Heat Transfer Mechanisms — Question 3

Question 3

Temperature-dependent conductivity creates a curved profile

A plane slab has length $L=.200$ m, area $.0100\,\mathrm{m^2}$, and conductivity $k(T)=k_0[1+b(T-300)]$, with $k_0=10.0$ W/(m K), $b=.0100\,\mathrm{K^{-1}}$. Its faces at $x=0,L$ are held at 400 K and 300 K. Use steady $P=-Ak(T)dT/dx$, with no generation or lateral loss.

1. Integrate the conduction law to find the heat rate.

2. Find the temperature at $x=L/2$ and explain why it is not the arithmetic mean of the face temperatures.

Question 3 - Solution

Temperature-dependent conductivity creates a curved profile

(a) Method and calculation

$$\frac{PL}{A}=\int_{300}^{400}k(T)dT=k_0\left[\Delta T+\frac b2(\Delta T)^2\right]=1500\,\mathrm{W/m}.$$$$P=\frac{.0100}{.200}(1500)=75.0\,\mathrm W.$$

(b) Interpretation and further calculation

Let $\theta=T-300$. Integrating from position $x$ to the cold face gives$$\theta+.005\theta^2=150(1-x/L).$$$$x=L/2:\quad .005\theta^2+\theta-75=0\ \Rightarrow\ \theta=-100+\sqrt{25000}=58.1139\,\mathrm K.$$$$T(L/2)=358.114\,\mathrm K.$$The hot region has larger conductivity and hence a smaller magnitude of temperature gradient for the same heat rate.

Physical check

The endpoint-averaged conductivity gives the exact total rate here because $k$ is linear in temperature, but it does not make $T(x)$ linear.