EEE 103 - Load Flow Analysis

4 Electrical and Electronics Engineering Institute University of the Philippines

RDDELMUNDO EEE 103 AY2010-11 S2

EEE 103 – Introduction to Power Systems

Power Flow Through Short Transmission Lines

Let us consider a short transmission line. The single-phase equivalent circuit is shown below:

• •

• • +

-

Vs = |Vs| ∠α VR = |VR| ∠0

+

-

Is IR

Z = (r + jxL )L = |Z| ∠θ

IS = IR = I VS = ZI + VR




We calculate for the current I and its conjugate I* : ( ) ( )

( )( ) ( )

( )

0

0

S R

S R

V VI

Z

V VI

Z

α

θ

α

θ∗ −

−

∠ − ∠=

∠

∠ − ∠=

∠

v

v

We calculate the single-phase complex power at the sending and receiving ends:

( )

( 0)S S

R R

S V IS V I

α ∗

∗

= ∠ ⋅

= ∠ ⋅

v vv v

The direction of power flow will be inherent in the direction of the current I, i.e., SS is the supplied power when positive, and SR is the load power when positive.




Looking at the sending-end complex power:

( )

( )

2

2

cos cos

sin sin

S S RS

S S RS

V V VPZ Z

V V VQZ Z

θ α θ

θ α θ

# $ ⋅# $= ⋅ − ⋅ +' ( ' () *) *

# $ ⋅# $= ⋅ − ⋅ +' ( ' () *) *

Getting the real and imaginary (reactive) components:

( ) ( )( )

( )2

( )0

( )

S S

S RS S

S S RS

S V IV V

S VZ

V V VSZ Z

α

αα

θ

θ α θ

∗

−

−

= ∠ ⋅

∠ − ∠= ∠ ⋅

∠

' ( ⋅' (= ∠ − ∠ +) * ) *+ ,+ ,

v v

v

v




If we assume the line reactance is much greater than the line resistance, i.e., xL >> rL, then we can neglect rL. This means θ = 90° and Z = X, which when we substitute in the previous equations yield:

( )

( )

2

2

2

cos90 cos 90

sin

sin 90 sin 90

cos

S S RS

S RS

S S RS

S S RS

V V VPX X

V VPX

V V VQX X

V V VQX X

α

α

α

α

" # ⋅" #= ⋅ ° − ⋅ + °& ' & '( )( )

⋅" #= ⋅& '( )

" # ⋅" #= ⋅ ° − ⋅ + °& ' & '( )( )

" # ⋅" #= − ⋅& ' & '

( )( )




Looking at the receiving-end complex power:

( )

( )

2

2

cos cos

sin sin

S R RR

S R RR

V V VPZ Z

V V VQZ Z

α θ θ

α θ θ

−

−

$ %⋅$ %= ⋅ + − ⋅' (' () * ) *

$ %⋅$ %= ⋅ + − ⋅' (' () * ) *

Getting the real and imaginary (reactive) components:

( ) ( )( )

( )2

( 0)0

( 0)

R R

S RR R

S R RR

S V IV V

S VZ

V V VSZ Z

α

θ

α θ θ

∗

−

−

−

= ∠ ⋅

∠ − ∠= ∠ ⋅

∠

' (⋅' (= ∠ + − ∠) *) *+ , + ,

v v

v

v




If we assume the line reactance is much greater than the line resistance, i.e., xL >> rL, then we can neglect rL. This means θ = 90° and Z = X, which when we substitute in the previous equations yield:

( )

( )

2

2

2

cos 90 cos90

sin

sin 90 sin90

cos

S R RR

S RR

S R RR

S R RR

V V VPX X

V VPX

V V VQX Z

V V VQX X

α

α

α

α

−

−

# $⋅# $= ⋅ + ° − ⋅ °& '& '( ) ( )

⋅# $= ⋅& '( )

# $⋅# $= ⋅ + ° − ⋅ °& '& '( ) ( )

# $⋅# $= ⋅ − & '& '( ) ( )




Observations 1.  Since we assumed that the transmission line consists

of pure reactance, real power is not dissipated in the line and PS = PR.

2.  If the transmission line resistance is non-negligible, we will have to use the “unsimplified” equations.

3.  Maximum real power transfer occurs when α = 90°. 4.  Real power transfer is more sensitive to the difference

between phase angles of the supply voltage and the load voltage.

5.  Reactive power transfer is more sensitive to the difference between magnitudes of the supply voltage and the load voltage.




The Load Flow Problem

How do you determine the voltage, current, power, and power factor at various points in a power system?

Sending End

Receiving End

VS = ?

Load 2 MVA, 3Ph

85%PF

VR = 13.2 kVLL

Line 1.1034 + j2.0856 ohms/phase

ISR = ?

VOLTAGE DROP = VS - VR

Solve for:· 1)  ISR = (SR/VR )*

2)  VD = ISRZL

3)  VS = VR + VD

4)  SS = VS·(ISR)*

Basic Electrical Engineering Solution




The Load Flow Problem Sending

End Receiving

End

VS = ?

Load 2 MVA, 3Ph

85%PF

VR = 13.2 kVLL

Line 1.1034 + j2.0856 ohms/phase

ISR = ?

Solve for: 1)  ISR = (SR/VR )*

2)  VD = ISRZL

3)  VS = VR + VD

4)  SS = VS·(ISR)*

( )( )( )

11

R

SR

S

S ( 2,000,000 / 3 ) cos (0.85 )

666,666.67 31.79 VA

V (13,200 / 3 ) 0 7621.02 0 V

666,666.67 31.79I 87.48 31.79 A

7621.02 0VD 87.48 31.79 1.1034 j2.0856 178.15 j104.23 V

V 7621.02 j0 178

φ−

∗

= ∠

= ∠

= ∠ = ∠

∠% &= = ∠−' (∠) *

= ∠− + = +

= + + ( )

S

.15 j104.23 7,799.87 0.77 V

V 7,799.87 0.77 /1000* 3 13.51 k V

+ = ∠

= ∠ =





Sending End

Receiving End

VS = 13.2 kVLL

Load 2 MVA, 3Ph

85%PF

VR = ?

Load Flow From the Real World Line

1.1034 + j2.0856 ohms/phase

ISR = ?

How do you solve for: 1)   ISR = ?

2)   VD = ?

3)   VR = ?

4)   SS = ?




The Load Flow Problem Load Flow of Distribution System

How do you solve for the Voltages, Currents, Power and Losses?

Bus1

Utility Grid

Bus2 Bus3

Bus4 V1 = 67 kV

Lumped Load A 2 MVA 85%PF

Lumped Load B 1 MVA 85%PF

V2 = ?

V4 = ?

V3 = ? I23 , Loss23 = ?

I24 , Loss24 = ?

I12 , Loss12 = ?

P1 , Q1 = ? P2 , Q2 = ?

P3 , Q3 = ?

P4 , Q4 = ?





Line 1

Line 3 Line 2

1 2

3

G G

How do you solve for the Voltages, Currents and Power of a LOOP power system?

Load Flow of Transmission and Subtransmission System




The Load Flow Problem !  How do you determine the voltage, current, and power

flows, at various points in the power system, under existing conditions of normal operations?

!  How do you determine the adequacy of the power system in meeting the demand during contingencies?

!  How about if there are contemplated changes in the power system? How will you determine in advance the effects of: !  Growth or Addition of loads !  Addition or Decommissioning of generating plants !  Expansion of the transmission and distribution systems

before the proposed changes are implemented?





Load Flow Analysis simulates (i.e., mathematically

determines) the performance of an electric power system under a given set of conditions.

Load Flow (also called Power Flow) takes a snapshot of the electric power system at a given point in time.

ANSWER: THE LOAD FLOW STUDY!




POWER SYSTEM MODELS FOR LOAD FLOW ANALYSIS




Network Models

!  The static components of the power system are modeled by the bus admittance matrix, [Ybus].

!!!!!!!!

"

#

$$$$$$$$

%

&

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

[YBUS] =

The number of buses (excluding the neutral bus) determines the dimension of the bus admittance, [Ybus].




Generator Models

1.  Voltage-controlled generating units to supply a scheduled active power P at a specified voltage magnitude V. The generators are equipped with voltage regulators to adjust the field excitation so that the units will supply or absorb a particular reactive power Q in order to maintain the voltage.

2.  Swing generating units to maintain the frequency at 60Hz in addition to the specified voltage. The generating unit is equipped with frequency following controller (quick-responding speed governor) and is assigned as the Swing Generator.




Bus Types

!  The power system is interconnected through the busses. The busses must therefore be identified in the load flow model. ! Generators, shunt admittances, and loads are

connected from their corresponding bus to the neutral bus.

! Transmission lines, transformers, and series impedances are connected from bus to bus.




Bus Types

!  To completely describe a particular bus, four quantities must be specified: ! Bus Voltage Magnitude, |VP| ! Bus Voltage Phase Angle, δP

! Bus Injected Active Power, PP

! Bus Injected Reactive Power, QP




The difference between the total load demand plus losses (both P and Q) and the scheduled generations is supplied by the swing bus. The voltage magnitude and phase angle are specified for the Swing Bus, also called the Slack Bus.

Swing Bus Specify: V, δ

Unknown: P, Q G

P,Q

δV∠+

-

Swing Bus Swing Bus or Slack Bus




Generator Bus (Voltage-Controlled) Bus or PV Bus The total real power Pp injected into the system through the bus is specified together with the magnitude of the voltage Vp at the bus. The bus voltage magnitude is maintained through reactive power injection.

G

P,Q

δV∠+

- Generator Bus

Specify: P, V

Unknown: Q, δ

Generator Bus




The total injected power Pp and the reactive power Qp at Bus P are specified and are assumed constant, independent of the small variations in bus voltage.

Load Bus or PQ Bus

P,Q

+

-

Load Bus Specify: P, Q

Unknown: V, δ δV∠

Load Bus




BBuuss TTyyppee

KKnnoowwnn QQuuaannttiittiieess

UUnnkknnoowwnn QQuuaannttiittiieess

SSwwiinngg

VVpp,, δδpp

PPpp,, QQpp

GGeenneerraattoorr

PPpp,, VVpp

QQpp,, δδpp

LLooaadd

PPpp,, QQpp

VVpp,, δδpp

Summary of Bus Types




Line 1

Line 3 Line 2

1 2

3

G G

Voltage Generation Load Bus No. V (p.u.) δ P Q P Q

Remarks

1 1.0 0.0 * * 0 0 Swing Bus 2 1.0 * 0.20 * 0 0 Gen Bus 3 * * 0 0 0.60 0.25 Load Bus

PBUS = PGEN – PLOAD

Bus Types

QBUS = QGEN – QLOAD

Injected Powers:




SOLUTIONS TO SIMULTANEOUS ALGEBRAIC EQUATIONS




Numerical Methods

!  Direct Methods ! Cramer’s Rule ! Matrix Inversion ! Gaussian Elimination Method ! Gauss-Jordan Reduction Method

!  Iterative Methods ! Gauss Iterative Method ! Gauss-Seidel Iterative Method ! Newton-Raphson Method




Iterative Methods

An iterative method (root word: iterate) is a repetitive process for obtaining the solution of an equation or a system of equations.

The solutions start from arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined.

Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some acceptable pre-specified precision index (the convergence index) for all variables.




Given the system of algebraic equations,

In the above equation, the x’s are unknown.

3nnn232131

2n2n222121

1n1n212111

yxaxaxa

yxaxa xay xaxaxa

=+++

↓↓↓↓

=+++

=+++

Gauss Iterative Method




In general, the jth equation may be written

as

)xab(a1

x i

n

1i jij

jj

jji

∑≠=

−=

n 2, 1, j …=

equation “a”





In general, the Gauss iterative estimates are:

where k is the iteration count


k

n

11

1nk

3

11

13k

2

11

12

11

11k

1 xaa

...xaa

xaa

ayx −−−−=+

xaa

...xaa

xaa

ay

x kn

22

2nk3

22

23k1

22

21

22

21k2 −−−−=+

k

1-n

nn

1-nn,k

2

nn

n2k

1

nn

n1

nn

n1k

n xaa

...xaa

xaa

ay

x −−−−=+




From an initial estimate of the unknowns (x10, x2

0,…xn

0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,

)xab(a1

x kn

1i jij

jj

1kj i

ji∑

≠=

+ −=

n 2, 1, j …=

equation “b”





A convergence check is conducted after each iteration. The latest values are compared with their values respectively. k

j1k

jk xxx −= +Δ

n 2, 1, j …=

equation “c”

The iteration process is terminated when:

t)(convergen |x |max kj εΔ <


k = maximum no. of iterations (non-convergent)




Example:

Assume a convergence index of ε = 0.001 and use the following initial estimates:

5 3x x x6 x 4x x

4 x x 4x

3 21

32 1

321

=++

=++

=+−

0.5 x x x b)

0.0 x x x a)0

30

20

1

0 3

0 2

0 1

===

===





Solution: a) The system of equation must be expressed

in standard form.

)x- x 4 (41x k

3k2

1k1 +=+


) x -x - 5(31x k

2k1

1k3 =+

) x - x - 6 ( 41x k

3k1

1k2 =+




Iteration 1 (k = 0):

1.6667 x max

6667.106667.1x5.105.1x

101x

1.6667 ) 0 -0 - 5(31

x

1.5 ) 0 - 0 - 6 ( 41

x

1.0 ) 0 - 0 4 (41x

03

03

02

01

1 3

1 2

1 1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ





0.83334 x max

83334.06667.1833333.0x66667.05.1833333.0x

041667.01958325.0x

0.833333 ) 1.5 -1.0 - 5(31

x

0.833333 ) 1.6667 - 1.0 - 6 ( 41

x

0.958333 ) 1.6667 - 1.5 4 (41x

13

13

12

11

2 3

2 2

2 1

=

−=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ







0.23617 x max

23617.08333.00695.1x21877.0833325.00521.1x

041667.0958325.01x

1.0695 ) 0.8333 -0.9583 - 5(31

x

1.0521 ) 0.8333 - 0.9583 - 6 ( 41

x

1.0 ) 0.8333 - 0.8333 4 (41x

23

23

22

21

3 3

3 2

3 1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ





0.0869 x max

0869.00695.19826.0x0695.00521.19826.0x

0044.019956.0x

0.9826 ) 1.0521 -1.0 - 5(31

x

0.9826 ) 1.0695 - 1.0 - 6 ( 41

x

0.9956 ) 1.0695 - 1.0521 4 (41x

33

33

32

31

4 3

4 2

4 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ







0.0247 x max

0247.09826.00073.1x0228.09826.00054.1x

0044.09956.01x

1.0073 0.9826) -0.9956 - 5(31

x

1.0054 ) 0.9826 - 0.9956 - 6 ( 41x

1.0 ) 0.9826 - 0.9826 4 (41

x

43

43

42

41

5 3

5 2

5 1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ





0.0091 x max

0091.00073.19982.ox0072.00054.19982.0x

0005.019995.0x

0.9982 ) 1.0054 -1.0 - 5(31

x

0.9982 ) 1.0071 - 1.0 - 6 ( 41

x

0.9995 ) 1.0073 - 1.0054 4 (41x

53

53

52

51

6 3

6 2

6 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ







0.0026 xmax

0026.09982.00008.1x0024.09982.00006.1x

0005.09995.01x

1.0008 0.9982) -0.9995 - 5(31

x

1.0006 ) 0.9982 - 0.9995 - 6 ( 41

x

1.0 ) 0.9982 - 0.9982 4 (41x

63

63

62

61

7 3

7 2

7 1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ





0.0010 x max

0010.00008.19998.0x0008.00006.19998.0x

0005.019995.0x

0.9998 ) 1.0008 -1.0 - 5(31

x

0.9998 ) 1.0008 - 1.0 - 6 ( 41x

0.9995 ) 1.0008 - 1.0006 4 (41

x

73

73

72

71

8 3

8 2

8 1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ






The Gauss iterative method has converged at iteration 8. The method yields the following solution:

0.9998 x0.9998 x

0.9995 x

3

2

1

=

=

=





GAUSS-SEIDEL ITERATIVE METHOD

FOR A SYSTEM OF EQUATIONS




The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jth equation may be written as follows.

)xab(a1

x i

n

1i jij

jj

jji

∑≠=

−= n 2, 1, j …=

From an initial estimates (x10, x2

0,…xn0), an updated value is

computed for x1 using the above equation with j set to 1.This new value replaces x1

0 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration.

Gauss-Seidel Iterative Method




Within an iteration, the most recent computed values are used in computing for the remaining unknowns. In general, at iteration k,

i j

nk+1j j ji i1

jj

1x (b a x )

a iα

≠=

= −∑

n 2, 1, j …=

j i if 1k j i if k where

<+=

>=α

After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method.





An improvement to the Gauss Iterative Method Gauss-Seidel Iterative Method

xaa

...xaa

ay

x kn

11

1nk2

11

12

11

11k

1−−−=

+

xaa

...xaa

ay

x kn

22

2n1k1

22

21

22

21k

2−−−= ++

1kn

ii

in1k1i

ii

1ii,1k1-i

ii

1-ii,1ki

ii

ij

ii

i xaa

xaa

xaa

...xaa

ay

x1k

i

+++

+++ −−−−−=+

xa

a...x

aa

ay

x 1k1-n

nn

1-nn,1k1

nn

n1

nn

n1k

n

++ −−−=+




Example: Solve the system of equations using the Gauss-Seidel method. Used a convergence index of ε = 0.001

5 3x x x6 x 4x x4 x x 4x

3 21

32 1

321

=++

=++

=+−

0.5 x x x 0 3

0 2

0 1 ===





Solution: a) The system of equation must be expressed

in standard form.

) x -x - 5(31x

) x - x - 6 ( 41x

)x- x 4 (41x

1k2

1k1

1k3

k3

1k1

1k2

k3

k2

1k1

+++

++

+

=

=

+=





Iteration 1 (k =0):

0.625 | x |max

4583.050.09583.0x625.050.0125.1x

50.05.01x

0.9583 ) 1.125 -1.0 - 5(31x

1.125 ) 0.5 - 1.0 - 6 ( 41

x

1.0 ) 0.5 - 0.5 4 (41

x

02

03

02

01

1 3

1 2

1 1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ

0.5 x x x with 0 3

0 2

0 1 ===





0.125 | x |max

0323.09583.09861.0x125.0125.11x0417.010417.1x

0.9861 ) 1.0 -1.0417 - 5(31x

1.0 ) 0.9583 - 1.0417 - 6 ( 41

x

1.0417 ) 0.9583 - 1.125 4 (41

x

12

13

12

11

2 3

2 2

2 1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ







0.0119 | x |max

0119.09861.09980.0x0026.010026.1x

0382.00417.10035.1x

0.9980 ) 1.0026 -1.0035 - 5(31x

1.0026 ) 0.9891 - 1.0035 - 6 ( 41

x

1.0035 ) 0.9861 - 1.0 4 (41x

23

23

22

21

3 3

3 2

3 1

=

=−=

=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ





0.0024 | x |max

0015.09980.09995.0x0024.00026.10002.1x

0023.00035.10012.1x

0.9995 1.0002) -1.0012 -1.0 - 5(31x

1.0002 0.9980) - 1.0012 - 6 ( 41

x

1.0012 )0.9980 -1.0026 4 (41

x

32

33

32

31

4 3

4 2

4 1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ







εΔ

Δ

Δ

Δ

0.001 | x|max

0004.09995.09999.0x0001.00002.10001.1x001.00012.10002.1x

0.9999 1.0001) -1.0002 - 5(31x

1.0001 0.9995) - 1.0002 - 6 ( 41

x

1.0002 )0.9995 - 1.0002 4 (41

x

4

43

42

41

5 3

5 2

5 1

<=

=−=

−=−=

−=−=

==

==

=+=





The Gauss-Seidel Iterative Method has converged after only 5 iterations with the following solutions:

0.9999 x1.0001 x1.0002 x

3

2

1

=

=

=





The real and reactive power into any bus P is:

Pp + jQp = Vp Ip*

where Pp = real power injected into bus P

Qp = reactive power injected into bus P

Vp = phasor voltage of bus P

Ip = current injected into bus P

Pp - jQp = Vp* Ip

or

Gauss-Seidel Load Flow

(1)

Linear Formulation of Load Flow Equations




Equation (1) may be rewritten as:

Ip = Pp - jQp _________

Vp*

From the Bus Admittance Matrix equation, the current injected into the bus are:

I1 = Y11V1 + Y12V2 + Y13V3

I2 = Y21V1 + Y22V2 + Y23V3

I3 = Y31V1 + Y32V2 + Y33V3


(2)

(3) Ip = Yp1V1 + Yp2V2 + … + YppVp + … + YpnVn




Substituting (3) into (2)

_________ Vp

*

Pp - jQp = Yp1V1 + Yp2V2 + … + YppVp + … + YpnVn


_________ V1

*

P1 – jQ1 = Y11V1 + Y12V2 + Y13V3

_________ V2

*

P2 – jQ2 = Y21V1 + Y22V2 + Y23V3

_________ V3

*

P3 – jQ3 = Y31V1 + Y32V2 + Y33V3

(4)




Solving for Vp in (4)

_______ Y11V1 = P1 – jQ1

V1*

- (___ + Y12V2 + Y13V3)


!"

#$%

&−−

−= 313212*

1

11

11

1 VYVYVjQP

Y1V

_______ Y22V2 = P2 – jQ2

V2*

- (Y12V2 + ___ + Y13V3)

!"

#$%

&−−

−= 313121*

2

22

22

2 VYVYVjQP

Y1V





!"

#$%

&−−

−= 232131*

3

33

33

3 VYVYVjQP

Y1V

_______ Y33V3 = P3 – jQ3

V3*

- (Y13V1 + Y23V2 + ___)

(5) Vp =

1 ___ Ypp

_______ Vp

*

Pp - jQp Σ - n

q=1 q≠p

YpqVq




Generalizing the Gauss-Seidel Load Flow, the estimate for the voltage Vp at bus p at the kth iteration is:

Vpk+1

=

1 ___ Ypp

_______ (Vp

k)*

Pp - jQp Σ - n

q=1 q≠p

YpqVqα

where, α = k if p < q α = k + 1 if p > q

Gauss-Seidel Load Flow Gauss-Seidel Load Flow Solution

(6)




!  Gauss-Seidel Voltage Equations of the form shown in (6) are written for all buses except for the swing bus. The solution proceeds iteratively from an estimate of all bus voltages

!  For a Load Bus (Type 3) whose real power and reactive power are specified, the G-S voltage equation is used directly to compute the next estimate of the bus voltage.

!  For a Generator Bus (Type 2) where the voltage magnitude is specified, an estimate of Qp must be determined first. This estimate is then compared with the reactive power limits of the generator. If it falls within the limits, the specified voltage is maintained and the computed Qp is inputted, in the Gauss-Seidel equation. Otherwise, the reactive power is set to an appropriate limit (Qmin or Qmax) and the bus is treated as a load bus in the current iteration.





Numerical Example Shown in the figure is a 3-bus power system. The line and bus data pertinent to the system are also given. The reactive limits of generator 2 are -50 MVARS and 50 MVARS. Use base power of 100 MVA. Solve the load flow problem using Gauss-Seidel iterative method with a 0.005 convergence index.

Line 1

Line 3 Line 2

1 2

3

G G





Branch Data

Line No. Bus Code Impedance Zpq (p.u.)

1 1 - 2 0.08 + j0.24 2 1 - 3 0.02 + j0.06 3 2 - 3 0.06 + j0.18

Bus Data Voltage Generation Load Bus

No. V (p.u.) δ P Q P Q Remarks

1 1.0 0.0 * * 0 0 Swing Bus 2 1.0 * 0.20 * 0 0 Gen Bus 3 * * 0 0 0.60 0.25 Load Bus





The Bus Admittance Matrix is:


Y11 = 6.25 - j18.75

Y21 = -1.25 + j3.75

Y31 = -5 + j15

Y12 = -1.25 + j3.75

Y22 = 2.9167 - j8.75

Y32 = -1.6667 + j5 Y33 = 6.6667 - j20

Y13 = -5 + j15

Y23 = -1.6667 + j5

[YBUS] =




Specified Variables: V1 = 1.0 δ1 = 0.0

V2 = 1.0 P2 = 0.2

P3 = -0.6 Q3 = -0.25

Initial Estimates of Unknown Variables:

δ20 = 0.0

V30 = 1.0

δ30 = 0.0

Note the negative sign of P and Q of the Load at Bus 3





Gauss-Seidel Load Flow Equations

!  Bus 1: Swing Bus

!  Bus 2: PV Bus

for all iterations ( )11 1.0 0kV + = ∠

We must first estimate the Q2 for Bus 2 by:

( ) ( )2

( 1) ( ) ( )2 2 21 1 22 2 23 3

k k k k kP jQ V Y V Y V Y V∗ +− = ⋅ ⋅ + ⋅ + ⋅

( )( 1) ( 1) ( )2 22 21 1 23 3( )

22 2

1 kk k k

k

P jQV Y V Y VY V

+ +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

Then we substitute the Q2 value (and only the Q2 value) to:




Gauss-Seidel Load Flow Equations

!  Bus 3: PQ Bus

( )( 1) ( 1) ( 1)3 33 13 1 23 3( )

33 3

1k k k

k

P jQV Y V Y VY V

+ + +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

The Gauss-Seidel Equation for PQ Bus is straightforward:




Gauss-Seidel Load Flow: Example Iteration 1: k = 0 Bus 1: Bus 2:

for all iterations 1 1.0 0kV = ∠

( ) ( )2

( 1) ( ) ( )2 2 21 1 22 2 23 3


02 0Q =

( )( ) ( )( ) ( )( ) ( )

2

02

1.25 3.75 1 0

1 0 2.9167 8.75 1 0 0 0

1.6667 5 1 0

j

P jQ j j

j

∗

− + ⋅ ∠% &' (

− = ∠ ⋅ + − ⋅ ∠ = −' (' (+ − + ⋅ ∠) *

02V

21Y

22Y

23Y

1kV02V

03V discard

02Q

(This value is within limits.)




Gauss-Seidel Load Flow: Example Bus 2: (iteration 1, continued)

1 12 21.0071 1.1705 1.0 1.1705aV V= ∠ → = ∠o o

( ) ( )( ) ( )

( ) ( )

12

0.2 0 1.25 3.75 1 011 0

2.9167 8.751.6667 5 1 0

j jV

jj

∗

" #$ %−' (− − + ⋅ ∠+ ,+ ,= ⋅ ∠' (- .− ' (

− − + ⋅ ∠' (/ 0

( )( 1) ( 1) ( )2 22 21 1 23 3( )

22 2

1 kk k k

k

P jQV Y V Y VY V

+ +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

22Y02V

2P 02Q 21Y 1

kV

23Y03V




Gauss-Seidel Load Flow: Example Bus 3: (iteration 1)

13 0.9799 1.0609V = ∠− o

( )( ) ( )

( ) ( )

13

0.60 0.25 5 15 1 01 1 0(6.6667 20)

1.6667 5 1.0 1.1705

j jV

jj

∗

" #$ %− +' (− − + ⋅ ∠+ ,+ ,∠' (= ⋅ - .− ' (

− − + ⋅ ∠' (/ 0o

( )( 1) ( 1) ( 1)3 33 31 1 32 2( )

33 3

1k k k

k

P jQV Y V Y VY V

+ + +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

33Y03V

3P 3Q 31Y 1kV

32Y12aV




Gauss-Seidel Load Flow: Example Check for convergence: (iteration 1) Action: Continue iterating.

( ) ( )1 1 02 2 2 1.0071 1.1705 1.0 0

0.0217 0.005

V V VΔ = − = ∠ − ∠

= >

o o

( ) ( )1 1 03 3 3 0.9799 1.0609 1.0 0

0.0272 0.005

V V VΔ = − = ∠− − ∠

= >

o o

12 1.0 1.1705aV = ∠ o 1

3 0.9799 1.0609V = ∠− o




Gauss-Seidel Load Flow: Example Iteration 2: k = 1

Bus 2: ( ) ( )2

( 1) ( ) ( )2 2 21 1 22 2 23 3


12 0.0160Q =

( )

( ) ( )( ) ( )( ) ( )

2

2

12

12

1.25 3.75 1 0

1.0 1.1705 2.9167 8.75 1.0 1.1705

1.6667 5 0.9799 1.0609

0.3024 0.0160

j

P jQ j

j

P jQ j

! "− + ⋅ ∠& '& '− = ∠− ⋅ + − ⋅ ∠& '& '+ − + ⋅ ∠−( )

− = −

o

o

o

1 *2aV

21Y

22Y

23Y

1kV12aV

13V


discard





2 22 20.9965 0.5648 1.0 0.5648aV V= ∠ → = ∠o o

( )( ) ( )

( ) ( )22

0.2 0.0160 1.25 3.75 1 01 1.0 1.17052.9167 8.75

1.6667 5 0.9799 1.0610

j jV

jj

! − #$ % − − + ⋅ ∠( )* +∠−, -= ⋅ * +− * +− − + ⋅ ∠−. /

oo

o

( )( 1) ( 1) ( )2 22 21 1 23 3( )

22 2

1 kk k k

k

P jQV Y V Y VY V

+ +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

22Y1 *2aV

2P 12Q 21Y 1

kV

23Y13V




Gauss-Seidel Load Flow: Example Bus 3: (iteration 2)

23 0.9791 1.2218V = ∠− o

( ) ( )

( ) ( )23

0.60 0.25 5 15 1 01 0.9799 1.0610(6.6667 20)

1.6667 5 1.0 0.5648

j jV

jj

! − + #$ % − − + ⋅ ∠( )* +∠, -= ⋅ * +− * +− − + ⋅ ∠. /

oo

o

( )( 1) ( 1) ( 1)3 33 31 1 32 2( )

33 3

1k k k

k

P jQV Y V Y VY V

+ + +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

33Y

1*3V

3P 3Q 31Y 1kV

32Y22aV




Gauss-Seidel Load Flow: Example Check for convergence: (iteration 2)

Action: Continue iterating.

( )( )

2 2 12 2 2

0.9965 0.5648

1.0 1.1705aV V V

∠Δ = − =

− ∠

o

o

( )( )

2 2 13 3 3

0.9791 1.2218

0.9799 1.0610V V V

∠−Δ = − =

− ∠−

o

o

22 1.0 0.5648aV = ∠ o 2

3 0.9791 1.2218V = ∠− o

0.0111 0.005= >

0.0029 0.005= <




Gauss-Seidel Load Flow: Example Iteration 3: k = 2

Bus 2: ( ) ( )2

( 1) ( ) ( )2 2 21 1 22 2 23 3


22 0.0438Q =

( )( ) ( )( ) ( )( ) ( )

2

2

22

22

1.25 3.75 1 0

1.0 0.5648 2.9167 8.75 1.0 0.5648

1.6667 5 0.9791 1.2218

0.2253 0.0438

j

P jQ j

j

P jQ j

! "− + ⋅ ∠& '& '− = ∠− ⋅ + − ⋅ ∠& '& '+ − + ⋅ ∠−( )

− = −

o

o

2 *2aV

21Y

22Y

23Y

1kV22aV

23V


discard





3 32 20.9991 0.4158 1.0 0.4158aV V= ∠ → = ∠o o

( )( ) ( )

( ) ( )32

0.2 0.0438 1.25 3.75 1 01 1.0 0.56482.9167 8.75

1.6667 5 0.9791 1.2218

j jV

jj

! − #$ % − − + ⋅ ∠( )* +∠−, -= ⋅ * +− * +− − + ⋅ ∠−. /

oo

o

( )( 1) ( 1) ( )2 22 21 1 23 3( )

22 2

1 kk k k

k

P jQV Y V Y VY V

+ +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

22Y2 *2aV

2P 12Q 21Y 1

kV

23Y23V




Gauss-Seidel Load Flow: Example Bus 3:

33 0.9790 1.2575V = ∠− o

( ) ( )

( ) ( )33

0.60 0.25 5 15 1 01 0.9799 1.0610(6.6667 20)

1.6667 5 1.0 0.4158

j jV

jj

! − + #$ % − − + ⋅ ∠( )* +∠, -= ⋅ * +− * +− − + ⋅ ∠. /

o

o

( )( 1) ( 1) ( 1)3 33 31 1 32 2( )

33 3

1k k k

k

P jQV Y V Y VY V

+ + +∗

" #$ %−' () *= ⋅ − ⋅ − ⋅' () *, -. /

33Y2*3V

3P 3Q 31Y 1kV

32Y32aV




Gauss-Seidel Load Flow: Example Check for convergence:

Action: Stop iterating. The solution has converged.

( )( )

3 3 22 2 2

0.9991 0.4158

1.0 0.5648aV V V

∠Δ = − =

− ∠

o

o

( )( )

3 3 23 3 3

0.9791 1.2575

0.9791 1.2218V V V

∠−Δ = − =

− ∠−

o

o

32 1.0 0.4158aV = ∠ o 3

3 0.9790 1.2575V = ∠− o

0.0027 0.005= <

0.0006 0.005= <




I

jX

Ei∠δ Vt∠0

Generator Voltage & Power Control

~

The complex power delivered to the bus (Generator Terminal) is

[ ] [ ] !"

#$%

& ∠−∠∠=∠=+

jXVEVIVjQP ti

tttt000 * δ

!"

#$%

&= δsinXVEP ti

t !"

#$%

&−=XV

XVEQ tti

t

2

cosδ

Principles of Load Flow Control





!"

#$%

&= δsinXVEP ti

t !"

#$%

&−=XV

XVEQ tti

t

2

cosδ

Observations: 1.  Real Power is injected into the bus (Generator Operation), δ must

be positive (Ei leads Vt)

2.  Real Power is drawn from the bus (Motor Operation), δ must be negative (Ei lags Vt)

3.  In actual operation, the numeric value of δ is small & since the slope of Sine function is maximum for small values, a minute change in δ can cause a substantial change in Pt






!"

#$%

&= δsinXVEP ti

t !"

#$%

&−=XV

XVEQ tti

t

2

cosδ

Observations: 4.  Reactive Power flow depends on relative values of EiCosδ and Vt

5.  Since the slope of Cosine function is minimum for small values of angle, Reactive Power is controlled by varying Ei •  Over-excitation (increasing Ei) will deliver Reactive Power into the Bus

•  Under-excitation (decreasing Ei) will absorb Reactive Power from the Bus





Capacitor Compensation

Ipq p

q

PL - jQL

+ jQc

~

p

qpq

p

qpqpq V

PXj

VQX

VE −−=

The voltage of bus q can be expressed as

Observations: 1.  The Reactive Power Qq causes a voltage drop and thus largely

affects the magnitude of Eq

2.  A capacitor bank connected to bus q will reduce Qq that will consequently reduce voltage drop





Tap-Changing Transformer a:1

q r

s p

The π equivalent circuit of transformer with the per unit transformation ratio:

pqyaa2

1−pqya

a 1−

pqya1

Observation: The voltage drop in the transformer is affected by the transformation ratio “a”





BASIC INFORMATION

"  Voltage Profile "   Injected Power (Pp and Qp) "   Line Currents (Ipq and Ipq) "  Power Flows (Ppq and Qpq) "   Line Losses (I2R and I2X)

Information from a Load Flow Study




The bus voltages are: V1 = 1.0∠0° V2 = 0.9990∠0.4129° V3 = 0.9788∠-1.2560°

The power injected into the buses are:

P1 - jQ1 = (1.0∠0) [(19.7642∠-71.5651°)(1.0∠0°)

+ (3.9528∠108.4349°)(0.9990∠0.4129°)

+ (15.8114∠108.4349°) (0.9788∠-1.25560°) = 0.4033 - j0.2272

P1 - jQ1 = V1* [Y11V1 + Y12V2 + Y13V3 ]





P2 - jQ2 = (0.999∠-0.4129°)[(3.9528∠108.4349°)(1.0∠0°) + (9.2233∠-71.5649°)(0.9990∠0.4129°) + (5.2705∠108.4349°)(0.9788∠-1.25560°)

= 0.2025 - j0.04286

P3 - jQ3 = (0.9788∠1.256°) [(15.8114∠108.4349°)(1.0∠0°) + (5.2705∠108.4349°)(0.9990∠0.4129°) + (21.0819 ∠ -71.5650°)(0.9788∠-1.25560° )

= -0.600 + j0.2498

P2 - jQ2 = V2* [Y21V1 + Y22V2 + Y23V3 ]

P3 - jQ3 = V3* [Y31V1 + Y32V2 + Y33V3 ]





p q ypo yqo

ypq Ipq Iqp Iline Vp Vq

ppoqppqpolinepqVy )VV(yI II +−=+=

qqopqpqqolineqpVy )VV(yI I I +−=+−=

The line current Ipq, measured at bus p is given by

Similarly, the line current Iqp, measured at bus q is

Line Currents





The branch currents are:

)VV(yII qppqlinepq −== )VV(yI I pqpqlineqp −=−=

I12 = y12 [V1 - V2] I21 = y12 [V2 – V1]

I13 = y13 [V1 – V3] I31 = y13 [V3 – V1]

I23 = y23 [V2 – V3] I32 = y23 [V3 – V2]





The power flow (Spq) from bus p to q is *

pqppqpqpqIVQj PS =+=

The power flow (Sqp) from bus q to p is *

qpqqpqpqpIVQj PS =+=

Power Flows





The branch power flows are:

P12 – jQ12 = V1* I12 P21 – jQ21 = V2

* I21

P13 – jQ13 = V1* I13 P31 – jQ31 = V3

* I31

P23 – jQ23 = V2* I23 P32 – jQ32 = V3

* I32





The power loss in line pq is the algebraic sum of the power flows Spq and Sqp

qppqlosslosslossSSQj PS +=+=

Line Losses


( ) *pqqp

*pqq

*pqp

IVVIVIV

+=

−=




The line losses are:

P12(Loss) – jQ12(Loss) = (P12 – jQ12) + (P21 – jQ21 )








Other Information:

"  Overvoltage and Undervoltage Buses "  Critical and Overloaded Transformers

and Lines "   Total System Losses




Sensitivity Analysis with Load Flow Study

9) Add or remove rotating or static var supply to buses.

8) Increase or decrease transformer size.

7) Change transformer taps.

6) Change bus voltages.

5) Increase conductor size on T&D lines.

4) Add new transmission or distribution lines.

3) Add, remove or shift generation to any bus.

2) Add, reduce or remove load to any or all buses. 1) Take any line, transformer or generator out of service.

Uses of Load Flow Studies




Uses of Load Flow Studies ! Sensitivity

Analysis Example

IEEE 14-Bus System





Analysis Example

Removal of Line 4-5





Analysis Example

IEEE 14-Bus System





Analysis Example

Removal of generator at Bus 2





Analysis Example

IEEE 14-Bus System





Analysis Example

Removal of rotating VAR supply at bus 3

From 1.010 pu





Analysis Example

IEEE 14-Bus System





Analysis Example

Increase in P and Q at bus 14

From 14.9 MW + 5.0 MVAR;

V: from 1.035 pu

Line 9-14: From 50% loading





!  Analysis of existing conditions: ! Check for voltage violations (undervoltage/overvoltage). ! Check for transformer overloading/line overloading. ! Check for system losses.

!  Analysis for correction of power quality issues: ! Voltage adjustment at the delivery points ! Transformer tap changing ! Capacitor compensation:

"  Compensation for Peak Loading "  Check for overvoltages during Off-Peak conditions "  Optimize capacitor allocation and capacitor switching





!  Analysis for Expansion Planning: ! Construction of new substation ! Addition of capacity to existing substation ! Construction of new feeder segment ! Extension of existing feeder segment ! Addition of parallel feeder segment ! Replacement of conductors in existing feeder segments ! Conversion of entire feeder circuits from one voltage

level to another voltage level ! Addition of generators





!  Contingency Analysis: ! Reliability of the Transmission, Subtransmission, and

Distribution Systems Reliability denotes that not only is the power system working, but that it is working properly. That is, no physical and technical constraints must be violated – i.e., voltage must be well regulated and within acceptable range, load limits of the transformers and the lines must not be exceeded, and power balance must be satisfied.

!  System Loss Analysis: ! Identification of lossy components in the power system.

EEE 103 - Load Flow Analysis

Documents