5 3i12 3 14 2 the no load reactance is 2 2 x nl z nl

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8.1432 − 2.52 = 17.97 Ω In the IEEE recommended equivalent circuit we assume that X 1 + X m = X NL = 17.97 Ω From the blocked-rotor test, the blocked-rotor resistance is: RBL = 7200 PBL = = 0.6667 Ω 2 3 * 60 2 3 I1 BL Note that in this example it does not mentioned the frequency of the blocked rotor test so we can use the rated frequency as the frequency of the blocked rotor test. The blocked-rotor impedance at frequency of blocked rotor test is: Z BL = V1 BL I1 BL = 163 / 3 = 1.5685 Ω 60 The blocked-rotor reactance is: 246 Chapter Four X BL = (Z 2 BL ) 2 − RBL = 1.56852 − 0.6667 2 = 1.42 Ω ′ X BL ≅ X 1 + X 2 = 1.42 Ω ′ Assume, X 1 = X 2 (at rated frequency) ′ then X 1 = X 2 = 0.71 Ω From no load test we know that X 1 + X m = X NL and X 1 = 0.71 Ω, then the magnetizing reactance is : X m = X NL − X 1 = 17.97 − 0.71 = 17.26 Ω Now R = RBL − R1 = 0.6667 − 0.25 = 0.4167 Ω. Form equation (4.43): 2 2 X′ + Xm 0.71 + 17.26 ′ R= R2 = 2 * 0.4167 = 0.4517 Ω Xm 17.26 Example 4.3 The following test results are obtained from a threephase 60 hp, 2200 V, six-pole, 60 Hz squirrel-cage induction motor. (1) No-load test: Supply frequency = 60 Hz, Line voltage = 2200 V Line current = 4.5 A, Input power = 1600 W (2) Blocked-rotor test: Frequency = 15 Hz, Line voltage = 270 V Line current = 25 A, Input power = 9000 W Three-Phase Induction Machine 247 (3) Average DC resistance per stator phase: R1 = 2.8 Ω (a) Determine the no-load rotational loss. (b) Determine the parameters of the IEEE-recommended equivalent circuit of Fig.4.9. (c) Determine the parameters (Vth , Rth , X th ) for the Thevenin equivalent circuit of Fig.4.10. Solution: (a) From the no-load test, the no-load power is: PNL = 1600 W The no-load rotational loss is: Prot = PNL − 3I12 R1 Prot = 1600 − 3 * 4.52 * 2.8 = 1429.9W (b) IEEE-recommended equivalent circuit. For the no-load ′ condition, R2 / s is very high. Therefore, in the equivalent circuit of Fig.4.9, the magnetizing reactance X m is shunted by a very high resistive branch representing the rotor circuit. The reactance of this parallel combination is almost the same as X m . Therefore the total reactance X NL , measured at no load at the stator terminals, is essentially X 1 + X m . The equivalent circuit at no load is shown in Fig.4.11a. 248 Chapter Four (a) No-load equivalent circuit (b) Blocked-rotor equivalent circuit. (c) Blocked-rotor equivalent circuit for improved value for R2 . Fig.4.13 V1 = 2200 = 1270.2 V / Phase 3 The no-load impedance is: Z NL = V1 1270.2 = = 282.27 Ω I1 4.5 The no-load resistance is: RNL = The no-load reactance is: PNL 1600 = = 26.34 Ω 3I12 3 * 4.52 249 Three-Phase Induction Machine 2 2 X NL = Z NL − RNL = 282.27 2 − 26.34 2 = 281Ω Thus, X 1 + X m = X NL = 281 Ω = 281.0 Ω. For the blocked-rotor test the slip is 1, the magnetizing reactance ′ ′ X m is shunted by the low-impedance branch R2 + jX 2 . Because ′ ′ X m >> R2 + jX 2 , the impedance X m can be neglected and the equivalent circuit for the blocked-rotor test reduces to the form shown in Fig.4.11b. From the blocked-rotor test, the blocked-rotor resistance is: RBL = 9000 PBL = = 4.8 Ω 3I12 3 * 252 ′ Therefore, R2 = RBL − R1 = 4.8 − 2.8 = 2Ω , The blocked-rotor impedance at 15 Hz is: Z BL = V1 270 = = 6.24 Ω I1 3 * 25 The blocked-rotor reactance at 15 Hz is X BL = (6.24 2 ) − 4.82 = 3.98 Ω Its value at 60 Hz is X BL = 3.98 * 60 = 15.92 Ω 15 ′ X BL ≅ X 1 + X 2 ′ Hence, X 1 = X 2 = 15.92 = 7.96 Ω (at 60 Hz) 2 The magnetizing reactance is therefore: 250 Chapter Four X m = 281 − 7.96 = 273.04 Ω Now R = RBL − R1 = 4.8 − 2.8 = 2 Ω . So, 2 7.96 + 273.04 ′ R2 = 2 = 2.12 Ω 273.04 (c) From Equation (4.31) Vth ≅ 273.04 V1 = 0.97 V1 7.96 + 273.04 From Equation (4.34) Rth ≅ 0.97 2 R1 = 0.97 2 * 2.8 = 2.63 Ω From Equation (4.35) X th ≅ X 1 = 7.96 Ω . 4.10 Performance Characteristics The equivalent circuits derived in the preceding section can be used to predict the performance characteristics of the induction machine. The important performance characteristics in the steady state are the efficiency, power factor, current, starting torque, maximum (or pull-out) torque, and so forth. The mechanical torque developed Tmech per phase is given by 2 Pmech = Tmechω mech = I 2 R2 (1 − s ) s (4.45) Three-Phase Induction Machine Where ω mech = 2πn 60 (4.46) ω mech is related to the synchronous speed The mechanical speed by: ω mech = (1 − s )ω syn ω mech = nsyn 60 and ω syn = 251 2π (1 − s ) 120 f 4π f1 * 2π = 60 P P (4.47) (4.48) (4.49) From Equations (4.45), (4.47), and (4.18) R2 = Pag s 2 Tmech ω syn = I 2 Then, Tmech = Tmech = Tmech = 1 ω syn 1 ω syn 1 ω syn 2 I2 ′ I 22 Pag (4.50) (4.51) R2 s (4.52) ′ R2 s (4.53) From the equivalent circuit of Fig.4.11 and Equation (4.53) Tmech 2 Vth R′ = * *2 ω syn (Rth + R2 / s )2 + ( X th + X 2 )2 s ′ ′ 1 (4.54) 252 Chapter Four Note that if the approximate equivalent c...
View Full Document

Ask a homework question - tutors are online