AC Drives

Dr. Adel A. El-Samahy

Department of Electrical Engineering

University of Helwan

Facts in Electrical Machines..

• If an e.m.f. source is applied to closed circuit, an electric current will pass through the circuit

• When electric current passes through a conductor or coil, a magnetic field is established (DC current produces DC magnetic field while AC current produces AC magnetic field). The direction of the magnetic field is determined by Right Hand Grip Rule .

Facts in Electrical Machines..

• When a magnetic field cuts a conductor/coil, an e.m.f. is induced in that conductor/coil. The magnitude of this e.m.f. proportional to the rate of cutting. The rate of cutting may be produced due to relative motion between conductors and magnetic field (DC or AC generators), or AC magnetic field and stationary winding (transformer)

• To produce electromagnetic torque, there must be two magnetic fields. These fields must have no relative speed and have an angle between them

Induction Motor Drives

.

.

Importance of Induction motors..

• The induction machine is used in a wide variety of applications as a means of converting electric power to mechanical work. It is without doubt the workhorse of the electric power industry. Pump, steel mill, and hoist drives are but a few applications of large multiphase induction motors.

• On a smaller scale, the 2-phase servomotor is used extensively in position-follow-up control systems and single-phase induction motors are widely used in household appliances as well as in hand and bench tools

Advantages of Induction motor Drives

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.

• AC motor are less expensive•  Ac motors have low maintenance• For the same rating, ac motors are

higher in weight as compared to dc motors.

•  AC motors can work in hazardous areas like chemical, petrochemical etc. whereas dc motors are unsuitable for such environments because of commutator sparking

Disadvantages of Induction motor Drives

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.

Power converters for 1. the control of ac motors are more

complex.2. ac motors are more expensive3. ac motors generate harmonics in the

supply system and load circuit. Hence AC motors gets derated.

Induction Motor Drives?Why does the rotor rotates?• When the stator winding is connected to 3

phase AC supply, a resultant rotating magnetic field with constant magnitude is established. This field rotates with synchronous speed

Where: f is the supply frequency, P is the number of poles

• This field cuts the rotor winding which is short circuited inducing an e.m.f. that causes current in the rotor winding which in turns produces another rotating magnetic fields.

• The interaction between the two fields produces an electromagnetic torque that forces the rotor to rotate

Induction Motor Drives?Definitions

ns: Speed of rotating magnetic field (Synchronous speed) rpmws : Speed of rotating magnetic field (Synchronous speed) rad/sn : motor speed rpmw : rotor speed rad/sf1: supply frequencyf2: frequency of the induced e.m.f in the rotor = sf1

s : slipns- n : Slip speeds

s

s

s

n

nns

p

fns

1120

60

2 ss

n

Induction Motor Drives?Example 1: . Calculate the frequency of the rotor current to produce an average torque at a speed of - 200 rpm. (p=2 and f1 =50 Hz) Solution

rpmp

fns 3000

120 1

1.06673000

)200(3000

s

s

n

nns

f2 = sf1 = 53.33 Hz

Induction Motor Equivalent Circuit

Induction Motor Equivalent Circuit

Induction Motor Equivalent Circuit

`21

2`2

1

`2

XXsR

R

VI ph

tan-1 ((X1+X2’)/R1+R2

’/s))

I1 =Io+ I2’ Im + I2

’

Where: Io is no load current

Power Flow GraphInput Power = Pin = 3VphIph cos f =Vl Il cos fStator copper losses = Pst = 3Iph

2 R1

Core losses = Pc = 3Vm2 /Rm

Air gap power = Pg = 3I22 R2/s

Mechanical Power (developed power) = Pm =Pg(1-s) = 3I2

2 R2/s(1-s)Pf =friction lossesOutput Power = P0 = Pm –Pf = Po/Pin

Power Flow GraphExample 2: . Find the efficiency of 3-phase, 50 Hp, 60 Hz, four-pole induction motor. The motor is star connected. nf.l=1755 rpm, VL = 440 V, Io = 18 A. at 0.085 Power factor lag. Also available from the manufacturer’s data sheet are the following design data: R1=0.1, R2

' =0.12 , X1=0.35 , X2' = 0.4 , Pcore=1200 W,

Prot=950 W.

Solution

=1800 rpmp

fns

1120

025.01800

)1755(1800

s

s

n

nns

Power Flow GraphExample 2:

Pin = 3 440 57.35 cos (26.7) = 39046 W

Pg= 3 I2' 2 r2

' /s =37896 W

Pout = Pm - Prot = Pg (1-s)-950 =36000 W

= Pout/ Pin = 92.2 %

Io= Im+ Irot =18 -850 Irot =950/(3

Im = 0.35 +j 17.95 I2' = Vph/Z2

’

I1= Im+ I2' = 57.35 -26.70

Torque Speed Characteristics

sXXsR

R

RVT

s

RIPT

s

ph

ss

g

2`21

2`2

1

`2

2

`2

2`2

`

3

3

Torque Speed CharacteristicsRegenerative braking

Pluging

Torque Speed CharacteristicsThere are three region of operation:

1.Motoring (0≤s≤1)The motor rotates in the same

direction as the field; as the slip increases, the torque increases, while air gap flux remains constant. Once the torque reaches its maximum value (s = scr), the torque decreases with the increase in the slip due to reduction in the air gap flux

Torque Speed CharacteristicsThere are three region of operation:2.Regeneration (s0)

The speed is greater than the synchronous speed, w and ws being in the same direction and the slip is negative. Therefore R2/s is negative which means the power is fed back from the shaft to the supply. The torque speed characteristics is similar to that of motoring, but having negative value

Torque Speed CharacteristicsThere are three region of operation:3.Plugging (1 s)

The speed is opposite to the direction of the field, this may happen if the sequence of the supply source is reversed, while motoring, therefore the direction of the field and the developed torque also reversed. The energy due to plugging must be dissipated within the motor and may cause excessive heating.

Torque Speed CharacteristicsCondition for Maximum Torque

dT/ds = 0

2`21

211

2

2`21

21

`2

2

3

XXRR

VT

XXR

Rs

s

phmm

cr

Torque Speed Characteristics

2`

212

11

2

2

3

XXRR

VT

s

phmr

The maximum regenerative torqueThe maximum regenerative torque may be obtained from the general torque equation by letting s= -scr

Torque Speed Characteristics

2211

2211

2

`2

1

211

12

2

12

cr

crmmmr

cr

crcrmmst

crcr

cr

crmm

XRR

XRRTT

as

ssaTT

R

Ra

sass

ss

asTT

`21 XXX cr

ss

ssT

Tcr

cr

max2

Torque Speed Characteristics

ss

ssT

Tcr

cr

mm

2

R1 is small compared to the other circuit impedances for motor with rating greater than 1 kW and my be neglected. In this case

`21

2

2

3

XX

VTT

s

phmrmm

`21

`2

XX

Rscr