Unformatted text preview: d the stator current
I1 will remain the same, and the machine will develop the same
torque (Equation (4.54)). Also, if the rotational losses are
neglected, the developed torque is the same as the load torque.
Therefore, for unity turns ratio, Example 4.6 The following test results are obtained from three
phase 100hp,460 V, eight pole star connected induction machine
Noload test : 460 V, 60 Hz, 40 A, 4.2 kW. Blocked rotor test is
100V, 60Hz, 140A 8kW. Average DC resistor between two stator
terminals is 0.152 Ω
(a) Determine the parameters of the equivalent circuit.
(b) The motor is connected to 3ϕ , 460 V, 60 Hz supply and runs
at 873 rpm. Determine the input current, input power, air
gap power, rotor cupper loss, mechanical power developed,
output power and efficiency of the motor. 268 Chapter Four
(c) Determine the speed of the rotor field relative to stator
structure and stator rotating field
Solution:
From no load test:
(a) Z NL = RNL = 460 / 3
= 6.64 Ω
40 PNL
4200
=
= 0.875 Ω\
2
3 * I1 3 * 40 2 X NL = 6.64 2 − 0.8752 = 6.58 Ω
Then, X 1 + X m = 6.58 Ω
From blocked rotor test: RBL =
R1 = 8000
= 0.136 Ω
3 *140 2 0.152
= 0.076 Ω (from resistance between two stator
2 terminals). Z BL = 100 / 3
= 0.412 Ω
140 X BL = 0.412 2 − 0.136 2 = 0.389 Ω
′
Then, X 1 + X 2 = 0.389 Ω 269 ThreePhase Induction Machine ′
Then, X 1 = X 2 = 0.389
= 0.1945 Ω
2 X m = 6.58 − 0.1945 = 6.3855 Ω
R = RBL − R1 = 0.136 − 0.076 = 0.06 Ω
2 0.1945 + 6.3855 ′
Then, R2 = * 0.06 = 0.0637 Ω
6.3855 0.076 Ω j0.195 Ω j0.195 Ω j6.386 Ω (b) ns = s= 0.0637
s 120 f 120 * 60
=
= 900rpm
8
P ns − n 900 − 873
=
= 0.03
900
ns ′
R2 0.0637
=
= 2.123 Ω
0.03
s
Input impedance Z1 Z1 = 0.076 + j 0.195 + I1 = ( j 6.386)(2.123 + j 0.195) = 2.121∠27.16o
2.123 + j (6.386 + 0.195) V1
460 / 3
=
= 125.22∠ − 27.16o
Z1 2.12∠27.16 270 Chapter Four
Input power: Pin = 3 * ( ) 460
*125.22 cos 27.16o = 88.767 kW
3 Stator CU losses: Pst = 3 *125.22 2 * 0.076 = 3.575 kW
Air gap power Pag = 88.767 − 3.575 = 85.192 kW
Rotor CU losses P2 = sPag = 0.03 * 85.192 = 2.556 kW
Mechanical power developed: Pmech = (1 − s ) Pag = (1 − 0.03) * 85.192 = 82.636 kW Pout = Pmech − Prot
From no load test: Prot = PNL − 3I12 * R1 = 4200 − 3 * 40 2 * 0.076 = 3835.2W
Pout = 82.636 *103 − 3835.2 = 78.8 kW
Then the efficiency of the motor is: η= Pout
78.8
*100 =
*100 = 88.77 %
88.767
Pin ThreePhase Induction Machine 271 Example 4.7 A three phase, 460 V 1450 rpm, 50 Hz, four pole
wound rotor induction motor has the following parameters per ′
′
phase ( R1 =0.2Ω, R2 =0.18 Ω, X 1 = X 2 =0.2Ω, X m =40Ω). The
rotational losses are 1500 W. Find,
(a) Starting current when started direct on full load voltage.
Also find starting torque. (b) (b) Slip, current, power factor, load torque and efficiency
at full load conditions. (c) Maximum torque and slip at which maximum torque will
be developed. (d) How much external resistance per phase should be
connected in the rotor circuit so that maximum torque
occurs at start? Solution:
(a) V1 = 460
= 265.6 V / phase
3 Z1 = 0.2 + j 0.2 +
Then, I st =
(b) s = j 40 * (0.18 + 0.2)
= 0.55∠46.59o Ω
0.18 + j 40.2 V1
265.6
=
== 482.91 ∠ − 46.3o Ω
o
I1 0.55∠46.59 1500 − 1450
= 0.0333
1500 272 Chapter Four ′
R2
0.18
=
= 5.4 Ω
s 0.0333
Z1 = 0.2 + j 0.2 +
Then I1 FL = j 40 * (5.4 + j 0.2)
= 4.959 ∠10.83o Ω
5.4 + j 45.4 265.6
= 53.56∠ − 10.83o A
4.959∠10.83o
o Then the power factor is: cos 10.83 = 0.9822 lag. ω sys =
Vth = 1500
* 2π = 157.08 rad / sec .
60 265.6 * ( j 40 )
= 264.275 ∠0.285o V
(0.2 + j 40.2) Then, Z th = j 40 * (0.2 + j 0.2 )
= 0.281432 ∠45.285o = 0.198 + j 0.2 Ω
0.2 + j 40.2 Then,
3 * (264.275) * 5.4
T=
= 228.68 Nm
2
2
157.08 * (0.198 + 5.4 ) + (0.2 + 0.2 )
2 Then, Pag = T * ω sys = 228.68 *157.08 = 35921.1W
Then, P2 = sPag = 0.0333 * 35921.1 = 1197 W
And, Pm = (1 − s )Pag = 34723.7W
Then, Pout = Pm − Prot = 34723.7 − 1500 = 33223.7W 273 ThreePhase Induction Machine Pin = 3 * 265.6 * 53.56 * 0.9822 = 41917 W
Then, η = Pout 33223.7
=
= 79.26 %
41914
Pin Then, Tm =
sTmax [ 3 * (264.275)2 ( 2 *188.5 0.198 + 0.198 + (0.2 + 0.2 )
0.18
=
= 0.4033
2 1/ 2
2
0.198 + (0.2 + 0.2 ) [ (d) sTmax = 1 = )] 2 1/ 2 2 = 862.56 Nm [0.198 ′
′
R2 + Rext
2 2 1/ 2 + (0.2 + 0.2 ) ′
′
Then, R2 + Rext = 0.446323
′
Then, Rext = 0.446323 − 0.18 = 0.26632 Ω
Example 4.8 The rotor current at start of a threephase, 460 volt,
1710 rpm, 60 Hz, four pole, squirrelcage induction motor is six
times the rotor current at full load.
(a) Determine the starting torque as percent of full load torque.
(b) Determine the slip and speed at which the motor develops
maximum torque.
(c) Determine the maximum torque developed by the motor as
percent of full load torque. 274 Chapter Four
Solution:
Note that the equivalent circuit parameters are not given.
Therefore equivalent circuit parameters cannot be used directly for
computation.
(a) The synchronous speed is From Equation (4.57)
2
2
I 2 R2 I 2 R2
α
, Thus,
T=
sω syn
s...
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 Electric motor, air gap, induction motor, Squirrelcage rotor, threephase induction machine

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