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ECE 3600 Induction Motor Examples A.Stolp 11/7/11 rev 11/12/14 Ex. 1 A 480-V. six-pole, 60-Hz, ∆-connected, 3-phase induction motor is rated at 60 hp. Its equivalent circuit components are R1 0.480. Ω R2 0.6. Ω RC ∞ X1 0.50. Ω X2 0.60. Ω XM 30. Ω 600 . W P mech N poles 6 480 . V Vφ For a slip of 0.04, find s 150 . W P misc 200 . W P core V φ = 480 V 4. % a) The line current (magnitude) R 1 = 0.48 Ω Z eq R1 I2 I1 1 j. X 1 j . X 1 = 0.5j Ω 1 R2 1 j. X M j. X 2 ∞ E1 RC j . X M = 30j Ω R2 s = 15 Ω j . X 2 = 0.6j Ω s Z eq = 12.104 + 6.786j Ω I1 Vφ I 1 = 30.171 Z eq I 1 = 34.59 A ∆-connected arg I 1 = 29.278 deg b) The stator copper losses c) The air-gap power P AG 16.916j A Z E1 3. P SCL 2. 1 P SCL = 1.723 kW E1 j. X 2 V φ. E1 1 j. X 2 I 1 . 3 = 59.91 A R1 1 j. X M I2 I1 IL = R2 Z E1 E 1 = 457.06 Z eq s P AG R2 3. I2 2. R2 P AG = 41.72 kW s P AG = 55.95 hp s d) The power converted from electrical to mechanical form j. X 1 I2 R1 R2 R2 I1 .( 1 s) s j. X M RC E1 6.966j V P conv = 3 . j. X 2 P conv I2 (1 2. R2 .( 1 s) = (1 s s ) . P AG s ) . P AG P conv = 40.05 kW P conv = 53.71 hp e) The motor speed in revolutions per minute and radians per second 7200. rpm n sync nm N poles (1 s ) . n sync n sync = 1200 rpm n m = 1152 rpm ECE 3600 ω sync ωm n sync . 2 . π. (1 rad . min rev 60. sec s ) . ω sync Induction Motor Examples ω sync = 125.7 ω m = 120.6 p1 rad sec rad sec ECE 3600 Induction Motor Examples P conv τ ind f) The induced torque p2 ωm P AG τ ind OR: τ ind = 332 N. m (easier) ω sync g) The load torque τ load Use Pcore here. Lump it in with the mechanical losses, Pmisc and Pmech. Read the last 2 paragraphs on p.302. P out P conv P core P mech P misc τ load h) The overall machine efficiency η P out = 39.105 kW P out τ load = 324.152 N. m ωm P out η = 90.006 % 3 . V φ. Re I 1 Ex. 2 For the motor in Example 1, a) what is the slip at the pullout torque? Create a Thevenin equivalent of the model with (R2/s + X2) as the load. j. X 1 R1 j. X M Vφ V φ. V TH j. X M j. X 1 R1 RC j. X M V TH = 472.014 + 7.428j V V TH = 472.073 V j. X 1 R1 1 Z TH j. X M 1 R1 RC 1 j. X 1 j. X M Z TH = 0.464 + 0.499j Ω Z TH = 0.6817 Ω At max Pout R2 = s j . X 2 = 1.193 Ω Z TH R2 s Z TH j. X 2 arg Z TH = 47.07 deg s = 50.287 % b) What is the pullout torque of this motor? nm s ) . n sync (1 V TH I2 Z TH j. X 2 R2 P AG 3. I2 s P conv τ max ECE 3600 rad . min n m. 2 . π. rev 60. sec ωm n m = 596.553 rpm Induction Motor Examples p2 (1 R2 s s ) . P AG P conv ωm 2. ω m = 62.471 rad sec P AG = 201.686 kW P conv = 100.264 kW τ max = 1605 N. m ECE 3600 Ex. 3 Induction Motor Examples p3 A 3-phase, Y-connected, induction motor has the following equivalent circuit components: 3500. rpm currently running at n current air-gap power R1 0.2. Ω 0.3. Ω X1 7.5. kW P AG I2 R2 0.13. Ω I1 R2 12. Ω XM E1 X2 RC RC a) Find I2 This has to be a 2-pole motor, so P AG 3 2 = I2 . 3600. rpm n sync R2 b) Find the rotor copper losses P RCL s s) 3600 3500 0.45. Ω ∞ s = 2.778 % 3600 P AG . s 3 R2 I2 s and .( 1 s I 2 = 23.113 A 2 3. I 2 . R 2 c) The output shaft torque is τ load 19. N. m Find the output power P out τ load . n . P RCL = 208.333 W 2 . π. rad . min rev 60. sec P out = 6.964 kW d) Find the mechanical power losses (all lumped together). P mech P AG P RCL P out P mech = 327.803 W e) Find the line current. Note: Don't try any shortcuts here. You need to do your math with full complex numbers. I advise you to assume the phase angle of I2 is 0o . IL =? E1 I 2. IL I2 Vφ E1 R2 s E1 j. X 2 E 1 = 108.167 + 10.401j V I L = 25.617 A j. X M I L. R 1 j. X 1 V φ = 116.7 V NOT asked for f) The stator copper losses P SCL g) The overall machine efficiency η = 2. 3. I L R1 P out P SCL P SCL = 393.75 W = 88.22 % P AG ECE 3600 Induction Motor Examples p3 ECE 3600 Ex. 4 Induction Motor Examples p4 2. (16 pts) A 208-V. four-pole, 60-Hz, Y-connected, induction motor is rated at 20 hp. Its equivalent circuit components are N poles 4 R1 0.20. Ω R2 0.12. Ω X1 0.40. Ω X2 0.40. Ω 300 . W P mech 0. W P misc 15. Ω XM P core 0. W For a slip of 0.06, the following values have been calculated for you: E 1 = 108.167 + 10.401j V The line current: E 1 = 108.665 V I L = 25.62 A Find the following: a) The stator copper losses 3. P SCL b) The air-gap P AG IL 2. R1 P SCL = 0.394 kW E1 I2 X2 2 R2 2 3. I 2 . P AG 2 R2 P AG = 8.13 kW s s c) The power converted from electrical to mechanical form P conv (1 s ) . P AG P conv = 7.905 kW d) The motor speed in revolutions per minute and radians per second n sync 7200. rpm N poles ω sync n sync = 1800 rpm rad . min n sync . 2 . π. rev 60. sec ω sync = 188.496 nm (1 s ) . n sync ωm n m = 1750 rpm rad OR sec 377 rad = 188.5 2 rad . min n m. 2 . π. rev 60. sec ω m = 183.26 sec rad sec e) The induced torque τ ind τ ind f) The load torque τ load P out P conv ωm P conv OR: P core τ ind P mech τ load P AG ω sync P misc P out ωm ECE 3600 τ ind = 43.133 N. m P out = 7.605 kW τ load = 41.496 N. m Induction Motor Examples p4
