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Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998.

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Presentation on theme: "Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998."— Presentation transcript:

1 Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998

2 Electric Drives2 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)

3 Electric Drives3 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)

4 Electric Drives THE CAGE AND WOUND ROTORS ARE EQUIVALENT Figure 8.3. Induction motor rotors a.) cage - type rotor b.) wound rotor

5 Electric Drives 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)

6 Electric Drives6 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

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

8 Electric Drives8 The inductances may be now assembled into the so called matrix inductance : (8.18) where:

9 Electric Drives 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.

10 Electric Drives THE PHASE COORDINATE MODEL GOES TO 8 th ORDER (8.23) (8.24) (8.25) (8.26) (8.27) Using (8.24) in (8.23) - with  er variable in time in any case - we obtain: (8.28) where: (8.29)

11 Electric Drives11 (8.30) (8.31) (8.32) The motion equation are: (8.33)

12 Electric Drives THE SPACE PHASOR MODEL (8.34) (8.35) (8.36) (8.37) (8.38)

13 Electric Drives13 (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, b r and c, c r equation (8.23) becomes: (8.44) (8.45) (8.46)

14 Electric Drives14 (8.48) (8.49) (8.50) (8.51) (8.52) (8.53) The torque should be calculated from (8.42) with the above notations:

15 Electric Drives15 We may now decompose in plane the space phasors along two orthogonal axes d and q moving at speed  b [5]: (8.55) (8.56)

16 Electric Drives16 Also from (8.49) - (8.50) and (8.47): (8.57) [P(  b )] is the Park transformation: (8.58) The inverse of Park transformation is: (8.59)

17 Electric Drives17 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)

18 Electric Drives18 The position of the space phasor for is shown in figure 8.7. Figure 8.7. The space phasor of sinusoidal three phase currents

19 Electric Drives 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  b =  1 (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: )

20 Electric Drives20 d/dt = j(  1 -  b ).(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)

21 Electric Drives ELECTRICAL TRANSIENTS WITH FLUX LINKAGES AS VARIABLES (8.69) (8.70) (8.71) equations (8.51) - (8.53) become: (8.72) (8.73)

22 Electric Drives22 with (8.74) While  s,  r are the stator and rotor time constants,  s ’ and  r ’ 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 (  b ) coordinates.

23 Electric Drives 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  b and on rotor speed  r, but their real part is, in general, negative suggesting attenuated periodic response.[4,6] Example 8.2. For an induction motor with the data r s = 0.5 , r r = 0.60 , L s = L r = 0.08H, L m = 0.075H, p = 2pole pairs calculate the complex eigen values for constant speed, in synchronous coordinates (  1 = 2  60) for n = 0 and n = 1800 rpm.

24 Electric Drives24 Solution: The value of  b =  1 = 2  60 rad/s. ,  s ’,  r ’, K s, K r are now calculated from (8.71) - (8.74): (8.76) (8.77) Also(8.78)

25 Electric Drives25 Now we rewrite (8.75) in a canonical form: (8.79) (8.80) For n = 0 (8.81) For n = 30rps (1800rpm): (8.82)

26 Electric Drives 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).

27 Electric Drives27 Figure Structural diagram of induction motor with constant rotor flux and speed in synchronous coordinates (  b =  1 )

28 Electric Drives 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)

29 Electric Drives29 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.

30 Electric Drives30 Figure Space phasor steady state equivalent circuit of induction machine The torque expression (8.54) with (8.91) and V r0 b = 0, yields: (8.94)

31 Electric Drives31 From (8.92): (8.95) Consequently the electromagnetic torque T e 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)

32 Electric Drives NO LOAD IDEAL SPEED MAY GO UNDER OR OVER CONVENTIONAL VALUE  1 The no load ideal speed (slip S 0 ) 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 S 0 = 0 and  ro =  1. 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  1 and S = The motor data are: r s = 0.5 , r r = 0.6 , L s = L r = 0.08H, L m = 0.075H, r0 = 0.8Wb,  1 = 2  60 rad/s, p = 2 pole pairs, V r b = 0 (cage rotor).

33 Electric Drives33 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 (T e ) is, (8.96),: (8.103) From (8.72) - (8.73) with d/dt = 0 and V r b = 0 we may now calculate the stator flux : (8.104) Notice that the motor parameters are as in example 8.2 and thus  r ’ = s,  s ’ = s, K s = K r = ,  = (8.105)

34 Electric Drives34 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)

35 Electric Drives35 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 The power factor angle  1 is: (8.110) Finally cos  1 = Figure Steady state space phasor diagram in synchronous coordinates (  b =  1 ): d.c. quantities

36 Electric Drives36 Example 8.4. Loss breakdown A high efficiency induction motor with cage rotor has the data: rated power P n = 5kW, rated line voltage (rms) V L = 220V (star connection), rated frequency f 1 = 60Hz, number of pole pairs p = 2, core loss (P iron ) = mechanical loss P mec = 1.5% of P n, additional losses p add = 1%P n, stator per rotor winding losses p cor /p cos = 2/3, rated efficiency  n = 0.9 and power factor cos  n = 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 Figure Induction motor energy conversion

37 Electric Drives37 The input power P in is: (8.111) The phase current I n (rms) is: (8.112) The total losses are: (8.113) Consequently: (8.114) (8.115) So (8.116) (8.117) (8.118)

38 Electric Drives38 The electromagnetic power P e is the active power that crosses the airgap: (8.119) (8.120) (8.121) The rotor winding loss p cor is:(8.122) and thus the rated slip is: (8.123) The rated speed n n is: (8.124) The shaft torque T n is calculated directly from mechanical power P n : (8.125) Finally the amplitude of the stator current, in synchronous coordinates (d.c. quantities) are:(8.126)

39 Electric Drives 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 P e ), is: (8.127) This expression is valid for the cage rotor (V r b = 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 (T e 0) but the energy transfer direction in the motor is reversed. The energy is pumped back into the power source through the stator.

40 Electric Drives40 Braking is obtained when again (T e > 0,  r 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)

41 Electric Drives41 (8.131) (8.132) The torque slip (speed) curve is shown in figure Figure Torque - speed curve of induction motors for constant voltage and frequency

42 Electric Drives 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 Figure Equivalent circuit for d.c. braking - stator coordinates - in space phasors; steady state

43 Electric Drives43 The electromagnetic torque is still computed from (8.94): (8.134) with (8.135) Finally: (8.136) The peak torque is obtained for  rk : (8.137) and its value is: (8.138)

44 Electric Drives44 (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  rk (which is fairly small in high efficiency motors) the torque is again rather small. Figure D.c. braking torque of induction motors

45 Electric Drives 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  r0 is ((8.100) with (8.99)): (8.140) Evidently for the cage rotor = 0 and thus or (8.141)  2 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 f 1 variation;  pole number (2p) changing;  wound rotor supply (or rotor frequency f 2 variation).

46 Electric Drives46 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 I m properly to avoid excessive magnetic saturation, while varying frequency f 1.  V 1 /f 1 - scalar control;  constant (controlled) rotor flux ( r ) vector control;  constant (controlled) stator flux ( s ) - 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  b.

47 Electric Drives V 1 /f 1 TORQUE - SPEED CURVES V 1 /f 1 control means that: (8.143) (8.144) Figure Torque / speed curves for (V 1 /f 1 control) a.) V 1 /f 1 dependences b.) T e /  r curves

48 Electric Drives ONLY FOR CONSTANT ROTOR FLUX TORQUE - SPEED CURVES ARE LINEAR (8.146) Figure Torque - speed curves for constant rotor flux r b = ct. up to base frequency  1b ; constant voltage and variable frequency above  1b

49 Electric Drives 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)

50 Electric Drives50 (8.151) Figure Torque - speed curves for constant stator flux amplitude s up to  1b and constant voltage above  1b


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