1 / 16

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.

mairi
Download Presentation

Synchronous Machines

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related