Power System Representation and Equations

A one-line diagram of a simple power system

Oil or liquid circuit breaker Rotating machine Two-winding power transformer Wye connection, neutral ground

Per-Phase, Per Unit System

( )base,3

2baseL,

base

base

baseL,base

baseL,

base,3base

3

3

Φ

Φ

=

=

=

SV

Z

I

VZ

V

SI

quantity of valuebase valueactualunitper in Quantity =

As we have seen in Chapter 2, many transformers and machines have their internal impedances specified as per unit resistances and reactances, using the voltage and apparent power ratings of the device itself as the base quantity. If these impedances are expressed in per unit to a base other than the one selected as a base for a power system, we must convert the impedances to per unit on the new base. This conversion could be done by using the original base impedance to convert the impedances back into ohms,

Bus 1 Bus 2

Load B

1 2

Load A

and then using the power system base impedance to convert the value on ohms to per unit on the new base. Alternatively, we may combine the two steps into a single equation:

−=−

given

new2

new

givengivennew V

Vunit perunit Per

SSZZ

Example A simple power system consisting of one synchronous generator and one synchronous motor connected by two transformers and a transmission line is shown in the following Figure. Develop a per-phase, per unit equivalent system for this power system using a base apparent power of 100 MVA and a base line voltage at generator G1 of 13.8 kV.

Solution:

kV 8.13base,1 =V Region 1

kV 110kV 8.13kV 110

base,1base,2 =

=VV Region 2

kV 2.13kV 120kV 4.14

base,2base,3 =

=VV Region 3

The corresponding base impedances in each region are:

( ) ( )Ω=== 904.1

MVA 100kV 8.13 2

base3φφ

2baseL,

base,1 SV

Z Region 1

( ) ( )Ω=== 121

MVA 100kV 110 2

base3φφ

2baseL,

base,2 SV

Z Region 2

1 2

Load A

G1 rating: 100 MVA 13.8 kV

R = 0.1 pu XS=0.9 pu

T1 ratings: 100 MVA

13.8/110 kV R = 0.01 pu X=0.05 pu

L1 R=15Ω X=75Ω

T2 ratings: 50 MVA

120/14.4 kV R = 0.01 pu

X=0.05 pu

M rating 50 MVA 13.8 kV

R = 0.1 pu X2=1.1 pu

Region 1 Region 2 Region 3

( ) ( )Ω=== 743.1

MVA 100kV 2.13 2

base3φφ

2baseL,

base,3 SV

Z Region 3

unitper 05.0

unitper 01.0

unitper 9.0

unitper 1.0

puT1,

puT1,

puG1,

puG1,

=

=

=

=

X

R

X

R

unitper 620.0 121 75

unitper 124.0 121 15

system line,

system line,

=

ΩΩ

=

=

ΩΩ

=

X

R

−=−

given

new2

new

givengivennew V

Vunit perunit Per

SSZZ

( )

( ) unitper 119.0MA 50MA 100

kV 2.13kV 4.14 05.0

unitper 238.0MA 50MA 100

kV 2.13kV 4.14 01.0

2

puT2,

2

puT2,

=

=

=

=

X

R

−=−

given

new2

new

givengivennew V

Vunit perunit Per

SSZZ

( )

( ) unitper 405.2MA 50MA 100

kV 2.13kV 8.13 05.0

unitper 219.0MA 50MA 100

kV 2.13kV 8.13 01.0

2

puM2,

2

puM2,

=

=

=

=

X

R

Per-phase, per unit equivalent circuit of the simple power system.

Writing Node Equations for Equivalent Circuit

Once the per-phase, per unit equivalent circuit of a power system is created, it may be used to find the voltages, currents, and powers present at various points in a power system. The most common technique used to solve such circuits is nodal analysis. In nodal analysis, we use Kirchhoff’s current law equations to determine the voltages at each node (each bus) in the power system, and then using the resulting voltages to calculate the currents and power flows at various points in the power system, and then use the resulting voltages to calculate the currents and power flows at various points in the system.Consider the following simple three-phase power system containing three busses connected by three transmission lines. The system includes a generator connected to bus 1, a load connected to bus 2, and a motor connected to bus 3.

+ -

+ - G1 M2

0.01 + j0.05 pu

0.1 + j0.9 pu

0.124 + j0.62 pu 0.023 + j0.119 pu

0.219 + j2.405 pu

A simple three-phase power system

G1 M3

Line 1

Line 2 Line 3

T1 T2

T3

T4 T5

T6

Load 2

Sum of currents out of a node = sum of currents into the node Apply KCL on node 1

( ) ( ) 113121 IV VV VV =+−+− dba YYY Apply KCL on node 2

( ) ( ) 223212 IV VV VV =+−+− eca YYY Apply KCL on node 3

( ) ( ) 332313 IV VV VV =+−+− fcb YYY Rearrange these equations to collect the terms in each voltage

( )( )

( ) 3321

2321

1321

IV V_V_IV_V V

IVVV

=+++=+++−=−−++

fcbcb

cecaa

badba

YYYYYYYYYY

YYYYY

The above equation can be expressed in matrix form

=

++−++−

++

3

2

1

3

2

1

III

VVV

- -

fcbc a

cecaa

badba

YY Y -YY-YYYYYY YYYY

This equation is of the form IVYbus =

Where Ybus is the self admittance of a system.

Ya Yc

Yb

Yd

I1 V1

1

Ye

I2

V3 V2

2

3

n

I3

Yf

There are many ways for solving systems of simultaneous linear equations, such as substitution, Gaussian elimination, LU factorization, and so forth! For us, MATLAB has very efficient equation solvers built directly into it! If a system of n simultaneous linear equations in n unknowns can be expressed in the form:

bAx = Where A is n × n matrix, and b is a n-element column vector. It may be expressed as

bAx 1−= Where A-1 is the inverse of a matrix!