Unformatted text preview: he power involved in the rotor circuit is:
2
P2 = I 2 R 2 (4.15) ThreePhase 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. ThreePhase 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 IEEERecommended Equivalent Circuit
In the induction machine, because of its air gap, the exciting
current 10 is highof the order of 30 to 50 percent of the fullload
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) ThreePhase 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:
Noload 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 noload 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
noload, 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 noload using rated linetoline voltage, VNL
(Fig.4.10). Measure the no load current I NL and the total
3phase active power PNL. Fig.4.10 A noload test permits the calculation of Rm and Rm
of the magnetizing branch.
IEEErecommended equivalent circuit. For the noload condition, ′
R2 / s is very high. Therefore, in the equivalent circuit of Fig.4.9, 239 ThreePhase 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.
 Spring '14

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