Electric Machines I - Philadelphia University · 2021. 3. 1. · 3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University General Principles Conversion of electrical power

Electric Machines I Three Phase Induction Motor

1

Dr. Firas Obeidat

2

Table of contents

1 • General Principles

2 • Construction

3 • Production of Rotating Field

4

• Why Does the Rotor Rotate

5

• The Slip and Rotor Current Frequency

6

• The Equivalent Circuit in an Induction Motor

7

• Losses and the Power Flow Diagram

8 • Torque-Speed Curve

Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University

General Principles

Conversion of electrical power into mechanical power takes

place in the rotating part of an electrical motor.

In AC motors, the rotor does not receive electrical power but

conduction by induction in the same way as the secondary of

2-winding transformer receives its power from the primary

winding.

Induction motor can be treated as a rotating transformer i.e.

one in which primary winding is stationary but the

secondary is free to rotate.

All of the ac motors, the polyphase induction motor is the

one which is extensively used for various kinds of industrial

drives.


General Principles

Advantages Disadvantages


Construction

An induction motor consists mainly of two main parts, Stator and Rotor

Stator

The stator of induction motor is made up of a number of stampings,

which are slotted to receive the windings.

The stator carries 3-phase winding and is fed from a 3-phase supply.

It is wound for definite number of poles, the exact number of poles being

determined by the requirements of speed.

When the stator winding supplied with 3-phase current, produce

magnetic flux, which is of constant magnitude but which revolves (or

rotates) at synchronous speed. This revolving magnetic flux induces an

emf in the rotor by mutual induction.


Construction

Most of the induction motors are squirrel cage type, because this type of

rotor has the simplest and most rugged construction imaginable and is

almost indestructible.

The rotor consists of cylindrical laminated core with parallel slots for

carrying the rotor conductors which are not wires but consist of heavy

bars of copper, aluminums or alloys. One bar is placed in each slot. The

rotor bars are brazed or electrically welded or bolted to two heavy and

stout short circuiting end ring.

Rotor

Squirrel-cage Rotor


Construction

A wound rotor has a complete set of three-phase

windings that are similar to the windings on the stator.

The three phases of the rotor windings are usually Y-

connected, and the ends of the three rotor wires are

tied to slip rings on the rotor's shaft. The rotor

windings are shorted through brushes riding on the

slip rings. These three brushes are further connected

externally to 3-phase star connected rheostat. This

make possible the introduction of additional resistance

in the rotor circuit during starting period for

increasing the starting torque of the motor.

Wound-rotor induction motors are more expensive

than cage induction motors, and they require much

more maintenance because of the wear associated with

their brushes and slip rings. As a result, wound-rotor

induction motors are rarely used.

Rotor

Phase-wound Rotor


Production of Rotating Field

When stationary coils wound for three phase are supplied by three phase

supply, a uniformly rotating (or revolving) magnetic flux of constant

value is produced.

When three phase winding displaced in space by 120o, are fed by three

phase current displaced in time by 120o, they produce a resultant

magnetic flux which rotates in space as if actual magnetic poles were

being rotated mechanically.

3-phase, 2-poles stator having three

identical windings places 120o space

degree

The flux due to three phase windings Positive direction of fluxes


Production of Rotating Field The maximum value of flux due to any one of the three phases is ϕm. The

resultant flux ϕr (at any instant) is given by the vector sum of the

individual fluxes ϕ1,ϕ2 , and ϕ3 due to three phases.

(1) When θ=0o

ϕ1=0o, 𝝓 𝟐= −

𝟑

𝟐𝝓

𝒎, 𝝓

𝟑=

𝟑

𝟐𝝓

𝒎

𝝓 𝒓= 𝟐 ×

𝟑

𝟐𝝓

𝒎𝒄𝒐𝒔

𝟔𝟎

𝟐= 𝟑 ×

𝟑

𝟐𝝓

𝒎=𝟑

𝟐𝝓

𝒎

(2) When θ=60o

𝝓 𝟏=

𝟑

𝟐𝝓

𝒎, 𝝓

𝟐= −

𝟑

𝟐𝝓

𝒎, 𝝓

𝟑= 𝟎

𝝓 𝒓= 𝟐 ×

𝟑

𝟐𝝓

𝒎𝒄𝒐𝒔

𝟔𝟎

𝟐= 𝟑 ×

𝟑

𝟐𝝓

𝒎=𝟑

𝟐𝝓

𝒎

Let

𝝓 𝟏= 𝝓

𝒎𝒔𝒊𝒏𝜔𝒕

𝝓 𝟐= 𝝓

𝒎𝒔𝒊𝒏(𝜔𝒕 − 𝟏𝟐𝟎)

𝝓 𝟑= 𝝓

𝒎𝒔𝒊𝒏(𝜔𝒕 − 𝟐𝟒𝟎)


Production of Rotating Field

(4) When θ=180o

ϕ1=0o, 𝝓 𝟐=

𝟑

𝟐𝝓

𝒎, 𝝓

𝟑= −

𝟑

𝟐𝝓

𝒎

𝝓 𝒓= 𝟐 ×

𝟑

𝟐𝝓

𝒎𝒄𝒐𝒔

𝟔𝟎

𝟐

𝝓 𝒓= 𝟑 ×

𝟑

𝟐𝝓

𝒎=𝟑

𝟐𝝓

𝒎

60o

ϕ3-ϕ2

Φr=1.5Φm

(1) θ=0o

60o

-ϕ2

ϕ1

Φr=1.5Φm

(2) θ=60o

Φr=1.5Φm

(4) θ=180o

ϕ2-ϕ360o

Φr=1.5Φm

(3) θ=120o

-ϕ3

ϕ160o

From the above four positions, it can

be concluded that:

1. The resultant flux is constant

value and equal to 1.5 ϕm.

2. The resultant flux rotates around

the stator at synchronous speed

given by Ns=120fs/p.

(3) When θ=120o

𝝓 𝟏=

𝟑

𝟐𝝓

𝒎, 𝝓

𝟐= 𝟎, 𝝓

𝟑= −

𝟑

𝟐𝝓

𝒎

𝝓 𝒓= 𝟐 ×

𝟑

𝟐𝝓

𝒎𝒄𝒐𝒔

𝟔𝟎

𝟐=𝟑

𝟐𝝓

𝒎

When 3-phase stator windings are fed by 3-phase supply, a magnetic

flux of constant magnitude, but rotating at synchronous speed, is set up.

The flux passes through the air gap, sweeps past the rotor surface and so

cuts the rotor conductors which, as yet, are stationary.

Due to relative speed between the rotating flux and the stationary

conductors an e.m.f. is induced in the conductors According to faraday’s

law.

The frequency of the induced e.m.f. is the same as the supply frequency.

The e.m.f. magnitude is proportional to the relative velocity between the

flux and the conductors, and its direction is given by Fleming’ right

hand rule.

Since the rotor bars or conductors form a closed circuit, rotor current is

produced whose direction, as given by Lenz’s law, is such as to oppose

the very cause producing it.

The cause which produces the rotor current is the relative velocity

between the rotating flux of the stator and the stationary rotor

conductors.

To reduce the relative speed, the rotor starts running in the same

direction as that of the flux and tries to catch up with the rotating flux.


Why Does the Rotor Rotate


Why Does the Rotor Rotate

The setting up of the torque for rotating the rotor

Figure (a) is shown the stator field which is assumed to be rotating

clockwise. The relative motion of the rotor with respect to the stator is

anticlockwise.

By applying right hand rule, the direction of the induced e.m.f. in the

rotor is found to be outwards.

The direction of the flux due to rotor current alone is as shown in figure

(b).

By applying left hand rule or by combined field as shown in figure ( c),

the rotor conductors experience a force tending to rotate them in

clockwise direction. So, the rotor is set into rotation in the same

direction as that of the stator flux.


The Slip and Rotor Current Frequency

The rotor never succeeds in catching up with the stator field. If it really

did, then there would be no relative speed between the two, hence no

rotor e.m.f., no rotor current and so no torque to maintain rotation.

The slip (s) is the difference between the synchronous speed Ns and the

actual speed N of the rotor.

𝒔 =𝑵𝒔 − 𝑵𝒎

𝑵𝒔× 𝟏𝟎𝟎%

𝑵𝒎 = (𝟏 − 𝒔)𝑵𝒔

When the rotor is stationary, the frequency of rotor current is the same

as the supply frequency.

When the rotor starts revolving, the frequency depends upon the

relative speed or on the slip speed.

𝑵𝒔𝒍𝒊𝒑 = 𝑵𝒔 −𝑵𝒎 is called slip speed

𝒔 =𝝎𝒔 −𝝎𝒎

𝝎𝒔× 𝟏𝟎𝟎% Where

Ns= synchronous speed in rpm

N m = rotor speed (mechanical

shaft speed) in rpm

𝜔s= synchronous angular

velocity (2π𝑵𝒔/𝟔𝟎) in rad/s

𝜔m= mechanical angular velocity

(2π𝑵𝒎/𝟔𝟎) in rad/s

𝝎𝒎 = (𝟏 − 𝒔)𝝎𝒔


The Slip and Rotor Current Frequency at any slip speed, the frequency of the rotor be fr

𝑵𝒔 − 𝑵𝒎 =𝟏𝟐𝟎𝒇𝒓𝑷

𝑵𝒔 =𝟏𝟐𝟎𝒇𝒔𝑷

Dividing the above equations one by other

𝑓𝑟𝑓𝑠=𝑁𝑠 − 𝑁𝑚

𝑁𝑠= 𝑠

𝒇𝒓 = 𝒔𝒇𝒔

Where

Ns= synchronous speed in rpm

N m = rotor speed (mechanical

shaft speed) in rpm

P=number of poles

fs=stator frequency in Hz

fr= rotor frequency in Hz

𝑓𝑟 = 𝑠𝑓𝑠 =𝑁𝑠 − 𝑁𝑚

𝑁𝑠𝑓𝑠 = (𝑁𝑠 − 𝑁𝑚)

𝑃

120𝑓𝑠𝑓𝑠

𝒇𝒓 =𝑷

𝟏𝟐𝟎(𝑵𝒔 − 𝑵𝒎)

Or

Substitute 𝑠 =𝑁𝑠−𝑁𝑚

𝑁𝑠 in the above equation gives


The Slip and Rotor Current Frequency

When the rotor stationary (at standstill) Nm= 0 rpm, the rotor frequency fr=fs

and the slip s=1. At Nm= Ns, the rotor frequency fr= 0 Hz, and the slip s=0.

Example: A 208-V, 10-hp, four-pole, 60 Hz, Y connected induction motor has

a full-load slip of 5 percent.

(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?

(a) 𝑁𝑠 =120𝑓𝑠𝑃

=120 × 60

4= 1800 𝑟𝑝𝑚

𝑁𝑚 = 1 − 𝑠 𝑁𝑠 = 1 − 0.05 1800 = 1710 𝑟𝑝𝑚 (b)

(c) 𝑓𝑟 = 𝑠𝑓𝑠 = 0.05 × 60 = 3 𝐻𝑧


The Equivalent Circuit in an Induction Motor

The largest relative motion occurs when the rotor is stationary, called the

locked-rotor or blocked-rotor condition, so the largest voltage and rotor

frequency are induced in the rotor at that condition.

The smallest voltage (0 V) and frequency (0 Hz) occur when the rotor

moves at the same speed as the stator magnetic field, resulting in no relative

motion.

The magnitude and frequency of the voltage induced in the rotor at any

speed between these extremes is directly proportional to the slip of the

rotor.

If the magnitude of the induced rotor voltage at locked-rotor conditions is

called E2 the magnitude of the induced voltage at any slip will be given by the

equation

𝐸𝑟 = 𝑠𝐸2

The frequency of the induced voltage at any slip will be given by the equation

𝒇𝒓 = 𝒔𝒇𝒔

The reactance of an induction motor rotor depends on the inductance of the

rotor and the frequency of the voltage and current in the rotor. With a rotor

inductance of Lr the rotor reactance is given by



Where X2 is the locked rotor reactance

𝑋𝑟 = 2𝜋𝑓𝑟𝐿𝑟 = 2𝜋𝑠𝑓𝑠𝐿𝑟 = 𝑠𝑋2

𝐼2 =𝐸𝑟

𝑅𝑟 + 𝑋𝑟=

𝑠𝐸2𝑅2 + 𝑠𝑋2

=𝐸2

𝑅2 𝑠 + 𝑋2

The rotor current flow is

To produce the final per-phase equivalent circuit for an induction motor, it is

necessary to refer the rotor parts of the model over to the stator side.

𝑎 =𝐸1𝐸2

The turn ratio of the induction motor is

𝐼2′ =

𝐼2𝑎

𝑹𝟐′ = 𝒂𝟐𝑹𝟐 𝑿𝟐

′ = 𝒂𝟐𝑿𝟐

So

𝐸2′ = 𝑎𝐸2



Part of the power coming across the air gap in an induction motor is

consumed in the rotor copper losses, and part of it is converted to

mechanical power to drive the motor's shaft. It is possible to separate the

two uses of the air-gap power and to indicate them separately on the motor

equivalent circuit.

Io

ImIc

XmRcV1

X1R1 X2'

E2'=E1

I1 I2'

R2's

1-s( )

R2'

In order to separate the rotor copper losses and the converted power to

mechanical power, the equivalent circuit of the induction motor as the

figure below


Losses and the Power Flow Diagram


The following power relations in an induction motor can be deduced


Input power 𝑷𝟏 = 𝟑𝑽𝟏𝑰𝟏𝒄𝒐𝒔𝜽𝟏 = 𝟑𝑽𝑳𝑰𝑳𝒄𝒐𝒔𝜽𝟏

Stator core losses 𝑷𝒇 = 𝑰𝑪𝟐𝑹𝑪

Stator copper losses 𝑷𝑪𝒔 = 𝟑𝑰𝟏𝟐𝑹𝟏

Power transferred to rotor (air gap power) 𝑷𝟐 = 𝟑𝑰𝟐′𝟐 𝑹𝟐

′

𝒔

Rotor copper losses 𝑷𝑪𝒓 = 𝟑𝑰𝟐′𝟐𝑹𝟐

′ = 𝒔𝑷𝟐

Mechanical power 𝑷𝒎 = 𝟑𝑰𝟐′𝟐𝑹𝟐

′ 𝟏−𝒔

𝒔

𝑷𝟐 = 𝑷𝟏 − 𝑷𝑪𝒔 − 𝑷𝒇

𝑷𝒎 = 𝑷𝟐 − 𝑷𝑪𝒓 = 𝑷𝟐 − 𝒔𝑷𝟐 = (𝟏 − 𝒔)𝑷𝟐

The gross torque developed by the

rotor (air gap torque) Tg is

𝑻𝒈 =𝑷𝒎

𝝎𝒎=𝟑𝑰𝟐

′𝟐𝑹𝟐′ 𝟏 − 𝒔

𝒔𝟐𝝅𝑵𝒎𝟔𝟎

𝐍𝐦



𝑻𝒈 =𝑷𝒎

𝝎𝒎=𝟑𝑰𝟐

′𝟐𝑹𝟐′ 𝟏 − 𝒔

𝒔𝟐𝝅(𝟏 − 𝒔)𝑵𝒔

𝟔𝟎

𝐍𝐦 =𝟑𝑰𝟐

′𝟐𝑹𝟐′

𝒔𝟐𝝅𝑵𝒔𝟔𝟎

𝐍𝐦 =𝑷𝟐

𝝎𝒔𝐍𝐦

The approximate circuit for the

induction motor as shown in the figure,

from this figure I2’ can be found as

Io

ImIc

XmRcV1

Xeq=X1+X2'

E2'=E1

I1 I2'

R2's

1-s( )

Req=R1+R2'

𝐼2′ =

𝑉1𝑅1 + 𝑅2

′ 𝑠 + 𝑗 𝑋1 + 𝑋2′

𝑁𝑚 = (1 − 𝑠)𝑁𝑠 But

𝐼2′ =

𝑉1

(𝑅1 + 𝑅2′ 𝑠 )2+(𝑋1 + 𝑋2

′)2

𝑻𝒈 =𝟑𝑽𝟏

𝟐

𝝎𝒔

𝑹𝟐′

𝒔(𝑹𝟏 + 𝑹𝟐

′ 𝒔 )𝟐+(𝑿𝟏 + 𝑿𝟐′)𝟐

𝐍𝐦

Output power= mechanical power –Rotational (Windage & Friction) loss

𝑷𝒐𝒖𝒕 = 𝑷𝒎 − 𝑷𝒘


Examples

(b) 𝑃𝑚 = 𝑃2 − 𝑃𝐶𝑟 = 38.6 − 0.7 = 37.9kW

(c) 𝑃𝑜𝑢𝑡 = 𝑃𝑚 − 𝑃𝑤 = 37.9 − 0.6 = 37.3kW

𝜂 =𝑃𝑜𝑢𝑡𝑃1

=37.3

42.4× 100% = 88% (d)

Examlpe: A 480-VL, 60-Hz, 50-hp, three-phase induction motor is drawing

60A at 0.85 pf lagging. The stator copper losses are 2 kW, and the rotor

copper losses are 700 W. The rotational losses are 600 W, the core losses are

1800 W. Find the following quantities:

(a) The air-gap power P2

(b) The power converted Pm

(c) The output power Pout

(d) The efficiency of the motor

𝑃1 = 3𝑉𝐿𝐼𝐿1𝑐𝑜𝑠θ1 = 3 × 480 × 60 × 0.85 = 42.4kW (a)

𝑃2 = 𝑃1 − 𝑃𝐶𝑠 − 𝑃𝑓 = 42.4 − 2 − 1.8 = 38.6kW

𝜼 =𝑷𝒐𝒖𝒕

𝑷𝟏× 𝟏𝟎𝟎%

The efficiency of the induction motor is The output torque

𝑻𝒐𝒖𝒕 =𝑷𝒐𝒖𝒕

𝝎𝒎


Examples

Examlpe: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected induction motor has

the following impedances in ohms per phase referred to the stator circuit:

R1=0.641Ω X1=1.106 Ω R2’=0.332Ω X2

’=0.464Ω Xm=26.3Ω

The total rotational (windage and friction) losses are 1100 W and are assumed

to be constant. The core loss is jumped in with the rotational losses. For a rotor

slip of 2.2 percent at the rated voltage and rated frequency, find the motor's

(a) Speed

(b) Stator current (c) Power factor (d) Air gap power and output power

(e) The air gap torque Tg and load torque Tout (f) Efficiency

(a) 𝑁𝑠 =120𝑓𝑠𝑃

=120 × 60

4= 1800 𝑟𝑝𝑚

𝑁𝑚 = 1 − 𝑠 𝑁𝑠 = 1 − 0.022 1800 = 1760.4 𝑟𝑝𝑚

(b) 𝑍2 =𝑅2

′

𝑠+ 𝑗𝑋2

′ =0.332

0.022+ 𝑗0.464 = 15.09 + 𝑗0.464 = 15.1∠1.76𝑜

Ω

𝑍𝑜 =1

1 𝑗𝑋𝑚 + 1 𝑍2 =

1

1 𝑗26.3 + 1 15.1∠1.76 =

1

−𝑗0.038 + 0.0662∠ − 1.76

𝑍𝑜 =1

0.0773∠ − 31.1= 12.94∠31.1𝑜 Ω


Examples

(c)

𝑍𝑇 = 𝑍1 + 𝑍𝑜 = 0.641 + 𝑗1.106 + 12.94∠31.1 = 11.72 + 𝑗7.79 = 14.07∠33.6𝑜 Ω

𝐼1 =𝑉1𝑍𝑇

=460/ 3

14.07∠33.6𝑜= 18.88∠ − 33.6 𝐴

𝑝𝑓 = 𝑐𝑜𝑠33.6 = 0.83 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

𝑃2 = 𝑃1 − 𝑃𝐶𝑠 − 𝑃𝑓 = 12.53 − 0.685 − 0.0 = 11.845 kW

𝑃1 = 3𝑉𝐿𝐼𝐿1𝑐𝑜𝑠θ1 = 3 × 460 × 18.88 × 0.83 = 12.53kW (d)

𝑃𝐶𝑠 = 3𝐼12𝑅1 = 3 × 18.882 × 0.641 = 0.685 kW

𝑃𝑚 = 𝑃2 − 𝑃𝐶𝑟 = 𝑃2 − 𝑠𝑃2 = 1 − 𝑠 𝑃2 = 1 − 0.022 11845 = 11585 W

𝑃𝑜𝑢𝑡 = 𝑃𝑚 − 𝑃𝑤 = 11585 − 1100 = 10485 W

(e)

𝑇𝑜𝑢𝑡 =𝑃𝑜𝑢𝑡𝜔𝑚

=10485

2𝜋1760.460

=10485

184.3= 56.9 Nm

𝑇𝑔 =𝑃2𝜔𝑠

=11845

2𝜋𝑁𝑠60

=11845

2𝜋180060

=11845

188.5= 62.8 Nm

𝜂 =𝑃𝑜𝑢𝑡𝑃1

=10485

12530× 100% = 83.7% (f)


Torque-Speed Curve

The torque speed (slip) curve

for an induction motor gives

us the information about the

variation of torque with the

slip.

At full load, the motor runs at speed of Nm. When mechanical load

increases, motor speed decreases tell the motor torque again becomes

equal to the load torque.

As long as the two torques are in balance, the motor will run at constant

(but lower) speed.

If the load torque exceeds the induction motor maximum torque, the

motor will suddenly stop.

To

rqu

e

Speed %

0

Nm

10080604020

Full Load

Torque

Maximum

Torque

Ns

0

Starting

Torque

Tfl

1.5Tfl

2.5Tfl

Slip %

0100 80 4060 20 0

When the rotor stationary (at

standstill) Nm= 0 rpm, the

rotor frequency fr=fs and the

slip s=1. At Nm= Ns, the rotor

frequency fr= 0 Hz, and the

slip s=0.


Comments on the Induction Motor Torque- Speed Curve

1. The induced torque of the motor is zero at synchronous speed.

2. 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.

3. 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.

4. 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.


Comments on the Induction Motor Torque- Speed Curve

5. 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.

6. If the rotor of the induction motor is driven faster than synchronous speed, then the direction of the induced torque in the machine reverses and the machine becomes a generator, converting mechanical power to electric power.

7. If the motor is turning backward relative to the direction of the magnetic fields, the induced torque in the machine will stop the machine very rapidly and will try to rotate it in the other direction. Since reversing the direction of magnetic field rotation is simply a matter of switching any two stator phases, this fact can be used as a way to very rapidly stop an induction motor. The act of switching two phases in order to stop the motor very rapidly is called plugging.


Maximum Torque in Induction Motor

𝑇𝑔 =3𝑉1

2

𝜔𝑠

𝑅2′

𝑠(𝑅1 + 𝑅2

′ 𝑠 )2+(𝑋1 + 𝑋2′)2

𝑁𝑚

𝑑𝑇𝑔

𝑑𝑠= 0

𝑑𝑇𝑔

𝑑𝑠=3𝑉1

2

𝜔𝑠

(𝑅1 + 𝑅2′ 𝑠 )2+(𝑋1 + 𝑋2

′)2−𝑅2

′

𝑠2− 2

𝑅2′

𝑠 𝑅1 + 𝑅2′ 𝑠

−𝑅2′

𝑠2

(𝑅1 + 𝑅2′ 𝑠 )2+(𝑋1 + 𝑋2

′)2 2= 0

(𝑅1 + 𝑅2′ 𝑠 )2+(𝑋1 + 𝑋2

′)2−𝑅2

′

𝑠2− 2

𝑅2′

𝑠𝑅1 + 𝑅2

′ 𝑠 −𝑅2

′

𝑠2= 0

(𝑅1 + 𝑅2′ 𝑠 )2+(𝑋1 + 𝑋2

′)2=2𝑅1𝑅2

′

𝑠+2𝑅2

′2

𝑠2

Since the induced torque is equal to P2/𝜔s. the maximum possible torque

occurs when the air-gap power is maximum. Since the air-gap power is

equal to the power consumed in the resistor R2’/s, the maximum induced

torque will occur when the power consumed by that resistor is maximum.



𝑅12 + (𝑋1 + 𝑋2

′)2=𝑅2

′2

𝑠2

𝑅2′

𝑠= ± 𝑅1

2 + (𝑋1 + 𝑋2′)2

𝒔𝒎𝒂𝒙 = ±𝑹𝟐

′

𝑹𝟏𝟐 + (𝑿𝟏 + 𝑿𝟐

′)𝟐

Substitute smax in torque equation to get the maximum torque equation

𝑻𝒈𝒎𝒂𝒙=𝟑𝑽𝟏

𝟐

𝟐𝝎𝒔

𝟏

𝑹𝟏 + 𝑹𝟏𝟐 + (𝑿𝟏 + 𝑿𝟐

′)𝟐𝑵𝒎

𝑅12 +

2𝑅1𝑅2′

𝑠+2𝑅2

′2

𝑠2+ (𝑋1 + 𝑋2

′)2=2𝑅1𝑅2

′

𝑠+2𝑅2

′2

𝑠2

𝑅12 + (𝑋1 + 𝑋2

′)2=𝑅2

′2

𝑠2

The plus (+) sign for motor.

The minus (-) sign for generator



The maximum torque is proportional to the square of the supply

voltage and is also inversely related to the size of the stator

impedances and the rotor reactance. The smaller a machine's

reactances, the larger the maximum torque it is capable of

achieving. slip at which the maximum torque occurs is directly

proportional to rotor resistance, but the value of the maximum

torque is independent of the value of rotor resistance.

It is possible to insert resistance into the rotor circuit of a wound

rotor because the rotor circuit is brought out to the stator through

slip rings. As the rotor resistance is increased, the pullout speed of

the motor decreases. but the maximum torque remains constant.

If a resistance is inserted into the rotor circuit, the maximum torque

can be adjusted to occur at starting conditions. Therefore. The

maximum possible torque would be available to start heavy loads.

On the other hand, once the load is turning, the extra resistance can

be removed from the circuit, and the maximum torque will move up

to near-synchronous speed for regular operation.



Example: A two-pole, 50-Hz induction motor supplies 15 kW to a load at a

speed of 2950 r/min. Neglecting the rotational losses.

(a) What is the motor 's slip?

(b) What is the induced torque in the motor in N.m under these conditions?

(c) How much power will be supplied by the motor when the torque is

doubled at the same speed?

𝑁𝑠 =120𝑓𝑠𝑃

=120 × 50

2= 3000 𝑟𝑝𝑚

(a)

𝑠 =𝑁𝑠 − 𝑁𝑚

𝑁𝑠× 100% =

3000 − 2950

3000× 100% = 1.66%

(b) 𝑇𝑔 =𝑃𝑚𝜔𝑚

=15000

2𝜋295060

= 48.6 Nm

(c) 𝑃𝑚 = 𝑇𝑔𝜔𝑚 = 48.6 × 22𝜋2950

60= 29.5 kW



Example: A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor

induction motor has the following impedances in ohms per phase referred to

the stator circuit:

R1=0.641Ω X1=1.106 Ω R2’=0.332Ω X2

’=0.464Ω Xm=26.3Ω

(a) What is the maximum torque of this motor? At what speed and slip does

it occur?

(b) What is the starting torque of this motor?

(c) When the rotor resistance is doubled, what is the speed at which the

maximum torque now occurs? What is the new starting torque of the

motor?

(a) 𝑇𝑔𝑚𝑎𝑥=3𝑉1

2

2𝜔𝑠

1

𝑅1 + 𝑅12 + (𝑋1 + 𝑋2

′)2

𝑁𝑠 =120𝑓𝑠𝑃

=120 × 60

4= 1800 𝑟𝑝𝑚

𝜔𝑠 =2𝜋𝑁𝑠

60=2𝜋1800

60= 188.5 𝑟𝑎𝑑/𝑠𝑒𝑐

𝑇𝑔𝑚𝑎𝑥=3(460/ 3)2

2 × 188.5

1

0.641 + 0.6412 + (1.106 + 0.464)2= 240𝑁𝑚



(b)

𝑠𝑚𝑎𝑥 =𝑅2

′

𝑅12 + (𝑋1 + 𝑋2

′)2=

0.332

0.6412 + (1.106 + 0.464)2= 0.196

𝑇𝑔𝑠𝑡𝑎𝑟𝑡=3𝑉1

2

𝜔𝑠

𝑅2′

𝑠(𝑅1 + 𝑅2

′ 𝑠 )2+(𝑋1 + 𝑋2′)2

𝑁𝑚

At standstill s=1

𝑇𝑔 =3(460/ 3)2

188.5

0.332

(0.641 + 0.332)2+(1.106 + 0.464)2= 109𝑁𝑚

(b) If the rotor resistance is doubled, then the slip at maximum torque doubles, too.

𝑠𝑚𝑎𝑥𝑛𝑒𝑤 = 0.196 × 2 = 0.392

𝑁𝑚𝑛𝑒𝑤 = 1 − 𝑠𝑚𝑎𝑥𝑛𝑒𝑤 𝑁𝑠 = 1 − 0.392 1800 = 1094.4 𝑟𝑝𝑚

𝑇𝑔𝑠𝑡𝑎𝑟𝑡=3𝑉1

2

𝜔𝑠

𝑅2′

𝑠(𝑅1 + 𝑅2

′ 𝑠 )2+(𝑋1 + 𝑋2′)2

𝑁𝑚

𝑇𝑔𝑠𝑡𝑎𝑟𝑡−𝑛𝑒𝑤=3(460/ 3)2

188.5

(2 × 0.332)

(0.641 + (2 × 0.332))2+(1.106 + 0.464)2= 170𝑁𝑚


Complete Torque- Speed Curve of a Three Phase Machine

Three phase machine can be run as motor when it takes electrical power and

supplies mechanical power. The direction of torque and rotor rotation are in the

same. For this case 0<Nm<Ns, 1<s<0.

The same machine can be used as an asynchronous generator when driven at

speed greater than the synchronous speed. In this case, it receives mechanical

energy from the stator. The torque is oppositely-directed. For this case Nm>Ns,

s<0.

The same machine can be used as a brake during the plugging period. For this

case Nm in opposite direction, s>1.

|← s>1 →|

|← 0 <s<1 →|

|← s<0 →|

s=1 s=0


Induction Motor Operating as Generator

When run faster than its synchronous speed, an induction motor runs as a

generator called induction generator.

The induction generator converts the mechanical power it receives into

electrical energy and this energy is released by the stator.

As soon as the motor speed exceeds its synchronous speed, it starts

delivering active power P to the 3-phase line. However, for creating its own

magnetic field, it absorbs reactive power Q from the line to which it

connected. Q flows in the opposite direction to P.


Plugging of an Induction Motor

An induction motor can be quickly stopped by inter-changing any of its two

stator leads. It reverses the direction of the revolving flux which produces a

torque in the reversed direction. Thus applying brake on the motor.

During this so-called plugging period, the motor acts as a brake. It absorbs

kinetic energy from the still revolving load causing its speed to fall.

The associated power Pm is dissipated as heat in the rotor. At the same time,

the rotor also continues to receive power P2 from the stator which also

dissipated as heat.

Plugging produces rotor I2R losses which even exceed those when the rotor

is locked.

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Electric Machines I - Philadelphia University · 2021. 3. 1. · 3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University General Principles Conversion of electrical power