Transcript 25471_energy_conversion_12..

ENERGY CONVERSION ONE
(Course 25741)
CHAPTER SIX (& S.G. parallel op. with Pow. Sys. of chapter 5)
SYNCHRONOUS MOTORS
Operation of AC Generators in Parallel
with Large Power Systems
• Isolated synchronous generator supplying its own load is very
rare (emergency generators)
• In general applications more than one generator operating in
parallel to supply loads
• In Iran national grid hundreds of generators share the load on
the system
• Advantages of generators operating in parallel:
1- several generators can supply a larger load
2- having many generators in parallel increase the
reliability of power system
3- having many generators operating in parallel allows
one or more of them to be removed for shutdown &
preventive maintenance
4- if only one generator employed & not operating near full load, it
will be relatively inefficient
Operation of AC Generators in Parallel
with Large Power Systems
INFINITE BUS
• When a Syn. Gen. connected to power system,
power sys. is so large that nothing operator of
generator does, have much effect on pwr. sys.
• Example: connection of a single generator to a
large power grid (i.e. Iran grid), no reasonable
action on part of one generator can cause an
observable change in overall grid frequency
• This idea belong to definition of “Infinite Bus”
which is: a so large power system, that its
voltage & frequency do not vary, (regardless of
amount of real and reactive power load)
Operation of AC Generators in Parallel
with Large Power Systems
• When a syn. Gen.
connected to a
power system:
1-The real power
versus frequency
characteristic of
such a system
2-And the reactive
power-voltage
characteristic
Operation of AC Generators in Parallel
with Large Power Systems
• Behavior of a generator
connected to a large
system
A generator connected in
parallel with a large
system as shown
• Frequency & voltage of
all machines must be the
same, their real powerfrequency (& reactive
power-voltage)
characteristics plotted
back to back

Operation of AC Generators in Parallel
with Large Power Systems
• Assume generator just been paralleled with
infinite bus, generator will be “floating” on the
line, supplying a small amount of real power
and little or no reactive power
• Suppose generator paralleled, however its
frequency being slightly lower than system’s
operating frequency
 At this frequency power supplied by
generator is less than system’s operating
frequency, generator will consume energy and
runs as motor
Operation of AC Generators in Parallel
with Large Power Systems
• In order that a generator comes on line and
supply power instead of consuming it, we
should ensure that oncoming machine’s
frequency is adjusted higher than running
system’s frequency
• Many generators have “reverse-power trip”
system
• And if such a generator ever starts to consume
power it will be automatically disconnected from
line
Synchronous Motors
• Synchronous machines
employed to convert
electric energy to
mechanical energy
• To present the principles
of Synchronous motor, a
2-pole synchronous
motor considered
 ind  kBR  BS  counterclockwise
• It has the same basic
speed, power, & torque
equations as Syn. Gen.
Synchronous Motors
Equivalent Circuit
• Syn. Motor is the same in all respects as Syn.
Gen., except than direction of power flow
• Since the direction of power flow reversed,
direction of current flow in stator of motor may
also be reversed
• Therefore its equivalent circuit is exactly as
Syn. Gen. equivalent circuit, except that the
reference direction of IA is reversed
Synchronous Motors
Equivalent Circuit
• 3 phase Eq. cct. 
• Per phase Eq. cct.
Synchronous Motors
Equivalent Circuit
• The related KVl equations:
Vφ=EA+jXS IA + RAIA
EA =Vφ-jXSIA –RAIA
• Operation From Magnetic
Field Perspective
• Considering a synchronous
generator connected to an
infinite bus
• Generator has a prime
mover turning its shaft,
causing it to rotate (rotation
in direction of Tapp)
FIGURE (1) 
for Generator
Synchronous Motors
Equivalent Circuit
• The phasor diagram of syn. Motor, & magnetic field diagram
FIGURE (2)
Synchronous Motors
Operation
• Induced torue is given by:
Tind=kBR x Bnet
(1)
Tind=kBR Bnet sinδ
(2)
• Note: from magnetic field diagram, induced
torque is clock wise, opposing direction of
rotation in Generator related diagram
• In other words; induced torque in generator is a
countertorque, opposing rotation caused by
external applied torque Tapp
Synchronous Motors
Operation
• Suppose, instead of turning shaft in direction of
motion, prime mover lose power & starts to
drag on machine’s shaft
• What happens to machine? Rotor slows down
because of drag on its shaft and falls behind net
magnetic field in machine  BR slows down &
falls behind Bnet , operation of machine
suddenly changes
• Using Equation (1), when BR behind Bnet ,
torque’s direction reverses & become
counterclockwise
Synchronous Motors
Operation
• Now, machine’s torque is in direction of motion
• Machine is acting as a motor
• With gradual increase of torque angle δ, larger
& larger torque develop in direction of rotation
until finally motor’s induced torque equals load
torque on its shaft
• Then machine will operate at steady state &
synchronous speed again, however as a motor
Synchronous Motors
Steady-state Operation
• Will study behavior of synchronous motors under
varying conditions of load & field current , also its
application to power-factor correction
• In discussions, armature resistance ignored for
simplicity
- Torque-Speed Characteristic
• Syn. Motors supply power to loads that are constant
speed devices
• Usually connected to power system, and power
systems appear as infinite buses to motors
• Means that terminal voltage & system frequency will
be constant regardless of amount of power drawn by
motor
Synchronous Motors
Steady-state Operation
• Speed of rotation is locked to
applied electrical frequency
• so speed of motor will be
constant regardless of the
load
• Resulting torque-speed
characteristic curve is shown
here 
• S.S. speed of motor is
constant from no-load up to
max. torque that motor can
supply (named : pullout
torque) so speed regulation
of motor is 0%
Synchronous Motors
Steady-state Operation
•
•
•
•
•
Torque equation:
Tind=kBRBnet sinδ
Tind = 3 Vφ EA sinδ /(ωm XS) (chapter 5, T=P/ωm)
Pullout torque occurs when δ=90◦
Full load torque is much less than that, may
typically be 1/3 of pullout torque
• When torque on shaft of syn. Motor exceeds
pullout torque, rotor can not remain locked to
stator & net magnetic fields. Instead rotor starts
to slip behind them.
Synchronous Motors
Steady-state Operation
• As rotor slows down, stator magnetic field “laps” it
repeatedly, and direction of induced torque in rotor
reverses with each pass
• Resulting huge torque surges, (which change direction
sequentially) cause whole motor to vibrate severely
• Loss of synchronization after pullout torque is
exceeded known as “slipping poles”
• Maximum or pullout torque of motor is:
Tmax=kBRBnet
Tmax=3VφEA/(ωm XS)
• From last equation, the larger the field current, larger
EA , the greater the torque of motor
• Therefore there is a stability advantage in operating
motor with large field current or EA
Synchronous Motors
Steady-state Operation
• Effect of load changes on motor operation
• when load attached to shaft, syn. Motor develop
enough torque to keep motor & its load turning
at syn. Speed
• Now if load changed on syn. motor, let examine
a syn. motor operating initially with a leading
power factor
Synchronous Motors
Steady-state Operation
• If load on shaft increased, rotor will initially slow down
• As it does, torque angle δ becomes larger & induced
torque increases
• Increase in induced torque speeds the rotor back up,
& rotor again turns at syn. Speed but with a larger
torque angle δ
• last figure show the phasor diagram before load
increased
• Internal induced voltage EA=Kφω depends on field
current & speed of machine
• Speed constrained to be constant by input power
supply, and since no one changed field current it is
also constant
Synchronous Motors
Steady-state Operation
• |EA| remain constant as load changes
• Distances proportional to power increase (EA sinδ or IA
cosθ) while EA must remain constant
• As load increases EA swings down as shown &
jXSIA has to
increase &
Consequently IA
Increase, Note: p.f.
angle θ change too,
causing less leading
& gradually lagging
Synchronous Motors
Steady-state Operation
• Example: A 208 V, 45 kVA, 0.8 PF leading, Δ
connected, 60 Hz synchronous machine has XS=2.5 Ω
and a negligible RA. Its friction and windage losses are
1.5 kW, and its core losses are 1.0 kW. Initially the
shaft is supplying a 15 hp load, and motor’s power
factor is 0.8 leading
(a) sketch phasor diagram of this motor, and find
values of IA, IL, and EA
(b) assume that shaft load is now increased to 30 hp
Sketch the behavior of the phasor diagram in
response to this change
(c ) Find IA, IL, and EA after load change. What is the
new motor power factor
Synchronous Motors
Steady-state Operation
• Solution:
(a) Pout=(15 hp)(0.746 kW/hp)=11.19 kW
electric power supplied to machine:
Pin=Pout+Pmech loss+Pcore loss+Pcopper loss=
= 11.19+1.5+1.0+0=13.69 kW
Since motor’s PF is 0.8 leading, the resulting
line current flow is:
IL=Pin/[√3 VT cos θ]=13.69/[√3 x 208 x 0.8]=47.5 A
IA=IL/√3  IA=27.4 /_36.78◦ A
Synchronous Motors
Steady-state Operation
• To find EA :
EA=Vφ-jXS IA =208 /_0◦ - (j2.5 Ω)(27.4/_36.78◦)=
208- 68/_126.87◦ =249.1 –j 54.8 V=255/_-12.4◦V
Synchronous Motors
Steady-state Operation
• (b) As load power on shaft increased to 30 hp,
shaft slows momentarily, & internal voltage EA
swings to a larger δ, maintaining its magnitude
Synchronous Motors
Steady-state Operation
• (c) After load changes, electric input power of
machine is:
• Pin=Pout + Pmech loss+Pcore loss+Pcopper =
(30)(0.746)+1.5+1.0+0=24.88 kW
since P= 3 Vφ EA sinδ /XS
torque angle determined:
δ=arcsin [XSP/(3VφEA)]
=arcsin[(2.5x24.88)/(3x208x255)]=
arcsin(0.391)=23◦
Synchronous Motors
Steady-state Operation
• EA=355/_-23◦ V
• IA= [Vφ-EA] / (jXS) = [208-255/_-23◦]/(j2.5)=
= [103/_105◦]/(j2.5) =41.2/_15◦ A
and IL :
IL= √3 IA =71.4 A
the final power factor is cos(-15)=0.966 leading
Synchronous Motors
Steady-state Operation
• Effect of Field current changes on a synchronous motor
• It was shown how change in shaft load affects motor torque
angle and the supply current
• Effect of field current change:
• Above phasor diagram shows a motor operating at a lagging
p.f.
• Now increase its IF & see what happens to motor
• This will increase EA, however don’t affect real mechanical
power supplied by motor. Since this power only changes when
shaft load torque change
Synchronous Motors
Steady-state Operation
• Since change in IF does not affect shaft speed nm
and, since load attached to shaft is unchanged, real
mechanical power supplied is unchanged
• VT is constant (by power source supply)
• power is proportional to following parameters in
phasor diagram : EAsinδ & IAcosθ and must be
constant
• When IF increased, EA must increase, however it can
be done along line of constant power as shown in next
slide
Synchronous Motors
Effect of an increase in field current
• Note: as EA increases first IA decreases and then increases
again
• At low EA armature current is lagging , and motor is an inductive
load, consuming reactive power Q
• As field current increased IA lines up with Vφ motor like a
resistor, & as IF increased further IA become leading and motor
become a capacitive load (capacitor-resistor) supplying reactive
power
Synchronous Motors
Steady-state Operation
• A plot of IA versus IF for a synchronous motor
shown below known as V curves in each curve
of constant P, minimum IA occur at unity P.F.