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8/10/2019 Electric Power Calculations-Rocky MT Electrical League
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PHILLIPS ENGINEERS+ CONSULTANTS, INC.www.phillipsengineers.com
ELECTRIC POWER CALCULATIONS
ROCKY MOUNTAINELECTRICAL LEAGUE
OCTOBER 24, 2002
K. James Phillips, Jr., P.E. [email protected]
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ELECTRIC POWER CALCULATIONS
• OVERVIEW
• PER PHASE
• SHORT CIRCUIT
• PER UNIT
• HARMONICS
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WHY ELECTRIC POWER CALCULATIONS
• USED TO PREDICTOUTCOMES
• SHORT CIRCUITS
• HARMONICS
• VOLTAGES
• LOADS
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THE BASICS
• ALL ELECTRICAL THEORY RELATESBACK TO THE BASICS
• Volts, Amps, Ohms
II
ZV
V = I * Z
Z = V / I
I = V / Z
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SERIES COMBINATIONS
Z1
Z2
Z
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COMPLEX IMPEDANCE
R
X
Z
0
X = Z * Sin 0
R = X
X/R
Z = R2
+X2
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SHORT CIRCUIT REQUIREMENTS
NEC 110-9 AND 110-10• Articles 110-9 and 110-10
• Equipment shall have adequateinterrupting rating
• Clear faults without extensive damage
• Implies must perform short circuit study
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AIC RATINGS
CIRCUIT BREAKER SHORT CIRCUIT
TYPE RATING
QOB 10,000
QOB-H 22,000
QOB-VH 42,000
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SHORT CIRCUIT AMPS (SCA)
Source
Circuit Breaker
Source
3 Phase
Circuit Breaker
Line-Ground
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PER PHASE ANALYSIS
~
~
~
~
A
B
C A
B
C
Ia
Ib
Ic
~
Three Phase Representation
Single Phase Representation
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PER PHASE ANALYSIS EXAMPLE
~
~
~
~A
B
C A
B
C
Ia
Ib
Ic
~
Three Phase Representation
Single Phase Representation
A 480Y / 277V source is serving a balanced three phase wye resistive load of 20ohms per phase. What is the current in phase A, B and C?
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PER PHASE ANALYSIS
SHORT CIRCUITS
~
~
~
~
A
B
C A
B
C
Ia
Ib
Ic
~
Three Phase Representation
Single Phase Representation
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DATA
SOURCE IMPEDANCE
• THEVENINEQUIVALANTIMPEDANCE
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DATA
TRANSFORMER IMPEDANCE
• TRANSFORMER
IMPEDANCE
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DATA
CONDUCTOR IMPEDANCE
• CONDUCTORIMPEDANCE
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BASIC SHORT CIRCUIT ANALYSIS
I = V / Z = 277 V / (.001+.01+.085 + 2.0)
I = 277 V / 2.096 ohms
I = 132.156 Amps of load current
I
V
Zsource
.001
Ztransformer
.01
Zconductor
.085 Zload
2.0
277 V
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BASIC SHORT CIRCUIT ANALYSIS
I = V / Z = 277 V / (.001+.01+.085)
I = 277 V / 0.096 ohms
I = 2885.4 Amps of short circuit current
IZsource
.001
Ztransformer
.01
Zconductor
.085 Zload
2.0
Short Circuit
V
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SCA < AIC
CALCULATEDSHORT CIRCUIT AMPS MUST BELESS THAN AICRATING
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CALCULATION TRICKTRANSFORMER IMPEDANCE
• VOLTAGE
• %Z
• KVA
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CALCULATION TRICKTRANSFORMER IMPEDANCE
Transformer
Variable voltage source
Short Circuit
A
Percent Impedance = Percent rated primaryvoltage that causes rated base/ambient fullload current to flow in the secondary of ashort circuited transformer.
i.e.. 5.75 percent primary voltage causesfull load current in short circuited secondary,the percent impedance is 5.75%
%Z = 100%
FLA SCA
SCA = (FLA * 100)
%Z
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SHORT CIRCUIT CALCULATION
EXAMPLE:
1500 KVA TRANSFORMER
5.75% IMPEDANCE
480 VOLT SECONDARY
1500 KVA
5.75%
480 VOLTS
SCAMPS?X
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SHORT CIRCUIT CALCULATION
STEP ONE:
FLA = (1500 KVA) / ( .48 KV * SQRT 3 )
FLA = 1804 AMPS
STEP TWO
SCA = (1804 AMPS * 100 ) / 5.75%
SCA = 31,374 AMPS
1500 KVA
5.75%
480 VOLTS
SCAMPS?X
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PROBLEM WITH IMPEDANCE ON TWOSIDES OF TRANSFORMER
10 : 1 RATIO
X
20 OHMS
0.8 OHMS
Z1(VIEWED FROM SECONDARY)
= Z1 * (1 / 10)2 = 20 Ù * .01 = .002 Ù
Z2(VIEWED FROM SECONDARY)
= Z2 * (10 / 1)2 = 0.8 Ù * 100 = 80 Ù
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WHY PER UNIT?
• Calculations that involve impedances atseveral voltage levels via transformers,
can become complicated due to thetransformer turns ratio.
• Per unit eliminates need to “reflect”
impedances to different voltages.
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PER UNIT EXAMPLE
Example:
A base number of 500 is used. The following is asummary of per unit values for various numbers:
Number Per Unit Per Cent
250 0.5 p.u. 50%
500 1.0 p.u. 100%
1000 2.0 p.u 200%
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BASE QUANTITIES
Current (p.u.) = Actual Current
Base Current
Voltage (p.u.) = Actual Voltage
Base Voltage
Impedance (p.u.) = Actual Impedance
Base Impedance
Power (p.u.) = Actual Power
Base Power
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BASE QUANTITIES
• The per unit system requires two basequantities to be selected: – Base kVA - Fixed Quantity
– Base Voltage - Variable Based on Voltage
• Two base quantities are calculated:
– Base Impedance - function of base voltage – Base Current - function of base voltage
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BASE KVA(SELECTED)
BASE KVA (MVA):
In electric power systems, the base quantity thatremains constant throughout the system is the basekVA. The kVA base is not affected by voltage levelsor transformer turns ratios. The amount of kVAentering the primary of a transformer is the same asthe kVA leaving the secondary of the transformer(neglecting transformer losses).
Any arbitrary number may be selected as the kVAbase, however, most electric utilities use a 100,000kVA base commonly referred to as a 100 MVA base.
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BASE VOLTAGE(SELECTED)
BASE VOLTAGE:
The base voltage will vary depending on the voltage
level of the system. The base voltage is generallythe nominal voltage of a particular voltage level.
PER UNIT VOLTAGE
V p.u. = V actual / V base V actual = V base * V p.u.
Example:
V base = 13.8 kV
V actual = 13.4 kV
V p.u. = 13.4 kV / 13.8 kV = 0.97 p.u.
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BASE CURRENT(CALCULATED)
• I base = MVAbase * 1000 / (sqrt 3 * kVbase)
• I base = kVAbase
/ (sqrt 3 * kVbase
)PER UNIT CURRENT
I p.u. = I actual / I base I actual = I base * I p.u.
Example:
Bases: 100 MVA, 13.8 kV
I base = (100 MVA * 1000) / (sqrt 3 * 13.8 kV) = 4183 Amps
Actual Amps = 2500 Amps
I p.u. = I actual / I base = 2500 / 4183 = 0.598 p.u.
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BASE IMPEDANCE(CALCULATED)
• Z base = kVbase2 / MVA base
Per Unit Impedance:Bases: 100 MVA, 13.8kV
Actual Impedance = 0.32 + j0.27 ohms
Z base = 13.8 kV2 / 100 MVA = 1.90 ohms
Z p.u. = 0.168 + j0.142 p.u.
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TABLE OF BASE VALUES100 MVA - NOMINAL VOLTAGES
Voltage (kV) Impedance
(Ohms)
Current
(Amps)
765 5852.25 75.47
500 2500.00 115.47345 1190.25 167.35
230 529.00 251.02
138 190.40 418.37
115 132.25 502.04
69 47.61 836.7434.5 11.90 1673.48
23 5.29 2510.22
13.8 1.90 4183.70
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BASIC SHORT CIRCUIT ANALYSISREVIEW
I = V / Z = 277 V / (.001+.01+.085)
I = 277 V / 0.096 ohms
I = 2885.4 Amps of short circuit current
IZsource
.001
Ztransformer
.01
Zconductor
.085 Zload
2.0
Short Circuit
V
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SHORT CIRCUIT ANALYSISPER UNIT
Zload
V base = 13.8 kV, 3 Phase
MVA base = 100
Assume impedances are all reactive i.e. “X” only. No “R” component.
I = V / Z = 1.0 p.u. / (.002+.03+.098 p.u.)
I = 1.0 p.u. / 0.13 p.u.
I = 7.69 p.u. short circuit current
I base = (MVA base * 1000) / (Sqrt 3 * kV base )I base = ( 100 MVA * 1000 ) / (Sqrt 3 * 13.8 kV)
I base = 4183.7 Amps
I actual = I p.u. * I base I actual = 7.69 p.u. * 4183.7 Amps I actual = 32,172.6 Amps
IVZsource
.002p.u.
Ztransformer
.03p.u.
Zconductor
.098p.u.1.0 p.u.
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HARMONICSDEFINITION
• Harmonic Order - integer multiple of thefundamental frequency. – Harmonic Order Frequency
• 1 ( fundamental) 60Hz
• 2 120Hz
• 3 180Hz
• 4 240Hz
• N N * Fundamental
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HARMONICS - 5TH
5th Harmonic (300 Hz)
Fundamental (60 Hz)
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NON LINEAR LOAD
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SOURCE OF HARMONICSTHE LOAD
• Solid State MotorDrives
• Rectifiers
• UPS systems
• Computer powersupplies
• Fluorescent lightingelectronic ballast's
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FREQUENCY SPECTRUM
1 3 5 7 9 11Harmonic Order
M a g n i t u d e
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HARMONIC RELATED PROBLEMS
• Blown Capacitors / Capacitors Fuses
• Transformer Overheating
• Neutral Overheating
• Motor / Generator Overheating
• Equipment Misoperation
• Circuit Breaker Misoperation
• Communication Interference
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EFFECTS OF CAPACITORS
• Without capacitors, thecircuit is predominantly
inductive.• When capacitors are
added, an L-C circuitresults.
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SYSTEM IMPEDANCEINDUCTIVE ONLY
Example:
Power factor correctionrequirements dictate that two 600kvar capacitor banks be installedat the substation bus. What doesthe system impedance look likebefore adding capacitors?
Short Circuit
Capacity = 30MVA
HarmonicSource
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SYSTEM IMPEDANCEINDUCTIVE ONLY
Frequency (harmonic order)
Z ( O h m s
)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Z = jwl, w = 2pi*f, f = frequency
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SIMPLIFIED RESONANCECALCULATIONS
Example:
Power factor correctionrequirements dictate that two 600kvar capacitor banks be installedat the substation bus. The utilityshort circuit current is 30 MVA(36,084 Amps @ 480V). What isthe resonance frequency whenthe 600 kvar bank is on line andwhat is the resonance frequency
when both 600 kvar banks are online.
Short Circuit
Capacity = 30MVA
2 x 600 kvarCapacitor Bank
Harmonic Source5TH, 7TH, 11TH, 13TH
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SIMPLIFIED RESONANCECALCULATIONS
MVAsc = short circuit MVA at the capacitorbank location.
Mvar cap = Mvar rating of the capacitor bank
SourceImpedance
PowerFactorCapacitor
HarmonicSource
XL XC
hr = MVAsc = Xc
Mvar cap Xsc
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SIMPLIFIED RESONANCECALCULATIONS
Frequency (harmonic order)
Z ( O h m s
)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1200 kvar 600 kvar
hr = MVAsc =
Mvar cap
30 MVAsc = 7
0.600 Mvar cap
30 MVAsc = 5
1.200 Mvar cap
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RESULTS
• PRODUCES SEVERE DISTORTION
• DESIGN CAPACITOR AS HARMONICFILTER
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IEEE-519
• Sets limits forvoltage and current
distortion at PCC• PCC is the point of
common coupling
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QUESTIONS ???
CONTACT INFORMATION:PHILLIPS ENGINEERS + CONSULTANTS, INC.4450 BELDEN VILLAGE ST., N.W. SUITE 309CANTON, OHIO 44718
Tel: 330.491.0261Fax: 330.491.0265 [email protected]
DESIGN ENGINEERING ANALYSIS
FEASABILITYMANAGEMENT
TRAINING