Power System-Per Unit Analysis

8/13/2019 Power System-Per Unit Analysis

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Power Systems I

Power System Analysis

Fundamentals of Power Systems (EEL 3216)

basic models of power apparatus,transformers, synchronous machines, transmission linessimple systems

one feeder radial to single load

What more is there?large interconnected systems

multiple loads; multiple generators

why have large interconnected systems?reliability; economics

analysis of the large systemflow of power and currents; control and stability of the systemproper handling of fault conditions; economic operation


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Power Systems I

Modern Power Systems

Power Producergeneration station

prime mover & generatorstep-up transformer

Transmission CompanyHV transmission linesswitching stations

circuit breakerstransformers

Distribution Utilitydistribution substations

step-down transformers

MV distribution feedersdistribution transformers


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Power Systems I

Network Layout

HV NetworksLarge quantities of powershipped over great distancesSharing of resources

Improved reliabilityEconomics of large scale

MV NetworksLocal distribution of powerNumerous systems

Economics of simplicity

Autonomous operationLoads

Industrial & CommercialResidential


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Power Systems I

System Control

Network ProtectionSwitchgear

instrumentation transformerscircuit breakersdisconnect switchesfuses

lightning arrestorsprotective relays

Energy ManagementSystems

Energy Control Centercomputer controlSCADA - Supervisory ControlAnd Data Acquisition


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Power Systems I

Computer Analysis

Practical power systemsmust be safereliableeconomical

System Analysisfor system planningfor system operationsrequires component modelingtypes of analysis

transmission line performancepower flow analysiseconomic generationschedulingfault and stability studies


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FAMU-FSU College of Engineering

Chapter 2

AC Power


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Power Systems I

Single-Phase Power Consumption

( )( )

( ) ( )

( ) ( )( ) ( ){ }

( ){ } ( )vvmmiv

ivivmm

ivmm

im

vm

t I V t I V t p

I I V V

t I V t p B A B A B A

t t I V t it vt p

t I t i

t V t v

++++=

===

+++=

++=++==

+=+=

2sinsin2cos1cos)(

22

2coscos)(coscoscoscos

coscos)()()(

cos)(

cos)(

21

21

21

energy flow intothe circuit

energy borrowed andreturned by the circuit

i(t )

v(t )


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Power Systems I

Average Active (Real) Power

( ){ } ( )

( ){ } ( )

( ) ( )

I V P

pf

I V P

dt t dt t

dt t t I V

dt t pP

t I V t I V t p

vv

vv

==

=

==

++++=

=

++++=

cos

cos

0sin0cos

sin2sincos2cos1

)(21

sin2sincos2cos1)(

2

0

2

0

2

0

2

0


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Power Systems I

Apparent Power

( ){ } ( )( ){ } ( ){ }

( ) ( )vv X vv R

vv

t S t I V t p

t Pt I V t pt I V t I V t p

I V S

I V P

+=+=

++=++=

++++=

=

=

2sinsinsin2sin)(

2cos1cos2cos1)(sin2sincos2cos1)(

cos


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Power Systems I

Reactive Power

for a pure resistorthe impedance angle is zero, power factor is unityapparent power and real power are equal

for a purely inductive circuit

the current lags the voltage by 90, average power is zerono transformation of energy

for a purely capacitive circuitthe current leads the voltage by 90, average power is zero

( ) ( )

( )v X

vv X

t Qt p I V S Q

t S t I V t p

+==

+=+=

2sin)(sinsin

2sinsin2sinsin)(


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Power Systems I

AC Power

Example

the supply voltage is given by v(t ) = 480 cos

t the load is inductive with impedance Z = 1.20 60determine the expression for the instantaneous current i(t ) andinstantaneous power p(t )

plot v(t ), i(t ), p(t ), pR(t ), pX(t ) over an interval of 0 to 2


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Power Systems I

Complex Power

Real Power, P

RMS based - thermally equivalent to DC powerReactive Power, QOscillating power into and out of the load because of its reactiveelement (L or C).

Positive value for inductive load (lagging pf)Complex Power, S

( )

22

*

sincos

QPS

jQP I V j I V S

S I V I V I V iv

+=

+=+=

===


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Power Systems I

Complex Power

P

Q

S

V

I v

i

Q

S

P

V I

v

i

Lagging Power Factor

Leading Power Factor


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Power Systems I

The Complex Power Balance

From the conservation of energy

Real power supplied by the source is equal to the sum of the realpowers absorbed by the load and the real losses in the systemReactive power must also be balanced

The balance is between the sum of leading and the sum of laggingreactive power producing elements

The total complex power delivered to the loads in parallel is thesum of the complex powers delivered to each

=

+=

=

lossesloadsgen

ind laggingcapsleading

lossesloadsgen

S S S

QQQQ

PPP

0

0

0


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Power Systems I

Complex Power

Example

in the circuit below, find the power absorbed by each load andthe total complex powerfind the capacitance of the capacitor to be connected across theloads to improve the overall power factor to 0.9 lagging

I

V

1200 V

I 2 I 3

Z 1=60+j0 Z 2=6+j12 Z 3=30-j30

I 1


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Power Systems I

Complex Power Flow

Consider the following circuit

For the assumed directionof current

The complex power

V 2

Z = R+j X =|Z|

V 1 I 12

( ) ( )

=

=

==

22

112211

12

222111

Z V

Z V

Z V V I

V V V V

( ) ( )

( )21212

1

221111*12112

+=

==

Z

V V

Z

V

Z

V

Z

V V I V S


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Power Systems I

Complex Power FlowThe real and reactive power at the sending end

Transmission lines have small resistance compared to thereactance. Often, it is assumed R = 0 ( Z = X 90)

( )

( )21212

112

2121

2

112

sinsin

coscos

+=

+=

Z

V V

Z

V Q

Z V V

Z V P

( ) ( )[ ]2121112212112 cossin == V V X V Q X V V P


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Power Systems I

Complex Power Flow

For a typical power system with small R / X ratio, thefollow observations are made

Small changes in 1 or 2 will have significant effect on the realpower flowSmall changes in voltage magnitude will not have appreciableeffect on the real power flowAssuming no resistance, the theoreticalmaximum power (static transmissioncapacity) occurs when the angulardifference, , is 90 and is given by:

For maintaining stability, the power system operates with smallload angle

The reactive power flow is determined by the magnitudedifference of the terminal voltages

X

V V P 21max =


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Power Systems I

Three-Phase Power

Balanced three-phase powerAssumes balanced loadsAssumes voltage and currents with phases that have 120 separation

==

==

==

L LL p p

L LL p p

L LL p p

I V I V S

I V I V Q

I V I V P

33

sin3sin3

cos3cos3

3

3

3


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FAMU-FSU College of Engineering

Chapter 3

Power Apparatus Modeling


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Power Systems I

System ModelingSystems are represented on a per-phase basis

A single-phase representation is used for a balanced systemthe system is modeled as one phase of a wye-connected network

Symmetrical components are used for unbalanced systemsunbalance systems may be caused by: generation, networkcomponents, loads, or unusual operating conditions such as faults

The per-unit system of measurements is usedReview of basic network component models

generators

transformersloadstransmission lines


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Power Systems I

Generator ModelsGenerator may be modeled in three different ways

Power Injection Model - the real, P, and reactive, Q, power of thegenerator is specified at the node that the generator is connected

either the voltage or injected current is specified at the connectednode, allowing the other quantity to be determined

Thevenin Model - induced AC voltage, E , behind the synchronousreactance, X d

Norton Model - injected AC current, I G, in parallel with thesynchronous reactance

E

X d

X d

I G

Node

Node


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Power Systems I

Transformer ModelEquivalent circuit of a two winding transformer

X m Rc

X 2 R2 X 1 R1

V 1 V 2 E 2

N 1 : N 2

E 1


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Power Systems I

Transformer ModelApproximate circuit referred to the primary

X EQ1 R EQ1

X m RcV 1 22

12 V N

N V =


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Power Systems I

Load ModelsModels are selected based on both the type of analysisand the load characteristicsConstant impedance, Z load

Load is made up of R, L, and C elements connected to a networknode and the ground (or neutral point of the system)

Constant current, I load The load has a constant current magnitude I , and a constantpower factor, independent of the nodal voltageAlso considered as a current injection into the network

Constant power, S load The load has a constant real, P , and reactive, Q , powercomponent independent of nodal voltage or current injectionAlso considered as a negative power injection into the network


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Power Systems I

Per Unit SystemAlmost all power system analyses are performed in per-units

Per unit system for power systemsBased on a per-phase, wye-connect, three-phase system3-phase power base, S 3

common power base is 100 MVA

Line-to-line voltage base, V LL

voltage base is usually selectedfrom the equipment rated voltage

Phase current base, I LPhase impedance base, Z

base LL

base

base L V

S

I

=

33

( ) ( )base

base LN

base

base LLbase S

V S

V Z

== 1

2

3

2

100%

).().()( )( x

unit engr xunit engr x pu x

base

actualgengineerinunit per ==


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Power Systems I

Per Unit SystemEquipment impedances are frequently given in per unitsor percentages of the impedance base

The impedance base for equipment is derived from the ratedpower and the rated voltageWhen modeling equipment in a system, the per unit impedancemust be converted so that the equipment and the system are ona common base

It is normal for the voltage bases to be the same:

( ) ( )

( ) ( ) 22

2

22

==

====

newbase

old base

old base

newbaseold

puold

puold base

old base

newbase

newbasenew

pu

newbase

newbase

newbase

new puold

base

old base

old base

old pu

V V

S S Z Z

S V

V S Z

V

S Z

Z Z

Z V

S Z

Z Z

Z

old

base

newbaseold

punew

pu S S

Z Z =


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Power Systems I

Per Unit SystemThe advantages of the per unit system for analysis

Gives a clear idea of relative magnitudes of various quantitiesThe per-unit impedance of equipment of the same general typebased upon their own ratings fall in a narrow range regardless ofthe rating of equipment.

Whereas their impedances in ohms vary greatly with the ratings.

The per-unit impedance, voltages, and currents of transformersare the same regardless of whether they are referred to theprimary or the secondary side.

Different voltage levels disappear across the entire system.

The system reduces to a system of simple impedancesThe circuit laws are valid in per-unit systems, and the power andvoltages equations are simplified since the factors of 3 and 3are eliminated in the per-unit system


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Power Systems I

Per Unit SystemExample

the one-line diagram of a three-phase power system is shownuse a common base of 100 MVA and 22 kV at the generator

draw an impedance diagram with all impedances marked in per-unitthe manufacturers data for each apparatus is given as follows

G: 90 MVA 22 kV 18%T1: 50 MVA 22/220 kV 10%L1: 48.4 ohmsT2: 40 MVA 220/11 kV 6%T3: 40 MVA 22/110 kV 6.4%

L2: 65.43 ohmsT4: 40 MVA 110/11 kV 8%M: 66.5 MVA 10.45 kV 18.5%Ld: 57 MVA 10.45 kV 0.6 pf lag

G

M

T2

T1

T4

T3

L1 L2

Ld

Power System-Per Unit Analysis

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