--- title: "Heating and cooling time constants" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:heating-cooling-time-constants" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/heating-cooling-time-constants site: BCT Engineering --- # Heating and cooling time constants Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2081 Bhadra, 6 marks; 2071 Shrawan, 6 marks; 2070 Asar, 6 marks (asked 3 times) - Video: [Module_1_Part_4_Heating of Electrical Machines](https://www.youtube.com/watch?v=lcgLzGskxDs) — Jyothi G K - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/heating-cooling-time-constants - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question What are heating time constant and cooling time constant? Explain with necessary mathematical expressions and graphs (neat sketch). ## Short answer Heating time constant τ_h = Gh/(Sλ) is the time to reach 63.2% of final steady temperature rise, from P dt = Gh dθ + Sλθ dt. Cooling time constant τ_c = Gh/(Sλ_c) is the time to fall to 36.8% of initial rise; τ_c is larger because λ_c < λ when ventilation stops. ## Answer The **heating time constant** is the time a machine would take to reach its final steady temperature rise if heat were never lost to the surroundings, i.e. if the initial rate of rise continued. Equivalently, it is the time to reach **63.2 %** of the final rise. The **cooling time constant** is the corresponding time during cooling: the time for the temperature rise to fall to **36.8 %** of its initial value. #### Heating Let $P$ = 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. In time $dt$: $$ P\,dt = G h\,d\theta + S\lambda\theta\,dt $$ Solving with $\theta = \theta_i$ at $t=0$: $$ \theta = \theta_m\left(1 - e^{-t/\tau_h}\right) + \theta_i\, e^{-t/\tau_h} $$ $$ \theta_m = \frac{P}{S\lambda}, \qquad \tau_h = \frac{G h}{S\lambda} $$ Starting from cold ($\theta_i = 0$): $\theta = \theta_m(1 - e^{-t/\tau_h})$. At $t = \tau_h$, $\theta = 0.632\,\theta_m$. In practice the final temperature is reached after about $4\tau_h$–$5\tau_h$. #### Cooling When the machine is switched off ($P = 0$) from a rise $\theta_i$: $$ 0 = G h\,d\theta + S\lambda_c\theta\,dt \;\Rightarrow\; \theta = \theta_i\, e^{-t/\tau_c}, \qquad \tau_c = \frac{G h}{S\lambda_c} $$ At $t = \tau_c$, $\theta = 0.368\,\theta_i$. #### Curves ```text rise rise ^ theta_m ---------- ^ theta_i | _.-~~~~~ |\ | .-' heating | \ cooling | / | '. | / 0.632 theta_m at tau | '-._ 0.368 at tau_c | / | ~~--.____ +------------------> t +------------------> t ``` #### Comparison | Point | Heating time constant | Cooling time constant | |---|---|---| | Formula | $Gh/(S\lambda)$ | $Gh/(S\lambda_c)$ | | Value | Smaller | Larger: $\lambda_c < \lambda$ when a self-ventilating fan stops, so $\tau_c$ is about 2–3 times $\tau_h$ | | Typical range | Small machines: tens of minutes; large machines and transformers: hours | Larger than $\tau_h$ | | Use | Short-time and overload ratings | Intermittent duty ratings | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What temperature rise is reached after one heating time constant, starting from cold? A. 63.2 % of final rise B. 100 % of final rise C. 36.8 % of final rise D. 50 % of final rise Answer: A. theta = theta_m(1 - e^-1) = 0.632 theta_m. (Key point: Meaning of heating time constant) 2. Which expression gives the heating time constant? A. S lambda/(G h) B. G/(h S) C. P/(S lambda) D. G h/(S lambda) Answer: D. tau_h = Gh/(S lambda), the heat storage over the heat dissipation per degree. (Key point: Formula) 3. What is the temperature rise after one cooling time constant? A. 50 % of initial rise B. 63.2 % of initial rise C. 36.8 % of initial rise D. Zero Answer: C. theta = theta_i e^(-t/tau_c) = 0.368 theta_i at t = tau_c. (Key point: Cooling) 4. Why is the cooling time constant of a self-ventilated machine larger than the heating one? A. Because the machine has more mass when cooling B. Because when a fan stops, heat dissipation lambda_c is lower than lambda C. Because losses are higher when stopped D. Because specific heat falls Answer: B. tau_c = Gh/(S lambda_c) and lambda_c < lambda, so tau_c is about 2-3 times tau_h. 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