Electric Machines I Three Phase Induction Motor Dr. Firas Obeidat 1
Table of contents 1 General Principles 2 Construction 3 Production of Rotating Field 4 Why Does the Rotor Rotate 5 The Slip and Rotor Current Frequency 6 The Equivalent Circuit in an Induction Motor 7 Losses and the Power Flow Diagram 8 Torque-Speed Curve 2
General Principles Conversion of electrical power into mechanical power takes place in the rotating part of an electrical motor. In AC motors, the rotor does not receive electrical power but conduction by induction in the same way as the secondary of 2-winding transformer receives its power from the primary winding. Induction motor can be treated as a rotating transformer i.e. one in which primary winding is stationary but the secondary is free to rotate. All of the ac motors, the polyphase induction motor is the one which is extensively used for various kinds of industrial drives. 3
General Principles Advantages Disadvantages 4
Construction An induction motor consists mainly of two main parts, Stator and Rotor Stator The stator of induction motor is made up of a number of stampings, which are slotted to receive the windings. The stator carries 3-phase winding and is fed from a 3-phase supply. It is wound for definite number of poles, the exact number of poles being determined by the requirements of speed. When the stator winding supplied with 3-phase current, produce magnetic flux, which is of constant magnitude but which revolves (or rotates) at synchronous speed. This revolving magnetic flux induces an emf in the rotor by mutual induction. 5
Construction Rotor Squirrel-cage Rotor Most of the induction motors are squirrel cage type, because this type of rotor has the simplest and most rugged construction imaginable and is almost indestructible. The rotor consists of cylindrical laminated core with parallel slots for carrying the rotor conductors which are not wires but consist of heavy bars of copper, aluminums or alloys. One bar is placed in each slot. The rotor bars are brazed or electrically welded or bolted to two heavy and stout short circuiting end ring. 6
Construction Rotor Phase-wound Rotor A wound rotor has a complete set of three-phase windings that are similar to the windings on the stator. The three phases of the rotor windings are usually Y- connected, and the ends of the three rotor wires are tied to slip rings on the rotor's shaft. The rotor windings are shorted through brushes riding on the slip rings. These three brushes are further connected externally to 3-phase star connected rheostat. This make possible the introduction of additional resistance in the rotor circuit during starting period for increasing the starting torque of the motor. Wound-rotor induction motors are more expensive than cage induction motors, and they require much more maintenance because of the wear associated with their brushes and slip rings. As a result, wound-rotor induction motors are rarely used. 7
Production of Rotating Field When stationary coils wound for three phase are supplied by three phase supply, a uniformly rotating (or revolving) magnetic flux of constant value is produced. When three phase winding displaced in space by 120 o, are fed by three phase current displaced in time by 120 o, they produce a resultant magnetic flux which rotates in space as if actual magnetic poles were being rotated mechanically. 3-phase, 2-poles stator having three identical windings places 120 o space degree The flux due to three phase windings Positive direction of fluxes 8
Production of Rotating Field The maximum value of flux due to any one of the three phases is ϕ m. The resultant flux ϕ r (at any instant) is given by the vector sum of the individual fluxes ϕ 1,ϕ 2, and ϕ 3 due to three phases. Let φ 1 = φ m sinωt φ 2 = φ m sin(ωt 120) φ 3 = φ m sin(ωt 240) (1) When θ=0 o ϕ 1 =0 o, φ 2 = 3 2 φ m, φ 3 = 3 2 φ m φ r = 2 3 2 φ 60 cos m 2 = 3 3 2 φ m = 3 2 φ m (2) When θ=60 o φ 1 = 3 φ 2 m, φ 2 = 3 φ 2 m, φ 3 = 0 φ r = 2 3 2 φ 60 cos m 2 = 3 3 2 φ m = 3 2 φ m 9
Production of Rotating Field (3) When θ=120 o 60 o ϕ 3 ϕ 1 φ 1 = φ r = 2 3 φ 2 m, φ 2 = 0, φ 3 = 3 φ 2 m 3 2 φ 60 cos m 2 = 3 2 φ m -ϕ 2 Φ r =1.5Φ m Φ r =1.5Φ m 60 o -ϕ 2 (4) When θ=180 o (1) θ=0 o (2) θ=60 o ϕ 1 =0 o, φ 2 = 3 φ 2 m, φ 3 = 3 φ 2 m Φ r =1.5Φ m Φ r =1.5Φ m φ r = 2 3 2 φ m cos 60 2 φ r = 3 3 2 φ m = 3 2 φ m From the above four positions, it can be concluded that: 1. The resultant flux is constant value and equal to 1.5 ϕ m. 2. The resultant flux rotates around the stator at synchronous speed given by N s =120f s /p. ϕ 1 (3) θ=120 o 60 o -ϕ 3 -ϕ 3 60 o (4) θ=180 o ϕ 2 10
Why Does the Rotor Rotate When 3-phase stator windings are fed by 3-phase supply, a magnetic flux of constant magnitude, but rotating at synchronous speed, is set up. The flux passes through the air gap, sweeps past the rotor surface and so cuts the rotor conductors which, as yet, are stationary. Due to relative speed between the rotating flux and the stationary conductors an e.m.f. is induced in the conductors According to faraday s law. The frequency of the induced e.m.f. is the same as the supply frequency. The e.m.f. magnitude is proportional to the relative velocity between the flux and the conductors, and its direction is given by Fleming right hand rule. Since the rotor bars or conductors form a closed circuit, rotor current is produced whose direction, as given by Lenz s law, is such as to oppose the very cause producing it. The cause which produces the rotor current is the relative velocity between the rotating flux of the stator and the stationary rotor conductors. To reduce the relative speed, the rotor starts running in the same direction as that of the flux and tries to catch up with the rotating flux. 11
Why Does the Rotor Rotate The setting up of the torque for rotating the rotor Figure (a) is shown the stator field which is assumed to be rotating clockwise. The relative motion of the rotor with respect to the stator is anticlockwise. By applying right hand rule, the direction of the induced e.m.f. in the rotor is found to be outwards. The direction of the flux due to rotor current alone is as shown in figure (b). By applying left hand rule or by combined field as shown in figure ( c), the rotor conductors experience a force tending to rotate them in clockwise direction. So, the rotor is set into rotation in the same direction as that of the stator flux. 12
The Slip and Rotor Current Frequency The rotor never succeeds in catching up with the stator field. If it really did, then there would be no relative speed between the two, hence no rotor e.m.f., no rotor current and so no torque to maintain rotation. The slip (s) is the difference between the synchronous speed N s and the actual speed N of the rotor. s = N s N m N s N m = (1 s)n s N slip = N s N m 100% s = ω s ω m ω s ω m = (1 s)ω s is called slip speed 100% Where N s = synchronous speed in rpm N m = rotor speed (mechanical shaft speed) in rpm ω s = synchronous angular velocity (2πN s /60) in rad/s ω m = mechanical angular velocity (2πN m /60) in rad/s When the rotor is stationary, the frequency of rotor current is the same as the supply frequency. When the rotor starts revolving, the frequency depends upon the relative speed or on the slip speed. 13
The Slip and Rotor Current Frequency at any slip speed, the frequency of the rotor be f r N s N m = 120f r P N s = 120f s P Dividing the above equations one by other Or f r f s = N s N m N s f r = sf s = s Where N s = synchronous speed in rpm N m = rotor speed (mechanical shaft speed) in rpm P=number of poles f s =stator frequency in Hz f r = rotor frequency in Hz Substitute s = N s N m N s in the above equation gives f r = sf s = N s N m P f N s = (N s N m ) f s 120f s s f r = P 120 (N s N m ) 14
The Slip and Rotor Current Frequency When the rotor stationary (at standstill) N m = 0 rpm, the rotor frequency f r =f s and the slip s=1. At N m = N s, the rotor frequency f r = 0 Hz, and the slip s=0. Example: A 208-V, 10-hp, four-pole, 60 Hz, Y connected induction motor has a full-load slip of 5 percent. (a) What is the synchronous speed of this motor? (b) What is the rotor speed of this motor at the rated load? (c) What is the rotor frequency of this motor at the rated load? (a) N s = 120f s P = 120 60 4 = 1800 rpm (b) N m = 1 s N s = 1 0.05 1800 = 1710 rpm (c) f r = sf s = 0.05 60 = 3 Hz 15
The Equivalent Circuit in an Induction Motor The largest relative motion occurs when the rotor is stationary, called the locked-rotor or blocked-rotor condition, so the largest voltage and rotor frequency are induced in the rotor at that condition. The smallest voltage (0 V) and frequency (0 Hz) occur when the rotor moves at the same speed as the stator magnetic field, resulting in no relative motion. The magnitude and frequency of the voltage induced in the rotor at any speed between these extremes is directly proportional to the slip of the rotor. If the magnitude of the induced rotor voltage at locked-rotor conditions is called E 2 the magnitude of the induced voltage at any slip will be given by the equation E r = se 2 The frequency of the induced voltage at any slip will be given by the equation f r = sf s The reactance of an induction motor rotor depends on the inductance of the rotor and the frequency of the voltage and current in the rotor. With a rotor inductance of L r the rotor reactance is given by 16
The Equivalent Circuit in an Induction Motor X r = 2πf r L r = 2πsf s L r = sx 2 Where X 2 is the locked rotor reactance The rotor current flow is I 2 = E r R r + X r = se 2 R 2 + sx 2 = E 2 R 2 s + X 2 To produce the final per-phase equivalent circuit for an induction motor, it is necessary to refer the rotor parts of the model over to the stator side. The turn ratio of the induction motor is a = E 1 E 2 So E 2 = ae 2 I 2 = I 2 a R 2 = a 2 R 2 X 2 = a 2 X 2 17
The Equivalent Circuit in an Induction Motor Part of the power coming across the air gap in an induction motor is consumed in the rotor copper losses, and part of it is converted to mechanical power to drive the motor's shaft. It is possible to separate the two uses of the air-gap power and to indicate them separately on the motor equivalent circuit. In order to separate the rotor copper losses and the converted power to mechanical power, the equivalent circuit of the induction motor as the figure below R 1 I 1 X 1 I o I 2 ' R 2 ' X 2 ' V 1 I c R c X m I m E 2 '=E 1 R 2 ' ( 1-s ) s 18
Losses and the Power Flow Diagram 19
Losses and the Power Flow Diagram The following power relations in an induction motor can be deduced Input power P 1 = 3V 1 I 1 cosθ 1 = 3V L I L cosθ 1 Stator copper losses P Cs = 3I 1 2 R 1 Stator core losses P f = I 2 C R C 2 R Power transferred to rotor (air gap power) P 2 = 3I 2 2 P 2 = P 1 P Cs P f s Rotor copper losses P Cr = 3I 2 2 R2 = sp 2 2 1 s Mechanical power P m = 3I 2 R2 s P m = P 2 P Cr = P 2 sp 2 = (1 s)p 2 The gross torque developed by the rotor (air gap torque) T g is T g = P m = 3I 2 2 R2 ω m 2πN m 60 1 s s Nm 20
Losses and the Power Flow Diagram But T g = P m = 3I 2 1 s 2 R2 s ω m 2π(1 s)n s 60 Nm = 3I 2 2 R 2 s 2πN s 60 The approximate circuit for the induction motor as shown in the figure, from this figure I 2 can be found as I 2 = N m = (1 s)n s R 1 + R 2 V 1 s + j X 1 + X 2 V 1 Nm = P 2 ω s Nm I 1 I c R c I o X m I 2 ' I m E 2 '=E 1 R eq =R 1 +R 2 ' X eq =X 1 +X 2 ' R 2 ' ( 1-s ) s I 2 = (R 1 + R 2 V 1 s) 2 +(X 1 + X 2 ) 2 T g = 3V 1 2 ω s (R 1 + R 2 R 2 s s) 2 +(X 1 + X Nm 2 ) 2 Output power= mechanical power Rotational (Windage & Friction) loss P out = P m P w 21
Examples The efficiency of the induction motor is The output torque η = P out P 1 100% T out = P out ω m Examlpe: A 480-V L, 60-Hz, 50-hp, three-phase induction motor is drawing 60A at 0.85 pf lagging. The stator copper losses are 2 kw, and the rotor copper losses are 700 W. The rotational losses are 600 W, the core losses are 1800 W. Find the following quantities: (a) The air-gap power P 2 (b) The power converted P m (c) The output power P out (d) The efficiency of the motor (a) P 1 = 3V L I L1 cosθ 1 = 3 480 60 0.85 = 42.4kW P 2 = P 1 P Cs P f = 42.4 2 1.8 = 38.6kW (b) P m = P 2 P Cr = 38.6 0.7 = 37.9kW (c) (d) P out = P m P w = 37.9 0.6 = 37.3kW η = P out = 37.3 100% = 88% P 1 42.4 22
Examples Examlpe: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R 1 =0.641Ω X 1 =1.106 Ω R 2 =0.332Ω X 2 =0.464Ω X m =26.3Ω The total rotational (windage and friction) losses are 1100 W and are assumed to be constant. The core loss is jumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor's (a) Speed (b) Stator current (c) Power factor (d) Air gap power and output power (e) The air gap torque T g and load torque T out (f) Efficiency (a) N s = 120f s 120 60 = = 1800 rpm P 4 N m = 1 s N s = 1 0.022 1800 = 1760.4 rpm (b) Z 2 = R 2 Z o = Z o = s + jx 2 = 0.332 0.022 + j0.464 = 15.09 + j0.464 = 15.1 1.76o Ω 1 1 jx m + 1 Z 2 = 1 0.0773 31.1 = 12.94 31.1o Ω 1 1 j26.3 + 1 15.1 1.76 = 1 j0.038 + 0.0662 1.76 23
Examples Z T = Z 1 + Z o = 0.641 + j1.106 + 12.94 31.1 = 11.72 + j7.79 = 14.07 33.6 o Ω (c) (d) I 1 = V 1 = 460/ 3 = 18.88 33.6 A Z T 14.07 33.6o pf = cos33.6 = 0.83 lagging P 1 = 3V L I L1 cosθ 1 = 3 460 18.88 0.83 = 12.53kW P Cs = 3I 1 2 R 1 = 3 18.88 2 0.641 = 0.685 kw P 2 = P 1 P Cs P f = 12.53 0.685 0.0 = 11.845 kw P m = P 2 P Cr = P 2 sp 2 = 1 s P 2 = 1 0.022 11845 = 11585 W P out = P m P w = 11585 1100 = 10485 W (e) (f) T g = P 2 ω s = 11845 2πN s 60 = 11845 2π1800 60 = 11845 188.5 = 62.8 Nm T out = P out = 10485 = 10485 = 56.9 Nm ω m 2π1760.4 184.3 60 η = P out = 10485 100% = 83.7% P 1 12530 24
Torque-Speed Curve The torque speed (slip) curve for an induction motor gives us the information about the variation of torque with the slip. When the rotor stationary (at standstill) N m = 0 rpm, the rotor frequency f r =f s and the slip s=1. At N m = N s, the rotor frequency f r = 0 Hz, and the slip s=0. Torque 2.5T fl 1.5T fl T fl 0 0 Starting Torque 20 40 60 Speed % Maximum Torque Full Load Torque 80 N m 100 0 100 80 60 40 20 0 Slip % N s At full load, the motor runs at speed of N m. When mechanical load increases, motor speed decreases tell the motor torque again becomes equal to the load torque. As long as the two torques are in balance, the motor will run at constant (but lower) speed. If the load torque exceeds the induction motor maximum torque, the motor will suddenly stop. 25
Comments on the Induction Motor Torque- Speed Curve 1. The induced torque of the motor is zero at synchronous speed. 2. The torque- speed curve is nearly linear between no load and full load. In this range, the rotor resistance is much larger than the rotor reactance, so the rotor current, the rotor magnetic field, and the induced torque increase linearly with increasing slip. 3. There is a maximum possible torque that cannot be exceeded. This torque, called the pullout torque or breakdown torque, is 2 to 3 times the rated full load torque of the motor. 4. The starting torque on the motor is slightly larger than its full-load torque. So this motor will start carrying any load that it can supply at full power. 26
Comments on the Induction Motor Torque- Speed Curve 5. The torque on the motor for a given slip varies as the square of the applied voltage. This fact is useful in one form of induction motor speed control. 6. If the rotor of the induction motor is driven faster than synchronous speed, then the direction of the induced torque in the machine reverses and the machine becomes a generator, converting mechanical power to electric power. 7. If the motor is turning backward relative to the direction of the magnetic fields, the induced torque in the machine will stop the machine very rapidly and will try to rotate it in the other direction. Since reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases, this fact can be used as a way to very rapidly stop an induction motor. The act of switching two phases in order to stop the motor very rapidly is called plugging. 27
Maximum Torque in Induction Motor Since the induced torque is equal to P 2 /ω s. the maximum possible torque occurs when the air-gap power is maximum. Since the air-gap power is equal to the power consumed in the resistor R 2 /s, the maximum induced torque will occur when the power consumed by that resistor is maximum. T g = 3V 1 2 ω s (R 1 + R 2 R 2 s s) 2 +(X 1 + X 2 ) 2 Nm dt g ds = 0 2 dt g ds = 3V (R 1 + R 2 1 ω s s) 2 +(X 1 + X 2 ) 2 R 2 s 2 2 R 2 R s R 1 + R 2 s 2 s 2 (R 1 + R 2 s) 2 +(X 1 + X 2 ) 2 2 = 0 (R 1 + R 2 s) 2 +(X 1 + X 2 ) 2 R 2 s 2 2 R 2 s R 1 + R 2 s R 2 s 2 = 0 (R 1 + R 2 s) 2 +(X 1 + X 2 ) 2 = 2R 1R 2 s + 2R 2 2 s 2 28
Maximum Torque in Induction Motor R 1 2 + 2R 1R 2 R 1 2 + (X 1 + X 2 ) 2 = R 2 2 R 2 s R 1 2 + (X 1 + X 2 ) 2 = R 2 2 + 2R 2 2 s 2 + (X 1 + X 2 ) 2 = 2R 1R 2 s s 2 s 2 s = ± R 1 2 + (X 1 + X 2 ) 2 + 2R 2 2 s 2 s max = ± R 2 R 1 2 + (X 1 + X 2 ) 2 The plus (+) sign for motor. The minus (-) sign for generator Substitute s max in torque equation to get the maximum torque equation T gmax = 3V 1 2 2ω s R 1 + 1 R 1 2 + (X 1 + X 2 ) 2 Nm 29
Maximum Torque in Induction Motor The maximum torque is proportional to the square of the supply voltage and is also inversely related to the size of the stator impedances and the rotor reactance. The smaller a machine's reactances, the larger the maximum torque it is capable of achieving. slip at which the maximum torque occurs is directly proportional to rotor resistance, but the value of the maximum torque is independent of the value of rotor resistance. It is possible to insert resistance into the rotor circuit of a wound rotor because the rotor circuit is brought out to the stator through slip rings. As the rotor resistance is increased, the pullout speed of the motor decreases. but the maximum torque remains constant. If a resistance is inserted into the rotor circuit, the maximum torque can be adjusted to occur at starting conditions. Therefore. The maximum possible torque would be available to start heavy loads. On the other hand, once the load is turning, the extra resistance can be removed from the circuit, and the maximum torque will move up to near-synchronous speed for regular operation. 30
Maximum Torque in Induction Motor Example: A two-pole, 50-Hz induction motor supplies 15 kw to a load at a speed of 2950 r/min. Neglecting the rotational losses. (a) What is the motor 's slip? (b) What is the induced torque in the motor in N.m under these conditions? (c) How much power will be supplied by the motor when the torque is doubled at the same speed? (a) N s = 120f s P = 120 50 2 = 3000 rpm s = N s N m N s 100% = 3000 2950 3000 100% = 1.66% (b) T g = P m ω m = 15000 2π2950 60 = 48.6 Nm (c) P m = T g ω m = 48.6 2 2π2950 60 = 29.5 kw 31
Maximum Torque in Induction Motor Example: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit: R 1 =0.641Ω X 1 =1.106 Ω R 2 =0.332Ω X 2 =0.464Ω X m =26.3Ω (a) What is the maximum torque of this motor? At what speed and slip does it occur? (b) What is the starting torque of this motor? (c) When the rotor resistance is doubled, what is the speed at which the maximum torque now occurs? What is the new starting torque of the motor? (a) T gmax = 3V 1 2 N s = 120f s P 2ω s 1 R 1 + R 1 2 + (X 1 + X 2 ) 2 = 120 60 4 = 1800 rpm ω s = 2πN s 60 = 2π1800 = 188.5 rad/sec 60 T gmax = 3(460/ 3)2 2 188.5 1 0.641 + 0.641 2 + (1.106 + 0.464) 2 = 240Nm 32
Maximum Torque in Induction Motor s max = R 2 R 1 2 + (X 1 + X 2 ) 2 = 0.332 0.641 2 + (1.106 + 0.464) 2 = 0.196 (b) At standstill s=1 T gstart = 3V 1 2 ω s (R 1 + R 2 R 2 s s) 2 +(X 1 + X 2 ) 2 Nm T g = 3(460/ 3)2 188.5 0.332 (0.641 + 0.332) 2 +(1.106 + 0.464) 2 = 109Nm (b) If the rotor resistance is doubled, then the slip at maximum torque doubles, too. s max new = 0.196 2 = 0.392 N m new = 1 s maxnew N s = 1 0.392 1800 = 1094.4 rpm T gstart = 3V 1 2 ω s T gstart new = (R 1 + R 2 3(460/ 3)2 188.5 R 2 s s) 2 +(X 1 + X 2 ) 2 Nm (2 0.332) (0.641 + (2 0.332)) 2 +(1.106 + 0.464) 2 = 170Nm 33
Complete Torque- Speed Curve of a Three Phase Machine Three phase machine can be run as motor when it takes electrical power and supplies mechanical power. The direction of torque and rotor rotation are in the same. For this case 0<N m <N s, 1<s<0. The same machine can be used as an asynchronous generator when driven at speed greater than the synchronous speed. In this case, it receives mechanical energy from the stator. The torque is oppositely-directed. For this case N m >N s, s<0. The same machine can be used as a brake during the plugging period. For this case N m in opposite direction, s>1. s>1 0 <s<1 s=1 s=0 s<0 34
Induction Motor Operating as Generator When run faster than its synchronous speed, an induction motor runs as a generator called induction generator. The induction generator converts the mechanical power it receives into electrical energy and this energy is released by the stator. As soon as the motor speed exceeds its synchronous speed, it starts delivering active power P to the 3-phase line. However, for creating its own magnetic field, it absorbs reactive power Q from the line to which it connected. Q flows in the opposite direction to P. 35
Plugging of an Induction Motor An induction motor can be quickly stopped by inter-changing any of its two stator leads. It reverses the direction of the revolving flux which produces a torque in the reversed direction. Thus applying brake on the motor. During this so-called plugging period, the motor acts as a brake. It absorbs kinetic energy from the still revolving load causing its speed to fall. The associated power P m is dissipated as heat in the rotor. At the same time, the rotor also continues to receive power P 2 from the stator which also dissipated as heat. Plugging produces rotor I 2 R losses which even exceed those when the rotor is locked. 36
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