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Synchronous Machines

Synchronous Machines. Example 1. A 150 kW, 460 V, 1200 r/min, Y-connected synchronous motor has a synchronous reactance of 0.8 W /phase. The internal voltage is 300 V. If the power angle is 30 o , determine the following The power The torque The pull-out torque of the motor.

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Synchronous Machines

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  1. Synchronous Machines

  2. Example 1 • A 150 kW, 460 V, 1200 r/min, Y-connected synchronous motor has a synchronous reactance of 0.8 W/phase. The internal voltage is 300 V. If the power angle is 30o, determine the following • The power • The torque • The pull-out torque of the motor

  3. Example 1 Solution

  4. Example 2 A 2000-hp, 2300-V, unity power factor, Y-connected, 30-pole, 60-Hz synchronous motor has a synchronous reactance of 1.95 W/phase. Neglect all losses. Compute the maximum power and torque which this motor can deliver if it is supplied with power directly from a 60-Hz, 2300-V supply. Assume field excitation is maintained constant at the value which will result in unity power factor at rated load.

  5. 374 A 1328 V 729 V 1515 V Example 2 Solution Rated kVA = 2000X0.746 = 1492 kVA, three phase = 497 kVA/phase Rated voltage = 2300/1.732=1328 V per phase Rated current = 497000/1328 = 374 A/phase-Y From the phasor diagram, The maximum power and torque,

  6. Example 3 • A three-phase, 225 r/min synchronous motor is connected to a 4-kV, 60-Hz line draws a current of 320 A and absorbs 2000 kW. Calculate • The apparent power supplied to the motor • The power factor • The reactive power absorbed • The number of poles on the rotor

  7. Example 3 Solution

  8. Example 4 • A three-phase synchronous motor rated at 800-hp, 2.4-kV, 60-Hz operates at unity power factor. The line voltage suddenly drops to 1.8 kV, but the exciting current remains unchanged. Explain how the following quantities are affected. • Motor speed and mechanical power output • Power angle, d • Position of the rotor poles • Power factor • Stator current

  9. The speed is constant, hence the load does not know that the line voltage has dropped. Therefore, the mechanical power will remain unchanged • P=(VtEf/Xs)sind, P, Ef and Xs are the same but Vt has fallen; consequently sind must increase, which means that d increases • The poles fall slightly behind their former position, because d increases • Terminal voltage is smaller than before, the motor internal voltage is bigger than the terminal voltage and as a result, the power factor will be less than unity and leading • As power factor is less than unity, apparent power S is greater now. The terminal voltage is smaller, • will increase Example 4 Solution

  10. Example 5 • A 4000-hp, 6.9-kV synchronous motor has a synchronous reactance of 10 W/phase. The stator is connected in wye, and the motor operates at full-load (4000 hp) with a leading power factor of 0.89. If the efficiency is 97%, calculate the following: • The apparent power • The line current • The internal voltage per phase with corresponding phasor diagram • The power angle • The total reactive power supplied to the system • The approximate maximum power [in hp] the motor can develop without pulling out of step

  11. Ia Vt 270 IaXs Ef Example 5 Solution

  12. Example 5 Solution (cont’d)

  13. Example 6 • A 1500-kW, 4600-V, 600 r/min, 60-Hz synchronous motor has a synchronous reactance of 16 W/phase and a stator resistance of 0.2 W/phase. The excitation voltage is 2400 V, and the moment of inertia of the motor and its load is 275 kg.m2. We wish to stop the motor by short-circuiting the armature while keeping the dc rotor current fixed. Calculate • The power dissipated in the armature at 600 r/min • The power dissipated in the armature at 150 r/min • The kinetic energy at 600 r/min • The kinetic energy at 150 r/min • The time required for the speed to fall from 600 r/min to 150 r/min

  14. jXs Ra + Ia + Ef Vt=0 Example 6 Solution

  15. Example 6 Solution (Cont’d)

  16. Example 6 Solution (Cont’d)

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