Lecture Notes on AC Machines PDF

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EE3010 ElectricalDevicesand Machines Dr.LeePengHin AssociateProfessor Office:EEE‐S1‐B1a‐09 Tel:67904474 Email:[email protected] 1

EE3010 ElectricalDevicesand Machines LECTURENO1

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Topics A.ACMachines i. InductionMotors B.DCMachines i. DCMachineryFundamentals ii. DCMotors iii. DCGenerators

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 Maintextbook:

StephenJ.Chapman ElectricMachinery Fundamentals, 5th Edition, McGrawHill, (TK2000.C4662012)

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 Referencebook: I.

Theodore Wildi, Electrical Machines, Drives and Power Systems, 6th Edition, Pearson (TK2182.W673 2006)

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Introduction to INDUCTION MOTORS • The induction motor was invented by Nikola Tesla (1856‐1943). Three‐phase induction motors are the motors most frequently encountered in the industry. • They are simple, rugged, easy to maintain and are less expensive than DC motors of equal power and speed ratings.

A three‐phase induction motor 6

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Introduction Stationary stator – consist essentially of a steel frame housing and the 3‐phase stator windings displaced from each other by 120 deg and are placed in slots cut on the inner surface of the stator frame. The windings are connected in either wye or delta configurations. Electrical power is supplied to the stator windings.

Stator winding 7

 Revolving rotor – composed of punched laminations which are carefully stacked to create a series of rotor slots to provide space for the rotor windings.  The type of rotor windings give rise to two main types of motors o squirrel‐cage induction motors o wound‐rotor induction motors.

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Squirrel‐cage – series of conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shorting rings.

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Wound‐rotor – consists of a set of 3‐phase windings similar to the stator windings and usually wye connected. The ends are tied to slip rings on the rotor’s shaft. The rotor windings are shorted through brushes riding on the slip rings. This allow external resistors to be connected in series with the rotor windings, to modify the torque‐speed characteristics.

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• The operating principle of the 3‐phase induction motor is very straight forward but unfortunately, the simplicity is obscured by the complicated construction of practical induction motors. • There is no electrical connections to the rotor. The 3‐phase supply is connected to the stator windings. • The transfer of energy from the stationary member to the rotating member is by means of electromagnetic induction. 11

• A rotating magnetic field, produced by the stator, induces an alternating emf and current in the rotor. The resultant interaction of the induced rotor current with the rotating magnetic field of the stator produces motor torque. • The torque‐speed characteristics is directly related to the resistance and reactance of the rotor. • Hence, different torque‐speed characteristics may be obtained by designing rotor circuits with different ratios of rotor resistance to rotor reactance. 12

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The rotating magnetic field • In a nutshell, the basic principle of all AC motor operation is as follows: If two magnetic fields are present in a machine, then a torque will be created which will tend to line up the two magnetic fields. • If one magnetic field is produced by the stator of an AC machine and the other one is produced by the rotor of the machine, then a torque will be induced in the rotor which will cause the rotor to turn and align itself with the stator magnetic field. 13

• If there are some way to make the stator magnetic field rotate, then the induced torque in the rotor would cause it to constantly “chase” the stator magnetic field around in a circle. • How can the stator magnetic field be made to rotate ? • The fundamental principle of ac machine operations : If a 3‐phase set of currents, each of equal magnitude and differing in phase by 120 degrees, flows in a three‐phase windings, then it will produce a rotating magnitude field of constant magnitude. 14

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The following demonstrates the above: • Three coils spaced 120 electrical degrees apart. • A balanced 3‐phase currents are applied to them.

ia (t )  I max sin(t ) A ib (t )  I max sin(t  1200 ) A ic ( t)  I max sin( t  240 0 ) A

  2 f s , fs is the frequency of the supply at the stator 15

• The currents will establish three time‐varying magnetic field intensity and hence magnetic flux density vectors given by B  Bˆ 00 T a

a

B b  Bˆ b120 0 T B c  Bˆ c 2400 T  Ba  Bmax sin( t )  Bb  Bmax sin( t  1200 )  Bc  Bmax sin( t  2400 )

The current flows into the a end and out of the a’ end of the coil. 16

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• The resultant net magnetic flux density in the stator is the sum of the individual flux density. B net  B a  B b  B c  1.5B max (t  900 ) T

• We can examine its value at specific instant of time. At t  0 0 ,B net  1.5 Bmax ( 90 0 ) T At t  90 0 , B net  1.5Bmax 00 T

The magnitude of the field is CONSTANT It rotates counterclockwise at an angular velocity. The speed of rotating flux - called synchronous speed, sync . 17

In rad/s,

 sync    2   n sync    2 f s  60  120 fs  n sync  2

(a) The rotating magnetic field in a stator represented as moving north and south stator poles – two‐pole (b) The poles complete one complete mechanical rotation around the stator surface for each electrical cycle of the applied current. (c) Therefore, the mechanical speed of rotation of the magnetic field in revs per sec is equal to the electric frequency in hertz. 18

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2sync    n sync 

120 fs 4

a) The four‐pole stator rotating magnetic field. The stator windings are doubled. In this case, each pole moves halfway around the stator surface in one electrical cycle. b) The electrical frequency of the current is twice the mechanical frequency of rotation.

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Therefore,

n sync 

120 f s p

where fs  freq of the 3-phase supply in Hz nsync  synchronous speed, i.e the speed of the rotating flux in rev/min p  number of poles formed by the stator winding

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Example. Determine the synchronous speed of a six‐pole 460V 60 Hz induction motor if the frequency is reduced to 85 percent of its rated value. 120 f s p 120(60)(0.85)  6  1020 r/min

nsync 

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Remarks • The rotor consists comprised of either conductors bars or windings. As stator magnetic field rotates, by Faraday’s law, a voltage is induced in the rotor conductors and a current flows in its closed circuit and an electromagnetic torque will be produced. As a result, the rotor starts rotating in the direction of the revolving magnetic field and ultimately, reaches a steady‐state speed which is close to the synchronous speed. • The rotor speed can never be equal to the synchronous speed. 22

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• 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. • An induction motor can thus speed up to near‐synchronous speed, but it can never exactly reach synchronous speed. 23

Concept of rotor slip • The voltage induced in the rotor bar depends on the speed of the rotor relative to the magnetic fields. Two terms are used to define the relative motion of the rotor and magnetic fields a) Slip speed : nslip  nsync  nm where nslip  slip speed of the machine, r/min nsync  speed of the magnetic field, sync speed, r/min nm  mechanical shaft speed of motor, r/min 24

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b) Slip slip  s  

nslip n sync

 x 100% 

nsync  nm n sync

 x 100%

sync  m  x 100%   sync

s  0  rotor turns at synchronous speed s  1  rotor is stationary nm  (1 s )nsync

m  (1 s )sync 25

The electrical frequency on the rotor • An induction motor works by inducing voltages and currents in the rotor of the machine – sometimes called a rotating transformer. • Unlike a transformer, the rotor (secondary) frequency is not the same as the stator (primary) frequency. s  0  n m  n sync , f r  0 s  1  nm  0,rotor is stationary, fr  f s . f r  sf s The rotor frequency

fr 

p nsync  nm  120

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Example. A 208‐V, 10 horse‐power (hp), four‐ pole, 60 Hz, wye‐connected induction motor has a full‐load slip of 5 percent. a) What is the synchronous speed of the 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? d) What is the shaft torque of this motor at the rated load? 27

a) n sync 

120 f s 120(60)   1800 r/min 4 p

b) slip  s 

nsync  nm nsync

 0.05 

1800  nm 1800

 nm  1710 r/min c) f r  sf s  0.05(60) 3 Hz d) Tload m  Pout  Tload (1710(2 ) / 60) 10(746 W/hp)  Tload  41.7 N.m 28

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EE3010 ElectricalDevicesand Machines LECTURENO2

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The equivalent circuit of an induction motor • Because of the induction of voltages and currents in the rotor circuit of an induction motor is essentially a transformer action, the equivalent circuit of an induction motor is very similar to that of a transformer.

Per phase equivalent circuit of an induction motor, with rotor and stator connectedbyanidealtransformerof turnsratioa_eff

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The circuit parameters per phase : R1 : stator winding resistance X1 : stator leakage reactance RR : rotor winding resistance XR : rotor leakage reactance aeff : effective turns ratio coupling E1 and ER Rc : resistance representing core loss Xm : magnetising reactance

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The rotor circuit model • In an induction motor, when a voltage is applied to the stator windings, a voltage is induced in the rotor windings. • The induced voltage is 0 V when the rotor moves at synchronous speed. • When the rotor is stationary, ‐ i.e. locked‐rotor or blocked‐rotor conditions, the largest voltage and rotor frequency are induced in rotor. At any other speed, 4

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The rms value of the induced voltage generated in the rotor as it is swept by the rotating flux is ER  4.44  f r  N r max  4.44  sf s  Nrmax  sERO where ERO  4.44 fs  Nr  max is the rotor induced voltage at locked-rotor conditions ( f r  f s , s  1 ) The reactance depends on the inductance and the freq of the voltage and current in the rotor. X R  r LR  2  fr  LR  2 sfs  LR  sXRO where X RO is the blocked-rotor reactance.

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Example. The frequency and induced voltage in the rotor of a certain six‐pole would rotor induction motor, whose shaft is blocked, are 60 Hz and 100V. Determine the corresponding values when the rotor is running at 1100 r/min. 120 fs 120(60)   1200 r/min 6 p n n s  sync m  0.0833 nsync

nsync 

f r  sf s  5 Hz ER  sERO  8.33V 6

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The rotor circuit model of aninduction motor

The rotor circuit model with all the frequency (slip) effects concentrated inresistor RR 7

Rotor currentas a functionof rotor speed

IR 

E RO R R / s  jX RO

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The final equivalent circuit of an induction motor

The per phase equivalent circuit of an induction motor referred to thestator.

• The core losses (W) are usually lumped together with the friction, windage and stray losses and called the rotational losses. 9

Hence, unless specified otherwise, RC will not appear in the equivalent circuit subsequently.

The referred rotor parameters to the stator side are: E1  aeff ERO , I2 

IR ERO where IR  aeff RR / s  jX RO

R2  a2eff RR ,

X 2  a2eff X RO

aeff  effective turns ratio 10

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Losses and Power‐Flow Diagram in Induction Motors

Prot (rotational losses) Powerflowdiagramof aninductionmotor 11

• Induction motor basically is a rotating transformer. • Input is 3‐phase supply. • Output is mechanical power. • Air gap power is the power transferred to the rotor of the machine across the air gap between the stator and the rotor. • The converted power is the power converted from electrical to mech form. • The core losses are usually lumped together with the friction, windage and stray load losses and called the rotational losses.

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Example. A 480‐V, 60 Hz, 50‐hp, three‐phase induction motor is drawing 60A at 0.85 pf lagging. The stator copper losses are 2kW, and the rotor copper losses are 700W. The total rotational losses which includes the friction, windage and core losses is 2400W, and the stray losses are negligible. Find, a)The air-gap power PAG b) The converted power Pconv c) The output power Pout d) The efficiency of the mot or  13

a ) Pin  3Vline I l ine cos  3 (480)(60)0.85  42.4 kW PAG  Pin  PSCL  40.4 kW b ) Pconv  PAG  PRCL  39700 W c) Pout  Pconv  Prot  37.3 kW d) The efficiency of the motor  

Pou t  100%   88% Pin 14

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EE3010 ElectricalDevicesand Machines LECTURENO3

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Power and Torque in induction motors From the per  phase equivalent circuit The stator copper losses PSCL  3 I12 R1 PAG  Pin  PSCL The air-gap power R  Since X 2 does not consumes real power, PAG  3 I 22  2   s  2 The rotor copper losses PRCL  3 I 2 R2 The converted power, also called developed power,   1 s   Pdev  Pconv  PAG  PRCL  3I 22  R 2     s  Also, PRCL  s PAG , Pconv  (1 s ) PAG , Pout  Pconv  Prot The induced torque T ind is the torque generated by the internal electric to mech power conversion. Also called developed torque Tdev Tind  Tdev 

Pconv

m



(1  s ) PAG P  AG (1  s) sync sync

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Separating Rotor Cu Losses and the Power Converted The per phase equivalent circuit with rotor copper losses and power convertedseparated R  T he air-gap power P AG  3I 22  2  . Rotor copper losses PRCL  3I 22 R2  s    1 s   T he converted power Pconv  PAG  PR CL  3 I 22  R 2     s  T hus,

1 s  R conv  R 2    s 

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Example. A 460‐V, 60 Hz, four‐pole induction motor is wye‐connected. The core losses are lumped together with the friction and windage losses and the stray losses to give the total rotational losses of 1100W and they are assumed to be constant. The impedances per phase referred to the stator circuit are R1  0.641 ,

R2  0.332 ,

X1  1.106 ,

X 2  0.464 , X M  26.3 For a rotor slip of 2.2 % at the rated voltage and rated frequency, find the motor's (a) Speed (b) Input stator curr ent (d) Pconv and Pout ( e)Tind and Tload

(c) Input power factor (f) Efficiency 4

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120 f s 120(60)   1800 r/min 4 p  nm  (1  s) nsync  1760 r/min (a) n sync 

(b) The total impedance per phase is calculated as Ztotal  11.72  j 7.79  14.0733.60   I1 

V Ztotal

266 00   18.88  33.60 A 0 14.07 33.6

(c) The pf is cos(33.60 )  0.833 lagging. 5

(d)

2 PAG  Pin  PSCL  3Vline Iline cos  3I1 R1

 3(460)(18.88)0.833 3(18.88)2 0.641 11845 W Pconv  (1  s) PAG  11585 W  Pout  Pconv  Prot  10485 W e) Tind 

Pconv

m



PAG

sync

 62.8 N.m. Tload 

f) The efficiency of the motor  

Pout

m

 56.9 N.m.

Pout x 100%  83.7% Pin 6

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Example. A three‐phase 230‐V, 60‐Hz, 100‐hp (1 hp = 746 W), 6‐pole induction motor operating at rated conditions has an efficiency of 91 % and draws a line current of 248 A. The stator copper and rotor copper losses are 2803 W and 1549 W, respectively. Determine (a) Power input (b) Total losses (c) Air‐gap power (d) Rotor speed (e) Input power factor (f) Rotational losses [Ans : 81978 W, 7378 W, 79175 W, 1176 rpm, 0.83 lagging, 3026 W] 7

(a) Eff  

P

out

 100x746   Pin  81978 W Pin

(b) Total losses = Pin  Pout  7378 W (c) PAG  Pin  PSCL  79175 W (d) From PRCL  sPAG  s  0.01956 fs  1200 rpm p n m  (1  s )nsync  1176 rpm nync  120

(e) From Pin  3VlineI line cos  cos  0.83 (f) Prot  Total losses  PSCL  PRCL  3026 W 8

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EE3010 ElectricalDevicesand Machines LECTURENO4

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Induction motor Torque‐Speed Characteristics • It is possible to use the equivalent circuit of an induction motor and the power flow diagram for the motor to derive a general expression for induced torque (developed torque) as a function of speed. Tind 

PAG

sync

,

2 R  PAG  3I 2  2   s 

The per phase equicircuitof aninductionmotor

 1  Tind    sync 

 2  R2  3I  2  s   2

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• We want to find an expression for the current I 2 . • One approach is to find the Thevenin’s equivalent to the left of the pts marked “x” in the figure. VTH 

jX M V R1  jX 1  jX M 

VTH  VTH VTH 

XM R  (X 1  X M )2 2 1

Z TH 

V

jX M (R1  jX 1 ) R1  jX 1  jX M

 RTH  jX TH 3

The resultant simplified equi circuit of an induction motor

I2 

VTH VTH  Z TH   R2 / s   jX 2  RTH  R2 / s   j  X TH  X 2 

I2 I2 

VTH

R

TH

 R 2 / s   X TH  X 2  2

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 1  2  R2   1   R2  VTH2 3  Tind   3I  s ync  2  s   sync   s   R  R / s 2   X  X 2     2 2 TH TH 4

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nm  (1 s )nsync

 m  (1  s) sync

Atypicalinduction motortorque‐speed characteristics curve

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Plugging (s  1)

Motoring (0  s  1)

Generating (s  0)

Induction motor torque‐speed characteristics curve showing the extendedoperating regions (braking andgenerator regions)

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Comments on the Induction Motor Torque‐Speed Characteristics Curve Motoring region • The motor rotates in same direction but slower than the rotating magnetic field. • The induced torque is zero at synchronous speed. • The starting torque is slightly higher than the full‐load torque. • There is a maximum possible developed torque – pullout torque 7

Generating region • If the rotor of the induction motor is driven faster than synchronous speed by an external prime mover, the direction of the induced torque reverses and the machine becomes a generator. • Induction generators in wind turbines are operated in this manner. As the torque applied to its shaft increases, the amount of power produced by the induction generator increases. 8