What is the minimum ampacity required for a feeder serving two motors at 15HP, one motor at 25HP, and one motor at 40HP?

To determine the minimum ampacity required for a feeder serving multiple motors, the National Electrical Code (NEC) provides guidelines for calculating the load. In this scenario, we need to calculate the total load from the motors at 15 HP, 15 HP, 25 HP, and 40 HP.

First, it's important to convert horsepower (HP) to kilowatts (kW) using the formula:

1 HP = 0.746 kW

Thus, we calculate:

• For two motors at 15 HP:

( 2 \times 15 \text{ HP} \times 0.746 = 22.38 \text{ kW} )

• One motor at 25 HP:

( 25 \text{ HP} \times 0.746 = 18.65 \text{ kW} )

• One motor at 40 HP:

( 40 \text{ HP} \times 0.746 = 29.84 \text{ kW} )

Next, we sum the outputs:

Total kW = 22.38 kW + 18.65 kW + 29.84 kW = 70