Heat generation in a current-carrying conductor: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2018, Q6

Question 6 — Heat generation in a current-carrying conductor (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. Solid cylinder, diameter 75 mm ($r_0=0.0375$ m), $k=70\ \text{W/m°C}$, uniform volumetric generation $\dot q'''$. Fluid $T_\infty=27$ °C, surface coefficient $h=568\ \text{W/m}^2\text{°C}$; peak (centreline) temperature limited to $T_\text{max}=540$ °C.

Find. (a) the maximum generation rate per unit length, (b) the surface temperature.

Figure 6 — Uniform generation gives a parabolic radial temperature profile. The centreline is hottest (limited to 540 °C); the surface sits above the fluid by the convective drop $\dot q'''r_0/2h$.

Approach. For a solid cylinder with uniform generation, the centre-to-surface rise is $\dot q'''r_0^2/4k$ and the surface-to-fluid rise is $\dot q'''r_0/2h$ (from a surface energy balance). Summing them to the 540 °C limit fixes $\dot q'''$; then the surface temperature follows.

Peak-temperature constraint. The centreline temperature is $$T_c=T_\infty+\underbrace{\frac{\dot q'''r_0}{2h}}_{\text{surface film}}+\underbrace{\frac{\dot q'''r_0^{2}}{4k}}_{\text{conduction in solid}}=540\ ^\circ\text{C}$$ Solving for the generation rate: $$\dot q'''=\frac{T_c-T_\infty}{\dfrac{r_0}{2h}+\dfrac{r_0^{2}}{4k}}=\frac{513}{3.30\times10^{-5}+5.02\times10^{-6}}=1.349\times10^{7}\ \text{W/m}^3$$
Generation per unit length. Multiply by the cross-sectional area $\pi r_0^2$: $$\dot q'=\dot q'''\,\pi r_0^{2}=1.349\times10^{7}\,\pi(0.0375)^{2}=5.96\times10^{4}\ \text{W/m}$$ $\dot q'\approx 59.6\ \text{kW/m}$
Surface temperature. From the surface film drop only: $$T_s=T_\infty+\frac{\dot q'''r_0}{2h}=27+\frac{1.349\times10^{7}(0.0375)}{2(568)}=27+445=472\ ^\circ\text{C}$$ $T_s\approx 472\ ^\circ\text{C}$ As a check, adding the conduction rise $\dot q'''r_0^2/4k=68$ °C recovers the 540 °C centreline. ✓

Quantity	Result
Volumetric generation $\dot q'''$	$1.35\times10^{7}$ W/m³
Generation per unit length $\dot q'$	59.6 kW/m
Surface temperature $T_s$	472 °C

Concept & Theory

Topic: steady one-dimensional conduction with uniform internal heat generation in a solid cylinder.

Key relations: temperature profile $T(r)=T_s+\dfrac{\dot q'''}{4k}(r_0^2-r^2)$; centre-to-surface rise $\dfrac{\dot q'''r_0^2}{4k}$; surface balance $\dot q'''\pi r_0^2=h\,2\pi r_0(T_s-T_\infty)\Rightarrow T_s-T_\infty=\dfrac{\dot q'''r_0}{2h}$.

Why this works: integrating the generation-conduction equation with symmetry at the centre gives the parabolic profile, whose maximum is at $r=0$. All the generated heat must ultimately convect off the surface, which fixes the surface temperature independently of $k$. The design limit is the centreline (the hottest point), so both drops add.

Common pitfall: applying the limit at the surface rather than the centre, or dropping one of the two temperature rises. Note $\dot q'$ (per length) $=\dot q'''\pi r_0^2$, not $\dot q'''\times$ perimeter.

Source: Incropera et al., Fundamentals of Heat and Mass Transfer 8th ed., §3.5 (conduction with generation).

Question 6 of 8: Heat generation in a current-carrying conductor

Question 6 — Heat generation in a current-carrying conductor (Part B, equal value)