---
title: "Plane wall with internal heat generation"
subject: "Heat Transfer"
chapter: 2
key: "semester_5/heat_transfer:plane-wall-with-heat-generation"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/plane-wall-with-heat-generation
site: BCT Engineering
---

# Plane wall with internal heat generation

Heat Transfer · Chapter 2: Conduction Heat Transfer

- Asked in: year not given, 8 marks
- Video: [One-Dimensional Heat Conduction with Generation](https://www.youtube.com/watch?v=Z4SuGpHcwJ4) — LearnChemE
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/plane-wall-with-heat-generation
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-2

## Question

A large plane plate of thickness 40 mm and thermal conductivity 20 W/m K generates heat uniformly at 5 x 10^6 W/m^3 (for example, an electrical heating element). Both faces are cooled by a fluid at 80 °C with a heat transfer coefficient of 500 W/m^2 K. Derive the temperature distribution for the plate and find (a) the surface temperature, (b) the maximum temperature and its location, and (c) the heat removed per square metre from each face.

## Short answer

For steady 1D conduction with uniform generation and symmetric convection cooling, the temperature distribution is T(x) = T_s + (q_g/(2k))(L^2 - x^2). Surface temperature T_s = 280 °C, maximum temperature at centre T_max = 330 °C, heat removed per face q'' = 100 kW/m^2.

## Answer

#### Derivation
Let the plate occupy $-L \le x \le L$ with $2L = 40$ mm, so $L = 0.02$ m. For steady one-dimensional conduction with generation $\dot q_g$:

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

Integrate: $T = -\dfrac{\dot q_g}{2k}x^2 + C_1x + C_2$. By symmetry $dT/dx = 0$ at $x = 0$, so $C_1 = 0$. At the surface $x = L$ the conduction equals convection:

$$
-k\frac{dT}{dx}\Big|_{L} = h\,(T_s - T_\infty)\;\Rightarrow\;\dot q_g L = h\,(T_s-T_\infty)
$$

so $T_s = T_\infty + \dfrac{\dot q_g L}{h}$ and

$$
T(x) = T_s + \frac{\dot q_g}{2k}\left(L^2 - x^2\right)
$$

The maximum is at the centre: $T_{max} = T_s + \dfrac{\dot q_g L^2}{2k}$.

#### (a) Surface temperature

$$
T_s = 80 + \frac{(5\times10^6)(0.02)}{500} = 80 + 200 = 280\ ^\circ\text{C}
$$

#### (b) Maximum temperature

$$
T_{max} = 280 + \frac{(5\times10^6)(0.02)^2}{2(20)} = 280 + 50 = 330\ ^\circ\text{C}
$$

It occurs at the mid-plane $x = 0$.

#### (c) Heat removed per face
By energy balance, heat generated in the half-plate equals heat leaving one face:

$$
q'' = \dot q_g L = (5\times10^6)(0.02) = 1\times10^5\ \text{W/m}^2 = 100\ \text{kW/m}^2
$$

Check: $h(T_s - T_\infty) = 500\times200 = 1\times10^5$ W/m$^2$.

**Answer:** $T_s = 280\ ^\circ$C; $T_{max} = 330\ ^\circ$C at the centre; $q'' = 100\ \text{kW/m}^2$ per face.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which equation governs steady 1-D conduction with generation?
   A. dT/dx + qg/k = 0
   B. d^2T/dx^2 + qg/k = 0
   C. d^2T/dx^2 = 0
   D. d^2T/dx^2 + qg/k = (1/alpha) dT/dt
   Answer: B. Steady state removes the time term but generation stays. (Key point: Governing equation)

2. What boundary condition fixes C1 = 0?
   A. T = Tinf at x = 0
   B. Q = 0 at the surface
   C. dT/dx = 0 at the mid-plane by symmetry
   D. T = Ts at x = 0
   Answer: C. Symmetric cooling means zero gradient at the centre. (Key point: Symmetry)

3. Surface temperature is found from:
   A. Ts = Tinf + qg L^2/(2k)
   B. Ts = Tinf - qg L/h
   C. Ts = Tinf + qg L/k
   D. Ts = Tinf + qg L/h
   Answer: D. Energy balance: qg L = h(Ts - Tinf); gives 280 C. (Key point: Surface temperature)

4. Where is the maximum temperature and how big?
   A. Centre plane, 330 C
   B. Surface, 280 C
   C. Centre plane, 280 C
   D. Surface, 330 C
   Answer: A. Tmax = Ts + qg L^2/(2k) = 280 + 50 at x = 0. (Key point: Maximum temperature)

## Related in Heat Transfer

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