A 1800 kW, 300 V variable-speed motor is driven by a 2200 kW generator, using a

Question

A 1800 kW, 300 V variable-speed motor is driven by a 2200 kW generator, using a Ward-Leonard
control system shown in the figure. The total resistance of the motor and generator armature circuit is
50 m?. The motor turns at a nominal speed of 500 rpm, when Eo is 300 V.

a. Calculate the motor torque and speed when Es = 150 V and Eo = 120 V.

b. Calculate the mechanical load torque attached to the shaft of the motor if the motor is spinning with
the same constant speed under conditions of part (a).

c. Assume that Es reduces to 100 V. What is the armature current, speed of rotation, and the braking
torque developed in the motor at the first moment while Eo is still unchanged.

Explanation / Answer

a. Speed N=500 rpm at Eo=300 V.

So, at 150V, N=250 rpm.

Es-Eo=Ia Ra

150-120=Ia(50 milliohm)

Ia= 30/(50 milliohm)= 600 A

Eo= Ka*phi*((2*pi*250)/60)

Ka*phi= (120*60)/(2*pi*250)

=4.58

Tm= 4.58*600=2748 Nm

b. Magnitude of mechanical load torque would be the same as motor produced torque in steady state condition just that it will be exerted in opposition to motor torque. So, Tl=2748Nm

c. Ia =20/(50 m OHm)= 400A

N=500 rpm

Braking torque = 4.58*400= 1832 NM

d. Ia= 600A, T=2748 Nm,

Ia*Ra =30V

Eo= Es-Ia*Ra

Eo= 100-30 =70V

&0= 4.58 * omega

omega= 70/4.58

N= omega /(2*pi)

N= 96 rpm (approx)

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