Magnetic field due to an electric current

When a conductor carries an electric current, a magnetic field is produced around that conductor.

Right Hand Rule

Review: Electromagnetism

Magnetic field of a solenoid

If a coil is wound on a steel rod and connected to a supply, the steel becomes magnetized and behaves like a permanent magnet.Grip Rule if solenoid is gripped with the right hand, with fingers pointing in the direction of current, then the thumb outdtretched parallel to the axis of the solenoid points in the direction of the magnetic field inside the solenoid.

Solenoid with a steel core

Force on a conductor carrying current across a magnetic field

F (force on conductor, N) = B (flux density, T) x L (length, m) x I (current through conductor, A)

Left Hand Rule: First finger: FluxSecond finger: CurrentThumb: Mechanical Force

BILF

Electromagnetic induction

When a conductor cuts or is cut by a magnetic flux, an emf is generated in the conductor and a magnitude of the generated emf is proportional to the rate at which the conductor cuts or is cut by the magnetic flux. Ex : transformer

Direction of induced e.m.f

Two methods:a) Fleming’s right hand rule - If the first finger of the right hand be

pointed in the direction of the magnetic flux and if the thumb be pointed in the direction of motion of the conductor relative to the magnetic field, then the second finger, held at right angles to both the thumb and the first finger, represents the direction of the emf.

b) Lenz’s law

Fleming’s right hand rule

The induction machine is the most rugged and the most widely used machine in industry. Like dc machine, the induction machine has a stator and a rotor mounted on bearings and separated from the stator by an air gap. However, in the induction machine both stator winding and rotor winding carry alternating currents. The induction machine can operate both as a motor and as generatorAs motors, they have many advantages. They are rugged, relatively inexpensive, and require very little maintenance. They range in size from a few watts to about 10,000 hp. The speed of an induction motor is nearly but not quite constant, dropping only a few percent in going from no load to full load.

The main disadvantages of induction motors are:· The speed is not easily controlled.· The starting current may be five to eight times full-load current.· The power factor is low and lagging when the machine is lightly loaded

INDUCTION MACHINE

INDUCTION MOTOR CONSTRUCTION

Two different types of induction motor which can be placed in stator:a) squirrel cage rotorb) wound rotor

Squirrel Cage rotor Wound rotor

Squirrel cage rotor – consists of conducting bars embedded in slots in the rotor magnetic core, and these bars are short circuited at each end by conducting end rings. The rotor bars and the rings are shaped like squirrel cage.

Wound rotor – carries three windings similar to the stator windings. The terminals of the rotor windings are connected to the insulated slip rings mounted on the rotor shaft. Carbon brushes bearing on these rings make the rotor terminals available to the user of the machine. For steady state operation, these terminals are short circuited.

Types of rotor

Rotor bars (slightly skewed)

End ring

Squirrel Cage Rotor

Wound Rotor

• Most motors use the squirrel-cage rotor because of the robust and maintenance-free construction.

• However, large, older motors use a wound rotor with three phase windings placed in the rotor slots.

• The windings are connected in a three-wire wye. • The ends of the windings are connected to three slip rings. • Resistors or power supplies are connected to the slip rings

through brushes for reduction of starting current and speed control

Induction Motor Components

BASIC INDUCTION MOTOR CONCEPT

A three phase set of voltages has been applied to the stator, and three phase set of stator currents is flowing. These produce a magnetic field Bs, which is rotating in a counterclockwise direction .

The speed of the magnetic field’s rotation is

P

fn e

sync

120

The rotating stator fields Bs induces a voltage in the rotor bars.

The rotor voltage produces a rotor current flow, which lags behind the voltage because of the inductance of the rotor.

The rotor current produces a rotor magnetic field BR lagging 900 behind itself, and BR interacts with Bnet to produce a counterclockwise torque in the machine

IvxBeind )(

sRind xBkB

THE CONCEPT OF ROTOR SLIPThe voltage induced in a rotor depends on the speed of the rotor relative to the magnetic field.

Slip speed is defined as the difference between synchronous speed and rotor speed

msyncslip n - n n

wherenslip = slip speed of the machinensync = speed of the magnetic fieldsnm = mechanical shaft speed of motor

Slip is the relative speed expressed on a per unit or a percentage basis

100% xn

n s

sync

slip 100% x n

n - n s

sync

msync

CONTINUED…

In term angular velocity (radians per second, rps)

100% x -

ssync

msync

If the rotor turns at synchronous speed, s = 0 while if the rotor is stationary/standstill, s = 1.

synxm s)n- (1 n

synxm s)- (1

THE ELECTRICAL FREQUENCY CONCEPT

Like a transformer, the primary (stator) induces a voltage in the secondary (rotor) but unlike a transformer, the secondary frequency is not necessary the same as the primary frequency.

If the rotor of a motor is locked, then the rotor will have same frequency as the stator.

The rotor frequency can be expressed

er sf f

)( msyncr n - n120

P f

EXERCISE 1

A 208V, 10hp, 4 pole, 60Hz, Y connected induction motor has full load slip of 5%.

Calculate,

i) nsync (Ans:1800rpm)

ii) nm (Ans: 1710rpm)

iii) fr at the rated load (Ans: 3 Hz)

iv) Shaft torque at the rated load (Ans: 41.7Nm)

THE EQUIVALENT CIRCUIT OF AN INDUCTION MOTOR

R1 = Per phase stator winding resistance

jX1 = Per phase stator leakage reactance

VP = Per phase terminal voltage

RC = Core resistance

jXM = Magnetizing reactance

ER = Rotor voltage

RR = Rotor resistance

jXR = Per phase rotor leakage reactance

THE ROTOR CIRCUIT MODEL

jXR0 = blocked rotor rotor resistance

ER0 = Locked rotor voltage

RR

RR jX R

E I

R0R

RR jsX R

E I

R0R

R0R

jX s

RE

I

The greater the relative motion between rotor and the stator magnetic fields, the greater the resulting rotor voltage and rotor frequency. The largest relative motion occurs when the rotor is stationary. (locked rotor or blocked rotor condition)

In an ordinary transformer, the voltages, currents, and impedances on the secondary of the device can be referred to the primary side by means of the turn ratio of the transformer:

THE FINAL EQUIVALENT CIRCUIT

SSP aVVV '

a

III S

SP 'SS ZaZ 2'

The transformed rotor voltage becomes

The rotor current becomes

The rotor impedances becomes

0'

1 ReffR EaEE

eff

R

a

II 2

0

22 R

Reff jX

s

RaZ

Reff RaR 22

02

2 Reff XaX

Referred rotor resistance and reactance

THE FINAL EQUIVALENT CIRCUIT

POWER FLOW DIAGRAM

When the secondary windings in an induction motor (rotor) are shorted out, so no electrical output exists from normal induction motors. Instead, the output is mechanical.

The relationship between input and output powers are shown below:

POWER AND TORQUE IN INDUCTION MOTOR

Stator Copper Loss,

Core Losses,

Air Gap Power,

Rotor Copper Loss,

Developed Mech. Power,

Output power,

Developed Torque,

12

13 RIPSCL

Ccore GEP 213

coreSCLinAG PPPP s

RIPAG

2223

AGRCL sP RIP 2223

RCLAGdevconv PPPP

s

s- R3I 2

2

12

m

convind

P

sync

AGP

miscWFconvout PPPP &

Separating the rotor copper losses and the power converted in Induction motor equivalent circuit

The derivation of the induction motor induced-torque equation

m

convind

P

sync

AGind

P

The induced torque in induction motor is:

s

RIPAG

222

s

RIPAG

2223

Air gap power:

Total Air gap power:

Air gap power is the power crossing the gap from the stator circuit to the rotor circuit. It is equal to the power absorbed in the resistance R2/s.

Continued…

To solve circuit above, we use thevenin’s theorem:

If I2 can be determined, then the air gap power and the induced torque will be known.

Continued…

1XX M 11 RXX M

2

11

M

MTH XX

XRR 1XX TH

)XX(jR

)jXR(jXjXRZ

M

MTHTHTH

11

11

THEVENIN’S EQUIVALENT CIRCUIT

22

2

jXjXs

RR

VI

THTH

TH

22

22

22

222

)()(

33

XXs

RR

s

RV

s

RIP

THTH

TH

AG

])()[(

3

22

22

22

XXs

RR

sR

VP

THTHsync

TH

sync

AGind