What is the full-load current for a 30hp, 460V, 3-phase induction motor?

To determine the full-load current for a 30 horsepower (hp), 460V, 3-phase induction motor, we can use the formula that relates horsepower, voltage, and full-load current in a three-phase electrical system:

[ \text{I} = \frac{P \times 746}{\sqrt{3} \times V \times \text{Efficiency} \times \text{Power Factor}} ]

In this formula:

• ( P ) is the power in horsepower,

• 746 is the number of watts in a horsepower,

• ( V ) is the line voltage,

• Efficiency and Power Factor can vary but are often approximated in practice.

For a typical induction motor, we might assume an efficiency of about 90% (0.9) and a power factor of about 0.8 for the calculations.

Plugging in the numbers for our specific motor:

• ( P ) = 30 hp (or 30 x 746 watts = 22,380 watts),

• ( V ) = 460V,

• Assuming efficiency = 0.9 and power factor = 0.8.

Using the formula to calculate the full-load current:

[ I = \frac