Transcript Chapter 4 : AC Machine

Chapter 4 :
AC Machine
Lecturer : Puan Nordiana Binti Mohamad Saaid
Slide prepared/edited by: Dr Rosemizi Abd Rahim/Puan Nordiana
Click here to watch the ac machine animation video
http://rmz4567.blogspot.my/2013/02/electrical-engineering.html
AC Machine
 Alternating current (ac) is the primary source of electrical
energy.
 It is less expensive to produce and transmit alternating
current (ac) than direct current (dc).
 For this reason, and because ac voltage is induced into the
armature of all generators, ac machines are generally more
practical.
 It is very easy to step up or step down the voltage for AC electricity
through the use of transformers. It is not very straight forward to
step up or step down the voltage for DC electricity- for power
loads, it is typically done through inversion to AC- stepping
up/down using a transformer- and rectification back to DC.
 May function as a generator (mechanical to electrical) or a
motor (electrical to mechanical)
DC Machine & AC machine
• DC motor - ends of the coil
connect to a split ring to 'rectify'
the emf produced
• AC motor - no need rectification,
so don't need split rings.
AC Motor
As in the DC motor case,
• a current is passed through
the coil, generating a torque
on the coil.
• Since the current is
alternating, the motor will
run smoothly only at the
frequency of the sine wave.
AC Generator
• This process can be described in terms of Faraday's law .
The rotation of the coil continually changes the magnetic
flux through the coil and therefore generates a voltage
Generator and Motor
How Does an Electric Generator Work?
Classification of AC Machines
 Two major classes of machines;
i) Synchronous Machines:
• Synchronous Generators: A primary source of
electrical energy.
• Synchronous Motors: Used as motors as well as
power factor compensators (synchronous condensers).
ii) Asynchronous (Induction) Machines:
• Induction Motors: Most widely used electrical motors in
both domestic and industrial applications.
• Induction Generators: Due to lack of a separate field
excitation, these machines are rarely used as generators.
Synchronous Machine
Synchronous Machine
 Origin of name: syn = equal, chronos = time
 Synchronous machines are called ‘synchronous’ because
their mechanical shaft speed is directly related to the power
system’s line frequency.
 the rotating air gap field and the rotor rotate at the same
speed, called the synchronous speed.
 Synchronous machines are ac machine that have a field
circuit supplied by an external dc source.
– DC field winding on the rotor,
– AC armature winding on the stator
Synchronous Machine
 Synchronous machines are used primarily as generators of
electrical power, called synchronous generators or
alternators.
 They are usually large machines generating electrical power
at hydro, nuclear, or thermal power stations.
 Synchronous motors are built in large units compare to
induction motors (Induction motors are cheaper for smaller
ratings) and synchronous motors are used for constant speed
industrial drives
 Application of synchronous machine as a motor: pumps in
generating stations, electric clocks, timers, and so forth
where constant speed is desired.
Synchronous Machine
The frequency of the induced voltage is related to the rotor speed by:
where ;
P = the number of magnetic poles
fe = the power line frequency.
Typical machines have two-poles, four-poles, six-poles etc.
Synchronous Machine
Construction
• Energy is stored in the inductance
• As the rotor moves, there is a change in
the energy stored
• Either energy is extracted from the
magnetic field (and becomes mechanical
energy – motor)
• Or energy is stored in the magnetic field
and eventually flows into the electrical
circuit that powers the stator – generator
Synchronous Machine
Construction
• DC field windings are mounted on the (rotating) rotor
- which is thus a rotating electromagnet
• AC windings are mounted on the (stationary) stator
resulting in three-phase AC stator voltages and
currents
 The main part in the synchronous machines are
i) Rotor
ii) Stator
Synchronous Machine
Rotor
 There are two types of rotors used in synchronous
machines:
i) cylindrical (or round) rotors
ii) salient pole rotors
 Machines with cylindrical rotors are typically found in
higher speed higher power applications such as
turbogenerators. Using 2 or 4 poles, these machines
rotate at 3600 or 1800 rpm (with 60hz systems).
 Salient pole machines are typically found in large (many
MW), low mechanical speed applications, including
hydrogenerators, or smaller higher speed machines (up to
1-2 MW).
 Salient pole rotors are less expensive than round rotors.
Synchronous Machine – Cylindrical rotor
D1m
Turbine
L  10 m
Steam

High speed

3600 r/min  2-pole

Stator
winding
N
Uniform air-gap
Stator
1800 r/min  4-pole
 Direct-conductor cooling (using
hydrogen or water as coolant)

d-axis
q-axis
Rotor
winding
Rotor
Rating up to 2000 MVA
S
Turbogenerator
Synchronous Machine – Cylindrical rotor
Stator
Cylindrical rotor
Synchronous Machine – Salient Pole
1. Most hydraulic turbines have to turn at low speeds
(between 50 and 300 r/min)
2. A large number of poles are required on the rotor
d-axis
Non-uniform
air-gap
N
D  10 m
q-axis
S
S
Turbine
Hydro (water)
Hydrogenerator
N
Synchronous Machine – Salient Pole
Stator
 Synchronous machine rotors are simply rotating electromagnets built
to have as many poles as are produced by the stator windings.
 Dc currents flowing in the field coils surrounding each pole magnetize
the rotor poles.
 The magnetic field produced by the rotor poles locks in with a rotating
stator field, so that the shaft and the stator field rotate in synchronism.
 Salient poles are too weak mechanically and develop too much wind
resistance and noise to be used in large, high-speed generators driven
by steam or gas turbines. For these big machines, the rotor must be a
solid, cylindrical steel forging to provide the necessary strength.
 Axial slots are cut in the surface of the cylinder to
accommodate the field windings.
 Since the rotor poles have constant polarity they must be
supplied with direct current.
 This current may be provided by an external dc generator or
by a rectifier.
 In this case the leads from the field winding are connected to insulated
rings mounted concentrically on the shaft.
 Stationary contacts called brushes ride on these slip rings to carry
current to the rotating field windings from the dc supply.
 The brushes are made of a carbon compound to provide a good
contact with low mechanical friction.
 An external dc generator used to provide current is called a “ brushless
exciter “.
Synchronous Machine
Stator
 The stator of a synchronous machine carries the armature or load
winding which is a three-phase winding.
 The armature winding is formed by interconnecting various
conductors in slots spread over the periphery of the machine’s stator.
Often, more than one independent three phase winding is on the
stator. An arrangement of a three-phase stator winding is shown in
Figure below. Notice that the windings of the three-phases are
displaced from each other in space.
Synchronous Machine
Construction
 Stator
Synchronous Machine
Magnetomotive Forces (MMF’s) and Fluxes Due to Armature
and Field Windings
Flux produced by a stator winding
Synchronous Machine
Magnetomotive Forces (MMF’s) and Fluxes Due to Armature
and Field Windings
Synchronous Machine
Magnetomotive Forces (MMF’s) and Fluxes Due to Armature
and Field Windings
Two Cycles of mmf around the Stator
Synchronous Generator
Equivalent circuit model – synchronous generator
 If the generator operates at a terminal voltage VT while supplying a
load corresponding to an armature current Ia, then;
 In an actual synchronous machine, the reactance is much greater
than the armature resistance, in which case;
 Among the steady-state characteristics of a
synchronous generator,
its voltage regulation and power-angle characteristics are the most
important ones. As for transformers, the voltage regulation of a
synchronous generator is defined at a given load as;
Synchronous Generator
Phasor diagram of a synchronous generator
The phasor diagram is to shows the relationship among the
voltages within a phase (Eφ,Vφ, jXSIA and RAIA) and the current IA in
the phase.
Unity P.F (1.0)
Synchronous Generator
Lagging P.F
Leading P.F.
Synchronous Generator
Power and Torque
In generators, not all the mechanical power going into a
synchronous generator becomes electric power out of the machine
The power losses in generator are represented by difference between
output power and input power shown in power flow diagram below
Synchronous Generator
Losses
Rotor
- resistance; iron parts moving in a magnetic field causing
currents to be generated in the rotor body
- resistance of connections to the rotor (slip rings)
Stator
- resistance; magnetic losses (e.g., hysteresis)
Mechanical
- friction at bearings, friction at slip rings
Stray load losses
- due to non-uniform current distribution
Synchronous Generator
The input mechanical power is the shaft power in the generator given by equation:
The power converted from mechanical to electrical form internally is given by
The real electric output power of the synchronous generator can be expressed in line
and phase quantities as
and reactive output power
Synchronous Generator
In real synchronous machines of any size, the armature resistance
RA is more than 10 times smaller than the synchronous reactance
XS (Xs >> RA). Therefore, RA can be ignored
Synchronous Motor
Synchronous Motor
Power Flow
Example : Synchronous Generator.

A three-phase, wye-connected 2500 kVA and 6.6 kV generator
operates at full-load. The per-phase armature resistance Ra and the
synchronous reactance, Xd, are (0.07+j10.4).
Calculate the percent voltage regulation at
(a) 0.8 power-factor lagging, and
(b) 0.8 power-factor leading.
Solution.
Induction Machine
 The machines are called induction machines because of the
rotor voltage which produces the rotor current and
the rotor magnetic field is induced in the rotor windings.
 Induction generator has many disadvantages and low
efficiency. Therefore induction machines are usually
referred to as induction motors.
Induction Machine
• Three-phase induction motors are the most common and
frequently encountered machines in industry
–
–
–
–
simple design, rugged, low-price, easy maintenance
wide range of power ratings: fractional horsepower to 10 MW
run essentially as constant speed from no-load to full load
Its speed depends on the frequency of the power source
• not easy to have variable speed control
• requires a variable-frequency power-electronic drive for optimal
speed control
Construction
Slip rings
Cutaway in a
typical woundrotor IM.
Notice the
brushes and
the slip rings
Brushes
Construction
• An induction motor has two main parts
i) a stationary stator
• consisting of a steel frame that supports a hollow,
cylindrical core
• core, constructed from stacked laminations (why?),
– having a number of evenly spaced slots, providing the space for
the stator winding
Stator of IM
Construction
ii) a revolving rotor
• composed of punched laminations, stacked to create a
series of rotor slots, providing space for the rotor winding
• conventional 3-phase windings made of insulated wire
(wound-rotor) » similar to the winding on the stator
• aluminum bus bars shorted together at the ends by two
aluminum rings, forming a squirrel-cage shaped circuit
(squirrel-cage)
Construction
• Two basic design types depending on the rotor design
– squirrel-cage: conducting bars laid into slots and
shorted at both ends by shorting rings.
– wound-rotor: complete set of three-phase windings
exactly as the stator. Usually Y-connected, the ends
of the three rotor wires are connected to 3 slip rings
on the rotor shaft. In this way, the rotor circuit is
accessible.
Construction
1. Squirrel cage – the conductors would look like one of
the exercise wheels that squirrel or hamsters run on.
Construction
2. Wound rotor – have brushes and slip rings at the end
of rotor
Notice the
slip rings
Rotating Magnetic Field
• Balanced three phase windings, i.e.
mechanically displaced 120 degrees form
each other, fed by balanced three phase
source
• A rotating magnetic field with constant
magnitude is produced, rotating with a speed
120 f e
nsync 
P
(rpm)
Where fe is the supply frequency and
P is the no. of poles and nsync is called
the synchronous speed in rpm
(revolutions per minute)
Synchronous speed
P
50 Hz
60 Hz
2
3000
3600
4
1500
1800
6
1000
1200
8
750
900
10
600
720
12
500
600
Principle of operation
• This rotating magnetic field cuts the rotor windings and produces an
induced voltage in the rotor windings
• Due to the fact that the rotor windings are short circuited, for both
squirrel cage and wound-rotor, and induced current flows in the rotor
windings
• The rotor current produces another magnetic field
• A torque is produced as a result of the interaction of those two magnetic
fields
 ind  kBR  Bs
Where ind is the induced torque and BR and BS are the
magnetic flux densities of the rotor and the stator respectively
Induction motor speed
• At what speed will the IM run?
– Can the IM run at the synchronous speed, why?
– If rotor runs at the synchronous speed, which is the same speed
of the rotating magnetic field, then the rotor will appear stationary
to the rotating magnetic field and the rotating magnetic field will
not cut the rotor. So, no induced current will flow in the rotor and
no rotor magnetic flux will be produced so no torque is generated
and the rotor speed will fall below the synchronous speed
– When the speed falls, the rotating magnetic field will cut the rotor
windings and a torque is produced
Induction motor speed
• So, the IM will always run at a speed lower than
the synchronous speed
• The difference between the motor speed and the
synchronous speed is called the Slip (s)
nslip  nsync  nm
Where nslip= slip speed
nsync= speed of the magnetic field
nm = mechanical shaft speed of the motor
The Slip
s
nsync  nm
nsync
Where s is the slip
Notice that : if the rotor runs at synchronous speed
s=0
if the rotor is stationary
s=1
Slip may be expressed as a percentage by multiplying the
above eq. by 100, notice that the slip is a ratio and doesn’t
have units
Induction Motors and Transformers
• Both IM and transformer works on the principle of
induced voltage
– Transformer: voltage applied to the primary windings produce an
induced voltage in the secondary windings
– Induction motor: voltage applied to the stator windings produce
an induced voltage in the rotor windings
– The difference is that, in the case of the induction motor, the
secondary windings can move
– Due to the rotation of the rotor (the secondary winding of the IM),
the induced voltage in it does not have the same frequency of
the stator (the primary) voltage
Frequency
• The frequency of the voltage induced in the rotor
is given by
Pn
fr 
120
Where fr = the rotor frequency (Hz)
P = number of stator poles
n = slip speed (rpm)
P  (ns  nm )
fr 
120
P  sns

 sf e
120
Frequency
• What would be the frequency of the rotor’s
induced voltage at any speed nm?
f r  sf e
• When the rotor is blocked (s=1) , the frequency
of the induced voltage is equal to the supply
frequency
• On the other hand, if the rotor runs at
synchronous speed (s = 0), the frequency will
be zero
Torque
• While the input to the induction motor is electrical power,
its output is mechanical power and for that we should
know some terms and quantities related to mechanical
power
• Any mechanical load applied to the motor shaft will
introduce a Torque on the motor shaft. This torque is
related to the motor output power and the rotor speed
 load 
Pout
m
( N .m) , where m 
2nm
60
(rad / s)
Horse power
• Another unit used to measure mechanical power is the
horse power
• It is used to refer to the mechanical output power of the
motor
• Since we, as an electrical engineers, deal with watts as a
unit to measure electrical power, there is a relation
between horse power and watts
1 hp  746 watts
Example
A 208-V, 10hp, four pole, 60 Hz, Y-connected
induction motor has a full-load slip of 5 percent
1. What is the synchronous speed of this motor?
2. What is the rotor speed of this motor at rated load?
3. What is the rotor frequency of this motor at rated
load?
4. What is the shaft torque of this motor at rated load?
Solution
120 f e 120(60)


 1800 rpm
P
4
1.
nsync
2.
nm  (1  s)ns
3.
f r  sfe  0.05  60  3Hz
4.
 (1  0.05) 1800  1710 rpm
Pout
m 2 nm
60
10 hp  746 watt / hp

 41.7 N .m
1710  2  (1/ 60)
 load 
Pout

Equivalent Circuit
• The induction motor is similar to the transformer with the
exception that its secondary windings are free to rotate
As we noticed in the transformer, it is easier if we can combine these
two circuits in one circuit but there are some difficulties
Equivalent Circuit
• When the rotor is locked (or blocked), i.e. s =1, the
largest voltage and rotor frequency are induced in the
rotor.
• On the other side, if the rotor rotates at synchronous
speed, i.e. s = 0, the induced voltage and frequency in
the rotor will be equal to zero.
ER  sE R 0
Where ER0 is the largest value of the rotor’s induced voltage
obtained at s = 1(locked rotor)
Equivalent Circuit
• The same is true for the frequency, i.e.
f r  sf e
• It is known that
X  L  2fL
• So, as the frequency of the induced voltage in
the rotor changes, the reactance of the rotor
circuit also changes
X r  r Lr  2f r Lr
Where Xr0 is the rotor reactance
at the supply frequency
(at blocked rotor)
 2sf e Lr
 sX r 0
Equivalent Circuit
• Then, we can draw the rotor equivalent circuit as
follows
Where ER is the induced voltage in the rotor and RR is
the rotor resistance
Equivalent Circuit
• Now we can calculate the rotor current as
ER
IR 
( RR  jX R )
sER 0

( RR  jsX R 0 )
• Dividing both the numerator and denominator by s so
nothing changes we get
ER 0
IR 
(
RR
 jX R 0 )
s
Where ER0 is the induced voltage and XR0 is the rotor reactance at
blocked rotor condition (s = 1)
Equivalent Circuit
• Now we can have the rotor equivalent
circuit
Equivalent Circuit
• Now as we managed to solve the induced voltage and
different frequency problems, we can combine the stator
and rotor circuits in one equivalent circuit
Where
2
X 2  aeff
X R0
2
R2  aeff
RR
I2 
IR
aeff
E1  aeff ER 0
aeff 
NS
NR
Power losses in Induction
machines
• Copper losses
– Copper loss in the stator (PSCL) = I12R1
– Copper loss in the rotor (PRCL) = I22R2
• Core loss (Pcore)
• Mechanical power loss due to friction and windage
• How this power flow in the motor?
Power flow in induction motor
Power relations
Pin  3 VL I L cos   3 Vph I ph cos 
PSCL  3 I12 R1
PAG  Pin  ( PSCL  Pcore )
PRCL  3I 22 R2
Pconv  PAG  PRCL
Pout  Pconv  ( Pf  w  Pstray )
 ind 
Pconv
m
Equivalent Circuit
• We can rearrange the equivalent circuit as
follows
Actual rotor
resistance
Resistance
equivalent to
mechanical load
Power relations
Pin  3 VL I L cos   3 Vph I ph cos 
PSCL  3 I12 R1
PAG  Pin  ( PSCL  Pcore )  Pconv  PRCL
R2
 3I
s
2
2
PRCL  3I 22 R2
Pconv  PAG  PRCL  3I 22
Pconv  (1  s) PAG
R2 (1  s )
s
Pout  Pconv  ( Pf  w  Pstray )
 ind 
PRCL

s
PRCL (1  s )

s
Pconv
m
(1  s ) PAG

(1  s )s
Power relations
PAG
Pconv
1
1-s
PRCL
s
PAG : PRCL : Pconv
1 : s : 1-s
Example
A 480-V, 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
friction and windage losses are 600 W, the core losses are
1800 W, and the stray losses are negligible. Find the
following quantities:
1.
2.
3.
4.
The air-gap power PAG.
The power converted Pconv.
The output power Pout.
The efficiency of the motor.
Solution
1.
Pin  3VL I L cos 
 3  480  60  0.85  42.4 kW
PAG  Pin  PSCL  Pcore
 42.4  2  1.8  38.6 kW
2.
Pconv  PAG  PRCL
700
 38.6 
 37.9 kW
1000
3.
Pout  Pconv  PF &W
600
 37.9 
 37.3 kW
1000
Solution
37.3
Pout 
 50 hp
0.746
4.

Pout
100%
Pin
37.3

100  88%
42.4
Example
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= 0.641 R2= 0.332
X1= 1.106  X2= 0.464  XM= 26.3 
The total rotational losses are 1100 W and are assumed to be constant.
The core loss is lumped in with the rotational losses. For a rotor slip of
2.2 percent at the rated voltage and rated frequency, find the motor’s
1.
2.
3.
Speed
Stator current
Power factor
4. Pconv and Pout
5. ind and load
6. Efficiency
Solution
1. n  120 f e  120  60  1800 rpm
sync
P
4
nm  (1  s)nsync  (1  0.022) 1800  1760 rpm
R2
0.332
 j 0.464
2. Z 2   jX 2 
s
0.022
 15.09  j 0.464  15.11.76 
1
1
Zf 

1/ jX M  1/ Z 2  j 0.038  0.0662  1.76
1

 12.9431.1 
0.0773  31.1
Solution
Ztot  Z stat  Z f
 0.641  j1.106  12.9431.1 
 11.72  j 7.79  14.0733.6 
4600
V
3
I1 

 18.88  33.6 A
Ztot 14.0733.6
3. PF  cos 33.6  0.833 lagging
4. Pin  3VL I L cos  3  460 18.88  0.833  12530 W
PSCL  3I12 R1  3(18.88)2  0.641  685 W
PAG  Pin  PSCL  12530  685  11845 W
Solution
Pconv  (1  s) PAG  (1  0.022)(11845)  11585 W
Pout  Pconv  PF &W  11585  1100  10485 W
5.  ind
 load
10485
=
 14.1 hp
746
PAG
11845


sync 2 1800
Pout
10485


m 2 1760
 62.8 N.m
60
 56.9 N.m
60
6.   Pout 100%  10485 100  83.7%
Pin
12530
Induction Motor – Power and Torque
The output power can be found as
Pout = Pconv – PF&W – Pmisc
The induced torque or developed torque:
Example
A two-pole, 50-Hz induction motor supplies 15kW to a
load at a speed of 2950 rpm.
1. What is the motor’s slip?
2. What is the induced torque in the motor in N.m under these
conditions?
3. What will be the operating speed of the motor if its torque is
doubled?
4. How much power will be supplied by the motor when the
torque is doubled?
Solution
1. n  120 f e  120  50  3000 rpm
sync
s
2.
P
nsync  nm
nsync
2
3000  2950

 0.0167 or 1.67%
3000
no Pf W given
 assume Pconv  Pload and  ind   load
 ind 
Pconv
m
15 103

 48.6 N.m
2
2950 
60
Solution
3. In the low-slip region, the torque-speed curve is
linear and the induced torque is direct proportional
to slip. So, if the torque is doubled the new slip will
be 3.33% and the motor speed will be
nm  (1  s )nsync  (1  0.0333)  3000  2900 rpm
4. Pconv   ind m
2
 (2  48.6)  (2900  )  29.5 kW
60
Torque-slip characteristics
• Slip,
s
nsync  nm
nsync
• Slip at maximum torque, SmaxT
• SmaxT =
𝑅2 ′
𝑋2 ′
=
𝑅2
𝑋2
=
𝑟𝑜𝑡𝑜𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑡𝑎𝑛𝑑𝑠𝑡𝑖𝑙𝑙 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒
Torque
• Maximum torque, 𝑇𝑚𝑎𝑥
• 𝑇𝑚𝑎𝑥 =
3 𝑉1 2
𝜔𝑠 2𝑋2
Torque
• Full load torque, Tfl
𝑉1
• Current, 𝐼2 =
• 𝑇𝑓𝑙 =
𝑃𝐺
𝜔𝑠
=
3
𝜔𝑠
𝑅2 /𝑠 2 +𝑋2 2
=
𝑉1 2 𝑅2 /𝑠
𝑅2 /𝑠 2 +𝑋2 2
Example
• A 5kV, 50Hz, 20-pole, 3-phase star-connected
induction motor has a rotor resistance of 0.15Ω
and standstill reactance of 1.15Ω. The motor has
a speed of 292.5rpm at full load. Calculate the
following:i) slip (slip at full load)
ii) slip at maximum torque
iii) the ratio of maximum torque to full-load
torque
Solution
• Slip, s 
nsync  nm
nsync
292.5
 1
 0.025
300
• Slip at maximum torque, SmaxT=
𝑅2
𝑋2
=
0.15
1.15
= 0.13
• Ratio of Tmax to Tfl
• 𝑇𝑚𝑎𝑥 =
• 𝑇𝑓𝑙 =
3 𝑉1 2
𝜔𝑠 2𝑋2
3
𝜔𝑠
• Ratio of
=
=
3
𝑉1 2
𝜔𝑠 2 1.15
𝑉1 2 𝑅2 /𝑠
𝑅2 /𝑠 2 +𝑋2 2
𝑇𝑚𝑎𝑥
𝑇𝑓𝑙
=
= 2.7045
3
𝑉1 2 0.15/0.025
𝜔𝑠 0.15/0.025 2 +1.152