conduction from axis to surface and convection from surface to fluid — act in series, so the maximum temperature is simply their sum above the fluid. Fixing that sum at the material limit is the binding constraint, and it uniquely sets the allowable generation.

Common pitfall: Using $r_0^2/(2k)$ instead of $r_0^2/(4k)$ for the internal rise (the factor is 4 for a cylinder, 6 for a sphere, 2 for a slab), or omitting the surface-convection rise so the whole 513 °C is attributed to conduction.

Source: Çengel & Ghajar, Heat and Mass Transfer 6th ed., §2-5 & §3-6 (heat generation in solids); Incropera 8th ed., §3.5.

Question 6 of 8: Current-carrying conductor with internal heat generation

Question 6 — Current-carrying conductor with internal heat generation (Part B, equal value)