Power Systems and Machines
Will McLennan
Based on lectures by J. Bialek
Short Description
This course aims to provide students with advanced knowledge and a thorough understanding of: the steady state performance of induction machines; the transient behaviour
of synchronous machines; power system protection equipment; power system and machine protection; operation and protection of embedded generators.
Summary of Intended Learning Outcomes
By the end of the course a student should be able to:
• Calculate and model the steady state operation and performance of induction
motors and generators;
• Represent the closed loop control models of synchronous generators using Laplace
techniques and determine their time performance;
• Estimate the steady state and transient fault levels in simple power systems containing embedded generation;
• Determine relay operating currents in overcurrent protection schemes and calculate
clearance times, incorporating amplitude and time discrimination;
Contents
1 Three-phase induction machines
1.1 Revision and fundamentals . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Induction motors . . . . . . . . . . . . . . . . . . . . . . . .
1.1.3 Production of a revolving field in the stator . . . . . . . . .
1.1.4 Induced EMFs and production of rotor current and torque
1.1.5 Synchronous speed, shaft speed and slip . . . . . . . . . . .
1.1.6 Torque/speed variation . . . . . . . . . . . . . . . . . . . .
1.2 Power flow and characteristics . . . . . . . . . . . . . . . . . . . . .
1.2.1 Apparent, active and reactive power. Power factor . . . . .
1.2.2 Power flow, losses and efficiency of induction motors . . . .
1.2.3 Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.4 Rotor power balance as a function of slip . . . . . . . . . .
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2 The equivalent circuit model of the induction machine
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Approximate model of induction machine . . . . . . . . .
2.2.1 Rotor . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Rotor and referred values . . . . . . . . . . . . . . . . . .
2.4 T-equivalent circuit . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Stator no-load circuit . . . . . . . . . . . . . . . .
2.4.2 Stator series impedance . . . . . . . . . . . . . . .
2.4.3 Circuit solution . . . . . . . . . . . . . . . . . . . .
2.4.4 Current and voltage phasor diagram . . . . . . . .
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3 Steady state behaviour of induction machines
3.1 Approximate single-phase equivalent circuit . .
3.1.1 Stator no-load circuit . . . . . . . . . .
3.1.2 Stator series impedance . . . . . . . . .
3.2 Calculation of machine performance . . . . . .
3.3 Performance operating as an induction motor .
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4 Operation of induction machines as motors or generators
4.1 Power flow reversibility . . . . . . . . . . . . . . . . . . . . .
4.2 Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 No load . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Phasor diagram operating as a motor and a generator . . .
4.6 Performance operating as an induction generator . . . . . .
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5 Determining equivalent circuit model impedance from machine test
results
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i
5.1
5.2
5.3
5.4
Introduction . . . . . . . . . . . . . . .
Electrical test method . . . . . . . . .
5.2.1 The no-load test - Rc and Xm
The DC test for stator resistance - R1
The locked-rotor test . . . . . . . . . .
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6 Operation of induction machines on the electricity
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . .
6.2 Connection and starting of induction motors . . . .
6.3 Connection and starting of induction motors . . . .
6.4 Power factor correction of induction motors . . . . .
supply network
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7 Steady state behaviour of synchronous machines
7.1 Revision of synchronous machine operation . . . .
7.2 DC injection . . . . . . . . . . . . . . . . . . . . .
7.3 Alignment of fields . . . . . . . . . . . . . . . . . .
7.4 Static excitation . . . . . . . . . . . . . . . . . . .
7.5 Brushless excitation . . . . . . . . . . . . . . . . .
7.6 Connection to infinite busbar . . . . . . . . . . . .
7.7 Airgap flux . . . . . . . . . . . . . . . . . . . . . .
7.8 Double excitation . . . . . . . . . . . . . . . . . . .
7.9 MMF phasor diagrams . . . . . . . . . . . . . . . .
7.9.1 Compensator mode - Figure 37i . . . . . . .
7.9.2 Motor mode - Figure 37ii . . . . . . . . . .
7.9.3 Generator mode - Figure 37iii . . . . . . . .
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8 Synchronous generators - voltage control
8.1 Introduction . . . . . . . . . . . . . . . . . .
8.2 Synchronous generators . . . . . . . . . . .
8.2.1 Isolated operation . . . . . . . . . .
8.3 Generator and AVR performance . . . . . .
8.3.1 Static excitation . . . . . . . . . . .
8.3.2 AVR . . . . . . . . . . . . . . . . . .
8.3.3 Main field . . . . . . . . . . . . . . .
8.4 Brushless excitation systems . . . . . . . . .
8.5 Generator modeling . . . . . . . . . . . . .
8.6 Steady state accuracy of excitation systems
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9 Synchronous generator modeling: control systems and stability
9.1 Time domain response - Laplace analysis . . . . . . . . . . . . . .
9.2 Generator/AVR root locus diagrams . . . . . . . . . . . . . . . . .
9.3 Improvement of AVR stability . . . . . . . . . . . . . . . . . . . . .
9.4 Time domain response of generator/AVR system . . . . . . . . . .
9.4.1 Impulse response . . . . . . . . . . . . . . . . . . . . . . . .
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9.4.2
Step changes in voltage . . . . . . . . . . . . . . . . . . . . . . . . 68
10 Operation of embedded generators
10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 Transmission and distribution networks . . . . . . . . . . .
10.3 Statutory requirements . . . . . . . . . . . . . . . . . . . . .
10.3.1 Applicable standards and codes of practice . . . . .
10.4 Integrating embedded generation into distribution networks
10.4.1 Power flow . . . . . . . . . . . . . . . . . . . . . . .
10.4.2 Thermal effects . . . . . . . . . . . . . . . . . . . . .
10.4.3 Voltage variation . . . . . . . . . . . . . . . . . . . .
10.4.4 Increased fault levels . . . . . . . . . . . . . . . . . .
10.4.5 Reduced stability . . . . . . . . . . . . . . . . . . . .
10.4.6 Compromised operation of protection equipment . .
10.4.7 Increased harmonic content . . . . . . . . . . . . . .
10.5 Synchronous or induction generators . . . . . . . . . . . . .
10.6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 Balanced per-unit fault analysis
11.1 The per-unit system . . . . . . . . . . . .
11.2 Choice of base . . . . . . . . . . . . . . .
11.3 Base MVA, voltage and impedance . . . .
11.4 Per unit reactances of transformers . . . .
11.5 Per-unit reactances of generators . . . . .
11.6 Per-unit reactances of synchronous motors
11.7 Conversions between bases . . . . . . . . .
11.8 Conversions between off-nominal voltages
11.9 Use in fault level calculations . . . . . . .
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12 Power system protection equipment
12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2 Switchgear and circuit breakers . . . . . . . . . . . . . . . . . . .
12.3 Protection components . . . . . . . . . . . . . . . . . . . . . . . .
12.3.1 Current transformer (CT) . . . . . . . . . . . . . . . . . .
12.3.2 Voltage transformer (VT) . . . . . . . . . . . . . . . . . .
12.4 Protection relays . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.5 Protection of embedded generators and the distribution network
12.6 General protection considerations . . . . . . . . . . . . . . . . . .
12.7 Neutral earthing . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.8 Distribution network protection . . . . . . . . . . . . . . . . . . .
12.9 DG protection schemes . . . . . . . . . . . . . . . . . . . . . . . .
12.9.1 Overcurrent protection - Relay designation 51 . . . . . . .
12.9.2 Earth fault protection - Relay designation 64 . . . . . . .
12.10Frequency operated protection - Relay designation 81 . . . . . .
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12.11Voltage operated protection - Relay designations 27 (under)/59 (over)
12.12Loss of Mains Protection - Relay designation 83 . . . . . . . . . . . . .
12.13Differential protection - Relay designation 87 . . . . . . . . . . . . . .
12.14Electro-mechanical protection . . . . . . . . . . . . . . . . . . . . . . .
12.14.1 Reverse power - Relay designation 32 . . . . . . . . . . . . . .
12.14.2 Overspeed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.14.3 Overtemperature . . . . . . . . . . . . . . . . . . . . . . . . . .
13 Overcurrent protection
13.1 Introduction to faults and overcurrent protection . . . . . . . . .
13.1.1 Fault levels in a typical system . . . . . . . . . . . . . . .
13.2 Overcurrent protection . . . . . . . . . . . . . . . . . . . . . . . .
13.2.1 Protection co-ordination . . . . . . . . . . . . . . . . . . .
13.3 Fuses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.3.1 An introduction to discrimination . . . . . . . . . . . . .
13.3.2 Disadvantages of fuses . . . . . . . . . . . . . . . . . . . .
13.4 MCBs, MCCBs . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.5 Overcurrent protection discrimination . . . . . . . . . . . . . . .
13.5.1 Discrimination by amplitude . . . . . . . . . . . . . . . .
13.5.2 Discrimination by time . . . . . . . . . . . . . . . . . . . .
13.5.3 Discrimination by time and current . . . . . . . . . . . . .
13.6 IDMT over-current relays . . . . . . . . . . . . . . . . . . . . . .
13.6.1 Current plug setting multiplier (PSM) . . . . . . . . . . .
13.6.2 Time multiplier setting (TMS) . . . . . . . . . . . . . . .
13.6.3 IDMT characteristics . . . . . . . . . . . . . . . . . . . . .
13.6.4 Factors influencing choice of PSM . . . . . . . . . . . . .
13.7 Overcurrent protection design . . . . . . . . . . . . . . . . . . . .
13.7.1 Discrimination time margin (grading margin) . . . . . . .
13.7.2 IDMT relay grading . . . . . . . . . . . . . . . . . . . . .
13.7.3 The graphical representation of time/current relationships
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14 Overcurrent protection - Case Study
14.1 Revision of Methodology for the Calculation of Balanced 3 phase Faults
14.2 Effects of location and impedance . . . . . . . . . . . . . . . . . . . . . .
14.3 Effects of type of EG and excitation . . . . . . . . . . . . . . . . . . . .
14.3.1 Fault contribution from generators . . . . . . . . . . . . . . . . .
14.4 Case study of a loaded radial feeder with new generation . . . . . . . . .
A Laplace transforms
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133
iv
Power Systems and Machines
1
1
Three-phase induction machines
1.1
1.1.1
Revision and fundamentals
Introduction
The basic concepts and analytical tools from the EPE2 (Power and Machines parts) and
EPE3 (Power Systems part) that are required for this course are listed below:
• Single-phase and three-phase supply voltages, currents and phasor diagrams.
• Resistance, reactance, impedance and phasor diagrams.
• Active (real), reactive (imaginary) and apparent/total (complex) power flow.
• Complex number arithmetic; rectangular to polar conversion.
• Operating principles of induction (or asynchronous) machines and synchronous
machines.
• Per-unit calculation of three-phase symmetrical short circuit fault levels.
It is assumed that you have the appropriate notes available, and that you will revise
the material from them. Tutorial 1 provides examples that will test how well you have
understood this material.
1.1.2
Induction motors
The electromechanical principles by which an induction motor is able to develop torque
and power are:
1. The creation of a revolving/rotating magnetic flux pattern in the bore of the stator;
2. The induction of electromotive forces in rotor conductors, linking with that rotating
magnetic field, and subsequent current flow in the rotor conductors;
3. The reaction between the rotating magnetic field from the stator and the current
in the rotor conductors, producing a force and torque on the rotor.
Some familiarity with these basic effects is assumed.
1.1.3
Production of a revolving field in the stator
A simplified three-phase stator is shown in figure 1. The phases are identified RED, YELLOW, and BLUE. For simplicity, the stator has been shown to have two distributed poles
in each phase. We have already seen in the EPE3 Power Systems that stators of induction machines are not wound with salient or concentrated poles, but have distributed
Power Systems and Machines
2
windings comprised of groups of diamond coils connected together to produce a revolving flux pattern which is sinusoidal in shape. Although there are six poles in absolute
terms, the stator is referred to as a two-pole stator (i.e. two per phase). The line to
neutral or phase voltages Vrn , Vyn , and Vbn are applied across each phase of the stator
winding, and cause nearly identical alternating phase currents to flow, since each phase
impedance is approximately equal. The phase currents are shown in figure 2.
!
Ir
R1
^FGG+06!/065!C:!
Y2
Φ
B1
Vyn
Current
12
b
Φy
Φ
r
2:!
":!
Y1
Iy
R2
Ib
!
B2
Φtotal
Vrn
Vbn
2<!
"<!
^FGG+06!5F6!5H!C<!
!
Figure 1.3.1a Three-phase two-pole stator
! stator
Figure 1: Three-phase two-pole
!
10
Φ
If we regard
the RED phase on its own initially and consider the phase winding
as a
8
r
6
set of coils
with “go” sides at the top and “return” sides at the bottom, Φwe can study
b
4
the variation
of the magnetic field between the RED phase poles R1 and R2.
When the
Φ
y
2
Φtotal
$!
! positive
!#$ is on a
%!#$half-cycle,
& t "!#$
RED phase
voltage
terminal
R1
is
positive
with
respect
to R2
0
Φatotal
0
90“into the
180 page” 270
360pole. This
450
and current
flows
at
the
R1
produces
clockwise
flux
pattern
-2
Φ
-4
that points
from right to left in the middle of the airgap. Φ
Current coming
out of the page b
Φ
Φ
Φy
b
r
at the -6R2 pole produces an anti-clockwise flux that also points
fromr right to left Φintotal
the
-8
middle of the airgap. The two fluxes add up to be a single magnetic field directed right
Φy
-10
to left-12between R1 and R2. On negative half-cycle of phase
voltage,
R2
is
positive
with
!
respect to R1, and current flows into the R2 end. This !produces a clockwise flux pattern
!that points from left to right in the middle of the airgap. Current coming out of the page
Figure 1.3.1b Stator phase currents (RED phase – solid line
Figure 1.3.1c Airgap flux vectors
at the R1 pole produces an anti-clockwise flux that also points from left to right in the
"#$$%&!'()*+!,!-)*(+-!./0+1!2$3#!'()*+!,!-)*(4-56!./0+7!
middle of the airgap. The two fluxes add up to be a single magnetic field directed left to
86! 69:;<==*9>?*! @!69 ! / 2 9A=B71! 6(+! C#D! EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?! K).F+!
right between R1 and R2. As time progresses, the magnetic field strength between R1
*/0@A=B79L:! )./M0+-! J+6N++0! C:! )0-! C<1! 6()6! HG5?! "#$$%&! /*! 0+M)6/K+! )0-! ()*! K).F+!
and R2 due to phase current IR changes magnitude sinusoidally, always in the horizontal
*/0@A=B!4!:<=B7!9!4=O>!)./M0+-!J+6N++0!"<!)0-!":1!)0-!6()6!HG5?!2$3#!/*!).*5!0+M)6/K+!)0-!()*!K).F+!
*/0@A=B!
4!<P=B7!
9! 4=O>!
)./M0+-!
J+6N++0! 2<!
)0-!
2:O! ! phases
Q(+! K+E65G!
*F?1!
5G! 656).! H/+.-1!
/*! -/G+E6+-!
direction.
On their
own,
the YELLOW,
and
BLUE
behave
identically,
but the
(5G/R506)..S!CTUVQ!65!$#WQ1!)./M0+-!J+6N++0!C:!65!C<1!)0-!()*!G+.)6/K+!K).F+!:O>!6/?+*!6(+!?)I/?F?!
◦
corresponding resulting fields are displaced mechanically (i.e., spatially) by -120 and
/0-/K/-F).!H.FI!HG5?!50+!'()*+!@:!L!=O>*/0X=B!L!=O>*/0X=B7O!
◦
!-240 from the RED phase.
!
86! 69Y;Z==*! @!69 7! / 6 9<:=B71! 6(+! "#$$%&! EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?! K).F+!
In addition, it is clear from the current waveforms that the maximum current in the
*/0@<:=B!4!:<=B7!9!L:! )./M0+-! J+6N++0! ":! )0-! "<1! 6()6! HG5?! 2$3#! /*! 0+M)6/K+! )0-! ()*! K).F+!
RED phase occurs earlier
in time
than the
current
the/*!YELLOW,
which in)0-! ()*!
*/0@<:=B!4!<P=B7!9!4=O>!
)./M0+-!
J+6N++0!
2<! maximum
)0-! 2:1! 6()6!
HG5?! in
C#D!
05N! ).*5! 0+M)6/K+!
turn occurs
earlier in time )./M0+-!
than that
in BLUE.
Electrically,
the yellow
and
blue
phase
K).F+!
*/0! @<:=B!4!=B7!9!4=O>!
J+6N++0!
C<! )0-!
C:O! Q(+! K+E65G!
*F?1! 5G!
656).!
H/+.-1!
/*! -/G+E6+-!
◦
◦
J+6N++0!":!65!"<1!/O+O!)6!L:<=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>!6/?+*!
currents are displaced in time by -120 , and -240 behind red current.
6(+!?)I/?F?!/0-/K/-F).!H.FI!HG5?!50+!'()*+O!
!
86! 6! 9! ::;Z==*! @!69 11! / 6 9XX=B71! 6(+! 2$3#! EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?! K).F+!
*/0@XX=B!4!<P=B7!9!L:! )./M0+-! J+6N++0! 2:! )0-! 2<1! 6()6! HG5?! C#D! /*! 0+M)6/K+! )0-! ()*! K).F+!
*/0@XX=B!4!=B7!9!4=O>!)./M0+-!J+6N++0!C<!)0-!C:1!)0-!6()6!HG5?!"#$$%&!/*!05N!).*5!0+M)6/K+!)0-!()*!
K).F+! */0@XX=B!4!:<=B7!9!4=O>! )./M0+-! J+6N++0! "<! )0-! ":O! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/G+E6+-!
J+6N++0!2:!)0-!2<1!/O+O!)6!L<P=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>!6/?+*!
6(+!?)I/?F?!/0-/K/-F).!H.FI!HG5?!50+!'()*+O!
!
86!6!9!:>;Z==*9:;P=*9<>?*1!5G!)H6+G!50+!HF..!ESE.+;'+G/5-!5H!<=?*1!6(+!?)M0+6/E!H/+.-!()*!G+6FG0+-!
65! C:! )0-! C<! 5G/+06)6/50O! T6! E)0! J+! E50E.F-+-! 6()6! 6(+! ?)M0+6/E! H/+.-! G56)6+*! N/6(! 6(+! *)?+! )0MF.)G!
K+.5E/6S;HG+[F+0ES!)*!6(+!+.+E6G/E).!*F''.S1!H5G!)!6N54'5.+!*6)65GO!!
!
1.4
INDUCED EMFS AND PRODUCTION OF ROTOR CURRENT AND TORQUE
!
Ir
R1
Power Systems and Machines
^FGG+06!/065!C:!
3
Y2
Φtotal
2<!
"<!
B2
Vrn
It is a propertyVbn
of magnetic
fields that theyΦmay be represented
and added vectorially. To
Φy Φr
b
assimilate the rotation of the magneticB1field in the stator bore, consider the time varying
2:!
":!
vector sum of the individual RED, YELLOW, and BLUEY1magnetic fields. At the times
Iy
when the magnetic flux
produced by the peak current in a phase is at the maximum
Vyn
R2
^FGG+06!5F6!5H!C<
!
between the respective RED, YELLOW and BLUE
poles, there are contributions to
the
total flux from the other two phases which resolve to strengthen/increase the total flux,
in the direction of the phase Ibwhich produces the maximum current.
The resolution of
!
Figure
1.3.1a
Three-phase
two-pole
stator
airgap flux vectors is shown in figure 3.
r
Current
!
R1
Y2
10
^FGG+06!/065!C:!
8
Φ
b
Φy
2
b
!#$
!
%!#$
90
180
270
$!
0
Φ
r
-2
-4
Y1
y
r
Φ
"<!
4
B2
Φ
2<!
6
Φtotal
B1
!
!
12
0
2:!
360
Φy
& t "!#$
Φtotal
450
Φ
":!
b
-6
-8
R2
-10
-12
Φtotal
Φy
Φ
r
Φtotal
^FGG+06!5F6!5H!C<!
!
!
!
! phase
FigureFigure
1.3.1a 2:
Three-phase
two-pole
stator
Stator
currents
(RED phase solid line, YELLOW phase dashed line,
Figure 1.3.1b Stator phase currents (RED phase – solid line
Figure 1.3.1c Airgap flux vector
BLUE phase dash-dot! line)
"#$$%&!'()*+!,!-)*(+-!./0+1!2$3#!'()*+!,!-)*(4-56!./0+7!
!
86! 69:;<==*9>?*! @!69 ! / 2 9A=B71!
6(+! C#D! EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?!
Φ
*/0@A=B79L:! )./M0+-! J+6N++0! rC:! )0-! C<1! 6()6! HG5?! "#$$%&! /*! 0+M)6/K+! )0-! ()*!
Φ
*/0@A=B!4!:<=B7!9!4=O>!)./M0+-!J+6N++0!"<!)0-!":1!)0-!6()6!HG5?!2$3#!/*!).*5!0+M)6/K+!)0-!()*
b
*/0@A=B! 4!<P=B7! 9! 4=O>! )./M0+-!
Φy J+6N++0! 2<! )0-! 2:O! ! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/
Φtotal
(5G/R506)..S!CTUVQ!65!$#WQ1!)./M0+-!J+6N++0!C:!65!C<1!)0-!()*!G+.)6/K+!K).F+!:O>!6/?+*!6(+!?)I
$!
!
!#$
%!#$
& t "!#$
/0-/K/-F).!H.FI!HG5?!50+!'()*+!@:!L!=O>*/0X=B!L!=O>*/0X=B7O!
Φtotal
90
180
270
360
450
!
Φ
b
86! 69Y;Z==*!
@!69 7!
/
6
9<:=B71! 6(+! "#$$%&!
Φ
Φ
Φ EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?!
Φy
b
r
Φtotal ":! )0-!
r "<1! 6()6! HG5?! 2$3#! /*! 0+M)6/K+! )0-! ()*!
*/0@<:=B!4!:<=B7!9!L:!
)./M0+-! J+6N++0!
*/0@<:=B!4!<P=B7!9!4=O>! )./M0+-! J+6N++0! 2<! )0-! 2:1! 6()6! HG5?! C#D! /*! 05N! ).*5! 0+M)6/K+! )0
Φ
K).F+! */0! @<:=B!4!=B7!9!4=O>! )./M0+-! J+6N++0! C<! y)0-! C:O! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/
!
J+6N++0!":!65!"<1!/O+O!)6!L:<=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>
!
6(+!?)I/?F?!/0-/K/-F).!H.FI!HG5?!50+!'()*+O!
Figure 3: Airgap flux vectors
!
ator phase currents (RED phase – solid
Figure@!69
1.3.1c
flux vectors
◦
86! 6!line
9! =
::;Z==*!
11!Airgap
/ 6 9XX=B71!
6(+! 2$3#!
EFGG+06;H.FI!
E506G/JF6/50! ()*! ?)I/?F?!
At t=1/200s=5ms
(ωt
π/2 = 90
), the
RED
current/flux
contribution
has maximum
,!-)*(+-!./0+1!2$3#!'()*+!,!-)*(4-56!./0+7!
*/0@XX=B!4!<P=B7!9!L:!
)./M0+-!
J+6N++0!
2:!
)0-!
2<1!
6()6!
HG5?!
C#D! /*! 0+M)6/K+! )0-! ()*!
value sin(90◦ )=+1 aligned between R1 and R2, that from YELLOW is negative and
has
*/0@XX=B!4!=B7!9!4=O>!)./M0+-!J+6N++0!C<!)0-!C:1!)0-!6()6!HG5?!"#$$%&!/*!05N!).*5!0+M)6/K+!)
?*! @!69 ! / 2 9A=B71! 6(+! C#D!
EFGG+06;H.FI!
E506G/JF6/50!
()*!
?)I/?F?!
K).F+!
◦ - 120◦ ) = -0.5 aligned between Y2 and Y1, and that from BLUE is also
value
sin(90
K).F+!
*/0@XX=B!4!:<=B7!9!4=O>!
)./M0+-!
"<! )0-! ":O! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/
/M0+-! J+6N++0! C:! )0-! C<1! 6()6!
HG5?!
"#$$%&! /*! 0+M)6/K+!
)0-!J+6N++0!
()*! K).F+!
negative and has
value sin(90◦ - 240◦ ) = -0.5 aligned between B2 and B1. The vector
J+6N++0!2:!)0-!2<1!/O+O!)6!L<P=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>
9!4=O>!)./M0+-!J+6N++0!"<!)0-!":1!)0-!6()6!HG5?!2$3#!/*!).*5!0+M)6/K+!)0-!()*!K).F+!
! 9! 4=O>! )./M0+-! J+6N++0!
)0-!6(+!?)I/?F?!/0-/K/-F).!H.FI!HG5?!50+!'()*+O!
2:O! !is
Q(+!
K+E65G!horizontally
*F?1! 5G! 656).!
H/+.-1! /*!
sum, or2<!
total
field,
directed
RIGHT
to -/G+E6+-!
LEFT, aligned between R1 to
!
VQ!65!$#WQ1!)./M0+-!J+6N++0!C:!65!C<1!)0-!()*!G+.)6/K+!K).F+!:O>!6/?+*!6(+!?)I/?F?!
R2, and has relative
value 1.5 times the maximum individual flux from one phase (1 +
86!6!9!:>;Z==*9:;P=*9<>?*1!5G!)H6+G!50+!HF..!ESE.+;'+G/5-!5H!<=?*1!6(+!?)M0+6/E!H/+.-!()*!G+
5?!50+!'()*+!@:!L!=O>*/0X=B!L!=O>*/0X=B7O!
◦ ).
0.5sin30◦ + 0.5sin30
65! C:! )0-!
C<! 5G/+06)6/50O! T6! E)0! J+! E50E.F-+-! 6()6! 6(+! ?)M0+6/E! H/+.-! G56)6+*! N/6(! 6(+! *)?+! )
K+.5E/6S;HG+[F+0ES!)*!6(+!+.+E6G/E).!*F''.S1!H5G!)!6N54'5.+!*6)65GO!!
◦ ), the YELLOW
!69 7! / 6 9<:=B71! At
6(+!
"#$$%&!
E506G/JF6/50!
()*! ?)I/?F?!
K).F+!
t=7/600s
(ωtEFGG+06;H.FI!
= 7π/6 = 210
current/flux
contribution has maximum
7!9!L:! )./M0+-! J+6N++0! ":! )0-! !"<1! 6()6! HG5?! 2$3#! /*! 0+M)6/K+! )0-! ()*! K).F+!
b
!
7!9!4=O>! )./M0+-! J+6N++0! 2<! )0-! 1.4
2:1! 6()6!
HG5?! C#D!
05N!
).*5! 0+M)6/K+!
()*!CURRENT AND TORQUE
INDUCED
EMFS/*!
AND
PRODUCTION
OF)0-!
ROTOR
!4!=B7!9!4=O>! )./M0+-! J+6N++0! C<! )0-!
C:O! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/G+E6+-!
!
<1!/O+O!)6!L:<=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>!6/?+*!
W/MFG+! :OPO! *(5N*! )! */?'./H/+-! E)M+! N/0-/0M1! N/6(!6(+! G565G! J5-S! G+?5K+-! H5G! E.)G/6SO! TH! 6(+! G
-/K/-F).!H.FI!HG5?!50+!'()*+O!
/0*+G6+-! /0! 6(+! J5G+! 5H! 6(+! ?)E(/0+! *6)65G! HG5?! )J5K+1! 6(+! G+K5.K/0M! ?)M0+6/E! H/+.-! N/..! \
6(G5FM(\! 5G! ./0]! N/6(! 6(+! J)G*! 5H! 6(+! E)M+! N/0-/0MO! ! 8*! .50M! )*! 6(+! ?)M0+6/E! H/+.-! G+K5.K+*
=*! @!69 11! / 6 9XX=B71! 6(+! 2$3#! EFGG+06;H.FI! E506G/JF6/50! ()*! ?)I/?F?! K).F+!
7!9!L:! )./M0+-! J+6N++0! 2:! )0-! 2<1! 6()6! HG5?! C#D! /*! 0+M)6/K+! )0-! ()*! K).F+!
Pag
!4=O>!)./M0+-!J+6N++0!C<!)0-!C:1!)0-!6()6!HG5?!"#$$%&!/*!05N!).*5!0+M)6/K+!)0-!()*!
!:<=B7!9!4=O>! )./M0+-! J+6N++0! "<! )0-! ":O! Q(+! K+E65G! *F?1! 5G! 656).! H/+.-1! /*! -/G+E6+-!
!2<1!/O+O!)6!L<P=B!N/6(!G+*'+E6!65!6()6!)6!6!9!=1!)0-!)M)/0!()*!G+.)6/K+!K).F+!:O>!6/?+*!
-/K/-F).!H.FI!HG5?!50+!'()*+O!
Power Systems and Machines
4
value sin(210◦ - 120◦ ) = +1 aligned between Y1 and Y2, that from BLUE is negative
and has value sin(210◦ - 240◦ ) = -0.5 aligned between B2 and B1, that from RED is
now also negative and has value sin (210◦ - 0◦ ) = -0.5 aligned between R2 and R1. The
vector sum, or total field, is directed between Y1 to Y2, i.e. at +120 with respect to
that at t = 0, and again has relative value 1.5 times the maximum individual flux from
one phase.
At t = 11/600s (ωt = 11π/6 = 330◦ ), the BLUE current/flux contribution has maximum
value sin(330◦ - 240◦ ) = +1 aligned between B1 and B2, that from RED is negative and
has value sin(330◦ - 0◦ ) = -0.5 aligned between R2 and R1, and that from YELLOW
is now also negative and has value sin(330◦ - 120◦ ) = -0.5 aligned between Y2 and Y1.
The vector sum, or total field, is directed between B1 and B2, i.e. at +240◦ with respect
to that at t = 0, and again has relative value 1.5 times the maximum individual flux
from one phase.
At t = 15/600s=1/40s=25ms, or after one full cycle/period of 20ms, the magnetic field
has returned to R1 and R2 orientation. It can be concluded that the magnetic field
rotates with the same angular velocity/frequency as the electrical supply, for a two-pole
stator.
1.1.4
Induced EMFs and production of rotor current and torque
Figure 4 shows a simplified cage winding, with the rotor body removed for clarity. If the
rotor is inserted in the bore of the machine stator from above, the revolving magnetic
field will “thread through” or link with the bars of the cage winding. As long as the
magnetic field revolves more quickly than the cage winding, there will be a rate of change
of the flux linking with each conductor and an electromotive force induced across the
ends of the conductors.
Faraday’s Law:
I
E2 .dl = = −
C
dφtotal
= φtotal (N − NR )
dt
(1.1)
shows that the rate of change of flux is directly proportional to the difference in speed
between the rotating flux pattern and the shaft speed of the rotor conductors. The rotor
conductors are made of copper, and exhibit very low impedance to the flow of current
under the action of induced emf, 2 . A current,
I2 =
2
Z2
(1.2)
flows in each conductor, which splits up in the rotor short circuiting rings and is returned by the other conductors in the cage winding. The effect is repeated in each bar
(conductor) of the rotor cage, as the flux from each pole of the stator field passes that
bar. The current in each rotor bar produces a separate rotor magnetic field which reacts
Power Systems and Machines
Rotating flux
N rev/min
5
ɸ
total
Rotor currents
I2 α E2/Z2
SC Ring
Induced emfs
E2 ɸ (N-Nr)
Shaft speed
Nr rev/min
F α ɸ I2
Figure 4: Simplified rotor cage winding
with that from the stator, and each bar experiences a force
F = BI2 l = φI2
(1.3)
which acts tangentially to the surface of the rotor and in the direction of the revolving
field. A torque is developed on the rotor of the machine, which accelerates the rotor
towards the speed of the rotating magnetic field in the stator bore. Provided there is
a net difference in speed with the rotor revolving more slowly than the stator field, the
induced emf and ensuing rotor current will create a torque on the rotor of the machine.
If this torque is transmitted to driven equipment, the machine operates as a motor. The
remainder of this lecture considers the machine operating as a motor.
1.1.5
Synchronous speed, shaft speed and slip
In the two-pole stator above (i.e., with one pair of poles per phase), the magnetic field
revolved through one mechanical revolution in the time taken for the electricity supply
to advance through one angular revolution. On a 50 Hz supply, the field will revolve
once every 1/50 sec, 50 times a second or at 3000 rev/min (for 60 Hz supply - 3600
rev/min). If the stator is wound to have twice as many poles, the magnetic field will
rotate from the first to the second pair of red poles in the same time. The stator must
be wound uniformly and in the same period of time the field rotates through half of the
“two-pole” angular displacement. The field revolves at half of the two-pole speed, 1500
rev/min. A six-pole winding (i.e., three pairs of poles per phase) would produce a field
revolving at one third of 3000 rev/min - 1000 rev/min.
The speed of the magnetic field produced by the stator is known as the SYNCHRONOUS
SPEED of the machine, denoted N. The synchronous speed is inversely proportional to
Power Systems and Machines
6
the number of POLE PAIRS, denoted p, and directly proportional to the supply frequency fsupply and this relationship determines the synchronous speed is given by,
N=
60.fsupply
rev/min.
p
(1.4)
Since this speed is determined solely by the stator winding configuration and supply
frequency, the synchronous speed of a given machine is fixed, and neither varies with
time nor with applied load. To produce torque, the rotor must revolve more slowly than
the synchronous speed of the magnetic field.
The SHAFT (i.e., ROTOR) SPEED NR is always less than synchronous in the case
of a motor, and is determined by the mechanical load applied to the shaft. Note that
applying load slows the rotor down and the machine develops more torque in response
to the added load.
The difference between synchronous speed and the shaft speed determines the torque
produced. It is therefore more convenient to work with this difference in speed as it is a
smaller figure. The SLIP is the difference between the synchronous speed and the shaft
speed, expressed as a percentage of synchronous, and is denoted s, where
s=
N − NR
N
NR = N (1 − s)
(1.5)
For a squirrel cage-type induction motor operating from no load up to full load, the slip
would typically increase from 1% to 10%, and the shaft speed would fall from 99% to
90% of synchronous.
The magnitude and frequency of the induced emf on the rotor of the machine is determined by shaft speed/slip. If the rotor were to revolve at exactly synchronous speed N,
there would be no rate of change of flux and no induced emf in the rotor. If the rotor was
locked and could not turn, the induced emf and frequency would be at the maximum,
since the rate of change of flux would be greatest. At standstill, the induced voltage on
the rotor E2 has maximum (open circuit) value Eoc and this reduces linearly with slip
as the speed increases.
At any speed NR , with slip s, the induced voltage on the rotor is given by
E2 = s.Eoc
At standstill, E2 = Eoc and at synchronous speed N, E2 = 0
At standstill, the induced frequency frotor has maximum value, just equal to the frequency of the ac supply on the stator fsupply, and this reduces linearly with slip s as
the speed increases. At any speed NR , with slip s, the induced frequency on the rotor
may be expressed as Frotor = s.fsupply .
Power Systems and Machines
1.1.6
7
Torque/speed variation
Figure 5 shows a typical torque-speed curve for an induction motor. When the machine
is started unloaded, the slip is instantaneously 100%, and the induced emf on the rotor is
at a maximum, since the induced emf is proportional to the rate at which the magnetic
field cuts the rotor conductors. The current in the rotor is also at the maximum at
start. At the instant of switch on, the stator winding has to be magnetised, requiring
a high magnetising current. Current at start may be up to 6 times full load current
(FLC).
The unloaded motor then accelerates up towards synchronous speed, and reaches a speed
corresponding to around 1% slip, where the induced emf, current and torque are just
sufficient to overcome the no-load losses incurred by friction and windage losses on the
rotor. The application of mechanical load to an unloaded machine, causes the rotor to
slow down. The slip increases, and the induced emf and rotor current rise. The torque
produced by the rotor rises to balance with that applied by the load. Full load torque
(FLT) would normally be developed around 90% of synchronous speed. Maximum torque
would be developed at around 80% of synchronous speed, at a value between 200 and
300% of full load torque.
Shaft torque TI
(% full load torque)
180
160
140
120
100
80
60
40
20
0
Full load
0 10 20 30 40 50
100 90 80 70 60 50
60 70
40 30
80 90 100 slip (%)
20 10 0 Nr (%)
Figure 5: Typical torque speed curve
Operating between no load and say twice full load torque, the machine behaves in a
stable manner. The torque produced varies almost linearly with slip, and machine
responds to added load by slowing down and producing more torque at the reduced
speed. Operation near the peak of the torque-speed curve, or beyond designed full load
torque is dangerously unstable. In that situation, added load may cause the rotor to
slow down to a speed below the tip of the torque curve. In this case, the output torque
starts to fall as speed reduces. Speed falls further and torque decreases again. The motor
Power Systems and Machines
8
stalls and returns to a locked rotor condition. This is an extremely dangerous situation,
since the ventilating air flow through the rotor and stator is removed and the electrical
losses may be increased by a factor of 36 (6xFLC)2 . The motor very quickly reaches
dangerous temperatures and will certainly burn out if it is not disconnected from the
ac supply. Separate overcurrent and stall protection is therefore desirable for induction
motors, and is discussed later on.
1.2
1.2.1
Power flow and characteristics
Apparent, active and reactive power. Power factor
An induction motor may be regarded as an inductive-resistive load on the electrical
supply, as shown in figure 6. Voltage and current phasor diagrams for a star connected
induction motor are shown in figure . The alternating current flowing in the stator
always flows later in time than the applied voltage causing the current flow, due to the
inductive effects of the stator winding. The line currents are shown at an angle, φ, to
the corresponding voltages below.
A voltmeter, connected between two line conductors, could be used to measure line
voltage Vline . An ammeter, inserted in a line conductor, could be used to measure line
current Iline . One wattmeter W (whose reading was multiplied by 3) could measure
active power input Pin , since the impedances and loads presented by each phase of the
motor should be balanced, i.e., equal.
Red
Yellow
Blue
Neutral
VI
Vline
MOTOR
A
V0
W
I phase = I line
V phase =
V line / 3
Star connected
Vphase =
Vline
I line
I phase =
I line / 3
Delta connected
Figure 6: Induction motor load
A supply voltage (usually obtained from a star connected secondary of a distribution
Power Systems and Machines
9
Vbn
ɸ
Vyb
ly
ɸ
lb
Vbr
ɸ lr
V phase
Vrn
Vry
120
Vline
Vyn
Figure 7: Voltage and current phasor diagrams
transformer) with line value Vline , has phase value:
Vline
Vphase = √ V
3
(1.6)
In a star connected machine, a line current of value Iline is equal to a phase current Iphase .
In a delta connected machine, the line and phase voltages are identical, and the phase
current is:
Iline
(1.7)
Iphase = √ V
3
Simply multiplying together the rms values of Vphase and Iphase would suggest the power
taken by one phase of the motor, but would not take account of the phase angle between current and voltage. The total power apparently taken by the machine would
be 3Vphase Iphase . Substituting line values, since they are more readily measurable as
neutral conductor may or may not be available, the total three-phase COMPLEX or
APPARENT POWER, in units of Volt-amperes (VA) may be calculated as:
√
Sin = 3Vline Iline V A
(1.8)
The component of line current which is out of phase with the applied voltage Vphase , Iline
sinφ, is that which magnetises and demagnetises the inductance of the motor stator every
successive half-cycle of the ac supply. If the inductance is considered pure and there was
no other circuit resistance, no power would be dissipated in the process. This component
of current contributes to the overall value of line current as a reactive component, and,
consequently, to the apparent power flow, but does not affect the real or active power
taken from the supply.
The total three-phase IMAGINARY or REACTIVE POWER, in units of volt-amperes
reactive (VAr) may be calculated as:
√
Qin = 3Vline Iline sin φ V Ar
(1.9)
Power Systems and Machines
10
The component of line current in phase with the applied voltage Vphase , Iline cos φ, is
that which performs perceptible “useful” work, either as machine output power or as
losses from the machine. This component of current contributes to the overall value
of line current as an active component, and consequently to the apparent power. This
component of current determines the useful power taken from the electrical supply.
The total three-phase REAL or ACTIVE POWER, in units Watts (W) may be calculated
as:
√
Pin = 3Vline Iline cos φ W
(1.10)
The power factor at which the motor operates is simply the ratio of the real power taken
(to produce output power and supply the losses) to the apparent power taken by the
motor (to both magnetise and supply the losses and output power from the machine).
Thus:
Pin
Power Factor =
= cos φ
(1.11)
Sin
Since in an inductive circuit the current always lags behind the applied voltage, the induction motor always operates at a lagging power factor, typically 0.9 on full load.
1.2.2
Power flow, losses and efficiency of induction motors
The principle of energy conservation enables the separation of the stator and rotor of
the motor into energy centres either side of the airgap of the machine. Figure 8 shows
the power flow diagram for the induction motor, with the rotor unthreaded.
In the
Power
out of
equals
steady state, the power flow through the machine and the losses are unvarying.
flow into and out of the stator can be balanced. Similarly, power flow into and
the rotor can be balanced. The power leaving the stator via the airgap exactly
that supplied to the rotor from the airgap.
The INPUT POWER to the stator is provided by the electricity supply, and is designated
Pin .
If the stator is fed from a three-phase supply at line voltage Vline , and draws line current
Iline at a power factor of cos φ, the active power taken into the stator is Pin .
The STATOR LOSSES in the machine arise in two ways. Due to the revolving magnetic
field, small emfs induced in the stator laminations circulate eddy currents in these laminations. The ohmic (I2 R) heating losses associated with these currents heat the stator
core and represent an IRON LOSS designated Pf . The iron loss is present under all
conditions of load between no load and full load, and is nearly constant. Also due to the
flow of supply current in the stator windings, there is a direct ohmic heating loss due
to the resistance of the stator windings. This is referred to as a COPPER LOSS and
2
is designated Psl . The copper loss is proportional to the 3.Iline
and hence varies with
load.
Power Systems and Machines
Psl Copper
loss Pf
Iron
loss
Pin
Active Power to stator
11
Airgap power
to rotor
Pr
Pc
Copper loss
Pw
Friction and windage loss
Pl
Shaft ouput power
to load
Rotor
Figure 8: Power flow and losses in a machine
The AIRGAP POWER may be established by considering the power balance on the stator, and represents that amount of power transferred to the rotor, designated PR.
PR = Pin − (Psl + Pf ) Watts
(1.12)
The ROTOR LOSSES in the machine arise due to the circulation of rotor currents
in the ohmic resistance of the rotor conductors. This loss, proportional to load, is a
COPPER LOSS designated PC . The copper losses on the rotor are determined by the
magnitude and frequency of the induced emf on the rotor, which is directly proportional
to slip.
If the rotor was to rotate at exactly synchronous speed (i.e., at s = 0), there would be
no induced rotor emf E2 or rotor current I2 . The rotor copper losses would be zero at
synchronous speed.
When the rotor is locked and NR = 0, s = 1, no power is developed or transmitted to
the load. In this situation, all of the rotor power PR is dissipated as rotor copper losses,
or PC = PR . A reasonable representation of this effect is to relate the losses to the slip
and total power on the rotor linearly as:
PC = s.PR
Watts
(1.13)
The FRICTION AND WINDAGE LOSSES on the rotor further subtract from the power
transferred across the airgap, and are incurred as the rotor overcomes bearing friction
forces and circulates cooling air in the machine. Designated PW , the friction and windage
losses may be assumed to remain almost constant from no load to full load conditions,
at about 1% of the input power.
Performing an energy balance on the rotor of the machine enables the calculation of
the shaft power or output power transferred to the load, denoted PL . Slip may be
used to estimate the relative amounts of: a) power dissipated as copper losses PC , and
Power Systems and Machines
12
b) developed gross mechanical power. GROSS MECHANICAL POWER (GMP) is the
sum of the mechanical power transmitted to the load and the friction and windage losses
being overcome
(PL + PW ) = PR − PC
and
PC = sPR
(1.14)
Thus, since
PL + PW
W
(1.15)
1−s
The associated output torque may also be determined if the shaft speed NR and torque
delivered to the load Tload is known:
PR =
PL = Pload = ωτ =
2πNR τload
W
60
(1.16)
Addition of the stator losses enables the calculation of the input power to the machine,
from which line current could be estimated at a given power factor.
Pin = Pr + Psl + Pf =
1.2.3
Pl + Pw
+ Psl + Pf W
1−s
(1.17)
Efficiency
It is now possible to calculate or verify machine efficiency based on shaft output power
and any data enabling the calculation of input power.
Efficiency = ξ = η =
Pin − (Pf + Psl + Pr + Pw )
Shaft output power
Pload
=
=
Electrical input power
Pin
Pin
(1.18)
Taking the input power as a nominal 100% and the efficiency of a typical well designed
motor as 94%, the losses in the machine might comprise, in percentage terms.
Input Power
Less
Stator iron losses
Stator copper losses
Rotor copper losses
Friction and windage
Leaving
Output power
100%
-1.0%
-1.5%
-2.5%
-1.0%
——
94%
With this information, it should now be possible to perform some first order assessments
of machine performance, based on the above theory, nameplate data, and electrical
measurements made at the supply.
Power Systems and Machines
1.2.4
13
Rotor power balance as a function of slip
As a prelude to future material, we need to note for the development of the induction
machine model, that since:
(PL + PW ) = PR − PC
and PC = sPR
(1.19)
(PL + PW ) = PC /s − PC = PC (1 − s)/s
(1.20)
(PL + PW ) = PR (1 − s)
(1.21)
The triple-ratio of (PL + PW ) : PC : PR = (1 − s) : s : 1 defines the proportions of gross
mechanical power : rotor copper losses : rotor power from the airgap.
Power Systems and Machines
2
14
The equivalent circuit model of the induction machine
2.1
Introduction
A three-phase induction machine could operate as a motor or as a generator when
connected to a synchronous three-phase electrical supply. The physical properties of
the machine do not vary from those designed and manufactured into the machine, and
remain independent of the mode of operation. Consequently, it is possible to represent
the machine by a simplified equivalent circuit, whose parameters are independent of
whether the machine operates as a motor or as a generator. The steady state behaviour
of the machine may be analysed or simulated using the equivalent circuit as a model of
the machine.
This section revises and develops the approximate model of the induction machine. A
later section goes on to illustrate how manufacturers’ test data may be used to estimate
component values in the model. From there, the steady state behaviour and validity of
the model may be assessed when the machine is connected to grid systems of varying
strength.
2.2
Approximate model of induction machine
The model is evolved by considering the machine as a motor initially, and then reconsidering it in a later lecture as a generator. All circuit diagrams represent one phase
of the machine (“per-phase” representation) and all parameters are phase values, unless
otherwise stated.
2.2.1
Rotor
Figure 9 shows the two components which may be used to represent the rotor of an
induction machine.
R2
E 2 = s.Eoc
sX2
I2
R2 /s
E 2 =Eoc
X2
I2
Figure 9: Rotor equivalent circuit (left) and simplified circuit (right)
The rotor conductors, being made of copper, exhibit an ohmic resistance to the flow
of rotor current. Resistance R2 represents the rotor winding resistance. The fact that
an alternating current flows in the rotor conductors, the proximity of the conductors
and their configuration dictates that the rotor circuit must be inductive in nature. The
Power Systems and Machines
15
inductive reactance of the rotor is frequency dependent, and since rotor frequency frotor
is slip dependent, the inductive behaviour of the rotor may be represented by an inductive
reactance of ohmic value s.X2 , where X2 is the standstill reactance of the rotor.
At standstill, the induced emf E2 on the rotor is at the maximum value EOC . At any
value of slip, the induced emf on the rotor E2 = s.EOC . In the model, E2 is applied
directly across Z2 , the series combination of R2 + jsX2 (figure 9):
q
sX2
Z2 = R2 + j(sX2 ) = R22 + (sX2 )2 6 + φ2 ohms/phase where φ2 = tan− 1
(2.1)
R2
Thus, the rotor current I2 , which flows at any value of slip, is
sEOC
I2 = p 2
R2 + (sX2 )2
6
− φ2
(2.2)
To simplify the circuit and reduce the number of slip dependent terms, it is more convenient to fix E2 = EOC , to fix the reactance as X2 and to make the resistive term slip
dependent, R2 /s, as shown in figure 9. The rotor current I2 becomes dependent on only
one ‘s’ term. There is now a single (slip-dependent) resistor R2 /s in the rotor circuit. It
is necessary to represent the rotor copper losses and gross mechanical power as ohmic
heating losses in the rotor, preferably still with one slip-dependent term.
Eliminating PR from equations 1.19 and 1.21 gives two useful expressions to work back
and forward from.
PC =
s(PL + PW )
(1 − s)
or
(PL + PW ) = PC
(1 − s)
s
(2.3)
To represent individually the rotor copper losses and gross mechanical power in the
correct ratio, the resistor R2/s may be separated into two components, R2 and R2 .(1 −
s)/s, since
R2
(1 − s)
= R2 + R2
(2.4)
s
s
• The total rotor power is now represented by the ohmic losses in resistor R2 /s.
• The ohmic heating loss in the model component R2 now represents the rotor copper
losses PC .
• The ohmic heating loss in the model component R2 (1-s)/s represents the gross
mechanical power (PL + PW ) developed by the rotor.
R2
s
2
PC = 3I2 R2
PR = 3I22
PL + PW
1−s
= 3I22 R2
s
(2.5)
(2.6)
(2.7)
Power Systems and Machines
16
R2
sX2
E 2 = Eoc
R2 (1-s)
s
I2
Figure 10: Rotor GMP resistor added
2.3
Rotor and referred values
Since emfs are induced across the airgap of the machine, the rotor and stator windings
may be regarded as those of a transformer linked by an air core. In this case, the
secondary winding rotates, but under a fixed load, the rotor revolves at a fixed speed
and the rate of change of flux linkage is constant. The rotating flux φ links both with
the stator and rotor windings and induces separate emfs in each; E1 in the stator and E2
in the rotor. The phase voltage E2 induced in the rotor conductors may be considered
as the secondary (internal) emf in the airgap transformer. The phase voltage E1 induced
in the stator windings may be regarded as the primary (internal) emf. In most cases, E2
is less than E1 and the transformer may be regarded as a step down transformer. Figure
11 shows the airgap transformer in circuit.
I2
R2
sX2
R2 (1-s)
E 2 = Eoc
E1
dɸ
dt
N1: N2
s
I2
Figure 11: Airgap transformer and rotor circuit
If we regard the transformer to have N1 turns in the primary (stator), N2 turns in
the secondary (rotor), ratio Neff and to be ideal: the same flux (and rate of change of
flux) links both windings, and Faradays law applies to both windings as follows: In the
Power Systems and Machines
17
primary winding
E1 = N1
dφ
dt
(2.8)
and for the secondary
dφ
(2.9)
dt
Therefore, we may equate volts per turn, and derive the voltage and turns ratio expression.
N1
E1
=
(2.10)
E2
N2
Since the induction machine is singly-excited from its stator supply, there is only one
(total) mmf raising this flux, and the ampere-turns in each winding must be equal,
E2 = N2
mmf = N1 I1 = N2 I2
and
I1
N2
=
I2
N1
(2.11)
The impedance of the rotor circuit in the secondary winding may be expressed as Z2 .
The impedance seen in the primary due to the current drawn in the primary to supply
the secondary may be expressed as Z1 , where
Z2 =
E2
I2
and Z1 =
E1
I1
(2.12)
Substituting secondary values for E1 and I1 ,
Z1 =
N1
E2 N
2
2
I2 N
N1
E2 N1 2
2
=
= Z2 Neff
I2 N2
(2.13)
Due to the step down characteristics of the transformer, looking into the rotor, or transformer secondary, “across the motor airgap” from the stator, all of the effects of the
rotor impedances will be increased by the square of the turns ratio, as evidenced in the
stator circuit.
The model of the machine is most useful if it represents the power flow from the stator
into the rotor, and must therefore correctly represent the stator and rotor circuits. It is
not convenient to represent the machine with a transformer in the model, and all of the
rotor impedances may be referred into the stator circuit by the square of the turns ratio
Neff . Referred values are denoted R2 ’ and X2 ’, where
2
R20 = Neff
R2
2
and X20 = Neff
X2
(2.14)
The transformer may now be removed from the equivalent circuit, its effects having been
absorbed and modelled. The rotor may be represented in the stator circuit as a series
impedance Z2’ (also denoted as “rotor branch impedance”, Zt ), where the prime or ’
denotes a referred value.
Zt = Z20 =
R20
+ jX20
s
or Z20 6 φ2
where
φ2 = tan−1
sX20
R20
(2.15)
Power Systems and Machines
18
The rotor model just developed may be incorporated into two different overall machine
models. The ‘Tee’ model in figure 12 is more accurate, but involves a lengthier solution.
The approximate model in section is more easily solved and sufficiently accurate for most
applications.
2.4
T-equivalent circuit
Figure 12 shows the remaining stator components of the T equivalent circuit model
which are added to the rotor of the machine. Each is a phase quantity and is explained
below.
R1
X1
E1
V1
X’2
R2
Xm
R’2 (1-s)
RC
s
I’2
I om
Figure 12: T equivalent circuit
2.4.1
Stator no-load circuit
RC - Since the revolving magnetic field induces eddy currents in the stator core, which in
turn incur iron losses, under all conditions of load, there must be an active component of
stator current which is nearly fixed, and which dissipates power in a resistor to represent
iron core losses. Resistance RC represents this loss as a core-loss resistor.
Xm - Under all loading conditions, a current must flow to magnetise the stator and the
airgap with the rotating field, and therefore a component Xm represents the magnetising
reactance of the machine.
The parallel combination of RC and Xm is sometimes referred to as the “no load shunt
circuit” of the machine, since on no load the slip, s, is very small, and consequently R02 /s
is very large. I2 ’ becomes small enough to be neglected. Hence, the rotor component
of the equivalent circuit may be neglected on no load. The no-load shunt impedance is
designated Z0 and the no load current I0 may be calculated from E1 and Z0 :
Z0 =
RC jXm
RC + jXm
or
Z0 6 φ0
where
φ0 = tan−1
Im Z0
Re Z0
(2.16)
Power Systems and Machines
2.4.2
19
Stator series impedance
R1 - The copper conductor of the stator windings has a finite resistance. The flow of
stator current I1 causes a stator copper heating loss in the machine, which is represented
as an ohmic power loss in R1 . In real terms, R1 is the series resistance of each phase
stator winding of the machine.
X1 - Not all of the revolving magnetic field links with the rotor. Some of the flux is
deemed to have leaked from the useful path. This is represented by the primary winding
of the airgap transformer having a number of turns that produce flux that cross the
airgap to the rotor, and a number of turns that produce the flux that does not. These
turns are separate from the effective number of turns, and are designated as the stator
winding leakage reactance. The series reactance of the stator becomes
Z1 = R1 + jX1
2.4.3
Z1 6 φ0
or
where
φ0 = tan−1
X1
Im Z1
= tan−1
(2.17)
Re Z1
R1
Circuit solution
E1 is an internal voltage that is not initially known, because of the as-yet uncalculated
voltage drop across Z1. The circuit has to be solved by calculating the total impedance
Zin out to the terminals, across which phase voltage V1 6 0◦ is known.
The combined impedance ZRS of the rotor and stator no-load circuit is calculated
from
ZRS =
Z0 Z20
Z0 + Z20
or
ZRS 6 φRS
where
φRS = tan−1
Im ZRS
Re ZRS
(2.18)
but this is needed in rectangular form for addition to Z1 , as
ZRS (cos φRS + j sin φRS )
(2.19)
The total input impedance of the circuit, Zin becomes
Zin = (R1 + ZRS cos φRS ) + j (X1 + ZRS sin φRS )
(2.20)
The phase current drawn into each phase of the machine Iin becomes
Iin = I1 =
V1
6 − φin
Zin
where
φin = tan−1
X1 + ZRS sin φRS
R1 + ZRS cos φRS
Internal voltage E1 is calculated by subtracting the voltage drop across Z1 .
E1 = V1 6 0 − I1 Z1 6 (0 − φ1 ) = V1 − I1 Z1 cos(−φ) − j (I1 Z1 sin(−φ1 ))
(2.21)
(2.22)
Power Systems and Machines
20
Current I2 ’ can only be calculated once E1 is known, because of the unknown voltage
drop across R1 + jX1 . Having the internal voltage E1 known, allows the calculation of
the no-load stator current I0 and the referred rotor current I2 ’.
E1
sX20
6 − φ2 where φ2 = tan−1
(2.23)
Z2
R20
E1
Im Z0
6 − φ0 where φ0 = tan−1
I00 =
(2.24)
Z0
Re Z0
The branch currents I1 = Iin , I2 ’ and I0 may now be used to calculate the machine
performance.
I20 =
2.4.4
Current and voltage phasor diagram
In the T equivalent circuit, the sum of the currents drawn by the no-load circuit Z0 and
the rotor referred impedance Z2 ’ has to be supplied through the stator windings, which
also have impedance. As a result of the phase current drawn, there are series voltage
drops across the stator series resistance R1 and leakage reactance X1 . V1 must exceed in
magnitude and lead in phase angle E1 . This is shown below in the current and voltage
phasor diagram for the above circuit.
The no-load circuit draws real and reactive components of current I0r and I0m to supply
iron losses in the stator and magnetise the stator and airgap, respectively. These two
quadrature components of current add at the T node to become I0 = I0r + jI0m . The
rotor circuit draws current I2 ’ from the supply as the load on the rotor increases. This
adds to the no load current I0 at the T node, to become the total current drawn into
each phase of the stator winding Ip , or I1 .
V1
I 1 jX1
E1
I 1R 1
I1
ɸ
1
ɸ
0
Io
I om
ɸ2
I or
I’2
ɸ axis
Figure 13: Current and voltage phasor diagram for T equivalent circuit
The T equivalent circuit solution is lengthy, but can be further simplified to become an
approximate single-phase equivalent circuit of an induction machine.
Power Systems and Machines
3
21
Steady state behaviour of induction machines
3.1
Approximate single-phase equivalent circuit
Since the no-load current is virtually constant, and much smaller than the current in the
stator due to the rotor current (called the “referred rotor current”), the no-load shunt
circuit is moved out to be directly across the terminals of each phase equivalent circuit,
as shown below in figure 14. The series voltage drops across the stator components are
also sufficiently small, so that it is valid to regard the phase voltage V1 as connected
across the no load circuit.
X’2
R2
V1 Xm
R’2 (1-s)
RC
s
I’2
I om
Figure 14: Approximate single-phase equivalent circuit
3.1.1
Stator no-load circuit
As above, but moved out to the terminals with Z0 connected across phase voltage V1 or
Vph , given by equation 2.16 from before.
3.1.2
Stator series impedance
As above, but now calculated to be in series with the rotor referred impedance Z2 .
Z20 =
R20
+ jX20
s
or Z20 6 φ2
R20
+ j X1 + X20
Zt = R1 +
s
s
R20 2
Zt =
R1 +
+ j (X1 + X20 )2 6 φt
s
where
φ2 = tan−1
φt = tan−1
sX20
R20
(3.1)
X1 + X20
R1 + R20 /s
(3.2)
V1
6 − φt
Zt
(3.3)
I20 =
The referred rotor current I2 ’ flowing in the series Zt branch is determined by Z1 and
Z2 ’. Note that since Z2 ’ has a slip dependent term, I2 ’ rises as the slip increases. This
Power Systems and Machines
22
correctly represents the increase in current as the machine is loaded. Refer to figure
13.
The current drawn into the terminals of the approximate equivalent circuit I1 or Iph is
the sum of I2 ’ and I0 , and is the phase current in each winding. Note that I2 ’ and I0
are unlikely to be in phase and the addition must be vectorial, in rectangular components.
I1 = I0 + I20
(3.4)
The input impedance of the machine may be calculated under all loading conditions as
the parallel combination of Z0 and Zt .
Zin =
Z0 Zt
Z0 6 + φ0 .Zt 6 + φt
=
Z0 + Zt
(Z0 cos φ0 + Zt cos φt ) + j(Z0 sin φ0 + Zt sin φt )
(3.5)
p
(3.6)
Then
(Z0 cos φ0 + Zt cos φt )2 + j(Z0 sin φ0 + Zt sin φt )2 6 + φC
Z0 sin φ0 + Zt sin φt
φC = tan−1
Z0 cos φ0 + Zt cos φt
ZC =
(3.7)
where φin is the phase angle of the overall impedance of the approximate equivalent
circuit, or machine input impedance. The input current I1 drawn by the machine and
its phase angle φin relative to the applied phase voltage V1 can now be calculated. The
operating power factor can be determined from the cosine of the phase angle between
the applied voltage and current drawn.
Iphase = I1 =
3.2
V1
6 − φin
Zin
and power factor = cos(−φin )
(3.8)
Calculation of machine performance
The apparent (or complex) input power to the machine can be calculated from
√
Pin = 3Vline Iline
(3.9)
The active (or real) input power is
Pin =
√
3Vline Iline cos(−φin )
(3.10)
The reactive (or imaginary) input power is
√
Pin = 3Vline Iline sin(−φin )
(3.11)
The efficiency of the machine becomes
ξ=η=
Pload
,
Pin
where
Pload = 3I22 R20
1−s
− PW
s
(3.12)
Power Systems and Machines
23
Note that the individual electrical losses in the machine are represented by the I2 R losses
in the corresponding resistance. The friction and windage might be estimated as about
1% of the shaft output. On this basis, the input current, power factor, and efficiency of
the machine at any given load is a function of the equivalent circuit component values
and slip.
The performance of the machine may now be modelled at any value of slip, particularly
within the stable areas of operation, using relatively simple mathematical analysis and
electrical techniques. For ranges of positive slip values, the machine may be modelled as
a motor. For ranges of negative slip values, the machine may be modelled as a generator.
The voltage and current phasor diagrams associated with the models correctly represent
the steady state behaviour of the machine in either mode.
Figure 13 shows the voltage and current phasor diagrams for the approximate equivalent
circuit with the induction machine motoring. The no load current I0 is heavily reactive,
at angle φ0 to the applied voltage, since it is fixed in magnitude and phase by the
real and imaginary currents required to provide iron losses and magnetise the machine
respectively. This current is independent of mechanical load on the machine.
On no load, the rotor and stator series current I2 ’ is negligible, and I1 , the current drawn,
is mainly reactive at a very lagging phase angle determined by the inductive reactance
of the magnetising branch.
I0 = I0R + jI0M
I0R = I0 cos φ0
I0M = I0 sin φ0
I1 = I0 + I20
(3.13)
As load on the machine increases and the slip rises from its no-load value to that corresponding to the increased load, I2 ’ starts to increase and contribute to the increase
in I1 . The angle φ1 between V1 and I1 also reduces. The active component of current
I1 cos φ1 also increases, improving the power factor and drawing more active power from
the three-phase supply.
3.3
Performance operating as an induction motor
The approximate model of the induction machine shown in figure 12 may be used to
predict the performance of the machine, operating either as a motor or as a generator,
if the parameters of the machine are known.
Earlier, it was shown that the gross mechanical power (PL + PW) could be represented
by the ohmic heating losses in the variable resistor (equation 2.7). The gross mechanical
power may be converted to an airgap torque, by taking account of shaft rotating speed.
Power Systems and Machines
24
Since
2πNR (TL + TW )
60
60(PL + PW )
(TL + TW ) =
2πNR (1 − s)
(PL + PW ) =
(3.14)
(3.15)
0
(TL + TW ) =
60.3.I22 R20 1 − s
180R20 20
=
I
2πN (1 − s) s
2πN s 2
(3.16)
The referred rotor current was derived earlier in equation 3.3. Substituting for I2 ’ gives
an expression for airgap torque in terms of machine constants/parameters and slip.
180R20
V12
2πN s (R1 + R2 /s)2 + (X1 + X20 )2
sR20
180V12
(TL + TW ) =
2πN (sR1 + R2 )2 + s2 (X1 + X20 )2
(TL + TW ) =
(3.17)
(3.18)
Plotting this relationship against slip in range from +100% for a motor at standstill
through no load at 0% slip to -100% for a generator driven out of synchronism at 200%
speed, produces the correct form of torque-slip curve, shown in figure 15. When s is
positive, a torque is produced and the machine operates as a motor. When s is negative,
a torque is absorbed and the machine operates as a generator. The stable region of small
slip either side of synchronous speed is properly represented.
The stationary points on the torque-slip curve may be determined by equating dT/ds =
0. When dT/ds = 0, a quadratic equation in s occurs with roots giving the stationary
points below, corresponding to maximum torque at two +/- values of slip, either side of
s = 0.
±R20
s= p 2
(3.19)
R1 + (X1 + X20 )2
The maximum torque produced or absorbed at these values of slip becomes
(TL + TW )max = ±
45V12
1
p
πN R1 ± R12 + (X1 + X20 )2
(3.20)
In practice, the stator series resistance and leakage reactance are much lower than the
rotor parameters R1 << R2 ’ and X1 <<X2 ’. Thus setting R1 = 0 and X1 = 0,
(TL + TW )max = ±
45V12 1
πN X20
when
s=±
R20
X20
(3.21)
I.e. maximum torque is developed for slip s = R2 ’/X2 ’.
Note that for a maximum torque around 80% speed, or 20% slip, R02 ≈ 0.2X2 ’. The values
of R1 , X1 , R2 ’ and X2 ’ determine the “steepness” of the stable area of the induction
Power Systems and Machines
25
Shaft torque Tl - produced
(% full load torque)
180
+Tmax
s = + R2/X2
160
140
120
100
Full load
80
60
Motor
40
slip (%)
-100
-90
-80
-70
-60
-50
-40
-30
-20
20
-20
20
30
40
50
60
70
80
90
100 slip (%)
-40
Generator
-60
-80
Full load
-100
-120
-140
-160
-180
-Tmax
s = -R2/X2
Shaft torque Tl absorbed
(% full load torque)
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Figure 15: Torque-speed curve
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motor/generator torque-speed curve, and the speed regulation of the machine. The
!
ratio of TMAX :TFULL-LOAD is also important,
since the ability of the machine to sustain
# / *)
!
0
$
mechanical overloads is determined )by the ratio* of) “Pull-out torque” : “Full load torque”.
!
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!
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+
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may now be modelled
.9
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variations
against
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B:$:!4'>(4/4!&)+./$!(%!6$3$1)-$6!A)+!%1(-!%!8!J!D
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Power Systems and Machines
4
26
Operation of induction machines as motors or generators
4.1
Power flow reversibility
Power Systems And Machines
Like any other electrical machine, the induction machine may convey energy in for-
5. ward
OPERATION
MACHINES
AS MOTORS
OR GENERATORS
(i.e., motor)OF
or INDUCTION
reverse (i.e., generator)
directions.
The machine
can be made to
from Flow
motoring
to generating by simply substituting the driven equipment with
5.1move
Power
Reversibility
Like
any
other equipment,
electrical machine,
the induction
machine
may
convey
energy
in forward
(i.e., motor)
the
driving
and accelerating
the rotor
from
positive
slip
as a motor,
through
or synchronous
reverse (i.e.,speed
generator)
directions.
The
machine
can
be
made
to
move
from
to
when slip is zero and the machine is on no load, to negative motoring
slip
generating by simply substituting the driven equipment with the driving equipment, and accelerating
the synchronous
speed,
whenthrough
the machine
will become
a generator.
Figure
16 machine
theabove
rotor from
positive slip as
a motor,
synchronous
speed when
slip is zero
and the
refers,
and
is
cited
a
few
times
in
these
notes.
is on no load, to negative slip – above the synchronous speed, when the machine will become a
generator. Figure 5.1 refers, and is cited a few times in these notes.
Psl + Pf
Pc
Pf
Pw
Psl + Pf
Pw
Pc
V
IL L
Pw
V
I0 L
V
IL L
P
Iom
in
P
Motor
in
P
No load
out
Generator
Nr < N
Nr = N
s +ve
P
s=0
Nr > N
L
P = Pw
s -ve
L
P = (Psl+Pf) +Pc+Pw+P
P
P
=
Pf
in
L
L
in
P = (Psl+Pf) +Pc+Pw+P + Pout
L
!"#$%&'()*'+',-".'/01'.23&%'4-23'+'5262%"0#'/01'#&0&%/6"0#'2.&%/6"0#'521&7'
5.2
Figure 16: Slip and power flow - motoring and generating operating modes
Motor
Operating as a motor, the rotor is loaded mechanically and revolves more slowly than the stator
magnetic field. The slip is typically 0.05 and is positive. Since the magnetic field from the stator
revolves
8 rev/min and the rotor at 89:8);*+7< rev/min, the stator magnetic field revolves around
4.2 at
Motor
the rotor at 8+89':'8);*+;*+7<<=7)8 rev/min. The induced emfs and currents on the rotor thus revolve
at 7)8 rev/min relative to the rotor, which rotates at 8;*+7< rev/min. The magnetic flux due to the
Operating
as ina the
motor,
rotor
is loaded
and revolves
morewords,
slowlyirrespective
than
current
induced
rotorthe
thus
revolves
at 89mechanically
'='7)8:8);*+7=7<:8.
In other
of
stator magnetic
The slip
is typically
0.05 and with
is positive.
Since
theand
magnetic
thethe
operating
slip, the field.
rotor field
revolves
in synchronism
the stator
field
the two fields
interact
to produce
resulting
torque
shown below
(left).
Even though the two fields
field from
the stator
revolves
at as
N rev/min
and in
theFigure
rotor 5.2
at N
R =N.(1-s) rev/min, the
revolve in synchronism, the stator field cuts the rotor conductors. The field from the stator is almost
stator magnetic field revolves around the rotor at N-N = N.(1-(1-s))=s.N rev/min.
linear, radially-inwards through the rotor; the field fromR the rotor current (around the rotor
The induced
emfs and currents
on the rotor
thus revolve
at s.N
rev/min relative
to thethe rotor
conductors)
is cylindrical.
The directions
of these
two fields
concentrate
flux behind
rotor, which
rotatesitatin N(1-s)
magnetic
due the
to the
induced a force
conductors
and reduce
front ofrev/min.
the rotor The
conductor.
As aflux
result,
rotorcurrent
bar experiences
driving
it in
the same
direction at
as N
the
field, and produces
in the
rotor
thus revolves
+ s.N=N.(1-s+s)=N.
In torque.
other words, irrespective of
R stator
the operating slip, the rotor field revolves in synchronism with the stator field and the
two fields interact to produce resulting torque as shown below in figure 17 (left). Even
though the two fields revolve in synchronism, the stator field cuts the rotor conductors.
'
!"#$%&'()>'+'?01$@&1'@$%%&067'/01'62%A$&7'42%'5262%"0#'/01'#&0&%/6"0#'521&7'
For example, a six pole machine on a 50Hz supply operating at 5% slip would have the rotor turning
at 950 rev/min and stator magnetic field cutting the rotor conductors at 50 rev/min. The induced
frequency 49 on the rotor would be 7)4,BCCDE = 2.5 Hz, and the flux pattern from the rotor currents
would revolve at FG49HIH9J. = 50 rev/min relative to the surface of the rotor. The rotor revolves at
Power Systems and Machines
27
The field from the stator is almost linear, radially-inwards through the rotor; the field
from the rotor current (around the rotor conductors) is cylindrical. The directions of
these two fields concentrate flux behind the rotor conductors and reduce it in front of
the rotor conductor. As a result, the rotor bar experiences a force driving it in the same
direction as the stator field, and produces torque.
N
N r/m
Nr
T prod
s +ve
motoring
N r/m
N
N-Nr r/m
+ve
+
Nr
T abs
N-Nr r/m
-ve
s -ve
generating
Figure 17: Induced currents and torques for motoring and generating modes
For example, a six pole machine on a 50Hz supply operating at 5% slip would have the
rotor turning at 950 rev/min and stator magnetic field cutting the rotor conductors at
50 rev/min. The induced frequency fR on the rotor would be s.fSUPPLY = 2.5 Hz, and
the flux pattern from the rotor currents would revolve at 60fROTOR /p = 50 rev/min
relative to the surface of the rotor. The rotor revolves at 950 rev/min, and the induced
magnetic flux on the rotor revolves at 950+50 = 1000 rev/min relative to the stator,
i.e., at the synchronous speed.
The electricity supply system provides all of the active power flowing into and through
the machine which comprises: stator iron losses PF ; rotor and stator copper losses, PC
and PSL ; friction and windage losses PW ; and the shaft output power PL . This is shown
in figure 16 (left).
4.3
No load
If a prime mover (e.g., water or wind turbine) accelerates the rotor up to synchronous
speed, the slip decreases to zero, the induced emf and frequency of the rotor and rotor
currents fall to zero, and the mechanical output power falls to zero. At the synchronous
speed, the engine supplies the shaft mechanical losses (friction and windage), and the
electricity supply provides the stator no load losses (iron losses only, since there are
virtually zero copper losses in the stator). Also, there are no copper losses in the rotor,
since the rotor emf and current is zero. Figure 16 (centre) refers.
Power Systems and Machines
4.4
28
Generator
In the case of the induction generator, the synchronous supply is still required to produce
the revolving stator magnetic field linking with the rotor. If the prime mover drives the
rotor at a speed greater than synchronous, the slip is typically -0.05 and is negative.
Since the magnetic field from the stator revolves at N rev/min and the rotor revolves
at NR =N.(1+s) rev/min, the stator magnetic field revolves around the rotor at N-NR =
N.(1-(1+s)) = -s.N rev/min.
The induced emfs and currents on the rotor thus revolve at -s.N rev/min relative to the
rotor, which rotates at N(1+s) rev/min. The magnetic flux due to the current induced
in the rotor thus revolves at NR - s.N=N(1+s-s)=N.
From the above example, the same six pole machine on a 50 Hz supply operating as a
generator at -5% slip would have the rotor turning at 1050 rev/min and rotor conductors
cutting the stator magnetic field at 50 rev/min. The induced frequency fR on the rotor
would be s.f = -2.5 Hz, and the flux pattern from the rotor currents would revolve at
60fROTOR /p = -50 rev/min relative to the surface of the rotor. The rotor revolves at
1050 rev/min, and the induced magnetic flux on the rotor revolves at 1050-50 = 1000
rev/min relative to the stator, again at the synchronous speed. Thus as a generator,
irrespective of the operating slip, the rotor field still revolves in synchronism with the
stator field and the two fields interact to absorb torque as shown above in 17 (right).
The two fields revolve in synchronism, but the stator field cuts the rotor conductors
in the reverse direction to that operating as a motor. The emf and current in each
rotor conductor reverses, and the directions of the fields concentrates flux ahead of the
rotor conductor and reduces it behind the rotor conductor. As a result, the rotor bar
experiences a force repelling it against the direction of rotation and the machine balances
and absorbs torque from the engine.
Operating as a generator connected to an ac supply, there is still only one source of
magnetising current, and the machine is singly excited. As a result, the magnetomotive
forces either side of the airgap must always balance (N1 I1 = N2 I2 ). Since the rotor
current I2 has changed direction, to maintain an mmf balance across the airgap, the
referred rotor current in the stator winding I2 ’ must also reverse and the phase current
I1 resolves to flow out of the stator.
In response to an input power and torque provided by the prime mover, the machine
absorbs torque and generates electrical power that flows out into the system, provided the stator is magnetised from a synchronous supply (i.e. the supply itself). In
this case, the engine supplies all of the mechanical power PL which is used to overcome friction and windage losses PW and, after conversion, supplies all of the electrical losses
√ in the machine PC , PSL , and PF and produces electrical output power
POUT = 3VL IL cos φWatts. Figure 16 (right) shows this above.
Power Systems and Machines
4.5
29
Phasor diagram operating as a motor and a generator
Figure 12 shows the T equivalent circuit of the induction motor, and figure 13 the current
and voltage phasor diagram for the induction machine motoring - including stator series
voltage drops.
In both cases, the no load current I0 is heavily reactive, at an angle φ0 to the applied
voltage, since it is fixed in magnitude and phase by the real and reactive currents required
to provide iron losses and magnetise the airgap respectively. This current is nearly
independent of mechanical load or input power to the machine. Operating as a motor,
as the mechanical load on the machine increases and as the slip rises, I2 ’ starts to increase
and contribute to the increase in I1 . The angle between V1 and I1 also reduces and the
power factor improves. Figure 18 shows the voltage and current phasor diagram for an
induction machine operating as a generator, based on changes to current flows in the
T equivalent circuit, which is also shown. The referred rotor current in the stator I2 ’
has reversed to become -I2 ’. The machine still requires a source of magnetising current,
and still draws magnetising reactive current I0 from the electricity supply. The output
current I1 from the generator is the vector sum of I0 and -I2 ’.
I1R1
I 1 jX1
E1
I’2
ɸ axis
I1
V1
ɸ1
ɸ2
Ior
Io
ɸo
om
Figure 18: Induction generator Tee equivalent current and voltage phasor diagram
Convention dictates that for operating as a generator the phasor diagram is drawn from
an internal viewpoint, and the stator emf and currents now lead the system voltage.
Note that since the supply current leads the system voltage, the machine generates
active power at a leading power factor.
4.6
Performance operating as an induction generator
As stated earlier, the induction machine will motor or generate either side of zero slip,
depending whether the slip is positive or negative. The machine does not change physically, as the shaft speed goes above the synchronous, and all of the model component
Power Systems and Machines
30
values remain fixed only slip changes. The model for the machine shown in figure 14
therefore continues to represent the machine in the generator mode. Operating as a
generator with negative slip, the expressions for torque, referred rotor current, and gross
mechanical power all change.
They shows that the airgap torque changes to an absorbed torque. The expression for
the rotor active power shows that the “loss” in the resistor representing gross mechanical
power becomes a “negative loss” or an effective power gain, which correctly represents
the machine generating. This being the case, I2 ’ reverses direction and active power
is exported from the machine, since the phase angle of Zt and I2 ’ changes. Note also
that the vector combination of I2 ’ and I0 to find I1 and the output power factor all still
hold, as do the scalar calculations of input power, losses, efficiencies etc. It was noted
earlier that after the rotor of the machine has been accelerated above the synchronous
speed, the prime mover had to supply the friction and windage losses in the machine,
and produce enough active power to supply the internal electrical losses. The increase
in speed results in an increase in friction and windage losses. Since the no load current
is largely independent of electrical load or speed of rotation, the magnetising current
and iron losses remain virtually unchanged. However, as a motor, I2 ’ supplied the stator
and rotor copper and iron losses in the stator and rotor, whereas in the generating mode
I2 ’ has to supply ALL of the electrical losses. I2 ’ is therefore higher and the rotor and
stator copper losses increase slightly (in R1 and R2 ’). The result is that as a generator
the induction machine is around 0.5% less efficient. To produce the same output power,
the shaft input torque has to be higher.
Power Systems and Machines
5
31
Determining equivalent circuit model impedance from
machine test results
5.1
Introduction
The equivalent circuit of an induction machine is a very useful tool for determining the
machine’s operating conditions and response to changes in load. However, if a model
is to represent a machine realistically, it is necessary to determine what the component
values are that go into the model. Values of R1 , RC , R2 , X1 , X2 and XM may be
estimated for a production motor or generator by two methods - either by analysis of
electrical test results, or interpolation between part load performance figures.
5.2
5.2.1
Electrical test method
The no-load test - Rc and Xm
The no-load test of an induction machine measures the friction and windage losses of the
motor, and also provides information about its magnetisation current, and iron losses.
The test circuit is shown in figure 19. Wattmeters W1 and W2 measure the total threephase power taken by the motor on test. Voltmeter V1 measures the line voltage applied,
and three ammeters A1 -A3 measure the RED, YELLOW and BLUE line currents. The
induction motor is not connected to an external mechanical load and is allowed to spin
freely. The only mechanical load on the motor is the friction and windage losses, so all
of the airgap power in the machine is converted to mechanical losses, and the slip of the
motor is very small (possibly as small as 0.001 or less). The equivalent circuit of this
motor operating on no load is also shown in figure 19.
Operating with very small slip, R2 ’.(1-s)/s is much larger than the resistance corresponding to the rotor copper losses R2 ’, and larger than the ohmic value of the rotor reactance
X2 .
R2 ’.(1-s)/s tends towards open circuit. I2 ’, reduces to virtually zero and the equivalent
circuit reduces approximately to the magnetising impedance RC in parallel with XM . At
no-load conditions, the input power measured by wattmeters W1 and W2 must represent
the losses in the motor. With the approximate model, the losses must be dissipated in
RC as stator iron losses. Even on no load, the rotating magnetic field links with and
causes hysteresis losses in the stator frame and laminations.
The total apparent power taken by the motor is:
Stotal = Vp (IR + IY + IB )V A
(5.1)
For a balanced three-phase ac voltage supply:
IR = IY = IB = Iline
(5.2)
Power Systems and Machines
32
Variable voltage and freq
Red
Yellow
Blue
Neutral
VI
A1 A2 A3 V0
W1
W2
Full voltage
I = Ir + Iy + Ib
3
W = W1 + W2
No-load
Nr -> N
Figure 19: No load test equivalent circuit
After substitution:
Ptotal
√
3Vline Iline V A
√
= W1 + W2 = 3Vline Iline cos φ0 W
Stotal = 3Vp Iline or
(5.3)
(5.4)
The no load power factor cos φ0 may be calculated from wattmeter, voltmeter, and
ammeter readings obtained on test and might typically be 0.1 lagging. Phase angle φ0
may be determined from the inverse cosine of no load power factor. The no load current
I0 has real and imaginary components I cos φ0 , and I sin φ0 .
Power Factor =
W1 + W2
No load active power
=√
No load apparent power
3Vline Iline
(5.5)
Since the no-load circuit is the parallel combination of RC and jXM .
Iline cos φ0 =
Vphase
RC
and Iline sin φ0 =
Vphase
jXm
(5.6)
This allows the estimation of RC and XM from the phase voltage and resolved components of no-load current.
RC =
5.3
Vphase
Iline cos φ0
and
Xm =
Vphase
jIline sin φ0
(5.7)
The DC test for stator resistance - R1
R1 must be determined independent of X1 , R2 , X2 . Manufacturers carry out a “dc test”,
where a dc voltage is applied to the stator windings of the induction motor. Because
Power Systems and Machines
33
the current is not alternating, there is no induced voltage in the rotor circuit, and no
resulting rotor current flow. Also, the leakage reactance X1 of the stator is zero under
the effects of direct current. Therefore, the only quantity limiting current flow in the
motor is the stator winding resistance R1 , and that resistance can be determined.
The circuit for the dc test is shown in figure 20. A dc power supply is connected to
two of the three line terminals of a star-connected induction motor. For the test, the
current in the stator windings is adjusted to the rated value, and the voltage between
the terminals is measured. The line current must be adjusted to the rms value of the full
load current in an attempt to heat the windings up to the same temperature they would
be during normal operation (since winding resistance is a function of temperature). The
direct current flows through two of the windings (RED and YELLOW in figure 20), so
the total winding resistance measured is 2xR1 , and R1 may be found from:
R1 =
Vdc
2Idc
(5.8)
Variable voltage dc supply
VI
A1
I dc ~ 1 fl
Current limit
resistor
Figure 20: DC test circuit
5.4
The locked-rotor test
The rotor resistance R2 plays an extremely critical role in the determination of the
performance of an induction motor. Among other things, R2 determines the shape of the
torque-speed curve, and the speed at which the pull-out torque occurs. Manufacturers
would normally carry out a standard motor test called the locked-rotor test, which, in
conjunction with the value of R1 obtained in the dc test, can be used to determine the
rotor circuit resistance and total reactance.
Power Systems and Machines
34
This test is similar to the short-circuit test on a transformer. In this test, the rotor is
locked or blocked, so that it cannot move. An ac voltage is applied to the motor and
the resulting voltage, current and power are measured, using the no-load test circuit as
shown in figure 21. Great care must be taken with the locked rotor test, since there is
no cooling effect from rotating shaft fans. Measurements must be made quickly, before
the rotor and stator windings overheat. To perform the locked-rotor test, a variable ac
voltage is applied to the stator, and the current flow is adjusted to approximately the
full-load value. The voltage, current, and power flowing into the motor are measured.
The equivalent circuit for this test is also shown in figure 21. In this case, since the
rotor is not moving, the slip s = 1. The resistor representing gross mechanical power
R02 .(1 − s)/s reduces in value to zero, since no mechanical power is being taken off the
rotor. All of the power induced into the rotor appears as copper loss, and is represented
by the ohmic losses in R2 ’.
Variable voltage and freq
Red
Yellow
Blue
Neutral
VI
A1 A2 A3 V0
W1
W2
Reduced voltage
I = Ir + Iy + Ib
3
W = W1 + W2
Locked rotor
Nr = 0
Figure 21: Locked rotor equivalent circuit
The total rotor resistance is just equal to R02 , which is quite a small value. Therefore,
the circuit under these conditions looks like a series combination of R1 , R02 , X1 , and
X2 . Value of R1 has already been determined by the dc test. The current taken by
the machine is so much greater than that taken by the machine on no-load, that the
magnetising branch of the equivalent circuit may be neglected when interpreting the test
results.
In normal operation, the stator frequency is the line frequency of the power system (50 or
60 Hz). During this test, the slip is 1, and since fROTOR = s.fSUPPLY , the rotor emfs are
also at line frequency. This determines X02 correctly as the standstill reactance, since the
rotor circuit was redrawn to leave the rotor reactance independent of induced frequency,
at the standstill value.
Power Systems and Machines
35
The input power to the motor can be described by:
√
PLR = 3VLR Iline cos φLR W
(5.9)
and measured by the two wattmeter test method as
PLR = W1 + W2 W
(5.10)
so the locked-rotor power factor, and phase angle φLR between line voltage and standstill
current can be found from the inverse cosine of the active:apparent powers measured as
cos−1 (PLR /SLR )
W1 + W2
−1
√
(5.11)
φLR = cos
3VLR Iline
The input impedance of the motor under locked rotor conditions is:
ZLR =
Vphase
Vline
=√
Iline
3Iline
(5.12)
and the angle of the total impedance is φLR . Therefore, the locked rotor input impedance
can be separated into active and reactive components as follows
Vline
Vline
cos φLR + j √
sin φLR
ZLR = RLR + jXLR = √
3Iline
3Iline
(5.13)
The total locked-rotor resistance RLR is equal to the sum of R1 and R2 ’. Hence R2 ’ may
be calculated as
Vline
R2 = √
cos φLR − R1
(5.14)
3Iline
The locked-rotor reactance XLR is equal to the sum of the stator leakage reactance X1
and the referred value of the rotor standstill reactance X2 ’. Thus
XLR = X1 + X20
(5.15)
This total reactance is the inductive component of the measured locked-rotor impedance,
so
Vline
X1 + X20 = √
sin φLR
(5.16)
3Iline
Unfortunately, there is no simple way to separate the stator and rotor reactances from
each other. In normal practice, it really does not matter just how XLR is broken down,
since the reactance appears as the sum of X1 and X2 ’ in all the torque equations. In most
industrial induction motors and generators, X1 and X2 ’ may be taken as equal.
The equivalent circuit component values determined as above may be substituted into
the expressions for torque, current, power factor, efficiency and power, and the variation
of each of these parameters evaluated as slip varies with load or output power.
Power Systems and Machines
6
36
Operation of induction machines on the electricity supply network
6.1
Introduction
While the induction machine is recognised as the simplest, most rugged and most reliable
rotating electrical machine, it requires to be selected carefully and its performance should
be analysed before it may be applied to a specific task. The approximate equivalent
circuit represents, with reasonable accuracy, how the machine behaves in the steady
state, and may be modified to consider particularly onerous conditions, such as starting
or overloading of the machine.
6.2
Connection and starting of induction motors
During the start-up, an induction motor may draw up to six times its full load current.
This effect is particularly serious for two related reasons:
(i) Connection to weak grid systems with high series impedance and reduced threephase symmetrical short circuit fault level will give rise to terminal voltage drop,
causing
(ii) Reduced torque margins at start (danger of failure to start) and rotor overheating
during stalled conditions.
Consider the equivalent circuit of the induction motor (as shown in figure 22) at standstill, suddenly connected to a grid system which may be considered to have infinite
strength (i.e. zero grid impedance).
I start
Z grid
Vinf
V1
reduced
R 1 +R’2
Z start
X 1 +X’2
R’2 (1-s)
s
I’ 2
Figure 22: Induction motor suddenly connected to infinite grid (Zgrid = 0)
At the instant of connection, the rotor is stationary, and s=1. The power taken off the
rotor is zero. The resistor in the equivalent circuit representing gross mechanical power
R02 (1 − s)/s is shorted out. The referred rotor impedance reduces from say (20.R02 +j.X02 )
when s=0.05 to (R02 +j.X02 ) at s=1. In the first few revolutions of the rotor, while s is
Power Systems and Machines
37
close to 1, I2 ’ is much greater than I0 , so the equivalent circuit reduces to that shown
above. The rotor current I02 may rise by a factor of 5 or 6, depending on the values of
R02 and X02 .
The input impedance at switch-on becomes Zstart = (R1 + R20 ) + j(X1 + X20 ) and the
instantaneous current drawn is given by:
Istart = p
V1
6
(R1 + R2 )2 + (X1 + X2 )2
tan−1
X1 + X20
R1 + R20
(6.1)
If the three-phase supply is very ‘stiff’ (i.e., strong) and has a high fault level, approaching
the capacity of infinite busbars, Zgrid ≈ 0, then V1 ≈ Vgrid at all times during starting,
irrespective of the current drawn. Since starting torque is proportional to V21 , and V1
has not fallen appreciably, the motor will accelerate up to speed, R2 ’/ s will increase, and
the current Istart will decrease to the full load value at the end of a successful start.
If the fault level of the three-phase supply is low, due to the point of connection being
geographically or electrically remote from a high voltage, then Zgrid may be high. All
transmission lines, transformers, and cables add their series impedances into Zgrid , and
particularly in rural areas with high series impedance, there may be problems starting
large induction machines. Under normal conditions, the full load voltage drop across
Zgrid must typically be less than 5% of the statutory supply voltage. Drawing motor full
load current from a weak grid supply should leave Vstart > 0.95Vgrid . The voltage drop
across the grid impedance in the equivalent circuit shown in figure 23 becomes
Vdrop = I1 Zgrid < 0.05Vgrid
Volts drop
Z grid
Vinf
(6.2)
I start
V1
reduced
R 1 +R’2
Z start
X 1 +X’2
R’2 (1-s)
s
I’ 2
Figure 23: Induction motor suddenly connected to a weak grid (Zgrid > 0)
When starting an induction motor on a weak grid, the increased voltage drop across
Zgrid associated with 6x full load current can be appreciable
Istart = 6.I1 and Volts drop = Istart Zgrid = 6.I1 .Zgrid ≈ 6.0.05.Vgrid ≈ 0.3.Vgrid
(6.3)
Vstart = Vgrid − Vdrop = Vgrid .(1 − Istart Zgrid ) = Vgrid .(1 − 0.3) = 0.7Vgrid
(6.4)
Power Systems and Machines
38
This would reduce the voltage at the motor terminals to Vstart = 0.7.Vgrid . Since the
torque produced is proportional to the square of the terminal voltage, the torque produced would fall to 0.49 of the normal starting torque, at a time when accelerating
torque is most needed. The motor would almost certainly stall, and would remain in
or will return to the locked rotor condition, or - equally serious - would run up to and
‘creep’ at a reduced speed.
Note that the effect of this individual machine would reduce the supply voltage to all
other machines on that voltage node, busbar or site - all of whose output torques would
reduce and they could also stall. Light intensities would also fall with the square of the
reduced voltage. Without correct power supply back-up, computer supplies may fail,
and process control may be lost. This voltage drop is also known as the “voltage sag”
(or “voltage dip”) due to motor starting.
For accurate estimation of expected voltage drops during normal operation and starting,
comprehensive load flow studies should be carried out using detailed system calculations,
usually in software packages. Initially, however, it is useful to make less accurate estimates of voltage drop to determine if more detailed studies are necessary. There are
three techniques: 1 is the best method, 2 and 3 are less accurate, but may be used for
quick checks.
1. Calculate series voltage drop, as above, where system impedance and motor starting current are known. This may also be done in per-unit.
Vstart = Vgrid − Vdrop = Vgrid .(1 − Istart Zgrid ), all as vectors.
2. At the start, the machine reactances dominate as the magnetic fields are set up
and reactive potential division may be used, which ignores the effects of rotor and
stator series resistance (since they are very low) and stator core loss resistance
(since it is high).
3. Where least of all information is available, an empirical ratio method may be used.
At the instant of switch-on, the magnetising reactance Xm of the stator has to be charged
up (Xm initially looks like a short circuit), to produce the rotating magnetic field, although this effect lasts only a few cycles of the supply waveform. The starting reactance
applicable during this time may be estimated from the formula below, which is derived
directly from the approximate model, where series and magnetising reactances are in
parallel.
(X1 + X20 )Xm
6 + 90◦
Xstart =
(6.5)
X1 + X20 + Xm
Assuming the grid impedance to be entirely reactive, the lowest voltage on the motor
during start may be estimated by reactive potential division as shown in figure 24.
Vstart = Vgrid
Xstart
Xstart + Xgrid
(6.6)
Power Systems and Machines
39
I1
Z grid
Vinf
V start
X 1 +X’2
Xm
I om
Figure 24: Estimated voltage drop - reactive potential division
For example, a 2500 HP (1865 kW) motor might operate with full load efficiency and
power factor of 0.945 and 0.84 lagging respectively. On a stiff 2300V supply, the full
load current taken would be 589 A, and the motor base MVA would be 2.35 MVA
(1.865x106/0.945/0.84). The machine is to be connected to the 2.3 kV grid at a point
where the fault level is around 20 MVA. The ohmic value of the reactance of the 20 MVA
grid reactance at that point may be calculated from
Xgrid
2
2
kVphase
Vphase
kVline
=
=
=
= 0.265 ohms/phase
Ifault
M V Afault ((phase)
M V Afault ((total)
(6.7)
The reactances suggested by analysis of the machine test results are X1 = 0.403, X02 =
0.35 and XM = 7.96 ohms/phase. The starting reactance Xstart could be estimated as
above to be 0.687 ohms.
√
Phase voltage of this machine is 2300/ 3 = 1328 V. If the machine was supplied from
the infinite grid, the starting current would be 1328/0.687 = 1933A (only 3.28 times full
load!). Due to the series impedance of the weak grid at 0.265 ohms/phase, the starting
current drawn is reduced to 1328/(0.687+0.265) = 1409 A (now 2.39 times full load).
Torque is also proportional to the square of the current drawn, and the starting torque
would fall to (2.39/3.28)2 or 0.732 = 53%. Reactive potential division suggests that the
starting voltage would fall to Vstart = 0.72Vgrid .
Calculating the series voltage drop due to the increased current at start would bring
about a voltage drop of Istart Zgrid = 1409.0.265 = 373V or down to 1328-373 = 954V
or 73% of nominal. This is only valid if we assume that Istart lags the phase voltage at
-90 degrees and Zgrid is wholly reactive at +90 degrees. Then, the voltage drop is in
phase with the phase voltage and may be deducted arithmetically. In reality, the network
impedance will have finite resistance and an identified X:R ratio, and the starting current
may not lag by as much as 90 degrees.
Finally - if only the nameplate details of the motor are available, another technique may
be used to estimate the voltage drop at start. The per-unit impedance of the three-phase
Power Systems and Machines
40
supply may be calculated to the base MVA of the grid
Zgrid (pu) =
2
kVgrid
where MVAgrid is fault level
M V Agrid
(6.8)
Since the full load current taken by the motor is
M V Amotor
Pload
= √
Ifull load = √
ξ 3Vline cos φ
3Vline
(6.9)
and at start the line current rises in the ratio SC/FLC, the starting MVA of the motor
approximates to
M V Astart =
√
3Vline
SC
SC
Iline =
M V Amotor
F LC
F LC
(6.10)
The full load, or motor base impedance may be calculated using equation 6.7. The
impedance of the motor, falls during start to the following value:
Zstart =
2
2
kVline
kVline
F LC
= Zmotor
=
SC
M V Astart
SC
M V Amotor
F LC
(6.11)
Thus, to its own base the per unit impedance of the motor falls to
F LC
SC
(6.12)
F LC M V Agrid
.
SC M V Amotor
(6.13)
Zstart (pu) =
or changing this to the base MVA of the grid
Zstart (pu) =
The voltage drop may now be estimated as a convenient rule-of-thumb by potential
division, only this time using equivalent impedances for the grid and the motor as it
starts.
Zstart
Zstart + Zgrid
1
Vstart (pu) = Vgrid
=
=
=
1+
1
Zstart
1+
SC M V Amotor
F LC M V Agrid
1
1
= 73%
1 + 3.28. 2.35
20
(6.14)
(6.15)
(6.16)
(6.17)
For the above induction motor, which might be expected to draw 3.28 x FLC on full
voltage and have base MVA = 2.35 on full voltage, the voltage drop predicted by this
method would be also be 73%.
%%
Z start
1
1
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!
start
grid
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1.0
SC MVA motor
Z start # Z grid
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%
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Power Systems and Machines 1
41
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/+$% 6'/'5%
7$=$9'4%
4$"R% /'5@,$% "/% "% -904% '(% % WETZE% !% Y% O"/% -/"27-/099.% '5% 7,502#%
Nominally
R2 ≈1099%
0.1X
2 , and the motor develops peak torque at a slip of R2 /X2 ≈
-/"5/?,4Q.%"27%0/%1099%-/"5/%6'5$%$(($8/0=$9>A%%\'1$=$5.%0(%/+$%"770/0'2"9%-/"5/02#%5$-0-/"28$%0-%5$7,8$7%"-%
0.1, or at a speed of 90% of synchronous. If additional resistance around 0.9X2 is added
/+$%6'/'5%5,2-%,4%/'%-4$$7.%4$"R%/'5@,$%8"2%3$%6"02/"02$7%"99%/+$%1">%,4%/'%2'56"9%-4$$7.%1+$2%/+$%
externally across the slip-rings in the form of variable rotor resistance, the motor will
-+'5/?8058,0/%-10/8+%0-%89'-$7%"27%/+$%6'/'5%3$+"=$-%"-%"%2'56"9%027,8/0'2%6"8+02$A%
develop peak torque at a slip of R2 /X2 ≈ 1 (at standstill, or during start-up), and it will
start more effectively. However, if the additional starting resistance is reduced as!"#$%&%'(%)%
the
motor runs up to speed, peak torque can be maintained all the way up to normal speed,
when the short-circuit switch is closed and the motor behaves as a normal induction
machine.
Power Systems and Machines
6.4
42
Power factor correction of induction motors
Since the induction motor must always draw its magnetising current from the ac supply
to produce the rotating magnetic field, there will always be a fixed amount of lagging
VArs taken from the supply. The current and voltage phasor diagram for the complete
machine (i.e., the rotor currents are not separated out) is shown below in figure 26,
together with the active and reactive power diagram.
V1
V1
ɸ
ɸ
I1
1
ɸ
ɸ
2
0
I0
V1
s
ɸ
req
I1
P
I’
I cap
I or
I om
I 1 sin ɸ
req
Q
1
I 1sin ɸ1
Figure 26: Phasor diagrams
On no load, IOM = I0 sin φ0 is the reactive component of current which magnetises the
machine
√ and IOR = I0 cos φ0 is the active component of current which produces real
power 3Vline I0 cos φ0 which drives the rotor at no-load speed, overcoming friction and
windage losses, and supplies the iron losses in the machine. Since I0 sin φ0 >> I0 cos φ0 ,
the power factor at no load is poor, sometimes as low as 0.1 lagging. As the machine is
loaded mechanically, it takes a larger active component of current. Iline cos φ1 increases
with load and the power factor improves as φ1 gets smaller.
Most induction motors are designed to operate at 0.9 or 0.95 power factor lagging. On
fractional/lower load, the power factor will reduce from 0.9. Electricity supply authorities prefer to sell active power (kW), and would prefer to only generate enough reactive
power (kVAr) for their own system compensation. Industrial plants with high induction
motor loads (operating at fractional output) present a large demand for magnetising
VArs. The supply authorities operate higher tariffs as the power factor of a consumer
falls from unity through 0.9 and below.
Where a number of machines, cables, and resistive loads exist on a supply, it is possible
to estimate the individual real and reactive power requirements. The overall site VAr
demand is simply the sum of the capacitive VArs for cables and transmission lines less
the sum of the inductive VArs for motors. The overall site Watt demand is simply the
sum of all of the active power requirements of the motors and connecting/transmission
losses. The total site apparent power requirement may be calculated from these figures
Power Systems and Machines
43
by rectangular resolution. The site power factor is then established as the ratio of
active/apparent power. If the VAr demand is on balance positive, the site power factor
will be capacitive or leading, but this is quite unusual. If the VAr demand is on balance
negative or inductive, the site power factor will be lagging. This is more usually the case
and improving the power factor of a site is economically prudent, and may be done in
two ways; by:
(i) static capacitor power factor correction
(ii) installing synchronous - induction motors (SIMs)
SIMs are synchronous induction motors with wound rotors, and slip-rings. They are
accelerated to near synchronous speed with rotor-resistor controllers, and run-up as
induction motors. Near synchronous speed, excitation is applied, and the machine synchronises (as a standard synchronous machine). Thereafter the unit, or site power factor
may be improved by overexciting the motor to export VArs to other inductive site loads
and to the grid.
Using static capacitors to correct individual machines, the magnetising current and VArs
for full load power is reduced by connecting power factor correction capacitors across
the machine terminals, either in star or delta form, as shown in figure 27.
Red
Yellow
Blue
Neutral
C
Xm
V1
Xc V
c
Correction
capacitors
Induction
motor
Figure 27: Power Factor Correction
The uncorrected power factor may be taken from the full load power factor stipulated
on the nameplate. Otherwise, the operating power factor and phase angle 1 is easily
determined from the machine details, provided efficiency ξ may be estimated or taken
Power Systems and Machines
44
from test certificates, using
PLOAD
φ1 = cos−1 √
ξ 3Vline Iline
(6.18)
The phase angle of the required corrected power factor is determined from
φreq = cos−1 (P Frequired )
(6.19)
Power factor correction will not adjust the real power taken by the motor, which is fixed
by the applied mechanical load and the full load efficiency, P1 and I1 cos φ1 are fixed,
but
Qreq
Qin
φ1 = tan−1
and
φreq = tan−1
(6.20)
Pin
Preq
Thus
Qin =
and
Qreq =
√
√
3Vline Iline sin φ1 = Pin tan φ1
(6.21)
3Vline Ireq sin φ1 = Pin tan φreq
(6.22)
From the active/reactive power diagram in figure 26 above, the reduction in lagging
VArs necessary to correct to the new power factor is provided by connecting capacitors
at the machine terminals to excite leading capacitive VArs
Capacitive VArs = Qin Qreq = Pin (tan φ1 − tan φreq )
(6.23)
For example, a 100 kW (mechanical power) motor operating at 92% efficiency, 0.9 power
factor on a 400 V supply would take a line current of 174 A, at a lagging phase angle of
25.8 degrees. This is a total magnetising demand of 52.5 kVArs (17.55 kVArs/phase),
and a fixed demand for electrical active power of 108.7 kW. To improve the power factor
to 0.95, the new phase angle would be 18.2 degrees, at the same active power. For this
new power factor the capacitive VArs = 16.8kVArs.
Thus, 5.6 kVArs of capacitive load must be connected at each of the motor terminals, per
phase. The capacitors would operate at 230 V, each taking a 90 degree leading current
of 24.3 A, with a reactance of j9.44Ω at 50 Hz. (30mF/phase).
There are two operational problems with power factor correction. If the power factor is
corrected for full load operation, say to 0.95, the combined power factor will go more
leading as the load on the motor reduces. To maintain 0.95 power factor, the capacitors
might need to be in switchable banks. Such equipment becomes complex and expensive if overvoltages and discharge transients are to be avoided. More significantly - if
the power factor was corrected to near unity, this implies the inductive and capacitive
reactances would be nearly equal, and resonance might occur. Large currents would
resonate between the capacitors and parallel stator inductance, inducing high voltages
on the stator. Self-excitation must be avoided on induction motors by fitting overvoltage
protection relays.
Power Systems and Machines
7
45
Steady state behaviour of synchronous machines
7.1
Revision of synchronous machine operation
Synchronous machines are doubly excited, as shown in figure 28, where the rotor body
is magnetised by the injection of direct current into the field winding, and it is possible
to control the behaviour of the machine by adjustment of the magnetic field in the
airgap.
Like induction and dc machines, the synchronous machine has the ability to accept or
produce mechanical power on the rotor and produce or consume electrical power at
the stator terminals. All synchronous machines will move from motoring to generating
conditions freely. Traditionally referred to as alternators (because of the dc on the rotor
and ac in the stator), synchronous generators rely on the following principles in the
production of electrical power at the stator terminals:
1. The creation of a magnetic flux pattern in the airgap of the generator, which links
with the stator winding,
andSYSTEMS
revolves AND
at synchronous
POWER
MACHINES speed;
2. The delivery
of input power to the rotor shaft from a prime mover power source;
Behaviour of Synchronous
Machines
3. The induction of a three-phase sinusoidal voltage waveform in the stator winding;
hronous Machine Operation
s are doubly excited, as shown in Figure 8.1, where the rotor body is magnetised
4. The connection of a three-phase load or grid to the stator terminals to which the
ct current into the field winding, and it is possible to control the behaviour of the
generated
t of the magnetic field in the
airgap. power is delivered
Stator field N rev/min
machines, the synchronous machine has the
Rotor
N
roduce mechanical power on the rotor and
electrical power at the stator terminals. All
s will move from motoring to generating
ditionally referred to as alternators (because
otor and ac in the stator), synchronous
he following principles in the production of
stator terminals:
a magnetic flux pattern in the airgap of the
h links with the stator winding, and revolves
speed;
input power to the rotor shaft from a prime
urce;
a three-phase sinusoidal voltage waveform
Rotor field - dc - magnetises
ding;
the rotor and two fields align
of a three-phase load or grid to the stator
in attraction
ch the generated power is delivered
Figure 8.1
- Synchronous
machine-doubly
excited
Figure
28: Synchronous
machine-doubly
and manufactured to have the same number of poles as the stator and its
ld. The field (i.e. rotor) coils are energised with direct current, and produce a
which is stationary relative to the rotor body. The rotor coils, which are
are wound consecutively clockwise and anticlockwise such that the flow of
duces alternate North and South poles around the circumference of the rotor, as
ds
the magnetic attraction of adjacent North and South poles in the revolving stator
nding S and N poles on the rotor. For this to occur, the rotor must always rotate
s speed (in the steady state). The torque ! produced (or absorbed) at this speed
anical output power of the machine operating as a motor, or absorbed as a
o-load conditions, the magnetic fields on the rotor and stator align exactly.
cal load to the rotor causes it to slow down momentarily and the rotor axis falls
s, to enter a steady state at the retarded load angle ". In this condition, the
the stator field produces a torque which just balances the applied load torque:
rque by backward adjustment of load angle ". If an engine contributes power to
ne operating as a generator, the rotor accelerates momentarily to forward load
excited
machine by adjustment of the magnetic field in the airgap.
Stator field N rev/min
Like induction and dc machines, the synchronous machine has the
Rotor
N
ability to accept or produce mechanical power on the rotor and
produce or consume electrical power at the stator terminals. All
synchronous machines will move from motoring to generating
and
Machines
46
conditionsPower
freely. Systems
Traditionally
referred
to as alternators (because
of the dc on the rotor and ac in the stator), synchronous
generators rely on the following principles in the production of
electrical power at the stator terminals:
7.2 DC injection
1.
The creation of a magnetic flux pattern in the airgap of the
generator, which links with the stator winding, and revolves
at synchronous
The rotor isspeed;
designed and manufactured to have the same number of poles as the stator
2.
The and
delivery
of input power
to the
rotor
shaft
from
its revolving
magnetic
field.
The
field
(i.e.a prime
rotor) coils are energised with direct
mover power source;
current, and produce a magnetic flux pattern which is stationary relative to the rotor
3.
The induction of a three-phase sinusoidal voltage waveform
Rotor field - dc - magnetises
body.
The
rotor coils, which are connected in series, are wound consecutively
clockwise
in the
stator
winding;
the rotor and two
fields align
anticlockwise
such that the
flow
excitation
current produces
alternate North and
4.
The and
connection
of a three-phase
load
or ofgrid
to the stator
in attraction
terminals
which
the generated
power is delivered
South to
poles
around
the circumference
of the rotor, as shown in figure 29.
Figure 8.1 - Synchronous machine-doubly excited
8.2
DC injection
The rotor 7.3
is designed
and manufactured
Alignment
of fields to have the same number of poles as the stator and its
revolving magnetic field. The field (i.e. rotor) coils are energised with direct current, and produce a
magnetic flux pattern which is stationary relative to the rotor body. The rotor coils, which are
Torque is produced by the magnetic attraction of adjacent North and South poles in the
connected in series, are wound consecutively clockwise and anticlockwise such that the flow of
stator field
with North
the corresponding
S and
N poles
on the rotor. ofFor
to as
excitation revolving
current produces
alternate
and South poles
around
the circumference
thethis
rotor,
occur, 8.3.
the rotor must always rotate at exactly synchronous speed (in the steady state).
shown in Figure
The torque
! produced (or absorbed) at this speed determines the mechanical output
8.3
Alignment
of fields
power of by
thethe
machine
operating
as ofa adjacent
motor, or
absorbed
as a poles
generator.
On true noTorque is produced
magnetic
attraction
North
and South
in the revolving
stator
field with the
S and
N polesfields
on the
For this
occur,align
the rotor
mustApplication
always rotate
loadcorresponding
conditions, the
magnetic
onrotor.
the rotor
andtostator
exactly.
at exactly of
synchronous
(in the
the rotor
steadycauses
state).it The
torque
! produced
(or absorbed)
thisaxis
speed
mechanicalspeed
load to
to slow
down
momentarily
and the at
rotor
determines
thebehind
mechanical
outputaxis,
power
of thea machine
operating
a motor,load
or absorbed
as a
falls
the stator
to enter
steady state
at theasretarded
angle ”. In
generator. On true no-load conditions, the magnetic fields on the rotor and stator align exactly.
this condition, the increased attraction to the stator field produces a torque which just
Application of mechanical load to the rotor causes it to slow down momentarily and the rotor axis falls
balances
the applied
torque:state
the at
motor
torque
by backward
behind the
stator axis,
to enterload
a steady
the produces
retarded load
angle
". In this adjustment
condition, the
of
load
angle.
If
an
engine
contributes
power
to
the
rotor
of
the
machine
operating
as
increased attraction to the stator field produces a torque which just balances the applied
load torque:
the motoraproduces
torque
by
backward
adjustment
of
load
angle
".
If
an
engine
contributes
power
generator, the rotor accelerates momentarily to forward load angle where the applied to
the rotor of
the machine
operating
a generator,
the in
rotor
momentarily
torque
is balanced
by theas
force
of attraction
theaccelerates
extended magnetic
field.to forward load
angle " where the applied torque is balanced by the force of attraction in the extended magnetic field.
No load - stator field and rotor
poles aligned - " = 0
"
N
N
S
S
NS
S N
S
N
S
N
N
S
S
N
On load as a motor - rotor poles
displaced backwards - " negative and
airgap torque produced
Figure 8.3 - Rotor and stator fields and airgap forces – for a 4-pole machine
Figure 29: Rotor and stator fields and airgap forces for a 4-pole machine
The analogy may be made that the force acting between the poles resembles a fictitious
Page 1 of 5
“spring”, whose restoring force is proportional to the extension from the no-load or
Power Systems and Machines
47
relaxed state. Figure 29 shows the stator and rotor fields linked and displaced.
7.4
Static excitation
In a direct or statically excited synchronous machine, the rotor field coils are connected
via slip-rings to the excitation supply. A solid state automatic voltage regulator (AVR)
acts directly into the main field connections, and controls the field current to control the
airgap magnetic flux. The excitation power is supplied from either the machine busbars
or the grid derived supplies. A controlled thyristor bridge produces direct current which
is fed to the rotor slip-rings. Adjustment of the firing angle of the thyristors in the
rectifier controls the excitation voltage and current, varying the flux in the machine
airgap and adjusting the terminal voltage of the machine. Most excitation systems
incorporate automatic voltage regulators which automatically control generated voltage
by adjustment of excitation current. A static excitation scheme is shown schematically in
30. Static excitation schemes suffer the disadvantage that they require auxiliary supplies
to establish excitation and start the machine. They require additional equipment for
“black-starting” the generating unit. An operational disadvantage of static excitation is
that the slip-rings and brushes require regular maintenance.
Main generator
Rotor
Sliprings
Power
transformer
Thyristor
converter
Figure 30: Static Excitation
7.5
Brushless excitation
In this case, permanent magnet generator (pmg) is driven by the shaft of the main generator. The permanent magnet rotor induces an emf in its wire wound stator, such that
by the time the generator has reached synchronous speed, full excitation voltage is available. The brushless system has the ability to “black-start” and energises busbars in an
Power Systems and Machines
48
isolated power supply system. The pmg output voltage is rectified by the AVR thyristor
bridge and applied to the stationary winding of the ac exciter, creating a stationary field
in the stator of the exciter. The rotor or armature of the exciter revolves within this
field and induces an alternating voltage on the rotor winding. Shaft mounted, rotating
diodes rectify this voltage and apply it to the main field as the excitation voltage. If
the statically excited system loses its power source, it cannot force excitation to maintain terminal voltage. During voltage dips on the connected electrical system, however,
the brushless system has capacity to boost excitation and maintain terminal voltage.
In automatic voltage regulation, an accurate voltage sensing transformer continuously
measures the terminal voltage of the generator. The output of the sensing VT is fed to
the AVR where it is compared with an adjustable reference voltage. The firing angle of
the thyristor bridge is controlled by the error between the reference and output voltage
signals. Adjustment of a voltage setting potentiometer varies the reference and provides
control of the set voltage. Figure 31 shows a brushless excitation system, including the
power flow from pmg to the main field winding.
Permanent
magent
generator
Exciter
Rotating
rectifier
~
Main generator
M
Manual
A
Auto
AVR
Figure 31: Brushless excitation system
7.6
Connection to infinite busbar
To begin to understand or analyse the operation of the synchronous machine, it must
be connected to an “infinite busbar”. Any grid system may be represented as a source
of emf E grid and finite grid and cable impedance up to the point of connection, Figure
32 refers.
d with an adjustable reference
e of the thyristor bridge is
between the reference and
djustment of a voltage setting
reference and provides control
gure 8.5 shows a brushless
ing the power flow Power
from pmg
Systems
and Machines
49
Figure 8.5 - Brushless excitation system
e busbar
d or analyse the operation of the synchronous
ected to an "infinite busbar". Any grid system may
ce of emf Egrid and finite grid and cable impedance
Xcable
Xgrid
Vgrid
tion, Figure 8.6a on the right refers.
Egrid
alised grid connection consists of an infinitely large
system, connected to a zero impedance busbar grid
. While an infinitely large number of synchronous
Figure 32: Connection to infinite busbar
Page 2 of 5
The infinite busbar or idealised grid connection consists of an infinitely large synchronous
generating system, connected to a zero impedance busbar grid by zero impedance links.
While an infinitely large number of synchronous machines is impossible in practice, a
large national grid, such as may be found in developed countries, has enough synchronous
generators operating in parallel that their capacities add to be several GVA. In addition,
their reactances are very low, connected to the busbar by low cable reactances. The
busbar connects the (generator+cable) reactances up to the busbar in parallel, and the
resulting overall impedance approaches zero. Figure 33 refers.
In a finite system, the addition of
a complex load demands additional active (kW) and
POWER SYSTEMS AND MACHINES
reactive (kVAr) power from the connected generators. Even correctly governed, the
ossible in practice, a large national grid, such as may be found in developed countries,
driving
the generators
will slow add
down
to GVA.
produce
chronous generatorsengines
operating
in parallel
that their capacities
to slightly
be several
In the additional active
power.
As a result,
the
generated
frequency
falls.
eactances are very low,
connected
to the
busbar
by low
cable reactances.
The busbar
generator+cable) reactances up to the busbar in parallel, and the resulting overall
oaches zero. Figure 8.6b refers.
tem, the addition of a complex load
onal active (kW) and reactive (kVAr)
e connected generators. Even correctly
engines driving the generators will slow
produce the additional active power. As a
rated frequency falls.
Figure 8.6b - Infinite grid
Xn
X3
X2
X1
Vgrid
Egrid
reactive power resolves with that for active power to required apparent power at the
Figure 33: Infinite grid
on. Under the applied voltage, the apparent power is provided by current flowing into
urrent must flow through the series reactance of the grid between the sources of emf
The
demand series
for reactive
power
withinternal
that forimpedance
active power
connection. There is
a resulting
voltage
drop,resolves
due to the
of to required apparent
s made up of generator,
transformer,
busbar,
cable
and
line
impedances
connected
in
power at the point of extraction. Under the applied voltage, the apparent power is
el.
provided by current flowing into the load. That current must flow through the series
reactance of the grid between the sources of emf and the point of connection. There is
bar system (in theory) contains an infinite number of synchronous machines, whose
a resulting
series
voltage drop,
due toup
thetointernal
impedance
of the grid, which is made
bine in parallel to become
zero. The
connecting
impedance
the point
of connection
generator,
transformer,
cable
and line
impedances
e grid can generate up
or of
absorb
an infinite
amount busbar,
of active
(Watts)
or reactive
(VAr)connected in series and
he voltage of the busbar
changing. The infinite number of generators also have infinite
parallel.
ed kinetic energy. The additionally demanded active power is instantaneously provided
an infinite
number
of synchronous makinetic energy, but,The
dueinfinite
to the busbar
infinite system
inertia, (in
the theory)
speed ofcontains
the generators
does
not
nite grid can therefore
produce
or reactances
absorb infinite
amounts
of active
power without
the connecting impedance
chines,
whose
combine
in parallel
to become
zero. The
grid changing.
ve of the active or reactive power drawn by a connected machine, the terminal voltage
t the point of connection remains fixed. The terminal voltage of the ideal synchronous
ted to the grid is fixed. The frequency and speed of rotation of the machine are also
first premise in the understanding of synchronous machines.
x
chine, the stator windings have negligible
eakage reactance. From Figure 8.7a this
vector sum of internal emfs and voltage
ays be exactly equal to Vp, and since the
ected to the infinite busbar, internal emf
be fixed since Vp cannot change. From
Vgrid
Vp
POWER SYSTEMS AND MACHINES
mpossible in practice, a large national grid, such as may be found in developed countries,
nchronous generators operating in parallel that their capacities add to be several GVA. In
reactances are very low, connected to the busbar by low cable reactances. The busbar
Power Systems
andbusbar
Machines
(generator+cable) reactances
up to the
in parallel, and the resulting overall
proaches zero. Figure 8.6b refers.
50
Xn
X3
ystem, the addition of a complex load
X2
X1
to the
point (kVAr)
of connection is zero, and the grid can generate or absorb an infinite
tional active (kW)upand
reactive
amount Even
of active
(Watts) or reactive (VAr) power, without Vgrid
the voltage of the busbar
he connected generators.
correctly
engines driving the
generators
will
slow
changing. The infinite number of generators also have infinite inertia and stored kinetic
o produce the additional
active
power.
As a
Egrid
energy.
The
additionally
demanded active power is instantaneously
provided from the
erated frequency falls.
stored
kinetic
energy,
but,
due
to
the
infinite
inertia,
the
speed
of
the
generators does
Figure 8.6b - Infinite grid
not change. The infinite grid can therefore produce or absorb infinite amounts of active
or reactive power resolves
with thatthe
for frequency
active power
to required
apparent power at the
power without
of the
grid changing.
tion. Under the applied voltage, the apparent power is provided by current flowing into
Thus, irrespective
of the active
orgrid
reactive
power
byofaemf
connected machine, the
current must flow through
the series reactance
of the
between
thedrawn
sources
of connection. Thereterminal
is a resulting
series
voltage
drop,
due
to
the
internal
impedance
of fixed. The terminal
voltage and frequency at the point of connection remains
h is made up of generator,
transformer,
busbar,
cable
and
line
impedances
connected
in
voltage of the ideal synchronous machine connected to the grid is fixed. The frequency
allel.
and speed of rotation of the machine are also fixed. This is the first premise in the
understanding
of synchronous
machines.
usbar system (in theory)
contains an
infinite number
of synchronous machines, whose
mbine in parallel to become zero. The connecting impedance up to the point of connection
he grid can generate or absorb an infinite amount of active (Watts) or reactive (VAr)
the voltage of the busbar
changing.flux
The infinite number of generators also have infinite
7.7 Airgap
red kinetic energy. The additionally demanded active power is instantaneously provided
d kinetic energy, but, due to the infinite inertia, the speed of the generators does not
In the ideal machine, the stator windings have negligible resistance and leakage reactance.
finite grid can therefore produce or absorb infinite amounts of active power without the
From figure 34 this means that the vector sum of internal emfs and voltage drops must
he grid changing.
always be exactly equal to Vp , and since the machine is connected to the infinite busbar,
Faradays Law
to the rate of
ive of the active or internal
reactive power
a connected
machine,
terminal
voltage From
emf Epdrawn
mustbyalways
be fixed
since Vthe
change.
p cannot
at the point of connection remains fixed. The terminal voltage of the ideal synchronous
(equation 1.1), the induced emf in the stator winding is proportional
ected to the grid is fixed. The frequency and speed of rotation of the machine are also
change
of airgapofflux
linkage. machines.
he first premise in the
understanding
synchronous
ux
achine, the stator windings have negligible
leakage reactance. From Figure 8.7a this
e vector sum of internal emfs and voltage
ways be exactly equal to Vp, and since the
nnected to the infinite busbar, internal emf
s be fixed since Vp cannot change. From
, the induced emf in the stator winding is
the rate of change of airgap flux linkage:
Vgrid
Egrid
Grid
d"g
dt
Vp
Ep
Sync machine
! N (rev/min)."g
Figure
34: Infinite grid and ideal sync machine
Figure 8.7a - Infinite grid and ideal sync machine
The total
flux in thespeed
airgapNφand
speed
N and
g revolves
n the airgap "g revolves
at synchronous
inducesat
Epsynchronous
, which is held
constant
induces Ep , which
held are
constant
by the
infinite
grid.
Frequency
and speed
are also
grid. Frequency andisspeed
also held
constant,
what
produces
the second
assertion
in held constant, what
By
ding of the synchronous
machine,
which is
that theintotal
flux "g isofconstant.
produces
the second
assertion
the airgap
understanding
the synchronous
machine, which
e total flux in the airgap
is constant,
magnetomotive
is that
the total the
airgap
flux φg is constant. By extension, if the total flux in the airgap
Ep
which excites this flux
is also fixed
constant.
is constant,
theand
magnetomotive
force Vp
(mmf)
Fo which excites this flux is also fixed and
constant. Ep is always in phase with Vp , since there is no intervening impedance to
phase with Vp, since there is no intervening impedance
displace it, but since the induced emf is the time derivative of a sinusoidally varying
but since the induced emf is the time derivative of a
flux pattern, the induced emf Ep is π/2 degrees ahead of φg . Thus, the constant airgap
arying flux pattern, the induced emf Ep is #/2 degrees
mmf Fo liesF 90lies
degrees
behind the induced emf and terminal voltage, as shown in figure
Thus, the constant airgap mmf
90 degrees behind
o
mf and terminal voltage, as shown in Figure 8.7b. Note
8.7b the voltages, flux and mmf are rotating vectors
venience, have been 'frozen' with Vp at 12 o'clock.
constant
"g
Fo
Figure 8.7b - Airgap flux and induced emf
Page 3 of 5
xed since Vp cannot change. From
Egrid
duced emf in the stator winding is
e of change of airgap flux linkage:
Ep
Sync machine
Grid
N (rev/min)."g
Power Systems and Machines
51
Figure 8.7a - Infinite grid and ideal sync machine
rgap "g revolves at synchronous speed N and induces Ep, which is held constant
35. also
Note
that
in figure
35 produces
the voltages,
flux and
mmf are
equency and speed are
held
constant,
what
the second
assertion
in
convenience,
been
with flux
Vp at
is o’clock.
constant. By
the synchronous machine,
which have
is that
the‘frozen’
total airgap
"g12
flux in the airgap is constant, the magnetomotive
Vp
excites this flux is also fixed and constant.
rotating vectors which, for
Ep
with Vp, since there is no intervening impedance
ce the induced emf is the time derivative of a
ux pattern, the induced emf Ep is #/2 degrees
e constant airgap mmf Fo lies 90 degrees behind
constant
terminal voltage, as shown in Figure 8.7b. Note
he voltages, flux and mmf are rotating vectors
e, have been 'frozen' with Vp at 12 o'clock.
"g
Fo
Figure 8.7b - Airgap flux and induced emf
Figure 35: Airgap flux and induced emf
Page 3 of 5
7.8
Double excitation
Remembering that the synchronous machine is doubly excited, the flux g in the airgap
is produced by
Fo which must be the sum of two separate mmfs (mmf = N.I ampere turns)
F1 is proportional to phase current Ip in the stator winding and number of turns in the
stator coils
F2 is proportional to the rotor current I2 injected into the rotor winding and the number
of turns in the rotor field coils.
SYSTEMS
AND MACHINES
The fluxes and mmfs mayPOWER
be treated
as vectors,
since they have magnitude and direction, and the vector identity in Figure 8.8 emerges as the third construction in this
synchronous machine is doubly excited, the flux !g in the airgap is produced by
understanding of the synchronous machine.
um of two separate mmfs (mmf = N.I ampere turns):
Fo ≡ F1 + F2
ase current Ip in the stator winding and number of turns in the stator coils, and
e rotor current I2 injected into the rotor winding and the number of turns in the
Vp
may be treated as vectors, since they have
on, and the vector identity in Figure 8.8 emerges Ip
tion in this understanding of the synchronous
Fo " F1 + F2
Figure 8.8 - MMF diagram
Ep
$
!g
#
Fo
F1
F2
y, where Fo is fixed in length and chosen to lie
vertical Ep. F1 and F2 are free to take up any magnitude
andMMF
direction
which
Figure 36:
diagram
The excitation control in a synchronous machine adjusts the field current I2 and
tude of vector F2. The direction of F2 (relative to Fo) is determined by the
and stator axes, or load angle # between them. Since the load angle # between
nd F2 in the rotor axis varies with mechanical load and Fo"F1+F2, F1 will always
h the adjustment in F2 brought about by applied load or adjustment of rotor
magnitude or direction of F1 necessary to complete the mmf triangle created
rrent Ip changing value and direction relative to Vp. The phase angle $ between
determines the operating power factor of the machine. Thus, Ip and power
onse to added load, but may also be modified by the field current setting I2.
p
(7.1)
Power Systems and Machines
52
This is a vector identity, where Fo is fixed in length and chosen to lie horizontally behind
a vertical Ep . F1 and F2 are free to take up any magnitude and direction which satisfies
this identity. The excitation control in a synchronous machine adjusts the field current
I2 and as a result the magnitude of vector F2 . The direction of F2 (relative to Fo ) is
determined by the alignment of the rotor and stator axes, or load angle δ between them.
Since the load angle δ between Fo in the stator axis and F2 in the rotor axis varies with
mechanical load and Fo ≡ F1 + F2 , F1 will always vary in sympathy with the adjustment
in F2 brought about by applied load or adjustment of rotor current I2 .
The adjustment of the magnitude or direction of F1 necessary to complete the mmf
triangle created results in the phase current Ip changing value and direction relative to
Vp . The phase angle φ between current and voltage Vp determines the operating power
factor of the machine. Thus, Ip and power factor will vary in response to added load,
but may also be modified by the field current setting I2 .
This is the crux of the understanding of the synchronous machine - the magnitude
and direction of the phase current in the stator may be adjusted by varying the rotor
excitation. The machine is controllable, and may be made to operate at desired power
factors.
7.9
MMF phasor diagrams
There are three load conditions to consider in the examination of the behaviour of the
synchronous machine, and for completeness there are two excitation conditions in each
of the machine states. If the machine is operating:
• On true no-load, the rotor and stator axes are aligned and δ = 0
• As a motor, the rotor will be dragged behind the stator axis and δ will be -ve
• As a generator, the rotor will be driven ahead of the stator axis and δ will be +ve
In each of these cases, the mmf vector identity Fo ≡ F1 + F2 must be maintained, with
F1 adjusting to compensate for the insufficiency or excess of excitation mmf F2 from the
rotor.
If F2 < Fo the airgap is under-excited with the implication that φg would be insufficient
to induce Ep . This can not be the case, and F1 has to become a magnetising mmf.
If F2 > Fo the airgap is over-excited with the implication that φg would induce an emf
greater than Ep . This can not be the case and F1 has to become a de-magnetising mmf
There are therefore six mmf diagrams to consider and six modes of operation to understand.
Power Systems and Machines
7.9.1
53
Compensator mode - Figure 37i
Operating on no-load, or as a synchronous compensator, no shaft mechanical load is applied or supplied, and no real power is exchanged. In the ideal machine, if no mechanical
power is taken off the rotor, coupling the rotor and stator fields, and axes, align perfectly
and load angle δ is zero. Airgap flux φg must induce Ep and must be π/2 behind the
applied voltage Vp . Thus, Fo is π/2 behind Vp and Ep . The vector sum of F1 + F2
must always remain π/2 behind Vp and Ep . Since, on no load the rotor and stator axes
remain aligned. F1 and F2 align, and add linearly to directly equal Fo . Stator current
Ip must always be in phase with F1 , since the mmf vector defines the magnitude and
direction of the current vector.
Under excitation F2 < Fo - If rotor excitation F2 is less than sufficient to magnetise
the airgap (and produce Ep ), the stator must produce F1 as a magnetising mmf, where
F1 lies π/2 behind Vp and the stator draws magnetising current Ip . Note that in this
condition Ip lags Vp by π/2, and the machine operates at zero lagging power factor,
looking inductive - (pf = watts/va).
Over excitation F2 > Fo - If the rotor excitation is increased, such that F2 exceeds the
Fo which would be required to magnetise the airgap (i.e. Ep > Vp ), the stator must
reduce overall flux and mmf and F1 becomes demagnetising. In this case, the stator
takes a demagnetising or leading current Ip , and operates at zero leading power factor
looking capacitive.
Note that in this mode, the machine runs unloaded, with the stator behaving as a
polyphase inductor (underexcited) or capacitor (overexcited). In weak transmission and
distribution systems, such machines can be used to “stiffen” the grid, trying to modify
the power factor of the receiving end of transmission lines. It is also interesting to reflect
on the condition where F2 = Fo in which case the machine would operate at unity power
factor, only drawing sufficient active power to overcome friction and windage losses, and
would revolve at synchronous speed with no load or engine connected.
7.9.2
Motor mode - Figure 37ii
If a mechanical load is applied to the rotor coupling, the rotor retards to load angle δ,
and revolves with the axes of F2 and Fo displaced by this angle, producing torque on
the rotor due to the force of attraction between the stator and rotor fields. In order
to maintain the vector equality Fo ≡ F1 + F2 advances, and stator current Ip advances
from being π/2 behind Vp to angle φ. Notice
now that the machine draws active current
√
Ip cos φ, per phase, and active power 3Vl Ip cos φ to balance the applied load power.
If the mechanical load is fixed, δ will be fixed, but depending on rotor excitation F2 ,
the stator can still be made to draw a magnetising component (+Ip sin φ) of current if
underexcited, or a demagnetising component (−Ip sin φ) of current if overexcited, operating at a lagging, or leading power factor respectively. Note that since Ip cos φ is very
Power Systems and Machines
N
54
S
N
S
N
δ
S
δ
V1 E1
V1 E1
ɸ1
ɸ1
I1F1 F2
F0
F2
F0
δ
I1F1 V1 E1
V1 E1
I1F1
F2
I1F1
λ
a) Under excited
ɸ1
V1 E1
δ
ɸ1
F0
λ
I1F1
V1 E1
F2
ɸ1
F0
F1
F2
λ
δ
λ
ɸ1
b) Over excited
F2
i
No load/compensator
δ
F0
ii
Motor
I1F1
iii
Generator
Figure 37: The mmf and voltage phasor diagrams
F0
Power Systems and Machines
55
much greater when the motor is overexcited than when underexcited, the motor will
produce greater output power and torque in this condition. It is for this reason that all
synchronous motors are designed to operated overexcited at a leading power factor of
around 0.9.
7.9.3
Generator mode - Figure 37iii
The last condition to understand is where a prime-mover, such as a diesel engine, steam
or gas turbine or water turbine imparts mechanical power to the shaft of the machine,
and causes load angle δ to advance. The rotor axis, and F2 advance on Fo , and F1
must retard to maintain Fo ≡ F1 + F2 . The
√ active component of current now becomes
−Ip cos φ and the machine delivers power 3Vl Ip cos φ into the electrical supply, to balance the mechanical input power. The machine generates active and reactive power
determined by the mechanical input power and operating power factor (rotor excitation). Again, the vector identity Fo ≡ F1 + F2 determines that F1 will vary in position
and magnitude as F2 is adjusted. If the generator is underexcited, the stator has to take
a magnetising component of current, (+Ip sin φ), which in this case represents operation
at a leading power factor. If the machine is overexcited, the stator produces a component of demagnetising current (−Ip sin φ) and in this case operates at a lagging power
factor.
Power Systems and Machines
8
56
Synchronous generators - voltage control
8.1
Introduction
This and the next lecture will develop analytical techniques for considering the transient
behaviour of a synchronous generator controlled by an automatic voltage regulator.
8.2
Synchronous generators
A synchronous machine would produce electrical power at the required supply frequency
if the prime mover is correctly governed to drive the machine at the synchronous speed.
The machine developed torque and produced power is determined/controlled by forward
adjustment of rotor load angle δ. Figure 38 shows the mmf and voltage phasor diagrams
for a synchronous generator in overexcited and underexcited operating conditions. Note
that the synchronous reactance of the generator Xs is added.
I1.jXs
V1
I1.jXs
Et
δ
I1
ɸ
V1
Et
F2 < F0
Underexcited
F2
δ
δ
F1
F0
ɸ
F2 > F0
Overexcited
I1
F2
δ
F1
F0
Figure 38: The mmf and voltage phasor diagrams
8.2.1
Isolated operation
Unlike a synchronous motor, the synchronous generator can produce electrical power
remote (i.e. isolated) from a synchronous grid. In this situation (since there is no synchronous supply to define system frequency), the governor on the prime-mover must
be set to control shaft speed and in that way frequency of the generated electricity.
Additionally, the excitation current on the rotor field windings must be regulated automatically by the AVR, in order to generate the correct terminal voltage as the electrical
load connected to the machine varies. Consider the machine to operate with a fixed excitation, which is more than sufficient to magnetise the airgap, (F1 + F2 > F0 required,
and Et > V1 ). On initial load, I1 is at phase angle φ1 and I1 Xs completes the voltage
Power Systems and Machines
57
phasor triangle. The governor on the engine will admit just enough fuel to meet the
initial load at synchronous speed, and the rotor will be advanced by initial load angle δ,
as shown below.
ΔEt
I1.jXs
Voltage drop
E’t
Et
I’1
ɸ’
ɸ
F’2
I1
δ
δ’
F2
F1
F’1
F0
Figure 39: Application of the electrical load isolated operation
8.3
Generator and AVR performance
The addition of electrical load to a generator causes an increased voltage drop across
the synchronous reactance of the stator, which reduces phase voltage V1 . The AVR
senses this change, and by comparison with a reference voltage V, uses the amplified
error signal to boost the rotor field current Et in the generator. This increases F2 , and
in turn Et and restores V1 to the desired value preset by Vref . Under the AVR control,
the combined voltage response is predicted using the forward transfer function of the
generator, and the transfer function of the AVR. There are two methods of providing
excitation power to and voltage control of synchronous generators, as discussed in the
previous section.
8.3.1
Static excitation
Statically excited generators, as shown in figure 40, have the excitation power taken from
the connected electrical system, or from the machine terminals. An excitation supply
transformer provides excitation power to the AVR at a reduced voltage. In smaller
generators, this voltage transformer (VT) may also provide the measurement, or “reset”
signals of V1 to the AVR. Neglecting saturation effects in the VT, the reset signal to
the AVR is directly proportional to the line or phase voltage Vl, and for the purposes
of the control system is also termed V1. Adjustment of a voltage setting potentiometer
Power Systems and Machines
58
produces a controllable reference voltage Vref , which ideally is independent of generated
voltage.
The reference voltage is usually created internally within the AVR by rectifying, filtering,
and zener-stabilising the power supply, to provide a fixed dc voltage (around 12 V). The
first stage of the AVR is a comparator, which rectifies the reset voltage V1 and compares
it with the reference voltage Vref set by the operator. Whilst both voltages are dc, the
internal value of V1 varies with generated voltage Vline , while Vref is constant. The error
output from the comparator is termed Verr .
Above the minimum field voltage required to establish open circuit voltage at the generator terminals, the output voltage of the AVR is directly proportional to the error signal
Verr derived by the comparator. In a statically excited generator, the output voltage
of the AVR is applied via sliprings directly to the main field of the generator as field
voltage Vf . Increased current If flows in the rotor, increasing F2 , and Et , to restore
V1 .
Grid system
Reset signal
Exn power
Voltage
setting
pot.
Vline
V1
Automicatic
Voltage
Regulator
Vref
Vf
Generator
If
Rotor
Sliprings and
brushes
Stator
Figure 40: Statically excited generator and AVR
8.3.2
AVR
Since the field voltage of the generator Vf will normally be much greater than the error
signal Verr detected by the AVR, the AVR includes an amplifier of gain KA. Thus:
∆Vf
∆Vf
=
= KA
∆Vref − ∆V1
∆Verr
(8.1)
Where ±∆V1 is brought about by changes in generator load and the voltage drop at the
Power Systems and Machines
59
generator terminals caused by the line current and the series regulation of the stator
winding. ±∆Vref is caused by manual adjustment of the voltage setting potentiometer
in the AVR, for synchronising or power factor control. During a voltage transient, at
least two of the above voltages vary with time, usually Vf and V1 .
Also, since the measured voltage V1 is rectified and filtered prior to comparison in the
AVR, there is a finite delay introduced into the transfer of the amplified error signal.
This delay introduces a time constant TA (or τA ) into the dynamic response of the
AVR controlling the generated voltage. Thus, taking the Laplaces transforms of the
two time variations, for transient analysis, the forward transfer function of the AVR
becomes
∆Vf (s)
KA
(8.2)
=
∆Verr (s)
1 + sτA
The denominator introduces the delay term in the time domain response.
8.3.3
Main field
Neglecting the effects of exciter field saturation, the generated emf Et in the generator is
a linear function of Vf , and If . To correctly consider the variation in terminal voltage,
the stimulus which brings about a system transient is a change in load, or line current,
since the current taken from the generator reduces Et to V1 . For small signal analysis,
it is reasonable to consider Et ≈ V1 , and regard V1 as directly proportional to Vf .
The terminal voltage of the machine will typically be 20 times greater than the main
field voltage Vf , and the relationship between field voltage, field current and airgap flux
linkage, which induces Et introduces a main field voltage gain KF , where
∆V1 (t)
= Kf
∆Vf (t)
and
Kf
∆V1 (s)
=
∆Vf (s)
(1 + sτf )
(8.3)
Remembering that the main field has inductance LF and resistance RF , and that any
changes in VF are simply steps up and down in dc levels, the field voltage and current
will vary exponentially with time constant TF (the open circuit field time constant).
The main field voltage/stator terminal voltage thus has the above transfer function.
The statically excited synchronous generator and AVR may thus be represented in block
diagram form as shown in 41 below.
W
Vref
+
V1
-
Verr
AVR
KA
Vf
(1+sTA)
Generator
KF
(1+sTF)
Figure 41: Block diagram of statically excited generator
Power Systems and Machines
8.4
60
Brushless excitation systems
In the brushless excitation system, as shown below in figure ??, all of the same gains
and time constants apply to the AVR and main field. To improve voltage build-up
at start, and provide a source of excitation power which is independent of terminal
voltage, a permanent magnet generator provides excitation supply. To circumvent the
need for brushes or slip-rings, an ac exciter and rotating rectifier are fitted to the rotor.
Neglecting any rectifier delays in the rotor, the ac exciter introduces another gain and
delay into the forward transfer function of a brushless synchronous generator. The
transient performance is reduced by the delay in the ac exciter. Time constant TE is
approximately the ratio of exciter inductance LE to resistance RE .
Exn power
Vref
AVR
ac
dc
Voltage
setting
pot.
Grid system
Vline
V1
Reset signal
dc
PMG
AC exciter
Rotor
Generator
Stator
Figure 42: Brushless excitation system
Now, the output of the AVR is termed VE , and is applied to the field of the ac exciter
as a variable dc voltage, which causes a current to flow in the stator of the ac exciter,
and set up a stationary field, in which the wound rotor of the ac exciter revolves. An
alternating voltage is induced in the exciter rotor, which is rectified and applied to the
main field as VF and IF .
The block diagram representation of a brushless excited synchronous generator is shown
below in figure 43. Typical values for the gains and time constants are:
KA
KE
KF
100 500
1.0 - 2.0
10 - 30
TA
TE
TF
0.02 - 0.1 sec
0.7 - 1.5 sec
0.7-1.5 sec
Power Systems and Machines
W
Vref
+
V1
AVR
Verr
61
KA
Ve
(1+sTA)
-
AC Exciter
KE
Vf
(1+sTE)
Generator
KF
(1+sTF)
Figure 43: Block diagram of brushless generator and excitation system
8.5
Generator modeling
The forward transfer functions for the AVR, exciter (for brushless machines), and main
field-stator of synchronous generators combine serially under open-loop conditions, the
open loop transfer function combines to become
G(s) =
KA
KE
KF
V1 (s)
=
+
+
Vref (s)
(1 + sτA ) (1 + sτE ) (1 + sτF )
(8.4)
where the open-loop forward gain of the system becomes the product of the individual
stage gains. K = KA .KE .KF . The machine is of course controlled in the closed loop
mode by the inclusion of the voltage feedback loop
The closed loop model can be reduced as follows
V1 (s) = G(s).Verr (s) = G(s)(Vref (s) − V1 (s))
(8.5)
V1 (s)(1 + G(s)) = G(s)Vref (s)
(8.6)
and
and
G(s)
V1 (s)
=
Vref (s)
1 + G(s)
(8.7)
K
(1 + sτA )(1 + sτE )(1 + sτF )
(8.8)
H(s) =
where
G(s) =
Note that for a statically excited generator, the exciter term is deleted from the transfer
function, and
1
KA KF
H(s) =
=
(8.9)
1
KA KF + (1 + sτA )(1 + sτF )
1+
G(s)
The statically excited generator system has a lower forward gain, since the electromagnetic gain of the ac exciter is not present. The dynamic response of the generator is
improved, however, since one time constant is removed from the forward transfer function. The AVR acts straight into the main field, and can force main field excitation much
more effectively than a brushless excited generator. For the brushless generator
H(s) =
KA KE KF
KA KE KF + (1 + sτA )(1 + sτE )(1 + sτF )
(8.10)
Power Systems and Machines
8.6
62
Steady state accuracy of excitation systems
The complete excitation system must be stable in response to transients in terminal
voltage, or reference voltage, which implies it must have adequate forcing margin and
speed of response. In the steady state, however, the AVR must regulate the terminal
voltage V1 accurately, to say within ±1/2% of the V1 prescribed by Vref . For there to
be an output voltage V1 , there must always be an error signal at the input to the feed
forward block. If V1 is less than Vref, the error voltage Verr is positive and the forward
gain K multiplies this to become the output voltage V1 . If the steady state error Verr
between V1 prescribed by Vref and the actual measured V1 is to be small, the forward
gain K must be high. Also, for fast response and high accuracy (V1 ≈ Vref ) the gain K
must be high.
The block diagram of the closed loop system has
Verr (s) = Vref (s) − V 1(s)
V 1(s) = G(s).Verr (s)
Thus
Vref = Verr (s)(1 + G(s))
and
Verr = Vref (s)
1
1 + G(s)
(8.11)
(8.12)
Since s = 0 during steady state conditions
Verr = Vref
1
1+K
(8.13)
If the acceptable error in the controlled voltage is 1%, the forward gain of the system
must exceed 99. If the gain falls to say 19, the percentage error rises from 1 to 5%.
Typically, if KA = 100, KE = 1, and KF = 10, the steady state error is 0.1%. The error
described above is the difference between the wholly accurate output voltage and the
actual output voltage, which, while very small, is inherent in this feed-forward controller.
The accuracy of the controller is the actual output voltage expressed as a fraction or
percentage of the error-free output, or the ratio of the steady state output voltage V1 to
Vref and is calculated as follows.
H(s) =
K
K +1
thus
V1 =
K
Vref
K +1
and
V1
K
=
Vref
K +1
(8.14)
To reformulate above mentioned discussion: If the accuracy required in the AVR system
is 99%, the forward gain of the system must exceed 99. If the gain falls to say 19, the
percentage accuracy falls from 99% to 95%. Typically if KA = 100, KE = 1, and KF =
10, the steady state accuracy is 99.9%
Note also that the sum of the accuracy and error must always equal 1.
Power Systems and Machines
63
Vref
Verr
V1
V1
K
1
Vref +
V
K+1
K+1 ref
time
Figure 44: Steady state error and accuracy
9
9.1
Synchronous generator modeling: control systems and
stability
Time domain response - Laplace analysis
See appendix A for a list of Laplace transforms.
The closed loop block diagram of a generator and automatic voltage regulator (AVR)
was shown before in figure 43. The Closed Loop Forward Transfer Function (CLFTF)
of the generator and AVR system is
H(s) =
V1 (s)
G(s)
=
Vref
1 + G(s)
(9.1)
where G(s) is the Open Loop Forward Transfer Function (OLFTF) of the generator and
AVR. If the reference voltage Vref changes with time as a function Vref (t), the combined
s-plane voltage response of the AVR system becomes
V1 (s) = Vref
G(s)
1 + G(s)
(9.2)
The time domain behaviour of the whole generating unit becomes the inverse Laplace
transform of the s-plane response.
G(s)
−1
V1 (t) = L
Vref
(9.3)
1 + G(s)
The time response of the system is a critical function of the characteristic equation,
1 + G(s), in the denominator of the closed loop transfer function. The response of the
system is determined by the eigenvalues, or closed loop poles, which are the values of s
that cause (1 + G(s)) in the denominator of the closed loop to be zero. For the brushless
generator in figure 43. the OLTF is given by equation 8.8. In the statically excited
generator, the exciter stage is absent.
Power Systems and Machines
64
Solution of the characteristic equation, from the denominator of equation 9.2, becomes
1 + G(s) = 0, which produces a polynomial in s and K, whose roots (or eigenvalues)
determine the time response of the system. For the statically excited system
1 + G(s) = 0
→1+
KA KF
=0
(1 + sτA )(1 + sτF )
(9.4)
and
(1 + sτA )(1 + sτF ) + KA KF = 0
(9.5)
s τA τF + s(τA + τF ) + (1 + KA KF ) = 0
(9.6)
2
This is a quadratic in s, which will have two eigenvalues or roots that vary as K is
adjusted. When K is low, the roots are wholly real because (b2 − 4ac > 0), and solutions
s1 and s2 will be real and distinct. The time response will be of the form
V1 (t) = A exp(s1 t) + B exp(s2 t)
(9.7)
As K is increased, the roots become imaginary (b2 − 4ac < 0), and take values (σ +
jω and σ −jω), corresponding to quadratic solutions to equation 9.6. The time response
takes the general form
V1 (t) = A exp(σ + jωt) + B exp(σ − jωt)
(9.8)
or for the statically excited machine
V1 (t) = C exp(−σt) sin(ωt + b)
(9.9)
This is now oscillatory, and the terminal voltage V1 will experience damped, or undamped oscillations, depending on the sign of σ. Since most of the time constants
are fixed, the loop gain K (between maximum and minimum values), inserted in the
characteristic equation determines a range of values of s for which the response of the
generator and AVR will be finite, and predictable. Between these two extremes, there
is a whole range of values of ‘s’ (eigenvalues), defined by the loop gain K, which satisfy
the characteristic equation.
Note that the loop gain K at which oscillation begins can be determined from the condition b2 < 4ac or
(τA + τF )2
KA KF >
−1
(9.10)
4τA τF
Since values of s may be real and imaginary, of the form σ ± jω, each point s may be
represented as a point in the s-plane, having a real x coordinate σ and an imaginary y
coordinate jω. As K is varied, the values of s trace out a succession of points (or locus)
in the s-plane.
Since every value of s defined by 1 + K.F (s) = 0 is a root of the characteristic equation,
the corresponding curve is called the root-locus of the system in the s-plane. The behaviour of the system is represented in the s-plane by the root-locus diagram, and the
dynamic, or time domain response, may be assessed by root-locus analysis.
Power Systems and Machines
9.2
65
Generator/AVR root locus diagrams
The s-plane has real and imaginary axes σ and jω, as shown below in figure 45. Remembering that the values of s which cause an infinite response are termed the open
loop poles, and when s has these values, K must be zero on the open loop poles if the
response is to remain bounded. Consider the brushless generator, with the characteristic
equation 8.8, where in the determination of roots
(1 + KF (s) = 0), F (s) = 1/[(1 + sTA )(1 + sTE )(1 + sTF )]
(9.11)
Since there are three terms in s, there will be three poles on the s-plane, located at
sA = −1/τA , sB = −1/τE and sC = −1/τF .
Typically, TA = 0.02 sec, TE = 0.8 sec and TF = 1.2 sec, and, for this case sA = (−50, 0);
sE = (−1.25, 0); and sF = (−0.83, 0).
+jω
Stable (-σ)
-σ
Locus of
roots
X
X
Unstable (+σ)
+σ
X
oscillatory
when locus
leaves σ axis
X Poles
-jω
Figure 45: Third order root locus diagram - Brushless machine and AVR
The poles are shown on the root locus diagram in figure 45. There are three loci beginning
at K = 0 on the poles shown. The gain of the ac exciter KE , and main field KF are
fixed at typically 1.5, and 20, but the AVR forward gain KA can vary from 0 to between
300 - 500, resulting in loop gains varying from 0 to 9000 - 15000. As the overall gain is
increased, for each new value of K, there are another three roots, or eigenvalues. The
eigenvalues move away from the poles as KA increases. One (sA ) locus moves from -1/TA
to the left, and will always be stable. The other two loci (sE ,sF ) move along the axis,
towards each other from -1/TE and -1/TF , until they meet. Further increase in KA
results in two of the eigenvalues becoming conjugate complex, when s = σ ± jω. As the
gain is increased, the system begins to oscillate, with an A exp(−σt) sin(ωt) term. If the
gain is increased further, the speed of response continues to improve because of the less
negative σ value in sA . The complex conjugate (σ ± jω) pair however, oscillate faster as
jω increases, and lose stability as σ becomes positive when the locus passes through the
jω axis. The gain when the system goes unstable is termed the critical gain.
Power Systems and Machines
9.3
66
Improvement of AVR stability
The combination of high loop gain and three cascaded time constants makes the brushless
AVR/generator combination inherently unstable. In practice, there are two ways of
improving stability. If a high dynamic performance is required, a statically excited
generator would be used, with the AVR acting straight into the main field. The exciter
gain and time constant would be deleted.
The open loop transfer function becomes that of the statically excited generator and the
closed loop transfer function has characteristic equation
F (s) = 1/(1 + sTA ).(1 + sTE )
(9.12)
The above forward transfer function F(s) defines that the root locus has only two poles.
Calculation of the two root loci would show that they remain σ -ve under all conditions
of gain. Whilst the voltage response would contain oscillations after the loci met, further
increase in gain would only increase the frequency of oscillation as shown in figure 46
below.
Stable (-σ)
-σ
X
s=-1/Ta
+jω
X
s=-1/Tf
+σ
-jω
Figure 46: Root locus for statically excited generator
If the generator must be brushless and equipped with an ac exciter, an additional compensation filter is included in the AVR, which introduces phase lead, shown in figure
47. This has the effect of introducing a zero into the numerator of the characteristic
equation.
The open loop transfer function becomes
G(S) =
KA KE KF KC (1 + sτC )
(1 + sτA )(1 + sτE )(1 + sτF )
(9.13)
Power Systems and Machines
W
Vref
+
V1
67
Comp
AVR
Verr
KA
(1+sTA)
-
KC(1+sTC)
AC Exciter
Ve
KE
Vf
(1+sTE)
Generator
KF
V1
(1+sTF)
Figure 47: Brushless system with compensated AVR
The time constant of the compensation filter is adjustable, and during commissioning of
the unit may be set such that TC = TE . In that way, the adjusted time constant cancels
the exciter time constant, and the pole and zero in the function cancel. The open loop
transfer function reduces to
G(S) =
K
(1 + sτA )(1 + sτF )
(9.14)
producing the root locus shown in figure 46, and, therefore, unconditional stability.
9.4
Time domain response of generator/AVR system
For any value of loop gain K, the solution of the characteristic equation 1 + G(s) = 0 will
determine either a single real value, or complex conjugate pair of values of s. Consider
the statically excited generator. If the roots of the equation defined by K are wholly
real, say −4 < σ < −2, the system response will be of the form
B
A
+
= A exp(±σ1 t) + B exp(±σ2 t)
(9.15)
V1 (t) = L−1
s ± σ1 s ± σ2
and the system voltage changes will be as shown in figure 48.
W
Vref
+
V1
-
Verr
AVR
KA
(1+sTA)
Generator
KF
V1
(1+sTF)
Figure 48: Overdamped response
If the roots of the equation defined by K are real and imaginary, as σ ± jω, the system
response will be of the form
A
B
−1
V1 (t) = L
+
= A exp(±σ1 t) sin(ωt)
(9.16)
s ± σ1 s2 ± ω 2
and there will be an oscillatory term in the response. For the AVR and generator to be
stable, the transient terms must vanish to zero with time. This means that the solutions
Power Systems and Machines
68
of the characteristic equation must all include a −σ term, and must all lie to the left
of the jω axis, for the presiding value of loop gain K. The more negative the −σ term
becomes, the faster will be the convergence of the system oscillation. The changes which
affect the behaviour of the AVR system most frequently are the impulse function, and
the step function.
9.4.1
Impulse response
A sudden short circuit near the terminals of an AVR controlled generator instantaneously
tries to reduce the terminal voltage to 0, and may then be cleared by system protection.
Relative to a fixed value of Vref , this creates a spike in the error signal driving the
feed-forward loop, which may be analysed as the impulse response. Switching transients
associated with inductive loads also look like impulse functions. Whilst the effect occurs
at the generator terminals, the actuating signal in the output of the comparator cannot
distinguish between −∆V1 and +∆Vref . ∆Verr takes on the error signal generated by
the transient. The Laplace transform of a true impulse waveform is 1, and the voltage
response of the AVR system becomes
G(s)
V1 (t) = L−1
(9.17)
1 + G(s)
9.4.2
Step changes in voltage
A more frequent condition arises as electrical load is switched onto the generator. In
isolated operation, the sudden increase in current causes a sudden drop in terminal
voltage, as the voltage drop across the machine reactance Xs reduces V1 . The step
reduction in V1 creates an error signal to the AVR of magnitude Vref − ∆V1 . The AVR
detects this and increases the voltage applied to the field of the ac exciter, which in turn
increases the main field voltage VF and restores V1 to the value prescribed by Vref .
W
Vref
+
V1
-
Verr
AVR
KA
(1+sTA)
Generator
KF
V1
(1+sTF)
Figure 49: Block diagram and step function response
The response of the AVR to such a step function may be simulated by a sudden increase in Vref by ∆Vref , as shown in figure 49 above. Consider a brushless ac generator
where a stabilising term has been included in the AVR such that the transfer function
becomes
KA KC KE KF
G(s) =
(9.18)
(1 + sτA )(1 + sτF )
Power Systems and Machines
69
By suitable adjustment to cancel TC =TE , the open loop transfer function is reduced
to a two term model, where K = KA .KC .KE .KF . The step response for the system is
defined by the combined s-plane response of the step function in Vref and the closed loop
transfer function.
G(s)
−1
Vref
V1 (t) = L
(9.19)
1 + G(s)
G(s)
−1 ∆Vref
(9.20)
=L
s 1 + G(s)






1

1
−1
= ∆Vref L
(9.21)
1 

s


 1+

G(s)
K
−1 1
(9.22)
= ∆Vref L
s K + (1 + sτA )(1 + sτF )
K
−1 1
= ∆Vref L
(9.23)
s s2 (τA τF ) + s(τA + τF ) + (K + 1)
The solution is now best made particular to a given set of machine parameters. For a
low voltage machine with the following parameters: KA = 2.5, KF = 4, TA = 0.02 sec,
and TF = 1 sec, the solution is
= ∆Vref L
−1
1
500
2
s s + 51s + 550
(9.24)
This may be solved in one of three ways.
1. The denominator may be turned into a more ”recognisable” (factorised) form by
solution of the quadratic equation in s to produce
500
−1 1
V1 (t) = ∆Vref L
(9.25)
s (s + 35.5)(s + 15.5)
At this stage, the inverse Laplace transform may be taken to produce the time
domain response. Standard inverses are to be found in tables, a copy of which is
appended at the end.
2. Alternatively this expression may be expanded by partial fractions to
B
C
−1 A
V1 (t) = 500∆Vref L
+
+
s
(s + 35.5) (s + 15.5)
(9.26)
Solving for A, B, and C gives, A = 0.00182, B = 0.001408, C = 0.0032267 for
which the inverse Laplace Transform is
V1 (t) = ∆Vref [0.91 − 0.704 exp(−35.5t) + 1.613 exp(−15.5t)]
(9.27)
Power Systems and Machines
70
This is a stable, overdamped double exponential response. This shows that the
output changes exponentially with time, but in a stable manner. This could have
been predicted, since the eigenvalues of the characteristic equation were at -35.5
and -15.5, with no imaginary terms. The overdamped voltage response is shown
below in figure 50. Previous sections showed that the steady state accuracy could
be calculated from K/(1+K) = 500/550 = 90.9% or error = 10.09%. This result
is obvious here, if time is allowed to go to infinity for steady state.
This error is too high. Increasing KA , to say 10, to drive up the loop gain will
increase the accuracy. Increasing the loop gain creates imaginary parts in the
eigenvalues indicating oscillation and possible instability in V1 .
40
−1 1
V1 (t) = ∆Vref L
(9.28)
+
s 0.02s2 + 1.02s + 41
1
2000
= ∆Vref L−1
(9.29)
+ 2
s s + 51s + 2050
Accuracy K/(1+K) = 2000/2050 = 97.5%, error = 2.5%
3. Completion of squares - Halve ‘b’ term to produce 25.5 and the first root. Squaring
this becomes 650.25, meaning that the second root is the square root of (2050650.25), or 37.41. The inverse transform can be re-written as
2000
−1 1
V1 (t) = ∆Vref L
+
(9.30)
s (s + 25.5)2 + 37.5
and this may be compared with a standard inverse transform to arrive at the
time-domain response
2000
45.2
V1 (t) =
Vref 1 +
exp(−25.5t) sin(37.4t + 0.97)
(9.31)
2050
37.4
This is now a damped oscillatory response. Critical damping is when roots start
to contain imaginary terms, defined on the limit when b2 = 4ac. For this to be the
case KA cannot exceed 3.0 and the accuracy is limited to 92.3%.
Figure 50 shows the variation of voltage responses as gain is adjusted
Power Systems and Machines
71
+jω
-σ
V1
X
X
2
t
K < (Ta+Tf) -1
4TaTf
stable overdamped
+jω
+σ -σ
V1
X
X
-jω
2
t
+σ -σ
V1
X
X
-jω
K = (Ta+Tf) -1
4TaTf
critically damped
+jω
+jω
2
t
K > (Ta+Tf) -1
4TaTf
damped oscillatory
-σ
+σ
V1
-jω
-jω
X
X
t
+σ
-jω
2
K >> (Ta+Tf) -1
4TaTf
stable overdamped
Figure 50: Voltage responses
10
10.1
Operation of embedded generators
Introduction
The UK Government commitment to the Kyoto protocol is to reduce CO2 emissions
by 20% by 2010, with the target that renewable energy generation will supply 10% of
UK electricity by then. Additionally there will be a significant increase in the capacity
of fossil fueled combined heat and power (CHP) and co-generation plants towards a
target of 10GW capacity. Projections for these increases in different types of embedded
generation are as follows
OCGT/Diesel and Small Hydro
Renewables
CHP
Total
GW (1999)
2.80
0.63
2.10
5.53
GW (2010)
3-4
7-8
7-8
17-20
Increase
1.4x
12.7x
3.8x
3.6x
In Scotland the aspirational targets are twice those for the UK; 20% by 2010 and 40% by
2020. Virtually all of this new plant will connect to the distribution network, but currently none will be centrally constrained or dispatched in the way that larger plants are
connected to transmission network. There is growing and legitimate concern at the possible prevalence and effects of up to 20GW capacity of embedded and renewable energy
generation (EG) on the management and operation of the UK distribution networks.
The generic term is Distributed Generation (DG).
Power Systems and Machines
10.2
72
Transmission and distribution networks
The distribution networks were installed to convey active and reactive power from the 132
kV transmission system through primary sub-stations at 33 and 11kV for distribution to
urban and most rural consumers at 400/235V. Power flow was uni-directional towards
the low-voltage edges of the system, and the system operated to meet demand in a largely
passive mode. At its geographical extremities the network tends to consist of long to
medium length radial circuits, with few, if any, ready opportunities to re-inforce the
mesh connection across the open ends of the feeders. Figure 51 suggests the complexity
and structure of the Scottish transmission network.
Transformers that connect between transmission busbars, and those supplying the 3311kV distribution network are fitted with on-load automatic tap changers (OLTCs)
for automatic voltage control (AVC). As the load on these transformers rises and falls
the uncorrected secondary voltage would fall and rise due to the voltage drop across
the transformer and upstream series impedance. The AVC system senses the changing
voltage and adjusts up or down the high voltage winding tap position to maintain the
secondary voltage substantially constant.
Local transformers that connect between 11kV and the distribution network loads at
400V are usually provided with off-load fixed taps that can only be set by manually
adjusting links in the transformer when the 11kV and 400V systems either side are deenergised. These transformers are normally tapped up to compensate for line voltage
drop to ensure that all consumers receive supplies whose voltage is within statutory
limits under maximum loading. The installation and connection of an unconstrained
ED at the remote end (or any intermediate point) can reduce or reverse load flow and
cause voltages along the feeder to vary, ultimately out of statutory limits. With the
connection and presence of the EG the distribution system becomes active with nonconstrained bi-directional power flow, and unpredictable voltage variations that cannot
be readily controlled. Distribution Network Operators (DNOs) are required to provide
connections for proposed ED schemes but must maintain the quality of supply to local
consumers ensuring that the integrity, safety, reliability and security of the supply does
not reduce.
10.3
Statutory requirements
It is a legal and professional requirement that embedded generating plants are designed,
manufactured and installed to operate in an inherently safe manner, under normal and
abnormal or fault conditions. In addition to complying with the normal requirements of
the Health and Safety at Work etc Act, all embedded generating plants must comply with
the Electricity Act and the Electricity Supply Regulations (1988). Under these regulations no person may operate generating plant in parallel with a DNO’s system without
their written agreement, and the plant must meet minimum standards of operation,
protection and disconnection in the event of a distribution system or EG failure.
Power Systems and Machines
10.3.1
73
Applicable standards and codes of practice
The equipment installed must individually comply with all relevant British or IEC standards (BS/IEV) but the overall scheme has to be designed to operate safely and integrate
successfully with the distribution network. DNO distribution codes and Electricity Association Engineering Recommendations (ER’s) and Technical Reports (ETR’s) define
design, operational and protection requirements applied to EG schemes including those
below:
• The Energy Act, 1983
• Electricity Supply Regulations, 1988
• The Electricity Act, 1989
• Health and Safety at Work etc Act, 1974
• The Electricity (Private Gen. Stations and Requests by Private Gens. and Supplier) Regulations, 1984
• BS EN 50160 - Voltage Characteristics of Electricity Supplied by Public Distribution Systems, 1995
• BS EN 7671 - Requirements for Electrical Installations, 1992
• BS EN 7430 - Code of Practise for Earthing, 1998
• ER S5/1 - Earthing Installations in Substations, 1966
• ER P2/5 - Security of Supply
• ER P25 - The short-circuit characteristics of Electricity Boards’ low voltage distribution networks and the co-ordiation of over-current protective devices on 240V
single phase supplies, 1985
• ER P26 - Planning Limits for Voltage Fluctuations Caused by Industrial, Commerical and Domestic Equipment in the UK, 1989
• Planning Limits for Voltage Unbalance in the UK for 132kV and below, 1990
• Limits for Harmonics in the UK Electricity Supply, 1976
• ER G59/1 - Recommendations for the Connection of Embedded Generating Plant
to the Electricity Companies’ Distribution Systems, Amendment 2, 1995
• ER G75 - Recommendations for Embedded Generation Plant Connected to the
Public Electricity Supplier’s Distribution Systems above 20kV, or with Outputs
over 5MW, 1995
• ER G74 - Procedure to Meet the Requirements of IEC909 for the Calculation of
Short Circuits Three Phase AC Power Systems, 1992
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able standards and codes of practice.
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the rural distribution network from which the surrounding load centres are also supplied.
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• Increased losses and thermal overloads.
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regulation.
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• Increases in short circuit fault levels.
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• Reductions in transient stability.
ating Embedded Generation Into Distribution Networks
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harmonic
content.
.%-0%,$36.%'(%'2$3",-'0"/%."($,;%,7$%JKL%7".%,'%1$,$36-0$%"01%"::$2,%E'3%0',H%,7$%-62":,%
,-'0%'(%,7$-3%1-.,3-85,-'0%0$,='3A9%2"3,-:5/"3/;%-0%,$36.%'(%3$15:,-'0.%'(%,7$%.522/;%45"/-,;%,'%
/$:,3-:"//;%:/'.$%,'%,7$%CO<%%!'..-8/$%-62":,.%'(%,7$%CO%-0:/51$`%
10.4.1 Power flow
0%-0%2'=$3%(/'=a%%
1%/'..$.%"01%,7$36"/%'N$3/'"1.a%
%,'%N'/,"#$%3$#5/",-'0a%%
Consider the effects of increasing EG capacity Pgen + jQgen connected to a radial
.%-0%.7'3,%:-3:5-,%("5/,%/$N$/.a%%
relative to the demand of a fixed local load Pload + jQload , as shown below in
0.%-0%,3"0.-$0,%.,"8-/-,;a%%
51.
6-.$1%1-.:3-6-0",-'0%'3%'2$3",-'0%'(%0$,='3A%23',$:,-'0a%
1%7"36'0-:%:'0,$0,<%
Q5.%CO%
!/-0$b%cd/-0$%
Flow
CO%
D%
ce%
%$(($:,.%'(%-0:3$".-0#%CO%:"2":-,;%
]-[$1%
:'00$:,$1% ,'% "% 3"1-"/% ($$1$3%
7$% 1$6"01% '(% "% (-[$1% /':"/% /'"1% ! b cd
#$0 %
#$0%
%".%.7'=0<%
%
Q5.%)%
)%Y%CO%:'00$:,$1%,'%3$6',$%$01%
3-85,-'0%.;.,$6%
\':"/%
/'"1%
!/'"1%b%cd/'"1%
+0(%
O3-1%
Z"3-"8/$%
P&%
Q5.%&%
D"1-"/%($$1$3%
Figure 51: EG connected to remote end of radial distribution system
!"#$%&%'(%)*%
feeder
figure
Power Systems and Machines
75
The apparent power in the radial feeder will be the difference between generator production and load demand. Note the conventions that the original direction of power flow
it towards the load and the power leaving the network is negative, power entering the
network is positive.
Pline +jQline = −(Pload +jQload )+(Pgen +jQgen ) = (Pgen −Pload )+j(Qgen −Qload ) (10.1)
There are 5 conditions to note.
1. Sgen = 0
When the generator is out of service the power flow is towards the load and has
magnitude Sline = −(Pload + jQload . The feeder losses will be the total I 2 R ohmic
losses in the feeder conductors and the copper losses in any transformers that are
in the supply circuit.
2. Sgen < Sload
If the capacity of the generator is less than the demand of the local load and it
operates at rated output, the power deliverd from the primary substation towards
the local busbar reduces, and since this is still a negative quantity, flowing towards
the load there is Sline = (Pgen − Pload ) + j(Qgen − Qload ) is less than the original
power flow Sline = −(Pload + jQload . This reduced current results in lower feeder
losses and reduced voltage drops along the line.
3. Sgen ≈ Sload
If the generator output increases until the kW and kVAr produced closely meets
that demanded by the local load, there is minimal load flow, Sline ≈ 0. Feeder
losses are minimised at this state and the voltage profile is virtually flat at the
nominal off-load voltage along the feeder.
4. 2 × Sload > Sgen > Sload
Where the power produced exceeds that demanded locally, but by less than a factor
of two, the excess is less than the orignal power flow and is exported back towards
the network, now a positive quantity. Sline = (Pgen − Pload ) + j(Qgen − Qload ) <
Pload + jQload . This re-establishes losses and local voltage variation (now a rise
above the off-load value) ≈ to 1.
5. Sgen > 2 × Sload
If the active and reactive power produced is more than twice the local demand,
the increase in power flow towards the network and feeder losses may exceed the
thermal limits of the line and the local voltage may exceed the statutory maximum,
leading to voltage violation. Sline = (2Pgen − Pload ) + j(2Qgen − Qload ) > Pload +
jQload . The net export of apparent power can lead to excess losses causing thermal
violations and to excursions in the voltage local to the EG that may be outside
the statutory limits.
Power Systems and Machines
10.4.2
76
Thermal effects
Network losses are proportional to
Losses = I 2 R =
P 2 + Q2
S2
R=
R
2
V
V2
(10.2)
Thus the transmission of additional active or reactive power - in either direction - increases feeder losses. If a large quantity of reactive power is carried then the maximum
real power that can be carried within the thermal limit of the line is reduces. A large
synchronous generator operated overexcited exports reactive power that can increase
voltage rise and system losses. The connection an operation of a large induction generator will add to the system losses as reactive power is supplied to magnetise it, but
a fortunate consequence is that absorbing power at the remote end voltage rise. If the
active power produced by the EG at smaller sites is absorbed by industrial load or local
consumers at the remote end of the line, this reduces the power transmitted down the
line, the losses in the line reduce, the voltage gradients in the line can reduce and the
effects are largely beneficial. New plants of capacity up to the demand of the local load
will integrate with few thermal problems, since this unloads the local transform. Plants
of capacity up to twice that of the load demanded locally (or the rating of the local
supply transformer) may export power at times of light load which exceeds the rating
of the transform and the radial feeder.
Only smaller plants, typically of capacity less than 500kW, are connected at low voltage
(400V, 3 phase 50 Hz). Transmitting electricity at high voltage reduces system losses
and allows the dispatch of larger amounts of power. Larger plants such as multi machine
wind farms or large-hydro power stations that are thermally limited by the capacity of
the low voltage rural network are therefore connected to the 11kV and 33kV networks.
The design, selection, and application of embedded generating plants is required to
be carried out carefully to ensure satisfactory operation of the plant and the network
under all conditions, including verifying the ability if the distribution network to absorb
thermally the resulting power flow.
The costs of up-grading the network to avoid thermal violations are borne by the developer and this can outweigh all the other equipment costs, depending on the distance to
the next voltage level in the network. Even if there are no problems brought about by the
thermal limits the most prevalent limit occurs due to increased voltage variation.
10.4.3
Voltage variation
Consider the two busbar network shown in figure 52. This is the simplest representation
of distributed generation being embedded in the distribution network.
The EG produces Pgen + jQgen into the network; a local load takes −(Pload + jQload )
from the network. The resulting flow into the network at the sending end, where the
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ge Variation
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52: Two bus system
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generator and load are located is
$,$.">/',% ",0% 1'"0% 0$1/D$.9% 7?..$,>% +1/,$% /,>'% >=$% 1/,$% ">% @="9$% ",#1$% !% >'% >=$% 9$,0/,#% $,0%
0I%/,%@$.%?,/>%>$.E9%O'.%!"#$%%D'1>"#$9P%>=$%D$7>'.%$J@.$99/',9%".$%
Pline +jQline = (Pgen +jQgen )−(Pload +jQload ) = (Pgen −Pload )+j(Qgen −Qload ) (10.3)
.$7%%2%%+1/,$KO:%2%3;P%%%%%%'.%%%%%%%8.$7%%%S%%%89$,0%%A%%+1/,$KO:%2%3;P%%%%%%8'1>9%%
Consider now the simple transmission line and phasor diagrams shown in figure 53,
"7>/D$%@'B$.%/,>'%>=$%1/,$%>=$%#$,$.">$0%D'1>"#$%8
where the combined generation
and load delivers
current Iline into the line at phase
9$,0%B/11%F$%"=$"0%'(%8
.$7%FG%D'1>"#$%",#1$%"K%
."11G%>"C$,%>'%F$%)K*%@$.A?,/>%('.%9$,0/,#%$,0%D'1>"#$%./9$%7"17?1">/',9K%%+(%>=$%#$,$.">'.%/9%
angle φ to the sending end voltage Vsend , in per unit terms (or phase voltages) the vector
% ",0% $J@'.>9% +1/,$% ">% "% 1"##/,#% @="9$% ",#1$% % >'% >=$% 89$,0% >=$% D'1>"#$% ./9$% "1',#% >=$% 1/,$%
expressions are
:% 2% 3;% B/11% 7"?9$% 89$,0% >'% $J7$$0% 8.$7% % S% % )K*% @?K% % +(% >=$% #$,$.">'.% /9% ?,0$.$J7/>$0% ",0%
Vsend = Vrec
Iline
(R + jX)
Volts >=$% 1/,$%
(10.4)
% ">% "% 1$"0/,#% @="9$% ",#1$% ! >=$% ",#?1".% 0/9@1"7$E$,>%
'(%+
>=$%
D'1>"#$%
./9$% "1',#%
:%2%3;%B/11%7"?9$%89$,0%>'%F$%1$99%>=",%8.$7K%%<=/9%/9%9='B,%/,%(/#?.$%))K&KMFK%
Delivering active power into the line the generated voltage Vsend will be ahead of Vrec by
a voltage angle δ. Vrec is generally taken 8to be% 1.0 per unit for sending end voltage rise
9$,0
calculations. If the generator is overexcited and exports j"V
Iline at a lagging phase angle to
δ%
the Vsend , the voltage rise along 8
the
line impedance R + jX will case Vsend to exceed Vrec
.$7% +:%
V
∆V
φ%
= rec
1.0 pu. If the generator
is under excited and +3;%
exports Iline at a leading phase angle φ,
+1/,$%of the
V/>=%8
the angular% displacement
voltage
rise along the line impedance R + jX will cause
.$7%"9%.$(YKK%
Vsned to be less than Vrec . This is shown in 53.
R + jX
The complex power received
may be used to calculate the voltage rise at the sending
Iline
end as follows
Vsend
δ
Vrec
∆V
-IR
-IjX
,0%"9%.$(%",0%1"##/,#%!Q%>=$.$%/9%
'@%"7.'99%W%",0%89$,0%X%8.$7%
Vsend
φ
Sline = Pline + jQ
δ line = Vsend Iline
j"V
Pline − jQline
Iline =
∗V
Vsend
rec
V/>=% 8
∆V
(10.6)
Pline
% "9% .$(% ",0% 1$"0/,#% !Q% D'1>9%
.$7
9$,0
RPline − XQline
Vsend
S line = Pline + jQ line = V send . I line
*
(10.5)
j"V
− jQline
-IR (R + jX)
8
Vsend
%
XPline − RQline
+j
∗
Vsend
9$,0
Vsend
=V
jX) =% Vrec +
rec + Iline (R +
%X%8
0.'@%"7.'99%W%F?>%8
Vsend = Vrec +
KMFK%A%<.",9E/99/',%1/,$%D'1>"#$%",0%7?..$,>%@="9'.%0/"#."E9K%
∗
'B$.% UG9>$E9%
@1$J%
@'B$.%
"G% F$% ?9$0% >'%
$% D'1>"#$% ./9$%
,0/,#% $,0% "9%
-IjX
(10.7)
(10.8)
thusThere are a number important features to note:
1. PMost−importantly,
this equation is non-linear. Since Vsend appears on both sides
jQ line
line
I line = and
is*determined by itself, it can only be solved by iteration from an initial guess.
V send
V send = V rec + I line .( R + jX ) = V rec +
V send = V rec +
( Pline − jQ line )
( R + jX )
*
V send
( RP line + XQ line )
( XP line − RQ line )
+ j
*
*
V send
V send
!"#$%&%'(%)*%
D'1>"#$%89$,0I%/,%@$.%?,/>%>$.E9%O'.%!"#$%%D'1>"#$9P%>=$%D$7>'.%$J@.$99/',9%".$%
%
89$,0%%%S%%%8.$7%%2%%+1/,$KO:%2%3;P%%%%%%'.%%%%%%%8.$7%%%S%%%89$,0%%A%%+1/,$KO:%2%3;P%%%%%%8'1>9%%
%%
T$1/D$./,#%"7>/D$%@'B$.%/,>'%>=$%1/,$%>=$%#$,$.">$0%D'1>"#$%89$,0%B/11%F$%"=$"0%'(%8.$7%FG%D'1>"#$%",#1$%"K%
8.$7%/9%#$,$."11G%>"C$,%>'%F$%)K*%@$.A?,/>%('.%9$,0/,#%$,0%D'1>"#$%./9$%7"17?1">/',9K%%+(%>=$%#$,$.">'.%/9%
'D$.$J7/>$0% ",0% $J@'.>9% +1/,$% ">% "% 1"##/,#% @="9$% ",#1$% % >'% >=$% 89$,0% >=$% D'1>"#$% ./9$% "1',#% >=$% 1/,$%
Power
Systems
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$J@'.>9% +1/,$% ">% "% 1$"0/,#% @="9$% ",#1$% ! >=$% ",#?1".% 0/9@1"7$E$,>% '(% >=$% D'1>"#$% ./9$% "1',#% >=$% 1/,$%
/E@$0",7$%:%2%3;%B/11%7"?9$%89$,0%>'%F$%1$99%>=",%8.$7K%%<=/9%/9%9='B,%/,%(/#?.$%))K&KMFK%
Iline
Vsend
R + jX
δ%
Vrec
φ%
%
%
%
89$,0%
8.$7% +:%
+1/,$%
j"V
+3;%
∆V
V/>=%8.$7%"9%.$(YKK%
Iline
Vsend
φ
Iline
δ
Vrec
∆V
-IR
-IjX
V/>=%89$,0%"9%.$(%",0%1"##/,#%!Q%>=$.$%/9%
D'1>9%0.'@%"7.'99%W%",0%89$,0%X%8.$7%
Vsend
φ
j"V
δ
-IjX
∆V
j"V
Vrec
V/>=% 89$,0% "9% .$(% ",0% 1$"0/,#% !Q% D'1>9%
0.'@%"7.'99%W%F?>%8.$7%X%89$,0%
-IR
%
Q/#?.$%))K&KMFK%A%<.",9E/99/',%1/,$%D'1>"#$%",0%7?..$,>%@="9'.%0/"#."E9K%
Figure 53: Transmission line voltage and current phasor diagrams
%
*
Q.'E%55M%!'B$.% UG9>$E9% S line = Pline + jQ line = V send . I line
>=$% Choosing
7'E@1$J% V @'B$.%
send as 1.0 per-unit voltage reference for phase and load angles means
.$7$/D$0% E"G% F$%∗ ?9$0% >'% thus ◦
1.0pu6 0 is a good starting point. This produces a new calculated
Vsend
=D'1>"#$%
Vsend =./9$%
7"17?1">$%
>=$%
Pline − jQ line
">% >=$%
9$,0/,#%
value
for V$,0%
send."9%AsI line
the =iteration
proceeds it should quickly converge in this two
*
('11'B9K%%
send .
bus sytems to a final value for V
Vsend
%
( Pline − jQ line )
%
.( R
( R + jX
V send = V rec + RP
I lineline
jXline) = V XP
++
XQ
− RQ*line
rec +
line
%
V send
Vsend (n + 1) = Vrec +
+j
(10.9)
∗
∗
Vsend (n)
Vsend (n)
%
( RP line + XQ line )
( XP line − RQ line )
%
V send = V rec +
+ j
*
*
%
V
sendand voltage at theVreceiving
send
end are
% 2. If, and only if, the complex power received
both known the equation is linear and can be solved without iteration.
Vsend
RPline + XQline
XPline − RQline
= Vrec +
+j
∗
∗
Vrec
Vrec
RPline + XQline
∗
Vsend
and
jδV = j
XPline − RQline
∗
Vsend
)
!"#$%&%'(%)*%
(10.10)
3. The in-phase and quadrature components of voltage rise across the line, given by
∆V =
%
(10.11)
4. In well designed transmission lines Rline << Xline and the following approximations
are fair
XQline
XPline
∆V =
and
jδV = j ∗
(10.12)
∗
Vsend
Vsend
The implication here is that, in the transmission network, voltage rise is largely determined by the conveyance of reactive power. However, in the distribution network,
conductor cross sectional areas are reduced towards the edges (demand and currents are
lower, so conductors can be lower rated). The network is know as “radially-tapered” and
Vsend (n + 1) = Vrec +
( RPline + XQline )
( XPline − RQline )
+j
%
*
*
Vsend
( n)
Vsend (n)
:@%2(;%4,$%<'03:$J%3'F$-%-$<$2A$>%".>%A':4"#$%"4%4,$%-$<$2A2.#%$.>%"-$%1'4,%G.'F.%4,$%
.%25%:2.$"-%".>%<".%1$%5':A$>%F24,'/4%24$-"42'.7%%%%
Power( RP
Systems
( XPrec − RQrec )
rec + XQand
rec ) Machines
Vsend = Vrec +
V
*
rec
+j
V
*
rec
79
%
"5$%".>%8/">-"4/-$%<'03'.$.45%'(%A':4"#$%-25$%"<-'55%%4,$%:2.$%25%#2A$.%1@%%
∆V =
increased
resistance towards
edges.) The line geometries do not vary to the same
+ XQ
− RQ
( RPline has
( XPlinethe
line )
line
δ
=
%
and
j
V
j
extent,
so
inductance
and
inductive
reactance
remain fairly constant. This means that
*
*
Vat
Vsend
sendthe edges of the network R rises
to be similar to, or exceed X. This introduces the like-
52#.$>%4-".502552'.%:2.$5%M
%NN%O
:2.$%".>%4,$%('::'F2.#%"33-'J20"42'.5%"-$%("2-7%
lihood :2.$
that
in the
distribution network, voltage rise will be
driven by the export of real
( XQlineand
) that real power (production
XPline )
power,
may
well
be
constrained
by line resistance.
%
and jδV = j *
∆V =
*
V
V
send
Where
X >> R a remote EG send
that produces real and reactive power (over excited syn-
,$-$% 25% 4,"4;% 2.% 4,$% 4-".502552'.% .$4F'-G;% A':4"#$% -25$% 25% :"-#$:@% >$4$-02.$>% 1@% 4,$%
chronous generator with +ve Q entering the network) causes a greater local voltage rise
$"<42A$% 3'F$-7% % P'F$A$-;% 2.% 4,$% >254-21/42'.% .$4F'-G;% <'.>/<4'-% <-'55% 5$<42'.":% "-$"5%
than it".>%
would
at unity
(Q=0) or<".%
if it1$%
were
under
excited. An EG that abF"->5% 4,$% $>#$5% Q>$0".>%
</--$.45%
"-$%power
:'F$-;%factor
5'% <'.>/<4'-5%
:'F$-%
-"4$>R7%%
G.'F.%"5%S-">2"::@94"3$-$>T%".>%,"5%2.<-$"5$>%-$5254".<$%4'F"->5%4,$%$>#$57%%+,$%:2.$%
sorbs reactive power (under excited synchronous generator or induction generator with
.'4% A"-@% 4'% 4,$% 5"0$%
2.>/<4".<$%
2.>/<42A$%
-$0"2.%
-ve Q$J4$.4;%
leaving5'%
the
network) ".>%
will reduce
the-$"<4".<$%
local voltage
rise.("2-:@%
0$".5% 4,"4% "4% 4,$% $>#$5% '(% 4,$% .$4F'-G% M% -25$5% 4'% 1$% 5202:"-% 4'% '-% $J<$$>% O7% % +,25%
2G$:2,''>% 4,"4% 2.% 4,$%
.$4F'-G%
-25$%
F2::%
>-2A$.% 1@%
$J3'-4%the
'(% -$":%
If >254-21/42'.%
the apparent
powerA':4"#$%
from the
EG
is 1$%
sufficiently
larger,
voltage rise can cause the
%-$":%3'F$-%3-'>/<42'.%0"@%F$::%1$%<'.54-"2.$>%1@%:2.$%-$5254".<$7%
generated voltage (and that of local consumers) to be out with statutory limits at the
sending end of the feeder, rising further at times of minimum local loading due to
"%-$0'4$%WX%4,"4%3-'>/<$5%-$":%".>%-$"<42A$%3'F$-%Q'A$-$J<24$>%5@.<,-'.'/5%#$.$-"4'-%
increase
in active
and reactive
power
being "4%
exported
to the("<4'-%
grid. Figure 54 shows
4,$% .$4F'-GR% <"/5$5%
"% #-$"4$-%
:'<":% A':4"#$%
-25$% 4,".%
24% F'/:>%
/.24@% 3'F$-%
F$-$% /.>$-$J<24$>7% voltage
% E.% WX%variation
4,"4% "15'-15%
3'F$-%
Q/.>$-9$J<24$>%
with -$"<42A$%
import and
export
of current. 5@.<,-'.'/5%
/<42'.%#$.$-"4'-%Y9A$%:$"A2.#%4,$%.$4F'-GR%F2::%-$>/<$%:'<":%A':4"#$%-25$7%%
3'F$-% (-'0% 4,$% WX% 25% 5/((2<2$.4:@%:"-#$% 4,$%
.%<"/5$%4,$%#$.$-"4$>%A':4"#$%Q".>%4,"4%'(%
5R% 4'% 1$% '/4F24,% 54"4/4'-@% :20245% "4% 4,$%
(% 4,$% ($$>$-;% -252.#% (/-4,$-% "4% 420$5% '(%
:'">2.#% >/$% 4'% 4,$% 2.<-$"5$% 2.% "<42A$% ".>%
1$2.#%$J3'-4$>%4'%4,$%#-2>7%%[2#/-$%))7L7K1%
"#$% A"-2"42'.% F24,% 203'-4% ".>% $J3'-4% '(%
the
the
=5$.>%
=-$<%
9Y%
ZY%
[2#/-$%))7L7K17%9%+-".502552'.%:2.$%A':4"#$%-25$%F24,%:2.$%</--$.4%
Figure 54: Transmission line voltage rise with line current.
$<:"-$>%A':4"#$%A"-2"42'.%('-%:'F%A':4"#$%>254-21/42'.%.$4F'-G5%2.%4,$%\]%25%Z)*^9_`%'(%"%
*=7%%+,$%A"-2"42'.%"4%a=%Q))9KKG=R%25%'.:@%"-'/.>%Z^9_`%1/4%4,$%3-'1:$05%F24,%A':4"#$%
The statutory
declared
voltage
variationF,$-$%
for lowM%voltage
'54% (-$8/$.4% ".>% '1A2'/5%
"4% b=7% % % I.%
"% F$"G%
-/-":% .$4F'-G;%
Z% cO% 25%distribution
-$:"42A$:@% networks in the UK
is +10/-6% of a nominal 400/230V. The variation at MV (11-33kV) is only around ±6%
$%<:$"-:@%203'5$5%"%:2024%'.%4,$%<"3"<24@%'(%4,$%WX%4,"4%<".%1$%<'..$<4$>%4'%4,$%5@54$0;%
"<24@%'3$-"42'.%:$">2.#%4'%A':4"#$%A2':"42'.57%%+,$-$%"-$%(2A$%<'.>242'.5%4'%.'4$7%
but the problems with voltage violation are most frequent and obvious at low voltage.
In a weak rural network, where R + jX is relatively high, voltage rise clearly imposes
a limit on the capacity of the EG that can be connected to the system, without full
-"4'-% 25% '/4% '(% 5$-A2<$% 4,$% "33"-$.4% 3'F$-% ".>% </--$.4% >-"F.% 1@% 4,$% :'">% F2::% <"/5$% "%
capacity $.>%
operation
leading
to voltage
violations.
There
are -$0'4$%
five conditions to note.
'0% 4,$% 3-20"-@% 5/154"42'.%
'(% 4,$%
>254-21/42'.%
.$4F'-G%
/3% 4'% 4,$%
0'54%
G$$3% 4,$% A':4"#$% "4%4,"4% <'.5/0$-% F24,2.% :20245;% 4,$% :'<":% '((9:'">%4-".5('-0$-% 0"@% 1$%
1. Sgen = 0
5$4%4,$%A':4"#$%>-'3%".>%4'%-$54'-$%4,$%A':4"#$%4,$-$%4'%"-'/.>%)7*%3$-9/.247%+,25%0"@%
When the generator is out of service the apparent power and
-".5('-0$-%'((9:'">%4"35%"-$%5$4%4'%>$:2A$-%)7*&%3/%F,$.%4,$-$%25%.'%:'">7%
current drawn by the
load will cause a voltage drop from the primary substation end of the distribution
network up to the most remote consumer. To keep the voltage at that consumer
<24@% -25$5% 4'F"->5% 4,"4% '(% 4,$% :'<":% :'">;% 4,$% .$4% 3'F$-% >-"F.% >'F.% 4,$% :2.$% (-'0% 4,$%
within limits, the local off-load transformer may be tapped up to offset the voltage
.%9%!:'">R%Z%cQY#$.9%Y:'">R%-$>/<$5%".>%4,$%4-".5('-0$-%25%/.:'">$>7%%+,$%A':4"#$%"4%4,$%
drop and to restore the voltage there to around 1.0 per unit. This may mean that
'.>"-@%-25$5%4'F"->5%4,$%'((9:'">%5$442.#7%
the transformer off-load taps are set to deliver 1.05
pu when there is no load.
!"#$%&%'(%)*%
2. Sgen < Sload
As the EG capacity rises towards that of the local load, the net power drawn down
Power Systems and Machines
80
the line from the transformer (Pgen − Pload ) + j(Qgen − Qload ) reduces and the
transformer is unloaded. The voltage at the transformer secondary rises towards
the off-load setting.
3. Sgen ≈ Sload
If the generator output increases until the kW and kVAr produced closely meets
that demanded by the local load, there is minimal load on the transformer and the
voltage all along the LV feeder will be around 1.05 pu.
4. 2 × Sload > Sgen > Sload
As the apparent power and current reverses, what was a voltage drop along the
feeder becomes a voltage rise, but above the 1.05 pu off-load value. If the power
produced by the EG causes the net power sent back to the transformer (Pgen −
Pload ) + j(Qgen − Qload ) to approach the original power drop delivered by the
transformer, the voltage rise above 1.05 pu will be approximately equal to the
voltage drop that was adjusted out of the taps, 0.05 pu. The voltage rises from
1.05 towards 1.1 pu, and statutory limit.
5. Sgen > 2 × Sload
If the active and reactive power produced is more than twice the local demand,
the increase in load f low towards the network can cause the voltage to rise on the
low voltage side of the transformer and for local consumers to exceed 1.1 pu. This
is a supply quality violation.
Depending on the capacity of the generator, it may be possible to tap back down one
or more of the fixed-tap transformers along the feeder to bring the full-output voltages
back below statutory limits. However, in the event that the EG trips or is shut down,
the voltage at the remote end supplying the local load can fall back below the statutory
minimum. At low voltage installations the supply transformers will be provided with
off-load taps and once set to accommodate the presence of the EG the resulting voltage
variation is unavoidable.
Differing voltage levels can also have undesirable effects on automatic voltage control
(AVC) equipment within distribution substations, such as auto-tap changing transformers, AVC relays and (negative reactance) line drop compensation schemes. This is because the AVC voltage reference is usually on the secondary winding of the supply transformer, and reverse power flow through the transformer (when the embedded generator
is exporting) causes erroneous operation of the voltage sensing relay and tap changing
mechanism. Time invariant energy sources such as small scale hydro plants or bio energy
schemes are in this category where excursions in real and reactive power flow may take
place on a weekly basis. Time-variant energy sources such as wind-farms or wave devices without storage produce power whose instantaneous value can change in seconds.
This is a much more serious operational problem because the voltage variation that this
causes can have AVC equipment operating continually rather than occasionally, leading
to increased wear and maintenance. There is however some more recent observation that
where generation follows demand such as at wind farms close to rural communities, the
Power Systems and Machines
81
net effect is actually to reduce the frequency of operation of the AVC equipment.
10.4.4
Increased fault levels
Prior to the installation of an EG the switchgear in the distribution network will have
been designed and specified based on the fault levels resulting from centralised generation
and the minimum system interconnected impedance. The presence of any EG in the
network will increase the fault levels all through the network, but most noticeably at the
point of connection of the EG. Synchronous and induction generators will both cause
an increase in the fault levels, but as a later section will show the main difference is the
peak asymmetric current and the duration of the sustained short circuit current.
In urban or industrial areas the proximity to reinforced stiff grids, with low intervening
impedance, means that prior to the installation of the EG the fault level will already
be high, and the new switchgear, which the developer will buy and own, will have to be
fully fault rated for the system fault level - perhaps 500MVA - even though the plant
to be connected may only have a small capacity, say 50kVA. In this case the cost of the
switchgear will be determined by the distribution network.
In rural areas the lower fault levels resulting from the higher intervening system impedance
may mean that the distribution switchgear, while adequately rater before the arrival of
the EG, will have to be upgraded at a cost borne by the developer.
Ironically, it is not only necessary to check the maximum fault level with and without
the EG at the design stage, it is just as necessary to check the minimum fault level at
the EG and out in the network. This is to ensure that in the event of a short circuit
fault close to or remote from the EG, enough fault current will flow and be detected to
cause the over current protection to operate and disconnect the sources of generation
feeding the fault.
10.4.5
Reduced stability
Steady-state and transient stability is necessary to determine the ability of the embedded synchronous generator to remain synchronised following small disturbances in the
grid system (e.g. gradual changes in load, excitation or prime over input) and large disturbances (e.g. severe load changes, switching operations and electrical faults). During
such disturbances there is a dynamic mismatch between mechanical power input into
the generator and that absorbed to produce the electrical power accepted by the grid
connection. This results in acceleration, dynamic change in shaft speed and rotor angle
and possible loss of synchronism.
In the event of a distribution network fault the impedance connecting the embedded
generator to the ‘infinite’ grid will change. In a radial feeder going open circuit to clear
a fault, the PES requires that the EG is disconnected immediately to prevent ‘islanded’
Power Systems and Machines
82
generation. Faults that occur further out in the grid can cause step changes in the
system reactance and load angle between the EG and the infinite grid. Depending on
the initial, intermediate and final load angles, the generator may not converge smoothly
on the new load angle and may lose stability.
The likelihood of loss of synchronism can be assessed using the graphical equal area
stability criterion. When a fault causes the rotor of the generator to swing such that the
area on the torque-angle plane represents the disturbing effort, exceeds that representing
the restoring effort, it is likely to lose synchronism. That angle is known as the critical
clearing angle. Dynamic stability studies estimate the time variation of rotor angles
following the application and clearance of faults at different point in the system, and
assesses wether the rotor will reach its critical clearing angle before the fault is cleared.
If the fault is cleared before the machine reaches the critical clearing angle, it will remain
synchronised. Low EG inertias, short transient time constraints, high machine or system
reactances, and long fault clearance times all reduce overall stability.
Smaller EG’s are unlikely to substantially change the stability of the distribution network
even when connected close-up, but the ability to remain synchronised (safely) through
system fault clearances will be of interest to the EG developer. As the point of connection
becomes more remote electrically, stability margins will be reduced until, in the extreme
case of an EG at the end of a long radial feeder, any fault in the feeder and subsequent
clearance will trip the generator on loss of mains or over current, depending on the
distribution of rural load. Individual stability of new EG’s will be of some interest to
the connecting DNO is terms of preserving local quality supply, but as the population of
EGs increases, overall system stability must be re-appraised. Demand in the UK is not
forecast to increase substantially over the next 10 years and a large population of of EGs
could displace ageing thermal plants. It will be important to ensure that loss of system
stored energy from displaced plants does not impair overall system stability.
10.4.6
Compromised operation of protection equipment
Design, calibration and setting of protection equipment installed in the EG’s synchronising or connection circuit breaker to protect the EG against the effects of the network
feeding a fault inside the EG boundary. It is more difficult to provide and set equipment
that will protect network equipment against the effects of the EG feeding a fault within
the network boundary. Differing nominal currents and directional sensitivity might be
used, but since the EG cannot be constrained, it is difficult to cost effectively provide
settings (since they cannot easily be changed) across all relays that provide close protection in the presence and absence of the EG. The 11 or 33kV circuit breakers within the
network may have current, or impedance operated protection relays set to protect the
network for uni-directional flow. Their operation may become ineffective or spurious as
a result of the redirection and changes to current flow in the system. Protection settings
for the under and over voltage relays that are appropriate for the condition where the
generator is out of service may cause the relays to operate and trip the DNO-owned cir-
Power Systems and Machines
83
cuit breakers when the generator is in service and loaded (or vice versa). That the EG
cannot be constrained is a problem for the DNO since they cannot (continually) adjust
their system voltage settings to compensate for the presence of the EG - having done
that the developer of the EG scheme may, without reference, shut the plant down.
10.4.7
Increased harmonic content
Synchronous generators that are designed to the appropriate British, IEC, or VDE standards and that are correctly installed will not add significantly to the harmonic content
of a balanced system of distribution network voltages. EGs that are connected to the
distribution network by power electronic converters such as photovoltaic cells producing
inverted direct current. Variable speed ac generators with rectifiers and inverters will,
at best, produce a re-assembled ac waveform that contains many higher order harmonics
that are not acceptable to the DNO since this will distort and contaminate the supply
to their consumers. It is possible to buy and fit harmonic filters but they are expensive
and their absorbance is limited.
Directly connected EGs can improve the harmonic content of the voltage waveform in the
distribution network because they generally have a low harmonic impedance that acts
as a relative short circuit to higher frequencies. Care has to be taken where induction
generators (or motors) are installed with power factor correction capacitors, because
the capacitors will shunt higher frequencies and can have appreciable high frequency
currents (and increased heating effects).
Induction generators can also be used to balance the voltages in a badly balanced rural
network. The induction generator has a low impedance to unbalanced voltages and tends
to draw off unbalanced currents. This can result in negative sequence currents, reverse
fluxes and increased heating on the rotor of the machine.
10.5
Synchronous or induction generators
Once the output power and speed of the prime mover are determined, a fundamental
choice that developers must make is whether to install synchronous or induction generator(s). Synchronous generators are normally selected when the delivery of mechanical
power has a low time variance, such as that from a water/steam/gas turbine. Induction
generators are also used to develop power from these prime-movers if they can be magnetised from the distribution grid to which they are connected. Where there is a high
time-variance such as with wind turbine, induction generators more readily absorb the
periodic changes in input power.
Induction generators are often initially selected because the cost per kW is lower, standard proprietary machines may be applied (overspeed permitting), there is no (apparent)
need for a governor or automatic voltage regulator, synchronising and protection equipment is much less sever than that from a synchronous generator. Where the grid is
Power Systems and Machines
84
‘weak’, at more remote points of the distribution system, the magnetising current demanded by the induction generator increases reactive power flow and this can locally
reduce system voltage.
Synchronous generators also cause load flow to vary and in weak areas of the distribution
grid can disturb unacceptably system voltage regulation. Exporting real and reactive
power from the remote end of a long radial feeder can cause excessive voltage rise at the
remote end. Indeed induction generators are now perceived to assist voltage regulation,
by absorbing reactive power and thus reducing voltage rise as real power is exported.
Some DNOs will now waive their right to charge for the supply of reactive power to some
EG sites.
Often, where the choice is left to the developer, induction generators are initially chosen because the cost per kW is lower, standard proprietary machines may be applied
(over-speed permitting), there is no (apparent) need for a governor or automatic voltage
regulator, synchronising and protection equipment is simpler and less expensive. The
additional expense of a synchronous generator and synchronising equipment is considerable and the economic justification for installing synchronous machines may be based on
comparing the sum of avoided kAVrh costs with the added capital costs of a synchronous
generator, governor and synchronising equipment.
10.6
Conclusions
As electricity demand continues to grow, along with environmental awareness and accountaibilty, the number of embedded small and mini hydro generators connected at
distribution voltage will continue to increase in industrialised, newly-industrialised and
less-developed countries.
To integrate around 2GW of new plant every year for the next 10 years will transform
the currently passive, radially tapered uni-directional distribution network into an omnidirectional system that will have to be managed actively. While generating plant and
the distribution network will have to be operated in a more flexible and innovative
way to accommodate a growing portfolio of generating plant, operating from diverse
energy sources, supply quality, integrity and safety cannot be compromised. Embedded
generator and distribution system protection will play an increasingly important role in
the definition and expansion of EG capacity and limits of operation while preserving
security of supply. The satisfactory electrical integration of EGs depends on the ability
of the distribution network to accept (within statutory voltage limits) the presence or
absence of the new plant and this is one of the main limiting factors to accommodating
large amounts of embedded renewable generation. Voltage and frequency limits at low
voltage have already been widened to accommodate and retain a large population of
EGs, and there is a need to assess whether these limits could be widened at higher
voltages to allow a greater transmission of energy from the rural distribution network
beyond the primary substation.
Power Systems and Machines
85
Many arguments are advanced about the perceived benefits of embedded and renewable
generation. To protect existing customers from the effects of local voltage variation
assessments of nearly all new embedded renewable energy schemes require that the local
distribution be upgraded, often at a cost that renders them unfeasible. This is because
the reverse load flow is determined when the EG is at full production and local demand
in the area affected is set at an arbitrary low value. Production and demand vary
independently with time, particularly from renewable EGs, and probabilistic analysis
of the net export of active and reactive power is likely to show that the actual extent
in time violations can be low. Constraining off the plant under these conditions may
be economically more acceptable than applying full system reinforcement costs. The
contribution that EGs make to local demand does not always allows deferral of renewal
or reinforcement of the network, which is tapered towards low voltages. Other than
in conditions 1 and 2 in section 10.4.1, which limit EG capacity, the need to upgrade
the network will always be a possibility. Even in these conditions, if local load drops
(due to population drift or industrial decline), it might become necessary to reinforce
retrospectively the distribution network to be able to continue accepting EG output. It
is already difficult to predict network load and consequently schedule constrained bulk
generators to provide a margin that covers the shutdown or tripping of unconstrained
EGs (added on top of capacity that may be constrained) to meet the same variability
of load, will make it more difficult to schedule confidently bulk generation. EGs usually
have lower inertia, stored energy and stability than the plan that they are displacing.
Their presence also reduces the extent of part-loaded plant that allows the system to
meet transients in a stable way. Additionally they will potentially displace plant that is
synchronised to absorb and regulate their fluctuating output. Overall system stability
will have to be preserved.
While the natural energy resource determine embedded generator rating, the capacity
at which the plant can operate depends on the characteristics of the rural network. Remoteness from the 11/33kV primary substation may restrict them thermally, or on the
basis of voltage regulation, the amount of power that the system can absorb. Larger machines should ideally be installed close to the primary substation but the location of the
energy source usually prevents this and requires the distribution to be upgraded.
Even at the remote ends of the rural network, embedded generators have to become fairly
large before overall fault levels are exceeded. Voltage rise on export of full generator
power is a serious limit but this may be mitigated by the installation of under-excited
synchronous generators or induction machines. Integrating a large capacity EG site at
a remote part of the system may be difficult, whichever type of generator is selected.
Voltage rise must always be considered carefully and is usually the limiting factor in
the development of a remote site. Initial consideration of the above criteria may lead
the DNOs to insist on up-grading some or all sections of their radial system up to the
point of connection. Once this has been designed and costed the remaining criteria may
still be significant. Large, low inertia synchronous machines, operating at lagging power
factors at remote point of the rural system may become unstable after transients such
Power Systems and Machines
86
as short circuit fault clearance.
The integration of a proposed plant into the network must be assessed carefully to
ensure satisfactory load flow, thermal acceptability, voltage regulation, final fault levels,
supply quality, transient stability and protection discrimination. The on and off load
voltages at the remote end of the system have to be identified under all directions
of real/reactive power flow. Protection co-ordination and transient stability must be
assessed to ensure that the embedded generator does not reduce overall system integrity
under fault conditions.
Power Systems and Machines
11
11.1
87
Balanced per-unit fault analysis
The per-unit system
In the network analysis of power systems, or in the modelling of rotating machines,
instead of using actual values of quantities it is usual to express them as fractions of
reference quantities, such as rated full-load values. These fractions are called per-unit
(denoted by pu) and the pu value of any quantity is defined as
Per-unit =
the actual value (in any unit)
the base or reference value (in the same unit)
(11.1)
For instance, to a base time of one minute, one second is 1/60 per unit. To a base power
of one kilowatt, a 60 W light bulb may be expressed as rated 0.06 pu, and a 100 W bulb
as 0.1 pu. Ten 60 W bulbs and twenty 100 W bulbs would collectively dissipate (10.0.06
+ 20.0.1) = 2.6 pu power. Referring back to the base power of 1000 W, this represents
a total power of 2600 W.
Although the use of pu values may seem a rather indirect method of expression, it allows
the expression of network or machine in unitary terms rather than as absolute values,
that may be in many different electrical units, such as volts, amps, ohms, or watts. In
the case of the above light bulbs, which had a base power of 1000 W, for a base voltage
of 250 V, base current would be 1000/250 = 4 A. Each of the 60 W bulbs (0.06 pu
power) takes 0.06 pu current, and each of the 100 W bulbs (0.1 pu power) would take
0.1 pu current when connected to 250 V (1.0 pu). Total current may now be calculated
in per units of base current as (10.0.06 + 20.0.1) = 2.6 pu current, or to a base of 4 A
= 10.4 A.
Neither base nor individual resistances have featured in this calculation, which demonstrates how a circuit may be solved without concern for absolute component values.
Base resistance would be base voltage/base current = 250/4 = 62.5 Ω and the respective
per unit and absolute resistances of the bulbs might be found from: R(pu) = V(pu)/I(pu)
= 1pu/0.06pu = 16.66 pu (1042 Ω) for the 60W bulbs and 1pu/0.1pu = 10 pu (625 Ω)
for the 100W bulbs.
A generator that was rated 5MVA would be expected to produce 1.0pu apparent power
to a base of 5 MVA, in this case its own rated capacity. If it is operated at 0.8 power
factor, 1.0pu real power would be 4 MW, and 3 MVAr would be
√ 1.0pu imaginary power.
On an 11kV network, this machine would deliver 262A, (S = 3Vl Il ), and this is 1.0pu
current.
Calculations of currents, impedances, powers, losses, etc., may become very cumbersome,
because there are numerous voltage levels linked by transformers. Without the use of the
per-unit system, it is necessary to convert power flows to different voltages and currents
on either side of transformers. In an ideal transformer, primary and secondary currents
Power Systems and Machines
88
are related by the inverse of voltage ratio, and impedances seen ‘through’ transformers
are related by the square of voltage ratio.
With the per-unit system, the whole transmission system for normal operating conditions
may be reduced to a homogeneous set of per-unit impedances operating entirely at 1.0
pu voltage. The pu impedances and their bases exhibit all the same relationships as
absolute values, and obey all the same circuit laws, such as Ohms law and Kirchhoffs
Law, but the arithmetic that is involved in the per-unit calculations is much simpler.
Within three-phase power systems, a per-unit phase voltage has the same numerical
value as the corresponding per-unit line voltage. In the per-unit system, three-phase
values of voltage, √
current and power can be used without undue anxiety about the result
being a factor of 3 incorrect. The actual values of resistance and reactance for lines,
cables and other apparatus are phase values.
The essence of per-unit calculations is
(a) Choose a base for all circuit parameters (NB one base may define others)
(b) Convert all absolute values into per-unit values
(c) Carry out the very much simpler calculations
(d) Convert all per-unit values back into absolute terms using base in (a)
11.2
Choice of base
All of per-unit analysis rests on the correct and consistent choice of a common base,
and accurate conversion of circuit parameters into and out of that base. In Power
Systems, four base quantities are required to completely define a per-unit system; these
are: voltage, current, complex power, and impedance.
Note that if two of them are set arbitrarily, then the other two are automatically fixed,
because complex power = voltage x current and impedance = voltage/current.
11.3
Base MVA, voltage and impedance
In a single-phase system, the following relationships hold:
Base apparent power (VA) = Base voltage (V) × Base current (A) = Vb Ib
(11.2)
Base MVA (MVAb ) = Base voltage (kV) × Base current (kA) = kVb kIb (11.3)
In a three-phase system, using phase voltage and current
Base MVA per phase = Phase voltage (kV) × Phase current (kA) = kVp .kIp
Using line voltage and phase current, as would be more normal
√
Base MVA total = 3xkVp .kIp = 3kVl .kIl
(11.4)
(11.5)
Power Systems and Machines
89
Iphase
Zbase
Vphase
Figure 55: Per Unit Impedance
In a circuit such as that shown in figure 55, if a current Ip = Iphase is measured in
the conductors under the effects of measured voltage Vp = Vphase , it seems as if there
exists an impedance equivalent to Vp /Ip somewhere to the right of the measured point
in the circuit. This is the base impedance of the circuit, determined from phase voltage
and current. It should be recognised that the actual impedance of a circuit component
plays no part in determining base impedance, which is determined solely by the voltage
at the point of interest and the current flowing at that point. The actual impedance
of the phase conductors is a separate entity, which may be measured (in ohms) and
subsequently related to the base impedance, Zbase in per-unit terms.
Base voltage (V)
Vb
=
Base current (A)
Ib
2
2
V
(kVb )
(kVb )2
= b =
=
Vb Ib
kVb kIb
MVAb
kVp2
=
MVAb (phase)
kVl2
=
MVAb (total)
Zbase =
(11.6)
(11.7)
(11.8)
(11.9)
All of the above calculations of base impedance produce the same ohmic value. Actual impedance, current, and voltage can now be converted to per-unit terms using the
respective base values.
Actual impedance(Ω)
Base impedance(Ω)
Actual current(A)
=
Base current(A)
Actual voltage(V )
=
Base voltage(V )
Per-unit impedance Zpu =
Per-unit current Ipu
Per-unit voltage Vpu
(11.10)
(11.11)
(11.12)
In the single-phase equivalent circuit
shown in figure 55, if the measured line voltage is
√
3.3 kV, the phase voltage 3300/ 3 = 1905V . The measured line=phase current is 953
A. If phase voltage and line current are taken as base voltage and base current, base
impedance is the ratio of base voltage to base current = 2Ω.
Power Systems and Machines
90
If the conductor has impedance 0.1 Ω at that point, the per unit impedance may be
calculated from the base impedance, as 0.05 pu. It would have been equivalent to
calculate base MVA using equation 11.9.
11.4
Per unit reactances of transformers
Transformer and generator reactances and resistances may be quoted in absolute ohms,
or in per unit terms. In either case, they will be required in per unit terms for efficient
computation. There is a straightforward convention that where impedances are quoted
in per unit terms this is always to the base MVA of the unit.
X
I1
nI2
I2
~
V1
nV1
V2
Figure 56: Transformer equivalent circuit
The 9 MVA transformer in figure 56 has a leakage reactance of 0.05 pu and steps 11 kV
down to 3.3 kV. The primary and secondary phase currents may be calculated (although
strictly, this is unnecessary).
MVA(total)
9 × 106
√
=√
= 472A/ph
3V1
3.11 × 103
MVA(total)
9 × 106
√
I2 =
=√
= 1575A/ph
3V2
3.3.3 × 103
I1 =
(11.13)
(11.14)
The base reactances may be calculated in both the high voltage winding
Bbase =
kVp2
6.352
=
= 13.44Ω/ph
MVAb (phase)
3
(11.15)
and in the low voltage winding
Bbase =
kVp2
1.952
=
= 1.21Ω/ph
MVAb (phase)
3
(11.16)
The leakage reactance is made up to represent total flux leakage in the primary and
secondary, but it is usually summed and expressed in either the primary or secondary.
Power Systems and Machines
91
The ohmic values of the leakage reactance may now be calculated from the base values
above.
X = Xbase .Xpu
(11.17)
which gives X1 = 0.672Ω/ph and X2 = 0.0605Ω/ph. While the ohmic values of leakage
reactance differ in the primary and secondary windings, the voltage drops across the
leakage reactances are V1 = 317V and V2 = 95V , which expressed as per-units of the
primary and secondary phase voltages are 317/6350 = 0.05 and 95/1905 = 0.05. This
illustrates the major virtue of the per-unit system. The per-unit voltage drop on 1.0 pu
voltage is exactly equal to the per-unit leakage reactance, either in the primary or in the
secondary, and the transformer may be represented as a single reactance. This is very
powerful in system analysis, because varying voltages are dealt with in the establishment
of base impedance.
11.5
Per-unit reactances of generators
A 30 MW, 0.8 lagging power factor synchronous generator has synchronous reactance
Xd of 1.24 pu and operates at 6.6 kV. If the generator is star connected, the line and
phase currents are equal and may be calculated (although again this is only included for
illustration).
MVA(total)
30 × 106
√
√
I1 =
=
= 3280A
(11.18)
3Vl
0.8 3.6.6 × 103
The base reactance may be calculated
Xbase =
kVl2
6.62
=
= 1.162Ω/phase
MVAb (total)
37.5
(11.19)
The ohmic value of the synchronous reactance may now be calculated from the base
values above. Xd = Xbase .Xpu = 1.162 × 1.24 = 1.44Ω/phase. The voltage drop across
the synchronous reactance at full load current would be 3280 x 1.44 = 4724V. Expressed
as a per-unit of phase voltage this is 4724/3810 = 1.24pu which is identically equal to
the per unit reactance multiplied by 1.0 pu current.
The fault contribution from a synchronous generator is a complex decaying exponential
function of time with three time constants. For the purposes of fault calculations, the
three-phase symmetrical short circuit contribution is determined from the sub-transient
reactance Xd00 . To the base reactance of the generator, Xd00 may lie between 0.06 and 0.2
per-unit. Based on 1.0 per-unit excitation in the generator, this means that the fault
current at the terminals could be as high as 1.0/0.06 = 16.66 pu, where 1.0 per unit current is rated current. Or, to the base MVA of the generator, the three-phase symmetrical
short circuit MVA would be 1.0 MVA/0.06 pu reactance = 16.66 pu MVA.
Power Systems and Machines
11.6
92
Per-unit reactances of synchronous motors
The calculations for the synchronous motor are identical in detail, but the major difference is that the full load MVA of the motor must be determined from the input
apparent power, which is determined from the shaft mechanical output power, overall
electromechanical efficiency and power factor.
MVA (total) =
11.7
Pout
η cos φ
(11.20)
Conversions between bases
In power system per-unit analysis, it is necessary to select (arbitrarily) a common base
MVA and express all per-unit resistances, reactances and impedances across the network
to this base. If given, absolute ohmic values will convert readily to the common base
MVA. Where component reactances have already been expressed to the base MVA of
the machine or transformer, the per unit reactance will have to be recalculated to the
network MVA base. Since
X(ohms)
MVAb
Xpu =
= X(ohms)
(11.21)
Xbase
kVl2
This formula may be used to calculate per-unit values of reactance, which are directly
proportional to base MVA. Thus per-unit reactances may be expressed to a new base
as
MVAb new
Xpu new = Xpu old ×
(11.22)
MVAb old
This is the most commonly used conversion in power system analysis.
11.8
Conversions between off-nominal voltages
In per-unit power systems analysis, per-unit reactances of transformers, motors, and
generators are expressed to the MVAb of the plant, assuming that the unit operates at
nominal 1.0 per unit voltage. In some cases, transformers are tapped-up to promote
the correct flow of reactive power and operate above the nominal voltage. To correctly
represent plant which is operating at off-nominal voltage, the per-unit reactance must be
re-expressed to the new voltage. Per-unit values of reactance are inversely proportional
to the square of base voltage, without absolute ohmic values or base MVA changing.
Thus per-unit reactances may be expressed to a new base voltage as follows.
kVl old2
Xpu new = Xpu old ×
(11.23)
kVl new2
This conversion is only used when a transformer, motor or generator is operating at
a voltage other than nominal 1.0 per unit, whereby the pu reactance was originally
calculated.
Power Systems and Machines
11.9
93
Use in fault level calculations
The advantage of the per-unit system for balanced fault analysis emerges in the solution
of even small power systems (such as that shown in figure 57), which may contain
many components at different voltages, passing widely varying currents through differing
impedances. Calculation of absolute values directly becomes complex, but provided that
the power system is balanced and may be represented by a single-phase equivalent circuit,
it may be analysed in per-unit terms. Once
• all of the line (and phase) voltages are identified,
• plant capacities (in MVA) are identified,
• and individual impedances are converted to per-unit (to a common base MVA),
the per-unit system of analysis reduces all the individual parameters into “per-units”
of nominal values and removes absolute terms. The complex circuit above reduces to a
single impedance to earth with 1.0 per unit voltage across it.
Since phase voltage has been set to 1.0 pu, the per-unit fault current may be calculated
as phase
If (pu) =
Vphase (pu)
Xfault (pu)
and If (A) = If (pu)xIbase
A/phase
(11.24)
The fault current emerges from this calculation as If pu. To convert back into absolute
current at any point in the circuit, the per-unit fault current at that part of the network is multiplied by base current. The chosen common base MVA applies across the
whole circuit, but -depending on voltage 1pu current will vary. Fault current is usually
expressed in A or kA, but an equally common unit for fault load flow is MVA.
MVAfault (pu) = Vphase (pu)xIphase (pu) =
2
Vphase
(pu)
Xfault (pu)
(11.25)
but since Vphase = 1.0pu
MVAfault (pu) =
1
(11.26)
Xfault (pu)
expressed back as fault MVA
Fault level (MVA) = MVAfault (pu).MVAbase =
MVAbase
Xfault (pu)
MVA/fault
Steps in the solution of the fault circuit are:
1. Draw single-phase equivalent circuit
2. Identify all system operating values
Voltages (kV)
Plant Capacities (MVA) - remembering to allow for motor efficiency
Impedances or Reactances (in Ω or in pu)
(11.27)
Power Systems and Machines
Overhead
line
Transformer
T1
33 kV Grid
Bus
section
circuit
(Normally closed)
breaker
Xt
11 kV Switchboard
XI
Cable
Hydrogenerator
94
Xc
~
T2
Xg
Xt
T3
If pu
Xt
~
415 V
Switchboard
Xf pu
V = 1.0 pu
Loads
Figure 57: Industrial power system with fault at 415V and single phase equivalent circuit
3. Consider position of fault (and impedance of fault - usually = 0).
4. Assert 1.0 pu voltage throughout system
5. Choose common MVA base for whole system
Usually three-phase capacity of largest transformer or generator, or multiple of 10.
6. Check per-unit reactances are specified based on operating voltage
Otherwise, adjust for change of voltage. For instance, a generator may have reactance 0.24 pu to its own base and 13.8 kV line voltage. In a system where it
is operated deliberately at 15 kV, Xpu will have to be adjusted. If it is necessary
to correctly represent plant which is operating at off-nominal voltage, the per-unit
reactance must be re-expressed to the new voltage
2 (old) kVline
Xpu (new) = Xpu (old)
(11.28)
2 (new)
kVline
7. Convert all ohmic and per-unit impedances to new pu value defined by base MVA
Where the ohmic values are given
Xbase =
2
Vphase
kVline
=
Iphase
MVAbase
and X(pu) =
X(Ω)
Xbase
(11.29)
Where per-unit values are given to the base MVA of the component, such as in
transformers, generators or motors, the per-unit value must be recalculated to the
common base MVA.
MVAbase (new)
Xpu (new) = Xpu (old)
(11.30)
MVAbase (old)
Power Systems and Machines
95
8. Combine reactances/impedances in series/parallel
To achieve single voltage source on left-hand side (LHS) feeding fault via single pu
reactance.
9. Calculate fault current in pu (Equation 11.24)
10. Calculate actual circuit fault currents (where required).
Power Systems and Machines
12
96
Power system protection equipment
12.1
Introduction
Electricity supply system is designed and equipped to distribute electricity to geographically separate customer loads in a manner that is safe, reliable and economic. Security
of supply is ensured by:
• Capacity margin of generation over peak load demand,
• Topography of the network providing alternative/back-up circuits to supply the
loads,
• Design and operation of the switchgear or circuit breakers that control the flow of
power to loads,
• Design and operation of the protection equipment that protect the network equipment and personnel.
The UK transmission system operates at two voltages 400 kV and 275 kV (also 132 kV
in Scotland). Bulk generation (¿1200 MW capacity) is connected to the transmission
system. The distribution system operates at lower voltages, from 132 kV through 66
kV, 33 kV, and 11 kV finally supplying domestic load at 400V/230V. The normal flow
of power is controlled by switchgear (collective term for circuit breakers arranged into
switchboards with a common busbar) owned by the network operator.
12.2
Switchgear and circuit breakers
Network circuit breakers are three-phase switches, with mechanically operated contacts
that may open and close in a controllable manner. Operators may open or close them
locally or from remote positions, or they may be activated by the local or remote protection equipment. They must open and close very quickly, under both normal and
abnormal operating conditions. The mechanical power that closes them may be from
motor, or (manually) charged spring mechanisms, or solenoids. Once closed, they must
be able to open without calling for additional motive power. The closing process charges
springs that remain compressed (by a latching mechanism), while the breaker is closed.
When the breaker is to be opened, or “tripped”, the latch is withdrawn and the energy
stored in the opening springs drives the contacts open very quickly. While in operation
and closed, the circuit breaker contacts must be able to conduct rated current without
exceeding rated temperature. While in operation and open, with one side live, the circuit breaker must be able to maintain the operating voltage across the contacts under
both clean and polluted conditions (such as degradation of the insulating medium which
may be oil or air). When called upon to open under normal or abnormal conditions,
the circuit breaker must be able to interrupt the flow of current safely and quickly. As
the circuit breaker contacts separate, an arc forms from the vaporised contact material.
Power Systems and Machines
97
The alternating arc current is naturally zero twice during a 50Hz cycle, aiding extinction. However, if the contacts do not open far enough by the next voltage peak (current
zero), which will be 10ms second later, the arc may re-strike. The arc is extinguished
through elongation and cooling. The magnetic forces of the arc current or an air blast
may be utilised to elongate the arc. Circuit breakers are classified based on the medium
used to extinguish the arc, i.e., air, vacuum, oil or SF6 gas. Additionally, the circuit
breakers must be able to withstand thermally and electromechanically any prospective
short circuit currents that may flow in the network.
Voltage
transformer
Secondary
wiring panel
Upper busbars
Current
transformers
Mechanism
Circuit
breaker
truck
Lower
busbars Vacuum
interupter
in SF6
Fixed contact stem
Sputter
shield
Bellows
Contacts
Retaining
bolt
Moving
contact
stem
Figure 58: Alstom HMX indoor circuit breaker (2000A, 36kV), the vacuum interrupter
in SF6
Indoor switchboards are made up of a row of separate circuit breakers in metal-clad
enclosures, shown end-on in Figure 13.2. This is a double busbar arrangement and the
upper and lower busbars are connected together along the length of the switchboard.
The incoming or outgoing circuits are connected at the rear of the switchboard. Protection and instrumentation current transformers (CTs) and voltage transformers (VTs)
are incorporated, as well as feeding relays and meters which may be mounted in the
secondary wiring panel. The circuit breaker truck may be isolated by horizontal withdrawal, and may connect to either the upper or lower busbars by raising or lowering the
mechanism.
Power Systems and Machines
12.3
98
Protection components
The simplest form of operation and protection system is a manually operated switch (or
circuit breaker) protected by one (or three) fuses. In distribution networks, switchboards
control the flow of power, wherein the 3 basic protection components are: circuit breakers and switchgear, sensing and measurement CTs and VTs and protection relays. The
sensing equipment monitors the system voltages, currents, directions and phase angles
continuously. Their outputs are permanently connected to voltage and current operated
protection relays, which continuously detect the state of the network. If the relays determine that the network is healthy, they remain reset. If a fault occurs or if the network
conditions become abnormal, the voltage and current signals fed to the protection relays
will be interpreted to determine whether and how quickly to operate the output of the
protection relays. If the protection relays determine that the circuit breaker is to be
tripped, their output contacts close in the circuit breaker trip circuit, the breaker opens
and the fault should be cleared. Figure 13.3 below shows the schematic of overcurrent
protection and manual tripping.
CT
Relay I’
operating
coil
Relay
contacts
Battery
Manual
trip
X
X
Breakers
Circuit breaker
I
=
R1
X
Trip coil
Figure 59: An overcurrent protection scheme.
Relays compare and react to current and voltage signals derived at the CTs and VTs.
The secondary currents and voltages are scaled measures of the primary values. In
addition to being easier to use at circuit board level, this allows the secondary circuits
to be isolated from the power system.
12.3.1
Current transformer (CT)
Current transformers (CTs) are used to produce a scaled measure of the primary currents
in the power system. The primary winding of a CT usually consists of a single turn (N1
= 1 turn) obtained by running the power systems primary conductors (or circuit side
busbars) through the CT core. CTs are either bar-type with the secondary winding
Power Systems and Machines
99
wound and encapsulated around a busbar, or ring type, wound in a toroid through
which one or more phase conductors can be passed. The normal current rating of a
CT secondary is normally 1A or 5A, although the first few cycles under short circuit
conditions CT currents are 10 to 20 times greater than the normal rating.
Specification of CTs
The CTs are specified to BS 3938 by their current ratio, the burden that they can supply
within their accuracy limits and their accuracy class.
Current ratio: (primary current:5 or primary current:1) say 1200:5 or 1200:1
Burden: may be between 5VA and 15VA
Accuracy class: 5P10, or 10P10
Protection CTs must function accurately under fault conditions, up to a fault current
known as the ‘accuracy limit current’. For a 10P10 CT, this current is 10 times the
normal current, and error is 10%. CTs that are used to detect nulls or balance between
opposing currents such as in zonal, or differential protection are required to be fully
matched and of much higher accuracy such as class 1.0 or class ‘X’.
Protection of CTs
The CTs are designed to operate with specified accuracy into a short circuit, when their
ratio limits the secondary nominal current to 1 A or 5 A. CTs are taken out of service
by shorting the secondary terminals with shorting links. If the secondary impedance
rises, the induced voltage in the secondary rises to try to maintain the correct secondary
current. For this reason, CT secondaries should never be left open-circuited. The induced
secondary voltage can damage the secondary insulation or be hazardous to personnel.
They should never have fuses in the secondary circuits.
12.3.2
Voltage transformer (VT)
These are used to produce a scaled measure of the primary voltages on the power system.
The VT may be modelled as an ideal transformer for power system protection purposes,
where Vin/Vout = N1 /N2 = n. Term ‘n.Vout ’ is a scaled version of Vin in phase with Vin
(by minimising the VT leakage impedance). Ideally, the VT output is connected to a
voltage sensing device of infinite impedance. Consider the 3-limb transformer in Figure
13.3.2 (only primary windings are shown). VA , VB , VC are the system voltages, line to
earth. Normally VA + VB + VC = 0.
When an earth fault occurs, one voltage is held to earth and VA + VB + VC 6= 0
but with only 3-limbs to P
carry the flux ΦA + ΦB + ΦC = 0, there is P
a conflict between
system voltage dictating
Φ 6= 0, and transformer design requiring
Φ = 0. For this
reason, a 5-limb transformer (figure 60) must be used if the secondary is to give a true
representation of the primary voltages under all circumstances. During earth faults, zero
sequence currents flow and produce zero sequence voltages. The zero sequence voltage
at the relay point is
VA + VB + VC = 3VAO
(12.1)
Power Systems and Machines
100
is called the residual voltage and is obtained from an open delta winding on a 5-limb
VT if it is required. The open delta winding sums vectorially the secondary voltages.
If the system is balanced, i.e., if there is no earth fault, VA + VB + VC = 0. If there is
an earth fault, this is no longer the case, and the residual voltage is therefore a way of
detecting and discriminating earth fault conditions.
A
N A
B
C
B
C
c
b
a
n
ɸc
ɸb
ɸa
Residual
voltage
Figure 60: A 3-limb and a 5-limb voltage transformer.
Primary
VA
Secondary
Residual
Va
VA
Vb
VC
VB
VC
VB
Vres Vc
Figure 61: Resolution of residual voltage
Specification of VTs
VTs are specified to BS3941 by their voltage ratio, the burden that they can supply
within their accuracy limits and their accuracy class.
Voltage ratio: line voltage:110 say 3300:110 or 11000:110
Burden: may be between 5VA and 50VA per phase
Accuracy class: Class 1 or 3
Protection of VTs
At voltages < 66kV, high rupture capacity fuses are utilised on the primary side to
protect the VT; however, fuses are not normally able to produce a sufficient interrupting
capacity for higher voltages. Secondary side fuses or miniature circuit breakers should
always be fitted to protect against short circuits on the secondary, which are likely to
produce many times the rated current in the secondary, and only a small current at the
primary due to the turns ratio.
Power Systems and Machines
12.4
101
Protection relays
Protection relays determine the health of the system being monitored and call for circuit
breaker operation in abnormal or unhealthy conditions. Their signals to operate are
derived from VT (110V) and/or CT (5A or 1A) secondary windings. The relay compares
this with a signal that is either derived externally or internally. There is a great number
of protection relays manufactured by different companies, but they must comply with
common operating, performance and accuracy standards. The forms of protection and
the types of relays necessary to provide this can be grouped as follows.
Unit (Magnitude)
Overcurrent
Earth fault
Under and overvoltage
Under and overfrequency
Loss of mains
Negative phase sequence
Directional (Mag & phase)
Overcurrent
Reverse power
Zonal(Mag & phase)
Differential
Merz Price
Distance
Unit protection schemes make measurements of magnitude or absolute values to protect
devices such as motors, generators, transformers or specific (unbounded) parts of the
supply network. Directional protection schemes sense the direction of current or power
flow and operate if directions change. Zonal protection schemes test for normality within
prescribed boundaries, e.g., across generators or over specific areas of the distribution
network bounded by distance measurements. It is thus possible to discriminate according
to either or both the type of fault and its location.
12.5
Protection of embedded generators and the distribution network
It is a legal and professional requirement that all embedded or distributed generating
plants are designed, manufactured and installed to operate in an inherently safe manner, under normal and abnormal or fault conditions. In addition to complying with
the requirements of the Health and Safety at Work etc Act (1974), all embedded generating plants must comply with the Electricity Act (1989) and the Electricity Supply
Regulations (1988). Under these Regulations no person may operate generating plant in
parallel with a Distribution Network Operators (DNO’s) system without their written
agreement, and the plant must meet minimum standards of operation, protection and
disconnection in the event of a distribution network or plant failure.
The presence of a distributed generator (DG) may cause protection relays to operate,
which were correctly set for the direction of active and reactive power flows and voltage gradients before the installation of the DG. Adjustment of protection settings to
accommodate the DG may leave the radial system less closely protected than before, in
the event that the DG trips or is shut down. For this reason, the minimum acceptable
Power Systems and Machines
102
generator protection is specified by the DNO. Voltage and frequency operated protection which complies with Engineering Recommendation G59/1 is required by the DNO
to be included in the generator protection. While not demanded by the DNO, current
operated protection is also required to protect the generator and the system against
overcurrents, short circuits, and earth faults.
The UK Energy Network Association documents G59/1, G75 and ETR113 specify the
nature and extent of protection recommended for DGs in the UK. The protection is
intended primarily to protect consumers, supply authority personnel and the network
by preventing:
• Connection of the DG to an unhealthy or faulty network;
• The DG from displacing voltage or frequency outside statutory limits;
• The DG from continuing to back-feed and energise part of the system.
The typical requirement is that over- and undervoltage relays operate within +10%-10%
of nominal 230V (253-207V) and over- and underfrequency relays set to operate within
+1%/-6% of nominal 50 Hz (50.5-47Hz). While the recommended time of operation is 0.5
seconds, mutual agreement with the supply authority may allow increased time settings.
This is especially useful where transient network disturbances such as autoreclosure
of network circuit breakers and motor starting may cause momentary voltage dips or
where loss of generation, or transmission circuit failure, can cause momentary national
frequency fluctuations. Figure 62 shows an embedded generator and local load connected
to the DNO network.
12.6
General protection considerations
Switchgear associated with new DGs has to include a prescribed minimum system of
protection relays that will operate to disconnect the generator from the distribution
network in the event that:
• A fault develops on either side of the generator circuit breaker;
• The system voltage or frequency goes out of agreed limits within the statutory
maxima and minima;
• The network becomes disconnected at a remote point.
In addition, the protection must ensure that the DG cannot be connected to, or continue
to energise, an unhealthy distribution network. Current, time and zonal discrimination
within the distribution network must be maintained through the adjustment of existing
device settings and/or careful selection of settings on new protection relays.
Power Systems and Machines
Embedded
Generator
103
CB Fuse
G
Radial
Feeder
Supply
Transformer
G
Grid
Local
load
Figure 62: Distributed generator and local load
12.7
Neutral earthing
The phase windings of the energy source (generator or transformer secondary) had to
be star-connected to form the neutral connection. In order to define the voltage at the
centre of the three-phase vector system, this star-point of the energy source is earthed. To
prevent large circulating currents caused by non-zero neutral voltages in an unbalanced
system, the neutral is never earthed anywhere else in the same area of the network
below the energy source. The neutral of the distribution system is likely to be earthed
at the star-point of the secondary winding of the distribution transformer, defining the
neutral voltage in the system connected to that transformer winding. If a distributed
generator is to be operated isolated from the distribution network, the star-point of
the stator winding must be earthed to define the neutral voltage of the isolated threephase system. Generators that are to be synchronised need their neutrals connected
to earth to define the incoming voltage system, prior to connection. After they are
synchronised to the network, neutral earth defines the voltage on the generator neutral,
and the generator neutral must be disconnected from earth at the instant of connection
to the network to prevent circulating current flow. If the generator neutral voltage
resolved (VN = VRN + VY N + VBN ) to be marginally above earth, there would be a
continual circulating current limited only by the resistance of the earth return path
to the substation containing the distribution transformer. Also, if there was a singlephase earth fault between the generator and the distribution transformer, the phase
voltage in one phase would collapse to a very low value. The voltage on the transformer
and generator neutral points would rise to full phase voltage. Very high neutral currents
would flow, limited only by the phase-earth loop impedances, until the phase overcurrent
protection or earth fault protection cleared the fault from both sides. There are a number
of methods of earthing generator neutral points illustrated in figure 63, but they must
be agreed with the (DNO).
Power Systems and Machines
104
Neutral
Earthing
Resistor
Solid
Earth
Load
R
Neutral
Earthing
Transformer
Figure 63: Different neutral earthing methods
12.8
Distribution network protection
There may be current- or impedance-operated protection relays within the DNO network
arranged to remain reset for prescribed unidirectional power flow, which may operate
or become unstable under the reverse power flow. Overvoltage relays that are reset
when the DG is out of service may operate when the DG reaches part or full load
export, and undervoltage relays may operate on loss of DG supply. Existing and new
network protection may have to be adjusted to remain effective while the generator is in
service, and yet be no less effective when the DG is shut down or is tripped by its own
protection. Adjustment of protection settings to accommodate the new plant may leave
the radial network less closely protected than before, while the DG is in service or out
of service.
12.9
DG protection schemes
Current and voltage operated protection relays are required specifically to protect the
DG. Choice and design of the DG protection system is at the discretion of the developer,
but the DNO will expect that their system is protected by this equipment when the DG
is in service. The DG protection may include:
• Inverse definite minimum time (IDMT) overcurrent, with instantaneous high-set
overcurrent,
• Instantaneous or time-graded earth fault,
• Neutral voltage displacement,
• Differential protection and reverse power.
12.9.1
Overcurrent protection - Relay designation 51
The protection listed above operates on voltage and frequency signals and, generally,
DNOs will require to witness the proof testing at site of the protective equipment.
Current operated equipment is of less interest to the DNOs, since its primary function is
Power Systems and Machines
105
250kVa, 415V
Embedded
Generator
Cable
Over
current
relay
400:1 CTs
Generator CB
to protect the plant items from the effects of an overcurrent or short circuit current. The
DNO system will be fully protected up to the point of common coupling, and the presence
of an embedded generator will not reduce the overcurrent protection provided by the
DNO circuit breakers or fuses. Current operated relays operate with sensing current
supplies of either 1A or 5A, and are fed from current transformers in the generator
circuit breaker, or neutral connections.
400A
Fuses
51a
51b
51c
64
Earth fault
relay
Figure 64: Generator IDMT OC& EF protection
There are two main types of current-operated protection systems:
• Magnitude or overcurrent: seeking to detect the excess currents;
• Zonal: seeking to detect abnormalities within a zone of the electrical system defined
by the position of the current transformers (see later section).
Suppose an embedded generator (250kVA, 415 V) feeds a distribution transformer where
the outgoing cable is protected by three 400A high rupturing capacity (HRC) fuses, and
the generator is protected by an IDMT overcurrent relay fed from 400:1 CTs, as shown
in Figure 13.9a. A fault in the cable leading to the distribution transformer could give
rise to a fault current from the generator of 7 x FLC = 7 x 348 = 2436A. From the BS88
fuse characteristics for a 400A fuse, the fuses should blow in 5 sec. Trying to preserve
the fuses by tripping the generator circuit breaker first, the circuit breaker must open
before 5 sec has elapsed, so deducting a grading margin of 0.4 sec the IDMT relay must
send out a trip signal in less than 4.6sec.
The relay should remain reset until a current of 387A (IFLC /Reset ratio = 348/0.9)
flows. In the event of the fault, the ratio of fault/reset current is 2436/387 = 6.3, and
the operating time would be 3.73 sec with a time multiplier of 1. The relay would
actually open the circuit breaker at 4.13sec which is 81% of the fuse operating time.
Power Systems and Machines
106
Setting the relay time multiplier to 0.5 reduces the circuit breaker operating time to
2.265sec and this would allow a 50% margin to time grade with the fuses.
12.9.2
Earth fault protection - Relay designation 64
Symmetrical components theory shows that when the system is balanced only positive
sequence components of current exist. When the loads are balanced and the system is
healthy, all phase currents are equal, resolve to zero and there is no neutral current.
When the loads are not balanced, or when there is an asymmetric fault, the phase currents are unequal and do not resolve to zero. The sum is the neutral current. When the
system becomes unbalanced, there are positive, negative and zero sequence components
of current. In = Ia + Ib + Ic .
Zero sequence currents can only flow if there is a neutral connection. In = Ia0 + Ib0 + Ic0
and Ia0 = Ib0 = Ic0 so I= 3I0 .
Earth path
Ig
a
Ia
c
Ic = 0
Ib = 0
b
Figure 65: Earth leakage fault
A phase to phase fault in an unearthed system will give rise to positive and negative
sequence currents, but there can be no zero sequence currents. Earth faults can only
occur if a live conductor(s) touches a conducting surface that is connected to earth, and
if there is a return path back to source of energy. If a phase becomes earthed, in an
earthed-neutral system, a path is provided and the zero-phase sequence components of
current are established. Since the zero sequence components are defined as having equal
magnitude and are in phase with each other
Ia = Ia0 + Ia1 + Ia2 = Ig
Iearth = Ia0 + Ib0 + Ic0
Ia0 = Ib0 = Ic0
Ia = Iearth = 3I0 .
Ib = 0, Ic = 0
(12.2)
Ia = Iearth
(12.3)
(12.4)
(12.5)
Power Systems and Machines
107
Representing the sequence components as vectors: those in phase a total Ia , or Ig ,
whilst those in phases b&c total zero. When the system is healthy, the sum of the 3
phase currents equals zero. As mentioned, in the event of an earth fault, the phase
currents do not resolve to zero, and the fault current returns from earth to the neutral
of the energy source. Any current in the earthed neutral has to indicate an earth fault.
If there are any zero sequence currents detected (in excess of healthy current unbalance)
this also indicates an earth fault in the system.
X
X
Battery
Circuit breaker
X
Trip coil
Figure 66: Core-balance protection
Earth fault protection is therefore designed to detect zero sequence currents, or nonresolution to zero of the phase currents. Phase currents may be resolved and added
using a core-balance CT system as shown in figure 66. All three phase conductors are
passed through a single current transformer core. The secondary winding is connected
to a single-pole relay. Under healthy conditions, the mmfs due to the 3 line currents
are balanced. Under line-to-line fault conditions, the mmfs due to the positive sequence
components balance, as do those due to the negative sequence components. There will
be no flux in the transformer core and therefore no emf or current in the secondary.
If an earth fault occurs on any conductor, the currents and mmfs no longer resolve to
zero, and there is residual flux in the transformer core that induces voltage and current
in the secondary winding. This current flows in the coil of the protection relay causing it
to operate. Any earth fault is potentially serious, and must be cleared by instantaneous
operation of the relay to trip the generator circuit breaker in minimum time. The earth
fault relay would be set to operate instantly if there was a resolved residual current in
excess of say 5% of nominal.
It is also possible to resolve the unbalanced current flow using three star-connected CTs
in an overcurrent and earth fault system, as shown in figure 67. If the generator is
Power Systems and Machines
108
operated synchronised to the network, the neutral may not be connected to earth, or
the network neutral. The action of the star point of the CT secondary windings is to
sum the secondary values of the line currents of the generator. If the system is healthy
and the phase currents balance, the CT secondary currents resolve to zero and there is
no current flow in earth fault element 64. An earth fault in the phase c would return via
the earthed transformer neutral, and the two healthy phases to the generator neutral.
The increased current in the phase c and the resulting imbalance would cause the CT
secondary currents to no longer resolve to zero. The unbalanced current flows in earth
fault element 64 and trips the generator circuit breaker. The system is illustrated in
figure 67.
250kVa, 415V
Embedded
Generator Cable
64
Earth
fault
(standby)
400:1 CTs
Generator CB
The residual systems have the advantage that they resolve the unbalance by creating an
electrically separate neutral (the core for core-balance and the CT secondary neutral in
13.9c). Where the network neutral is connected to earth, it is possible to back up the
protection with a standby earth fault scheme. A CT in the earthed neutral will detect
any earth fault current and operate a standby earth fault relay to trip the circuit breaker
to isolate the energy source.
Transformer
Cable
400A
Fuses
Fault
Ief
51a
51b
51c
Over
current
relay
64
Earth fault
relay (residual)
Figure 67: Earth protection and fault path
12.10
Frequency operated protection - Relay designation 81
A requirement of G59/1, that the DNOs insist on, is that the frequency at the point
of connection of the DG must always be between 94 - 101% of nominal (47-50.5Hz).
Power Systems and Machines
109
Underfrequency protection relays monitor the frequency in one or all three phases of
the busbar (network) side of the generator circuit breaker. If it falls by more than 6%,
the relay will prevent connection or trip the generator from the network. Similarly, the
generator will be tripped if the frequency rises by more than 1%. In both cases, the
relays contain adjustable timers, which may be adjusted to allow the relays to remain
reset for transient frequency changes of less than 0.5 seconds in duration. Loss of mains
(LOM) protection also relies on detecting and reacting to a change in system frequency,
and is discussed below.
12.11
Voltage operated protection - Relay designations 27 (under)/59
(over)
It must not be possible to connect a generator to a network that exhibits conditions of
a fault - since the fault might be under repair and the incoming generator will energise
the network including the faulty area. Once synchronised to a healthy network, the
protection ensures that the generator cannot locally displace the voltage or frequency of
the supply. The DNOs also impose, through G59/1, that the voltage on the busbar side
of the DG must be within healthy limits before the plant can be started for connection.
Further, the voltage of the network and generator during operation must be maintained
within these limits. Under- and overvoltage relays must be fitted to monitor the busbar
(network) voltages in all three phases. The relays are set (inside) 90% and 110% of
nominal voltage, and contain adjustable timers, normally set at 0.5 seconds to avoid
disconnection of plant during short transients in voltage. Most voltage-operated relays
operate from 110V ac sensing supplies, and are fed from voltage transformers in the
busbar or generator side of the generator circuit breaker, of suitable ratio. Neutral
voltage displacement protection can be used to provide back up to detect phase voltage
failure and assist overcurrent protection that might not be sensitive to an overcurrent
or earth fault remote in the distribution network. Frequency and voltage operated
protection relays (and Loss of Mains) are mandatory components of DG protection.
They must be included in the circuit breaker that connects the DG to the network to
ensure that all phases of the network are healthy (i.e., within voltage and frequency
limits) prior to connection of the DG.
12.12
Loss of Mains Protection - Relay designation 83
There is a statutory requirement to disconnect DGs from the network using loss of mains
(LOM) protection if one or more phases of the network supply to the plant become unhealthy. This ensures that the generator does not continue to operate and energise any
part of the local network up to where the fault was cleared by network protection operating, which would be dangerous to consumers and DNO personnel. The loss of mains
protection is also intended to disconnect any embedded generator so that autoreclosing
Power Systems and Machines
110
110V
250kVa, 415V
Embedded
Generator
Δ
83 81 27 59 Relays
Cable
X
X
X
Generator CB
Figure 68: Voltage operated protection
circuit breakers cannot re-connect the network out-of-synchronism to a generator that
has continued to operate.
Loss of mains protection is normally provided by detecting a change in the generator
frequency, since when the network fails the power which was being exported out of
that area of the network increases kinetic energy of rotation. The turbine governor
will sense the speed change and reduce the input power to the generator very rapidly,
but at very least there will be a momentary change in frequency. The most common
LOM protection is provided by rate-of-change-of-frequency (ROCOF) relays, or change
of phase-angle relays that respectively detect the momentary acceleration (as a df/dt)
or change in voltage phase angle associated with loss of load. On sensing the change
of frequency (or faster still, the rate of frequency rise), the loss of mains relay will trip
the generator circuit breaker and shut the plant down. If the embedded generator also
supplies some local load, there is a problem condition when the generated power just
balances the demand of the local load. No power will be exported to the network in this
balance and loss of mains will have no effect on power flow - as such, the frequency will
not change and the mains failure could remain undetected. Currently, the microprocessor
technology in loss of mains relays will sense and react to a load change of about 1%, and
load is seldom balanced to that extent.
Distribution systems that utilise slow-speed fault clearance devices and autoreclosers can
give rise to instability of the DG, due to the critical clearance times of the generators
being significantly lower than protection operating times. To protect adequately the DG
and the rural system, the DG protection must be co-ordinated to current-, voltage- and
time-discriminate with the existing system protection.
Network supply failure would normally occur if, say, one phase of an overhead line broke,
or a pole-mounted fuse blew in the utility network. More commonly, in bad weather,
Power Systems and Machines
111
pole-mounted auto-reclosers (PMARs) in the utility network will trip to clear transient
faults or overcurrents. If the recloser is the last in the radial feeder to the DG, the
network connection to the synchronised generator will be cut off. It should be noted
that the PMAR will close back up automatically (e.g., after 15 seconds), and if the DG
has not been tripped, the network will be forcibly superimposed back onto the generator,
i.e., connecting the DG out of synchronism. This is potentially very dangerous, and loss
of mains protection must be fitted to trip the DG, which otherwise could keep running
isolated from the network, maybe carrying some local load. Most LOM relays offer
consistency and repeatability of operation. However, they can be prone to nuisance
tripping if set too finely, and the system frequency varies suddenly either when distant
loads change or when faults in the network cause recoverable frequency excursions. Work
is necessary to establish coarser LOM settings that do not compromise safety but make
DGs less vulnerable to the effects of distant switching operations and remain connected,
particularly if they can contribute to overall stability.
12.13
Differential protection - Relay designation 87
Differential protection compares the currents entering and leaving a protected zone, and
operates when the difference between these currents exceeds the pre-set magnitude. Current transformers can be arranged to protect plant inside zones defined by the position of
the CTs, either in the form of differential protection or restricted earth fault protection.
In the case of differential protection, the relay compares very accurately the currents in
each CT primary in each phase as measured by the CTs upstream and downstream of
the protected zone.
The protected zone is bounded by matching highly accurate (usually class X) CTs. When
the network is healthy, the current that enters the zone from the left exactly equals the
current leaving the zone to the right. Differential protection deems the network healthy
when CT secondary currents are equal, or their difference is zero. It can therefore be set
to detect very small imbalances (or “spill currents”) and is highly sensitive, nominally
set to 0.025, or 0.05 of nominal current setting. Suppose the CTs are 400:5 A, and the
healthy current is 320A. Then, 4A will flow in each relay secondary, in the directions
shown, and this will resolve to zero, since the current demanded by CT1 is just provided
by CT2, and the current demanded by CT2 is produced by CT1. There is no operating
current in the relay operating coil, set to operate instantaneously at say 0.05*5A, or
0.25A. A fault outside the zone could increase the current through the zone, and up to
the accuracy limit of the CTs, the secondary currents from CT1 and CT2 still resolve
to prevent current flow through the relay coil. This is known as a “through fault” and
the differential protection is designed to remain stable to through faults.
In the case of synchronous generators, there would be three line-side and three neutral
CTs (see Figure 13.13), and the differential protection relay would remain reset as long
as the secondary CT currents matched exactly in line and neutral sides of each phase.
In the event that there was an earth fault in the protected zone, say in the phase a
Matching
CT’s
112
Fault
400:1
CT’s
Generator CB
Power Systems and Machines
Cable
87a
Ia+ If
If
87b
87c
Ia
Figure 69: Differential protection
above the generator, some of the current leaving the generator would leak to earth, and
would not pass through the primary winding of the line-side CT. The secondary current
in the neutral CT would exceed that in the line-side CT, and the unbalanced secondary
current caused by the earth fault would flow through the 87a relay element causing it to
operate, tripping the generator circuit breaker.
12.14
Electro-mechanical protection
Other generator protection such as overspeed and vibration detection must be installed
to protect the plant against mechanical failure, together with devices that will detect
winding, bearing, or coolant overtemperature.
12.14.1
Reverse power - Relay designation 32
Since a synchronous generator can take active power from the network and operate as a
motor, it is necessary to trip the generator circuit breaker if the direction of power flow
reverses. This may result in mechanical damage to the turbine or engine. Such relays can
be set from 1% (water turbine) to around 10% (steam turbine or diesel engine).
Power Systems and Machines
12.14.2
113
Overspeed
If the governor or power control fails, it is necessary to ensure that the unregulated flow
of input power does not cause the generator to speed up to an injurious level. The runaway speed of the turbine or engine may be up to twice synchronous, and a mechanical
overspeed switch is usually arranged to trip the generator circuit breaker and cause a
shutdown of the plant.
12.14.3
Overtemperature
All areas of embedded generators and their power sources, which may overheat in operation, should be monitored, such as bearings and stator windings. Before they become
dangerously hot, temperature relays are used to initiate a shutdown.
Power Systems and Machines
13
13.1
114
Overcurrent protection
Introduction to faults and overcurrent protection
Previous lectures developed a per-unit methodology for calculating short circuit fault
levels and currents. Equipment must be specified to safely withstand (carry), or in the
case of switchgear, interrupt these currents. Protection engineers use the calculated
fault currents to ensure that overcurrent protection will operate circuit breakers in a
safe, controllable and systematic way. This lecture revises and examines overcurrent
protection and discrimination tools.
13.1.1
Fault levels in a typical system
The fault levels may be ascertained for each voltage level on a typical power system
(refer to figure 70) when given the reactances. Although the short circuit fault level
changes with the network conditions, there are two characteristic cases in analysis: 1)
all plants/generators connected, 2) minimum plants/generators connected.
Voltage
275 kV
132 kV
33 kV
11 kV
415 V
Fault level
15000 MVA
3500 MVA
750/1000 MVA
150/250 MVA
30 MVA
The table above outlines the typical maximum fault levels expected in the UK. As the
voltages increase, so does the possible short circuit currents. As parallel paths greatly
reduce the fault impedance (hence increasing fault current) sectionalising is used to
reduce fault levels. Sectionalising normally opens a circuit breaker to disconnect parallel
impedances and create higher impedance routes to the then separate sections of the
network.
When considering the connection of new distributed generators (DGs), the distribution
network operators (DNOs) are required to advise the three-phase symmetrical short
circuit fault level at the proposed point of connection of the DG, which will be higher
at locations electrically ’closer’ to the primary station than at the remote end of a rural
radial feeder. Where the fault level (and equipment fault rating) of the DNO network is
high, it is unlikely that small to medium capacity DGs will bring about an overall excess
fault level. The new DG switchgear has to be adequately rated for the total prospective
(and future) network fault current and the contribution from the generator. It should
be noted that while the DNO may quote a fault level at the point of connection, they
reserve the right to reinforce fully the network up to this point, and this future fault level
may exceed actual (i.e. current) values and must be taken into account. Some typical
circuit breaker ratings are shown below.
Power Systems and Machines
115
Infinite Busbar
275 kV
132 kV
X = 0.155 pu
2 x 150 MVA G 33 kV
11 kV
415V
Max plant 0.0107 pu
Min plant 0.0143 pu
Transformers
240 MVA
X = 0.062 pu
Transformers
90 MVA
X = 0.244 pu
Transformers
15 MVA
X = 0.666 pu
1 MVA
X = 4.75 pu
Figure 70: Typical Power System Fault Levels & Reactances
13.2
Overcurrent protection
Generally, the worst-case fault (where the fault currents are highest) is the three-phase
short circuit fault. Overcurrent protection is designed to open or “trip” circuit breakers
that supply increased current to the fault. Sustained overloads (say 1.2-2 times normal
current) may be accommodated for a modest period, limited by the temperature rise
and final temperature of the network conductors and equipment. For lower values of
overcurrent that are short-lived (say 2-5 times normal), the generators, transformers
and cables have sufficiently high thermal capacity and long thermal time constants to
enable them to pass these overcurrents without the temperature becoming excessive.
This thermal capacity is an advantage in a system with active loads, such as induction
motors or transformers, since the (short-lived) starting currents or inrush currents may
be easily accommodated. Higher over-currents (say 5-10 times normal, or greater) can
cause electro-mechanical stress between conductors and lead to insulation failure and/or
mechanical damage, which is likely to occur very rapidly. Continuing high current will
lead to thermal damage due to the I2.t ohmic heating effect. Clearly, such faults must
be detected and immediately interrupted by the over-current protection and subsequent
tripping of the circuit breakers.
Power Systems and Machines
13.2.1
116
Protection co-ordination
Protection devices are said to be co-ordinated when a protection scheme correctly distinguishes and isolates most promptly and only the part of the power system at fault.
• Faults should be cleared fastest by the circuit breakers closest to the fault, leaving
as much as possible of the healthy network in service. Faults at electrically remote
parts of the network should be disconnected by the nearest circuit breakers. This is
geographical discrimination that may be achieved by time-grading relay operation.
• Faults leading to high currents should be disconnected more rapidly that faults
with lower currents. This is achieved by amplitude discrimination. Fault currents
are lowest at the network edges and highest closest to the source of energy, so
amplitude discrimination is easiest in areas with high fault levels, farther towards
the centre of the network.
The term ‘grading margin’ refers to the required difference in relay operating times
(time discrimination) or difference in relay “setting” or operating currents (amplitude
discrimination) required between protection devices to produce a co-ordinated protection
scheme. To correctly co-ordinate overcurrent protection requires the ability to calculate
likely fault currents at each protected point in the circuit. The data required for a relay
setting study are:
• A single-line diagram of the power system, including the rating and setting of the
protection devices and their associated current transformers.
• The per-unit fault impedances of all rotating machines, power transformers, lines
and feeders.
• The maximum and minimum short circuit currents expected at each protective
device.
• The maximum/peak loading current through protective devices.
• Motor starting currents and stalling times for induction motors.
• Generator decrement curves taking account of sub-transient and transient effects
(for fault currents).
• Performance curves for the current transformers.
In this course, we are primarily concerned with the first 4 of these factors.
13.3
Fuses
The simplest form of overcurrent protection for LV and MV schemes is the fuse, and
in industrial systems high-rupturing capacity fuses are used for high-speed short circuit
protection. A fuse acts as both fault detector (only faults associated with overcurrent)
Power Systems and Machines
117
and interrupter (although the time/current relationship, once selected, cannot be readjusted). The two main applications of fuses are:
• Overload protection - in circuits where load does not vary widely outside a
normal value. For overcurrents in the range of 2-5x normal current, fuses will take
a relatively long time to blow.
• Short circuit protection - in circuits where loads vary considerably outside a
normal value. For the range >5x normal current, fuses will take a much shorter
time to blow, figure 72.
The fusing element melts under the effects of the short circuit current and requires a
fixed amount of energy to cause the fuse to ‘blow’, in order to interrupt the current. The
I 2 .t = K characteristic of fuses provides a measure of time discrimination: the fuses have
inverse time:current characteristics. In fuse characteristics, the operating time t may be
plotted against fault current I, and from the above equation the family of rectangular
hyperbolae is produced for the BS range of fuse ratings. Plotting t against I 2 would
produce a straight line characteristic, but it is more normal to recognise the range of
currents & times by plotting each on logarithmic scales. Cartridge type fuses are most
common, as they are sealed and therefore: maintain a reliable grading characteristic,
contain the arc and fault energy and are protected against operator abuse (e.g., inserting
copper wire into a re-wireable fuse).
13.3.1
An introduction to discrimination
The nominal rating current of a fuse may be used to co-ordinate the order in which a
string of series-connected fuses would blow, but it should be noted that fuse manufacturers will normally only guarantee co-ordination (for over-load protection) between fuses
whose ratings are in the ratio 2:1. For instance, in figure 71 below, a fault current of 100A
should cause the 20A fuse to clear in 1.5 sec (value taken from figure 72) and the 25A
fuse to clear in 5 sec, but the nature of the rupturing process and tolerances in manufacture could lead to a 32A fuse blowing first, what will isolate the fault in a non-selective
manner, as a healthy section of the system and its load is also isolated.
13.3.2
Disadvantages of fuses
Fuses are cheap and easily replaced, but they self-destruct on operation and can not be
reset. The fuses are also rated to just exceed the normal full-load current of a circuit,
and therefore provide limited protection against earth faults that account for over 90%
of electrical faults. Earth faults usually flow in higher resistance fault paths and create
lower currents, so they are harder to detect and interrupt. A further disadvantage of
fuses in a three-phase system is that a fault in one phase would not result in all three
phases tripping. As a consequence, connected three-phase loads could continue operating
(until damaged) on the remaining two healthy phases.
Power Systems and Machines
Mains
Supply
118
32A
20A
25A
~
t = 15s
t= 5s
t= 1.5s
100A
Fault
Figure 71: Discrimination in a fuse circuit
10000
1000
100
Time (seconds)
10
1
200A
0.1
6A
20A 32A 50A 80A
25A
0.01
10
100
1000
Prospective Current, r.m.s (A)
Figure 72: Fuse characteristics to BS88
125A
Power Systems and Machines
13.4
119
MCBs, MCCBs
Miniature circuit breakers (MCBs) are now used to provide domestic and light industrial
over-current protection, and come in single-, two- or three-pole configurations. They operate with a thermal characteristic from 1-5 x rated current and instantaneously between
5-10x rated current. They offer the advantage that they may operate and be reset many
thousands of times.
Moulded case circuit breakers (MCCBs) are power switches with built in protective
functions featuring:
• Normal load open and close functions
• Protection functions under thermal overload (matched closely to the system being
protected) or short circuit conditions.
The MCCB case is of resin/glass fibre, providing a rugged and insulated device. Thermalmagnetic MCCBs use a bi-metallic element heated by the load current to sense thermal
overload (refer to Figure 14.4), but the response of a bi-metallic strip is too slow to
provide adequate short circuit protection. Therefore, a magnetic trip action is used
where the short circuit current energises an electromagnet operating the trip mechanism, typically within a time of 1 cycle. Electronic MCCBs are now produced, which
retain a similar characteristic to electro-mechanical MCCBs; however, the characteristic is adjustable to the desired application (the arrows in Figure 73 illustrate such
flexibility).
MCBs and MCCBs may provide a reasonable degree of current grading when used in
series, by simply applying a higher rated MCCB upstream of a given MCCB. Usually,
co-ordination selectivity is maintained in the thermal overload and partial high current
region, but it is lost in the short circuit pick-up region of the up-stream device. A short
delay may be applied to the upstream device to maintain co-ordination.
13.5
Overcurrent protection discrimination
A radial feeder supplied from a generating station is shown in figure 74. The feeder
supplies a number of buses: A-D, and has over-current relays fitted to the circuit breakers A-C located on the lines feeding busbars B-D. There is also a 5 MVA distributed
generator connected at the bus B.
13.5.1
Discrimination by amplitude
A fault within the shown network will be fed from the transmission network and the
distributed generator. Fault currents will flow from all sources of energy, and should
Power Systems and Machines
120
ra
tur
Instantaneous
(short-circuit)
operating area
re
pe
tu
tem
ra
gh
pe
Hi
m
te
Operating time
w
Lo
= Adjustable with
electronic MCCB
e
Long delay
(thermal)
operating area
Current
Figure 73: Discrimination with a MCCB
66 kV
T1
11 kV
1.2 Ω
20 MVA 12.5%
T2
11 kV
11 kV
300A
20 MVA 12.5%
Transmission
A
Network 200 MVA
Relay C
0.8 Ω
Relay A
B
Relay B
5 MVA 15%
Figure 74: Radial feeder system
1.5 Ω 11 kV
250A
C
150A
D
Power Systems and Machines
121
be sensed by corresponding relays in the network. If the protection system is to discriminate satisfactorily, then, for a fault on any particular section of the feeder, only
the circuit breaker(s) nearest the fault should operate to disconnect the faulty section.
Since the fault current is limited by the impedance between the source and the fault,
it will gradually reduce as the point of detection gets farther away from the source. In
theory, therefore, it is possible to obtain discrimination by arranging for the various
overcurrent relays to operate at different currents, i.e. for relay C to operate at a lower
current than relay B and so on. Standard relay characteristics, such as that in figure
75, result in faster clearing of higher fault currents. Unfortunately, network impedance
increases and the fault currents reduce anyway towards the edges of the network, making
amplitude discrimination more difficult. In addition, the actual value of fault current
is also dependent on the amount of generating plant actually in service at the time the
fault occurs.
13.5.2
Discrimination by time
An alternative is to arrange for the overcurrent relays to have different time delays
between the moment of energising the overcurrent relay coils (which is the moment of
fault occurrence) and moment of closure of the output contacts in the tripping circuit
of the circuit breaker (which is set by adjusted time delay). The time taken to operate
relay C must therefore be shorter than that of B, which must be shorter than A. Discrete
time discrimination has the disadvantage that it is not sensitive to the severity of the
fault.
13.5.3
Discrimination by time and current
To combine amplitude (severity) and time (location) discrimination, the current:time
characteristic is made inverse, and programmable time delays are built in. The family of
current:time characteristics in figure 75 is typical. These are inverse definite minimum
time (IDMT) characteristics, and IDMT overcurrent relays can be made to operate on
curves 1-5 by choosing an appropriate “time multiplier setting” (TMS).
In the circuit shown in figure 74, identical relays A-C could be used, with TMS selected
to give the inverse current characteristic of Curve 3 at C, Curve 2 at B, and Curve 1 at
A. To keep the characteristics separate at all values of fault current, it is necessary that
there is a definite minimum time of operation.
13.6
IDMT over-current relays
The most common over-current protection relay has inverse definite minimum time
(IDMT) characteristic. Most systems still utilise an IDMT characteristic derived from
the induction disk or cup relay, shown in figure 76, or its analogue or digital equivalent.
Power Systems and Machines
122
10.0
7.0
Time (seconds)
4.0
2.0
Curve 1
(TMS = 1.0)
Curve 2
(TMS = 0.8)
1.0
0.7
Curve 3
(TMS = 0.6)
0.4
Curve 4
(TMS = 0.4)
0.2
Curve 5
(TMS = 0.1)
4
6
8 10
Current (multiples of plug setting or PSM)
Figure 75: Inverse time characteristics
The fluxes Φ1 and Φ2 are out of phase due to the difference in turns on the lower electromagnet and thus a torque is produced on the disk causing it to rotate through the
distance necessary to bring the trip contacts together.
Time to trip = distance/disc speed
Disk speed ∝ I 2
Time to trip = distance/I
(13.1)
(13.2)
2
(13.3)
These properties produce the characteristics such as those in figure 75, where the operating time is proportional to the inverse square of the current applied, for example,
Curve 4. Increasing the distance to trip increases the time taken, and creates another
Curve such as 3.
13.6.1
Current plug setting multiplier (PSM)
This adjusts the setting current by means of a plug setting bridge, which varies the
effective turns on the upper electromagnet. A typical time-current IDMT characteristic
plot uses the fault current plug setting multiplier (PSM - the ratio of fault current to
operate current defined by plug setting) as the x-axis scale.
Power Systems and Machines
Input
123
Plug
Tapping bridge
Input
Upper
magnet
Disk
carrying
contacts
Lower
magnet
Input
a) Wattmetric induction disk
b)Separately fed induction disk
Figure 76: Induction-disk IDMT relay
13.6.2
Time multiplier setting (TMS)
This adjusts the operating time at a given multiple of the current plug setting multiplier,
that is, the distance through which the disk must travel before the contacts make. Figure
75 illustrates various TMS for an IDMT relay, for example a fault current of 1200A,
detected by 100:5 CTs would have 60A flowing through the relay coil. If the PSM is
150%, the relay will begin to operate when 7.5 A flows in the coil, or 150A flows in the
faulty circuit.
The fault current is eight times that prescribed by the plug, so as a multiple of the
plug setting this is PSM=8 on x-axis. For a TMS of 1, the operating time would be 2
seconds. If this had to be shortened for discrimination, a TMS of 0.6 would give a lower
characteristic and an operate time of 1.2 seconds.
13.6.3
IDMT characteristics
IDMT characteristics are available in a number of shapes, e.g., in accordance to IEC
and British Standards as illustrated below in figure 77.
Long time standby earth fault, t =
Standard inverse, t =
Very inverse, t =
0.14
−1
I 0.02
13.5
I −1
Extremely inverse, t =
120
I −1
(13.4)
(13.5)
(13.6)
80
−1
I2
(13.7)
Power Systems and Machines
124
Operating time, t (seconds)
100
10
4
Longtime standby
earth fault
2
Standard inverse
1.0
Very inverse
Extremely inverse
0.1
1
10
100
I(xIs) Current (Multiple of setting)
Figure 77: Time delayed over-current element - operation time characteristics
Power Systems and Machines
13.6.4
125
Factors influencing choice of PSM
• Load conditions: relay must not trip during healthy conditions, i.e. for full load
current or starting/ inrush current.
• Fault currents: high fault currents may cause saturation of CTs, therefore PSM
ratios important.
• CT performance: magnetisation and internal resistance.
• Relay burden (ZBURDEN ): CT must be chosen carefully (use excitation curves) for
low tap settings.
• Relay accuracy: better at the top end of the characteristic.
13.7
Overcurrent protection design
Most of the discrimination problems in protection systems which engineers have to solve
involve only a portion of the whole supply network, from generators to consumers. This
is usually a middle portion, with a supply network on one side and a load distribution
network on the other. The problem of discrimination is to decide the number and extent
of the stages required, and whether the characteristics and settings of overcurrent relays
can be arranged to give the necessary amplitude and time margins.
13.7.1
Discrimination time margin (grading margin)
To limit the extent of outage due to a fault to a minimum, it is necessary for the relay
and circuit breaker nearest to the fault to operate first. The difference between the
operating time of this relay and the one above it, which is to remain reset, is called
the discriminating time. The time difference between the operation of two adjacent
overcurrent protection relays should account for
• Errors: CTs, VTs, and the relays themselves have error tolerances dependent upon
the operating conditions.
• Overshoot time of the relay: When relay is de-energised, operation may continue
for longer due to energy stored as inertia of the induction disk or as current in the
capacitance of an analogue device.
• The fault current interruption time of the circuit breaker: The circuit breaker must
completely interrupt the fault current before the discriminating relay ceases to be
energised.
• Final margin on completion of operation: An additional safety margin is added to
the previous error factors to guarantee correct protection operation.
Power Systems and Machines
126
Assessments can be made for all error factors in any particular case and an overall
allowance made. In practice, a grading margin of 0.5 seconds is sought for electromechanical relays, but margins above 0.4 seconds may still provide discrimination. Margins below 0.4 seconds do not guarantee discrimination with electro-mechanical relays.
Analogue or digital relays offer better tolerances and repeatability over their entire characteristic and may enable a typical grading margin of 0.3 sec.
13.7.2
IDMT relay grading
While the necessity for time to discrimination tends to increase the time delay required on
relays successively nearer the energy source or supply network, this can be alleviated to
some extent by taking advantage of the increase in the maximum fault current available.
This is where time/current coordination schemes become invaluable. Correct grading
is preserved when IDMT characteristics keep 0.5 sec discrimination over the range of
expected plug setting multiples. If they cross, an upstream relay may operate early.
13.7.3
The graphical representation of time/current relationships
Manufacturers publish the time/current characteristics of their overcurrent relays in
graphical form. Users employ graphical methods of determining the settings for relays.
The use of the logarithmic scale for current has the advantage that variation of CT ratio
or current setting leaves the shape of the characteristic unchanged and merely moves it
bodily along the scale. This permits a template to be used to draw the characteristics.
The logarithmic time scale permits the IDMT template to be moved vertically to simulate
variation of the time multiplier setting. Percentage tolerances are also easily drawn by
displacement of the template.
T
Incorrect co-ordination
Same setting despite
different PSM and TMS
I
T
Correct co-ordination
at every current, grading
margin maintained
Grading margins > 0.5s
Min
Max
I
Figure 78: Erroneous and correct co-ordination of IDMT setting curves
Power Systems and Machines
14
127
Overcurrent protection - Case Study
14.1
Revision of Methodology for the Calculation of Balanced 3 phase
Faults
The important assumptions were:
• The fault is balanced across all 3 phases.
• The impedance of the fault is negligible.
• Generated voltages remain constant and balanced during the fault.
• Generators are modelled by an EMF and the sub-transient reactance only.
• Generators operate at rated open-circuit terminal voltage (assumed as Et = 1pu).
• Busbar voltages are 1pu before the fault, and transformers are set at nominal tap
position.
• The network impedance is asserted to be purely reactive (valid until at the edges
of the distribution network where X:R ratios are lower).
• The healthy load currents that meet demand are small and may be ignored.
• Overhead lines or cables are represented simply by their series reactances.
• Transformers are represented by their leakage reactances.
The methodology was:
1. Draw single-phase equivalent circuit.
2. Identify all system operating values (kV, MVA, X).
3. Assert 1.0 pu voltage throughout the system.
4. Choose common MVA base for the system.
5. Check per-unit reactances are specified based on operating voltage.
6. Convert all ohmic and per-unit impedances to new pu values defined by common
base MVA.
7. Combine reactances/impedances in series/parallel to achieve the equivalent single
voltage source feeding the fault via a single pu reactance.
8. Calculate the fault current in pu.
9. Divide the fault current back through the circuit in pu terms.
10. Calculate the actual circuit fault currents (where required).
Power Systems and Machines
14.2
128
Effects of location and impedance
The three-phase symmetrical short circuit fault level of the distribution network will
be increased at all points due to the fault contribution made by the connection of an
embedded generator (EG). The impedance between the EG and the point of the fault will
limit the fault contribution: as the fault takes place further and further away from the
generator, contribution will be reduced at the edges of the network; contribution from
the reinforced areas of the network (above the bulk supply points) will increase.
14.3
Effects of type of EG and excitation
The extent and duration of the fault contribution is also influenced by the type of generator. Self-exciting synchronous generators with low reactances contribute the highest
sub-transient, transient and sustained short circuit currents. Separately or statically
excited synchronous generators offer reduced sustained short circuit currents, since the
short circuit causes the excitation supply voltage to collapse.
Induction generators supply least sustained and transient contributions, since they are
excited from the distribution network and after the sub-transient period the fault current
has decayed/decreased, the machine is largely demagnetised and cannot continue to
supply the fault.
The effects of a short circuit fault generally occur so quickly that the generator AVR
and prime-mover governor can not react quickly enough to have an effect on the machines behaviour at the inception and during the early stages of the fault. The inertia
of the revolving shaft system is usually sufficiently high that, even subjected to the high
electromechanical torques developed, the speed of the generator may be treated as constant in fault studies to determine the currents on inception and clearance of the fault.
Machines that are connected to the network via long radial feeders that are tripped
on overcurrent or loss of mains to clear an intervening fault are required to be shut
down. This is the requirements of G59/1 and G75 ENA recommendations, which are
formulated in order to prevent islanded operation and avoid safety hazards arising from
forcible reconnection to the restored network.
14.3.1
Fault contribution from generators
As a ”rule of thumb”, the fault contribution from a synchronous machine may be assessed
using machine MVA and information about its per-unit sub-transient reactance:
MVAfault =
MVAgen
Xd00 pu
(14.1)
Power Systems and Machines
129
Depending on the type of generator and rotor winding, the sub-transient reactance may
be as low as 0.05 p.u., resulting in the sub-transient fault current of up to 20x the full load
current of the machine. The initial fault contribution from an induction generator may
be evaluated in a similar way, but is likely to be lower than that from a synchronous
generator of equivalent capacity. For a more accurate computation, the fault current
must be calculated at specific time (say at t = 0.06 seconds), which correspond to the
moment when the circuit breaker contacts would start to separate and clear the fault.
To check that the generator and other protection grades fully and properly, the full
current-time decrement curve for the generator would be plotted and compared with the
time current characteristics of the overcurrent protection.
14.4
Case study of a loaded radial feeder with new generation
Using figure 74 from the last section, this refers to a system where a new distributed
generator rated 4MW, 0.8 power factor is to be connected at 11kV. Supply to the 11kV
radial feeder is by two parallel transformers from the 66kV network as shown. The
ampere loadings on each busbar are noted, and the interconnecting impedances are as
indicated. The current plug setting multipliers (PSM) for relays A-C are 150, 150 and
125%. Relays A-C are connected to CTs of ratios 800:5, 400:5 and 200:5, respectively.
The minimum relay time multiplier setting increment is 0.025, beginning at 0.05. The
task is to determine suitable IDMT relay settings that will provide a grading margin of
0.5 seconds along the radial feeder. In this example, this margin is also applied to relay
C, but relay C might have a minimum TMS of 0.05 to provide the shortest clearing time
(see example in Tutorial 7, where reworking of this case study is performed).
General methodology
First, the maximum and minimum fault currents at each busbar should be determined.
Maximum values are with both transformers and the generator in service; minimum will
be with only one transformer in operation and the generator out of service. The standing
current loads on busbars B, C and D are of no consequence in the fault calculation, but
need to be verified to be below overcurrent protection settings. To a base of 10MVA
(Xbase = 12.1 ohms at 11kV, Ibase = 525A), the maximum impedance up to and
minimum fault currents at each busbar are
Busbar
Max Z (pu)
Min If (pu)
Min If (A)
A
0.1125
8.88
4666
B
0.2125
4.705
2470
C
0.2785
3.59
1884
D
0.4024
2.485
1304
The minimum impedance up to and maximum fault currents at each busbar are
Busbar
Max Z (pu)
Min If (pu)
Min If (A)
A
0.0675
14.81
7775
B
0.1129
8.857
4648
C
0.1789
5.59
2933
D
0.3028
3.303
1733
Power Systems and Machines
130
Secondly start at the most remote end of the system and work back to the source,
i.e.:
1. Consider relay C for the maximum fault at D and apply the shortest TMS setting
for relay C.
2. Consider relay B for the maximum fault at D and determine the correct TMS
setting for relay B which will produce sufficient grading margin with relay C.
3. Consider relay A for the maximum fault at D and determine the correct TMS
setting for relay A, again producing sufficient grading margin with relay B.
4. Consider relay B for the maximum fault at C and determine the relay B operating
time using the TMS already set in step c.
5. Consider relay A for the maximum fault at C and determine the correct relay A
TMS producing sufficient grading margin with relay B.
6. Consider relay A for the maximum fault at B and determine the relay A operating
time using the TMS from point e. above.
Fault at D - Consider relay C
125% current setting on 200:5 current transformers. Primary setting current = 250A
Normal load current = 150 A, relay will remain reset during the peak loading conditions.
PSM (Fault current as multiple of plug setting) = Fault Current (D)/Current Setting
(C) = 1733/250= 6.93
Time of operation (TC) for IDMT relay from curve in Figure 15.4b is ≈ 3.5s.
Time multiplier setting (TMS) required to give minimum delay is 0.05, and actual operating time (TA) = 3.5 x 0.05 = 0.175s
Fault at D - Consider relay B
150% current setting on 400:5 current transformers. Setting current = 600A
Normal load current = 250+150 A = 400A, so relay will remain reset.
PSM = Fault Current (D)/Current Setting (B) = 1733/600 = 2.89
Time of operation (TC) for IDMT relay from curve = 7s.
Time required for discrimination = above (TA) of IDMT relay at C + required margin=0.175+0.5= 0.675s
Time multiplier setting (TMS) required for 0.675s delay is = 0.675/7 = 0.096, say 0.1
Actual operating time (TA) = 7 x 0.1 = 0.7s
Fault at D - Consider relay A
150% current setting on 800:5 current transformers. Setting current = 1200A
Normal load current = (300-262)+250+150 A, = 438A so relay will remain reset.
PSM = Fault Current (D)/Current Setting (A) = 1733/1200 = 1.444
Time of operation (TC) for IDMT relay from curve = 18s.
Time required to give discrimination = above (TA) of IDMT relay at B + required
margin = 0.7+0.5= 1.2s
Power Systems and Machines
131
Time multiplier setting (TMS) required to give 1.2s delay is = 1.2/18 = 0.066, say 0.075
Actual operating time (TA) = 18 x 0.075 = 1.35s
Fault at C - Consider relay B
Fault current (multiple of plug setting): Fault Current (C)/Current Setting (B) =
2933/600 = 4.88
Time of operation (TC) for IDMT relay from curve = 4.4s.
Time multiplier setting (TMS) from previous grading = 0.1
Actual operating time (TA) = 4.4 x 0.1 = 0.44s
Fault at C - Consider relay A
Fault current (multiple of plug setting): Fault Current (C)/Current Setting (A) =
2933/1200 = 2.44
Time of operation (TC) for IDMT relay from curve = 8s.
Time multiplier setting (TMS) from previous grading = 0.075
Actual operating time (TA) = 8 x 0.075 = 0.6s
This does not give the required 0.5 sec grading clearance with relay B.0.6-0.44 = 0.16sec
Time required for discrimination = above (TA) of IDMT relay at B + required margin=0.44+0.5= 0.94s
Time multiplier setting (TMS) required on relay A to give 0.94s delay = 0.94/8 = 0.1175,
say 0.125
Actual operating time (TA) = 8 x 0.125 = 1s now grades with B (1-0.44 = 0.56sec)
But Relay A must be checked for fault at D again
Actual operating time (TA) = 18 x 0.125 = 2.25s
With a new TMS of 0.125 the operating time of relay A becomes 2.25 seconds
Fault at B - Consider relay A
Fault current (multiple of plug setting): Fault Current (B)/Current Setting (A) =
4648/1200 = 3.87
Time of operation (TC) for IDMT relay from curve = 5.2s.
Time multiplier setting (TMS) = 0.125
Actual operating time (TA) = 5.2 x 0.125 = 0.65s
These settings give adequate discrimination between each relay, perhaps at the disadvantage of a longer (2.25 sec) clearance time on relay A for a fault at D. If thought
excessive, this could be shortened by reducing the current plug setting of relay A from
150 to 125%. This gives a greater PSM and shorter Tc.
Fault at D - reconsider relay A
125% current setting on 800:5 current transformers. Setting current = 1000A
Normal load current = (300-262)+250+150 A, = 438A, so relay will still remain reset.
Fault current (multiple of plug setting): Fault Current (D)/Current Setting (A) =
1733/1000 = 1.733
Time of operation (TC) for IDMT relay from curve = 12s. (previously was 18s!)
Time multiplier setting (TMS) = 0.125
Actual operating time (T) = 12 x 0.125 = 1.5s still grades with B (1.5-0.7 = 0.8s)
Power Systems and Machines
132
Fault at C - reconsider relay A
125% current setting on 800:5 current transformers. Setting current = 1000A
Fault current (multiple of plug setting): Fault Current (C)/Current Setting (A) =
2933/1000 = 2.93
Time of operation (TC) for IDMT relay from curve = 6.5s. (previously 8s)
Time multiplier setting (TMS) = 0.125
Actual operating time (TA) = 6.5 x 0.125 = 0.8125s loses grading with B! (0.81250.44 = 0.3725s), but close to the 0.4s, which is a low-range limit for time grading of
electro-mechanical relays.
Fault at B - reconsider relay A
Fault current (multiple of plug setting): Fault Current (B)/Current Setting (A) =
4648/1000 = 4.65
Time of operation (TC) for IDMT relay from curve = 4.8s.
Time multiplier setting (TMS) = 0.125
Actual operating time (TA) = 4.8 x 0.125 = 0.6s
C
B
A
Fault
PSM
6.93
2.89
1.73
at D
Tc
3.5
7.0
12
TMS
0.05
0.1
0.125
Ta
0.175
0.7
1.5
Fault at C
PSM Tc
TMS
Ta
Fault at B
PSM Tc
TMS
Ta
4.88
2.93
0.1
0.125
0.44
0.8125
4.65
0.125
0.6
4.4
6.5
4.8
This shows that the clearance time of relay A for a fault at D can be reduced from 2.25
to 1.5 seconds, but now at the expense of a slightly reduced grading margin between
relays A and B for a fault at C.
Power Systems and Machines
A
133
Laplace transforms
Laplace transform integral:
L [f (t)] =
L−1 [F1 (s)F2 (s)] =
Convolution integral
Final value theorem
Rt
0
f (t)e−st dt
f1 (t − τ )f2 (τ ) dt
limt→∞ f (t) = lims→0 sf (s)
Initial value theorem
f (0+) = lims→∞ sf (s)
2nd -order transfer function
f (s) =
Signal
Impulse
Waveform
δ(t)
Step function
u(t)
Ramp
tu(t)
Decaying exponential
e−αt
Increasing exponential
1 − e−αt
Damped ramp
te−αt
Sin
[sin(βt)]u(t)
Cos
[cos(βt)]u(t)
Derivative
Integral
f 0 (t)
dn
f (t)
dtn
Rt
0 f (x) dx
Damped sin
[e−αt sin(βt)]u(t)
Damped cos
[e−αt cos(βt)]u(t)
T translation
S translation
[f (t − a)]u(t − a)
e−αt f (t)
Scaling
f (at)
i
1 h
ω −at
1
+
e
sin(bt
−
φ)
ω2
b
nth derivative
0
R∞
where ω 2 = a2 + b2
ωn2
s2 + 2ζωn s + ωn2
Laplace transform
1
1
s
1
s2
1
s+α
α
s(s + α)
1
(s + α)2
β
2
s + β2
s
s2 + β 2
sF (s) − f (0+ )
P
dr−1
sn F (s) − nr=1 r−1 f (0)sn−r
dt
F (s)
s
β
(s + a)2 + β 2
(s + α)
(s + a)2 + β 2
e−as F (s)
F (s + α)
1 hsi
F
a
a
1
s[(s + a)2 + b2 ]
−b
and φ = tan−1
a