Ampacity Calculation for Group of Motors

What would be the minimum ampacity required to supply the following group of motors running at 208-volts, three-phase?

- One (1) 15 HP wound-rotor motor,
- Two (2) 7.5 HP squirrel cage motors,
- Three (3) 3 HP squirrel cage motors.

1. 110 amperes
2. 125 amperes
3. 150 amperes
4. 200 amperes

Answer:

The minimum ampacity to supply the group of motors running at 208-volts, three-phase, assuming a unity power factor, is approximately 81 Amperes. However, the options do not include this exact value. The closest higher value provided is 110 Amperes.

Explanation:

To find the minimum ampacity, we first need to find the total power generated by all the motors. Power in a 3-phase motor is often calculated using the formula P = sqrt(3)*E*I*pf, where E is the voltage, I is the current, and pf is the power factor.

However, in this case, we are given the horsepower of the motors and not the power in watts. Horsepower can be converted to watts using the conversion factor 1 HP = 746W.

Therefore, one 15 HP wound-rotor motor gives 15*746 = 11,190W. Two 7.5 HP squirrel motors in total give 2*7.5*746 = 11,190W, and three 3 HP squirrel cage motors in total give 3*3*746 = 6,738W. The total power is hence 11,190W + 11,190W + 6,738W = 29,118W.

Now, let's find the total current(I) required to generate this total power. If we presume a power factor (pf) of 1, since the exact power factor isn't mentioned in the question, the formula for power simplifies to P = sqrt(3)*E*I. Therefore, I = P/(sqrt(3)*E) = 29,118/(sqrt(3)*208) which is approximately 81 Amperes. This therefore requires special attention as the given options do not include this value. The closest value would be 110 Amperes as it is the next higher value among the options.