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phase, \( 400 \mathrm{~V}, 37 \mathrm{~kW} \) motor runs at a power factor of 0.87 . Determine: the apparent power drawn by the motor (Ans: 42.529 kVA ) 37 the reactive power generated by the motor. (Ans: 20.969 kVAr ) the current drawn by the motor. (Ans: 61.3849 A )

Ask by Crawford Stanley. in Singapore
Dec 10,2024

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1. The apparent power drawn by the motor is 42.529 kVA. 2. The reactive power generated by the motor is 20.969 kVAr. 3. The current drawn by the motor is 61.3849 A.

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Bonus Knowledge

To determine the apparent power (S) of the motor, we use the formula: \[ S = \frac{P}{\text{Power Factor}} \] So, with \( P = 37 \, \text{kW} \) and a power factor of 0.87, \[ S = \frac{37}{0.87} \approx 42.529 \, \text{kVA} \]. Next, to find the reactive power (Q), we can use the formula: \[ Q = \sqrt{S^2 - P^2} \] Calculating: \[ S^2 = (42.529)^2 \text{ and } P^2 = (37)^2 \] leads to \[ Q = \sqrt{(42.529)^2 - (37)^2} \approx 20.969 \, \text{kVAr} \]. Finally, the current (I) drawn by the motor can be determined using the formula: \[ I = \frac{S}{\sqrt{3} \times V} \] Where \( V = 400 \, V \), thus: \[ I = \frac{42.529 \, \text{kVA}}{\sqrt{3} \times 0.4} \approx 61.3849 \, A \]. Now you've got all the key figures for that motor! Keep the calculations rolling!

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