electrical power systems induction motor

Mubarek Kurt

INDUCTION MOTOR BY: MUBAREK KURT

Mubarek Kurt

Induction Motor

Scope of discussion■ Structure■ Basic concept■ Equivalent Circuit■ Power and Torque■ Speed Control■ Testing

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Introduction

■ The AC induction motor (IM) is well suited to application requiring constant speed operation.

■ In general, the induction motor is cheaper and easier to maintain compared to other alternatives.

■ The IM is made up of the stator, or stationary windings, and the rotor.

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Induction Motor- Structure -

Squirrel Cage – no winding and slip ring

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Wound Rotor – the winding connected to slip rings

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Synchronous Speed

■ The speed with which the stator magnetic field rotates, which will determine the speed of the rotor, is called the synchronous speed (SS).

■ The relationship to calculate the SS of an induction motor is

SS=(120 X f)/PWhere SS=Syn. Speed (RPM), f=frequency (Hz) and P number of poles.

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Example 1

■ Determine syn. Speed for 2 pole & 8 pole motor. Assume f=50Hz.

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Motor/Rotor Speed

■ The rotor in an induction motor can not turn at the syn. Speed. In order to induce an emf in the rotor, the rotor must slower than the Syn. Speed (SS).

■ The relationship between the rotor speed and the SS is called the Slip (S). Typically, the slip expressed as a percentage of the SS.

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Slip■ The equation for the motor slip is:

■ If you were sitting on the rotor, you would find that the rotor was slipping behind the rotating field by the slip rpm = SS-RS=S(SS).

■ At stand still, S=1, that is RS=0. At the synchronous speed, RS=SS, S=0.

■ The mechanical speed of the rotor, in terms of slip and syn. speed.

%100% xSSRSSSS

SSRS SSSSRS )1()1(

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Example 2

■ Find Slip in percentage for SS=900 RPM and RS 885 RPM.

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The Electrical Freq. on the Rotor

■ An induction motor works by inducing voltages and currents in the rotor of the machine, and for that reason it has sometimes been called a rotating transformer.

■ If the rotor of a motor is locked so that it cannot move, then the rotor will have the same freq. as the stator.

■ On the other hand, if the rotor turns at sync. Speed, the freq. on the rotor will be zero.

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Motor/Rotor Speed cont.■ The frequency fre of the induced voltage and current in the

rotor circuit will correspond to this slip rpm, because this is the relatives speed between the rotating field and the rotor winding. Thus,

sere Sff

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Example 3

■ A 3phase, 20hp, 208 V, 60Hz, 6 pole, wye connected induction motor delivers 15kW at a slip 5%. Determines

a) Syn. Speedb) Rotor Speedc) Freq. of the rotor current

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Example 4

■ A 3 phase, 460 V, 100 hp, 60 Hz, four pole induction machine delivers rated output power at a slip of 0.05. Determine the

a) Syn. Speed and motor speed.b) Speed of the rotating air gap field. c) Frequency of the rotor circuit.d) Slip RPM.

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Example 5

A 208V, 10hp, 4 pole, 50 Hz, Y-connected induction motor has a full-load slip of 5%.

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?d) What is the shaft torque of this motor at the rated

load?

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Example 6

A 220-V, three-phase, two-pole, 50-Hz induction motor is running at a slip of 5 percent. Find:(a) The speed of the magnetic fields in revolutions per minute(b) The speed of the rotor in revolutions per minute(c) The slip speed of the rotor(d) The rotor frequency in hertz

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Example 7

A 480-V, three-phase, four-pole, 60-Hz induction motor running at a slip of 0.035. Find;(a) The speed of the magnetic fields in revolutions per minute(b) The speed of the rotor in revolutions per minute(c) The slip speed of the rotor(d) The rotor frequency in hertz

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Example 8

A three-phase, 60-Hz induction motor runs at 890 r/min at no load and at 840 r/min at full load.(a) How many poles does this motor have?(b) What is the slip at rated load?(c) What is the speed at one-quarter of the rated load?(d) What is the rotor’s electrical frequency at one-quarter of the rated load?

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Equivalent Circuit■ An induction motor relies for its operation on the induction

of voltages and currents in its rotor circuit from stator circuit (transformer action).

■ Because the induction of voltages and currents in the rotor circuit of an induction motor is essentially a transformer operation, the equivalent circuit of an induction motor will turn out to very similar to the equivalent circuit of transformer.

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Induction Motor- Equivalent Circuit -

Rotor voltage – Locked-rotor voltage

Rotor frequency – Frequency of the induced voltage

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RR is constant but XR depend on S

Rotor current flow

Overall rotor impedance considered SReferring to stator side, considered turn ratio – new Rotor voltage

Induction Motor – Equi. Circuit

XR is known as blocked-rotor rotor reactance

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Induction Motor – Basic Concept

Rotor current

Rotor impedance

If,

Final per-phase equivalent circuit

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Induction Motor- Power and Torque -

Pconv=Pmech

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Relationship Pag, Pconv & Pmech

■ Power across the air gap is includes the rotor copper loss (RCL) and mechanical power developed.

SSPPP

SPRIP

PSSP

PSP

SSRIP

SRIP

SSRRIP

mechRCLag

agRCL

RCLmech

agmech

mech

ag

ag

1::1::

)1(

)1(

)1(

)1(

222

222

222

22

22

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Example 9■A 480 V, 60 Hz, 50 Hp, 3 phase induction motor

is drawing 60 A at 0.85 PF lagging. The stator copper losses are 2 kW and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities:

a) The air gap powerb) The converted power c) The output power d) The efficiency of the motor.

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Example 10

■ A 3phase, 15 hp, 460 V, 4 pole, 60 Hz, 1728 rpm induction motor delivers full output power to a load connected to its shaft. The windage and friction loss of the motor is 750 W. Determine the

a) Mechanical power developed (=Output Shaft power +Windage and friction loss)

b) Air gap powerc) Rotor copper loss.

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Air gap needs R2/S and rotor copper loss needs R2

Stator Core Losses Rotor Core Losses

Mechanical @ Converted Power

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Example 11

A 480V, 60Hz, 50hp, 3 phase induction motor is drawing 60A at 0.85 Pf lagging. The stator copper losses are 2 kW, and the rotor losses are 700W. The friction and windage losses is 600W, the core losses are 1800W, and the stray losses are negligible. Find:a) The air gap power PAG

b) The power converted Pconv

c) The output power Pout

d) The efficiency of the motor

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Example 11A 460V, 25hp, 50Hz, 4 pole, Y-connected induction motor has the following impedances in ohms per phase referred to stator circuit:

The total rotational losses are 1100W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2% at rated voltage and rated frequency, find the motor’sa) speed d) Pconv and Pout

b) stator current e) ind and load

c) power factor f ) efficiency

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Example 12A 208-V, two-pole, 60-Hz Y-connected wound-rotor induction motor is rated at 15 hp. Its equivalent circuit components areR1 = 0.200 Ω R2 = 0.120 Ω XM = 15.0 ΩX1= 0.410 Ω X2 = 0.410 ΩPmech = 250 W Pmisc ≈ 0 Pcore = 180 WFor a slip of 0.05, find(a) The line current IL(b) The stator copper losses PSCL

(c) The air-gap power PAG

(d) The power converted from electrical to mechanical form Pconv

(e) The induced torque τind

(f) The load torque load τload

(g) The overall machine efficiency(h) The motor speed in revolutions per minute and radians per second

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Example 13

A 50-kW, 440-V, 50-Hz, six-pole induction motor has a slip of 6 percent when operating at full-load conditions. At full-load conditions, the friction and windage losses are 300 W, and the core losses are 600 W. Find the following values for full-load conditions:(a) The shaft speed nm

(b) The output power in watts(c) The load torque τload in newton-meters(d) The induced torque τind in newton-meters(e) The rotor frequency in hertz

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Induction Motor- Speed Control - ■ Two methodsa) Varying stator and rotor magnetic field speed

- electrical frequency or changing the number of poles

b) Varying slip- varying rotor resistance or terminal voltage

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Created two poles, N and S

Created an extra poles by changing current polarity (only 2 speeds)

Induction Motor- Speed Control (pole changing)-

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Induction Motor- Speed Control (Frequency) - With or without adjustment to the terminal voltage:a) Vary frequency, stator voltage adjusted –

generally vary speed and maintain operating torque.

b) Vary frequency, stator voltage maintained – able to achieve higher speeds but a reduction of torque.

Could be combined together, easy way by introducing power electronic fields.

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Induction Motor- Speed Control (terminal voltage) -

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Induction Motor- Speed Control (rotor resistance) -

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Testing of Induction Motor■ The parameters of the equivalent circuit Rc, Xm, R1, X1, R2

and X2, can be determined from the results of a no-load test, a blocked-rotor test and from measurement of the DC resistance of the stator winding.

■ No-load test is like the open-circuit test on a transformer, gives information about exciting current and rotational losses.

■ The small power loss in the machine at no-load is due to the core loss and the friction and windage loss.

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Testing of Induction Motor cont.

■ The blocked-rotor test on an induction machine, like the short-circuit test on transformer, gives information about leakage impedances.

■ In this test the rotor is blocked so that so that the motor cannot rotate.

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Induction Motor- Testing (no load Test) -

Measuring rotational losses and providing info about its magnetizing current (X1 + XM)

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Due to the rotor speed close to sync speed, S small and higher rotational losses

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Induction Motor- Testing (no load test) -

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Induction Motor- Testing (DC test) -

To find R1 (stator resistance).a) Dc voltage is applied and adjusted to rated

condition.b) Voltage and current flow are recorded.

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Induction Motor- Testing (Locked rotor)-

Steps:1- locked2- AC voltage is applied and currents flow is adjusted to full load condition3- Measure voltage, current and power flow

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Induction Motor- Testing (Locked rotor)-

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Example 14The following test data were taken on a 7.5 hp, 4-pole, 208V, 60Hz, design star connected induction motor having a rated current of 28A

(a)Sketch the per-phase equivalent circuit for this motor.

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Example 151) No Load TestSupply Freq. =60HzLine Voltage = 2200 VLine Current= 4.5 AInput Power=1600 W

2) Blocked-rotor TestFreq. =15HzLine Voltage = 270 VLine Current= 25 AInput Power=9000 W

3) Average DC resistance per stator phaseR1=2.8 ohma) Determine the no-load rotational lossb) Determine all parameters

electrical power systems induction motor

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