# Download Example 5 Solution

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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 30o, determine the following a) The power b) The torque c) The pull-out torque of the motor Example 1 Solution a ) The phase voltage, Vt 460 3 266 V Output power : Vt E f 266 300 P sin sin 30 o 50 k W Xs 0.8 b) Output torque: 50,000 T 400 N .m 2 1200 60 c) Pull out torque: 266 300 sin 90 o 0.8 Tmax 800 N .m 2 1200 60 Example 2 A 2000-hp, 2300-V, unity power factor, Y-connected, 30pole, 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. 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 374 A 1515 V 1328 V 729 V From the phasor diagram, E f Vt 2 I a X s 1328 2 374 1.95 1515 V 2 2 The maximum power and torque, 1328 1515 3096 kW three phase 1.95 2 s 2 60 8 rad / sec 30 Tmax 3096 123 .2 kN.m 8 Pmax 3 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 a. b. c. d. The apparent power supplied to the motor The power factor The reactive power absorbed The number of poles on the rotor Example 3 Solution a ) S 3 4000 320 2217 k VA b) PF cos 2000 2217 0.902 c) Q 2217 2 2000 2 957 k VAR d ) No. of poles 120 60 225 32 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. a. b. c. d. e. Motor speed and mechanical power output Power angle, Position of the rotor poles Power factor Stator current Example 4 Solution a. The speed is constant, hence the load does not know that the line voltage has dropped. Therefore, the mechanical power will remain unchanged b. P=(VtEf/Xs)sin, P, Ef and Xs are the same but Vt has fallen; consequently sin must increase, which means that increases c. The poles fall slightly behind their former position, because increases d. 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 e. As power factor is less than unity, apparent power S is greater now. The terminal voltage is smaller, I S 3Vt will increase 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: a. The apparent power b. The line current c. The internal voltage per phase with corresponding phasor diagram d. The power angle e. The total reactive power supplied to the system f. The approximate maximum power [in hp] the motor can develop without pulling out of step Example 5 Solution a ) P0 4000 0.746 2984 k W Pi 2984 0.97 3076 k W S 3076 3457 k VA 0.89 b) I a 3457 289 .3 A 3 6 .9 c) Vt 6900 3984 V ; cos 0.89 27 0 3 Ia Vt 270 IaXs Ef E f Vt 2 I a X s 2 E f Vt cos 90 0 27 0 5889 V 2 Example 5 Solution (cont’d) d ) P0 _ per phase Vt E f Xs sin 3076000 25.9 0 3 e) Q 3457 2 3076 2 1578 kVAR f ) Pmax 3 3984 5889 sin 90 0 7039 k W 9436 hp 10 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 a. b. c. d. e. 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 Example 6 Solution a ) Z Ra2 X s2 0.2 2 16 2 16 W Current per phase Ef Ia 2400 150 A Z 16 Power dissipated in all three phases jXs + Ef Ia Ra + Vt=0 Pdissipated 3 I a2 Ra 3 150 2 0.2 13.5 kW b) Becausethe exciting currentis fixed, the int ernal voltageis proportional to speed Consequently , when the speed dropsto 150 r / min E f 2400 150 600 V 600 The frequencyis also proportional to the speed, and so f 60 150 15 Hz 600 The synchronous reac tan ce is also proportional to the speed, and so X s 16 150 4W 600 Example 6 Solution (Cont’d) The newimpedance per phaseat 150 r / min is Z Ra2 X s2 0.2 2 4 2 4 W Current per phase Ia Ef 600 150 A Z 4 Power dissipated in all three phases Pdissipated 3 I a2 Ra 3 150 2 0.2 13.5 k W As the short circuit currentremains unchanged, power dissipated in three phasesis the same. c) The k ineticenergyat 600 r / min is K E1 5.48 10 3 Jn 2 5.48 10 3 275 600 2 542 .5 k J d ) The k ineticenergyat 150 r / min is K E 2 5.48 10 3 275 150 2 33.9 k J Example 6 Solution (Cont’d) e) The loss in kineticenergyin deceleration from600 r / min to 150 r / min is W K E1 K E 2 542 .5 33.9 508 .6 kJ The energyis lost as heat in the armatureresis tan ce. The time for the speed to drop from600 r / min to 150 r / min is P W 13.5 508 .6 t 37.7 sec . t t The motor would stop much soonerif externalresistors were conected acrossthe stator ter min als.