9.1 A 4-pole wound-rotor induction motor is used as a frequency changer. The starter is connected to a 50 Hz, 3-phase supply. The load is connected to the rotor slip rings. What are the possible speeds at which the rotor can supply power to this load at 25 Hz? What would be the ratio of voltages at load terminals at these speeds? Assume the rotor impedance to be negligible.
Solution
sf = 25
\ s = 25
50 = 0.5 We can get 25 Hz voltage at slip = ± 0.5
or n n
n
s s
- = ± 0.5
where ns= 120 50
4
¥ = 1,500 rpm
\ n = 750 rpm for s = +0.5
= 2,250 rpm for s = –0.5 The rotor voltage at any slip s is
= s ¥ rotor voltage at standstill Therefore the rotor voltage is the same for both 0.5 and –0.5 slip. Therefore,
ratio of voltages = 1
9.2 A 6-pole, 50 Hz, 3-phase induction motor running on full-load develops a useful torque of 160 Nm and the rotor emf is absorbed to make 120 cycles/min. Calculate the net mechanical power devel-oped. If the torque loss in windage and friction is 12 Nm, find the copper loss in the rotor windings, the input to the motor and efficiency. Given: starter losses = 200 W (inclusive of core loss).
Solution
ns= 1,000 rpm, ws = 104.7 rad/s f2= sf
120
60 = s ¥ 50 or s =
¥ 120
60 50 = 0.04
w = (1 – s) ws = (1 – 0.04) ¥ 104.7 = 100.5 rad/s Mechanical power developed (net) = 160 ¥ 100.5 = 16.08 kW
Mechanical power developed (gross) = (160 + 12) ¥ 100.5
= 17.29 kW 3I2¢r2¢ 1 s 1 Ê - ˆ
Á ˜
Ë ¯ = Pm
Rotor copper loss = 17,290 ¥ 0.04 1 0.04
Ê ˆ
Á - ˜
Ë ¯ = 720 W
Motor input = 17.29 + 0.72 + 0.8 = 18.81 kW h = 16 08
18 81 .
. = 85.5%
9.3 A 12-pole, 3-phase, 50 Hz induction motor draws 2.80 A and 110 kW under the blocked-rotor test.
Find the starting torque when switched on direct to rated voltage and frequency supply. Assume the starter and rotor copper losses to be equal under the blocked-rotor test. What would be the starting Solution Phase windings connected in star
¢
9.4 A 3.3 kV, 20-pole, 50 Hz, 3-phase, star-connected induction motor has a slip ring rotor of resistance 0.025 W and standstill reactance of 0.28 W per phase. The motor has a speed of 294 rpm when full-load torque is applied. Compute: (a) slip at maximum torque, and (b) ratio of maximum to full-full-load torque. Neglect stator impedance.
Solution
torque if the motor is started by connecting the phase windings in star. (Try this part after studying Sec. 9.8) 3 ¥ (280)2 (r1 + r2¢) = 110 ¥ 103
Dividing Eq. (i) by (ii)
9.5 An 8-pole, 3-phase, 50 Hz induction motor runs at a speed of 710 rpm with an input power of 35 kW. The starter copper loss at this operating condition is 1200 W while the rotational losses are 600 W. Find: (a) rotor copper loss, (b) gross torque developed, (c) gross mechanical power devel-oped, and (d) net torque and mechanical power output.
Solution Electrical power input = 35 kW
Stator copper loss = 12 kW
\ Power across air-gap, PG = 35 – 1.2 = 33.8 kW
(c) Mechanical power developed (gross) = (1 – s)PG
= (1 – 0.053) ¥ 33.8 = 32 kW
9.6 A 7.5 kW, 440 V, 3-phase, star-connected, 50 Hz, 4-pole squirrel-cage induction motor develops full-load torque at a slip of 5% when operated at rated voltage and frequency. Rotational losses (core, windage and friction) are to be neglected. Motor impedance data is as follows:
r1= 1.32 W x1= x¢2 = 1.46 W xm= 22.7 W
Determine the maximum motor torque at rated voltage and the slip at which it will occur. Also calculate the starting torque.
Solution
Taking the Thevenin equivalent in Fig. P9.6.
VTH= 254 22.7
Full-load output = 7.5 kW ws = 1 500 2
Substituting the values corresponding to full-load 50.27 = 3
= 3
9.7 The motor of Prob 9.6 is fed through a feeder from 440 V, 50 Hz mains. The feeder has an impedance of (1.8 + j 1.2) W /phase. Find the maximum torque that the motor can deliver and corresponding slip, stator current and terminal voltage.
Solution
The per phase circuit model of the motor and feeder is drawn in Fig. P9.7. Taking the Thevenin equivalent,
Notice the reduction in Tmax.
For the remaining part it is convenient to proceed by finding Zf. r¢
Notice that the voltage of the stator terminals is considerably reduced because of the voltage drop in feeder impedance.
9.8 A 400 V, 3-phase, star-connected induction motor gave the following test results:
No-load 400 V 8.5 A 1,100 W
Blocked-rotor 180 V 45 A 5,700 W
Determine the ohmic values of the components in the circuit model and calculate the line current and power factor when the motor is operating at 5% slip. The stator resistance per phase is 1.5 W and the standstill leakage reactance of the rotor winding referred to the stator is equal to that of the stator winding. Note: ri here accounts for rotational losses.
Blocked-rotor test
X = ( .2 31)2 -( .0 94)2 = 2.11 W Now
R = r1 + r2¢ 0.938 = 0.5 + r2¢
\ r2¢ = 0.438
also,
x1= x¢2 = 2 11 2
. = 1.055 W Performance calculations
s = 0.05 0 438.
S = 0 438 0 05
.
. = 8.76 W In Fig. P9.8,
1 ZF
= (0.0062 – j 0.0363) + 1 8 76. + j1 055.
= (0.0062 – j 0.0363) + (0.112 – j 0.0135)
= (0.118 – j 0.05) = 0.128 ––23°
or Zf = 7.8 –23° = 7.18 + j 3.05
Z(total) = (0.5 + j 1.055) + (7.18 + j 3.05)
= (7.68 + j 4.105) = 8.71 –28.1°
I1= 231
8 71. = 26.5 A
pf = cos 28.1° = 0.882 lagging
l0
27.55 Zf
l1 0.5 1.055 1.055 l2
V/Phase 161.6 0.438/s
Circuit Model
9.9 A 15 kW, 415 V, 4-pole, 50 Hz delta-connected motor gave the following results on test (voltages and currents are in line values):
No-load test 415 V 10.5 A 1,510 W
Blocked-rotor test 105 V 28 A 2,040 W Using the approximate circuit model, determine:
(a) the line current and power factor for rated output,
+
–
Fig. P9.8
(b) the maximum torque, and
(c) the starting torque and line current if the motor is started with the stator star-connected.
Assume that the stator and rotor copper losses are equal at standstill.
Hint Part (a) is best attempted by means of a circle diagram. For proceeding computationally from the circuit model, we have to compute the complete output-slip curve and then read the slip for rated output.
Solution
OP0= 10.5 A cos f0= 1 510
3 415 10 5 ,
¥ ¥ . = 0.2
or f0= 78.5°
OPSC= 28 415 105
¥ = 110.7 A (at 415 V)
cos fSC= 2 040 3 105 28
,
¥ ¥ = 0.4
fSC= 66.4°
Rated output = 15 kW = 3 V(I cos f)rated
(I cos f)rated= 15 000 3 415
,
¥ = 20.87 A
(a) The circle diagram is drawn in Fig. P9.9. PP¢ is drawn parallel to the output line P0PSC at a vertical distance 20.87A above P0PSC. The point P pertains to this point (P¢ is the second but unaccept-able solution—too large a current):
Then,
Il= OP = 30.5 A
pf = cos f = cos 30° = 0.866 lagging
(b) Torque line P0F is drawn by locating F as the midpoint of PSCG (equal stator and rotor, losses).
The maximum torque point is located by drawing a tangent to the circle parallel to P0F. The maximum torque is given as
RS = 44 A Tmax= 3 ¥ 415
3 ¥ 44 = 31,626 syn watts ns= 1,500 rpm, ws = 157.1 rad/s Tmax= 31 626
157 1 ,
. = 201.3 Nm
(c) If the motor is started to delta (this is the connection for which test data are given) Is(delta) = OPSC = 110.7 A
PSCF = 22 A
Ts(delta) = ( ) . 3 415 22
157 1
¥ ¥
= 100.7 Nm
Star connection
Is(star) = 110 7 3
. = 36.9 A
Ts(star) = 100 73. = 33.57 Nm
9.10 A 400 V, 3-phase, 6-pole, 50 Hz induction motor gave the following test results:
No-load 400 V 8 V 0.16 power factor
Blocked-rotor 200 V 39 A 0.36 power factor
Determine the mechanical output, torque and slip when the motor draws a current of 30 A from the mains. Assume the stator and rotor copper losses to be equal.
Solution
cos f0 = 0.16; f0 = 80.8°
cos fSC = 0.36; fSC = 69°
The circle diagram is drawn in Fig. P9.10. From the circle diagram we get the following results.
Pm= 3 ¥ 400 ¥ PB(= 10.75 A)
= 7.45 kW Slip s = BC
PC = 3 5 13 5
. .
A
A = 0.26 Torque T = 3¥400¥PC (=13 5. A)
ws
and ws= 120 50
6 2 60
¥ p
= 104.7 rad/s
\ T = 3 400 13 5
104 7
¥ ¥ .
. = 89.33 Nm V
O P
R
S
F
G H C
P¢
X 20.9 A
PSC
78.5°
66.4°
f = 30°
P0
Fig. P9.9
9.11 A 4-pole, 3-phase, 400 V, 50 Hz induction motor has the following parameters for its circuit model (referred to the stator side on equivalent-star basis)
r1= 1.2 W x1= 1.16 W
¢
r2 = 0.4 W x2¢ = 1.16 W xm= 35 W
Rotational losses are 800 W.
(a) For a speed of 1,440 rpm, calculate the input current, power factor, net mechanical power and torque and efficiency.
(b) Calculate the maximum torque and the slip at which it occurs.
Solution
(a) ns = 1,500 rpm n = 1,440 rpm or w = 2 1 440 60 p ¥ ,
= 150.8 rad/s s = 60
1 500, = 0.04 r¢
s
2 = 0 040 4.. = 10 W
In this part it is convenient to proceed by finding Zf (Fig. P9.11), in order that the identity of the input current is preserved.
V
O
P
B
E
F
H G C
D
PSC
P0
I1 1.2 1.16 a 1.16 I¢2
–
400 3 = 231 V Zf 35 0.4/s = 10W
b Fig. P9.10
Fig. P9.11
Zf = j j Rotational losses = 0.8 kW
Mechanical power output (net) = 12.71 – 0.8 = 11.91 kW
n = 1,440 rpm or 150.8 rad/s
(b) It will be necessary here to use Thevenin theorem. Finding the Thevenin equivalent to the left of ab in Fig. P9.9, we get According to Eq. 9.10,
Tmax= 3 0 5
= 3
9.12 A 3-phase, 3.3 kV, 50 Hz, 10-pole, star-connected induction motor has a no-load magnetizing current of 45 A and a core loss of 3.5 kW. The stator and referred rotor standstill leakage imped-ances are respectively (0.2 + j 1.8) W /phase. The motor is supplied from 3.3 kV mains through a line of reactance 0.5 W /phase. Use the approximate circuit model:
(a) The motor is running at 0.03 slip. Estimate the gross torque, stator current and power factor.
Assume voltage at motor terminals to be 3.3 kV.
(b) Calculate the starting torque and current when the motor is switched on direct to line with voltage at the far end of the line being 3.3 kV.
Solution pf = 0.863 lagging ns= 120 50 (b) We shall neglect magnetizing current.
Z(total)|s=1= j 0.5 + (0.2 + j 1.8) + (0.45 + j 1.8)
9.13 A 6-pole, 440 V, 3-phase, 50 Hz induction motor has the following parameters of its circuit model (referred to the stator or equivalent star basis):
r1 = 0.0 W (stator copper loss negligible); x1 = 0.7 W r2¢ = 0.3 W x2¢ = 0.7 W
xm = 35 W
Rotational loss = 750 W
Calculate the net mechanical power output, stator current and power factor when the motor runs at a speed of 950 rpm.
Solution
The circuit model of the motor is drawn in Fig. P9.13(a) and its thevenin equivalent is given in Fig.
P9.10(b).
= 5.84 –16.2° = 5.61 + j 163
Z(total) = 5.61 + j(0.7 + 1.63) = 5.61 + j 2.33
= 6.07 –22.6°
I1= 254
6 07. = 41.8 A pf = cos 22.6° = 0.923 lagging
9.14 A 75 kW, 440 V, 3-phase, 6-pole, 50 Hz, wound-rotor induction motor has a full-load slip of 0.04 and a slip at maximum torque of 0.2 when operating at rated voltage and frequency with rotor winding short-circuited at the slip-rings. Assume the stator resistance and rotational losses to be negligible. Find:
(a) Maximum torque (b) Starting torque
(c) Full-load rotor copper loss.
The rotor resistance is now doubled by adding an external series resistance.
Determine:
(d) Slip at full-load (e) Full-load torque
(f) Slip at maximum torque.
Solution
If the stator resistance and rotational losses are neglected, the circuit model on the Thevenin basis is as in Fig. P9.14.
Dividing Eq. (iii) by Eq. (ii) T
(a) Smax,T = 0.2
Substituting in Eq. (iv) T
\ Tmax= 1,940.1 Nm (remains unchanged) Substituting in Eq. (iv)
1 940 1
Multiplying Eq. (vi) by Eq. (vii)
104.7 ¥ 1,940.1 (1 – sfl¢) = 75,000 ¥ 0.5 ¥ ( . )
0.02 0.89 I¢2
(e) 104.7 (1 – 0.083)Tfl¢ = 75,000 Tfl¢ = 781.2 Nm (f) s¢max , T = 2 ¥ 0.2 = 0.4
9.15 A 3-phase induction motor has a 4-pole, star-connected stator winding and runs on 50 Hz with 400 V between lines. The rotor resistance and standstill reactance per phase are 0.4 W and 3.6 W respectively. The effective ratio of rotor to stator turns is 0.67. Calculate: (a) the gross torque at 4%
slip, (b) the gross mechanical power at 4% slip, (c) maximum torque, (d) speed at maximum torque, and (e) maximum mechanical power (gross). Neglect stator impedance.
Solution
Rotor impedance (standstill) referred to stator
= 1 (e) Refer to Fig. P9.15
For Pm(max)
I2¢ = |( .0 89 0 1/ .231+ j8 02. ) | = 19.28 A
9.16 A 30 kW, 440 V, 50 Hz, 3-phase, 10-pole, delta-connected squirrel-cage induction motor has the following parameters referred to a stator phase:
r1 = 0.54 W r2 ¢= 0.81 W x1 + ¢x2 = 6.48 W
ri = 414 W xm = 48.6 W
Calculate the machine performance (input current, power, power factor, mechanical output (gross) and torque developed (gross) for the following conditions:
(a) as a motor at a slip of 0.025. The equivalent circuit is drawn in Fig. P9.16.
(a) Motoring
Pm= 1 1
-= –338 Nm (in direction opposite to which the rotor is running) (c) Braking
I1 = (16.8 – j 115.28) + (184 – j 15.67)
= 87.49 Nm (in the direction of field but opposite to the direction in which the rotor is rotating)
Torque is developed in the same direction in which the rotor is running. The field rotates in an opposite direction to the rotor (s = 2). The total electrical and mechanical power input (14.21 + 5.5
= 19.71 kW) is consumed in stator and rotor copper loss and in core loss. Braking, therefore, can only be a short-time operation.
9.17 The following test results were obtained on a 7.5 kW, 400 V, 4-pole, 50 Hz, delta-connected induc-tion motor with a stator resistance of 2.1 W /phase:
No-load 400 V 5.5 A 410 W
Rotor-blocked 140 V 20 A 1,550 W
Obtain the approximate equivalent circuit modal.
Estimate the braking torque developed when the motor, running with a slip of 0.05 has two of its supply terminals suddenly interchanged.
Solution
ri= 390 W (includes windage and friction loss) 400
R = 1.29 W X = 3.83 W
r1= 2 13. = 0.7 W (eqv. star) r2¢ = 1.29 – 0.7 = 0.59 Circuit model (approximate) (Fig. P9.17) Braking torque
9.18 A 3-phase wound-rotor induction motor has star-connected rotor winding with a rotor resistance of 0.12 W/phase. With the slip rings shorted the motor develops rated torque at a slip of 0.04 and a line current of 100 A. What external resistance must be inserted in each rotor phase to limit the starting current to 100 A? What pu torque will be developed with rotor resistance starting ? Solution
We must assume here that the magnetizing current can be neglected.
¢
9.19 In Prob 9.18 what external resistance must be inserted per rotor phase to develop full load torque at 3/4th synchronous speed with a line current of 100 A?
Solution
9.20 A 4-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 4.5 W/phase and a standstill reactance of 8.5 W/phase. With no external resistance in the rotor circuit, the starting torque of the motor is 85 Nm.
(a) What is the rotor voltage at standstill ?
(b) What would be the starting torque if a 3 W resistance were added in each rotor phase ? (c) Neglecting stator voltage drop, what would be the induced rotor voltage and the torque at a slip
of 0.03?
(c) Induced rotor voltage at slip s = s ¥ standstill voltage
= 0.03 ¥ 302.5
9.21 Calculate the ratio of transformation of an auto transformer starter for a 25 kW, 400 V, 3-phase induction motor if the starting torque is to be 75% of full-load torque. Assume the slip at full-load to be 3.5 % and the short-circuit current to be six times full-load current. Ignore the magnetizing current of the transformer and of the motor.
Solution
9.22 With reference to the circuit model of Fig. P9.13 (as reduced by the Thevenin theorem) show that T
For the circuit model of Fig. 9.3 we have the following relationships
T = 3 2 Dividing (i) by (ii), we have
T
= 1 1
9.23 A 3-phase, 50 Hz, 75 kW induction motor develops its rated power at a rotor slip of 2%. The maximum torque is 250% of rated torque (that is, the torque developed at rated power). The motor has a K-ratio (defined in Prob 9.22) of K = 4.33. Find:
(a) smax,T at maximum torque
(b) rotor current referred to the stator at maximum torque, (c) starting torque, and
(d) starting current.
The answers to parts (b), (c), and (d) should be expressed in terms of current and torque at full-load speed. (a) We use the result of Prob 9.22.
T
Here smax, T = 2.79 ¥ 10–4 means the machine has an unrealistically small resistance and therefore the answer is
We know that
Using (ii), (i) can be rewritten as
2
The reader should note that the actual value of referred rotor current cannot be computed.
(c) T = 3 2 (d) Similar to part (b), we have
2
9.24 A 3-phase induction motor is wound for P poles. If the modulation poles are PM, obtain the general condition to suppress P2 = (P + PM) poles. Under this condition show that the angle between the phase axes for P1 = (P – PM) poles is r(2p /3), where r = integer non-multiple of 3. If P = 10, find
where r = integer non-multiple of 3
9.25 The two cages of a 3-phase, 50 Hz, 4-pole, delta-connected induction motor have respective standstill leakage impedances of (2 + j 8) and (9 + j2) W/phase. Estimate the gross torque developed;
(i) at standstill, the effective rotor voltage being 230 V/phase.
(ii) at 1450 rpm when the effective rotor voltage is 400 V/phase. The rotor quantities given are all referred to the stator; the stator impedances are negligible.
What is the gross starting torque if a star-delta starter is used ? Solution
(ii) s = 1 500 1 450 1 500
1 30
, ,
,
- =
T1= 3 157
400 2 30
2 30 8
2
2 2
( ) ( )
( ) ( )
¥ ¥
¥ + = 50.1 Nm
T2= 3 157
400 9 30
9 30 2
2
2 2
( ) ( )
( ) ( )
¥ ¥
¥ + = 11.32 Nm
T(total) = 50.1 + 11.3 = 61.4 Nm Star-delta starting:
In star-delta starting, starting voltage/phase is 400
3 = 230 V Hence,
Starting torque = 136.7 Nm