Question

A motor has to exert power starting from zero and rising uniformly to 150 HP in 8 mins after which it works at constant power of 100 HP for 5 mins. The motor remains on no load for next 5 mins. The cycle again starts and repeats indefinitely. Determine suitable size of motor.

Answer 1

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### What We Need to Do:
We need to find the **equivalent power** of the motor. This is the steady power that the motor would need to operate continuously, while doing the same total work as in the cycle described.

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### Steps to Solve:

1. **Break the cycle into parts:**
   - **Ramp-up phase (0 to 150 HP in 8 minutes):**
     Power increases from 0 to 150 HP. The average power during this phase is:
     \[
     \text{Average power} = \frac{0 + 150}{2} = 75 \, \text{HP}.
     \]
     The work done during this phase is:
     \[
     \text{Work} = \text{Average power} \times \text{Time}.
     \]
     Time = 8 minutes = \( \frac{8}{60} = 0.133 \, \text{hours} \), so:
     \[
     \text{Work} = 75 \times 0.133 = 10 \, \text{HP-hours}.
     \]

   - **Constant power phase (100 HP for 5 minutes):**
     Power is constant at 100 HP. The work done is:
     \[
     \text{Work} = \text{Power} \times \text{Time}.
     \]
     Time = 5 minutes = \( \frac{5}{60} = 0.083 \, \text{hours} \), so:
     \[
     \text{Work} = 100 \times 0.083 = 8.33 \, \text{HP-hours}.
     \]

   - **No-load phase (0 HP for 5 minutes):**
     No power is used, so:
     \[
     \text{Work} = 0 \, \text{HP-hours}.
     \]

2. **Total work in one cycle:**
   Add up the work from all phases:
   \[
   \text{Total work} = 10 + 8.33 + 0 = 18.33 \, \text{HP-hours}.
   \]

3. **Find equivalent power:**
   The cycle takes 18 minutes total, which is \( 18 / 60 = 0.3 \, \text{hours} \). The equivalent power is:
   \[
   \text{Equivalent power} = \frac{\text{Total work}}{\text{Total time}}.
   \]
   \[
   \text{Equivalent power} = \frac{18.33}{0.3} = 61.1 \, \text{HP}.
   \]

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### Final Answer:
The motor should have a size of **61.1 HP**.