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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