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© 2012, Siemens Industry Inc., all rights reserved
Tab 18 -
Apparent Power, Active Power, Reactive Power Distribution Systems Engineering -
Course 1
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Some Basic Concepts
Review of some important definitions:
Peak Value of a sinusoidal current wave (Ipeak
)•
Occurs when the sinusoidal current wave is at its maximum amplitude
RMS Value (also called effective value) of a sinusoidal current wave
Average Value of a sinusoidal current wave
02sin10
T
peakaverage dttTI
TI
peakpeakT
peakRMS II
dttT
IT
I 707.02
2sin10
22
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Voltage and current relationship
)90sin(sin2)sin(cos2)sin(2)sin()( oRMSRMSRMSMAX tItItItIti
)sin(2)sin()( tVtVtv RMSMAX
)sin(cos2)( tIti RMSR
v(t)
i(t) LOADiR(t) iX(t)
• COMPONENTS OF TOTAL CURRENT
• TOTAL CURRENT
)90sin(sin2)( oRMSX tIti
• SYSTEM VOLTAGE v(t)
• COMPONENT OF CURRENT IN-PHASE WITH VOLTAGE (REAL COMPONENT)Real component of current supplies a NET energy to the LOAD
• COMPONENT OF CURRENT 90O
OUT-OF-PHASE WITH VOLTAGE (IMAGINARY COMPONENT)Imaginary component of current supplies no NET energy to the load
θ
= Angle by which voltage v(t) leads total current i(t)
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Voltage and current relationship (continued)
)90sin(sin2)sin(cos2)sin(2)( oRMSRMSRMS tItItIti
)sin(cos2)( tIti RMSR
v(t)
i(t) LOADiR(t) iX(t)
• TOTAL CURRENT
)90sin(sin2)( oRMSX tIti
• REAL COMPONENT OF CURRENT • IMAGINARY COMPONENT OF CURRENT
• RMS VALUE (IR
) OF REAL COMPONENT OF CURRENT:
• RMS VALUE (IX
) OF IMAGINARY COMPONENT OF CURRENT:
cosRMSR II
sinRMSX II
• RMS VALUE (IRMS
) OF TOTAL CURRENT: 22 XRRMSTOTAL IIII
IR
IX
ITOTAL
RELATIONSHIP BETWEEN RMS VALUE OF REAL COMPONENT, IMAGINARY COMPONENT, AND TOTAL CURRENT WHEN ANGLEΘ
IS POSITIVE:
θ
= Angle by which voltage v(t) leads current i(t)
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Voltage, current, & power relationship
)sin()()sin()( tItiandtVtv MAXMAX
)sin()sin()()()( ttIVtitvtp MAXMAX
)2sin(sin22
)2cos(1cos22
)( tIVtIVtp MAXMAXMAXMAX
• INSTANTANEOUS AND AVERAGE POWER
• USING TRIGONOMETRIC IDENTITIES, THE EXPRESSION FOR INSTANTANEOUS POWER p(t) IS:
)2sin(sin)2cos(1cos)( tIVtIVtp RMSRMSRMSRMS
• INSTANTANEOUS POWER, p(t), IS THE RATE AT WHICH ENERGY IS SUPPLIED TO THE LOAD
“+”
p(t) means system supplies energy to load, “-”
p(t) means load supplies energy back to system. Frequency of p(t) is twice that of system
Note that the real component of current, IRMS cosθ, produces one component of the instantaneous power thathas a non-zero average value. The imaginary component of current, IRMS sinθ, produces the second componentof instantaneous power that has an average value of zero.
v(t)
i(t)
LOAD
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Voltage, current, & power relationship (continued)
)2sin(sin)2cos(1cos)( tIVtIVtp RMSRMSRMSRMS
)2sin()2sin(sin)( tQtIVtp RMSRMSQ )2cos(1)2cos(1cos)( tPtIVtp RMSRMSP
• EXPRESSION FOR INSTANTANEOUS POWER
• THE INSTANTANEOUS POWER CAN ARBITRARILY BE SPLIT INTO TWO COMPONENTS CALLED pP
(t) AND pQ
(t)
(ACTIVE
AND REACTIVE
POWER RESPECTIVELY)
ACTIVE POWER , pp
(t) REACTIVE POWER , pQ
(t)
WattsinIVdttpT
P RMSRMST
AV cos)(1
0
• AVERAGE POWER SUPPLIED TO THE LOAD OVER INTEGER MULTIPLES OF PERIOD T IS:
• AVERAGE VALUE OF ACTIVE POWER = P • PEAK VALUE OF REACTIVE POWER = QcosRMSRMS IVP sinRMSRMS IVQ
• THE APPARENT POWER
IN THE CIRCUIT, S, IS THE PRODUCT OF THE RMS VALUE OF THE VOLTAGE,VRMS
,
AND THE RMS VALUE OF THE CURRENT, IRMS
:
RMSRMS IVS
22 QPS P
QS
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Voltage, current, & power relationship (continued)
•
EXAMPLE CALCULATION 1A SINGLE-PHASE TWO WIRE CIRCUIT OPERATES AT 7620 VOLTS RMS BETWEEN THE TWO WIRES.THE CURRENT IN THE PHASE WIRE IS MEASURED AT 25 AMPERES RMS. A METER CONNECTEDTO THE CIRCUIT SHOWS THE ACTIVE (REAL) POWER SUPPLIED IS 150 KW
1.
WHAT IS THE APPARENT
POWER SUPPLIED BY THE CIRCUIT?2.
WHAT IS THE REACTIVE
POWER SUPPLIED BY THE CIRCUIT?
LOADVRMS = 7620 VOLTS
IRMS = 25 AMPS
kVAAMPERESVOLTIVS RMSRMS 5.190500,19025*7620.1 kWPGIVEN 150:
kVArQPSQorQPS 43.1171505.190,,.2 222222
IF THE LOAD WERE MODIFIED IN SOME MANNER SUCH THAT IT DRAWS ONLY
150 kW OF REAL POWER (P) AND NO REACTIVE POWER (Q = 0), WHAT WOULD THE LINE CURRENT BE IN
AMPERES?
RMSRMS IVPSQWITHTHEN :0
AMPERESV
PIRMS
RMS 69.19620,7000,150
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Apparent Power (complex number)
The combination of active and reactive power is referred to as apparent power, defined as follows with complex number notation:
S
= P
+ jQS
= VI
cos
+ j VI
sin
whereS is the apparent power (VA)P is the active power (W)Q is the reactive power (VAR)θ
is the angle between the voltage and the current (voltageangle –
current angle in this definition)P is related to energy that becomes heat, light, mechanical motion,etc.Q is related to energy that is stored in an inductor in ½
cycle, and then returned to the system in the next ½
cycle.
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Apparent Power Equations
• APPARENT POWER “S”
IN TERMS OF VOLTAGE (V) AND CURRENT (I) AS COMPLEX NUMBERS
XRXR VjVVandVjVV *
XRXR IjIIandIjII *
XjRZ
V = VOLTAGE PHASOR (A COMPLEX NUMBER) –
RMS VALUEVR
= REAL
PART OF VOLTAGE PHASOR V –
RMS VALUEVX
= IMAGINARY
PART OF VOLTAGE PHASOR –
RMS VALUEV*
= COMPLEX CONJUGATE OF V
I = CURRENT PHASOR (A COMPLEX NUMBER) –
RMS IR
= REAL
PART OF CURRENT PHASOR I –
RMSIX
= IMAGINARY
PART OF CURRENT PHASOR I –
RMSI*
= COMPLEX CONJUGATE OF I -
RMS
*
2
*
***
ZV
ZVV
ZVVIVjQPS
ZIZIIIZIIVjQPS
2***
XRRXXXRRXRXR IVIVjIVIVIjIVjVIVQjPS *
XXRR IVIVP XRRX IVIVQ
• APPARENT POWER S
•
ACTIVE (REAL) POWER
• REACTIVE (IMAGINARY) POWER
V VX
VR
I
IXIR
ZV0o
I
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Apparent Power Equations (continued)
• EXAMPLE CALCULATION 2
THE OUTPUT OF A COMPUTER PROGRAM GIVES THE FOLLOWING FOR THE VOLTAGE AND CURRENTIN A SINGLE-PHASE LINE (SEE GENERALIZED SKETCH ABOVE)
VOLTAGE “V”
IS 7620 VOLTS AT AN ANGLE OF 10 DEGREESLINE CURRENT “I”
IS 25 AMPERES AT AN ANGLE OF -28.057 DEGREESWHAT IS THE ACTIVE (REAL) POWER AND REACTIVE (IMAGINARY) POWER SUPPLIED TO THE LOAD?
WATTSIVIVP XXRR 1.998,149759.11*20.1323062.22*24.7504 VARSIVIVQ XRRX 434,117)759.11(*24.7504062.22*20.1323
• REAL AND IMAGINARY PART OF VOLTAGE PHASOR:
VoltsVandVoltsV oXo
R 20.1323)0.10(sin762024.7504)0.10(cos7620
AmpsIandAmpsI oXo
R 759.11)057.28(sin0.25062.22)057.28(cos0.25
• REAL AND IMAGINARY PART OF LINE CURRENT PHASOR:
• ACTIVE (REAL) AND REACTIVE (IMAGINARY) POWER TO LOAD
Note: The V, I, P, and Q values in this example are the same as
in example calculation 1
V VX
VR
I
IX
IRZV 0o
I
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Apparent Power Equations (continued)
cosRMSRMS IVP
• EXAMPLE CALCULATION 2 (continued)
The active and reactive power, for this example, also can be calculated with the relationships below.
sinRMSRMS IVQ
θ
= Angle by which voltage v(t) leads current i(t)
VoltsVRMS 7620 AmperesIRMS 25
0057.38)057.28(0.10 ooIV
kWWattsP o 0.1503.999,149)057.38cos(25*7620
kVArVarsQ o 43.1178.432,117)057.38sin(25*7620
GIVENS:
θV
= VOLTAGE ANGLE
θI
= CURRENT ANGLE
ACTIVE (REAL) POWER:
REACTIVE (IMAGINARY) POWER:
© 2012, Siemens Industry Inc., all rights reserved
The Power Triangle Revisited
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Phasor Diagram of the Power Triangle
kVAkVAR
kWO
O
kVAkVAR
kW
INDUCTIVE LOAD CAPACITIVE LOAD
POWER FACTOR IS THE COSINE OF THE ANGLE BETWEEN THE APPARENT POWER
(kVA)
ANDTHE ACTIVE POWER (kW).
FOR INDUCTIVE LOAD THE REACTIVE POWER (VARS) IS “+”
IN SIGN
FOR CAPACITIVE LOAD THE REACTIVE POWER (VARS) IS “-”
IN SIGN
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Power Equations Revisited (Again!)
kVAkVAR
kWO
INDUCTIVE LOAD
• RELATIONSHIPS BETWEEN POWER FACTOR AND REACTIVE FACTOR:
sinkVAkVAR
coskVAkW
22 kVARkWkVA
kVAkVARFACTORREACTIVERF
kVAkWFACTORPOWERPF
220.1 RFPF
• RELATIONSHIPS BETWEEN KVA, KW, AND KVAR:
© 2012, Siemens Industry Inc., all rights reserved
Energy –
Billing and Load Composition
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15-16Siemens Industry Inc., Siemens Power Technologies International
Customers and Loads
As engineers we think of the customers as electrical loads that the power delivery system is designed to serve
Generation
Transmission
Primary Distribution
Secondary Distribution
The accountants see the electrical loads as customers (sources of revenue)
The customer has a meter that measures what they use, and the Utility sends them monthly bills that recover the cost of supplying the power
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The kWh Meter –
Residential Customers
An electro-mechanical type of kilowatt-hour meter to measure energy consumption
Flux from current and voltage coils interact producing a torque on the disc. The rotational speed of the disk is proportional to the real power (kW). Time integration of the power thru a gear mechanism is the energy consumed, as displayed on the dials.
A kWh meter only measures the active power. The amount of reactive power consumed (VARs) are not measured.
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The kVA Meter –
Industrial and Commercial Customers
Used to meter commercial and industrial loads
Monitors the kW and kVAR as well as the usage in time periods.
Digital readouts
Meter data can be downloaded to hand held computer
Meter can be read remotely
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What Are Electrical Loads?
The electric power system elements have both a resistive and reactive component
Resistive Components (R)•
Overhead line & Underground cable circuit conductor resistance•
Transformer winding resistance
Reactive Components (XL
, XC
)•
Overhead line & Underground cable circuit reactance•
Transformer winding leakage reactance•
Phase reactor reactance for limiting short circuit current •
Series capacitors to cancel line inductive reactance
Loads are comprised of
Resistive elements •
I2
* R = kW
Reactive elements•
I2
* X = kVAR
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Load Composition
Constant Power Loads•
Demand the same amount of kVA
regardless of the voltage supplied to them
Constant Current Loads•
Demand the same current regardless of the voltage applied
Constant Impedance Loads•
At all times, present the same impedance to the system, irrespective of voltage applied to them
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Load Categories
Residential
Single and multiple phases; seasonal
Commercial
Light to heavy
Industrial
Light to heavy with multiple shifts
Agricultural
Short burst of high demand
Lighting
Changes with time of year
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15-22Siemens Industry
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Distribution Circuits
They are an aggregate of the different types of loads categories
The demands fluctuate
The power factor for each load type is not the same
Each load has a different requirement with respect to the quality of power supply
© 2012, Siemens Industry Inc., all rights reserved
Tab 19 -
Power Quality Overview Distribution Systems Engineering –
Course 1
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Siemens Energy Inc., Power Technologies International 16-2
What is power quality?
Voltage disturbances related to power quality
Causes of PQ problems
Power quality standards
Solutions
Topics
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Siemens Energy
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What Is Good Power Quality?
Utility perspective:
The relative absence of utility-caused voltage variations at the point of common coupling (PCC).
Customer perspective:
Electric power which supports their operations with minimal power induced equipment disturbances and failures.
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Ideal Voltage Conditions
Voltage magnitude is well within ANSI C84.1 Range A (normal operation) limits
Sags, swells, or transient voltages are non-existent or very minor
No momentary or permanent interruptions (outages)
Harmonic distortion and noise are well within specified limits
No observable flicker
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Power Quality Disturbances
Long duration voltage variations
steady-state overvoltages
& undervoltages
sustained interruptions (permanent faults)
Short duration voltage variations
momentary interruptions (clearing temporary faults and reclosing)
sags & swells (from faults on system)
Frequency variations
Transients
impulsive
oscillatory
Voltage unbalance (imbalance)
Voltage fluctuations
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Power Quality Disturbances (continued)
Waveform distortion
harmonics & interharmonics
noise
notching
Reliability issues
momentary outages (interruptions)
permanent outages (interruptions)
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Siemens Energy
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VOLTAGE SAGS ARE THE MOST COMMON POWER QUALITY DISTURBANCE EXPERIENCED BY END USERS
ITIC CURVE PROVIDES GUIDANCE TO EQUIPMENT MANUFACTURERS ON VOLTAGE SUSCEPTIBILITY THAT THEY SHOULD DESIGN INTO THEIR EQUIPMENT
POINTS ON ITIC CURVE ARE AN AGGREGATION OF OVER 3000 VOLTAGE EVENTS RECORDED AT 100 LARGE MANUFACTURING PLANTS IN THE USA OVER A PERIOD OF ONE YEAR
Power Quality Disturbances (continued) Voltage Sags
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ens
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Siemens Energy
Inc., Power Technologies International 16-8
Substation Bus Voltage Sag for SLG Fault on Radial Feeder
32" 32"
48"
Oa Oc
Ob
n
DISC: CE Dist Course, 2006, #1Con Ed OH Line Z1 Z0.FCW
R1 = 0.198 Ohms / mileX 1 = 0.583 Ohms / mile
R0 = 0.485 Ohms / mileX 0 = 1.901 Ohms / mile
477 MCM AL PHASE4/0 CU NEUTRAL
• FACTORS WITH SIGNIFICANT IMPACT ON SUBSTATION BUS VOLTAGE
1.
AVAILABLE 3-PHASE AND SLG FAULT CURRENT ON SUB BUS (Z0
& Z1
)
2.
IMPEDANCE OF FAULTED FEEDER IN OHMS PER UNIT LENGTH
3.
DISTANCE FROM SUBSTATION TO FAULT (L)
LINE CONFIGURATIONAND IMPEDANCES
FOR PLOTS
SUB 13.2 KVBUS
FAULTED FEEDER (SLG PHASE A)
L
UNFAULTED FEEDERS
Z0Z1
Disk: CE Dist Course 2006, #1Sub Bus V, SLG FLT.FCW
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Siemens Energy
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Substation Bus Voltage Sag for SLG Fault on Radial Feeder (continued)
• SUBSTATION LINE-TO-GROUND BUS VOLTAGES –
13.2 kV SYSTEMFEEDER FAULTED PHASE = A, UNFAULTED PHASES = B & C
Notes:Substation bus sag andswell applied to all unfaulted
feeders untilfaulted feeder breakeropens.
For selected configuration, unfaulted
phase C bus voltage swell is less than unfaulted
phase B bus voltage swell.
In 27 kV system, voltage sag on substation bus (with same available fault current as at 13.2 kV), will be greater than sag at 13.2 kV.
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Siemens Energy
Inc., Power Technologies International 16-10
Power Quality Disturbances (continued)
• EXAMPLE OF CONDITION CREATING VOLTAGE UNBALANCE FORSMALL THREE-PHASE PUMPING APPLICATION
FLOATING-WYE DELTA TRANSFORMER BANK FOR 4-WIRE DELTA SERVICE
POWER LEG TRANSFORMERS = 10 KVA
LIGHTING LEG TRANSFORMER = 15 KVA
OPEN FUSE CUTOUTON ONE OF THE POWERLEG TRANSFORMER
4-WIRE DELTA SERVICE SUPPLYING A 3-PHASE PUMPING APPLICATION
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Siemens Energy
Inc., Power Technologies International 16-11
Then Versus Now
Power quality today is not the same as power quality in 1950
With the increase in nonlinear devices, the power has become ’dirty’
with lots of voltage and current distortion
It can present a localized problem that results in overheating of transformers and fuses due mainly to the presence of high harmonic currents
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Siemens Energy
Inc., Power Technologies International 16-12
Customers -
Sensitive Equipment
Industrial equipment failures (Power supplies & motors)
One study suggests that current surges following a voltage sag is more responsible for failures than are surge overvoltages
Industrial/commercial concerns
computer drive manufacturing systems
PCs for data processing and management application
adjustable speed drives/motor driven assembly systems
Programmable logic controllers (PLC)
Computer numerical control (CNC) machines
Servo Drives
Robots
the increasingly competitive nature of manufacturing
the effect of non-linear loads on utility system, other customers and plant equipment
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ens
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Siemens Energy
Inc., Power Technologies International 16-13
Customers -
Sensitive Equipment (continued)
Residential concerns
VCRs
digital clocks
home PCs
various other sags and outage sensitive electronic devices
Residential concern, open neutral in 120/240-volt service
Causes steady state overvoltages
/undervoltages
Disk: CE Dist Course 2006, #1Open Neutral Secondary.FCW
X1
X3
X2
H1
H2
120 V LOAD
120 V LOAD240 V LOAD
DISTRIBUTIONTRANSFORMER
O
OPENNEUTRAL
RESISTANCE TO GROUND AT SERVICE (ROD, H20 PIPE)
120 V
120 V
1
O2
V1
V2
HV O
POLE GROUNDCONNECTION
1-PHASE 3-WIRE SERVICE 3-WIRE LOAD
PN
SN
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Siemens Energy
Inc., Power Technologies International 16-14
Origins of Power Quality Problems
Utility problems
weak system (High Impedance with load injecting high harmonic currents)
single phasing in three-phase lines
open neutral on primary or secondary system
poor system design
fault induced momentary outages, voltage sags and voltage swells
equipment failures (transformers, splices, etc)
switching surges
capacitor switching transients
lightning surges
harmonic resonance
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Siemens Energy
Inc., Power Technologies International 16-15
Origins of Power Quality Problems (continued)
Customer caused problems
poor system design or defective wiring
interaction of loads
grounding problems and loops
electromagnetic compatibility problems
Harmonics
switching of large loads (e.g. motors, arc furnaces) producing flicker
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Siemens Energy
Inc., Power Technologies International 16-16
Origins of Power Quality Problems (continued)
Manufacturers of Utilization Equipment
inadequate design of utilization equipment
cost-cutting measures which make equipment more sensitive
improper installation or application
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Siemens Energy
Inc., Power Technologies International 16-17
Who is responsible for PQ?
Utilities
to provide reliable service and voltage within specified limits
Customers
to buy less sensitive electronic devices
commercial and industrial customers should understand their power environment and purchase equipment from manufacturers who are aware of PQ issues
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Siemens Energy
Inc., Power Technologies International 16-18
Who is responsible for PQ? (continued)
Manufacturers
to produce products which will operate properly on the electromagnetic environment of a typical power system
Derating
required of induction motors in presence of voltage unbalance
0.70
0.75
0.80
0.85
0.90
0.95
1.00
0 1 2 3 4 5
OPERATION ABOVE 5 % VOLTAGE UNBALANCE IS NOTRECOMMENDED BY NEMA STANDARDS
CE 2006 #1, NEMA DERATE.EP
PERCENT VOLTAGE UNBALANCE
DE
RA
TIN
G F
AC
TOR
Voltage unbalance at many pointson radial distribution circuits will exceed 1 %.
In old days, many inductionmotors would operate satisfactorily with voltage unbalances up to 3.0 %.
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Siemens Energy
Inc., Power Technologies International 16-19
Utility Power Quality Programs
1996 PQ survey of utilities done by PTI:
80% of utilities had PQ programs for commercial and industrial customers, 62% also had programs for residential customer
29% of utilities charged for PQ consulting
PQ programs were usually reported as effective
PQ programs
monitoring (mainly, large customer loads)
educational programs
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ens
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Siemens Energy
Inc., Power Technologies International 16-20
Power Quality Standards
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Siemens Energy
Inc., Power Technologies International 16-21
Power Quality Standards (continued)
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Siemens Energy
Inc., Power Technologies International 16-22
Power Quality Standards (continued)
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Siemens Energy
Inc., Power Technologies International 16-23
Some Disturbances and Corresponding Standards
Transients
ANSI/IEEE C62
IEEE Std 1100
Harmonics
IEEE Std 519
Voltage sags
IEEE Std P1346
CBEMA / ITIC tolerance curves
Voltage Flicker
IEEE Std 519
IEEE Std 141
GE Flicker Curve
© 2012, Siemens Industry Inc., all rights reserved
Tab 20 -
Voltage Unbalance, Flicker and Transients
Distribution Systems Engineering –
Course 1
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Siemens Energy Inc., Power Technologies International 17-2
Voltage Unbalance Under Steady-State Conditions
Voltage levels are not the same on all three phases of a feeder at the same moment in time
Distribution systems are typically unbalanced•
The load demand at each phase is not the same
•
The current flow in the phase conductors is not the same
•
Line is not symmetrical•
The voltage drop is different for each phase
•
The voltage on the substation bus is unbalanced
This results in a voltage unbalance between the phases, and between the phases and ground
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Siemens Energy
Inc., Power Technologies International 17-3
Unbalanced transformerbank (open-wye
open-delta) causes unbalanced line currents on primary side as well as current in the primary multi-grounded neutral conductor
Voltage Unbalance Under Steady-State Conditions (continued)
LIGHTING LEG TRANSFORMER
POWER LEG TRANSFORMERPRIMARY NEUTRAL
PRIMARY PHASEWIRES
Note:
Transformer bank is on a two-phase tap line supplied from a three-phase four-wire multi-
grounded neutral circuit.
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Siemens Energy
Inc., Power Technologies International 17-4
Voltage Unbalance Under Steady-State Conditions (continued)
Basic Definition
Voltage Unbalance In percent
AVE
max
VVx 100
Where: MAXVΔ = The maximum deviation from the average phase-to-phase voltage
AVEV = The average phase-to-phase voltage 3
|V||V||V| cabcab
Vab
, Vbc
, and Vca
are the three phase-to-phase voltages
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Siemens Energy
Inc., Power Technologies International 17-5
Voltage Unbalance Under Steady-State Conditions (continued)
Example: Suppose a customer connected to the end of a 240 V three-phase service and assume that at the PCC the phase-to-
phase voltages are
|Vab
| = 230 volts|Vbc
| = 232 volts|Vca
| = 225 volts
Then: VAVE = (230 + 232 + 225) / 3 = 229 volts
∆V1
= |230 –
229| = 1 volt∆V2
= |232 –
229| = 3 volts∆V3
= |225 –
229| = 4 volts∆VMAX = 4 volts
Consequently: Voltage unbalance = (4/229)·100 = 1.75 %
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Siemens Energy
Inc., Power Technologies International 17-6
ANSI C84.1 Voltage Standards
Recommendation: Voltage unbalance at the PCC or service entrance shall not exceed 3% under no load conditions
The PCC is the point of common coupling, typically at the meter or service entrance equipment
The electric utilities are responsible only for satisfying the service entrance or PCC voltage requirements.
Customers are responsible for maintaining proper voltage downstream of the service entrance or PCC!
Classification Range A Range B Service Voltage 114 to 126 volts 110 to 127 volts
Utilization Voltage 110 to 125 volts 106 to 127 volts
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Siemens Energy
Inc., Power Technologies International 17-7
Voltage Unbalance –
Alternate Definition and Limit
CBA VaVaVV 21 31
CBA VaVaVV 22 31 2
321120 jea
Oj
100*%1
22 V
Vd
%2% 2 d
• NEGATIVE-SEQUENCE VOLTAGE UNBALANCE (d2
)
V2
= MAGNITUDE OF NEGATIVE-SEQUENCE VOLTAGE
V1
= MAGNITUDE OF POSITIVE-SEQUENCE VOLTAGE
• GIVEN THREE PHASE-TO-PHASE
VOLTAGES
VA
, VB
, AND VC (PHASORS WITHBOTH MAGNITUDE AND ANGLE):
• RECOMMENDED LIMIT FOR d2
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Siemens Energy
Inc., Power Technologies International 17-8
By knowing Ea
, Eb
and Ec
it is possible with “chart”
to determine the angle θ
and the unbalance factor d2
. Chart is applicable only when there are no zero-sequence components in the three voltages. From Westinghouse T&D Reference Book
Voltage Unbalance –
Alternate Calculation Method From Chart
EXAMPLE:Ea
= 235.0, Eb
= 230.0, Ec
= 222.0Eb
/Ea
= 0.979 Ec
/Ea = 0.945FROM CHART:
d2
= |V2
| / |V1
| = 0.034
Θ
= 37 Degrees
FROM EXACT CALCULATIONS:|V2
|| / |V1
| = 0.0330Θ
= 37.54 Degrees
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Siemens Energy
Inc., Power Technologies International 17-9
Voltage Flicker
VOLTAGE FLICKER IS:
Voltage Drop as Seen By Customers
Visible With Incandescent lights and some CFL’s
Complaints Normally From Residential Customers
Much Lower in Magnitude Than A Fault Caused Voltage Sag –
Usually Only a Few Volts on a 120-Volt Bas
Noticeable at Low Levels and Annoying at Higher Levels
VOLTAGE FLICKER DOES NOT:
Normally Does Not Cause Equipment Failure or Downtime
Does Not Damage Other Customers Equipment
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Siemens Energy
Inc., Power Technologies International 17-10
Voltage Flicker
• CAUSE-FLUCTUATING LOADS1.
ARC FURNACES2.
WELDERS3.
RECIPROCATING COMPRESSORS4.
ROCK CRUSHERS5.
SAWMILLS6.
CAR SHREDDERS
Note:1.
Flicker is worst at fluctuating load, anddecreases at points up stream from theflicker source.
• EFFECT OF VOLTAGE FLICKERCAN CAUSE VARIATION IN LAMPLIGHT OUTPUT THAT CAN BE EITHER PERCEPTIBLE OR ANNOYING TO THEEND USER
• RAPID CHANGES IN THE RMS VOLTAGE
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Siemens Energy
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Fluctuations and Dips
• EXAMPLE OF WAVEFORM PRODUCING VERY OBJECTIONABLE LAMP FLICKER
ΔV = 1.10 -
0.90 = 0.20 PER UNIT
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Siemens Energy
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GE Flicker Curve
Notes:1.
Based upon square-wave changes to supply voltages at indicated frequencies (see previous chart).
2.
From tests run by GE and utilities in 1930 with incandescent bulbs.3.
Used by 69 % of utilities based on 1985 IEEE survey.
• DEFINES BASED ON TESTS BORDERLINES OF VISIBILITY AND IRRITATION WITHINCANDESCENT LIGHTS
FUNCTION OF PERCENT VOLTAGE DIP AND FREQUENCY OF DIPS
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Siemens Energy
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Permissible Flicker Limits (Con Edison)
• FROM 1958 EEI T&D COMMITTEE MEETING
Notes:Upper stair case curves apply to radial services and underground
networks.Lower stair case curve applies to primary distribution lines.
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Measuring Voltage Flicker The Flicker Meter
Siemens Energy
Inc., Power Technologies International 17-14
The IEC 61000-4-15 flickermeter
standard was developed in Europe and later adopted in the U.S. as IEEE 1453-2004. The measurement methodology described in the IEC standard more accurately accounts for complex voltage fluctuations encountered in actual practice by including the effects of multiple flicker sources, frequencies and varying voltage modulation waveforms. In addition, the flickermeter
approach standardizes flicker monitoring across different manufacturers–IEC 61000-4-15 compliant instruments should all produce the same results for a given flicker excitation.
The real advantage of the flickermeter
method is inherent in its ability to accurately model the human flicker perception. This rather complex modeling is accomplished by five signal processing blocks described in the IEC standard, which represent the lamp-eye-brain response to light flicker –
the response of a lamp to supply voltage variations, the perception of the human eye and the memory characteristics of the human brain.
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Measuring Voltage Flicker The Flicker Meter
Siemens Energy
Inc., Power Technologies International 17-15
The flickermeter
system provides several measurement outputs. IFL, or instantaneous flicker level, represents the real time voltage modulation modified by the lamp-eye-brain response. This output can be plotted as a time interval graph and is useful for tracking down sources of voltage fluctuations.A statistical analysis block completes the human perception system by providing short and long term flicker severity indexes. Short term flicker severity Pst
is evaluated over a 10 minute observation period and is used to
evaluate disturbances caused by flicker sources with short duty cycles.
According to the IEEE 1453 standard, a Pst
value of 1.0 represents the system compatibility level, the level below which customer complaints are not likely to occur. Pst
is therefore commonly used to evaluate whether the measured voltage fluctuations are severe enough to cause flicker complaints.Long term flicker Plt
is derived from 12 successive Pst
values, or two hours, and is more suitable for evaluating the combined effect of several randomly operating loads such as welders or motors over longer periods of
time. A Plt
value of 0.8 is considered the system compatibility limit according to IEEE 1453.
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Oscillatory Transient -
Capacitor Switching
Closing capacitor switch produceshigh-frequency inrush current andassociated high frequency componentin the system voltages, that decay exponentially with time
L
CVrms
SUBSTATIONBUS FEEDER
IMPEDANCE
Vbus
CAPBANK
CE 2006 D#2, Circuit Cap Sw Transient.FCW
S
LFRF
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Siemens Energy
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Oscillatory Transient –
Inductive Load Switching
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Impulsive Transient -
Lightning Surges
BOTH MEASUREMENTSMADE ON CIRCUIT OR PTSECONDARY WITH A NOMINAL VOLTAGE OF120 VOLTS RMS OR 169.7VOLTS CREST
© 2012, Siemens Industry Inc., all rights reserved
Tab 21 -
Harmonics, Notching and Noise Distribution Systems Engineering –
Course 1
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Voltage Notching
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Electrical Noise
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Harmonics
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Siemens Industry
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Harmonics (continued)
Harmonics are multiples of the fundamental frequency
Like the waves in a pond, they will add or subtract to the fundamental
Under and Over voltageFUNDAMENTAL
TOTAL
THIRD HARMONIC
FIFTH HARMONIC
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Harmonics (continued)
• COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS
unitperOrderHarmonic
PeakHarmonic 1
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Harmonics (continued)
• COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS-
Harmonics lagged by 30 degrees from zero of fundamentalunitper
OrderHarmonicPeakHarmonic 1
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Harmonics (continued)
• COMBINATION OF FUNDAMENTAL AND FIRST THREE ODD HARMONICS-
Harmonics lagged by 60 degrees from zero of fundamental
unitperOrderHarmonic
PeakHarmonic 1
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Siemens Industry
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Harmonic Distortion Factor
Distortion Factor in %
Each harmonic component contributes to the overall distortion of
the fundamental frequency wave
Each harmonic component can be expressed in percent of the amplitude of the fundamental frequency wave
Total harmonic distortion (THD) is the distortion factor including all relevant harmonic components (usually from 2nd to 50th)
100
%
wavelfundamentatheofAmplitudeAmplitudesComponentHarmonictheofSquarestheofSum
THD
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Calculation of THD
Example: Consider the voltage waveshape
shown below
Voltage Waveform at F9: phase A
5 10 15 20 25 30 35 40 45 50
0
0.5
1
0
-0.5
-1
Time (ms)
Voltage (pu)
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Calculation of THD (continued)
A harmonic spectral analysis of this wave indicates the following components:-
a fundamental frequency wave (60Hz) of magnitude 1 pu-
a 5th harmonic (300Hz) wave of magnitude 0.14 pu-
a 7th harmonic (420Hz) wave of magnitude 0.12 pu-
a 11th harmonic (660Hz) wave of magnitude 0.03 pu-
a 13th harmonic (780Hz) wave of magnitude 0.03 pu
Then:
%9.18100
103.0)03.0(12.014.0
%2222
THD
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Calculation of THD (continued)
• THEORETICAL TOTAL WAVE CONSIDERING FUNDAMENTAL, 5TH, 7TH, AND 11TH
HARMONICFROM PREVIOUS EXAMPLE
FUNDAMENTAL = 1.0 PU5TH
HARMONIC = 0.14 PU7TH
HARMONIC = 0.12 PU11TH
HARMONIC = 0.03 PU
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TVs
Computers
Fluorescent lights
Dimmers for incandescent lights
HVDC terminals
UPS Systems
AC adjustable speed motor drives
Battery Chargers (Electric Cars)
Transformers (magnetizing current)
Harmonic Sources
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Fluorescent Lighting System V-I Waves
OLD STYLE WITH FERRORESONANTLIGHTING BALLAST
TYPICAL SCREW-IN COMPACT FLUORESCENT LAMP
VOLTAGE: BLUE CURVECURRENT: RED CURVE
DISPLACEMENT POWER FACTOR ≈
40 % Total Harmonic Distortion = 19 %
VOLTAGE: BLUE CURVECURRENT: RED CURVE
DISPLACEMENT POWER FACTOR ≈
SLIGHTLY LEADTotal Harmonic Distortion = 120 %
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Fluorescent Lighting System V-I Waves (continued)
FLUORESCENT LAMPS WITH ELECTRONICBALLASTS
VOLTAGE: BLUE CURVECURRENT: RED CURVE
DISPLACEMENT POWER FACTOR ≈
UNITY Total Harmonic Distortion = 49 %
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Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems
Balanced Three-Phase System
Current waveform in each phase is identical
Fundamental component of current in each phase is displaced by 120 degrees from that in any other phase
5th, 7th, 11th, 13th, 17th, 19th, etc
Harmonic currents are either positive- or negative-sequence
No harmonic currents in neutral return path (multi-grounded neutral of primary system and earth)
3rd, 9th, 15th, 21st, etc
Harmonic currents are zero-sequence
At each harmonic, current in neutral return path is three-times that in the phase
Do not pass from secondary to primary of delta (primary) wye-grounded (secondary) transformer
Do pass from secondary to primary of wye-grounded wye-grounded transformer (flow in neutral conductor of primary system)
Siemens Industry
Inc., Siemens Power Technologies International 18-16
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Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems
Siemens Industry
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ttIA 3sin1.0)sin(0.1
)
32(3sin1.0)
32sin(0.1 ttIB
)
32(3sin1.0)
32sin(0.1 ttIC
ϕA
ϕB
ϕC tIRESIDUAL 3sin3.0
RESIDUAL IS 3RD
HARMONIC
100 % Fundamental, 10 % Third Harmonic
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Harmonics in Balanced Three-Phase Four- Wire Multi-Grounded Neutral Systems
• ZERO-SEQUENCE HARMONIC TRANFER THROUGH DISTRIBUTION TRANSFORMERS
I3RD3
I3RD
I3RD
I3RD
N 1
I3RDN
NO ZERO-SEQUENCE HARMONICSIN PRIMARY PHASE CONDUCTORS OR IN PRIMARY NEUTRAL CONDUCTOR.
LOADS ZERO-SEQUENCE HARMONICSCIRCULATE IN DELTA WINDING.
I3RD3
I3RD
I3RD
I3RD
1
I3RD3
I3RD3
n
I3RDn
I3RDn
I3RDn
I3RDn3
I3RDn3
ZERO-SEQUENCE HARMONICS INPRIMARY PHASE CONDUCTORS AND IN PRIMARY NEUTRALCONDUCTOR (3 TIMES THAT INPRIMARY PHASE CONDUCTORS)
PTI 2010, D#1, Harmonic Transfer Thru Dist Xfrs.FCW
LOADSIDE
LOADSIDE
0
00
0
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Harmonics Produced Problems
Lighting Systems
High intensity discharge lamps outside voltage threshold can shut down
Transformers and Capacitor Banks
Resonances
Transformer derating
required with high harmonics (K Factor)
kVAr
rating of capacitor bank exceeded due to harmonic current
Other problems
Nuisance fuse operations
Device missoperations
Telephone interference (rural systems-old days)
Equipment heating•
Eventual Component failure
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Siemens Industry
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IEEE 519
IEEE established a Standard for harmonic measurement and control
IEEE Std. 519
It specifically set
Requirements for Utilities•
voltage
Requirements for Customers•
current
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IEEE Std 519 Voltage Requirements
For periods less than one hour, limits may be increased by 50%.
Bus Voltage at PCC
Individual Distortion
(%)
Total Distortion THD (%)
138kV 1 1.5
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IEEE Std 519 Current Requirements
AS THE STIFFNESS OF THE UTILITY SUPPLY SYSTEM INCREASES, THE LOAD IS ALLOWED TO GENERATE MORE HARMONIC CURRENTS.
© 2012, Siemens Industry Inc., all rights reserved
Tab 22 –
Voltage Sags, Voltage Swells, Momentary Interruptions
Distribution Systems Engineering –
Course 1
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Voltage Swell
This can occur, for instance, on the healthy (unfaulted) phases of a multi-grounded neutral distribution system during a single-line-to-ground fault
Duration of Swell
is typically between ½
cycle to 1 second
LINE-TO-GROUND VOLTAGE
1.0 PER UNIT1.25 PER UNIT
1.0 PER UNIT
CE DE 2010, Voltage Wave Sag.FCW
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Siemens Industry
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Voltage Swell –
Unfaulted
Phase-to-
Ground Voltages
For Single Phase-to-Ground Fault on Phase A
Z1
= POSITIVE-SEQUENCE IMPEDANCEAT FAULT POINT
Z0
= ZERO-SEQUENCE IMPEDANCE ATFAULT POINT
Θ1 = POSITIVE-SEQUENCE IMPEDANCEANGLE AT FAULT POINT
Θ0 = ZERO-SEQUENCE IMPEDANCEANGLE AT FAULT POINT
unitperZZ
ZZaVB01
012
2
unitperZZ
ZZaVC01
01
2
23
21
23
21
1202
120
jea
jea
o
o
j
j
• UNFAULTED Φ-TO-GROUNDVOLTAGE AT FAULT POINT
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Siemens Industry
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Momentary Interruption
Time in Cycles forLoss of voltage
Typically Caused By a Circuit Recloser
or Circuit Breaker Opening and Successfully Reclosing Due to a Temporary (Transient) Fault
LINE-TO-NEUTRAL VOLTAGE
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Momentary Interruption (continued)
• PERCENTAGE OF DEVICES THAT WERE ABLE TO SUCCESSFULLY RIDE THROUGH A MOMENTARY INTERRUPTION OF THE GIVEN
DURATION
DEVICEMOMENTARY INTERRUPTION DURATION
(SECONDS / CYCLES)0.5 / 30 2.0 / 120 16.7 / 1000
Digital Clocks 70 % 60 % 0
Microwave Ovens 60 % 0 0
VCR 50 % 37.5 % 0
Computer 0 0 0
FROM: POWER QUALITY ASSURANCE MAGAZINE, PP 296-310, 1990
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Momentary Interruptions (continued)
Equipment
is sensitive and can not ‘ride through’
resulting in the machine shutting down
System solutions
faster reclosing
fewer faults (tree trimming, animal & bird guards, better lightning protection)
elimination of fuse saving with feeder breaker instantaneous tripping in substation
installation of circuit reclosers
in main feeder and on major branches
closed bus tie breakers in substation (prevents momentary interruptions from loss of subtransmission
sources)
Customer solutions
static transfer switch, UPS
employ spot network systems
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Momentary Interruptions (continued)
• SYSTEM MEASURES TO REDUCE NUMBER OF MOMENTARY INTERRUPTIONS:1.
ELIMINATE INSTANTANEOUS TRIPPING OF FEEDER BREAKER 52-1 IN ATTEMPTINGTO PREVENT FUSE BLOWING FOR TEMPORARY FAULTS ON LATERALS. THIS MAY RESULT IN MORE PERMANENT OUTAGES FOR CUSTOMERS SERVED FROM FUSEDLATERALS.
2.
INSTALL RECLOSER IN MAIN FEEDER SO THAT TEMPORARY AND PERMANENTFAULTS ON “2ND
HALF OF FEEDER 2
DO NOT CAUSE A MOMENTARY INTERRUPTIONTO CUSTOMERS ON FIRST HALF OF FEEDER.
3.
USE SINGLE-PHASE RATHER THAN THREE-PHASE RECLOSERS IN 3-PHASE LINESSUBSTATION
BUS
Disk: CE Dist Course 2006, #2, Momentary Reduction.FCW
52-1 FUSE
LATERAL
RECLOSER52-2
52-3
FEEDER 1
FEEDER 2
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Voltage Sag
An RMS reduction in the nominal AC voltage, at the power frequency, to between 10% and 90% of nominal, for durations from 1/2 cycle to 1 minute.
IEEE 1250: Guide on Service to Equipment Sensitive to Momentary Voltage Disturbances
In three-phase systems, voltage sags may be same in each phase or different, depending on fault type or disturbance
In the multi-grounded neutral system, the single line-to-ground (neutral) fault can cause voltage sags in the faulted phase, and
voltage swells in the unfaulted
phases
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Examples of Voltage Sag
Even when there isn’t a complete loss of voltage, customer equipment (industrial motors, ASD) are still impacted
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Causes of Voltage Sags In Distribution Systems
Faults in sub transmission system
Faults on adjacent distribution circuits supplied from same bus section in the distribution substation
Starting of large motors
Transfer of large loads
EXAMPLES:
84% OF SAG EVENTS RESULT IN VOLTAGE LESS THAN 82.5%
40% OF SAG EVENTS RESULT IN VOLTAGE LESS 70%
16% OF SAG EVENTS RESULT IN VOLTAGE LESS THAN 40%
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Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit
FAULTED FEEDER (SLG PHASE A)
UNFAULTED FEEDERS
Z0SUBZ1SUB
Disk: CE Dist Course 2006, #1SUB SAG SLG FLT.FCW
Z0FDRZ1FDRVASUB
Z1SUB
= POSITIVE-SEQUENCE IMPEDANCE LOOKING INTO SUBSTATION BUS
Z0SUB
= ZERO-SEQUENCE IMPEDANCE LOOKING INTO SUBSTATION BUS
Z1FDR
= POSITIVE-SEQUENCE IMPEDANCE OF FEEDER BETWEEN SUBSTATION & FAULT
Z0FDR
= ZERO-SEQUENCE IMPEDANCE OF FEEDER BETWEEN SUBSTATION AND FAULT
unitperZZZZ
ZZV
FDRFDRSUBSUB
FDRFDRASUB
0101
01
222
• VOLTAGE ON FAULTED PHASE AT SUBSTATION BUS, VASUB
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Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit (continued)
• EXAMPLE CALCULATION: 13.2 KV SYSTEM, ELN
= 7620 VOLTSOH LINE, SLG FAULT ON ΦA 1.25 MILES FROM SUBSTATION
• SYSTEM DATA:AVAILABLE 3-Φ
FAULT CURRENT ON SUB BUS = I3P
= 15,000 AMP, X/R = 20AVAILABLE SLG FAULT CURRENT ON SUB BUS = ISLG
= 15,750 AMP, X/R = 20
Z1FDR
= (0.198 + j 0.583 Ω/MILE)*1.25 MILE = 0.2475 + j 0.7288 ΩZ0FDR
=(0.485 + j 1.901 Ω/MILE)*1.25 MILE = 0.6063 + j 2.3763 Ω
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Voltage Sag on Substation Bus Single Line-to-Ground Fault on Adjacent Circuit
(continued)
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))/tan(sin())/tan(cos(3
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