What is the minimum ampacity permitted for the feeder supplying a group of single-phase squirrel-cage induction motors totaling 32.5 HP?

To determine the minimum ampacity required for the feeder supplying a group of single-phase squirrel-cage induction motors totaling 32.5 horsepower (HP), it is essential to convert the horsepower into current using the proper formulas.

First, you can convert horsepower to kVA using the formula:

1 HP = 0.746 kW. Thus, 32.5 HP converts to kilowatts as follows:

32.5 HP × 0.746 kW/HP = 24.295 kW.

Next, for single-phase motors, the formula to calculate current (amps) is:

Current (I) = Power (kW) / (Voltage (V) × Power Factor (PF)).

Assuming a standard voltage of 240 volts (which is common in many single-phase applications in residential and commercial environments) and a typical power factor of 0.9:

I = 24.295 kW / (240 V × 0.9)

I = 24.295 kW / 216 V

I ≈ 112.5 A.

In practical applications, when sizing conductors and choosing a breaker, the calculated current is typically increased to allow for continuous loads and to comply with NEC guidelines