A 3-phase AC induction motor (25 kW capacity) is used for pumping operation. Electrical parameters such as current, volt, and power factor were measured with a power analyzer. Find the energy consumption of motor in one hour? (Volts. = 415 V, current = 20 amps and PF = 0.85).

Solution: To find the energy consumption of the 3-phase AC induction motor in one hour, we'll first calculate the real power consumed by the motor, then determine the energy consumption per hour.

Given:

• Power of the motor ($𝑃$) = 25 kW = 25,000 W
• Supply voltage ($𝑉$) = 415 V
• Current ($𝐼$) = 20 A
• Power factor ($𝑃𝐹$) = 0.85
• Operating time per hour = 1 hour

First, let's calculate the real power consumed by the motor using the formula:

$\text{Real power (W)}=𝑉\times 𝐼\times 𝑃𝐹$

$\text{Real power (W)}=415\text{V}\times 20\text{A}\times 0.85$

$\text{Real power (W)}=7030\text{W}$

Now, to find the energy consumption per hour, we multiply the real power consumed by the operating time per hour:

$\text{Energy consumption per hour (Wh)}=\text{Real power (W)}\times \text{Operating time per hour (hours)}$

$\text{Energy consumption per hour (Wh)}=7030\text{W}\times 1\text{hour}$

$\text{Energy consumption per hour (Wh)}=7030\text{Wh}$

So, the energy consumption of the 3-phase AC induction motor in one hour is 7030 watt-hours (Wh).