Load Flow Analysissmart.tabrizu.ac.ir/Files/Content/Smart/Load-Flow-Analysis-6th-Batch.pdf · Training Course in Load Flow Analysis The Load Flow Problem Load Flow Analysis simulates

NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER

Certificate inPower System Modeling and Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

U. P. NATIONAL ENGINEERING CENTER

Training Course in

Load Flow Analysis

2

Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Load Flow Analysis

Course Outline

1. The Load Flow Problem

2. Power System Models for Load Flow Analysis

3. Gauss-Seidel Load Flow

4. Newton-Raphson Load Flow

5. Backward/Forward Sweep Load Flow

6. Principles of Load Flow Control

7. Uses of Load Flow Studies

3




The Load Flow Problem

Basic Electrical Engineering Solution

Load Flow of Distribution System

Load Flow of Transmission and Subtransmission System

Load Flow of a Contemplated System

Load Flow of a Single Line

4





Basic Electrical Engineering Solution

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

Sending End

Receiving End

VS = ?

Load2 MVA, 3Ph

85%PF

VR = 13.2 kVLL

Line1.1034 + j2.0856 ohms/phase

ISR = ?

VOLTAGE DROP = VS - VR

Solve for:

1) ISR = (SR/VR )*

2) VD = ISRZL

3) VS = VR + VD

4) SS = VSx(ISR)*

5




The Load Flow ProblemSending

EndReceiving

End

VS = ?

Load2 MVA, 3Ph

85%PF

VR = 13.2 kVLL


ISR = ?

Solve for:

1) ISR = (SR/VR )*

2) VD = ISRZL

3) VS = VR + VD

4) SS = VSx(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 A7621.02 0

VD 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

+ = ∠

= ∠ =

6





Load Flow From the Real World

Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?

Load2 MVA, 3Ph

85%PFHow do you solve for:

1) ISR = ?

2) VD = ?

3) VR = ?

4) SS = ?

7





Load Flow of Distribution System

Bus1Bus2

Bus3

Bus4V1 = 67 kVV2 = ?

V4 = ?

V3 = ?I23 , Loss23 = ?

I24 , Loss24 = ?

I12 , Loss12 = ?

P1 , Q1 = ?P2 , Q2 = ?

P3 , Q3 = ?

P4 , Q4 = ?

Utility Grid

Lumped Load A2 MVA 85%PF

Lumped Load B1 MVA 85%PFHow do you solve

for the Voltages, Currents, Power and Losses?

8





Load Flow of Transmission and Subtransmission System

Line 1Line 1

Line 3Line 3Line 2Line 2

11 22

33

G G

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

9





Load Flow of a Contemplated System

How about if there are contemplated changes in the System?

How will you determine in advance the effects of:• Growth or addition of new loads• Addition of generating plants• Upgrading of Substation• Expansion of distribution lines

before the proposed changes are implemented?

Answer: LOAD FLOW ANALYSIS

10





Load Flow Analysis simulates (i.e., mathematically determine) the performance of an electric power system under a given set of conditions.

Load Flow (also called Power Flow) is a snapshot picture of the power system at a given point.

11






Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?

Load2 MVA, 3Ph 85%PF

SR = VR x (ISR)*

Injected Power at Receiving End

VS = VR + Z x ISR

Voltage at Sending End

ISR = (SR / VR)*

Solving for the Current

VR = VS - Z x SR*/VR*

Voltage at Receiving End

12






Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?


Converting Quantities in Per Unit

VS(pu) = 13.2 /13.2 = 1/0Base Power = 1 MVA

Base Voltage = 13.2 kV

Base Impedance = [13.2]2/1

= 174.24 ohms

SR(pu) = 2/cos-1(0.85) / 1

Zpu = (1.1034 + j2.0856)/174.24

= 0.00633 + j0.01197

13





Load Flow of a Single LineSending

EndReceiving

End

VS = 13.2 kVLL VR = ?


ISR = ?


VR(k) = VS - Z x [SR]* / [VR

(k-1) ]*

Let VR(0) = 1/0

For k = 1VR

(1) = __________

∆V(1) = __________

For k = 2VR

(2) = __________

∆V(2) = __________

14






Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?


VR(k) = VS - Z x [SR]* / [VR

(k-1) ]*

For k = 3VR

(3) = __________

∆V(3) = __________

For k = 4VR

(4) = __________

∆V(4) = __________

VR(2) = __________

15






Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?


ISR = __________

SR = __________

SS = VS x [ISR]*

SS = __________

VS = __________

VR = __________

VD = VS – VR

VD = __________

16






Sending End

Receiving End

VS = 13.2 kVLL VR = ?


ISR = ?


SLoss = PLoss + QLoss

SLoss = SS - SR

PLoss = _________

QLoss = _________

PFR = PR / SR

PFR = _________

PFS = PS / SS

PFS = _________

17




Power System Models for Load flow Analysis

Bus Admittance Matrix, Ybus

Network Models

Generator Models

Bus Types for Load Flow Analysis

18




Power System Models for Load Flow Analysis

The power system components are interconnected through the buses. The buses must therefore be identified in the load flow model.

Generators and loads are connected from bus to neutral.Transmission lines and transformers are connected from one bus to another bus.

19




⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

[YBUS] =

The static components (transformers and lines) are represented by the bus admittance matrix, Ybus

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

Network Models


20





Network ModelsLine 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

Line 1Line 1


11 22

33

Set-up the Ybus

21





Network Models

Compute the branch admittances to set up Ybus:

y12 = ____1z12

= ______________10.08 + j0.24

= 1.25 - j3.75

y13 = ____1z13

= ______________10.02 + j0.06

= 5 - j15

y23 = ____1z23

= ______________10.06 + j0.18

= 1.667 - j5

22




Set-up the bus admittance matrix:


Y11 = y12 + y13

= (1.25 - j3.75) + (5 - j15)= 6.25 - j18.75 = 19.7642 ∠ -71.5651°

Y12 = -y12

= -1.25 + j3.75 = 3.9528 ∠ 108.4349°

Y13 = -y13= -5 + j15 = 15.8114 ∠ 108.4349°

Y21 = Y12 = -y12= -1.25 + j3.75 = 3.9528 ∠ 108.4349°

23





Y22 = y12 + y23

= (1.25 - j3.75) + (1.6667 - j5)= 2.9167 - j8.75 = 9.2233 ∠ -71.5649°

Y23 = -y23= -1.6667 + j5 = 5.2705 ∠ 108.4349°

Y31 = Y13 = -y13= -5 + j15 = 15.8114 ∠ 108.4349°

Y32 = Y23 = -y23

= -1.6667 + j5 = 5.2705 ∠ 108.4349°= (5 - j15) + (1.6667 - j5)Y33 = y13 + y23

= 6.6667 - j20 = 21.0819 ∠ -71.5650°

24





Voltage-controlled generating units to supply a scheduled active power (P) at a specified voltage (V). The generating units are equipped with voltage regulator to adjust the field excitation so that the units will operate at particular reactive power (Q) in order to maintain the voltage. Swing generating units to maintain the frequency at 60Hz in addition to maintaining the specified voltage. The generating unit is equipped with frequency-following controller (very fast speed governor) and is assigned as Swing generator

Generator Models

25





Bus Types for Load Flow

Generators and loads are connected from bus to neutral.

Four quantities must be specified to completely describe a bus. These are:

Bus voltage magnitude, VpBus voltage phase angle, δpBus injected active power, PpBus injected reactive power, Qp

26




Swing Bus or Slack BusThe 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.

Type 1: Swing Bus

Specify: V, δ

Unknown: P, QG

P,QP,Q

δV∠++

--


27




Generator Bus (Voltage-Controlled) Bus or PV BusThe 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,QP,Q

δV∠++

--

Specify: P, VType 2: Generator Bus

Unknown: Q, δ


28





Load Bus or PQ 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.

P,QP,Q

++Type 3: Load Bus

Specify: P, Q

--δV∠

Unknown: V, δ

29





SUMMARY OF BUS TYPES

BB uu ss TT yy pp ee

KK nn oo ww nn QQ uu aa nn tt iitt iiee ss

UU nn kk nn oo ww nn QQ uu aa nn tt iitt iiee ss

TT yy pp ee 11 :: SS ww iinn gg

VV pp ,, δδ pp

PP pp ,, QQ pp

TT yy pp ee 22 :: GG ee nn ee rraa ttoo rr

PP pp ,, VV pp

QQ pp ,, δδ pp

TT yy pp ee 33 :: LL oo aa dd

PP pp ,, QQ pp

VV pp ,, δδ pp

30




Line 1Line 1


11 22

33

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

Bus Types

31




Gauss-Seidel Load Flow

Linear Formulation of Load Flow Equations

Gauss-Seidel Load Flow Solution

Numerical Example

32





The real and reactive power into any bus P is:

Linear Formulation of Load Flow Equations

Pp + jQp = Vp Ip*

or(1)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

33




Equation (1) may be rewritten as:

Ip = Pp - jQp_________

Vp*


(2)

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

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

I1 = Y11V1 + Y12V2 + Y13V3

I2 = Y21V1 + Y22V2 + Y23V3

I3 = Y31V1 + Y32V2 + Y33V3

34





Substituting (3) into (2)

_________Vp

*

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

_________V1

*

P1 – jQ1= Y11V1 + Y12V2 + Y13V3

(4)

_________V2

*

P2 – jQ2= Y21V1 + Y22V2 + Y23V3

_________V3

*

P3 – jQ3= Y31V1 + Y32V2 + Y33V3

35




Gauss-Seidel Load FlowSolving for Vp in (4)

_______Y11V1 = P1 – jQ1

V1*

- (___ + Y12V2 + Y13V3)

⎥⎦

⎤⎢⎣

⎡−−

−= 313212*

1

11

11

1 VYVYV

jQPY1V

_______Y22V2 = P2 – jQ2

V2*

- (Y12V2 + ___ + Y13V3)

⎥⎦

⎤⎢⎣

⎡−−

−= 313121*

2

22

22

2 VYVYV

jQPY1V

36





⎥⎦

⎤⎢⎣

⎡−−

−= 232131*

3

33

33

3 VYVYV

jQPY1V

_______Y33V3 = P3 – jQ3

V3*

- (Y13V1 + Y23V2 + ___)

Vp =1___

Ypp

_______ Vp

*

Pp - jQp Σ-n

q=1q≠p

YpqVq (5)

37





Gauss-Seidel Load Flow Solution

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=1q≠p

YpqVqα

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

(6)

38




Gauss-Seidel Load FlowGauss-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.

39




Gauss-Seidel Load FlowNumerical ExampleShown 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 zero and 50 MVARS, respectively. Base power used is 100 MVA. Solve the load flow problem using Gauss-Seidel iterative method assuming a 0.005 convergence index.

Line 1Line 1


11 22

33

G G

40





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 DataVoltage Generation Load Bus

No. V (p.u.) δ P Q P Q Remarks


41





Specified Variables:V1 = 1.0 δ1 = 0.0

V2 = 1.0 P2 = 0.2 Note the negative sign of P and Q of the Load at Bus 3P3 = -0.6 Q3 = -0.25

Initial Estimates of Unknown Variables:

δ20 = 0.0

V30 = 1.0

δ30 = 0.0

42





The Bus Admittance Matrix elements are:Y11 = 6.25 - j18.75 = 19.7642 ∠ -71.5651°Y12 = -1.25 + j3.75 = 3.9528 ∠ 108.4349°Y13 = -5 + j15 = 15.8114 ∠ 108.4349°Y21 = -1.25 + j3.75 = 3.9528 ∠ 108.4349°Y22 = 2.9167 - j8.75 = 9.2233 ∠ -71.5649°Y23 = -1.6667 + j5 = 5.2705 ∠ 108.4349°Y31 = -5 + j15 = 15.8114 ∠ 108.4349°Y32 = -1.6667 + j5 = 5.2705 ∠ 108.4349°Y33 = 6.6667 - j20 = 21.0819 ∠ -71.5650°

43




Gauss-Seidel Load FlowGauss-Seidel EquationsBus 1: Swing Bus

( ) 01V 1k1 ∠=+

for all iterations

Bus 2: Generator BusQ2 must first be determined from:

P2 - jQ2(k+1) = (V2

(k))* [Y21V1(k+1) + Y22V2

(k) + Y23V3(k)]

then substitute it to:

( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQPY

1V

44





Bus 3: Load Bus

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQPY1V

Iteration 1 (k = 0): V1 (1) = 1.0∠0°

P2 - jQ2(1) = (1.0∠0°) [(-1.25 + j3.75)(1.0∠0°)

+ (2.9167 - j8.75)(1.0∠0°)

+ (-1.6667 + j5)(1.0∠0°)

= 0.0 + j0.0

Q2(1) = 0.0 [This value is within the limits.]

45





( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQPY1V

( )1k22 jQP +−

V2(1) =

1___________________9.2233∠-71.5650

0.2 - j0.0

1.0∠0°___________

- (-1.25 +j3.75) (1.0∠0°)

- (-1.6667 + j5) (1.0∠0°)

21Y

23Y

( )1k1V +

( )k3V

( )( )*k2V22Y

= 1.0071∠1.1705°

46





V31 =

1_____________________21.0819∠-71.5650

-0.6 + j0.25

1.0∠0°____________

- (-5 +j15) (1.0∠0°)

- (5.2705∠108.4349°)(1.0071∠1.1705°)

33Y31Y

32Y

33 jQP −

( )*k3V

( )1k1V +

( )1k2V +

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQPY1V

= 0.9816 ∠-1.0570°

47





∆V2 = V2(1) - V2

(0)

= 1.0071∠1.1705° - 1.0∠0°

∆V2 = 0.0217

∆V3 = V3(1) - V3

(0)

= 0.9816∠-1.0570° - 1.0∠0°

∆V3 = 0.0259

48





Iteration 2 (k = 1): V1(2) = 1.0∠0°

Let, V2(1) = 1.0∠1.1705°

P2 - jQ2(2) = (1.0∠-1.1705°)[(-1.25 + j3.75)(1.0∠0°)

+ (9.2233∠-71.5649°)(1.0∠1.1705°)

+ (5.2705∠108.4349° )(0.9816∠-1.0570°)

= 0.2995 - j0.0073

Q2 (2) = 0.0073 [This value is within the limits.]

49





( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQPY1V

0.2 - j0.0073

1.0 ∠ -1.1705°______________V2

(2) =1___________________

9.2233 ∠ -71.5650

- (-1.25 +j3.75) (1.0 ∠ 0°)

- (5.2705 ∠ 108.4349° ) (0.9816 ∠ -1.0570°)

= 0.9966 ∠ 0.5819°

50





( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQPY1V

-0.6 + j0.25

0.9816 ∠ 1.0570°___________________V3

(2) =1_____________________

21.0819 ∠ -71.5650

- (-5 +j15) (1.0 ∠ 0°)

- (5.2705 ∠ 108.4349°) (0.9966 ∠ 0.5819° )

= 0.9783 ∠ -1.2166°

51





∆V2 = V2(2) - V2

(1)

= 0.9966 ∠ 0.5819° - 1.0071 ∠ 1.1705°

∆V2 = 0.0125

∆V3 = V3(2) - V3

(1)

= 0.9783 ∠ -1.2166° - 0.9816 ∠ -1.0570°

∆V3 = 0.004

52





Iteration 3 (k = 2): V1(2) = 1.0∠0°

Let, V22 = 1.0 ∠ 0.5819°

P2 - jQ22 = (1.0 ∠-0.5819°) [(-1.25 + j3.75)(1.0 ∠ 0°)

+ (9.2233 ∠ -71.5649° ) (1.0 ∠ 0.5819°)

+ (5.2705 ∠ 108.4349° ) (0.9783 ∠ -1.2166° )

= 0.2287 - j0.0472

Q22 = 0.0472 [This value is within the limits.]

53





( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQPY1V

V23 =

1___________________9.2233 ∠ -71.5650

0.2 - j0.0472

1.0 ∠ -0.5819°______________

- (-1.25 +j3.75) (1.0 ∠ 0°)

- (5.2705 ∠ 108.4349° ) (0.9783 ∠ -1.2166° )

= 0.9990 ∠ 0.4129°

54





( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQPY1V

-0.6 + j0.25

0.9783 ∠ 1.2166°

___________________V33 =

1_____________________21.0819 ∠-71.5650

- (-5 +j15)(1.0∠0°)

- (5.2705∠108.4349°)(0.9990∠0.4129°)

= 0.9788∠-1.2560°

55





∆V2 = V2(3) - V2

(2)

= 0.9990∠0.4129° - 1.0∠0.5819°

∆V2 = 0.003 < 0.005

∆V3 = V3(3) - V3

(2)

= 0.9788∠-1.2560° - 0.9783∠-1.2166°

∆V3 = 0.0008 < 0.005

The solution has converged.

56




Newton-Raphson Load Flow

Non-Linear Formulation of Load Flow Equations

Newton-Raphson Load Flow Solution

Numerical Example

57





Non-Linear Formulation of Load Flow Equations

The complex power injected into Bus p is*

p p p pP jQ E I− =and the current equation may be written as

(1)

n

p pq qq 1

I Y E=

= ∑Substituting (2) into (1)

(2)

(3)q

n

1qpq

*ppp EYEjQP ∑

=

•=−

58




Newton-Raphson Load FlowLet

p p pE V δ= ∠ q q qE V δ= ∠

p q p q p qY Y θ= ∠Substituting into equation (3),

n

p p p q pq pq q pq 1

P jQ V V Y ( )θ δ δ=

− = ∠ + −∑

n

(4)

Separating the real and imaginary components

p p q pq pq q pq 1

P V V Y co s( )θ δ δ=

= + −∑n

p p q pq pq q pq 1

Q V V Y sin( )θ δ δ=

= − + −∑

(5)

(6)

59




Newton-Raphson Load FlowThe formulation results in a set of non-linear equations, two for each Bus of the system.

Equations Pp are written for all Buses except the Swing Bus.

Equations Qp are written for Load Buses only

The system of equations may be written for

i number of buses minus the swing bus (n-1)

j number of load buses

60




Newton-Raphson Load FlowThe system of equations may be written as

1 1 1 2 i 1 2 jP P ( , ,..., ,v ,v ....,v )δ δ δ=

2 2 1 2 i 1 2 jP P ( , ,..., ,v ,v ....,v )δ δ δ=

i i 1 2 i 1 2 jP P ( , ,..., ,v ,v ....,v )δ δ δ=

1 1 1 2 i 1 2 jQ Q ( , ,..., ,v ,v ....,v )δ δ δ=

2 2 1 2 i 1 2 jQ Q ( , ,..., ,v ,v ....,v )δ δ δ=

jj 1 2 i 1 2 jQ Q ( , ,..., ,v ,v ....,v )δ δ δ=

M M

M M

(7)

61




Newton-Raphson Load FlowEquation (7) may be linearized using a First-Order Taylor-Series Expansion

spec calc 1 1 1 1 1 11 1 1 2 i 1 2 j

1 2 i 1 2 j

spec calc 2 2 2 2 2 22 2 1 2 i 1 2 j

1 2 i 1 2 j

spec calc i ii i 1 2

1 2

P P P P P PP P ... V V ... V

V V V

P P P P P PP P ... V V ... V

V V V

P PP P ...

∆δ ∆δ ∆δ ∆ ∆ ∆δ δ δ

∆δ ∆δ ∆δ ∆ ∆ ∆δ δ δ

∆δ ∆δδ δ

∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂= + + + +

∂ ∂i i i i

i 1 2 j2 1 2 j

spec calc 1 1 1 1 1 11 1 1 2 i 1 2 j

1 2 i 1 2 j

spec calc 2 2 2 2 2 22 2 1 2 i 1 2

1 2 i 1 2

P P P PV V ... V

V V V

Q Q Q Q Q QQ Q ... V V ... V

V V V

Q Q Q Q Q QQ Q ... V V ...

V V

∆δ ∆ ∆ ∆δ

∆δ ∆δ ∆δ ∆ ∆ ∆δ δ δ

∆δ ∆δ ∆δ ∆ ∆δ δ δ

∂ ∂ ∂ ∂+ + + +

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂ jj

j j j j j jspec calcj j 1 2 i 1 2 j

1 2 i 1 2 j

VV

Q Q Q Q Q QQ Q ... V V ... V

V V V

∆

∆δ ∆δ ∆δ ∆ ∆ ∆δ δ δ

∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂

M M M M M M M M M

M M M M M M M M M

62



Training Course in Load Flow Analysis1 1 1 1 1 1s p e c c a lc

1 11 2 i 1 2 j

2 2 2 2s p e c c a lc2 2

1 2 i 1

s p e c c a lci i

s p e c c a lc1 1

s p e c c a lc2 2

s p e c c a lcj j

P P P P P PP P

V V V

P P P PP P

V

P P

Q Q

Q Q

Q Q

δ δ δ

δ δ δ

∂ ∂ ∂ ∂ ∂ ∂⎡ ⎤−

∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥⎢ ⎥

∂ ∂ ∂ ∂⎢ ⎥−⎢ ⎥ ∂ ∂ ∂ ∂⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥

=⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

L L

L

M

M

2 2

2 j

i i i i i i

1 2 2 1 2 j

1 1 1 1 1 1

1 2 i 1 2 j

2 2 2 2 2 2

1 2 i 1 2 j

j j j j j j

1 2 i 1 2 j

P PV V

P P P P P PV V V

Q Q Q Q Q QV V V

Q Q Q Q Q QV V V

Q Q Q Q Q QV V V

δ δ δ

δ δ δ

δ δ δ

δ δ δ

⎡⎢⎢⎢ ∂ ∂⎢⎢ ∂ ∂⎢⎢⎢⎢⎢∂ ∂ ∂ ∂ ∂ ∂⎢

⎢ ∂ ∂ ∂ ∂ ∂ ∂⎢⎢ ∂ ∂ ∂ ∂ ∂ ∂⎢⎢ ∂ ∂ ∂ ∂ ∂ ∂⎢⎢ ∂ ∂ ∂ ∂ ∂ ∂⎢∂ ∂ ∂ ∂ ∂ ∂⎢

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂⎣

L

M M M M M M

L L

L L

L L

M M M M M M M

L L

1

2

i

1

2

j

V

V

V

∆ δ

∆ δ

∆ δ

∆

∆

∆

⎤⎡ ⎤⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎦

M

M

63




Newton-Raphson Load Flowor simply

⎥⎦

⎤⎢⎣

⎡∆∆

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡∆∆

VVQQVPP

QP δ

δ

δ

⎥⎥⎦

⎤

⎢⎢⎣

⎡∆∆

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡∆∆

VV

VQVQVPVP

QP δ

δ

δ

64




Newton-Raphson Load FlowNewton-Raphson Load Flow Solution

1 2

3 4

J JP

VJ JQ V

∆δ∆

∆∆

⎡ ⎤⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

np

p q p q p q q pq 1 ,q pp

1p

p q p q p q q pq

PV V Y sin ( )

JP

V V Y sin ( )

θ δ δδ

θ δ δδ

= ≠

∂⎧= + −⎪ ∂⎪

⎨ ∂⎪ = − + −⎪ ∂⎩

∑

65





p 2p p p p p p p

p2

pq p q p q p q q p

q

np

p q p q p q q pq 1 ,q pp

3p

p q p q p q q pq

PV P V Y c o s

VJ

PV V V Y c o s ( )

V

QV V Y c o s ( )

JQ

V V Y c o s ( )

θ

θ δ δ

θ δ δδ

θ δ δδ

= ≠

∂⎧= +⎪ ∂⎪

⎨ ∂⎪ = + −⎪ ∂⎩∂⎧

= + −⎪ ∂⎪⎨ ∂⎪ = − + −⎪ ∂⎩

∑

66





p 2p p p p p p q

p4

pq p q p q p q q p

q

QV Q V Y sin

VJ

QV V V Y sin ( )

V

θ

θ δ δ

∂⎧= −⎪ ∂⎪

⎨ ∂⎪ = − + −⎪ ∂⎩

The solution of the load flow equations proceeds iteratively from the set of initial estimates. These estimates are updated after evaluating the Jacobianmatrix.

67




Newton-Raphson Load FlowAt the kth iteration,

( k 1 ) ( k ) ( k )p p p

( k 1 ) ( k ) ( k )p p pV V V

δ δ ∆δ

∆

+

+

= +

= +

The process is terminated once convergence is achieved whrein

( k ) ( k )p qMAX P and MAX Q∆ ε ∆ ε≤ ≤

68




Numerical ExampleShown 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 zero and 50 MVARS, respectively. Base power used is 100 MVA. Solve the load flow problem using Gauss-Seidel iterative method assuming a 0.005 convergence index.

Line 1Line 1


11 22

33

G G


69





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 DataVoltage Generation Load Bus

No. V (p.u.) δ P Q P Q Remarks


70




Newton-Raphson Load FlowThe elements of the Bus Admittance Matrix are:

11

12

13

21

22

23

31

Y 6.25 j18.75 19.7642 71.5651Y 1,25 j3.75 3.9528 108.4349Y 5 j15 15.8114 108.4349Y 1.25 j .375 3.9528 108.4349Y 2.9167 j8.75 9.2233 71.5649Y 1.6667 j5 5.2705 108.4349Y 5 j15 15.811

= − = ∠ −= − + = ∠= − + = ∠

= − + = ∠= − = ∠ −= − + = ∠

= − + =

32

33

4 108.4349Y 1.6667 j5 5.2705 108.4349Y 6.6667 j 20 21.0819 71.5650

∠

= − + = ∠

= − = ∠ −

71




Newton-Raphson Load FlowBus 1: Swing Bus (Not included)

Bus 2: Generator Bus (Compute for P2)

Bus 32: Generator Bus (Compute for P2 and Q2)

2 2 22 3 2

2 3 3

3 3 33 3 3

2 3 3

3 3 3 33 3

2 3 3 3

P P PP VV

P P PP VV

Q Q Q VQ VV V

∆ ∆δδ δ

∆ ∆δδ δ

∆∆δ δ

⎡ ⎤⎡ ⎤∂ ∂ ∂⎡ ⎤ ⎢ ⎥⎢ ⎥∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ∂ ∂ ∂⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

72





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

73




Newton-Raphson Load FlowCompute Initial Power Estimates

02 2 1 21 21 1 2

2 2 22 22 2 3 23 23 3 2

P V V Y cos( ) V V Y cos V V Y cos( ) ( 1.0 )( 1.0 )( 3.9528 )cos( 108.4349 0.0 0.0 ) ( 1.0 )( 1.0 )( 9.2233 )cos( 71.5649 ) ( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 ) 0.0

θ δ δθ θ δ δ

= + −+ + + −

= + ++ −+

=

74





03 3 1 31 31 1 3

3 2 32 22 2 3 3 3 33 33

P V V Y cos( ) V V Y cos( ) V V Y cos ( 1.0 )(1.0 )(15.8114 )cos(108.4349 0.0 0.0 ) ( 1.0 )(1.0 )( 5.2705 )cos(108.4349 0.0 0.0 ) ( 1.0 )(1.0 )( 21.0819 )cos( 71.5650 )

θ δ δθ δ δ θ

= + −

+ + − +

= + −+ + ++ −

0.0=

75





03 3 1 31 31 1 3

3 2 32 32 2 3 3 3 33 33

Q V V Y sin( ) V V Y sin( ) V V Y sin ( 1.0 )( 1.0 )( 15.8114 ) sin( 108.4349 0.0 0.0 ) ( 1.0 )( 1.0 )( 5.2705 ) sin( 108.4349 0.0 0.0 ) ( 1.0 )( 1.0 )( 21.0819 ) sin( 71.5650 )

θ δ δθ δ δ θ

= + −+ + − +

= + −+ + ++ −

0.0=

76




Newton-Raphson Load FlowCompute Power Mismatch

02P 0 .2 0 .0 0 .2∆ = − =

03P 0 .6 0 .0 0 .6∆ = − − = −03Q 0 .2 5 0 .0 0 .2 5∆ = − − = −

Evaluate elements of Jacobian Matrix

P PV

JQ QV

V

δ δ

δ

∂ ∂⎡ ⎤⎢ ⎥∂ ∂⎢ ⎥=⎢ ⎥∂ ∂⎢ ⎥⎢ ⎥∂ ∂⎣ ⎦

77




Newton-Raphson Load FlowElements of J1:

22 1 21 21 1 2

2

2 3 23 23 3 2

P V V Y sin( )

V V Y sin( ) ( 1.0 )(1.0 )( 3.9528 )sin(108.4349 0.0 0.0 ) ( 1.0 )(1.0 )( 5.2705 )sin(108.4349 0.0 0.0 ) 8.75

θ δ δδ

θ δ δ

∂= + −

∂+ + −

= + −+ + −

=

78




Newton-Raphson Load FlowNewton-Raphson Load FlowElements of J1:

22 3 23 23 3 2

3

PV V Y sin( )

( 1.0 )(1.0 )( 5.2705 )sin(108.4349 0.0 0.0 ) 5

θ δ δδ∂

= − + −∂

= − + −= −

33 2 32 32 2 3

2

PV V Y sin( )

( 1.0 )(1.0 )( 5.2705 )sin(108.4349 0.0 0.0 ) 5

θ δ δδ∂

= − + −∂

= − + −= −

79





33 1 31 31 1 3

3

3 2 23 32 2 3

P V V Y sin( )

V V Y sin( ) ( 1.0 )(1.0 )(15.8114 )sin(108.4349 0.0 0.0 ) ( 1.0 )(1.0 )( 5.2705 )sin(108.4349 0.0 0.0 ) 20

θ δ δδ

θ δ δ

∂= + −

∂+ + −

= + −+ + −

=

80





( )232323323

23 cos δδθ ++=∂∂ YVVVPV

( )( )( ) ( )6667.1

0.00.04349.108cos2705.50.10.1−=

−+=

233 3 3 33 33

32

PV P V Y cosV

0.0 ( 1.0 ) ( 8.2233 )cos( 71.5649 ) 2.9167

θ∂= +

∂

= + −=

81





33 2 32 32 2 3

2

QV V Y cos( )

( 1.0 )(1.0 )( 5.2705 )cos(108.4349 0.0 0.0 ) 1.6667

θ δ δδ∂

= − + −∂

= − + −=

33 1 31 31 1 3

3

3 2 32 32 2 3

Q V V Y cos( )

V Y Y cos( ) ( 1.0 )( 1.0 )( 15.8114 )cos( 108.4349 0.0 0.0 ) ( 1.0 )( 1.0 )( 5.2705 )cos( 108.4349 0.0 0.0 ) 6.6667

θ δ δδ

θ δ δ

∂= + −

∂+ + −

= + −+ + −

= −

82





233 3 3 33 33

32

PV Q V Y sin

0.0 (1.0 ) ( 21.0819 )sin( 71.5649 ) 20

θδ∂

= −∂

= − −=

In Matrix Form,

8.75 5 1.66675 20 2.9167

1.6667 6.6667 20

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

83




Newton-Raphson Load FlowSolving for Gradients,

2

3

3 3

0.2 8.75 5 1.66670.6 5 20 2.91670.25 1.6667 6.6667 20 V /V

∆δ∆δ

∆

− −⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥− = −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦

02 0.003984rad. 0.2283deg∆δ = =

03

3

V 0.02145V∆

= − 03V 0.02145∆ = −

deg4822.1.rad02587.003 −=−=δ∆

84




Newton-Raphson Load FlowUpdate Initial Estimates

1 0 02 2 2δ δ ∆δ= +121 0 03 3 3

131 0 0

3 3 31

3

0.0 0.2283 0.2283

0.0 1.4822 1.4822

V V V

V 1.0 0.02145 0.97855

δ

δ δ ∆δ

δ

∆

= + =

= +

= − = −

= +

= − =

Specified Variables1

1111

2

V 1.0

0.0

V 1.0

δ

=

=

=

85




Newton-Raphson Load FlowUpdate Estimates of Injected Power

12 2 1 21 31 1 3

2 2 22 22

2 3 23 23 3 2

P VVY cos( ) VV Y cos( ) VV Y cos( ) = (1.0)(1.0)(3.9528)cos(108.4349 0.0 0.2283) (1.0)(1.0)(9.2233)cos( 71.5649) (1.0)(0.97855)(5.2705)cos(10

θ δ δθθ δ δ

= + −++ + −

+ −+ −+ 8.4349 1.4822 0.2283)

0.1975− −

=

4822.1+

86





13 3 1 31 31 1 3

3 2 32 32 2 3

3 3 33 33

P V V Y cos( ) V V Y cos( ) V V Y cos( ) = (0.97855 )(1.0 )(15.8114 )cos(108.4349 0.0 1.4822 ) (0.97855 )(1.0 )( 5.2705 )cos(108.4349 0.2283 1.4822 ) (0.97

θ δ δθ δ δθ

= + −

+ + −

++ +

+ + ++ 855 )(0.97855 )( 21.0819 )cos( 71.5650 )

0.66633−

= −

87





[

][

13 3 1 31 31 1 3

3 2 32 32 2 3

3 3 33 33

Q V V Y sin( ) V V Y sin( ) V V Y sin( )

= - (0.97855)(1.0 )(15.8114 )sin(108.4349 0.0 1.4822) (0.97855)(1.0 )(5.2705)sin(108.4349 0.2283 1.4822) (0.

θ δ δθ δ δθ

= − + −

+ + −

+

+ +

+ + +

+ ]97855)(0.97855)( 21.0819 )sin( 71.5650 ) 0.2375

−

= −

88





2 2 ,sp 2 ,ca lc

3 3 ,sp 3 ,ca lc

3 3 ,sp 3 ,ca lc

P P P

0 .2 0 .1 9 7 5 0 .0 0 2 5

P P P

0 .6 0 .6 6 3 3 0 .0 6 3 3

Q Q Q

.0 .2 5 0 .2 3 7 5 0 .0 1 2 5

∆

∆

∆

= −

= −

=

= −

= − +

=

= −

= − +

= Proceed to Iteration 2

89




Newton-Raphson Load FlowEvaluate Elements of Jacobian Matrix

Elements of J1:

22 1 21 21 1 2

2

2 3 23 23 3 2

P VVY sin( )

VVY sin( ) (1.0)(1.0)(3.9528)sin(108.4349 0.0 0.2283) (1.0)(0.97855)(5.2705)sin(108.4349 1.4822 0.2283) 8.6942

θ δ δδ

θ δ δ

∂= + −

∂+ + −

= + −+ − −

=

90




Newton-Raphson Load FlowNewton-Raphson Load FlowElements of J1:

22 3 23 23 3 2

3

P VVY sin( )

(1.0)(0.97855)(5.2705)sin(108.4349 1.4822 0.2283) 4.9393

θ δ δδ∂

=− + −∂

=− − −=−

33 2 32 32 2 3

2

P VV Y sin( )

(0.97855)(1.0)(5.2705)sin(108.4349 0.2283 1.4822) 4.8419

θ δ δδ∂

=− + −∂

=− + +=−

91





33 1 31 31 1 3

3

3 2 23 32 2 3

P V VY sin( )

V V Y sin( ) (0.97855)(1.0 )(15.8114 )sin(108.4349 0.0 1.4822) (0.97855)(1.0 )(5.2705)sin(108.4349 0.2283 1.4822) 19.3887

θ δ δδ

θ δ δ

∂= + −

∂+ + −

= + ++ + +

=

92





( )232323323

23 cos δδθ ++=∂∂ YVVVPV

( )( )( ) ( )4842.1

2283.04822.14349.108cos2705.5097855.00.1−=

−+=

233 3 3 33 33

32

PV P V Y cosV

0.6633 (0.97855 ) ( 21.0819 )cos( 71.5650 ) 5.7205

θ∂= +

∂

= − + −=

93





33 2 32 32 2 3

2

Q VV Y cos( )

(0.97855)(1.0)(5.2705)cos(108.4349 0.2283 1.4822) 1.7762

θ δ δδ∂

=− + −∂

=− + +=3

3 1 31 31 1 33

3 2 32 32 2 3

Q V V Y cos( )

V Y Y cos( ) (0.97855 )(1.0 )(15.8114 )cos(108.4349 0.0 1.4822 ) (0.97855 )(1.0 )( 5.2705 )cos(108.4349 0.2283 1.4822 ) 7.0470

θ δ δδ

θ δ δ

∂= + −

∂

+ + −= + ++ + +

= −

94





233 3 3 33 33

32

PV Q V Y sin

0.2375 (0.97855 ) ( 21.0819 )sin( 71.565 ) 18.9137

θδ∂

= −∂

= − − −=

In Matrix Form,

8.6942 4.9393 1.48424.8419 19.3887 5.7205

1.7762 7.0470 18.9137

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

95




Newton-Raphson Load FlowSolving for Gradients,

2

3

3 3

0.0025 8.6942 4.9393 1.48420.0633 4.8419 19.3887 5.72050.0125 1.7762 7.0470 18.9137 V / V

∆δ∆δ

∆

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

0012 1458.0rad/180 x rad0025.0 == πδ∆

0013 2150.0rad/180 x rad0038.0 == πδ∆

0005.0VV

3

03 =

∆ ( )( ) 0005.097855.00005.0V 13 ==∆

96




Newton-Raphson Load FlowUpdate Previous Estimates

11111

2

V 1.0

0.0

V 1.0

δ

=

=

=

Specified Variables

2 1 12 2 2

0

2 1 13 3 3

0

2 1 13 3 3

0.2283 0.1458 0.3741

1.4822 0.2150 1.2672V V V 0.97855 0.0005 0.9791

δ δ ∆δ

δ δ ∆δ

∆

= +

= + =

= +

= − + = −

= +

= + =

97




Newton-Raphson Load FlowUpdate Previous Estimates of Injected Power

22 2 1 21 21 1 2

2 2 22 22

2 3 23 23 3 2

P V V Y cos( ) V V Y cos( ) V V Y cos( ) = (1.0 )(1.0 )( 3.9528 )cos(108.4349 0.0 0.3741) (1.0 )(1.0 )( 9.2233 )cos( 71.5649 ) (1.0 )(0.9791)( 5.2705 )cos(108

θ δ δθθ δ δ

= + −++ + −

+ −+ −+ .4349 1.2672 0.3741)

0.2018− −

=

98





23 3 1 31 31 1 3

3 2 32 22 2 3 3 3 33 33

P V V Y cos( ) V V Y cos( ) V V Y cos (0.9791)(1.0 )(15.8114 )cos(108.4349 0.0 1.2672 ) ( 0.9791)(1.0 )( 5.2705 )cos(108.4349 0.3741 1.2672 ) ( 0.9791)(0.9791)(

θ δ δθ δ δ θ

= + −+ + − +

= + −+ + ++ 21.0819 )cos( 71.5650 )

0.5995−

= −

99





[]

[

23 3 1 31 31 1 3

3 2 32 32 2 3 3 3 33 33

Q V V Y sin( )

V V Y sin( ) V V Y sin

(0.9791)(1.0 )(15.8114)sin(108.4349 0.0 1.2672) (0.9791)(1.0 )(5.2705)sin(108.4349 0.37411 1.2672) (0.9791)(0.979

θ δ δ

θ δ δ θ

= − + −

+ + − +

= − + −

+ + +

+ ]1)(21.0819)sin( 71.5650) 0.2487

−

= −

100





The solution has converged

2 2 ,s p 2 ,c a lc

3 3 ,s p 3 ,c a lc

3 3 ,s p 3 ,c a lc

P P P

0 .2 0 .2 0 1 8 0 .0 0 1 8

P P P

0 .6 0 .5 9 9 5 0 .0 0 0 5

Q Q Q

.0 .2 5 0 .2 4 8 7 0 .0 0 1 3

∆

∆

∆

= −

= −

=

= −

= − +

=

= −

= − +

=

101




Newton-Raphson Load FlowThe solution of the Load Flow Problem is

03

02

01

2672.19791.0V3741.00.1V

00.1V

−∠=

∠=

∠=

102




The bus voltages are:

Information from a Load Flow Study

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°)P1 - jQ1 = V1

* [Y11V1 + Y12V2 + Y13V3 ]

+ (3.9528∠108.4349°)(0.9990∠0.4129°)

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

103




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


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 = V3* [Y31V1 + Y32V2 + Y33V3 ]

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

104





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]

105




p qypo yqo

ypq VqIpq IlineVp

ppoqppqpolinepqVy )VV(yI II +

Iqp

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

Line Currents


−=+=

qqopqpqqolineqpVy )VV(yI I I +−=+−=

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

106





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

107





Power FLOWS

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

pqppqpqpq IVQj PS *=−=

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

qpqqpqpqp IVQj PS *=−=

108





The line losses are:

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



109





Line Losses

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

qppqlosslosslossSSQj PS +=+=

( ) *pqqp

*pqq

*pqp

IVV

IVIV

+=

−=

110





BASIC INFORMATION

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

111





OTHER INFORMATION

Overvoltage and Undervoltage BusesCritical and Overloaded Transformers and LinesTotal System Losses

112




Backward/Forward Sweep Load Flow

Load Flow for Radial Distribution System

Procedure: Iterative Solution

Initialization

Solving for Injected Currents through the nodes

Backward Sweep

Forward Sweep

Solving for Injected Power

Solving for Voltage Mismatch

113





Bus1Bus2

Bus3

Bus4V1 = 67 kVV2 = ?

V4 = ?

V3 = ?

I23 , Loss23 = ?

I24 , Loss24 = ?

I12 , Loss12 = ?

P1 , Q1 = ?P2 , Q2 = ?

P3 , Q3 = ?

P4 , Q4 = ?

0.635 + j1.970 Ω

0.4223 + j0.7980 Ω0.131 + j1.595 Ω

Utility Grid


Lumped Load B1 MVA 85%PF

Load Flow for Radial Distribution System

114





Equivalent Circuit

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

Base Values

Sbase = 10 MVA

Vbase1 = 67 kV

Vbase2 = 13.2 kV

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

Base Z =13.22/10 =17.424Ω

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458pu

115





Iterative Solution1. Solve Injected Currents by Loads

2. Solve Line Currents (Backward Sweep)

3. Update Voltages (Forward Sweep)

4. Solve for Injected Power

5. Solve for Power MismatchContinue iteration by Backward-Forward Sweep until convergence is achieved

After convergence, solve Iinj, Pinj, Qinj, PF, PLoss, QLoss

116





Initialization

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

Initialize, V1(0) = 1/0

V2(0) = 1/0

V3(0) = 1/0

V4(0) = 1/0

117





Solving for Injected Currents

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

I1(0) = 0

I2(0) = 0

I3(0) = S3* /[V3

(0)]* = __________

I4(0) = S4* /[V4

(0)]* = __________

I1(0) = 0

I2(0) = 0

I3(0) = S3* /[V3

(0)]* = __________

I4(0) = S4* /[V4

(0)]* = __________

Solve Injected Currents by Loads

118





Backward Sweep

Bus1Utility

Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

I24(0) = I4

(0) = _______

I23(0) = I3

(0) = _______

I12(0) = 0 + I23

(0) + I24(0) = _______

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

Solve Line Currents

(Backward Sweep)

119





Forward Sweep

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

Update Voltages

(Forward Sweep)

V1(1) = 1/0

V2(1) = V1

(0) – [I12(0)][Z12] = ________

V3(1) = V2

(1) – [I23(0)][Z23] = ________

V4(1) = V2

(1) – [I24(0)][Z24] = ________

120





Solving for Injected Power

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

1 + j0 pu 0.085+ j0.05267 pu

0.17 + j0.10536 pu

Solve Injected Power

S1(1) = [V1

(1)][I1(0)]* = ___________

S2(1) = [V2

(1)][I2(0)]* = ___________

S3(1) = [V3

(1)][I3(0)]* = ___________

S4(1) = [V4

(1)][I4(0)]* = ___________

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

121





Solving for Power Mismatch

Bus1Utility

Grid

Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Bus2

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

Solve Power Mismatch

∆S1(1) = S1

(sp) - S1(calc) = ____________

∆S2(1) = S2

(sp) – S2(calc) = ____________

∆S3(1) = S3

(sp) – S3(calc) = ____________

∆S4(1) = S4

(sp) – S4(calc) = ____________

0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

122





Iterative Solution

Iteration 2:

I1(1) = 0

I2(1) = 0

I3(1) = S3* /[V3

(1)]* = __________

I4(1) = S4* /[V4

(1)]* = __________


I24(1) = I4

(1) = _______

I23(1) = I3

(1) = _______

I12(1) = 0 + I23

(1) + I24(1) = _______

Solve Line Currents

(Backward Sweep)

123





Iterative Solution

Update Voltages

(Forward Sweep)

V1(2) = 1/0

V2(2) = V1

(1) – [I12(1)][Z12] = ________

V3(2) = V2

(1) – [I23(1)][Z23] = ________

V4(2) = V2

(1) – [I24(1)][Z24] = ________

S1(2) = [V1

(2)][I1(1)]* = ___________

S2(2) = [V2

(2)][I2(1)]* = ___________

S3(2) = [V3

(2)][I3(1)]* = ___________

S4(2) = [V4

(2)][I4(1)]* = ___________


124





Iterative Solution


∆S1(2) = S1

(sp) - S1(calc) = ____________

∆S2(2) = S2

(sp) – S2(calc) = ____________

∆S3(2) = S3

(sp) – S3(calc) = ____________

∆S4(2) = S4

(sp) – S4(calc) = ____________

If Mismatch is higher than set convergence index, repeat the procedure (Backward-Forward Sweep) [Iteration 3]

125





Iterative Solution

Iteration 3:

I1(2) = 0

I2(2) = 0

I3(2) = S3* /[V3

(2)]* = __________

I4(2) = S4* /[V4

(2)]* = __________


I24(2) = I4

(2) = _______

I23(2) = I3

(2) = _______

I12(2) = 0 + I23

(2) + I24(2) = _______

Solve Line Currents

(Backward Sweep)

126





Iterative Solution

Update Voltages

(Forward Sweep)

V1(3) = 1/0

V2(3) = V1

(2) – [I12(2)][Z12] = ________

V3(3) = V2

(2) – [I23(2)][Z23] = ________

V4(3) = V2

(2) – [I24(2)][Z24] = ________

S1(3) = [V1

(3)][I1(2)]* = ___________

S2(3) = [V2

(3)][I2(2)]* = ___________

S3(3) = [V3

(3)][I3(2)]* = ___________

S4(3) = [V4

(3)][I4(2)]* = ___________


127





Iterative Solution


∆S1(3) = S1

(3) - S1(2)

∆S2(3) = S2

(3) – S2(2) = ____________

∆S3(3) = S3

(3) – S3(2) = ____________

∆S4(3) = S4

(3) – S4(2) = ____________

If Mismatch is lower than set convergence index, compute power flows

128




Bus1

Utility Grid

Bus2Bus3

Bus4


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu0.0242+j0.0458

Lumped Load A2 MVA 85%PFVOLTAGE PROFILE

V1 = ________

V2 = ________

V3 = ________

V4 = ________


INJECTED POWER

P1 + jQ1 = ________ + j ________

P2 + jQ2 = ________ + j ________

P3 + jQ3 = ________ + j ________

P4 + jQ4 = ________ + j ________

129




Bus2Bus3

Bus1

Utility Grid


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu0.0242+j0.0458 pu

Bus4POWER FLOW (P-Q)

P12 + jQ12 = ________ + j ________

P23 + jQ23 = ________ + j ________

P24 + jQ24 = ________ + j ________



POWER FLOW (Q-P)

P21 + jQ21 = ________ + j ________

P32 + jQ32 = ________ + j ________

P42 + jQ42 = ________ + j ________

130





Bus2Bus3

Bus1 0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu0.0242+j0.0458 pu

Utility Grid

Bus4Branch Currents Lumped Load A

2 MVA 85%PFI12 = ________

I23 = ________

I24 = ________


POWER LOSSES

I2R12 + jI2X12 = ________ + j ________

I2R23 + jI2X24 = ________ + j ________

I2R24 + jI2X24 = ________ + j ________

131





Line sections in the radial network are ordered by layers away from the root node (substation bus).

1 2 3

4 5 6

7 89 10 1211

13 14 15 1617 18 19 20

21 22 2324 25 26

3130292827

32 33 34

35

Layer 1

Layer 4

Layer 5

Layer 6

Layer 7

Layer 8

8

Layer 2

Layer 3

132




Three-Phase Forward/ Backward Sweep Method

The iterative algorithm for solving the radial system consists of three steps. At iteration k:

( )( )( )

)1k(

ic

ib

ia

*ic

*ib

*ia

)1k(icic

)1k(ibib

)1k(iaia

)k(

ic

ib

ia

VVV

YY

Y

V/SV/SV/S

III −

∗−

∗−

∗−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

Step 1: Nodal current calculation

icibia

icibia

icibia

icibia

Y,Y,YV,V,VS,S,S

I,I,IWhere, Current injections at node i

Scheduled power injections at node i

Voltages at node i

Admittances of all shunt elements at node i

133





Step 2: Backward Sweep to sum up line section current

∑∈ ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

Mm

)k(

mc

mb

ma

)k(

jc

jb

ja)k(

lc

lb

la

JJJ

III

JJJ

Starting from the line section in the last layer and moving towards the root node. The current in the line section l is:

jclbla J,J,JWhere, are the current flows on line section l

M and l Is the set of line sections connected to node j

134





Step 3: Forward Sweep to update nodal voltageStarting from the first layer and moving towards the last layer, the voltage at node j is:

)k(

lc

lb

la

l,ccl,bcl,ac

l,bcl,bbl,ab

l,acl,abl,aa)k(

ic

ib

ia

)k(

jc

jb

ja

JJJ

zzzzzzzzz

VVV

VVV

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

135





After the three steps are executed in one iteration, the power mismatches at each node for all phases are calculated:

( )( )( ) ic

2

ic*

ic)k(

ic)k(

ic)k(

ic

ib

2

ib*

ia)k(

ib)k(

ib)k(

ib

ia

2

ia*

ia)k(

ia)k(

ia)k(

ia

SVYIVS

SVYIVS

SVYIVS

−−=

−−=

−−=

∗

∗

∗

∆

∆

∆

If the real and imaginary part (real and reactive power) of any of these power mismatches is greater than a convergence criterion, steps 1, 2 & 3 are repeated until convergence is achieved.

136




Principles of Load Flow Control

Prime mover and excitation control of generatorsReactive Var Compensation (e.g., Capacitors)Control of tap-changing and voltage regulating transformers

137





Generator Voltage & Power Control

jXjX

IIEEii∠δ∠δ

VVtt∠∠00

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

~ [ ] [ ] ⎥⎦

⎤⎢⎣

⎡ ∠−∠∠=∠=+

jXVEVIVjQP ti

tttt000 * δ

⎥⎦⎤

⎢⎣⎡= δsin

XVEP ti

t ⎥⎦

⎤⎢⎣

⎡−=

XV

XVEQ tti

t

2

cosδ

138






⎥⎦⎤

⎢⎣⎡= δsin

XVEP 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

139






⎥⎦⎤

⎢⎣⎡= δsin

XVEP 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

140





Capacitor Compensation

Ipqp

PL - jQL

q

+ 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

141





Tap-Changing Transformera:1

q r

s p

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

pqya

a2

1−pqy

aa 1−

pqya1

Observation:

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

142




Uses of Load Flow Study

Sensitivity Analysis with Load Flow Study

Analysis of Existing Conditions

Analysis for Correcting PQ Problems

Expansion Planning

Contingency Analysis

System Loss Analysis

143




Uses of Load Flow StudiesSensitivity Analysis with Load Flow Study1) Take any line, transformer or generator out of service.2) Add, reduce or remove load to any or all buses.

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

4) Add new transmission or distribution lines.

5) Increase conductor size on T&D lines.

6) Change bus voltages.

7) Change transformer taps.

8) Increase or decrease transformer size.

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

144




Uses of Load Flow Studies

1) ANALYSIS OF EXISTING CONDITIONS• Check for voltage violations

PGC: 0.95 – 1.05 p.u. (For Transmission)PDC: 0.90 – 1.10 p.u (For Distribution)*

*Recommended 0.95 – 1.05 p.u.• Check for branch power flow violations

Transformer OverloadsLine Overloads

• Check for system lossesCaps on Segregated DSL

145





2) ANALYSIS FOR CORRECTING PQ PROBLEMS

• Voltage adjustment by utility at delivery pointRequest TransCo to improve voltage at connection pointTransCo as System Operator will determine feasibility based on Economic Dispatch and other adjustments such as transformer tap changing and reactive power compensation

146






• Transformer tap changingAvailable Taps

At Primary SideAt Secondary SideBoth Sides

Typical Taps Tap 1: +5%Tap 2: +2.5%Tap 3: 0% (Rated Voltage)Tap 4: -2.5%Tap 5: -5%

147






•Capacitor compensation• Compensate for Peak Loading• Check overvoltages during Off-Peak• Optimize Capacitor Plan

• System configuration improvement

148





3) EXPANSION PLANNING• New substation construction• Substation capacity expansion• New feeder segment construction / extension• Addition of parallel feeder segment• Reconducting of existing feeder segment/ circuit • Circuit conversion to higher voltage• Generator addition

149





4) CONTINGENCY ANALYSIS

Reliability analysis of the Transmission (Grid) and Subtransmission System

5) SYSTEM LOSS ANALYSIS

Segregation of System Losses

150




Load Flow Analysissmart.tabrizu.ac.ir/Files/Content/Smart/Load-Flow-Analysis-6th-Batch.pdf · Training Course in Load Flow Analysis The Load Flow Problem Load Flow Analysis simulates

Documents