1D conduction with generation using FDM

Governing equation and discretisation

Steady 1D conduction with generation:

$$k\frac{d^2T}{dx^2}+\dot q=0$$

Central difference at an interior node $i$:

$$k\frac{T_{i+1}-2T_i+T_{i-1}}{\Delta x^2}+\dot q=0\ \Rightarrow\ -T_{i-1}+2T_i-T_{i+1}=\frac{\dot q\Delta x^2}{k}$$

 x=0     0.025    0.05    0.075    0.1
 (0)------(1)------(2)------(3)------(4)
 100 C                               200 C

Data: $\Delta x=0.1/4=0.025$ m.

$$\frac{\dot q\Delta x^2}{k}=\frac{10^6\times(0.025)^2}{20}=31.25\ \text{K}$$

with $T_0=100$ and $T_4=200$.

Nodal equations

• Node 1: $2T_1-T_2=31.25+100=131.25$
• Node 2: $-T_1+2T_2-T_3=31.25$
• Node 3: $-T_2+2T_3=31.25+200=231.25$

In matrix form:

$$\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}\begin{bmatrix}T_1\\T_2\\T_3\end{bmatrix}=\begin{bmatrix}131.25\\31.25\\231.25\end{bmatrix}$$

Solution (elimination)

From node 1: $T_1=(131.25+T_2)/2$. From node 3: $T_3=(231.25+T_2)/2$. Substitute in node 2:

$$-\frac{131.25+T_2}{2}+2T_2-\frac{231.25+T_2}{2}=31.25\ \Rightarrow\ T_2-181.25=31.25$$ $$T_2=212.5\ ^\circ\text{C},\quad T_1=\frac{131.25+212.5}{2}=171.875\ ^\circ\text{C},\quad T_3=\frac{231.25+212.5}{2}=221.875\ ^\circ\text{C}$$

Exact solution

$$T(x)=T_0+(T_L-T_0)\frac xL+\frac{\dot q}{2k}x(L-x)$$

At $x=0.025$: $100+25+25000\times0.025\times0.075=171.875$. Similarly 212.5 and 221.875 at 0.05 and 0.075.

The two agree exactly because the exact solution is a quadratic, and the central difference formula for $d^2T/dx^2$ has an error proportional to $T''''$, which is zero here.

The maximum temperature is found from $dT/dx=1000+25000(0.1-2x)=0$, giving $x=0.07$ m, between nodes 2 and 3 (just before node 3), with $T_{max}=100+70+25000\times0.07\times0.03=222.5$ °C.

Answer: $T_1=171.9$ °C, $T_2=212.5$ °C, $T_3=221.9$ °C (equal to the exact values); $T_{max}=222.5$ °C at $x=0.07$ m.