Presentation on theme: "Three phase induction motors"— Presentation transcript:
1 Three phase induction motors Three-phase induction motors are the motors most frequently encountered in industry. They are simple, low-priced, and easy to maintainSuch machines are called induction machines because the rotor voltage (which produces the rotor current and the rotor magnetic field) is induced in the rotor windings rather than being physically connected by wires.In this chapter we cover the basic principles of the 3-phase induction motor and develop the fundamental equations describing its behavior
3 Stator It consists of a core of stacked, insulated, iron laminations, with windings of insulated copper wire filling the slots in the core
4 Squirrel-Cage Rotor2aleb inside aluminium liquidThe rotor consists of a shaft, a steel laminated rotor, and an embedded copper or aluminum squirrel cage
5 Wound RotorA wound rotor has a 3-phase winding, similar to the one on the stator. The winding is uniformly distributed in the slots and is usually connected in 3- wire wye. The terminals are connected to three slip-rings, which turn with the rotor. The revolving slip-rings and associated stationary brushes enable us to connect external resistors in series with the rotor winding.Typical wound rotors for induction motors. Notice the slip rings and the bars connecting the rotor windings to the slip rings.Ghale, este3mel 5ass jedan, machine connecting to other machineUsage less than 5%The external resistors are mainly used during the start-up period; under normal running conditions, the three brushes are short-circuited.
6 a. Molten aluminum is poured into a cylindrical cavity a. Molten aluminum is poured into a cylindrical cavity. The laminated rotor stacking is firmly held betweentwo molds.b. Compressed air rams the mold assembly into the cavity. Molten aluminum is forced upward through therotor bar holes and into the upper mold.c. Compressed air withdraws the mold assembly, now completely filled with hot (but hardened) aluminum.d. The upper and lower molds are pulled away, revealing the die-cast rotor. The cross section view showsthat the upper and lower end-rings are joined by the rotor bars.
7 Three phase induction motors An induction motor has 2 main parts; the Stator and Rotor. The Stator is the stationary part and the rotor is the rotating part. The Rotor sits inside the Stator. There will be a small gap between rotor and stator, known as air-gap. The value of the radial air-gap may vary from 0.5 to 2 mm.
8 Principle of operation Example to understand the behavior of a three phase induction motorsConsider a series of conductors of length l, whose extremities are short-circuited by two bars A and B. A permanent magnet placed above this conducting ladder, moves rapidly to the right at a speed; so that its magnetic field B sweeps across the conductors.
9 Principle of operation (Faraday's law):A voltage E = Blv is induced in each conductor while it is being cut by the flux2. The induced voltage immediately produces a current I, which flows down the conductor underneath the pole-face, through the end-bars, and back through the other conductors.3. Because the current-carrying conductor lies in the magnetic field of the permanent magnet, it experiences a mechanical force (Lorentz force).4. The force always acts in a direction to drag the conductor along with the magnetic field “Lenz law”However, as it picks up speed, the conductors will be cut less rapidly by the moving magnet with the result that the induced voltage E and the current I will diminish. Consequently, the force acting on the conductors will also decrease5. If the conducting ladder is free to move, it will accelerate toward the right.
10 Principle of operation In an induction motor the ladder is closed upon itself to form a squirrel-cage and the moving magnet is replaced by a rotating field. The field is produced by the 3-phase currents that flow in the stator windings, as we will now explain.
11 Rotating fieldConsider a simple stator having 6 salient poles. Each of which carries a coil, Coils that are diametrically opposite are connected in series.This creates three identical sets of windings AN, BN, CN, that are mechanically spaced at 120° to each other. The three sets of windings are connected in wyeCurrents flowing from line to neutral are considered to be positive.
12 Rotating fieldAs time goes by, we will consider the instantaneous value and direction of the current in each winding and establish the successive flux patternsIf we connect a 3-phase source to terminals A, B. C, alternating currents 𝐼 𝑎 , 𝐼 𝑏 and 𝐼 𝑐 will flow in the windings. The currents will have the same value but will be displaced in time by an angle of 120°. These currents produce magnetomotive forces which, in turn, create a magnetic flux. It is this flux we are interested in.
13 Rotating field at instant 1 At instant 1, current 𝐼 𝑎 is positive whereas 𝐼 𝑏 and 𝐼 𝑐 both have a negative value. The direction of the mmf depends upon the instantaneous current flows and, using the right-hand rule, we find that the direction of the resulting magnetic field is as shown here.NNSSSFlux pattern at instant 1
14 Rotating field at instant 2 Flux pattern at instant 2We discover that the new field has the same shape as before, except that it has moved clockwise by an angle of 60°. In other words, the flux makes 1/6 of a turn between instants 1 and 2.
15 Proceeding in this way for each of the successive instants 3, 4,5,6, and 7, separated by intervals of 1/6 cycle, we find that the magnetic field makes one complete turn during one cycle
16 Rotating fieldThe rotational speed of the field depends, therefore, upon the duration of one cycle, which in turn depends on the frequency of the source.If the frequency is 50 Hz, the resulting field makes one turn in 1/50 s, that is, 3000 revolutions per minute.Because the speed of the rotating field is necessarily synchronized with the frequency of the source, it is called synchronous speed.
17 Rotating field direction When the positive crests of the currents follow each other in the order A-B-C, this phase sequence produces a field that rotates clockwise. If we interchange any two of the lines connected to the stator, we find that the field now revolves at synchronous speed in the opposite, or counterclockwise direction.Interchanging any two lines of a 3-phase motor will, therefore, reverse its direction of rotation.
18 Number of poles- synchronous speed How to construct a 4-pole stator??NSSNSoon after the invention of the induction motor, it was found that the speed of the revolving flux could be reduced by increasing the number of poles2 pole stator4 pole statorThe four identical coils of phase A now span spaced 90° instead of 180 °The 4 coils are connected to each others in such a way that when a current 𝐼 𝑎 flows in the stator winding of phase A, it creates four alternate N-S poles.
19 Rotating fieldIn comparing the two figures, it is clear that the entire magnetic field has shifted by an angle of 45°-and this gives us the clue to finding the speed of rotation. The flux moves 45° in one half cycle and so it takes 8 half-cycles (= 4 cycles) to make a complete turn. On a 50 Hz system the time to make one turn is therefore 4 x 1/50 = 0.08 s. Consequently, the flux turns at the rate of 12.5 r/s or 750 r/min.
20 Rotating fieldThe speed of a rotating field depends therefore upon the frequency of the source and the number of poles on the stator.𝑛 𝑠 = synchronous speed [r/min]f= frequency of the source [Hz]p= number of poles𝒏 𝒔 = 𝟏𝟐𝟎𝒇 𝒑This equation shows that the synchronous speed increases with frequency and decreases with the number of poles.
21 Starting characteristics of a squirrel-cage motor ”rotor locked” A 3-phase voltage is applied to the stator of an induction motor.2. This 3 phase voltage creates a three phase current which creates a revolving magnetic field3. The revolving field induces a voltage in the rotor bars.4. The induced voltage creates large circulating currents which flow in the rotor bars and end-rings.5. The current-carrying rotor bars are immersed in the magnetic field created by the stator; they are therefore subjected to a strong mechanical force. ”Lorentz force”6. The sum of the mechanical forces on all the rotor bars produces a torque which tends to drag the rotor along in the same direction as the revolving field. “LENZ Law”
22 Acceleration of the rotor-slip As soon as the rotor is released, it rapidly accelerates in the direction of the rotating field. “Lenz law”The speed will continue to increase, but it will never catch up with the revolving field, In effect, if the rotor did turn at the same speed as the field (synchronous speed), the flux would no longer cut the rotor bars and the induced voltage and current would fall to zero. Under these conditions the force acting on the rotor bars would also become zero and the friction and windage would immediately cause the rotor to slow down.Pour annuler la vitesse, s’oppose a la causeThe rotor speed is always slightly less than the synchronous speed
23 Slip 𝑠 =slip (%) 𝑛 𝑠 = synchronous speed [r/min] The slip s of an induction motor is the difference between the synchronous speed and the rotor speed, expressed as a percent (or per-unit) of synchronous speed. The per-unit slip is given by the equation𝑠= 𝑛 𝑠 −𝑛 𝑛 𝑠𝑠 =slip (%)𝑛 𝑠 = synchronous speed [r/min]𝑛 = rotor speed [r/min]The slip is less than 0.1%at no-load and is equal to 1 (or 100%) when the rotor is locked.Revolution per minute
24 Frequency of the voltage induced in the rotor The frequency induced in the rotor depend upon the slip. It is given by the following equation:𝒇 𝟐 = 𝐬𝐟Wheref 2 =Frequency of the voltage and current in the rotor [Hz]f = frequency of the source connected to the stator [Hz]s =slip
25 Equivalent Circuit of a squirrel cage Induction Motor at standstill At standstill, it acts exactly like a conventional transformer and so its equivalent circuit is the same as that of a transformer.On standstillA 3-phase wound-rotor induction motor is very similar in construction to a 3-phase transformer. Thus, the motor has 3 identical primary windings and 3 identical secondary windings-one set for each phase. On account of the perfect symmetry, we can consider a single primary winding and a single secondary winding in analyzing the behavior of the motor.
26 Equivalent Circuit of the Induction Motor This equivalent circuit of an induction motor is so similar to that of a transformer that it is not surprising that the induction motor is sometimes called a rotary transformer.
27 Can we remove the magnetizing branch composed of j 𝑿 𝒎 and 𝑹 𝒎 under load for simplifying? In the case of a conventional 3-phase transformer, we would be justified in removing the magnetizing branch composed of j 𝑋 𝑚 and 𝑅 𝑚 because the exciting current 𝐼 0 is negligible compared to the load current 𝐼 𝑝 . However, in a motor this is no longer true: 𝐼 0 may be as high as 40 percent of 𝐼 𝑝 because of the air gap.!
28 Simplifying the equivalent circuit However for motors exceeding 2 hp, we can shift the magnetizing branch to the input terminals. This greatly simplifies the equations that describe the behavior of the motor, without compromising accuracy
29 Equivalent Circuit of the Induction Motor when the motor starts turning at slip s the frequency in the secondary winding will become 𝑠𝑓 the secondary leakage reactance will change from 𝑗 𝑥 2 to 𝑗 𝑠𝑥 2 . 𝐸 2 =𝑠 𝐸 1
30 Equivalent Circuit of the Wound Induction Motor In practice, to construct a final simplified equivalent diagram, we divide the secondary mesh equation by s, which shows an inductance equivalent leakage at frequency f. The frequencies of the primary and secondary then being identical with this manipulation, the elements are than shifted to the transformer primary.
31 Equivalent Circuit of the Wound Induction Motor In this diagram, the circuit elements are fixed, except for the resistance 𝑟 2 /𝑠 . Its value depends upon the slip and hence upon the speed of the motor. Thus, the value of 𝑟 2 /𝑠 will vary from 𝑟 2 to infinity as the motor goes from start-up (s= 1) to synchronous speed (s=0).
32 Active power flowIt is easy to see how electrical energy is converted into mechanical energy b y following the active power as it flows through the machine.
33 active power P c flows from the line into the 3-phase stator. Active power flowactive power P c flows from the line into the 3-phase stator.P js is dissipated as heat in the windings.P f is dissipated as heat in the stator coreThe remaining active power P r is carried across the air gap and transferred to the rotor by electromagnetic induction.P jr is dissipated as heat in the rotor windingsThe remainder is finally available in the form of mechanical power P m
34 Active power flowThe power flow diagram enables us to identify and to calculate three important properties of the induction motor:its efficiencyits powerits torque
35 Efficiency of an induction motor By definition:the efficiency of a motor is the ratio of the output power to the input power:Efficiency (η) = 𝑷 𝑳 / 𝑷 𝒆
36 𝐈 𝟐 𝐑 losses in the rotor1. Active power absorbed by the motor is 𝑷= 𝑬 𝒈 𝟐 𝑹 𝒎 + 𝑰 𝟏 𝟐 𝒓 𝟏 + 𝑰 𝟏 𝟐 𝒓 𝟐 ′/𝒔.2. Reactive power absorbed by the motor is 𝐐= 𝐄 𝐠 𝟐 𝐗 𝐦 + 𝐈 𝟏 𝟐 𝐱 (x= 𝒙 𝟏 + 𝒙 𝟐 ′)3. Apparent power absorbed by the motor is 𝑺= 𝑷 𝟐 + 𝑸 𝟐4. Power factor of the motor is 𝒄𝒐𝒔 𝜽 = 𝑷/𝑺5. Line current is 𝑰 𝑷 = 𝑺/ 𝑬 𝒈6. Active power supplied to the rotor is 𝑷 𝒓 = 𝑰 𝟏 𝟐 𝒓 𝟐 ′ 𝒔7. Power dissipated as 𝐼 2 𝑅 losses in the rotor circuit is 𝑷 𝒋𝒓 = 𝑰 𝟏 𝟐 𝒓 𝟐 ′=𝒔 𝑷 𝒓Per phase quantities
37 𝐈 𝟐 𝐑 losses in the rotorP jr = s P rEquation above shows that as the slip increases, the rotor I 2 R losses consume a larger and larger proportion of the power P r transmitted across the air gap to the rotor. A rotor turning at half synchronous speed (s= 0.5) dissipates in the form of heat 50 percent of the active power it receives. When the rotor is locked (s = 1), all the power transmitted to the rotor is dissipated as heat.
38 Mechanical power developed by the motor is 𝑷 𝒎 = 𝑷 𝒓 − 𝑷 𝒋𝒓 = 𝑷 𝒓 (𝟏−𝒔)
39 ExampleShow that the efficiency can be written: η= 1−𝑠 𝑃 𝑒 − 𝑃 𝑓 − 𝑃 𝐽𝑆 − 𝑃 𝑚𝑒𝑐 𝑃 𝑒
40 Motor torqueThe torque Tm developed by the motor at n speed is given byT m = 60 𝑃 𝑚 2𝜋 𝑛 = 60 𝑃 𝑟 1−𝑠 2𝜋 𝑛 𝑠 1−𝑠 =9.5 5𝑃 𝑟 / 𝑛 𝑠Therefore,T m =9.55 P r / n sWhereT m = torque developed by the motor at any speed [N. m]P r = power transmitted to the rotor [W]n s = synchronous speed [r/min]Tm=60 Pm/2pi*nEquation above shows that the torque is directly proportional to the active power transmitted to the rotor. Thus, to develop a high locked-rotor torque, the rotor must absorb a large amount of active power. The latter is dissipated in the form of heat consequently; the temperature of the rotor rises very rapidly.
41 Torque versus speed curve Nominal full-load torque is T.The starting torque is 1.5 TThe maximum torque (called breakdown torque) is 2.5 T.Pull-up torque is the minimum torque developed by the motor while it is accelerating from rest to the breakdown torque.torque developed by a motor depends upon its speed, but the relationship between the two cannot be expressed by a simple equation. Consequently, we prefer to show the relationship in the form of a curve.
42 Torque versus speed curve The induced torque of the motor is zero at synchronous speed. This fact hasbeen discussed previously.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.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.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.Notice that 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 that will be described later.
43 Torque versus speed curve At full-load the motor runs at a speed n. If the mechanical load increases slightly, the speed will drop until the motor torque is again equal to the load torque. As soon as the two torques are in balance. The motor will turn at a constant but slightly lower speed.However, if the load torque exceeds 2.5 T (the breakdown torque), the motor will quickly stop.
44 Motor under loadThe motor and the mechanical load will reach a state of equilibrium when the motor torque is exactly equal to the load torque. When this state is reached, the speed will cease to drop and the motor will turn at a constant rate.It is very important to understand that a motor only turns at constant speed when its torque is exactly equal to the torque exerted by the mechanical load. The moment this state of equilibrium is upset, the motor speed will start to change.
45 Effect of rotor resistance the starting torque doublesbreakdown torque remain unchangedlocked-rotor current dcreasesThe motor develops its breakdown torque at a speed N d of 500 r/min. compared to the original breakdown speed of 800 r/min.
46 Effect of rotor resistance If we again double the rotor resistance so that it becomes 5 R, the locked-rotor torque attains a maximum value of 250 N m for a corresponding current of 70 A
47 Effect of rotor resistance In summary:a high rotor resistance is desirable because it produces a high starting torque and a relatively low starting current. Unfortunately, it also produces a rapid fall-off in speed with increasing load. Furthermore, because the slip at rated torque is high, the motor I 2 R losses are high. The efficiency is therefore low and the motor tends to overheat.Under running conditions it is preferable to have a low rotor resistance. The speed decreases much less with increasing load, and the slip at rated torque is small. Consequently, the efficiency is high and the motor tends to run cool.
48 Wound-rotor vs squirrel rotor Although a wound-rotor motor costs more than a squirrel-cage motor, it offers the following advantages:1. The locked-rotor current can be drastically reduced by inserting three external resistors in series with the rotor. Nevertheless, the locked-rotor torque will still be as high as that of a squirrel-cage motor.2. The speed can be varied by varying the external rotor resistors.3. The motor is ideally suited to accelerate high-inertia loads, which require a long time to bring up to speed.
50 ExampleThe stator of an induction motor is connected in wye during startup, and then in delta for normal operation. 1- Show that the line current consumed in wye connection is three times smaller that in delta connection. 2- It is assumed that the engine output torque is proportional to the square of the voltage. Show that the output torque is divided by three during the starting phase. 3- What is the advantage of starting “wye - delta"? What is it’s disadvantage?
51 Speed control of induction motors It can accomplished by :Changing the number of poles on the machineChanging the applied electrical frequencyChanging the applied terminal voltage “torque proportional to the square of the applied voltage”Changing the rotor resistant in the case of a wound-rotor induction motor
52 Induction motor operating as a generator We can make an asynchronous generator by connecting an ordinary squirrel-cage motor to a 3- phase line and coupling it to a gasoline engine. As soon as the engine speed exceeds the synchronous speed, the motor becomes a generator, delivering active power P to the electrical system to which it is connected. However, to create its magnetic field, the motor has to absorb reactive power Q. This power can only come from the ac line. With the result that the reactive power Q flows in the opposite direction to the active power PI
53 Complete torque-speed characteristic of an induction machine
54 Induction motor operating as a generator Induction generators are usually rather small machines and are used principally with alternative energy sources, such as windmills, or with energy recovery systems. Almost all the really large generators in use are synchronous generators
55 Tests to determine the equivalent circuit No-load testNo load s<<< 𝑟 2 ′ 𝑠 >>>> 𝐼 1 is negligible compared to 𝐼 0‘
56 Tests to determine the equivalent circuit No-load testa. Measure the stator resistance 𝑅 𝐿𝐿 between any two terminals. Assuming a wye connection, the value of 𝑟 1 is𝑟 1 = 𝑅 𝐿𝐿 /2b. Run the motor at no-load using rated line-to line voltage 𝐸 𝑁𝐿 . Measure the no load current 𝐼 𝑁𝐿 and the total 3-phase active power 𝑃 𝑁𝐿 .
57 No-load testThe following calculations of total apparent power 𝑆 𝑁𝐿 and total reactive power 𝑄 𝑁𝐿 are then made:𝑆 𝑁𝐿 = 𝐸 𝑁𝐿 𝐼 𝑁𝐿 3𝑄 𝑁𝐿 = 𝑆 𝑁𝐿 2 − 𝑃 𝑁𝐿 2𝑃 𝑓 + 𝑃 𝑣 = windage, friction, and iron losses= 𝑃 𝑁𝐿 −3 𝐼 𝑁𝐿 2 𝑟 1The resistance 𝑅 𝑚 representing 𝑃 𝑓 + 𝑃 𝑣 , is𝑅 𝑚 = 𝐸 𝑁𝐿 2 /( 𝑃 𝑓 + 𝑃 𝑣 )The magnetizing reactance is:𝑋 𝑚 = 𝐸 𝑁𝐿 2 / 𝑄 𝑁𝐿
58 Tests to determine the equivalent circuit Locked rotor testUnder rated line voltage, when the rotor of an induction motor is locked, the stator current 𝐼 𝑃 is almost six times its rated value. Furthermore, the slip s is equal to one. This means that 𝑟 2 ′/𝑠 is equal to 𝑟 2 ′ where 𝑟 2 ′ is the resistance of the rotor reflected into the stator. Because 𝐼 𝑃 is much greater than the exciting current 𝐼 0 , we can neglect the magnetizing branch
60 Tests to determine the equivalent circuit Locked rotor testApply reduced 3-phase voltage to the stator so that the stator current is about equal to its rated valueb. Take readings of 𝐸 𝐿𝑅 (line-to-line) 𝐼 𝐿𝑅 and the total 3-phase power 𝑃 𝐿𝑅
61 The following calculations are then made: 𝑆 𝐿𝑅 = 𝐸 𝐿𝑅 𝐼 𝐿𝑅 3 Locked rotor testThe following calculations are then made:𝑆 𝐿𝑅 = 𝐸 𝐿𝑅 𝐼 𝐿𝑅 3𝑄 𝐿𝑅 = 𝑆 𝐿𝑅 2 − 𝑃 𝐿𝑅 2𝑥= 𝑄 𝐿𝑅 /3 𝐼 𝐿𝑅 23 𝐼 𝐿𝑅 2 𝑟 1 + 𝑟 2 = 𝑃 𝐿𝑅Hence,𝑟 2 = 𝑃 𝐿𝑅 3 𝐼 𝐿𝑅 2 − 𝑟 1