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Electric power calculations
1. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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ELECTRIC POWER CALCULATIONS
ROCKY MOUNTAIN
ELECTRICAL LEAGUE
OCTOBER 24, 2002
K. James Phillips, Jr., P.E.
jphillips@phillipsengineers.com
2. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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ELECTRIC POWER CALCULATIONS
• OVERVIEW
• PER PHASE
• SHORT CIRCUIT
• PER UNIT
• HARMONICS
3. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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WHY ELECTRIC POWER CALCULATIONS
• USED TO PREDICT
OUTCOMES
• SHORT CIRCUITS
• HARMONICS
• VOLTAGES
• LOADS
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THE BASICS
• ALL ELECTRICAL THEORY RELATES
BACK TO THE BASICS
• Volts, Amps, Ohms
II
Z
V
V = I * Z
Z = V / I
I = V / Z
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SHORT CIRCUIT REQUIREMENTS
NEC 110-9 AND 110-10
• Articles 110-9 and 110-10
• Equipment shall have adequate
interrupting 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 20
ohms 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
16. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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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
CALCULATED
SHORT CIRCUIT
AMPS MUST BE
LESS THAN AIC
RATING
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CALCULATION TRICK
TRANSFORMER IMPEDANCE
Transformer
Variable voltage source
Short Circuit
A
Percent Impedance = Percent rated primary
voltage that causes rated base/ambient full
load current to flow in the secondary of a
short circuited transformer.
i.e.. 5.75 percent primary voltage causes
full 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 TWO
SIDES 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 at
several voltage levels via transformers,
can become complicated due to the
transformer 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 a
summary 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 base
quantities 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 that
remains constant throughout the system is the base
kVA. The kVA base is not affected by voltage levels
or transformer turns ratios. The amount of kVA
entering the primary of a transformer is the same as
the kVA leaving the secondary of the transformer
(neglecting transformer losses).
Any arbitrary number may be selected as the kVA
base, however, most electric utilities use a 100,000
kVA 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 generally
the 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.
31. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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BASE IMPEDANCE
(CALCULATED)
• Z base = kVbase
2 / 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.
33. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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BASIC SHORT CIRCUIT ANALYSIS
REVIEW
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
34. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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SHORT CIRCUIT ANALYSIS
PER 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
IV
Zsource
.002p.u.
Ztransformer
.03p.u.
Zconductor
.098p.u.
1.0 p.u.
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HARMONICS
DEFINITION
• Harmonic Order - integer multiple of the
fundamental frequency.
– Harmonic Order Frequency
• 1 ( fundamental) 60Hz
• 2 120Hz
• 3 180Hz
• 4 240Hz
• N N * Fundamental
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SOURCE OF HARMONICS
THE LOAD
• Solid State Motor
Drives
• Rectifiers
• UPS systems
• Computer power
supplies
• Fluorescent lighting
electronic ballast's
40. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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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
41. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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EFFECTS OF CAPACITORS
• Without capacitors, the
circuit is predominantly
inductive.
• When capacitors are
added, an L-C circuit
results.
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SYSTEM IMPEDANCE
INDUCTIVE ONLY
Example:
Power factor correction
requirements dictate that two 600
kvar capacitor banks be installed
at the substation bus. What does
the system impedance look like
before adding capacitors?
Short Circuit
Capacity = 30MVA
Harmonic
Source
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SYSTEM IMPEDANCE
INDUCTIVE ONLY
Frequency (harmonic order)
Z
(Ohms
)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Z = jwl, w = 2pi*f, f = frequency
44. PHILLIPS ENGINEERS+ CONSULTANTS, INC.
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SIMPLIFIED RESONANCE
CALCULATIONS
Example:
Power factor correction
requirements dictate that two 600
kvar capacitor banks be installed
at the substation bus. The utility
short circuit current is 30 MVA
(36,084 Amps @ 480V). What is
the resonance frequency when
the 600 kvar bank is on line and
what is the resonance frequency
when both 600 kvar banks are on
line.
Short Circuit
Capacity = 30MVA
2 x 600 kvar
Capacitor BankHarmonic Source
5TH, 7TH, 11TH, 13TH
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SIMPLIFIED RESONANCE
CALCULATIONS
MVAsc = short circuit MVA at the capacitor
bank location.
Mvarcap = Mvar rating of the capacitor bank
Source
Impedance
Power
Factor
Capacitor
Harmonic
Source
XL XC
hr = MVAsc = Xc
Mvarcap Xsc
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RESULTS
• PRODUCES SEVERE DISTORTION
• DESIGN CAPACITOR AS HARMONIC
FILTER
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IEEE-519
• Sets limits for
voltage 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 309
CANTON, OHIO 44718
Tel: 330.491.0261
Fax: 330.491.0265
jphillips@phillipsengineers.com
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