Solution of Tutorial 3 Synchronous Motor Drives

Transcription

1 chool of lectrical ngineering & Telecommunications, UNW lectric Drive ystems olution of Tutorial 3 ynchronous Motor Drives Question. R 0, XO 72at the base speed corresponding to 50 Hz operation fo 3200 /phase b 380 /phase 3 (i) (ii) fo 50 N 2. 5 rev/sec rad/sec. p 8/ rev/min 3 f P sin sin 300, X 72 0 sin jx s fo jxs = j72 q 380 fo = ~ 3200 d (iii) cos 00, W However, and are not known. From the phasor diagram, olution of Tutorial 3 M drive F. Rahman/8-ep-3/2:33 PM

2 lectric Drive ystems sin ; cos ; d Also sin qx sin q 3. 22A X 72 And cos fo dx q dX d 69A. 72 d tan tan ( ) q PF cos , lagging. Now P 3 cos , 3. 33A Note also that, 3.33 A 2 2 d q Alternative solution: o 0 fo jx 7290 so A PF cos , lagging. (iv). The imum power the motor can develop at base speed ( R 0) 3 fo P for 90 X , Watts 508 kw 72 The imum torque the motor can develop at base speed T P 4 p 508, 6, 47 Nm 34 olution of Tutorial 3 M drive 2 F. Rahman/8-ep-3/2:33 PM

3 lectric Drive ystems Question 2 The excitation is increased, so that f 0 is now larger. The motor operates at unity power factor, i.e., the phase angle between the input voltage and current phasors is zero degree. 0 ; jx s fo jx s q 380 fo ~ d (i) P 300, W; 3 cos 300, W 300, A or, 3 fo sin 300, (2.) X and are both unknown. Now developed Power / phase 00, W q q or, 00, cos 00, cos (2.2) From (2.2) and (2.), 00, 3 sin 300 or tan X cos X olution of Tutorial 3 M drive 3 F. Rahman/8-ep-3/2:33 PM

4 lectric Drive ystems or X tan Alternative solution: cos 0 00, ; Hence fo 3 f0 (ii) Now P = sin 300, W X 00, 72 f0 4, olts ( rms) 380 sin 00, 380 fo A A (iii) P 3 f kW X 72 P 4 T Nm,, / p 34 Question 3 PF = 0.8 leading cos cos 300, (i) 32. 8A cos j q X s fo X s = j72 j d X s q ~ 380 fo d f olution of Tutorial 3 M drive 4 F. Rahman/8-ep-3/2:33 PM

5 lectric Drive ystems (ii) From the phasor diagram, sin qx (3.) cos f 0 dx (3.2) cos( ) cos cos sinsin cos 9. 6 sin q From (3.) 380 sin ( cos 9. 6 sin ) X 380sin 890 cos 47. 4sin or, ( )sin 890 cos 890 or, tan Alternative solution: cos X 380 cos sin( ) 5, 558 f0 d cos , so that = 32.8 A 380 fo Thus, fo (iii) P 882, 332 W 882 kw X 72 Question 4 p 4 T P 882,, 239 Nm 34 Given R. 5 / phase, X 5 / base speed 230, 5 A, b raated rated 0 N f / p 50 / 2 rev /sec 500 rev /min b 0 fo = induced voltage per phase (due to rotor excitation) at base speed = 200. Constant v/f operation i.e. f. olution of Tutorial 3 M drive 5 F. Rahman/8-ep-3/2:33 PM

6 lectric Drive ystems 3 f0 (i) P = sin W X = torque angle for operation at imum power or or imum torque = 90 regardless of speed (input frequency) if R 0. (ii) 3 fo P 9200 Watts at base speed X 5 p 2 T Nm At N = 750 rev/min, = = 0.5, f Hz f 34 / 2 57 rad/ sec P 3 f W X ; f , X 0. 5X P Watts p 2 T P Nm. 57 At N = 250 rev/min, = 250/500 = 0.67, f Hz As before, P 533 Watts 2 T Nm 57 olution of Tutorial 3 M drive 6 F. Rahman/8-ep-3/2:33 PM

7 lectric Drive ystems Question 5 Given R. 5 /phase, X 5 base speed 230, 200 / phase o raated f 0 0 At N = 250 rev/min, =250/500 = 0.67, f Hz , X X For imum torque, X tan R 5. T 3p 0 f0 X0 sin R(cos ) f R ( X0) Nm sin 5. cos Nm compared to 58.6 Nm when R is neglected. Question 6 Given R 0, X 7. base speed; d0 X 5. base speed q0 The base speed is given by N N f / p 50 /( 4 / 2) = 25 rev/ sec 500 rev/min b 0 olution of Tutorial 3 M drive 7 F. Rahman/8-ep-3/2:33 PM

8 lectric Drive ystems f 50 at 200 rev/ min 50 f /phase; /phase; = For operation at 200 rev/min, o 2 3 fo X do X qo T sin sin 2 2 o / p Xdo 2 XdoX qo Nm For 20 6 T [ ] Nm 34 Question 7 p 4/ 2 2 ; f 50 Hz ; R. 5 /phase X 0 5 base speed 230 /phase 0 raated ; 5 0 raated A/phase 200 f0 3 f (i) T cos, Nm m For a supply frequency of f, T 3 f cos. f The ratio remains constant with frequency, since both vary proportionately with frequency. Thus, f /rad/sec. 34 olution of Tutorial 3 M drive 8 F. Rahman/8-ep-3/2:33 PM

9 lectric Drive ystems For the imum torque/ampere (MTPA) characteristic, 0 i.e. f and are in phase. T Thus K 382. in this case. To develop 60 Nm, T A. K T T The amplitude of the stator current for this condition of operation (60 Nm, 0 ) is, A m (ii) The developed power for this case is: 750 P T kw 60 (iii) Power factor R =.5 jx s = j7.5 jx s R f f Clearly, the power factor, PF cos, is lagging. Taking f as the reference phasor at 0, R jx A, R. 5, X 5 / 2 f At 750 rev/min, f is 25 Hz, f 00. olution of Tutorial 3 M drive 9 F. Rahman/8-ep-3/2:33 PM

10 lectric Drive ystems 00c j j j ; PF cos , lagging (iv) 30 T cos A in order to develop 60 Nm A m.5 j7.5 q-axis f = jx s R fo d-axis j j j ; PF cos , lagging olution of Tutorial 3 M drive 0 F. Rahman/8-ep-3/2:33 PM

11 lectric Drive ystems Question 8 DC DC Link Reactor ~ M 20 DC DC Dotted trace represents the fundamental of the input phase current waveform. n this case, it is also in phase with the back emf waveform. f0 50 Base speed 25 rev/se c p 4/ 2 rev/min / 2 / 2 m DC m DC Now, f f T 3p cos 3p Nm olution of Tutorial 3 M drive F. Rahman/8-ep-3/2:33 PM

12 lectric Drive ystems and f T Nm 34 With the rated output power, the developed torque is 6 Power. 5 0 T 9, 554 Nm peed 34 / A A DC olution of Tutorial 3 M drive 2 F. Rahman/8-ep-3/2:33 PM