Download presentation

Published bySeamus Screen Modified over 3 years ago

1
ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998 Electric Drives

2
**8. INDUCTION MOTORS FOR DRIVES**

8.1. THE STATOR AND ITS TRAVELING FIELD Figure 8.1. Cross section of an induction motor with two poles (8.1) Electric Drives

3
**Figure 8.2. The m.m.f. and airgap flux density of phase a**

(8.2) (8.3) (8.6) (8.8) (8.9) (8.11) Electric Drives

4
**8.2. THE CAGE AND WOUND ROTORS ARE EQUIVALENT**

Figure 8.3. Induction motor rotors a.) cage - type rotor b.) wound rotor Electric Drives

5
**8.3. SLOT SHAPING DEPENDS ON APPLICATION AND POWER LEVEL**

Figure 8.4. Stator or wound rotor slots a.) for low power (semiclosed slots) b.) for high power (open slots) Electric Drives

6
**a.) semiclosed - round for low power and variable frequency motors **

Figure 8.5. Rotor slots a.) semiclosed - round for low power and variable frequency motors b.) closed - for low noise or high speed motors c.) semiclosed with moderate skin effect and starting torque - constant frequency d.) semiclosed with high skin effect for high starting torque - constant frequency e,f.) double cage - for superhigh starting torque - constant frequency g.) for high speed inverter - fed motors Electric Drives

7
**Figure 8.6. Three phase induction motor with equivalent wound rotor**

8.4. THE INDUCTANCE MATRIX Figure 8.6. Three phase induction motor with equivalent wound rotor (8.12) (8.13) (8.14) Electric Drives

8
**The inductances may be now assembled into the so called matrix inductance**

: (8.18) where: Electric Drives

9
**8.5. REDUCING THE ROTOR TO STATOR**

(8.19) (8.20) (8.21) (8.22) So the new inductance matrix is similar to that in (8.18), but with Electric Drives

10
**8.6. THE PHASE COORDINATE MODEL GOES TO 8th ORDER**

(8.23) (8.24) (8.25) (8.26) (8.27) Using (8.24) in (8.23) - with qer variable in time in any case - we obtain: (8.28) where: (8.29) Electric Drives

11
**The motion equation are: (8.33)**

(8.30) (8.31) (8.32) The motion equation are: (8.33) Electric Drives

12
**8.7. THE SPACE PHASOR MODEL (8.34) (8.35) (8.36) (8.37) (8.38)**

Electric Drives

13
**With definitions (8.37) - (8.38), equations (8.35) - (8.36) become: **

(8.41) With definitions (8.37) - (8.38), equations (8.35) - (8.36) become: (8.42) (8.43) If to (8.42) - (8.43) we add similar equations for phases b, br and c, cr equation (8.23) becomes: (8.44) (8.45) (8.46) Electric Drives

14
**The torque should be calculated from (8.42) with the above notations:**

(8.48) (8.49) (8.50) (8.51) (8.52) (8.53) The torque should be calculated from (8.42) with the above notations: Electric Drives

15
We may now decompose in plane the space phasors along two orthogonal axes d and q moving at speed wb [5]: (8.55) (8.56) Electric Drives

16
**[P(qb)] is the Park transformation:**

Also from (8.49) - (8.50) and (8.47): (8.57) [P(qb)] is the Park transformation: (8.58) The inverse of Park transformation is: (8.59) Electric Drives

17
**Example 8.1. The space phasor of sinusoidal symmetric currents. **

Consider three symmetrical sinusoidal currents and show how their complex space phasor in time through 6 instants. Give a graphical description of this process in time. Solution: The three phase currents may be written as: (8.61) The space phasor in stator coordinates is, (8.37),: (8.62) with (8.63) (8.62) becomes: (8.64) Electric Drives

18
**The position of the space phasor for is shown in figure 8.7.**

Figure 8.7. The space phasor of sinusoidal three phase currents Electric Drives

19
**8.8. THE SPACE PHASOR DIAGRAM FOR ELECTRICAL TRANSIENTS**

The space phasor model equations (8.51) - (8.53) may be represented on a space phasor diagram in the d - q plane with axes d and q rotating at speed wb = w1 (figure 8.8). Let us consider Figure 8.8. The space phasor diagram of induction motor valid for transients (for steady d/dt = 0) in synchronous coordinates (cage rotor: ) Electric Drives

20
**For this case the torque expression (8.54) becomes: (8.68)**

d/dt = j(w1 - wb). (8.66) Also for steady state of rotor flux the rotor flux equation (8.52) becomes: (8.67) Only for (short - circuited, cage, rotor) the rotor current and flux space phasors are orthogonal to each other. For this case the torque expression (8.54) becomes: (8.68) Electric Drives

21
**8.9. ELECTRICAL TRANSIENTS WITH FLUX LINKAGES AS VARIABLES**

(8.69) (8.70) (8.71) equations (8.51) - (8.53) become: (8.72) (8.73) Electric Drives

22
**and as variables; random speed (wb) coordinates.**

with (8.74) While ts, tr are the stator and rotor time constants, ts’ and tr’ are transient time constants of rotor and stator.The structural diagram corresponding to (8.72) - (8.73) is shown on figure 8.9. Figure 8.9. Structural diagram of induction motor with stator and rotor flux and as variables; random speed (wb) coordinates. Electric Drives

23
**8.10. COMPLEX EIGEN VALUES FOR ELECTRICAL TRANSIENTS**

Equations (8.72) - (8.73) imply a second order complex - variable system with only two complex eigen values corresponding to the determinant: (8.75) The complex eigen values from (8.75) depend on the reference system speed wb and on rotor speed wr, but their real part is, in general, negative suggesting attenuated periodic response.[4,6] Example 8.2. For an induction motor with the data rs = 0.5W, rr = 0.60W, Ls = Lr = 0.08H, Lm = 0.075H, p = 2pole pairs calculate the complex eigen values for constant speed, in synchronous coordinates (w1 = 2p60) for n = 0 and n = 1800 rpm. Electric Drives

24
Solution: The value of wb = w1 = 2p60 rad/s. s, ts’, tr’, Ks, Kr are now calculated from (8.71) - (8.74): (8.76) (8.77) Also (8.78) Electric Drives

25
**Now we rewrite (8.75) in a canonical form: (8.79)**

(8.80) For n = 0 (8.81) For n = 30rps (1800rpm): (8.82) Electric Drives

26
**8.11 ELECTRICAL TRANSIENTS FOR CONSTANT ROTOR FLUX**

(8.83) Using (8.83) in (8.73): (8.84) It is more convenient to use synchronous coordinates for equations (8.72): (8.85) Equations (8.84) - (8.85) lead to a simplification in the structural diagram of figure 8.9. as the derivative term in the rotor disappears (figure 8.10). Electric Drives

27
Figure Structural diagram of induction motor with constant rotor flux and speed in synchronous coordinates (wb = w1) Electric Drives

28
**8.12. STEADY STATE: IT IS D.C. IN SYNCHRONOUS COORDINATES**

Steady state means in general that the three phase voltages are symmetric and sinusoidal: ; i = 1, 2, 3 (8.86) The voltage space phasor in random coordinates is: (8.87) with (8.86): (8.88) For steady state: (8.89) Consequently: (8.90) Electric Drives

29
It is obvious that for steady state the currents in the model have to have the voltage frequency, which is: . Once again for steady state we use (8.51), with d/dt = as inferred in (8.66). Consequently from (8.51): (8.91) In the flux equations (8.52) - (8.53) we may separate the main (airgap) flux linkage : (8.92) (8.93) Equations (8.91) - (8.93) lead to the standard equivalent circuit of figure 8.11. Electric Drives

30
**The torque expression (8.54) with (8.91) and Vr0b = 0, yields: (8.94)**

Figure Space phasor steady state equivalent circuit of induction machine The torque expression (8.54) with (8.91) and Vr0b = 0, yields: (8.94) Electric Drives

31
**Consequently the electromagnetic torque Te is: (8.96)**

From (8.92): (8.95) Consequently the electromagnetic torque Te is: (8.96) Using the equivalent circuit of figure (8.11) and (8.95) we may obtain the conventional torque expression: (8.97) with (8.98) Electric Drives

32
**8.13. NO LOAD IDEAL SPEED MAY GO UNDER OR OVER CONVENTIONAL VALUE w1**

The no load ideal speed (slip S0) corresponds to zero torque, that is, (8.91), zero rotor current: (8.99) Only for short - circuited rotor windings (or passive impedance at rotor terminals) ( ) the ideal no load slip S0 = 0 and wro = w1. When the induction machine is doubly fed ( ) the ideal no load slip is different from zero and the no load ideal speed is, in general,: (8.100) Example 8.3. For steady state, calculate the stator voltage, stator flux, current, power factor, torque of an induction motor at 10% ideal no load (synchronous) speed w1 and S = 0.02. The motor data are: rs = 0.5W, rr = 0.6W, Ls = Lr = 0.08H, Lm = 0.075H, lr0 = 0.8Wb, w1 = 2p60 rad/s, p = 2 pole pairs, Vrb = 0 (cage rotor). Electric Drives

33
Solution: First we have to calculate the initial conditions, which are implicitly steady state. Let us use synchronous coordinates: (8.101) So (8.102) The torque (Te) is, (8.96),: (8.103) From (8.72) - (8.73) with d/dt = 0 and Vrb = 0 we may now calculate the stator flux : (8.104) Notice that the motor parameters are as in example 8.2 and thus tr’ = s, ts’ = s, Ks = Kr = , s = (8.105) Electric Drives

34
**Let us consider the d axis along the rotor flux and thus = . **

Now from (8.72) with we may find the stator voltage : (8.106) The motor phase voltage (r.m.s. value): (8.107) The stator current is obtained from (8.69): (8.108) Electric Drives

35
**The rotor space phasor is (8.95): (8.109)**

The amplitude of the stator current The results are illustrated by the space phasor diagram in figure 8.12. The power factor angle j1 is: (8.110) Finally cosj1 = Figure Steady state space phasor diagram in synchronous coordinates (wb = w1): d.c. quantities Electric Drives

36
**Figure 8.13. Induction motor energy conversion**

Example 8.4. Loss breakdown A high efficiency induction motor with cage rotor has the data: rated power Pn = 5kW, rated line voltage (rms) VL = 220V (star connection), rated frequency f1 = 60Hz, number of pole pairs p = 2, core loss (Piron) = mechanical loss Pmec = 1.5% of Pn, additional losses padd = 1%Pn, stator per rotor winding losses pcor/pcos = 2/3, rated efficiency hn = 0.9 and power factor cosjn = 0.88. Calculate all loss components, phase current (rms), then rated slip, speed, electromagnetic torque, shaft torque and stator current (as space phasors). Solution: The loss breakdown diagram of the induction motor is shown in figure 8.13. Figure Induction motor energy conversion Electric Drives

37
**The phase current In (rms) is: (8.112)**

The input power Pin is: (8.111) The phase current In (rms) is: (8.112) The total losses are: (8.113) Consequently: (8.114) (8.115) So (8.116) (8.117) (8.118) Electric Drives

38
**The rotor winding loss pcor is: (8.122) and thus the rated slip is: **

The electromagnetic power Pe is the active power that crosses the airgap: (8.119) (8.120) (8.121) The rotor winding loss pcor is: (8.122) and thus the rated slip is: (8.123) The rated speed nn is: (8.124) The shaft torque Tn is calculated directly from mechanical power Pn: (8.125) Finally the amplitude of the stator current , in synchronous coordinates (d.c. quantities) are: (8.126) Electric Drives

39
**8.14. MOTORING, GENERATING, A.C. BRAKING**

The equivalent circuit for steady state (figure 8.11) shows that the active power in the rotor (the electromagnetic power Pe), is: (8.127) This expression is valid for the cage rotor (Vrb = 0). Motoring mode is defined as the situation when the torque has the same sign as the speed: (8.128) For generating the torque is negative (S<0) in (8.127) but the speed is positive: (8.129) The generator produces braking (Te < 0, wr > 0) but the energy transfer direction in the motor is reversed. The energy is pumped back into the power source through the stator. Electric Drives

40
**We may synthesize the results on operation modes as in table 8.1. **

Braking is obtained when again (Te > 0, wr < 0) (or Te < 0, wr > 0) but still the electromagnetic power is positive: (8.130) We may synthesize the results on operation modes as in table 8.1. Table 8.1. Operation modes (cage rotor) Electric Drives

41
**The torque slip (speed) curve is shown in figure 8.14.**

(8.131) (8.132) The torque slip (speed) curve is shown in figure 8.14. Figure Torque - speed curve of induction motors for constant voltage and frequency Electric Drives

42
**8.15. D.C. BRAKING: ZERO BRAKING TORQUE AT ZERO SPEED**

For moderate braking requirements d.c. braking is commonly used in modern electrical drives. To calculate the d.c. braking torque we redraw the space phasor equivalent circuit in figure 8.11, but with a d.c. stator current source space phasor in figure 8.15. Figure Equivalent circuit for d.c. braking - stator coordinates - in space phasors; steady state Electric Drives

43
**The electromagnetic torque is still computed from (8.94): (8.134)**

with (8.135) Finally: (8.136) The peak torque is obtained for wrk: (8.137) and its value is: (8.138) Electric Drives

44
**Figure 8.16. D.c. braking torque of induction motors**

(8.139) The torque speed curve for d.c. braking is shown in figure Notice also that the rotor kinetic energy is dumped into the rotor resistor and that for zero speed the braking torque is zero. Also above wrk (which is fairly small in high efficiency motors) the torque is again rather small. Figure D.c. braking torque of induction motors Electric Drives

45
8.16. SPEED CONTROL METHODS Variable speed is required in many applications. It has to be performed at high energy conversion rates. The no load ideal speed wr0 is ((8.100) with (8.99)): (8.140) Evidently for the cage rotor = 0 and thus or (8.141) w2 is the frequency of the rotor current (or of rotor P.E.C.). There are three essential methods to vary speed by varying the no load ideal speed (as the rated slip is small) as suggested by (8.140): stator frequency f1 variation; pole number (2p) changing; wound rotor supply (or rotor frequency f2 variation). Electric Drives

46
**constant (controlled) rotor flux (lr) vector control; **

Stator frequency control is far more frequently used especially for wide speed control range. However the level of flux depends on the current in the machine. Especially on the magnetisation current (8.142) Consequently it is crucial to control Im properly to avoid excessive magnetic saturation, while varying frequency f1. V1/f1 - scalar control; constant (controlled) rotor flux (lr) vector control; constant (controlled) stator flux (ls) - vector control. Here only the torque / speed curves obtainable with the above methods are given. Notice that in all these cases the P.E.C. voltage supplying the induction motor is voltage limited with the maximum voltage reached at base speed wb. Electric Drives

47
**8.17. V1/f1 TORQUE - SPEED CURVES**

V1/f1 control means that: (8.143) (8.144) Figure 8.17. Torque / speed curves for (V1/f1 control) a.) V1/f1 dependences b.) Te/wr curves Electric Drives

48
**8.18. ONLY FOR CONSTANT ROTOR FLUX TORQUE - SPEED CURVES ARE LINEAR**

(8.146) Figure Torque - speed curves for constant rotor flux lrb = ct. up to base frequency w1b; constant voltage and variable frequency above w1b Electric Drives

49
**8.19. CONSTANT STATOR FLUX TORQUE - SPEED CURVES HAVE**

TWO BREAKDOWN POINTS (8.147) (8.148) This expression has extreme (critical) values for: (8.149) (8.150) Electric Drives

50
(8.151) Figure Torque - speed curves for constant stator flux amplitude ls up to w1b and constant voltage above w1b Electric Drives

Similar presentations

Presentation is loading. Please wait....

OK

Lesson 12a: Three Phase Induction Motors

Lesson 12a: Three Phase Induction Motors

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on introduction to object-oriented programming language Ppt online exam form Ppt on bionics medical Ppt on south african culture and food Download ppt on data handling class 8 Ppt on history of badminton game Ppt on resistance temperature detector probe Ppt on principles of object-oriented programming in c++ Ppt on international financial markets Ppt on community based disaster management for class 9