ONLINE GATE COACHING CLASSES BY Prof. Ch. SAI BABU Professor of Electrical & Electronics Engineering JNTUK Kakinada Lecture 8 of Electrical Machines Topic: Three phase induction motors
Jawaharlal Nehru Technological University, Kakinada-533003, Andhra Pradesh
1
Number Syllabus Detailed Syllabus
Lecture No Load and Blocked Rotor No Load and Blocked Tests on Induction Motor, Rotor Tests on Induction VIII Equivalent Circuit, Motor, Equivalent Calculation of Various Circuit, Calculation of Parameters. Various Parameters.
2 Equivalent Circuit of Induction Motor
The equivalent circuit of an induction motor is similar to that of the transformer.
The stator model of an induction motor consists of a stator phase
winding resistance R1, a stator phase winding leakage reactance X1.
The no-load current I0 is simulated by a pure inductive reactor
X0 taking the magnetizing component Iµ and a non-inductive resistor
R0 carrying the core-loss current Iω.
Thus, 푰ퟎ = 푰풘 + 푰흁
3 Equivalent Circuit of Induction Motor
Stator model of an induction motor Rotor circuit of an induction motor
Basic Equivalent Circuit
4 Approximate Equivalent Circuit of an Induction Motor The equivalent circuit is further simplified by shifting the
shunt impedance branches R0 and X0 to the input terminals as shown in the circuit diagram.
5 No Load Test on Induction Motor
In this test, the motor is made to run without any load i.e no- load condition. The rated voltage is applied to the stator. The input line current and total input power is measured.
6 No Load Test on Induction Motor
The two wattmeter method is used to measure the total input power.
The total power input W0 is the algebraic sum of the two wattmeter readings.
The Power input W0 consists of stator copper loss, stator core loss, friction and windage loss. The observation table is
7 No Load Test on Induction Motor Calculations:
푾ퟎ = ퟑ푽ퟎ푰ퟎ cos ∅0 푊0 cos ∅0 = 3푉0퐼0 Equivalent circuit parameters:
Active component of no-load current,푰풄 = 푰ퟎ cos ∅0
Magnetizing component of no-load current,푰풎 = 푰ퟎ sin ∅0 푽ퟎ No-load branch Resistance, 푹ퟎ = 푰풄 푽ퟎ No-load branch Reactance, 푿ퟎ = 푰풎
8 Blocked Rotor Test on Induction Motor In this test, the rotor is locked and it is not allowed to rotate. Thus the slip, S=1. If the motor is slip ring induction motor then the windings are short circuited at the slip rings. The reduced voltage such that stator carries rated current is applied.
Now the applied voltage (Vsc), the input power (Wsc) and a short
circuit current (Isc) are measured.
9 Blocked Rotor Test on Induction Motor
10 Blocked Rotor Test on Induction Motor Calculations:
푾풔풄 = ퟑ푽풔풄푰풔풄 cos ∅푠푐 푊푠푐 cos ∅푠푐 = 3푉푠푐퐼푠푐 Equivalent circuit parameters: ퟐ 푾풔풄 = ퟑ 푰풔풄 푹ퟏ풆 푾풔풄 Equivalent resistance referred to stator, 푹ퟏ풆 = ퟐ ퟑ 푰풔풄 푽풔풄 Equivalent impedance referred to stator, 풁ퟏ풆 = 푰풔풄 Equivalent reactance referred to stator, ퟐ ퟐ ∴ 푿ퟏ풆 = 풁ퟏ풆 − 푹ퟏ풆
11 Power flow in Induction Motor
12 Power flow in Induction Motor There are two types of losses in induction motor Constant or fixed losses, Variable losses. Constant or Fixed Losses The fixed losses can be obtained by performing no-load test on the three phase induction motor. These losses are further classified as 1. Iron or core losses 2. Mechanical losses 3. Friction losses
13 Power flow in Induction Motor Iron or Core Losses: Iron or core losses are further divided into hysteresis and eddy current losses. Eddy current losses are minimized by using lamination on core. Hysteresis losses are minimized by using high grade silicon steel.
Mechanical and Brush Friction Losses: Mechanical losses occur at the bearing and brush friction loss occurs in wound rotor induction motor.
14 Power flow in Induction Motor
Variable Losses
15 Power flow in Induction Motor Variable Losses:
These losses are also called copper losses. These losses occur due to current flowing in stator and rotor windings. As the load changes, the current flowing in rotor and stator winding also changes and hence these losses also changes. Therefore these losses are called variable losses. The copper losses are obtained by performing blocked rotor test on three phase induction motor.
16 Power flow in Induction Motor
Efficiency of Three Phase Induction Motor:
Efficiency is defined as the ratio of the output to that of input,
풐풖풕풑풖풕 풑풐풘풆풓 푬풇풇풊풄풊풆풏풄풚 = 풊풏풑풖풕 풑풐풘풆풓
17 Power flow in Induction Motor
푹풐풕풐풓 품풓풐풔풔 풐풖풕풑풖풕, 푷풎 = ퟐ흅푵 × 푻품
푹풐풕풐풓 풊풏풑풖풕, 푷ퟐ = ퟐ흅푵풔 × 푻품
푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔, 푷풄 = 푷ퟐ − 푷풎, ∴ 푷풄 = ퟐ흅(푵풔 − 푵) × 푻품
푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 푵 − 푵 = 풔 = 풔 풓풐풕풐풓 풊풏풑풖풕 푵풔
∴ 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕 = 풔푷ퟐ
18 Power flow in Induction Motor
풓풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 푹풐풕풐풓 풊풏풑풖풕 = 풔 Rotor gross output, Pm = input (P2) − rotor Cu loss = P2 − s × rotor input = (1 − s)*P2
∴ rotor gross output, Pm = (1 − s) *rotor input
19 Power flow in Induction Motor
풓풐풕풐풓 품풓풐풔풔 풐풖풕풑풖풕, 푷 푵 풎 = ퟏ − 풔 = ; 풓풐풕풐풓 풊풏풑풖풕, 푷ퟐ 푵풔
푷 푵 풎 = 푷ퟐ 푵풔 푵 ∴ 풓풐풕풐풓 풆풇풇풊풄풊풆풏풄풚 = 푵풔 풓풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 풔 = 풓풐풕풐풓 품풓풐풔풔 풐풖풕풑풖풕 ퟏ − 풔
∴ 푷ퟐ: 푷풎: 푷풄 ∷ ퟏ: ퟏ − 풔 : 풔
20 Multiple Choice Questions 1. The term „cogging‟ is associated with (a) Three-phase transformer (b)Compound Generator (c) Induction motor (d) D.C series motor Ans: (c) Explanation:
During the starting of squirrel cage induction motor if the number of stator slots is equal to the number of rotor slots or an integral multiple of rotor slot then it causes magnetic locking between stator and rotor slot. The upper and lower attracting force between the stator and rotor slot becomes more prominent than the tangential force, as a result, the rotation of the motor is stopped. 21 2. The crawling in the induction motor is caused by (a) High Loads (b) Low Voltage supply (c) Harmonic developed in the motor (d) Improper design of machine Ans: (c) Explanation:
The crawling word it self-suggest crawl means moving with low speed. This characteristic is the result of improper functioning of the motor that means either motor is running at very slow speed or it is not taking the load. The resultant speed is nearly 1/7th of its synchronous speed.
22 3. No load test of 3-phase induction motor used to determine (a) Variable loss (b) Constant loss (c) Eddy current loss only (d) Hysteresis loss only
Ans: (b)
Explanation:
This test is similar to the open circuit test of a transformer. The motor is uncoupled from its load and the rated voltage and frequency is applied to the stator. Since the motor runs at no load, the total input power is equal to constant iron-loss, friction loss, and windage losses of the motor.
23 4. Blocked rotor test in an induction motor is used to determine (a) Leakage impedance (b) Copper loss (c) Both 1 & 2 (d) None of the above Ans: (c) Explanation:
In blocked rotor test, the induction motor is locked, so it can‟t be moved therefore rotor will draw more current from supply because of the load demand on it.
Copper losses occur due to the current flowing in the rotor and stator windings.
Therefore blocked rotor test is used to determine copper losses and leakage impedance.
24 5. Which of the following losses are negligible in blocked rotor test? (a) Mechanical losses (b) Iron losses (c) Both 1 & 2 (d) None of the above Ans: (c) Explanation:
Since the rotor is locked in blocked rotor test, therefore, the mechanical loss is negligible. The voltage applied on the stator to perform blocked rotor test is low since the high voltage can damage the stator winding, therefore, it has negligible iron losses.
25 6. The rotor power output of a 3-phase induction motor is 30 KW and corresponding slip is 4%. The rotor copper loss will be (a) 625 Watt (b) 250 Watt (c) 1000 Watt (d) 1250 Watt
Ans: (d) Explanation:
Rotor copper losses = rotor input- rotor output and rotor output power = (1-s) rotor input power ∴ rotor Input power = output/(1 – s)= 30000 /(1 – 0.04) = 31250 Rotor copper losses = 31250 -30000 = 1250 watt
26 7. what is the ratio of rotor input power to rotor copper loss in an induction motor? (a) 1/(1 – S) (b) 1 – S (c) 1/S (d) S
Ans: (c)
Explanation:
Rotor copper loss = S × Rotor input power ∴ Rotor input / Rotor copper loss = 1/S.
27 8. What happens if fifth harmonics is given to induction motor? (a) Short-circuit the motor (b) Motor will rotate in reverse direction (c) Motor will rotate in the same direction (d) None of the above
Ans: (b)
Explanation:
The fifth harmonic is negative harmonic having the phase displacement of 120 degrees having negative phase sequence with the reference to the motor phase sequence. Hence induction motor will rotate in the reverse direction.
28 9. In a three phase induction motor, the electrical representation of the variable mechanical load is a resistance of ퟏ (a) 푹 (풔 − ퟏ) (b) 푹 − ퟏ ퟐ ퟐ 풔 풑 (c) 품 (d) 풑 (ퟏ − 풔) 풔 품 Ans: (b) Explanation: In the induction motor equivalent circuit, r2 is the actual rotor winding resistance whereas r2[(1-s)/s)] is the electrical analogue of mechanical load.
29 10. The power input to the rotor of a 3-phase, 50Hz, 6-pole induction motor is 80kW at a slip of 3.3%. The rotor copper loss per phase is (a) 2.64 kW (b) 880 W (c) 77.36 kW (d) 25.8 kW
Ans: (a)
Explanation:
∴ 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕
∴ 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = ퟎ. ퟎퟑퟑ × ퟖퟎ = 2.64 kW
30 11. In a three phase induction motor, the rotor input power per phase is 6 kW. The rotor is running at 5% slip. The rotor copper loss per phase is equal to (a) 300 W (b) 600 W (c) 5 W (d) 100 W
Ans: (a)
Explanation:
∴ 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕 ∴ 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = ퟎ. ퟎퟓ × ퟔퟎퟎퟎ = 300 W
31 12. The rotor power output of 3-phase induction motor is 15 KW. The rotor copper losses at a slip of 4% will be (a) 700 W (b) 650 W (c) 625 W (d) 600 W Ans: (d)
Explanation:
푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓
푹풐풕풐풓 풐풖풕풑풖풕 풑풐풘풆풓 = (ퟏ − 풔) × 풓풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓 ퟏퟓퟎퟎퟎ 풓풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓 = = ퟏퟓퟔퟐퟓ 푾 (ퟏ − ퟎ. ퟎퟒ) 푹풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = ퟎ. ퟎퟒ × ퟏퟓퟔퟐퟓ = ퟔퟐퟓ 푾
32 13. The rotor input of a motor is 20 kW and slip is 4%. The rotor output will be (a) 18 kW (b) 190 kW (c) 19.6 kW (d) 19.2kW Ans: (d) Explanation: rotor output, Pm = (1 − s) *rotor input
rotor output, Pm = (1 − 0.04) *20 = 19.2 kW
33 14. If the rotor input of an induction motor running with slip of 10% is 100 kW. The gross power developed by the rotor is (a) 10 kW (b) 90 kW (c) 99 kW (d) 80 kW
Ans: (b)
Explanation: rotor output, Pm = (1 − s) *rotor input
rotor output, Pm = (1 − 0.1) *100 = 90 kW
34 15. When the rotor of three phase induction motor is blocked then the slip is (a) 1 (b) 0 (c) 0.5 (d) 0.1 Ans: (a) Explanation:
In blocked rotor condition, Speed (N) =0
푵 −푵 푵 −ퟎ Slip, 퐬 = 풔 = 풔 푵풔 푵풔 ∴ 풔 = ퟏ
35 16. The power supplied to a three-phase induction motor is 32 kW and the stator losses are 1200 Watt. If the slip is 5%. determine the rotor copper loss
(a) 2.8 kW (b) 3.5 kW (c) 4 kW (d) 1.54 kW
Ans: (d)
Explanation: Input Power of rotor = Stator Input Power – Stator Losses = 32000-1200=30.8 kW
36 17. A 3-phase induction motor is running at 2% slip. If the input to rotor is 1000 W, then mechanical power developed by the motor is (a) 500 W (b) 200 W (c) 20 W (d) 980 W Ans: (d) Explanation:
Mechanical power developed in 3-phase motor = (1 − s) × power input to rotor = (1 – 0.02) x 1000 = 980 W
37 18. The approximate efficiency of a 3-phase, 50 Hz, 4-pole induction motor running at 1350 r.p.m. is (a) 90% (b) 60% (c) 45% (d) 100%
Ans: (a)
Explanation:
38 19. The conditions of an induction motor at no-load resemble those of a transformer whose secondary is (a) Short-circuited (b) Open-circuited (c) Supplying a variable resistive load (d) None of the above Ans: (b)
Explanation:
The operation of an induction motor under the no-load condition is similar to a transformer under open circuit condition.
39 20. In case of 3-phase induction motor, shaft power is 2700W and mechanical losses are 180W. At a slip of 4%, the rotor ohmic losses are (a) 115.2 W (b) 120 W (c) 108 W (d) 105 W Ans: (b) Explanation: Mechanical power developed in 3-phase motor = (1 − s)*power input to rotor
Mechanical power developed in 3−phase motor 푷풐풘풆풓 풊풏풑풖풕 풕풐 풓풐풕풐풓 = (ퟏ − 풔) (ퟐퟕퟎퟎ+ퟏퟖퟎ) = = ퟑퟎퟎퟎ푾 (ퟏ−ퟎ.ퟎퟒ)
∴ 푹풐풕풐풓 풐풉풎풊풄 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕 ∴ 푹풐풕풐풓 풐풉풎풊풄 풍풐풔풔 = ퟎ. ퟎퟒ × ퟑퟎퟎퟎ = 120 W 40 21. In an induction motor, air gap power is 10 kW and mechanically developed power is 8 kW. What are the rotor ohmic losses? (a) 3KW (b) 1.5KW (c) 2KW (d) 0.5Kw Ans: (c)
Explanation: Rotor ohmic loss = air gap power or rotor input power - mechanical developed power by rotor (∵ Rotor core losses are neglected due to low rotor frequency) Rotor ohmic loss = 10-8 = 2 kW.
41 22. A 3-phase, 4-pole, 50 Hz induction motor takes a power input of 30 kW at its full load speed of 1440 rpm. Total stator losses are 1 kW. The slip and rotor ohmic losses at full load are (a) 0.02, 600W (b) 0.04, 580W (c) 0.04, 1160W (d) 0.04, 1200W Ans: (c) Explanation: ퟏퟐퟎ × 풇 ퟏퟐퟎ ∗ ퟓퟎ ∴ 푹풐풕풐풓 풐풉풎풊풄 풍풐풔풔 푵 = = = ퟏퟓퟎퟎ 풓풑풎 풔 푷 ퟒ = 풔 × 풓풐풕풐풓 풊풏풑풖풕
푵 − 푵 ퟏퟓퟎퟎ − ퟏퟒퟒퟎ ∴ 푹풐풕풐풓 풐풉풎풊풄 풍풐풔풔 푺풍풊풑, 풔 = 풔 = = ퟎ. ퟎퟒ = ퟎ. ퟎퟒ × ퟐퟗ, ퟎퟎퟎ 푵풔 ퟏퟓퟎퟎ = 1160 W
풓풐풕풐풓 풊풏풑풖풕 = 풎풐풕풐풓 풊풏풑풖풕 − 풔풕풂풕풐풓 풍풐풔풔풆풔 Rotor input= 30000-1000=29,000W
42 23. During the no-load test, the power input to a 3-phase induction motor has to supply (a) core loss, friction and windage (FW) loss (b) small ohmic loss and FW loss (c) small ohmic loss, core loss and FW loss (d) small ohmic loss, FW loss and stray load loss Ans: (a) Explanation:
As the motor is running at no load, the total input power is equal to the constant iron loss, friction and windage losses of the motor.
43 24. During blocked rotor test on a 3-phase induction motor, the power input to a 3-phase induction motor has to supply (a) ohmic loss (b) ohmic loss and core loss (c) core loss (d) ohmic loss, core loss, friction and windage loss Ans: (a) Explanation: In block rotor test, the low voltage is applied so that the rotor does not rotate and its speed becomes zero and full load current passes through the stator winding. The constant losses are depends on voltage and ohmic loss are depends on full load current. Hence, the power input to a 3-phase induction motor has to supply ohmic loss
44 25. The power input to a 415V, 50Hz, 6-pole, 3-phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction and windage losses are 2 kW. The efficiency of the motor is
(a) 92.5% (b) 91% (c) 90.06% (d) 88% Ans: (c) Explanation:
ퟏퟐퟎ×ퟓퟎ power i/p = 40 kW , 푵 = = ퟏퟎퟎퟎ풓풑풎, 풔 ퟔ
ퟏퟎퟎퟎ−ퟗퟕퟓ 풔 = = ퟎ. ퟎퟐퟓ, Rotor i/p = 40 – 1 = 39 kW ퟏퟎퟎퟎ Gross mechanical power = (1 - S)*39 = 38.025 kW, Shaft o/p = 38.025 – 2 = 36.025 kW ∴ η = 36.025/40 = 90%
45 26. The approximate value of efficiency of 3-phase induction motor running at slip „s‟ is given by
ퟏ 풔 ퟏ−풔 풔 (a) (b) (c) (d) ퟏ+풔 ퟏ+풔 ퟏ+풔 ퟏ−풔
Ans: (c) Explanation:
풓풐풕풐풓 품풓풐풔풔 풐풖풕풑풖풕,푷 Approximate value of efficiency = 풎 = ퟏ − 풔 풓풐풕풐풓 풊풏풑풖풕,푷ퟐ
1−푠 1 − 푠 푖푠 푎푙푠표 푤푟푖푡푡푒푛 푎푠 (푠푖푛푐푒 푣푎푙푢푒 표푓 푠푙푖푝 푖푠 푣푒푟푦 푙표푤 ) 1+푠
46 27. A 3-phase slip-ring induction motor is fed from the rotor side with stator winding short circuited. The frequency of the currents in the short-circuited stator is (a) slip frequency (b) supply frequency (c) frequency corresponding to rotor speed (d) zero Ans: (a)
Explanation:
풇풓 = 풔 × 풇풓풆풒풖풆풏풄풚
47 28. A 3-phase star connected slip ring induction motor is fed from 400V, 50 Hz, source. Stator to rotor effective turns ratio is 2, at rotor speed of 1440 rpm, rotor induced emf per phase would be
(a) 4.62V (b) 46.2V (c) 8.0V (d) 9.24V Ans: (a) Explanation: ퟒퟎퟎ ퟏ ퟏퟐퟎ×ퟓퟎ 푬 = × = ퟏퟏퟓ. ퟒퟕퟑ, 푵 = = ퟏퟓퟎퟎ풓풑풎, ퟐ ퟑ ퟐ 풔 ퟒ ퟏퟓퟎퟎ−ퟏퟒퟒퟎ 풔 = = ퟎ. ퟎퟒ, ퟏퟓퟎퟎ 푹풐풕풐풓 풊풏풅풖풄풆풅 풆풎풇 풑풆풓 풑풉풂풔풆 = 풔 × 푬ퟐ
푹풐풕풐풓 풊풏풅풖풄풆풅 풆풎풇 풑풆풓 풑풉풂풔풆 = ퟎ. ퟎퟒ × ퟏퟏퟓ. ퟒퟕퟑ = 4.62 V
48 29. The efficiency and power factor of an induction motor increases in proportion to its (a) speed (b) mechanical load (c) voltage (d) rotor torque Ans: (b) Explanation: The efficiency of the induction increases when the mechanical load increases because as the motors load increases, its slip increases, and the rotor speed falls.
The input power factor of the induction increases when the mechanical load increases because in general, the higher the resistance (a load), the higher the power factor.
49 30. The 3-phase, 4-pole 5Hz induction motor is running at 0.05 slip. The supply voltage per phase is 230V and the ratio of stator to rotor turns is 2. The induced emf in the rotor per phase will be (a) 5.75 V (b) 5.8 V (c) 5.6 V (d) 6.0 V Ans: (a) Explanation: ퟏ 푬 = ퟐퟑퟎ × = ퟏퟏퟓ. ퟒퟕퟑ, ퟐ ퟐ
푹풐풕풐풓 풊풏풅풖풄풆풅 풆풎풇 풑풆풓 풑풉풂풔풆 = 풔 × 푬ퟐ
푹풐풕풐풓 풊풏풅풖풄풆풅 풆풎풇 풑풆풓 풑풉풂풔풆 = ퟎ. ퟎퟓ × ퟏퟏퟓ. ퟒퟕퟑ = 5.77 V
50 31. The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the rms value of the supply voltage at the rated frequency is GATE 2019 (a) magnetizing reactance (b) rotor leakage reactance (c) rotor resistance (d) stator resistance Ans: (a) Explanation:
51 32. A delta-connected, 3.7 kW, 400 V (line), three phase, 4-pole, 50-Hz squirrel-cage induction motor has the following equivalent circuit parameters per phase referred to the stator:푹ퟏ = ퟓ. ퟑퟗΩ, 푹ퟐ = ퟓ. ퟕퟐΩ, 푿ퟏ = 푿ퟐ = ퟖ. ퟐퟐΩ. Neglect shunt branch in the equivalent circuit. The starting line current in amperes (round off to two decimal places) when it is a connected to a 100V (line), 10 Hz, three-phase AC source is GATE 2019
Ans: (14.95) Explanation:
52 33. A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has following equivalent circuit parameters per phase referred to the stator: 푹ퟏ = ퟎ. ퟑΩ, 푹ퟐ = ퟎ. ퟑΩ, 푿ퟏ = 푿ퟐ = ퟎ. ퟒퟏΩ Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3-phase AC source is ______. GATE 2017 Ans: (71.4) Explanation:
53 34. The figure shows the per-phase equivalent circuit of a two- pole three-phase induction motor operating at 50 Hz. The “air-gap” voltage, Vg across the magnetizing inductance, is 210 V rms, and the slip, is 0.005. The torque (in Nm) produced by the motor is______GATE 2015
Ans: 401.69
Explanation:
54 35. The locked rotor current in a 3-phase, star connected 15 kW, 4 pole, 230 V, 50 Hz induction motor at rated conditions is 50A. Neglecting losses and magnetizing current, the approximate locked rotor line current drawn when the motor is connected to a 236V, 57 Hz supply is GATE 2012
(a) 58.5A (b) 45.0 A (c) 42.7 A (d) 55.6 A
Ans: (b)
Explanation:
55 36. A 400 V, 50 Hz 30 hp, three-phase induction motor is drawing 50A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be GATE 2008 (a) 23.06 kW (b) 24.11 kW (c) 25.01 kW (d) 26.21 kW Ans : (c) Explanation:
56 37. A 3-phase, 10 kW, 400 V, 4-pole, 50Hz, star connected induction motor draws 20A on full load. Its no load and blocked rotor test data are given below. GATE 2006 No Load Test : 400 V 6 A 1002 W Blocked Rotor Test : 90 V 15 A 762 W Neglecting copper loss in no load test and core loss in blocked rotor test, estimate motor‟s full load efficiency (a) 76% (b) 81% (c) 82.4% (d) 85% Ans: (b) Explanation:
57 38. For an induction motor, operation at a slip s, the ratio of gross power output to air gap power is equal to (a) ퟏ − 풔 ퟐ (b) ퟏ − 풔
(c) ퟏ − 풔 (d) ퟏ − 풔 GATE 2005 Ans: (b) Explanation:
58 39. The synchronous speed for the seventh space harmonic mmf wave of a 3-phase, 8-pole, 50 Hz induction machine is (a) 107.14 rpm in forward direction GATE 2004 (b) 107.14 rpm in reverse direction (c) 5250 rpm in forward direction (d) 5250 rpm in reverse direction Ans : (a) Explanation:
59 40. No-load test on a 3-phase induction motor was conducted at different supply voltage and a plot of input power versus voltage was drawn. This curve was extrapolated to intersect the y-axis. The intersection point yields (a) Core loss (b) Stator copper loss (c) Stray load loss (d) Friction and windage loss Ans: (d) GATE 2003 Explanation:
60 41. A three phase wound rotor induction motor is to be operated with slip energy recovery in the constant torque mode, when delivers an output power P0 at slip s. Then theoretically, the maximum power that is available for recovery at the rotor terminals is equal to GATE 2000 푷 푷 .풔 (a) 푷 (b) 푷 . 풔 (c) ퟎ (d) ퟎ ퟎ ퟎ ퟏ−풔 ퟏ−풔 Ans: (d) Explanation:
푳풆풕, 풕풉풆 풊풏풑풖풕 풑풐풘풆풓 풃풆 푷풊풏 푮풓풐풔풔 풑풐풘풆풓, 푷ퟎ = (ퟏ − 풔) × 푷풊풏 풔 풔풍풊풑 풑풐풘풆풓 풘풉풊풄풉 풄풂풏 풃풆 풓풆풄풐풗풆풓풆풅 = 풔푷 = × 푷 풊풏 ퟏ − 풔 ퟎ
61 42. A 3-phase squirrel cage induction motor has full load efficiency of 0.8 and a maximum efficiency of 0.9. It is operated at a slip of 0.6 by applying a reduced voltage. The efficiency of the motor at this operating point is (a) less than 0.4 GATE 1998 (b) greater than 0.6 (c) in the range of 0.8±0.1 (d) none of the above
Ans: (a)
Explanation: 푴풐풕풐풓 풆풇풇풊풄풊풆풏풄풚 = ퟏ − 풔 =1-0.6 =0.4
62 43. If an induction motor, if the air gap is increased (a) speed will reduce (b) Efficiency will improve GATE 1996 (c) Power factor will be lowered (d) break down torque will reduce Ans: (c) Explanation:
The magnetizing reactance Xm of the air gap is inversely related to air gap length. As air gap increases, Xm decreases & so magnetizing current drawn by the motor increases.
Operating power factor is improved as the reactive component decreases & stability also increases. The regulation of induction generator will improve as the reactive part decreases & power factor improves.
63 44. Cogging in an induction motor is caused IES/ESE 2019 (a) If the number of stator slots are unequal to number of rotor slots (b) If the number of stator slots are an integral multiple of rotor slots (c) If the motor is running at fraction of its rated speed (d) Due to ퟓ풕풉 harmonic Ans: (b)
Explanation:
Cogging is a phenomenon in squirrel cage induction motors which results from magnetic locking between stator core and rotor core and prevents the motor from running.
It happens if the stator number of slots are equal to rotor number of slots or the ratio between the two is an integer.
64 45. A 400V, 50Hz, 30hp, three phase induction motor is drawing 50A, current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5kW and 900W respectively. The friction and windage losses are 1050W and core losses are 1200W. The air gap power of the motor will be, nearly
(a) 15 kW (b) 20 kW (c) 25 kW (d) 30 kW Ans: (c) IES/ESE 2018
Explanation:
65 46. A 3-phase, 37 kW, induction motor has an efficiency of 90% when delivering full load. At this load the stator copper losses and rotor copper losses are equal and are equal to stator iron losses. The mechanical losses are one-third of no-load losses. then the motor runs at a slip of IES/ESE 2018
(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.04 Ans: (c) Explanation:
66 47. A 3-phase induction motor operating at a slip of 5% develops 20 kW rotor power output. What is the corresponding rotor copper loss in this operating condition? IES/ESE 2016
(a) 750 W (b) 900 W (c) 1050 W (d) 1200 W Ans: (c) Explanation:
67 48. The stator loss of a 3-phase induction motor is 2 KW. If the motor is running with a slip of 4% and power input of 90 kW, then what is the rotor mechanical power developed? (a) 84.48 kW (b) 86.35 kW (c) 89.72 kW (d) 90.52 kW IES/ESE 2016 Ans: (a) Explanation:
68 49. A 3–phase induction motor draws 50 kW from a 220 V, 50 Hz mains. The rotor emf makes 100 oscillations/minute. If the stator losses are 2 kW the rotor copper loss would be
(a) 0.16 kW (b) 0.32 kW (c) 1.6 kW (d) 3.2 kW
Ans: (c) IES/ESE 2014 Explanation:
69 50. What is the rotor copper loss of a 3 phase 550 Volt, 50 Hz, 6 poles induction motor developing 4·1 kW at the shaft with mechanical loss of 750 W at 970rpm? IES/ESE 2009 (a) 175 W (b) 150 W (c) 100 W (d) 250 W Ans: (b) Explanation:
70 51. The power input to a 415 V, 50 Hz, 6 pole 3·phase induction motor running at 975 rpm is 40 kW. The stator losses are 1 kW and friction and windage losses total 2 kW. What is the efficiency of motor? IES/ESE 2009
(a) 92.5% (b) 92% (c) 90% (d) 88% Ans: (c) Explanation: power i/p = 40 kw Slip = 0.025 Rotor i/p = 40 – 1 = 39 kw Pm = (1 - S) 39 = 38.025 kw Shaft o/p = 38.025 – 2 = 36.025 kw 71 ∴ η = 36.025/40 = 90% 52. An induction motor when started on load does not accelerate up ퟏ 풕풉 to full speed but runs at of the rated speed . The motor is ퟕ said to be IES/ESE 2001 (a) Locking (b) plugging (c) Crawling (d) Cogging Ans: (c) Explanation:
Induction motor runs nearer to synchronous speed but this is a
peculiar behaviour when the motor runs at a speed near to Ns/7 falling below the rated torque and unable to accelerate to its full speed. This is called crawling. Crawling is due the presence of slot harmonics mainly due to the dominant presence of 7th harmonic.
72 53. If the rotor power factor of a 3-phase induction motor is 0.866, the spatial displacement between the stator magnetic field and rotor magnetic field will be IES/ESE 2000 (a) ퟑퟎퟎ (b) ퟗퟎퟎ (c) ퟏퟐퟎퟎ (d)ퟏퟓퟎퟎ Ans: (c) Explanation:
Power factor, cos ∅ = ퟎ. ퟖퟔퟔ ∅ = ퟑퟎퟎ 푺풑풂풕풊풂풍 풅풊풔풕풂풏풄풆(휹) = ퟗퟎퟎ + ∅ 푺풑풂풕풊풂풍 풅풊풔풕풂풏풄풆 = ퟗퟎퟎ + ퟑퟎퟎ 휹 = ퟏퟐퟎퟎ
73 54. The stator of a 6–pole, 3-phase induction motor is fed from a 3–phase, 50 Hz supply which contains a pronounced fifth time harmonic. The speed of the fifth space harmonic field produced by the fifth time harmonic in stator supply will be (a) 200 rpm (b) 1500 rpm (c) 1000 rpm (d) 5000 rpm IES/ESE 2000 Ans: (a) Explanation:
ퟏ 푺풑풆풆풅 풐풇 ퟓ풕풉 풉풂풓풎풐풏풊풄 풇풊풆풍풅 = × 푵 ퟓ 풔 ퟏퟐퟎ×ퟓퟎ 푵 = = ퟏퟎퟎퟎ 풓풑풎 풔 ퟔ ퟏ 푺풑풆풆풅 풐풇 ퟓ풕풉 풉풂풓풎풐풏풊풄 풇풊풆풍풅 = × ퟏퟎퟎퟎ = ퟐퟎퟎ풓풑풎 ퟓ
74 55. Crawling in an induction motor is due to (a) time harmonics in supply IES/ESE 1999 (b) slip ring rotor (c) space harmonics produced by winding currents (d) insufficient starting torque
Ans: (c) Explanation:
There will be a maximum torque at a speed just below ( 1 / 7)Ns, and if this is high enough, the net torque produced can be higher than the torque due to the line frequency even though speed is ( 1 / 7)Ns.
The slip all this while remaining high. This can cause the motor to crawl at a speed just below (1 / 7) of synchronous speed. Thus crawling of IM is due to space harmonics.
75 56. A 3-phase slip-ring induction motor when started picks up speed but runs stably at about half the normal speed. This is because of IES/ESE 1998 (a) unbalance in the supply voltages (b) non-sinusoidal nature of the supply voltage (c) stator circuit asymmetry (d) rotor circuit asymmetry Ans: (b) Explanation:
76 57. The phenomenon of crawling in a 3-phase cage induction
motor may be due to IES/ESE 1998 (a) unbalanced supply voltage (b) 7th space harmonics of air-gap voltage (c) 7th time harmonics of voltage wave (d) 5th space harmonics
Ans: (b) Explanation: squirrel cage induction motors exhibits a tendency to run at very slow speeds (as low as one-seventh of their synchronous speed). This phenomenon is called as crawling of an induction motor.
This action is due to the fact that, flux wave produced by a stator winding is not purely sine wave.
77 58. For a more accurate analysis of the torque slip relationship, the equivalent circuit of a 3-phase induction motor for running condition is to be derived from the given standstill condition: K= ratio of transformation, E1=V1(applied voltage- 푰흁푿푳 ), at running condition, if the two reactance in the equivalent circuit are denoted by A and B, then the values of A and B will be ퟐ (a) 풌 푹ퟏ풂풏풅 푹ퟐ ퟐ (b) 풔풌 푹ퟏ풂풏풅 푹ퟐ ퟐ (c) 풌 푹ퟏ풂풏풅 풔푹ퟐ ퟐ (d) 풔풌 푹ퟏ풂풏풅 풔푹ퟐ Ans: (a) IES/ESE 1998 Explanation:
78 59. The rotor power output of a 3-phase induction motor is 15kW. The corresponding slip is 4%. The rotor copper loss will be (a) 600W (b) 625W (c) 650W (d) 700W
Ans: (b) IES/ESE 1997 Explanation:
Rotor copper losses = rotor input- rotor output and rotor output power = (1-s) rotor input power ∴ rotor Input power = output/(1 – s)= 15000 /(1 – 0.04) = 15625 Rotor copper losses = 15625-15000= 625 W
79 60. A 3-phase induction motor has rotor resistance R2, standstill rotor induced emf E2, and stator to rotor effective turns ratio of m. In an equivalent circuit of this machine, the rotor circuit 푹 resistance is shown as 풎ퟐ. ퟐ, where „s‟ is the slip. This 풔 implies that the value of the equivalent rotor circuit voltage will be
(a) E2 (b) s E2 (c) m E2 (d) m sE2 IES/ESE 1993 Ans: (c)
80 61. The presence of a dominant 7th harmonic in the winding distribution of a 3-phase, 6-pole, 50Hz induction motor may cause the motor to crawl at a speed of about (a) 750 rpm (b) 500 rpm (c) 242 rpm (d) 143 rpm IES/ESE 1993 Ans: (d)
Explanation:
ퟏퟐퟎ×ퟓퟎ 푵 = = ퟏퟎퟎퟎ 풓풑풎 풔 ퟔ
ퟏ ퟏ ퟕ풕풉 풉풂풓풎풐풏풊풄 푺풑풆풆풅 = × 푵 = × ퟏퟎퟎퟎ = ퟏퟒퟑ 풓풑풎 풔 풔 ퟕ
81 62. Cogging of induction motor occurs when (a) number of stator teeth-number of rotor teeth =odd number (b) number of stator teeth-number of rotor teeth =even number (c) number of stator teeth-number of rotor teeth =0 (d) number of stator teeth-number of rotor teeth =negative number Ans: (c) IES/ESE 1992 Explanation:
Squirrel cage rotors are not designed for any number of poles on the rotor. As the poles are automatically formed in equal number, due to low starting if the rotor poles and stator poles are quite horizontal to each other or they are even multiple to each other, a strong alignment force occurs. Due to this locking tendency the rotor refuses to start which is called as cogging.
Slip ring rotor design is quite complicated compared to squirrel cage with respect to cogging effect. The number of poles on the rotor should not be equal to stator poles. To achieve this rotor slots are always less than stator slots. The rotor poles also should not be in integral multiples of stator slots.
82 Assignment questions
1. An induction motor is, in general analogous to (a) two winding transformer with secondary short circuited. (b) two winding transformer with secondary open circuited. (c) auto transformer. (d) none of the above.
83 2. The power factor of squirrel cage induction motor is (a) low at light loads only (b) low at heavy loads only (c) both A and B (d) low at rated load only
84 3. A 3-phase, 50 Hz, 4 pole induction motor is running at 1440 rpm and rotor input power is 2KW. Find the rotor copper loss? (a) 40W (b) 60W (c) 80W (d) 100W
85 4. A 3-phase, 50Hz, 4 pole induction motor is running at 1440 rpm and rotor input power is 2KW. Find the rotor output power? (a) 2000W (b) 1450W (c) 1820W (d) 1920W
86 5. A 3-phase, 50 Hz, 4 pole induction motor is running at 1440 rpm. Find the rotor efficiency? (a) 96% (b) 92% (c) 94% (d) 95%
87 For any queries/clarifications/suggestions...feel free to contact through... [email protected]
88
Solutions for Assignment questions
1. An induction motor is, in general analogous to (a) two winding transformer with secondary short circuited. (b) two winding transformer with secondary open circuited. (c) auto transformer. (d) none of the above. Ans: (a) Explanation:
The rotor consists of thick laminations, made up of silicon steel with skewed slots which has less depth. Solid copper conductors are placed which are short circuited at both ends using end rings made up of copper.
This gives a completely closed rotor circuit. Induction motor stator winding is equivalent to primary of a transformer and its rotor winding is equivalent to secondary of a transformer but it is short circuited with
a rotating mechanical equivalent of r2 / s. 91 2. The power factor of squirrel cage induction motor is (a) low at light loads only (b) low at heavy loads only (c) both A and B (d) low at rated load only Ans: (a)
Explanation:
The operating power factor on no load or light loads is very low, because of magnetizing current.
The losses on no load are only core losses which are negligible.
Therefore magnetizing current is greater than core loss component current and the no load power factor is very low around 0.2 lagging.
As the load increases the core loss component current will increase and the power factor will improve.
92 3. A 3-phase, 50 Hz, 4 pole induction motor is running at 1440 rpm and rotor input power is 2KW. Find the rotor copper loss? (a) 40W (b) 60W (c) 80W (d) 100W
Ans: (c) Explanation:
풓풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = 풔 × 풓풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓
ퟏퟐퟎ × ퟓퟎ 푵 = = ퟏퟓퟎퟎ 풔 ퟒ
푵 −푵 ퟏퟓퟎퟎ−ퟏퟒퟒퟎ 푺풍풊풑, 풔 = 풔 = = ퟎ. ퟎퟒ 푵 ퟏퟓퟎퟎ
풓풐풕풐풓 풄풐풑풑풆풓 풍풐풔풔 = ퟎ. ퟎퟒ × ퟐퟎퟎퟎ = ퟖퟎ 푾
93 4. A 3-phase, 50Hz, 4 pole induction motor is running at 1440 rpm and rotor input power is 2KW. Find the rotor output power? (a) 2000W (b) 1450W (c) 1820W (d) 1920W
Ans: (d)
Explanation: ퟏퟐퟎ × ퟓퟎ 푵 = = ퟏퟓퟎퟎ 풔 ퟒ
푵 −푵 ퟏퟓퟎퟎ−ퟏퟒퟒퟎ 푺풍풊풑, 풔 = 풔 = = ퟎ. ퟎퟒ 푵 ퟏퟓퟎퟎ 풓풐풕풐풓 풐풖풕풑풖풕 풑풐풘풆풓 = ퟏ − 풔 × 풓풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓 = ퟏ − ퟎ. ퟎퟒ × ퟐퟎퟎퟎ = ퟏퟗퟐퟎ푾
94 5. A 3-phase, 50 Hz, 4 pole induction motor is running at 1440 rpm. Find the rotor efficiency? (a) 96% (b) 92% (c) 94% (d) 95% Ans: (a) Explanation: 푹풐풕풐풓 풐풖풕풑풖풕 풑풐풘풆풓 푵 푹풐풕풐풓 풆풇풇풊풄풊풆풏풄풚 = = ퟏ − 풔 = 푹풐풕풐풓 풊풏풑풖풕 풑풐풘풆풓 푵풔
ퟏퟐퟎ × ퟓퟎ 푵 = = ퟏퟓퟎퟎ 풔 ퟒ
ퟏퟒퟒퟎ 푹풐풕풐풓 풆풇풇풊풄풊풆풏풄풚 = × ퟏퟎퟎ = ퟗퟔ% ퟏퟓퟎퟎ
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