A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator, using a Ward-Leonard control system shown in Fig. 6. The total resistance of the motor and generator armature circuit is 10 m. The motor turns at a nominal speed of 300 r/min, when Eo is 500 V.

Question

A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator, using a Ward-Leonard control system shown in Fig. 6. The total resistance of the motor and generator armature circuit is 10 m. The motor turns at a nominal speed of 300 r/min, when $E_{ o }$ is 500 V.

Calculate

a. The motor torque and speed when

E_{ s }=400 V \text { and } E_{ o }=380 V

b. The motor torque and speed when

E_{ s }=350 V \text { and } E_{ o }=380 V

Calculate

a. The motor torque and speed when

[latex]E_{ s }=400 V ext { and } E_{ o }=380 V[/latex]

b. The motor torque and speed when

[latex]E_{ s }=350 V ext { and } E_{ o }=380 V[/latex]

Accepted Answer

a. The armature current is

[latex]\begin{aligned}I &=\left(E_{ s }-E_{ o } ight) / R=(400-380) / 0.01 \&=2000 A\end{aligned}[/latex]

The power to the motor armature is

[latex]P=E_{ o } I=380 imes 2000=760 kW[/latex]

The motor speed is

[latex]n=(380 V / 500 V ) imes 300=228 r / min[/latex]

The motor torque is

[latex]\begin{aligned}T &=9.55 P / n \&=(9.55 imes 760000) / 228 \&=31.8 kN \cdot m\end{aligned}[/latex]

b. Because [latex]E_{ o }[/latex] = 380 V, the motor speed is still 228 r/min.

The armature current is

[latex]\begin{aligned}I &=\left(E_{ s }-E_{ o } ight) / R=(350-380) / 0.01 \&=-3000 A\end{aligned}[/latex]

The current is negative and so it flows in reverse; consequently, the motor torque also reverses.

Power returned by the motor to the generator and the 10 mΩ resistance:

[latex]P=E_{ o } I=380 imes 3000=1140 kW[/latex]

Braking torque developed by the motor:

[latex]\begin{aligned}T &=9.55 P / n \&=(9.55 imes 1140000) / 228 \&=47.8 kN \cdot m\end{aligned}[/latex]

Question 5.3: A 2000 kW, 500 V, variable-speed motor is driven by a 2500 k...

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