Answer:
I = Line Current = 242.58 A
Q = Reactive Power = 41.5 kVAr
Explanation:
Firstly, converting 100 hp to kW.
Since, 1 hp = 0.746 kW,
100 hp = 0.746 kW x 100
100 hp = 74.6 kW
The power of a three phase induction motor can be given as:
[tex]P_{in} = \sqrt{3} VI Cos\alpha\\[/tex]
where,
P in = Input Power required by the motor
V = Line Voltage
I = Line Current
Cosα = Power Factor
Now, calculating Pin:
[tex]efficiency = \frac{{P_{out}} }{P_{in} }\\0.97 = \frac{74.6}{P_{in} } \\P_{in} = \frac{74.6}{0.97}\\ P_{in} = 76.9 kW[/tex]
a) Calculating the line current:
[tex]P_{in} = \sqrt{3}VICos\alpha \\76.9 * 1000= \sqrt{3}*208*I*0.88\\I = \frac{76.9*1000}{\sqrt{3}*208*0.88 }\\I = 242.58 A[/tex]
b) Calculating Reactive Power:
The reactive power can be calculated as:
Q = P tanα
where,
Q = Reactive power
P = Active Power
α = power factor angle
Since,
[tex]Cos\alpha =0.88\\\alpha =Cos^{-1}(0.88)\\\alpha=28.36[/tex]
Therefore,
[tex]Q = 76.9 * tan (28.36)\\Q = 76.9 * (0.5397)\\Q = 41. 5 kVAr[/tex]
