Question 3 of 6: B1 — Conduction with temperature-dependent conductivity

Question 3: B1 — Conduction with temperature-dependent conductivity

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. One-dimensional steady conduction, no generation. Conductivity $k=k_0(1+\beta T)$. Surfaces held at $T_0$ (hot) and $T_L$ (cold). Compare against the constant-$k$ (reference) profile, which is linear in $x$ for the wall and logarithmic in $r$ for the cylinder.

Find. The location of maximum deviation between the actual and constant-$k$ temperature profiles, (a) in the plane wall, (b) in the hollow cylinder.

Approach. The deviation is greatest where its $x$- (or $r$-) derivative vanishes, i.e. where the actual profile’s slope equals the reference slope; because the constant flux equals the mean conductivity times the reference slope, that condition pins the local temperature to the mean $T_m$.

1. Constant flux integral (plane wall). With $q=-k\,dT/dx$ constant, $-k_0(1+\beta T)\,dT=q\,dx$; integrating, $$k_0\!\left(T+\tfrac{\beta}{2}T^2\right)=D-qx .$$ The reference (constant-$k$) profile is the straight line $T_{ref}(x)=T_0+(T_L-T_0)\,x/L$ with constant slope $(T_L-T_0)/L$.
2. Mean-conductivity form of the flux. Integrating $q$ across the wall gives $q=\dfrac{k_0\left[1+\tfrac{\beta}{2}(T_0+T_L)\right](T_0-T_L)}{L}=\bar k\,\dfrac{T_0-T_L}{L}$, where $\bar k=k_0\!\left[1+\tfrac{\beta}{2}(T_0+T_L)\right]$ is $k$ evaluated at the mean temperature.
3. Locate maximum deviation. The gap $\Delta(x)=T(x)-T_{ref}(x)$ is stationary where $dT/dx=(T_L-T_0)/L$. Substituting this slope into $-k_0(1+\beta T)\,dT/dx=q=\bar k(T_0-T_L)/L$ forces $1+\beta T=1+\tfrac{\beta}{2}(T_0+T_L)$, i.e. $$\boxed{\,T^\ast=\frac{T_0+T_L}{2}\equiv T_m\,}\qquad(\text{independent of geometry}).$$
4. Position in the wall (part a). Setting $T(x^\ast)=T_m$ in the integral of Step 1 and simplifying gives $$\frac{x^\ast}{L}=\frac{1+\tfrac{\beta}{4}(3T_0+T_L)}{2\left[1+\tfrac{\beta}{2}(T_0+T_L)\right]}\ \xrightarrow{\ \beta\to0\ }\ \tfrac12 .$$ Since $T_0>T_L$ gives $3T_0+T_L>2(T_0+T_L)$, the ratio exceeds $\tfrac12$ for $\beta>0$: the higher conductivity near the hot face flattens the gradient there, so the actual profile bows above the straight line and does not fall to $T_m$ until past the mid-plane — the peak-deviation plane sits toward the cold face (toward the hot face if $\beta<0$).
5. Hollow cylinder (part b). The same argument with $q'=-k\,2\pi r\,dT/dr$ constant gives $k_0(T+\tfrac{\beta}{2}T^2)=D-\dfrac{q'}{2\pi}\ln r$; the reference profile is logarithmic. The maximum deviation is again at $T=T_m$ (the slope-matching argument is unchanged, since $k\,r\,dT/dr$ is constant and the reference has $r\,dT_{ref}/dr$ constant). Because $k_0(T+\tfrac{\beta}{2}T^2)$ is now linear in $\ln r$ exactly as it was linear in $x$ for the wall, the same fraction applies to $\ln r$: $$\boxed{\,\frac{\ln(r^\ast/R_0)}{\ln[(R_0+L)/R_0]}=\frac{1+\tfrac{\beta}{4}(3T_0+T_L)}{2\left[1+\tfrac{\beta}{2}(T_0+T_L)\right]},\qquad r^\ast=R_0\left(\frac{R_0+L}{R_0}\right)^{x^\ast/L}\,}$$ As $\beta\to0$ this tends to the geometric-mean radius $\sqrt{R_0(R_0+L)}$; for $\beta>0$ it lies outside it, toward the cold outer surface.