One hour rating of a 20 hp induction motor

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.)