--- title: "Temperature rise-time curve under heating" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:heating-curve-time-constant" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/heating-curve-time-constant site: BCT Engineering --- # Temperature rise-time curve under heating Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2074 Chaitra, 8 marks - Video: [5. Temperature rise - time curve](https://www.youtube.com/watch?v=lIH7tLwNEmE) — Nisha Prakash - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/heating-curve-time-constant - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question Derive the temperature rise-time curve of the machine under heating condition and also define heating time constant. ## Short answer Under constant losses and Newtonian cooling, temperature rise follows θ = θ_m (1 - e^{-t/τ}), with θ_m = P/(Sλ) and τ = Gh/(Sλ). The heating time constant τ is the time to reach 63.2% of final rise, and is the ratio of thermal capacity to dissipation. ## Answer A machine heats up because its losses produce heat faster than it can be dissipated at first. The temperature rises **exponentially** towards a final steady value, at which all the heat produced is dissipated. #### Assumptions - The machine is a homogeneous body with uniform temperature. - Heat produced $P$ (W) is constant; heat dissipation is proportional to the temperature rise (Newton's law of cooling). #### Notation $P$ = losses (W), $G$ = mass (kg), $h$ = specific heat (J/kg°C), $S$ = cooling surface (m²), $\lambda$ = specific heat dissipation (W/m²°C), $\theta$ = temperature rise above ambient at time $t$. #### Derivation In a small time $dt$ the temperature rises by $d\theta$. Energy balance: $$ \underbrace{P\,dt}_{\text{produced}} = \underbrace{G h\,d\theta}_{\text{stored}} + \underbrace{S\lambda\theta\,dt}_{\text{dissipated}} $$ Rearranging: $$ \begin{aligned} \frac{dt}{Gh} &= \frac{d\theta}{P - S\lambda\theta} \\ \frac{S\lambda}{Gh}\,dt &= \frac{d\theta}{\dfrac{P}{S\lambda} - \theta} \end{aligned} $$ Let $\theta_m = \dfrac{P}{S\lambda}$ and $\tau = \dfrac{Gh}{S\lambda}$. Integrating, with $\theta = \theta_i$ at $t = 0$: $$ \begin{aligned} \frac{t}{\tau} &= \ln\frac{\theta_m - \theta_i}{\theta_m - \theta} \\ \theta &= \theta_m\left(1 - e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau} \end{aligned} $$ Starting cold ($\theta_i = 0$): $$ \theta = \theta_m\left(1 - e^{-t/\tau}\right) $$ As $t \to \infty$, $\theta \to \theta_m = P/(S\lambda)$, the **final steady temperature rise**: all heat produced is then dissipated. ```text rise ^ theta_m --------------------- | _..--~~~~~~ | _.-' | .-' <- 0.632 theta_m at t = tau | / | / initial slope = theta_m / tau | / +---|------|------|------|------> t tau 2tau 3tau 4tau ``` #### Heating time constant $$ \tau_h = \frac{G h}{S\lambda} $$ It is defined as: 1. the time to reach the final rise $\theta_m$ if the initial rate of rise continued unchanged (no heat dissipated); or 2. the time to reach **63.2 %** of the final rise, since $1 - e^{-1} = 0.632$. In practice the machine is considered thermally steady after about $4\tau$–$5\tau$ (98–99 %). The time constant is large for large, heavy machines with poor cooling, and is reduced by better ventilation (larger $\lambda$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In deriving the heating curve, which energy balance is used for a small time dt? A. Heat produced = heat stored in the body + heat dissipated B. Heat produced = heat dissipated only C. Heat stored = heat dissipated minus heat produced D. Heat produced = heat stored only Answer: A. Losses P dt go partly into raising the temperature (G h dθ) and partly leave the surface (Sλθ dt). (Key point: Energy balance) 2. What is the final steady temperature rise θm in terms of the losses and cooling? A. Gh/(Sλ) B. P/(Sλ) C. PSλ D. P/(Gh) Answer: B. At steady state nothing is stored, so P = Sλθm, giving θm = P/(Sλ). (Key point: Final steady rise) 3. For a machine starting cold, the temperature rise at time t is: A. θ = θm e^(-t/τ) B. θ = θm t/τ C. θ = θm(1 - e^(-t/τ)) D. θ = θm(1 + e^(-t/τ)) Answer: C. Integrating the balance from θ = 0 gives an exponential rise towards θm. (Key point: Heating curve) 4. The heating time constant τ = Gh/(Sλ) is the time to reach what fraction of the final rise? A. 50 % B. 95 % C. 100 % D. 63.2 % Answer: D. At t = τ, 1 - e^-1 = 0.632 of θm; equally it is the time to reach θm if the initial rate continued. 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