Author: Ion Boldea, S.A.Nasar………… ……… Chapter 7 STEADY STATE EQUIVALENT CIRCUIT AND PERFORMANCE 7.1 BASIC STEADY-STATE EQUIVALENT CIRCUIT When the IM is fed in the stator from a three-phase balanced power source, the three-phase currents produce a traveling field in the airgap. This field produces emfs both in the stator and in the rotor windings: E 1 and E 2s . A symmetrical cage in the rotor may be reduced to an equivalent three-phase winding. The frequency of E 2s is f 2 . 111 1 1 1 1 1 112 Sfpn n nn pn p f npff = − =         −=−= (7.1) This is so because the stator mmf travels around the airgap with a speed n 1 = f 1 /p 1 while the rotor travels with a speed of n. Consequently, the relative speed of the mmf wave with respect to rotor conductors is (n 1 – n) and thus the rotor frequency f 2 in (7.1) is obtained. Now the emf in the short-circuited rotor “acts upon” the rotor resistance R r and leakage inductance L rl : () r rl1r 2s2 ILjSRESE ω+== (7.2) r rl1 r 2 s2 ILj S R E S E       ω+== (7.3) If Equation (7.2) includes variables at rotor frequency Sω 1 , with the rotor in motion, Equation (7.3) refers to a circuit at stator frequency ω 1 , that is with a “fictious” rotor at standstill. Now after reducing E 2 , I r , R r ,and L rl to the stator by the procedure shown in Chapter 6, Equation (7.3) yields 'I'Lj1 S 1 'R'RE'E r rl1rr 12       ω+       −+== (7.4) rotors cagefor 2/1 W,1K ;K WK WK E E 2w2E 11w 22w 1 2 ==== © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… rotors cagefor Nm ,2/1 W;K WKm WKm 'I I r22I 22w2 11w1 r r ==== (7.5) I E rl rl r r K K 'L L 'R R == (7.6) skew2y2q2w1y1q1w KKKK ;KKK ⋅⋅=⋅= (7.7) W 1 , W 2 are turns per phase (or per current path) K w1 , K w2 are winding factors for the fundamental mmf waves m 1 , m 2 are the numbers of stator and rotor phases, N r is the number of rotor slots The stator phase equation is easily written: () sl1s s s 1 LjRIVE ω+−=− (7.8) because in addition to the emf, there is only the stator resistance and leakage inductance voltage drop. Finally, as there is current (mmf) in the rotor, the emf E 1 is produced concurrently by the two mmfs (I s , I r ′). () 'IILj dt d E rs m11 m1 1 +ω−= ψ −= (7.9) If the rotor is not short-circuited, Equation (7.4) becomes 'I'Lj1 S 1 'R'R S 'V E r rl1rr r 1       ω+       −+=− (7.10) The division of V r (rotor applied voltage) by slip (S) comes into place as the derivation of (7.10) starts in (7.2) where () r rl1r r 2 ILjSRVES ω+=− (7.11) The rotor circuit is considered as a source, while the stator circuit is a sink. Now Equations (7.8) – (7.11) constitute the IM equations per phase reduced to the stator for the rotor circuit. Notice that in these equations there is only one frequency, the stator frequency ω 1 , which means that they refer to an equivalent rotor at standstill, but with an additional “virtual” rotor resistance per phase R r (1/S−1) dependent on slip (speed). It is now evident that the active power in this additional resistance is in fact the electro-mechanical power of the actual motor () 2 r rem 'I1 S 1 'R3n2TP       −=π⋅= (7.12) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… with () S1 p f n 1 1 −= (7.13) () elm 2 r r e 1 1 P S 'I'R3 T p == ω (7.14) P elm is called the electromagnetic power, the active power which crosses the airgap, from stator to rotor for motoring and vice versa for generating. Equation (7.14) provides a definition of slip which is very useful for design purposes: () elm Cor elm 2 r r P P P 'I'R3 S == (7.15) Equation (7.15) signifies that, for a given electromagnetic power P elm (or torque, for given frequency), the slip is proportional to rotor winding losses. I s V s E 1 R s j L ω 1sl E 1 R’ r j L’ ω 1rl R’ r (1-S) S V S r I s V s R s j L =jX ω 1sl sl R 1m I 0a j L =jX ω 11m 1m I m R’ r R’ r (1-S) S V S r + - j L’ =jX’ ω 1rl rl E =+j L ( + ’)=j L ωω II I 111msr1m1m a.) b.) Figure 7.1 The equivalent circuit © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… Equations (7.8) – (7.11) lead progressively to the ideal equivalent circuit in Figure 7.1. Still missing in Figure 7.1a are the parameters to account for core losses, additional losses (in the cores and windings due to harmonics), and the mechanical losses. The additional losses P ad will be left out and considered separately in Chapter 11 as they amount, in general, to up to 3% of rated power in well- designed IM. The mechanical and fundamental core losses may be combined in a resistance R 1m in parallel with X 1m in Figure 7.1b, as at least core losses are produced by the main path flux (and magnetization current I m ). R 1m may also be combined as a resistance in series with X 1m , for convenience in constant frequency IMs. For variable frequency IMs, however, the parallel resistance R 1m varies only slightly with frequency as the power in it (mainly core losses) is proportional to E 1 2 = ω 1 2 K w1 2 Φ 1 2 , which is consistent to eddy current core loss variation with frequency and flux squared. R 1m may be calculated in the design stage or may be found through standard measurements. moa iron 2 m 2 m1 iron 2 1 m1 II ; P IX3 P E3 R <<== (7.16) 7.2 CLASSIFICATION OF OPERATION MODES The electromagnetic (active) power crossing the airgap P elm (7.14) is positive for S > 0 and negative for S < 0. That is, for S < 0, the electromagnetic power flows from the rotor to the stator. After covering the stator losses, the rest of it is sent back to the power source. For ω 1 > 0 (7.14) S < 0 means negative torque T e . Also, S < 0 means n > n 1 = f 1 /p 1 . For S > 1 from the slip definition, S = (n 1 – n)/n 1 , it means that either n < 0 and n 1 (f 1 ) > 0 or n > 0 and n 1 (f 1 ) < 0. In both cases, as S > 1 (S > 0), the electromagnetic power P elm > 0 and thus flows from the power source into the machine. On the other hand, with n > 0, n 1 (ω 1 ) < 0, the torque T e is negative; it is opposite to motion direction. That is braking. The same is true for n < 0 and n 1 (ω 1 ) > 0. In this case, the machine absorbs electric power through the stator and mechanical power from the shaft and transforms them into heat in the rotor circuit total resistances. Now for 0 < S < 1, T e > 0, 0 < n < n 1 , ω 1 > 0, the IM is motoring as the torque acts along the direction of motion. The above reasoning is summarized in Table 7.1. Positive ω 1 (f 1 ) means positive sequence-forward mmf traveling wave. For negative ω 1 (f 1 ), a companion table for reverse motion may be obtained. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… Table 7.1 Operation modes (f 1 /p 1 > 0) S - - - - 0 + + + + 1 + + + + n + + + + f 1 /p 1 + + + + 0 T e P elm 0 - - - - 0 0 + + + + + + + + + + + + + + + + 0 0 Operation mode Generator Motor Braking 7.3 IDEAL NO-LOAD OPERATION The ideal no-load operation mode corresponds to zero rotor current. From (7.11), for I r0 = 0 we obtain 2 R 0 R 2 0 E V S ;0VES ==− (7.17) The slip S 0 for ideal no-load depends on the value and phase of the rotor applied voltage V R . For V R in phase with E 2 : S 0 > 0 and, with them in opposite phase, S 0 < 0. The conventional ideal no – load-synchronism, for the short-circuited rotor (V R = 0) corresponds to S 0 = 0, n 0 = f 1 /p 1 . If the rotor windings (in a wound rotor) are supplied with a forward voltage sequence of adequate frequency f 2 = Sf 1 (f 1 > 0, f 2 > 0), subsynchronous operation (motoring and generating) may be obtained. If the rotor voltage sequence is changed, f 2 = sf 1 < 0 (f 1 > 0), supersynchronous operation may be obtained. This is the case of the doubly fed induction machine. For the time being we will deal, however, with the conventional ideal no-load (conventional synchronism) for which S 0 = 0. The equivalent circuit degenerates into the one in Figure 7.2a (rotor circuit is open). Building the phasor diagram (Figure 7.2b) starts with I m , continues with jX 1m I m , then I 0a m1 m m1 oa R IjX I = (7.18) and moas0 III += (7.19) Finally, the stator phase voltage V s (Figure 7.2b) is 0s sl s0 s m m1 s IjXIRIjXV ++= (7.20) The input (active) power P s0 is dissipated into electromagnetic loss, fundamental and harmonics stator core losses, and stator windings and space harmonics caused rotor core and cage losses. The driving motor covers the mechanical losses and whatever losses would occur in the rotor core and squirrel cage due to space harmonics fields and hysteresis. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… I s0 V s R s jX sl R 1m I 0a I m jX 1m a.) jX 1m I m V s jX sl I s0 R s I s0 I 0a I m I s0 ϕ oi b.) P s0 p iron Co p c.) synchronous motor 2p poles 3~ f 1 Power analyser VA R I A C 3~ f 1 d.) n=f /p 11 2p poles 1 1 Figure 7.2 Ideal no-load operation (V R = 0): a.) equivalent circuit b.) phasor diagram c.) power balance d.) test rig For the time being, when doing the measurements, we will consider only stator core and winding losses. 2 0ssiron 2 0ss 2 m1 2 m1 2 m1 m1 2 0ss 2 a0m10s IR3p IR3 RX X R3IR3IR3P += =         + + =+≈ (7.21) From d.c. measurements, we may determine the stator resistance R s . Then, from (7.21), with P s0 , I s0 measured with a power analyzer, we may calculate the iron losses p iron for given stator voltage V s and frequency f 1 . We may repeat the testing for variable f 1 and V 1 (or variable V 1 /f 1 ) to determine the core loss dependence on frequency and voltage. The input reactive power is 2 0ssl 2 m1 2 m1 2 m1 m10s IX3 RX R X3Q         + + = (7.22) From (7.21)-(7.22), with R s known, Q s0 , I s0 , P s0 measured, we may calculate only two out of the three unknowns (parameters): X 1m , R 1m , X sl . We know that R 1m >> X 1m >> X sl . However, X sl may be taken by the design value, or the value measured with the rotor out or half the stall rotor (S = 1) reactance X sc , as shown later in this chapter. Consequently, X 1m and R 1m may be found with good precision from the ideal no-load test (Figure 7.2d). Unfortunately, a synchronous motor with the © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… same number of poles is needed to provide driving at synchronism. This is why the no-load motoring operation mode has become standard for industrial use, while the ideal no-load test is mainly used for prototyping. Example 7.1 Ideal no-load parameters An induction motor driven at synchronism (n = n 1 = 1800rpm, f 1 = 60Hz, p 1 = 2) is fed at rated voltage V 1 = 240 V (phase RMS) and draws a phase current I s0 = 3 A, the power analyzer measures P s0 = 36 W, Q s0 = 700 VAR, the stator resistance R s = 0.1 Ω, X sl = 0.3 Ω. Let us calculate the core loss p iron , X 1m , R 1m . Solution From (7.21), the core loss p iron is W3.3331.0336IR3Pp 2 2 0ss0siron =⋅⋅−=−= (7.23) Now, from (7.21) and (7.22), we get Ω== ⋅ ⋅⋅− = − = + 233.1 27 3.33 33 31.0336 I3 IR3P RX XR 2 2 2 0s 2 0ss0s 2 m1 2 m1 2 m1m1 (7.24) Ω= ⋅ ⋅⋅− = − = + 626.25 33 33.03700 I3 IX3Q RX XR 2 2 2 0s 2 0ssl0s 2 m1 2 m1 m1 2 m1 (7.25) Dividing (7.25) by (7.26) we get 78.20 233.1 626.25 X R m1 m1 == (7.26) From (7.25), () 685.25X626.25 178.20 X ;626.25 1 R X X m1 2 m1 2 m1 m1 m1 =⇒= + = +         (7.27) R 1m is calculated from (7.26), Ω=⋅=⋅= 74.53378.20626.2578.20XR m1m1 (7.28) By doing experiments for various frequencies f 1 (and V s /f 1 ratios), the slight dependence of R 1m on frequency f1 and on the magnetization current I m may be proved. As a bonus, the power factor cos ϕ oi may be obtained as 05136.0 P Q tancoscos 0s 0s 1 i0 =                 =ϕ − (7.29) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… The lower the power factor at ideal no-load, the lower the core loss in the machine (the winding losses are low in this case). In general, when the machine is driven under load, the value of emf (E 1 = X 1m I m ) does not vary notably up to rated load and thus the core loss found from ideal no-load testing may be used for assessing performance during loading, through the loss segregation procedure. Note however that, on load, additional losses, produced by space field harmonics,occur. For a precise efficiency computation, these “stray load losses” have to be added to the core loss measured under ideal no-load or even for no-load motoring operation. 7.4 SHORT-CIRCUIT (ZERO SPEED) OPERATION At start, the IM speed is zero (S = 1), but the electromagnetic torque is positive (Table 7.1), so when three-phase fed, the IM tends to start (rotate); to prevent this, the rotor has to be stalled. First, we adapt the equivalent circuit by letting S = 1 and R r ′ and X sl , X rl ′ be replaced by their values as affected by skin effect and magnetic saturation (mainly leakage saturation as shown in Chapter 6): X slstart , R′ rstart , X′ rlstart (Figure 7.3). I sc V ssc R s jX slstart R 1m 1m I m R’ rstart jX I’ rsc jX’ rlstart V ssc R =R +R’ sc sc jX srstart sc X =X +X’ slstart rlstart a.) b.) I sc mandatory at low frequency © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… P sc p iron (very small) p =3(R +R’ )I Cosc c.) R I sc sc sc j X I sc V ssc ϕ sc I sc d. ) srstart sc 2 Three phase power analyser VA R I A C 3~ f 1 e.) a b c n=0 S=1 Z - impedance per phase sc Single phase power analyser VA R I A C 1~ f 1 f.) a b c n=0 S=1 Z - impedance per phase sc 3 2 Figure 7.3 Short-circuit (zero speed) operation: a.) complete equivalent circuit at S = 1, b.) simplified equivalent circuit S = 1, c.) power balance, d.) phasor diagram, e.) three-phase zero speed testing, f.) single-phase supply zero speed testing For standard frequencies of 50(60) Hz, and above, X 1m >> R′ rstart . Also, X 1m >> X′ rlstart , so it is acceptable to neglect it in the classical short-circuit equivalent circuit (Figure 7.3b). For low frequencies, however, this is not so; that is, X 1m <> R ′ rstart , so the complete equivalent circuit (Figure 7.3a) is mandatory. The power balance and the phasor diagram (for the simplified circuit of Figure 7.3b) are shown in Figure 7.3c and d. The test rigs for three-phase and single-phase supply testing are presented in Figure 7.3e and f. It is evident that for single-phase supply, there is no starting torque as we have a non-traveling stator mmf aligned to phase a. Consequently, no rotor stalling is required. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… The equivalent impedance is now (3/2)Z sc because phase a is in series with phases b and c in parallel. The simplified equivalent circuit (Figure 7.3b) may be used consistently for supply frequencies above 50(60) Hz. For lower frequencies, the complete equivalent circuit has to be known. Still, the core loss may be neglected (R 1m ≈ ∞ ) but, from ideal no-load test at various voltages, we have to have L 1m (I m ) function. A rather cumbersome iterative (optimization) procedure is then required to find R′ rstart , X′ rlstart , and X slstart with only two equations from measurements of R sc , V sc /I sc . 2 rscrstart 2 scssc 'I'R3IR3P += (7.30) () () rlstartm1rlstart rlstartrlstartm1 slstarts sc 'XXj'R 'jX'RjX jXRZ ++ + ++= (7.31) This particular case, so typical with variable frequency IM supplies, will not be pursued further. For frequencies above 50(60) Hz the short-circuit impedance is simply rlstartslstartscrlstartsscscsc sc 'XXX ;'RRR ;jXRZ +=+=+≈ (7.32) and with P sc , V ssc , I sc measured, we may calculate 2 sc 2 sc ssc sc 2 sc sc sc R I V X ; I3 P R −         == (7.33) for three phase zero speed testing and 2 sc 2 ~sc ~ssc sc 2 ~sc ~sc sc R 2 3 I V 3 2 X ; I P 3 2 R       −         == (7.34) for single phase zero speed testing. If the test is done at rated voltage, the starting current I start (I sc ) Vsn is much larger than the rated current, 0.85.4 I I n start ÷≈ (7.35) for cage rotors, larger for high efficiency motors, and 1210 I I n start ÷≈ (7.36) for short-circuited rotor windings. The starting torque T es is: © 2002 by CRC Press LLC [...]... is applied to the shaft, the IM works on no-load In terms of energy conversion, the IM input power has to cover the core, winding, and mechanical losses The IM has to develop some torque to cover the mechanical losses So there are some currents in the rotor However, they tend to be small and, consequently, they are usually neglected The equivalent circuit for this case is similar to the case of ideal... stator (Peml < 0) and thus, after covering the stator losses, the rest of it is sent back to the power grid As expected, the machine has to be driven at the shaft at a speed n > f1/p1 as the electromagnetic torque Te (and Pelm) is negative (Figure 7.8) Driving motor n>f1/p1 f1 Induction generator 3~ power grid Figure 7.8 Induction generator at power grid The driving motor could be a wind turbine, a... an a.c emf in the stator phases Then three-phase emfs of frequency f1 = p1⋅n cause currents to flow in the stator phases and capacitors Their phase angle is such that they are producing an airgap field that always increases the remanent field; then again, this field produces a higher stator emf and so on until the machine settles at a certain voltage Vs and frequency f1 ≈ p1n Changing the speed will... explore the principle of IG capacitor excitation on no-load The machine is driven at a speed n n>f1/p1 3~ f1 Driving motor 3 phase load 2p1 poles C∆ Figure 7.12 Autonomous induction generator (IG) with capacitor magnetization The d.c remanent magnetization in the rotor, if any (if none, d.c magnetization may be provided as a few d.c current surges through the stator with one phase in series with the other... reactive power to produce and maintain the magnetic field in the machine As known, this reactive power may be “produced” with synchronous condensers (or capacitors)–Figure 7.12 The capacitors are ∆ connected to reduce their capacitance as they are supplied by line voltage Now the voltage Vs and frequency f1 of the IG on noload and on load depend essentially on machine parameters, capacitors C∆, and... reactances in the machine The equivalent circuit degenerates into that in Figure 7.13 The presence of rotor remanent flux density (from prior action) is depicted by the resultant small emf Erem (Erem = 2 to 4 V) whose frequency is f1 = np1 The frequency f1 is essentially imposed by speed The machine equation becomes simply Vs 0 = jX1m I m + E rem = − j 1 I m = Vs 0 (I m ) ω1C Y (7.59) As we know, the magnetization... application software such Matlab, etc Above a certain load, the machine voltage drops gradually to zero as there will be a deficit of capacitor energy to produce machine magnetization Point A on the magnetization curve will drop gradually to zero As the load increases, the slip (negative) increases and, for given speed n, the frequency decreases So the IG can produce power above a certain level of magnetic... regulate the voltage in the power grid Example 7.4 Generator at power grid A squirrel cage IM with the parameters Rs = Rr′ = 0.6 Ω, Xsl = Xrl′ = 2 Ω, X1ms = 60 Ω, and R1ms = 3 Ω (the equivalent series resistance to cover the core losses, instead of a parallel one)–Figure 7.11 – works as a generator at the power grid Let us find the two slip values S0g1 and S0g2 between which it delivers power to the grid... and in the rotor In general, due to difficulties in computation, the stray load losses are still assigned a constant value in some standards (0.5 or 1% of rated power) More on stray losses in Chapter 11 The slip definition (7.15) is a bit confusing as Pelm is defined as active power crossing the airgap As the stray load losses occur both in the stator and rotor, part of them should be counted in the stator... P2 electric + p Cos + piron + p Cor + ps + p mec Above the speed n max = ( ) f1 1 − S0 g 2 ; S0 g 2 < 0 , P1 (7.54) as evident in Figure 7.9, the IM remains in the generator mode but all the electric power produced is dissipated as loss in the machine itself Induction generators are used more and more for industrial generation to produce part of the plant energy at convenient timing and costs However, . The lower the power factor at ideal no-load, the lower the core loss in the machine (the winding losses are low in this case). In general, when the machine. reduced to the stator for the rotor circuit. Notice that in these equations there is only one frequency, the stator frequency ω 1 , which means that they