Let’s solve the problem step by step: ## **Given Duty Cycle** | Interval | Load (kW) | Time (min) | |----------|-------------------------|------------| | i | 0 → 400 (rising) | 5 | | ii | 500 (constant) | 5 | | iii | -400 (regenerative) | 4 | | iv | 0 (idle) | 2 | Assume losses are proportional to (Power)². We’ll estimate the **equivalent (rms) power** for power rating (since losses ∝ P²). --- ## **Step 1: Break Down Each Interval** ### i) Load rises from 0 to 400 kW in 5 min - Power profile: Linear increase from 0 to 400 kW. ### ii) Uniform load of 500 kW for 5 min - Power profile: Constant at 500 kW. ### iii) Regenerative 400 kW for 4 min - Power profile: -400 kW (negative sign for regeneration). ### iv) Remains idle for 2 min - Power profile: 0 kW. --- ## **Step 2: Find Equivalent (rms) Power** The **rms (root mean square) power** is found by: \[ P_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (P(t))^2 dt} \] Where \( T = 16 \) min (total cycle). Let’s compute \( \sum (P^2 \times \Delta t) \) for each interval. --- ### **Interval i: 0 to 400 kW, 5 min** Power increases linearly: \( P(t) = \frac{400}{5}t = 80t \), with \( t \) in minutes, \( t = 0 \) to \( 5 \). \[ \int_{0}^{5} [P(t)]^2 dt = \int_0^5 (80t)^2 dt = 6400 \int_0^5 t^2 dt = 6400 \left[ \frac{t^3}{3} \right]_0^5 = 6400 \times \frac{125}{3} = 6400 \times 41.6667 = 266,666.7 \] --- ### **Interval ii: 500 kW, 5 min** \[ P^2 = 500^2 = 250,000 \] \[ \text{Contribution} = 250,000 \times 5 = 1,250,000 \] --- ### **Interval iii: -400 kW, 4 min** \[ P^2 = (-400)^2 = 160,000 \] \[ \text{Contribution} = 160,000 \times 4 = 640,000 \] --- ### **Interval iv: 0 kW, 2 min** \[ P^2 = 0 \] \[ \text{Contribution} = 0 \] --- ## **Step 3: Total and Average** Total \(P^2 \times t\) sum: \[ = 266,666.7 + 1,250,000 + 640,000 + 0 = 2,156,666.7 \] Total time \(T = 5 + 5 + 4 + 2 = 16\) min --- ## **Step 4: Calculate rms Power** \[ P_{\text{rms}} = \sqrt{\frac{2,156,666.7}{16}} \] \[ = \sqrt{134,791.67} \] \[ \approx 367\, \text{kW} \] --- ## **Step 5: Final Answer** **The power rating of the motor should be about \( \boxed{367 \text{ kW}} \).** --- ### **Summary Table** | Interval | Power (kW) | Time (min) | \(P^2 \times t\) | |----------|---------------|------------|--------------------| | i | 0→400 (linear)| 5 | 266,667 | | ii | 500 | 5 | 1,250,000 | | iii | -400 | 4 | 640,000 | | iv | 0 | 2 | 0 | | **Total**| | **16** | **2,156,667** | --- **Power rating of the motor ≈ 367 kW** (rounded to nearest kW).