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 blockedrotor test, the blockedrotor 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 blockedrotor impedance at frequency of blocked rotor test
is: Z BL = V1 BL
I1 BL = 163 / 3
= 1.5685 Ω
60 The blockedrotor 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, sixpole, 60 Hz squirrelcage induction
motor.
(1) Noload test:
Supply frequency = 60 Hz, Line voltage = 2200 V Line current = 4.5 A, Input power = 1600 W
(2) Blockedrotor test:
Frequency = 15 Hz, Line voltage = 270 V
Line current = 25 A, Input power = 9000 W ThreePhase Induction Machine 247 (3) Average DC resistance per stator phase: R1 = 2.8 Ω
(a) Determine the noload rotational loss.
(b) Determine the parameters of the IEEErecommended
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 noload test, the noload power is: PNL = 1600 W
The noload rotational loss is: Prot = PNL − 3I12 R1
Prot = 1600 − 3 * 4.52 * 2.8 = 1429.9W
(b) IEEErecommended equivalent circuit. For the noload ′
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) Noload equivalent circuit (b) Blockedrotor equivalent circuit. (c) Blockedrotor equivalent circuit for improved value for R2 .
Fig.4.13 V1 = 2200
= 1270.2 V / Phase
3 The noload impedance is: Z NL = V1 1270.2
=
= 282.27 Ω
I1
4.5 The noload resistance is: RNL =
The noload reactance is: PNL
1600
=
= 26.34 Ω
3I12 3 * 4.52 249 ThreePhase 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 blockedrotor test the slip is 1, the magnetizing reactance ′
′
X m is shunted by the lowimpedance branch R2 + jX 2 . Because
′
′
X m >> R2 + jX 2 , the impedance X m can be neglected and the
equivalent circuit for the blockedrotor test reduces to the form
shown in Fig.4.11b. From the blockedrotor test, the blockedrotor
resistance is: RBL = 9000
PBL
=
= 4.8 Ω
3I12 3 * 252 ′
Therefore, R2 = RBL − R1 = 4.8 − 2.8 = 2Ω , The blockedrotor
impedance at 15 Hz is: Z BL = V1
270
=
= 6.24 Ω
I1
3 * 25 The blockedrotor 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 pullout) 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) ThreePhase 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...
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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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