ME241A — HW2 P3 Corner Node FDM (Laid Out • No Toggles • No Scrolls)

ME241A — HW 2, Problem 3: Corner Node Finite-Difference Equations

Everything is expanded and in-flow. No toggles, no internal scroll panes.

Problem 3 official solution schematic and equations — Official solution figure for Problem 3 (corner control volume and derived equations).

Setup (applies to all three cases)

Node & neighbors: \(P\equiv T_{m,n}\). Neighbors \(W=T_{m-1,n},\;E=T_{m+1,n},\;S=T_{m,n-1},\;N=T_{m,n+1}\).

Grid & CV: spacing \(\Delta x\) (horizontal), \(\Delta y\) (vertical). Corner CV uses half-faces on the sides touching the boundaries.

CV face	face length	distance to neighbor
West \(\to W\)	\(\Delta y\)	\(\Delta x\)
South \(\to S\)	\(\Delta x/2\)	\(\Delta y\)
East (lower half) \(\to E\)	\(\Delta y/2\)	\(\Delta x\)
North (left half) \(\to N\) or insulated/convective	\(\Delta x/2\)	\(\Delta y\)

Steady CV energy balance (outward positive): \(q_W + q_S + q_E^{(\text{cond})} + q_N^{(\text{cond})} + q_E^{(\text{bc})} + q_N^{(\text{bc})} = 0\). Conduction: \(q=-k\,\partial T/\partial n \cdot A\). Convection: \(q=hA\,(T_P-T_\infty)\). No volumetric generation.

(a) Top insulated, right convective \((T_\infty,h)\)

Step 1 — face fluxes

\( q_W = k\,\dfrac{\Delta y}{\Delta x}\,(T_P-T_W) \)

\( q_S = k\,\dfrac{\Delta x/2}{\Delta y}\,(T_P-T_S) \)

\( q_E^{\rm cond} = k\,\dfrac{\Delta y/2}{\Delta x}\,(T_P-T_E) \)

\( q_N^{\rm cond} = k\,\dfrac{\Delta x/2}{\Delta y}\,(T_P-T_N) \)

\( q_E^{\rm bc} = h\,\dfrac{\Delta y}{2}\,(T_P-T_\infty) \) (right)

\( q_N^{\rm bc} = 0 \) (top insulated)

Step 2 — assemble

\[ \Bigg[ k\frac{\Delta y}{\Delta x} +k\frac{\Delta x}{2\Delta y} +k\frac{\Delta y}{2\Delta x} +k\frac{\Delta x}{2\Delta y} +h\frac{\Delta y}{2} \Bigg]T_P = k\frac{\Delta y}{\Delta x}T_W +k\frac{\Delta x}{2\Delta y}T_S +k\frac{\Delta y}{2\Delta x}T_E +k\frac{\Delta x}{2\Delta y}T_N +h\frac{\Delta y}{2}T_\infty. \]

Step 3 — square grid \( \Delta x=\Delta y\equiv \Delta \)

\[ \begin{aligned} a_P &= 3k + \tfrac{h\Delta}{2} \\ a_W &= k \\ a_S &= \tfrac{k}{2} \\ a_E &= \tfrac{k}{2} \\ a_N &= k \\ S_u &= \left(\tfrac{h\Delta}{2}\right) T_\infty \end{aligned} \]

Step 4 — scale to match solution sheet

\[ \boxed{\,2(W+N)+(E+S)+\frac{h\Delta}{k}T_\infty-\Big(6+\frac{h\Delta}{k}\Big)P=0\,} \]

(b) Both boundaries insulated

Set \(h=0\). With \(\Delta x=\Delta y=\Delta\) and multiplying by \(2/k\):

\[ \boxed{\,2(W+N)+(E+S)-6P=0\,} \]

(c) Top & right both convective \((T_\infty,h)\)

Add \(+\,h(\Delta x/2)(T_P-T_\infty)\) on the north half-face:

\[ \boxed{\,2(W+N)+(E+S)+\dfrac{2h\Delta}{k}\,T_\infty-\Big(6+\dfrac{2h\Delta}{k}\Big)P=0\,} \]

What changed from line to line?

Face lengths (full vs half) → the 2 vs 1 coefficients after scaling by \(2/k\).
Convection adds to the diagonal and adds a \(T_\infty\) source term.
Setting \(\Delta x=\Delta y=\Delta\) and scaling by \(2/k\) yields tidy integer weights.

Problem 4.57 — Beginner-Friendly Step-by-Step (Print-Ready)

Problem 4.57 — Beginner-Friendly Step-by-Step (Print)

0) The Big Picture

2-D steady conduction in a wall with uniform volumetric generation, \( \dot{q} = 1\times10^{6}\ \text{W/m}^{3} \). Grid: \( \Delta x = \Delta y = 0.025\ \text{m} \). Thermal conductivity: \( k = 10\ \text{W}/(\text{m}\cdot\text{K}) \).

Given nodal temperatures (°C)	Convection boundaries
\(T_2=95.47,\; T_3=117.3,\; T_5=79.79,\) \(T_6=77.29,\; T_8=87.28,\; T_{10}=77.65\)	Inner (Surface B): \(h_i=500\), \(T_{\infty,i}=50^{\circ}\text{C}\) Outer (Surface A): \(h_o=250\), \(T_{\infty,o}=25^{\circ}\text{C}\)

Unknowns: \(T_1, T_4, T_7, T_9\). Then compute \(q'_A\) (outer) and \(q'_B\) (inner), and verify overall energy balance.

Mindset: treat each unknown as the center of a tiny control volume (CV). At steady state: energy in − energy out + generation = 0.

1) Universal Recipe (works at any node)

\[ \sum (\text{conduction to neighbors}) + \sum (\text{convection on boundary faces}) + \dot{q}\, A_{\text{CV}} = 0 \]

Conduction through a face: \( k\, \dfrac{\Delta T}{\Delta n} \times L \)
Convection on a boundary face: \( h\, (T_{\text{face}} - T_{\infty}) \times L \)
CV area (per unit depth): interior \( \Delta x \Delta y \); edges use half-faces; corners use quarter-faces.

2a) Node 1 — Corner (insulated top & left)

Fluxes present: conduct east to \(T_2\), south to \(T_3\), plus generation. No flux through insulated west/top.

\[ 0 = k\,\frac{(\Delta y/2)(T_2 - T_1)}{\Delta x} + k\,\frac{(\Delta x/2)(T_3 - T_1)}{\Delta y} + \dot{q}\,\frac{\Delta x\,\Delta y}{4} \] With \( \Delta x=\Delta y \): \[ \Rightarrow\; T_1 = \frac{T_2 + T_3}{2} + \frac{\dot{q}\,\Delta x^{2}}{4k} \]

Result: \(T_1 \approx 122.0^{\circ}\text{C}\)

2b) Node 4 — Interior corner on Surface B (convection inside)

Conducts to nodes \(2,5,8,3\) and convects to inner fluid at 50 °C along two CV edges; includes generation in an interior CV.

\[ 0 = k\frac{(\Delta x/2)(T_2 - T_4)}{\Delta y} + h_i\frac{(\Delta y/2)}{}(T_{\infty,i} - T_4) + h_i\frac{(\Delta x/2)}{}(T_{\infty,i} - T_4) + k\frac{(\Delta y)(T_5 - T_4)}{\Delta x} + k\frac{(\Delta y)(T_8 - T_4)}{\Delta x} + k\frac{(\Delta y)(T_3 - T_4)}{\Delta x} + \dot{q}\,\frac{3\,\Delta x\,\Delta y}{4} \] Rearranged (with \( \Delta x=\Delta y \)): \[ T_4 = \frac{ \big[T_2 + 2T_3 + T_5 + 2T_8 \big] + 2\,(h_i\Delta x/k)\,T_{\infty,i} + \left(3\,\dot{q}\,\Delta x^2 / 2k\right)} { \big[6 + 2\,(h_i\Delta x/k)\big] } \]

Result: \(T_4 \approx 94.50^{\circ}\text{C}\)

2c) Node 7 — Outer Surface A (convection outside)

Fluxes: conduct north to \(T_3\), east to \(T_8\), convection to outside fluid across a half-face, generation in a corner CV.

\[ 0 = k\frac{(\Delta x/2)(T_3 - T_7)}{\Delta y} + k\frac{(\Delta y)(T_8 - T_7)}{\Delta x} + h_o\,(\Delta x)\,(T_{\infty,o} - T_7) + \dot{q}\,\frac{\Delta x\,\Delta y}{4} \] \[ \Rightarrow\; T_7 = \frac{ T_3 + T_8 + (h_o\Delta x/k)\,T_{\infty,o} + \dot{q}\,\Delta x^2/(2k) } { 2 + (h_o\Delta x/k) } \]

Result: \(T_7 \approx 95.80^{\circ}\text{C}\)

2d) Node 9 — Outer Surface A (convection outside)

Fluxes: conduct north to \(T_5\), east to \(T_{10}\), convection to outside fluid across a whole face, generation in a edge CV.

\[ 0 = k\frac{(\Delta x)(T_5 - T_9)}{\Delta y} + k\frac{(\Delta y/2)(T_{10} - T_9)}{\Delta y} + h_o\,(\Delta x)\,(T_{\infty,o} - T_9) + k\frac{(\Delta y/2)(T_8 - T_9)}{\Delta x} + \dot{q}\,(\Delta x\,\Delta y/2) \] \[ \Rightarrow\; T_9 = \frac{ T_5 + 0.5\,T_8 + 0.5\,T_{10} + (h_o\Delta x/k)\,T_{\infty,o} + \dot{q}\,\Delta x^2/(2k) } { 2 + (h_o\Delta x/k) } \]

Result: \(T_9 \approx 79.67^{\circ}\text{C}\)

3) Heat rate per unit length from outer surface A to the fluid, \(q'_A\)

Sum convection from the outer faces of nodes 7, 8, 9 and 10 (each vertical face length = \( \Delta x \); corner half-faces use \( \Delta x/2 \)).

\[ q'_A = h_o\Big[ (\tfrac{\Delta x}{2})(T_7 - T_{\infty,o}) + \Delta x (T_8 - T_{\infty,o}) + \Delta x (T_9 - T_{\infty,o}) + (\tfrac{\Delta x}{2})(T_{10} - T_{\infty,o}) \Big] \]

Numerical result: \(q'_A \approx 1117\ \text{W/m}\)

4) Heat rate per unit length from inner fluid to Surface B, \(q'_B\)

Sum convection into nodes 2, 4, 5, and 6 along inner faces (use half-/quarter-lengths as appropriate).

\[ q'_B = h_i\Big[ (\tfrac{\Delta y}{2})(T_{\infty,i} - T_2) + (\tfrac{\Delta y}{2}+\tfrac{\Delta x}{2})(T_{\infty,i} - T_4) + \Delta x\,(T_{\infty,i} - T_5) + (\tfrac{\Delta x}{2})(T_{\infty,i} - T_6) \Big] \]

Numerical result: \(q'_B \approx -1383\ \text{W/m}\)

5) Energy Balance Check

\[ \dot{E}'_{\text{in}} - \dot{E}'_{\text{out}} + \dot{E}'_{\text{gen}} = 0 \quad\Rightarrow\quad - q'_A + q'_B + \dot{Q}'_{\text{gen}} = 0 \] with \( \dot{Q}'_{\text{gen}} = \dot{q}\,V' = 10^{6}\ \text{W/m}^{3}\times (25\times 50 + 25\times 50)\times 10^{-6}\ \text{m}^{2} = 2500\ \text{W/m} \).

Balance: \(-1117 - 1383 + 2500 = 0\;\checkmark\)

Quick Reminders (for interviews)

Every unknown node = one tiny CV energy balance. State what faces are active.
Edges use half-faces; corners use quarter-faces for \(L\) and \(A_{CV}\).
When asked for heat rate to a fluid, use only the convection faces that touch that fluid.
Always finish with a 10-second whole-section energy-balance check.

2-D Conduction with Generation & Convection — FDM Walkthrough

2-D Steady Conduction with Internal Generation & Convection — Finite-Difference Method

This problem involves a 2-D steady-state heat conduction analysis of a wall with internal heat generation and convection boundaries. The solution uses the finite-difference method (FDM), which approximates a continuous body with a grid of discrete nodes. Steps: 1) set up, 2) solve unknown nodal temperatures, 3) compute surface heat rates, 4) energy balance.

1) Set up 2) Nodal temperatures 3) Surface heat rates 4) Energy balance

1. Setting Up the Problem 📝

L-shaped wall, grid spacing \( \Delta x = \Delta y = \Delta = \textbf{0.025\ \mathrm{m}} \); some nodal temperatures are known from a table.

Given Information

Volumetric generation \( \dot q \) \( 1\times 10^{6}\ \mathrm{W/m^3} \)

Thermal conductivity \( k \) \( 10\ \mathrm{W/(m\cdot K)} \)

Grid spacing \( \Delta=0.025\ \mathrm{m} \)

Inner convection \( h_i,\, T_{\infty,i} \) \( 500\ \mathrm{W/(m^2K)},\ 50^{\circ}\mathrm{C} \)

Outer convection \( h_o,\, T_{\infty,o} \) \( 250\ \mathrm{W/(m^2K)},\ 25^{\circ}\mathrm{C} \)

Known nodal \( T \) \( T_2, T_3, T_5, T_6, T_8, T_{10} \) given

Goal

Unknown temperatures: \( T_1, T_4, T_7, T_9 \).
Heat rates per unit length: outer \( q'_A \), inner \( q'_B \).
Verify with an overall energy balance.

Node energy balance (control volume)

Conduction from neighbors + convection from fluid + internal generation = 0

\[ \sum_{\text{faces}} k\frac{A}{L_n}\,(T_N - T_P) \;+\; \sum_{\text{conv}} hA\,(T_{\infty}-T_P) \;+\; \dot q\,V \;=\; 0 \]

Corner CV: \(V=\tfrac{1}{4}\Delta^2\); Edge CV: \(V=\tfrac{1}{2}\Delta^2\); Re-entrant (Node 4): \(V=\tfrac{3}{4}\Delta^2\).

2. Calculating Nodal Temperatures 🌡️

Node 1 — Insulated Corner

Top & left insulated; conduction from Nodes 2 and 3; generation in a quarter CV:

\[ 0 = k\frac{\Delta}{\Delta/2}(T_2 - T_1) + k\frac{\Delta}{\Delta/2}(T_3 - T_1) + \dot q\left(\tfrac{1}{4}\Delta^2\right) \] \[ \Rightarrow\; T_1 = \frac{2T_2 + T_3}{3} + \frac{\dot q\,\Delta^2}{12k} \quad\Rightarrow\quad T_1 \approx 122.0^\circ\mathrm{C} \]

Node 4 — Re-entrant Corner

Three-quarter CV; conduction from 2,3,5,8 and inner convection:

\[ T_4 = \frac{6 + 2\!\left(\frac{k}{h_i\,\Delta}\right)}{6}\,T_2 + \frac{2}{6}\,T_3 + \frac{1}{6}\,T_5 + \frac{2}{6}\,T_8 + \frac{2\!\left(\frac{k}{h_i\,\Delta}\right)}{6}\,T_{\infty,i} + \frac{2k}{3}\,\frac{\dot q\,\Delta^2}{6k} \;\;\Rightarrow\;\; T_4 \approx 94.50^\circ\mathrm{C} \]

Node 7 — Outer Corner with Convection

\[ T_7 = \frac{2 + \left(\frac{k}{h_o\,\Delta}\right)}{2}\,T_3 + \frac{1}{2}\,T_8 + \frac{\left(\frac{k}{h_o\,\Delta}\right)}{2}\,T_{\infty,o} + \frac{2k}{2}\,\frac{\dot q\,\Delta^2}{2k} \;\;\Rightarrow\;\; T_7 \approx 95.80^\circ\mathrm{C} \]

Node 9 — Outer Edge with Convection

\[ T_9 = \frac{2 + \left(\frac{k}{h_o\,\Delta}\right)}{2}\,T_5 + \frac{1}{4}\,T_8 + \frac{1}{4}\,T_{10} + \frac{\left(\frac{k}{h_o\,\Delta}\right)}{2}\,T_{\infty,o} + \frac{2k}{2}\,\frac{\dot q\,\Delta^2}{2k} \;\;\Rightarrow\;\; T_9 \approx 79.67^\circ\mathrm{C} \]

3. Calculating Surface Heat Rates 🔥

Per unit out-of-plane length (\(\mathrm{W/m}\)). Use exposed boundary face lengths.

Outer Surface \( A \): nodes 7, 8, 9, 10

\[ q_A' \;=\; h_o \Big[ \tfrac{2}{\Delta}\,\Delta\,(T_7 - T_{\infty,o}) \;+\; \Delta\,(T_8 - T_{\infty,o}) \;+\; \Delta\,(T_9 - T_{\infty,o}) \;+\; \tfrac{2}{\Delta}\,\Delta\,(T_{10} - T_{\infty,o}) \Big] \quad\Rightarrow\quad q_A' \approx 1117\ \mathrm{W/m} \]

Inner Surface \( B \): nodes 2, 4, 5, 6

Area note: Use full exposed face for each node — nodes 2 & 6: length \(2\Delta\); node 4: re-entrant faces sum to \(2\Delta\); node 5: \( \Delta \). This matches the second document’s approach.

\[ q_B' \;=\; h_i \Big[ \tfrac{2}{\Delta}\,\Delta\,(T_{\infty,i}-T_2) \;+\; \Delta\,(T_{\infty,i}-T_4) \;+\; \Delta\,(T_{\infty,i}-T_5) \;+\; \tfrac{2}{\Delta}\,\Delta\,(T_{\infty,i}-T_6) \Big] \quad\Rightarrow\quad q_B' \approx -1383\ \mathrm{W/m} \quad (\text{negative } \Rightarrow \text{heat leaves fluid into wall}) \]

4. Overall Energy Balance ⚖️

\[ \dot E'_{\text{in}} - \dot E'_{\text{out}} + \dot E'_{\text{gen}} = 0 \quad\Rightarrow\quad (-q_A') + q_B' + \dot q\,V' = 0 \]

L-section area \(V'\)

Add CV pieces / two rectangles: \[ V' = 2.5\times 10^{-3}\ \mathrm{m^2} \]

Generation term

\[ \dot E'_{\text{gen}} = \dot q\,V' = (1\times 10^{6}) (2.5\times 10^{-3}) = 2500\ \mathrm{W/m} \]

\[ -1117 + (-1383) + 2500 = -2500 + 2500 = 0 \]

Energy balance satisfied — results are self-consistent.

Legend / Quick Keys

\(T_P\): node of interest \(T_N\): neighbor node \(h\): convection coefficient \(k\): conductivity \(\Delta\): grid spacing \(A\), \(V\): face area, control-volume area