Analysis of Three-Phase Induction Motors: Power and Torque Insights

THREE-PHASE INDUCTION MOTORS Analyze The Power And Torque

Development of IM Equivalent Circuit When a three-phase induction motor is excited by a balanced three-phase source, the currents in the phase windings must be equal in magnitude and 120o electrical apart in phase. The same must be true for the currents in the rotor windings as the energy is transferred across the air-gap from the stator to the rotor by induction. The frequency of the induced emf in the rotor is proportional to its slip [ fr = s fe]. Balanced 3-phase source Time-domain Phasor-domain 2

Development of IM Equivalent Circuit This induction is essentially a transformer operation, hence the equivalent circuit of a three-phase induction motor is similar to the equivalent circuitof a three-phase transformer. A transformer per-phase equivalent circuit, representing the operation ofan induction motor is shown in Figure 1-1 below: Figure 1-1 Per-phase equivalent circuit of a balanced three-phase induction motor. Also, the stator and the rotor windings are coupled inductively, thenan induction motor resembles a three-phase transformer with a rotating secondary winding. The similarity becomes even more striking when the rotor is at rest, s =1. 3

Development of IM Equivalent Circuit As in any transformer, there is certain resistance and self-inductance inthe primary (stator) windings, which must be represented in the equivalent circuit of the machine. They are R1 stator resistance and X1 stator leakage reactance. Also, like any transformer with an iron core, the flux in the machine is related to the integral of the applied voltage E1. The curve of mmf vs flux (magnetization curve) for this machine is compared to a similar curve fora transformer, as shown in Figure 1-2: The slope of the induction motor s mmf-flux curve is much lower than the curve of agood transformer. This is because there must be an air gap in an induction motor, which greatly increases the reluctance of the flux path and thus reduces the coupling between stator(primary) and rotor (secondary) windings. Figure 1-2 Magnetizing curve of a three-phase induction motor. 4

Development of IM Equivalent Circuit The higher reluctance caused by the air gap means that a higher magnetizing current is required to obtain a given flux level. Therefore, the magnetizing reactance Xm in the equivalent circuit will have a much smaller value than it would in a transformer. The primary internal stator voltage is E1 is coupled to the secondary Er by an ideal transformer with an effective turns ratio a. The turns ratio for a wound rotor is basically the ratio of the conductors per phase on the stator to the conductors per phase on the rotor. It is rather difficult tosee a clearly in the cage rotor because there are no distinct windings on the cage rotor. Erin the rotor produces current flow in the shorted rotor (or secondary) circuit of the machine. Figure1-1 The primary impedances and the magnetization current of theinduction motor are very similar to the corresponding components in atransformer equivalent circuit. 5

Development of IM Equivalent Circuit In this figure, V1= applied voltage on a per-phase basis R1= per-phase stator winding resistance L1 = per-phase stator winding leakage inductance X1= 2 f L1 = per-phase stator winding leakagereactance Rr = per-phase rotor winding resistance Lb = per-phase rotor winding leakage inductance Xb Figure1-1 = 2 f Lb = per-phase rotor winding leakage reactance under blocked-rotorcondition (s = 1) Xr= 2 sf Lb = sXb = per-phase rotor winding leakage reactance at slip s. Xm = per-phase magnetizationreactance Rc = per-phase equivalent core-loss resistance N1 = actual turns per phase of the stator winding N2 = actual turns per phase of the rotor winding 6

Development of IM Equivalent Circuit From the per-phase equivalent circuit (Figure 1-1), it is evident that the current in the rotor circuit is E~ E sE = b = r = b Figure1-1 I (R / s) + jX r r R + jX R + j s X b r r r b Based upon the above equation, we can develop another circuit ofan induction motor as given in Figure 1-3. Figure 1-3 Modified equivalent circuit of a balanced three-phase motor on a per-phase basis. 7

Development of IM Equivalent Circuit The resistance Rr/s in the rotor circuit is called the effectiveresistance. The effective resistance is the same as the actual rotor resistancewhen the rotor is at rest (standstill or blocked-rotor condition; s =1). On the other hand, when the slip approaches zero underno-load condition, the effective resistance is very high (Rr/s ). By defining the ratio of transformation, thea-ratio, as a =N1k 1 N2k 2 we can represent the induction motor by its per-phase equivalent circuitas referred to the stator. Such an equivalent circuit is shown in Figure1-4, where R = a2R 2 r X 2= a Xb 2 I~ a ~ 2I = r Figure 1- 4 Per-phase equivalent circuit of a balanced three- phase induction motor as referred to the stator side. 8

Development of IM Equivalent Circuit For this equivalent circuit I~R 1 s I~= I~ + I~ c ~ ~ E1 Ic =R ~ E = ~ + jI X 2 2 2 2 m Figure1-4 E jX = 1 I where , and m m c The per-phase stator winding current and the applied voltageare I~ = I~+ I~ 1 2 and V~ = E~+ I~(R + j X ) 1 1 1 1 1 The equivalent circuit of the rotor in Figure 1-4 is in terms ofthe hypothetical resistance R2/s. 9

Development of IM Equivalent Circuit From Power perspective: In this circuit, 2 I R / s However, the per-phase copper loss in the rotor must be I 2 R. 2 represents the per-phase power delivered to therotor. 2 2 2 Thus, the per-phase power developed by the motoris 1 s s = R + R 1 s R2 R2 s I R = I R 2I2 2 2 2 2 or 2 2 2 2 s s The above equation establishes the fact that the hypotheticalresistance R2/s can be divided into two series components: the actual resistance of the rotor R2and an additional resistance R2[(1 - s)/s]. R2/s R2 R2[(1 - s)/s] The additional resistance R2[(1 - s)/s] is called the load resistance orthe dynamic resistance. 10

Development of IM Equivalent Circuit The load resistance depends upon the speed of the motor and is said to represent the load on the motor because the mechanical power developed by the motor is proportional to it. In other words, the load resistance is the electrical equivalent ofa mechanical load on the motor. An equivalent circuit of an induction motor in terms of the loadresistance is given in Figure 1-5. This circuit is proclaimed as the exact equivalent circuit of balanced three- phase induction motor on a per-phasebasis. Figure 1-5 The equivalent circuit of Figure 1-4 modified to show the rotor and the load resistance. 11

Power Relations For 3-Phase IM Since the load resistance varies with the slip and the slip adjusts itself to the mechanical load on the motor, the power delivered to the load resistance is equivalent to the power developed by the motor. Thus, the performance of the motor at any slip can be determined fromits equivalent circuit, as given in Figure 1-5. Figure1-4 For a balanced three-phase induction motor the total input poweris Pin= 3V1 I1 cos V1 Power factor: pf = cos( v- i) pf = cos( ) where is the phase difference between the applied voltage and the stator winding current. I1 12

Power Relations For 3-Phase IM Since the power input is electrical in nature, we must account for the electrical losses (i.e. copper losses) first. The immediate electrical loss that must be taken into consideration is the stator copper loss. The total stator copper lossis P = 3I 2R scl 1 1 If the core loss is modeled by an equivalent core-loss resistance Rc, as shown in the figure, we must also take into account the total core loss (magnetic loss) as P = 3I2R or P = 3E2 /R m c c 1 m c The net power that is crossing the air-gap and is transported to the rotor by electromagnetic induction is called the air-gap power. In this case,the air-gap power is Pag Pscl = Pin Pm The air-gap power must also equal the power delivered tothe hypothetical resistance R2/s. Thatis, Pag= 3I2 R2 s 2 13

Power Relations For 3-Phase IM The electrical power loss in the rotor circuit (total rotor copper loss)is P = 3I2 R = sP rcl 2 Hence, the power developed by the motoris 2 ag 3I2(1 s)R = 22 Pd= Pag Prcl = (1 s)Pag s The electromagnetic torque developed by the motoris =(1 s)Pag=Pag T m By subtracting the rotational loss from the power developed, we obtain the power output of the motora Po = Pd Pr =Pd R2 s =3I2 (1 s) s s 2 d s The load torque or the shaft torqueis T =Po o m 14

Power Relations For 3-Phase IM Since the core loss has already been accounted for, the rotational loss Pr includes the friction and windage loss Pf and the stray-load loss Pst. The corresponding power-flow diagram is given in Figure 1-6a. When the core loss Pm is also considered a part of the rotational loss, the core-loss resistance Rc in Figure 1-5 must be omitted, the air-gap power becomes Pag = Pin - Pscl. The power-flow diagramwhen the core loss is a part of the rotational loss is given in Figure 1-6b. Figure 1-6 Power-flow diagram when the core loss is (a) simulated by Rc, and (b) treated as a part of the rotational loss. 15