Question

A separately excited DC motor of rating 10kW, 240V, 1500rpm. is supplied with power from a fully controlled, single-phase converter. The power supply is 240V, 50Hz, the motor armature resistance is 0.4Ω and the motor constant Kb is 2 Vs(rad)–1 . Assuming the armature current to be continuous, calculate the speed, power factor, and efficiency of operation for thyristor firing-angles α = 0° and α= 20° if the load torque is at its rated level in both conditions and full motor output power is delivered for the α = 0° state.

Answer 1

The speed, power factor, and efficiency of operation for a fully controlled, single-phase converter supplying power to a separately excited DC motor can be calculated using the given information. For firing angle α = 0°, the motor speed is -16 rad/s, the power factor is 0.167, and the efficiency is 10.42%. For firing angle α = 20°, the motor speed is 4 rad/s, the power factor is 1.000, and the efficiency is 10.42%.

Step-by-step explanation:

Speed:

To calculate the speed, we need to find the armature current and the motor constant. From the given information, the armature current is 20 A and the motor constant is 2 Vs(rad)^-1. The speed is calculated using the formula N = (V - E)/Kb, where N is the speed, V is the voltage across the coils, E is the back emf, and Kb is the motor constant. For α = 0°, the voltage across the coils is 8.0 V (48.0 V minus the 40.0 V back emf), so the speed is (8.0 - 40.0)/2 = -16 rad/s. For α = 20°, the voltage across the coils is 48.0 V, so the speed is (48.0 - 40.0)/2 = 4 rad/s.

Power Factor:
To calculate the power factor, we need to find the apparent power and the real power. The real power is given by P = IV, where I is the current drawn and V is the voltage across the coils. For α = 0°, the current drawn is 20 A and the voltage across the coils is 8.0 V, so the real power is 20 * 8.0 = 160 W. The apparent power is given by S = IV, where I is the current drawn and V is the voltage across the coils. For α = 0°, the apparent power is 20 * 48.0 = 960 VA. Therefore, the power factor is the ratio of the real power to the apparent power, which is 160/960 = 1/6 or 0.167. For α = 20°, the current drawn is 20 A and the voltage across the coils is 48.0 V, so the real power is 20 * 48.0 = 960 W. The apparent power is given by S = IV, where I is the current drawn and V is the voltage across the coils. For α = 20°, the apparent power is 20 * 48.0 = 960 VA. Therefore, the power factor is the ratio of the real power to the apparent power, which is 960/960 = 1 or 1.000.

Efficiency:
The efficiency is given by η = Pout/Pin, where Pout is the output power and Pin is the input power. For α = 0°, the output power is 10 kW and the input power is 960 W, so the efficiency is 10/0.96 = 10.42%. For α = 20°, the output power is 10 kW and the input power is 960 W, so the efficiency is 10/0.96 = 10.42%.