A 10 H.P., 4 pole, 25 Hz, 3-phase, wound rotor induction motor is taking 9100 watt from the line. Core loss is 290 watt, stator copper loss is 568 watt, rotor copper loss in 445 watt, friction and windage losses are 100 watt. Determine; (a) power transferred across air gap; (b) mechanical power

Question

A 10 H.P., 4 pole, 25 Hz, 3-phase, wound rotor induction motor is taking 9100 watt from the line. Core loss is 290 watt, stator copper loss is 568 watt, rotor copper loss in 445 watt, friction and windage losses are 100 watt. Determine; (a) power transferred across air gap; (b) mechanical power in watt developed by rotor; (c) mechanical power output in watt; (d) efficiency; (e) slip.

Accepted Answer

[latex]Power input to motor or stator = 9100 watt[/latex]

[latex]Power transferred across air gap = Stator input – Stator core loss – Stator copper loss \[0.5cm] \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \hspace{30 pt} \quad \qquad \; = 9100 – 290 – 568 = \mathbf{8242 W} [/latex]

[latex]Mechanical power developed in rotor = rotor input – Rotor copper loss = 8242 – 445 = \mathbf{7797} [/latex]

[latex]Rotor output = Mechanical power developed – Mechanical loss \[0.5cm] \hspace{30 pt} \quad \quad = 7797 – 100 = \mathbf{7697 W} [/latex]

[latex]Motor efficiency = \frac{Output}{input} \times 100 = \frac{7697}{9100} \times 100 = \mathbf{ 84.58 \% } [/latex]

Slip, [latex] S = \frac{Rotor copper loss}{Rotor input} = \frac{445}{8242} = \mathbf{ 0.05399 } [/latex]

Question 9.12: A 10 H.P., 4 pole, 25 Hz, 3-phase, wound rotor induction mot...

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