--- title: "Finite difference nodal temperatures in a square plate" subject: "Heat Transfer" chapter: 2 key: "semester_5/heat_transfer:two-dimensional-conduction-nodal-temperatures" source: https://www.bctengineer.com/q/semester_5/heat_transfer/two-dimensional-conduction-nodal-temperatures site: BCT Engineering --- # Finite difference nodal temperatures in a square plate Heat Transfer · Chapter 2: Conduction Heat Transfer - Asked in: year not given, 8 marks - Video: [Finite Difference Method/Heat Transfer/Simple Node Problem](https://www.youtube.com/watch?v=6BwnF9gGQOM) — The Chemical Engineering Tutor - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/heat_transfer/two-dimensional-conduction-nodal-temperatures - Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-2 ## Question Explain the finite-difference method for two-dimensional steady-state conduction and obtain the nodal equation for an interior node. A long square bar has its cross-section divided into a square grid of mesh size dx = dy with four interior nodes arranged 2 x 2 [nodes 1 and 2 in the top row, nodes 3 and 4 in the bottom row]. The top surface of the bar is held at 500 °C, and the left, right and bottom surfaces at 100 °C. Find the four nodal temperatures. ## Short answer Using finite-difference method with square grid, interior node temperature is the average of its four neighbours. Solving the nodal equations for the given boundary conditions yields T1 = T2 = 250 °C and T3 = T4 = 150 °C. ## Answer #### Finite-difference method The region is covered by a grid of nodes spaced $\Delta x$ and $\Delta y$. The Laplace equation $\dfrac{\partial^2T}{\partial x^2}+\dfrac{\partial^2T}{\partial y^2}=0$ is replaced using central differences: $$ \frac{\partial^2T}{\partial x^2}\approx\frac{T_{m+1,n}-2T_{m,n}+T_{m-1,n}}{\Delta x^2},\qquad \frac{\partial^2T}{\partial y^2}\approx\frac{T_{m,n+1}-2T_{m,n}+T_{m,n-1}}{\Delta y^2} $$ For $\Delta x = \Delta y$ this gives the interior-node equation $$ T_{m,n} = \frac{1}{4}\left(T_{m+1,n}+T_{m-1,n}+T_{m,n+1}+T_{m,n-1}\right) $$ so each node is the average of its four neighbours. Writing it for every unknown node gives a set of linear equations, solved by matrix inversion or iteration (Gauss-Seidel). #### Nodal equations ```text 500 500 500 500 100 [ 1 ]-[ 2 ] 100 | | 100 [ 3 ]-[ 4 ] 100 100 100 100 100 ``` Node 1 neighbours: top 500, left 100, node 2, node 3. $$ \begin{aligned} 4T_1 &= 500 + 100 + T_2 + T_3\\ 4T_2 &= 500 + 100 + T_1 + T_4\\ 4T_3 &= 100 + 100 + T_1 + T_4\\ 4T_4 &= 100 + 100 + T_2 + T_3 \end{aligned} $$ #### Solution The conditions are symmetric about the vertical centre line, so $T_1 = T_2$ and $T_3 = T_4$: $$ \begin{aligned} 3T_1 - T_3 &= 600\\ 3T_3 - T_1 &= 200 \end{aligned} $$ From the second, $T_1 = 3T_3 - 200$. Substitute in the first: $9T_3 - 600 - T_3 = 600$, so $T_3 = 150$ °C and $T_1 = 250$ °C. Check in node 1 equation: $(500+100+250+150)/4 = 250$. Correct. **Answer:** $T_1 = T_2 = 250\ ^\circ$C; $T_3 = T_4 = 150\ ^\circ$C. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For dx = dy, an interior node temperature equals: A. Sum of its four neighbours B. Average of two neighbours C. Half of the hottest neighbour D. Average of its four neighbours Answer: D. Central differences on Laplace's equation give T = (T_E+T_W+T_N+T_S)/4. (Key point: Interior node equation) 2. Which is the correct equation for node 1? A. 4T1 = 500 + 100 + T2 + T3 B. 4T1 = 100 + 100 + T2 + T3 C. 4T1 = 500 + 500 + T2 + T3 D. T1 = 500 + 100 + T2 + T3 Answer: A. Node 1 touches the top (500), left (100), node 2 and node 3. (Key point: Node 1 equation) 3. Why can T1 = T2 and T3 = T4 be assumed? A. All surfaces are at the same temperature B. Conditions are symmetric about the vertical centre line C. The grid is only 2 x 2 D. Heat generation is uniform Answer: B. Left and right sides are both 100 C, so the problem is symmetric. (Key point: Symmetry) 4. The nodal temperatures are: A. T1 = T2 = 150 C, T3 = T4 = 250 C B. All 200 C C. T1 = T2 = 250 C, T3 = T4 = 150 C D. T1 = T2 = 300 C, T3 = T4 = 100 C Answer: C. Solving 3T1 - T3 = 600, 3T3 - T1 = 200 gives T3 = 150, T1 = 250. 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