---
title: "Internal temperature of a homogeneous material"
subject: "Electric Machine Design"
chapter: 2
key: "semester_5/electric_machine_design:internal-temperature-homogeneous-material"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/internal-temperature-homogeneous-material
site: BCT Engineering
---

# Internal temperature of a homogeneous material

Electric Machine Design · Chapter 2: Heating and cooling of electric machine

- Asked in: 2074 Asoj, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/internal-temperature-homogeneous-material
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2

## Question

Derive the expression for calculation of internal temperature of a homogenous material of thickness 't' length 'l' and width 'w'. Other necessary data's can be assumed.

## Short answer

Uniform heat p flows across thickness t to two faces. Surface rise is p t/(2 lambda). Internal conduction adds p rho_t t^2/8. The hot-spot temperature is theta_a + p t/(2 lambda) + p rho_t t^2/8.

## Answer

When losses are produced inside a body, heat must flow through the material to reach its cooling surfaces. The interior is therefore hotter than the surface. The highest (internal or hot-spot) temperature is found as follows.

#### Assumptions
- A block of homogeneous material of thickness $t$, length $l$ and width $w$, with $l$ and $w$ much larger than $t$. Heat therefore flows only across the thickness ($x$ direction) and leaves equally from the two large faces.
- Uniform loss $p$ W/m³; thermal resistivity $\rho$ °C·m/W (constant).
- Surface heat dissipation coefficient $\lambda$ W/m²°C; ambient temperature $\theta_a$.

```text
          face 1 (area l x w)
   +--------------------------------+   ^
   |   heat <---  centre  ---> heat |   | t
   +--------------------------------+   v
          face 2 (area l x w)
```

#### Step 1: surface temperature
Total heat produced $= p\,l\,w\,t$. It leaves from two faces of total area $2lw$:

$$
\begin{aligned}
p\,l\,w\,t &= 2\,l\,w\,\lambda\,(\theta_s - \theta_a) \\
\theta_s &= \theta_a + \frac{p\,t}{2\lambda}
\end{aligned}
$$

#### Step 2: temperature rise inside
Take $x$ from the mid-plane. Heat produced between the mid-plane and a plane at $x$ crosses that plane. Its flow per unit area is $p\,x$. By Fourier's law:

$$
\begin{aligned}
\frac{d\theta}{dx} &= -\rho\,p\,x \\
\theta(x) &= \theta_h - \frac{\rho\,p\,x^2}{2}
\end{aligned}
$$

At $x = t/2$, $\theta = \theta_s$:

$$
\theta_h - \theta_s = \frac{p\,\rho\,t^2}{8}
$$

#### Step 3: internal (hot-spot) temperature
$$
\theta_h = \theta_a + \frac{p\,t}{2\lambda} + \frac{p\,\rho\,t^2}{8}
$$

```text
 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>
```

#### Remarks
- The first term is the rise of the surface above ambient, set by the cooling. The second is the internal rise, set by conduction.
- The internal rise varies as $t^2$. Dividing the block with a cooling duct at mid-thickness reduces it to one-quarter, which is why cores are split into packets.
- If heat flows in two directions (thickness $t$ and width $w$, with resistivities $\rho_t$, $\rho_w$), the internal rise becomes
$$
\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_t t^2} + \dfrac{1}{\rho_w w^2}\right]}
$$

## Check yourself

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

1. Heat leaves from two faces of area 2lw with coefficient lambda. What is the surface temperature?
   A. theta_a + p t^2/8
   B. theta_a + p t/(2 lambda)
   C. theta_a + p t lambda
   D. theta_a + 2 p t/lambda
   Answer: B. p l w t = 2 l w lambda (theta_s - theta_a). (Key point: Surface temperature)

2. What is the internal rise of the centre over the surface?
   A. p rho t^2 / 8
   B. p rho t / 2
   C. p t^2/(8 rho)
   D. 2 p rho t
   Answer: A. Integrating d theta/dx = -rho p x from the centre to x = t/2. (Key point: Internal rise)

3. What is the hot-spot temperature?
   A. theta_a + p rho t^2/8
   B. theta_a + p t/lambda
   C. theta_a + p rho t/2
   D. theta_a + p t/(2 lambda) + p rho t^2/8
   Answer: D. It is ambient plus surface rise (cooling) plus internal rise (conduction). (Key point: Final result)

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