Three Phase Induction Motors

# I2 se2 r2 jsx 2 414 the power involved in the rotor

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Unformatted text preview: he power involved in the rotor circuit is: 2 P2 = I 2 R 2 (4.15) Three-Phase Induction Machine 233 Which represents the rotor copper loss per phase. Equation (4.14) can be rewritten as: I2 = E2 (R2 / s ) + jX 2 (4.16) Equation (4.16) suggests the rotor equivalent circuit of Fig.4.7b. Although the magnitude and phase angle of I 2 are the same in Equations (4.14) and (4.16), there is a significant difference between these two equations and the circuits (Figs.4.7b and 4.14c) they represent. The current I 2 in Equation (4.14) is at slip frequency f 2 , whereas I 2 in Equation (4.16) is at line frequency f1 . In Equation (4.14) the rotor leakage reactance sX 2 varies with speed but resistance R2 remains fixed, whereas in Equation (4.16) the resistance R2 / s varies with speed but the leakage reactance X 2 remains unaltered. The per phase power associated with the equivalent circuit of Fig.4.7c is: 2 P = I2 R2 P2 = s s (4.17) Because induction machines are operated at low slips (typical values of slip s are 0.01 to 0.05) the power associated with Fig.4.7c is considerably larger. Note that the equivalent circuit of Fig.4.7c is at the stator frequency, and therefore this is the rotor 234 Chapter Four equivalent circuit as seen from the stator. The power in Equation (4.17) therefore represents the power that crosses the air gap and thus includes the rotor copper loss as well as the mechanical power developed. Equation (4.17) can be rewritten as: R 2 2R P = Pag = I 2 R2 + 2 (1 − s ) = I 2 2 s s (4.18) The corresponding equivalent circuit is shown in Fig.4.7d. The speed dependent resistance R2 (1 − s ) / s represents the mechanical power developed by the induction machine. 2 Pmech = I 2 R2 (1 − s ) s Pmech = (1 − s ) * Pag Pmech = (1 − s ) P s 2 2 (4.19) (4.20) (4.21) and, P2 = I 2 R2 = sPag (4.22) Thus, Pag : P2 : Pmech = 1 : s : (1 − s ) (4.23) Equation (4.23) indicates that, of the total power input to the rotor (i.e., power crossing the air gap, Pag ), a fraction is dissipated in the resistance of the rotor circuit (known as rotor copper loss) and the fraction 1 - s is converted into mechanical power. Three-Phase Induction Machine 235 Therefore, for efficient operation of the induction machine, it should operate at a low slip so that more of the air gap power is converted into mechanical power. Part of the mechanical power will be lost to overcome the windage and friction. The remainder of the mechanical power will be available as output shaft power. 4.7 IEEE-Recommended Equivalent Circuit In the induction machine, because of its air gap, the exciting current 10 is high-of the order of 30 to 50 percent of the full-load current. The leakage reactance X l is also high. The IEEE recommends that, in such a situation, the magnetizing reactance X m not be moved to the machine terminals but be retained at its appropriate place, as shown in Fig.4.8. The resistance Rc is, however, omitted, and the core loss is lumped with the windage and friction losses. This equivalent circuit (Fig.4.8) is to be preferred for situations in which the induced voltage E1 differs appreciably from the terminal voltageV1 . 236 Chapter Four Fig.4.8 IEEE recommended equivalent circuit. 4.8 Thevenin Equivalent Circuit In order to simplify computations, V1 , R1 , X 1 , and X m can be replaced by the Thevenin equivalent circuit values Vth , Rth , and X th , as shown in Fig.4.9, where: Vth = Xm R12 + ( X1 + X m ) 2 V1 (4.24) If R12 << ( X 1 + X m )2 as is usually case, Vth ≈ Xm V1 = K thV1 X1 + X m (4.25) The Thevenin impedance is: Z th = jX m (R1 + jX 1 ) = Rth + jX th R1 + j ( X 1 + X m ) (4.26) Three-Phase Induction Machine 237 If R1 << ( X 1 + X m ) , then, 2 2 2 Xm 2 Rth ≅ X + X R1 = K th R1 1 m (4.27) and since X 1 << X m , then, X th ≈ X 1 (4.28) Fig.4.9 Thevenin equivalent circuit. 4.9 Tests To Determine The Equivalent Circuit The approximate values of R1 , R2 , X m , Rm and X in the equivalent circuit can be found by means of the following tests: No-load test When an induction motor runs at no load, the slip is exceedingly small. Referring to Fig.4.7, this means that the value of R2 / s is very high and so current I1 is negligible compared to I o . Thus, at no-load the circuit consists essentially 238 Chapter Four of the magnetizing branch X m , Rm . Their values can be determined by measuring the voltage, current, and power at no-load, as follows: a. Measure the stator resistance RLL between any two terminals. Assuming a wye connection, the value of R1 is: R1 = RLL / 2 b. (4.29) Run the motor at no-load using rated line-to-line voltage, VNL (Fig.4.10). Measure the no load current I NL and the total 3-phase active power PNL. Fig.4.10 A no-load test permits the calculation of Rm and Rm of the magnetizing branch. IEEE-recommended equivalent circuit. For the no-load condition, ′ R2 / s is very high. Therefore, in the equivalent circuit of Fig.4.9, 239 Three-Phase Induction Machine the magnetizing reactance X m is shun...
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## This document was uploaded on 03/12/2014 for the course ENGINEERIN electrical at University of Manchester.

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