--- title: "One hour rating of a 20 hp induction motor" subject: "Electric Machine Design" chapter: 2 key: "semester_5/electric_machine_design:induction-motor-one-hour-rating" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/induction-motor-one-hour-rating site: BCT Engineering --- # One hour rating of a 20 hp induction motor Electric Machine Design · Chapter 2: Heating and cooling of electric machine - Asked in: 2082 Baishakh, 8 marks; 2076 Chaitra, 8 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/induction-motor-one-hour-rating - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-2 ## Question A 20 h.p., 400V, 3 phase, 50 Hz, induction motor has a final steady state temperature rise of 40°C when running at its rated output. Calculate its one hour rating for the same temperature rise, if the heating time constant is 180 min. The ratio of conductor losses to constant losses can be assumed as 1.25 and the total losses of the machine at full load is 1800 W. ## Short answer The motor's constant loss is 800 W, copper loss 1000 W. For a 40°C rise after 1 h, the equivalent steady rise is 141.1°C. Losses rise to 6349.9 W, giving a load factor of sqrt((6349.9 - 800)/1000) = 2.356, so one-hour rating ≈ 47.1 hp. ## Answer The **one-hour rating** is the output the motor can deliver for one hour, starting from cold, without its temperature rise exceeding the permitted value (here 40 °C, the same as the continuous rating). #### Given - Rated output = 20 hp, final steady rise at rated output $\theta_m = 40$ °C - Heating time constant $\tau = 180$ min, time $t = 60$ min - Total full-load losses = 1800 W, with $\dfrac{\text{copper loss}}{\text{constant loss}} = 1.25$ #### Step 1: split the losses $$ \begin{aligned} P_i + 1.25P_i &= 1800 \\ P_i\ (\text{constant}) &= \frac{1800}{2.25} = 800\ \text{W} \\ P_{cu}\ (\text{full-load copper}) &= 1800 - 800 = 1000\ \text{W} \end{aligned} $$ #### Step 2: final temperature rise allowed for the one-hour load Let $\theta_m'$ be the final steady rise the motor would reach at the one-hour load. For the rise to be just 40 °C after 60 min from cold: $$ \begin{aligned} 40 &= \theta_m'\left(1 - e^{-60/180}\right) \\ \theta_m' &= \frac{40}{1 - e^{-1/3}} = \frac{40}{1 - 0.7165} = \frac{40}{0.2835} = 141.11\ ^\circ\text{C} \end{aligned} $$ #### Step 3: losses at the one-hour load The final temperature rise is proportional to the losses ($\theta_m = P/S\lambda$, with the same cooling): $$ \begin{aligned} P' &= 1800 \times \frac{141.11}{40} = 1800 \times 3.5277 = 6349.9\ \text{W} \end{aligned} $$ #### Step 4: load corresponding to these losses Let the one-hour output be $x$ times the rated output. Constant loss stays at 800 W, and copper loss varies as $x^2$: $$ \begin{aligned} 800 + 1000\,x^2 &= 6349.9 \\ x^2 &= 5.5499 \\ x &= 2.356 \end{aligned} $$ $$ \text{One-hour rating} = 2.356 \times 20 = 47.12\ \text{hp}\ \ (\approx 35.1\ \text{kW}) $$ **Answer: one-hour rating ≈ 47.1 hp (about 2.36 times the continuous rating), for a 40 °C temperature rise.** (The motor would also need enough pull-out torque to carry this load. Thermally, this is the limit.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With copper to constant loss ratio of 1.25 and total loss 1800 W, what are the constant and copper losses? A. 720 W and 1080 W B. 800 W and 1000 W C. 1000 W and 800 W D. 900 W and 900 W Answer: B. Pi(1 + 1.25) = 1800 gives Pi = 800 W, and copper loss 1000 W. (Key point: Loss split) 2. The rise must be 40 °C after 60 min with tau = 180 min. What final rise theta_m' would the one-hour load give? A. About 141 °C B. About 56 °C C. About 400 °C D. 40 °C Answer: A. theta_m' = 40/(1 - e^(-1/3)) = 40/0.2835 = 141.1 °C. (Key point: Final rise at 1-hour load) 3. How are the losses at the one-hour load found? A. P' = 1800 x 40/141.11 B. P' equals 1800 W C. P' = 1800 x 60/180 D. Final rise is proportional to losses, so P' = 1800 x 141.11/40 = 6350 W Answer: D. theta_m = P/(S lambda) with the same cooling. (Key point: Loss proportionality) 4. Which relation gives the one-hour output multiple x? A. 1800 x^2 = 6350 B. 800 x^2 + 1000 = 6350 C. 800 + 1000 x^2 = 6350, so x = 2.36 D. 800 x + 1000 = 6350 Answer: C. Constant loss is fixed and copper loss varies as the square of load, so one-hour rating is about 47.1 hp. 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