What is the maximum setting for a non-time delay fuse providing protection for a 3-phase, 25 hp, 460 volt motor?

To determine the maximum setting for a non-time delay fuse protecting a 3-phase, 25 hp motor running at 460 volts, it is essential to calculate the full-load current of the motor. The full-load current can be derived using the formula:

[ \text{I} = \frac{\text{P} \times 1000}{\sqrt{3} \times \text{V} \times \text{PF}} ]

where:

• I = full-load current in amperes

• P = power in horsepower (hp)

• V = voltage in volts

• PF = power factor (typically assumed to be 1 for calculations unless stated otherwise)

A 25 hp motor at 460 volts would have a full-load current approximately calculated as follows:

1. Convert horsepower to watts: 25 hp × 746 watts/hp = 18,650 watts.

2. Use the formula:

[ \text{I} = \frac{18650 \text{ watts}}{\sqrt{3} \times 460 \text{ volts} \times 0.9 \text{ (assuming a power factor of 0.9)}} ]

3. Simplifying gives

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