Electric Machines I - Philadelphia University · 2021. 3. 1. · 3 Dr. Firas Obeidat Faculty of...
Transcript of Electric Machines I - Philadelphia University · 2021. 3. 1. · 3 Dr. Firas Obeidat Faculty of...
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
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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.
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General Principles
Advantages Disadvantages
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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.
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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
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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
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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
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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)
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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.
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Why Does the Rotor Rotate
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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.
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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
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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
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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
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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
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The Equivalent Circuit in an Induction Motor
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
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The Equivalent Circuit in an Induction Motor
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
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Losses and the Power Flow Diagram
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The following power relations in an induction motor can be deduced
Losses and the Power Flow Diagram
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
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Losses and the Power Flow Diagram
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
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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
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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
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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)
25 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
27 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
28 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
29 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Maximum Torque in Induction Motor
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
30 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Maximum Torque in Induction Motor
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.
31 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Maximum Torque in Induction Motor
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
32 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Maximum Torque in Induction Motor
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
33 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Maximum Torque in Induction Motor
(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
34 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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
35 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
36 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
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.
37