A DC separately excited motor is running 800 rpm driving a load whose torque is constant. Motor armature current is 500 A. The armature resistance drop and rotational losses are negligible. Magnetic circuit can be assumed to be linear. Calculate motor speed and armature current if terminal voltage is reduced to 50 % and field current is reduced to 80 %.?

Question

Answer

Problem Statement

A separately excited DC motor is running at 800 rpm with a constant load torque. The armature resistance drop and rotational losses are negligible. The motor armature current is 500 A. The magnetic circuit can be assumed to be linear. Calculate the motor speed and armature current if the terminal voltage is reduced to 50% and the field current is reduced to 80%.

Solution

Step 1: Calculate the back EMF and armature voltage

At the given speed of 800 rpm, the back EMF of the motor is given by:

EMF = K * ω = K * (2 * π * N / 60)

where K is the motor constant, N is the speed in rpm, and ω is the speed in radians per second.

Assuming the motor constant, K is 0.1 V/rpm, the back EMF is:

EMF = 0.1 * 2 * π * 800 / 60 = 10.5 V

The armature voltage is given by:

V = E + IaRa

where E is the back EMF, Ia is the armature current, and Ra is the armature resistance. Since the resistance drop is negligible, Ra can be assumed to be zero. Therefore,

V = E + IaRa = E = 10.5 V

Step 2: Calculate the new back EMF and armature voltage

If the terminal voltage is reduced to 50%, the new armature voltage is:

V' = 0.5 * V = 5.25 V

Assuming that the field current is reduced to 80%, the new back EMF is:

EMF' = 0.1 * 2 * π * N' / 60

where N' is the new speed of the motor. Since the load torque is constant, the new speed can be calculated using the torque-speed equation:

T = Kt * Ia

where Kt is the torque constant. Assuming the torque constant is 0.2 Nm/A, the torque is:

T = 0.2 * 500 = 100 Nm

The torque-speed equation can be rearranged to solve for the new speed:

N' = (V' - IaKt) / (0.1 * π)

Substituting V' = 5.25 V and T = 100 Nm, the new speed is:

N' = (5.25 - 100 * 0.2) / (0.1 * π) = 359.3 rpm

Step 3: Calculate the new armature current

The new armature current can be calculated using the torque-speed equation:

Ia' = T / Kt = 100 / 0.2 = 500 A

Therefore, the new motor speed is 359.3 rpm, and the new armature current is 500 A.

Conclusion

When the terminal voltage is reduced to 50% and the field current is reduced to 80%, the motor speed decreases to 359.3 rpm, and the armature current remains the same at 500 A.

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