Unformatted text preview: From Equation(4.72) ThreePhase Induction Machine 275 From Equation (4.72) Example 4.9 A 4 pole 50 Hz 20 hp motor has, at rated voltage
and frequency a starting torque of 150% and a maximum torque of
200 % of full load torque. Determine (i) full load speed (ii) speed
at maximum torque. 276 Chapter Four
Solution: T
1.5
Tst
T
= 1.5 and max = 2 then, st =
= 0.75
TFL
TFL
Tmax
2
From above and equation (4.72) 2 sTmax
Tst
=
= 0.75
2
Tmax 1 + sTmax 2 Then, 0.75 sTmax − 2 sTmax + 0.75 = 0
Then sTmax = 2.21525 (unacceptable)
Or sTmax = 0.451416
Also from Equation 4.72
2
2
Tmax sTmax + s FL
=
=2
TFL 2sTmax * s FL But sTmax = 0.451416
2
Tmax
0.4514162 + s FL
Then
=
=2
TFL 2 * 0.451416 * s FL
2
s FL − 4 * 0.451416 s FL + 0.451416 2 = 0
2
s FL − 1.80566 s FL + 0.203777 = 0 s FL = 1.6847 (unacceptable)
or s FL = 0.120957 ns = 120 * 50
= 1500 rpm
4 ThreePhase Induction Machine 277 then (a) nFL = (1 − s FL ) * ns nFL = (1 − 0.120957 ) *1500 = 1319 rpm
(b) ( ) nTmax = 1 − sTmax * ns = (1 − 0.451416) *1500 = 823 rpm
Example 4.10 A 3φ, 280 V, 60 Hz, 20 hp, fourpole induction
motor has the following equivalent circuit parameters. ′
′
R1 = 0.12 Ω, R2 = 0.1 Ω, X 1 = X 2 = 0.25 Ω, and X m = 10 Ω
The rotational loss is 400 W. For 5% slip, determine (a) The
motor speed in rpm and radians per sec. (b) The motor current. (c)
The stator culoss. (d) The air gap power. (e) The rotor culoss. (f)
The shaft power. (g) The developed torque and the shaft torque.
(h) The efficiency.
Solution: 120 * 60
= 1800 rpm
4
1800
ωs =
* 2π = 188.5 rad / sec
60
ns = 278 Chapter Four
0.12 Ω j0.25 Ω j0.25 Ω 0.1
=2
0.05 j10 Ω Z1 = 0.12 + j 0.25 + Re + X e
Z1 = 0.12 + j 0.25 +
V1 = I1 = 208 j10 * (2 + j 0.25)
= 2.1314∠23.55o Ω
2 + j10.25 = 120.1 V 3 120.1
= 2.1314∠ − 23.55o A
o
2.1314∠23.55
2 (c) P = 3 * 56.3479 * 0.12 = 1143.031W
1
(d) ( ) Ps = 3 *120.1 * 56.3479 * cos − 23.55o = 18610.9794 W
Pag = Ps − P = 17467.9485 W
1
(e) P2 = sPag = 0.05 *17467.9785 = 873.3974 W
(f) Pm = (1 − s ) Pag = 16594.5511W
(g) T = Pag
188.5 = 17467.9485
= 92.6682 N .m
188.5 ThreePhase Induction Machine Tshaft =
(h) η = Pshaft
188.5
Pshaft
Ps = 279 16194.5511
= 85.9127 Nm
188.5 *100 = 87.02% For operation at low slip, the motor torque can be considered
proportional to slip.
Example 4.11 A 30, 100 WA, 460 V, 60 Hz, eightpole induction
machine has the following
equivalent circuit parameters: ′
′
R1 = 0.07 Ω, R2 = 0.05 Ω, X 1 = X 2 = 0.2 Ω, and X m = 6.5 Ω
(a) Derive the Thevenin equivalent circuit for the induction machine.
(b) If the machine is connected to a 30, 460 V, 60 Hz supply,
determine the starting torque, the maximum torque the machine
can develop, and the speed at which the maximum torque is
developed.
(c) If the maximum torque is to occur at start, determine the
external resistance required in each rotor phase. Assume a
turns ratio (stator to rotor) of 1.2. 280 Chapter Four
Solution: Xm
6.5
* V1 =
* 265.6 = 257.7 V
X1 + X m
0.2 + 6.5 Vth = Rth + jX th = ( j 6.5) * ( j 0.2 + 0.07 ) = 0.06589 +
0.07 + j 0.2 + j 6.5 Ω
0.06589 j0.1947 Ω j 0.1947 Ω j0.2 Ω 0.05
s 257.7V (b) Tst = 3 * 257.7 2 * 0.05 [ 94.25 (0.06589 + 0.05) + (0.1947 + 0.2 ) Tmax = 2 2 = 624.7 Nm 3 * 257.7 2 [ 2 * 94.25 0.06589 + 0.06589 2 + (0.1947 + 0.2 )2 = 2267.8 Nm
sTmax =
Speed ( 0.05
0.06589 + (0.1947 + 0.2 ) 2 2 in ) rpm for which = 0.1249
max torque = 1 − sTmax * ns = (1 − 0.1249 ) * 900 = 787.5 rpm
(c) sTmax = ′
R2 R12 ′
+ ( X1 + X 2 ) 2 ′
α R2 occurs 281 ThreePhase Induction Machine ′
or R2 start = sstart = 1
sTmax ′
* R2 = 1
0.1249 * 0.05 = 0.4 Ω Then Rext = (0.4 − 0.05) / 1.2 2 = 0.243 Ω 1speed characteristic of an induction machine
We have seen that a 3phase squirrelcage induction motor can
also function as a generator or as a brake. These three modes of
operationmotor, generator, and brakemerge into each other, as
can be seen from the torquespeed curve of Fig.4.18. This curve,
together with the adjoining power flow diagrams, illustrates the
overall properties of a 3phase squirrelcage induction machine.
We see, for example, that when the shaft turns in the same
direction as the revolving field, the induction machine operates in 282 Chapter Four
either the motor or the generator mode. But to operate in the
generator mode, the shaft must turn faster than synchronous speed.
Similarly, to operate as a motor, the shaft must turn at less than
synchronous speed.
Finally, in order to operate as a brake, the shaft must turn in the
opposite direction to the revolving Fig.4.18 Complete torque speed curve of a 3phase induction
machine...
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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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