Exercise 01

Let's consider a DC motor with seperate excitation in which the flux is maintained constant. The back e.m.f (f.c.e.m : E) is proportional to the angular speed Ω: E = k.Ω with: k = 1.53 V.rad^-1.s

The armature resistance: R = 1.2Ω. The rated value of the armature voltage: U = 180V. The rated armature current i = 10A. The inductor absorbs a current J = 0.5A and is supplied with a voltage U_ex equal to 200V.

1. Calculate for nominal operation:

            1.1 The back e.m.f.

            1.2 The angular speed Ω and the rotor speed n.

            1.3 The electromagnetic power P_e and the moment of the electromagnetic torque T_e.

2. Demonstrate that the electromagnetic torque is proportional to the intensity i and give the value of the coefficient of proportionality.

3. Calculate the power absorbed by the motor P_a.

4. Calculate the copper losses in the armature.

5. Calculate the copper losses in the inductor.

6. Calculate the useful power of the motor if the collective losses are estimated at 200W.

7. Determine the useful torque of the motor.

8. Determine the moment of the loss torque C_p of the motor.

9. Calculate the motor efficiency.

 

Exercise 02

A 12-pole three-phase induction motor is supplied with a phase-to-phase voltage equal to 220V/60Hz. The stator windings are triangle connected. At nominal load, it rotates at a speed of 570 rpm, giving a shaft power of 2500W with an efficiency of 75% and a power factor of 0.8. If the mechanical losses pm = 200W, Calculate:

1. The intensity of the line current and that of the stator winding.

2. Synchronism speed, motor slip, frequency of rotor currents.

3. Rotor copper losses and torque on the shaft at nominal load.

 

Exercise 03

A three-phase turbo-alternator, star-coupled, with a nominal power of 200MVA, delivers a line voltage of 11kV at 50Hz. We neglect the phase resistance. The no-load excitation current is J = 100A. For the same excitation current (J = 100A), the short-circuit test gave a current of 6351A.

This alternator fed an induction motor which provides a useful power P_U = 90MW with an efficiency of 0.9.

1. Calculate the current absorbed by the motor for a power factor equal to 1 then to 0.8 inductive.

2. Calculate the excitation current corresponding to the motor power with a power factor equal to 1 then 0.8 capacitive then 0.8 inductive.