--- title: "Hot spot temperature of a plate" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:hot-spot-plate" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/hot-spot-plate site: BCT Engineering --- # Hot spot temperature of a plate Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2076 Asoj, 8 marks; 2082 Chaitra (new course), 5 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/hot-spot-plate - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question What is hot spot? Taking the example of a plate of thickness t, derive the expression to find the hottest spot temperature in terms of surface temperature and thickness of plate. ## Short answer For a plate of thickness t with uniform heat generation p and both faces cooled equally, the hot-spot temperature at the centre is θ_h = θ_s + pρt²/8, where ρ is thermal resistivity. The temperature distribution is parabolic, and the rise depends on t². ## Answer A **hot spot** is the point of highest temperature inside a body in which heat is produced, such as a core, a coil or a conductor. Heat produced inside must flow outwards to the cooling surfaces, so the interior is hotter than the surface. Insulation near the hot spot ages fastest, so the hot-spot temperature, not the average, must be kept within the limit of the insulation class. #### Derivation for a plate of thickness $t$ Consider a plate whose length and width are large compared with its thickness $t$, cooled equally on both faces. Then heat flows only across the thickness. Assume: - $p$ = heat (loss) produced per unit volume, W/m³, uniform - $\rho$ = thermal resistivity of the material, °C·m/W ($\rho = 1/k$) - heat flows only in the $x$ direction (the other dimensions are large), and both faces are at the same surface temperature $\theta_s$. By symmetry the hottest point is at the centre plane, $x = 0$, where no heat crosses. Consider a plane at distance $x$ from the centre. All heat produced between the centre and this plane must cross it. Per unit area: $$ Q_x = p\,x \quad \text{W/m}^2 $$ By Fourier's law, the temperature gradient is $$ \frac{d\theta}{dx} = -\rho\, Q_x = -\rho\, p\, x $$ Integrating from the centre (temperature $\theta_h$): $$ \begin{aligned} \theta(x) &= \theta_h - \frac{\rho\, p\, x^2}{2} \end{aligned} $$ So the temperature is **parabolic** across the thickness. At the surface, $x = t/2$, $\theta = \theta_s$: $$ \theta_h - \theta_s = \frac{\rho\, p}{2}\left(\frac{t}{2}\right)^2 = \frac{p\,\rho\, t^2}{8} $$ ```text temperature ^ theta_h (hot spot) | _..--.._ | .-' '-. | .' '. | / \ | theta_s theta_s +--|---------+---------|--> x -t/2 0 +t/2 <------ thickness t ------> ``` #### Result $$ \theta_h = \theta_s + \frac{p\,\rho\,t^2}{8} $$ where $\theta_h$ is the hottest-spot temperature (at the mid-plane) and $\theta_s$ the surface temperature. #### Points to note - The temperature profile across the plate is a **parabola** with its peak at the centre. - The hot-spot rise above the surface is proportional to **$t^2$**. Halving the thickness, for example by a cooling duct in the middle, reduces it to one-quarter. - It is also proportional to the loss density $p$ and to the thermal resistivity $\rho$. Materials with poor thermal conductivity (insulation, or laminated iron across the sheets) produce large internal gradients. - If only one face is cooled, the hot spot is at the insulated face, and $\theta_h - \theta_s = p\rho t^2/2$ (the same formula with thickness $2t$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is a hot spot? A. The point of highest temperature inside a body where heat is produced B. The average temperature of a coil C. The ambient temperature D. The point of lowest temperature on the surface Answer: A. Heat from inside must flow out, so the interior is hotter; the hottest point is the hot spot. (Key point: Definition) 2. What is the hot-spot temperature of a plate of thickness t cooled on both faces? A. theta_s + p rho t / 8 B. theta_s + p rho t^2 / 2 C. theta_s - p rho t^2 / 8 D. theta_s + p rho t^2 / 8 Answer: D. theta_h - theta_s = p rho t^2/8 from integrating the parabolic profile to x = t/2. (Key point: Result) 3. If a cooling duct splits the plate thickness in half, how does the hot-spot rise change? A. It is unchanged B. It halves C. It falls to one-quarter D. It doubles Answer: C. The rise varies as t^2. (Key point: Effect of thickness) 4. What is the shape of the temperature profile across the plate? A. Linear B. Parabolic with its peak at the centre C. Exponential D. Constant Answer: B. Heat crossing a plane grows with x, giving d theta/dx proportional to x and a parabola. (Key point: Profile) ## Related in Electric Machine Design - [Internal (hot spot) temperature of core](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ahot-spot-core-temperature) — Electric Machine Design, chapter 2, asked 6× ## Related in other subjects - [Hot and cold working; temperature and strain rate](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Ahot-cold-working-friction-strain-rate) — Manufacturing and Production Processes, chapter 5 - [Plate with convection and radiation: inside temperature](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aplate-convection-radiation-inside-temp) — Fundamental of Thermodynamics & Heat Transfer, chapter 7 - [Thickness of doubly refracting plate](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adoubly-refracting-plate-thickness) — Engineering Physics, chapter 4 - [Quarter wave plate thickness, quartz](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aquarter-wave-plate-quartz) — Engineering Physics, chapter 4