What is the total power consumption in kW of two dehumidification equipment units drawing 105 amps and 75 amps at 480/3/60?

What is the total power consumption in kW of two dehumidification equipment units drawing 105 amps and 75 amps at 480/3/60?

• A. 105 kW
• B. 149 kW
• C. 180 kW
• D. 135 kW

Answer: (B)

To determine the total power consumption in kilowatts (kW) for the two dehumidification equipment units, it is necessary to use the formula for three-phase power, which is: \[ \text{Power (kW)} = \frac{\sqrt{3} \times V \times I \times \text{Power Factor}}{1000} \] Here, \( V \) is the line voltage, \( I \) is the current in amps, and the power factor is typically assumed to be 1.0 for simplicity unless stated otherwise. In this question, both units operate at a voltage of 480 volts (V) and have currents of 105 amps and 75 amps, respectively. First, calculate the power for each unit: 1. For the unit drawing 105 amps: \[ \text{Power}_1 = \frac{\sqrt{3} \times 480 \, \text{V} \times 105 \, \text{A} \times 1.0}{1000} \] \[ \text{Power}_1 \approx \frac{1.732 \times 480 \times 105}{1000} \approx 87.29 \, \

To determine the total power consumption in kilowatts (kW) for the two dehumidification equipment units, it is necessary to use the formula for three-phase power, which is:

[ \text{Power (kW)} = \frac{\sqrt{3} \times V \times I \times \text{Power Factor}}{1000} ]

Here, ( V ) is the line voltage, ( I ) is the current in amps, and the power factor is typically assumed to be 1.0 for simplicity unless stated otherwise. In this question, both units operate at a voltage of 480 volts (V) and have currents of 105 amps and 75 amps, respectively.

First, calculate the power for each unit:

1. For the unit drawing 105 amps:

[ \text{Power}_1 = \frac{\sqrt{3} \times 480 , \text{V} \times 105 , \text{A} \times 1.0}{1000} ]

[ \text{Power}_1 \approx \frac{1.732 \times 480 \times 105}{1000} \approx 87.29 , \