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]




ELECTRIC POWER CALCULATIONS

• OVERVIEW

• PER PHASE

• SHORT CIRCUIT

• PER UNIT

• HARMONICS




WHY ELECTRIC POWER CALCULATIONS

• USED TO PREDICTOUTCOMES

• SHORT CIRCUITS

• HARMONICS

• VOLTAGES

• LOADS




THE BASICS

• ALL ELECTRICAL THEORY RELATESBACK TO THE BASICS

• Volts, Amps, Ohms

II

ZV

V = I * Z

Z = V / I

I = V / Z




SERIES COMBINATIONS

Z1

Z2

Z




COMPLEX IMPEDANCE

R

X

Z

0

X = Z * Sin 0

R = X

X/R

Z = R2

+X2




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




AIC RATINGS

CIRCUIT BREAKER SHORT CIRCUIT

TYPE RATING

QOB 10,000

QOB-H 22,000

QOB-VH 42,000




SHORT CIRCUIT AMPS (SCA)

Source

Circuit Breaker

Source

3 Phase

Circuit Breaker

Line-Ground




PER PHASE ANALYSIS

~

~

~

~

A

B

C A

B

C

Ia

Ib

Ic

~

Three Phase Representation

Single Phase Representation




PER PHASE ANALYSIS EXAMPLE

~

~

~

~A

B

C A

B

C

Ia

Ib

Ic

~



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?




PER PHASE ANALYSIS

SHORT CIRCUITS

~

~

~

~

A

B

C A

B

C

Ia

Ib

Ic

~






DATA

SOURCE IMPEDANCE

• THEVENINEQUIVALANTIMPEDANCE




DATA

TRANSFORMER IMPEDANCE

• TRANSFORMER

IMPEDANCE




DATA

CONDUCTOR IMPEDANCE

• CONDUCTORIMPEDANCE




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




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




SCA < AIC

CALCULATEDSHORT CIRCUIT AMPS MUST BELESS THAN AICRATING




CALCULATION TRICKTRANSFORMER IMPEDANCE

• VOLTAGE

• %Z

• KVA




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




SHORT CIRCUIT CALCULATION

EXAMPLE:

1500 KVA TRANSFORMER

5.75% IMPEDANCE

480 VOLT SECONDARY

1500 KVA

5.75%

480 VOLTS

SCAMPS?X




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




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




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.




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%




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




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




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.




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.




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.




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.




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




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




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.




HARMONICSDEFINITION

• Harmonic Order - integer multiple of thefundamental frequency. – Harmonic Order Frequency

• 1 ( fundamental) 60Hz

• 2 120Hz

• 3 180Hz

• 4 240Hz

• N N * Fundamental




HARMONICS - 5TH

5th Harmonic (300 Hz)

Fundamental (60 Hz)




NON LINEAR LOAD




SOURCE OF HARMONICSTHE LOAD

• Solid State MotorDrives

• Rectifiers

• UPS systems

• Computer powersupplies

• Fluorescent lightingelectronic ballast's




FREQUENCY SPECTRUM

1 3 5 7 9 11Harmonic Order

M a g n i t u d e




HARMONIC RELATED PROBLEMS

• Blown Capacitors / Capacitors Fuses

• Transformer Overheating

• Neutral Overheating

• Motor / Generator Overheating

• Equipment Misoperation

• Circuit Breaker Misoperation

• Communication Interference




EFFECTS OF CAPACITORS

• Without capacitors, thecircuit is predominantly

inductive.• When capacitors are

added, an L-C circuitresults.




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




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




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





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





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




RESULTS

• PRODUCES SEVERE DISTORTION

• DESIGN CAPACITOR AS HARMONICFILTER




IEEE-519

• Sets limits forvoltage and current

distortion at PCC• PCC is the point of

common coupling




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

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