Load Flow Analysis

NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER

Certificate in

Power System Modeling and Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

U. P. NATIONAL ENGINEERING CENTER

Load Flow Analysis

Training Course in

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




� 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

The Load Flow Problem

4





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)*

Basic Electrical Engineering Solution

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 A

7621.02 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





Sending End

Receiving End

VS = 13.2 kVLL

Load2 MVA, 3Ph

85%PF

VR = ?

Load Flow From the Real World


ISR = ?

How do you solve for:

1) ISR = ?

2) VD = ?

3) VR = ?

4) SS = ?

7





Load Flow of Distribution System

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

Bus1

Utility Grid

Bus2Bus3

Bus4V1 = 67 kV

Lumped Load A2 MVA 85%PF

Lumped Load B1 MVA 85%PF

V2 = ?

V4 = ?

V3 = ?I23 , Loss23 = ?

I24 , Loss24 = ?

I12 , Loss12 = ?

P1 , Q1 = ?P2 , Q2 = ?

P3 , Q3 = ?

P4 , Q4 = ?

8





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?

Load Flow of Transmission and Subtransmission System

9





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

Load Flow of a Contemplated System

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

Load2 MVA, 3Ph 85%PFVR = ?

Load Flow of a Single Line


ISR = ?

SR = VR x (ISR)*

Injected Power at Receiving End

VS = VR + Z x ISR

Voltage at Sending End

VR = VS - Z x SR*/VR*

Voltage at Receiving End

ISR = (SR / VR)*

Solving for the Current

12





Sending End

Receiving End

VS = 13.2 kVLL



ISR = ?

Converting Quantities in Per Unit

Base Power = 1 MVA

Base Voltage = 13.2 kV

Base Impedance = [13.2]2/1

= 174.24 ohms

VS(pu) = 13.2 /13.2 = 1/0

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

Zpu = (1.1034 + j2.0856)/174.24

= 0.00633 + j0.01197


13





Sending End

Receiving End

VS = 13.2 kVLL



ISR = ?

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

(k-1) ]*

Let VR(0) = 1/0

For k = 1VR

(1) = __________

ΔV(1) = __________

For k = 2VR

(2) = __________

ΔV(2) = __________


14





Sending End

Receiving End

VS = 13.2 kVLL



ISR = ?

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

(k-1) ]*

For k = 3VR

(3) = __________

ΔV(3) = __________

For k = 4VR

(4) = __________

ΔV(4) = __________

VR(2) = __________


15





Sending End

Receiving End

VS = 13.2 kVLL



ISR = ?

VS = __________

VR = __________

VD = VS – VR

VD = __________

ISR = __________

SR = __________

SS = VS x [ISR]*

SS = __________


16




� Bus Admittance Matrix, Ybus

� Network Models

� Generator Models

� Bus Types for Load Flow Analysis

Power System Models for Load flow Analysis

17




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.

Power System Models for Load Flow Analysis

18




⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

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


19




Line 1Line 1


11 22

33

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

Set-up the Ybus

Network Models


20




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

Network Models


21




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°


22




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°

Y33 = y13 + y23 = (5 - j15) + (1.6667 - j5)= 6.6667 - j20 = 21.0819 ∠ -71.5650°


23





� 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

24




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


� Bus voltage magnitude, Vp� Bus voltage phase angle, δδδδp� Bus injected active power, Pp� Bus injected reactive power, Qp

Bus Types for Load Flow

� Generators and loads are connected from bus to neutral.

25




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∠++

--


26




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∠++

--

Type 2: Generator Bus

Specify: P, V

Unknown: Q, δ


27




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,QP,Q

++

--

Type 3: Load Bus

Specify: P, Q

Unknown: V, δδV∠


28




SUMMARY OF BUS TYPES

BB uu ss TT yy pp ee

KK nn oo ww nn QQ uu aa nn tt ii tt 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


29




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


30




� Linear Formulation of Load Flow Equations

� Gauss-Seidel Load Flow Solution

� Numerical Example

Gauss-Seidel Load Flow

31




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(1)

Linear Formulation of Load Flow Equations


32




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


33




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)


34




Solving for Vp in (4)

_______Y11V1 = P1 – jQ1

V1*

- (___ + Y12V2 + Y13V3)

⎥⎦

⎤⎢⎣

⎡−−

−= 313212*

1

11

11

1 VYVYV

jQP

Y

1V

_______Y22V2 = P2 – jQ2

V2*

- (Y12V2 + ___ + Y13V3)

⎥⎦

⎤⎢⎣

⎡−−

−= 313121*

2

22

22

2 VYVYV

jQP

Y

1V


35




⎥⎦

⎤⎢⎣

⎡−−

−= 232131*

3

33

33

3 VYVYV

jQP

Y

1V

_______Y33V3 = P3 – jQ3

V3*

- (Y13V1 + Y23V2 + ___)

(5)Vp =

1___Ypp

_______

Vp*

Pp - jQp Σ-n

q=1q≠≠≠≠p

YpqVq


36




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

iteration is:

Vpk+1 =

1___Ypp

_______

(Vpk)*

Pp - jQp Σ-n

q=1q≠≠≠≠p

YpqVqαααα

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

Gauss-Seidel Load Flow Solution

(6)


37




Gauss-Seidel Load Flow� 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.

38




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


39




Branch Data


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



40




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


41




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°

Y33 = 6.6667 - j20 = 21.0819 ∠ -71.5650°

Y31 = -5 + j15 = 15.8114 ∠ 108.4349°Y32 = -1.6667 + j5 = 5.2705 ∠ 108.4349°


42




Gauss-Seidel EquationsBus 1: Swing Bus

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

jQP

Y

1V

( ) 01V 1k1 ∠=+


43




Bus 3: Load Bus

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQP

Y

1V

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.]


44




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°)

= 1.0071∠1.1705°

22Y

( )1k22 jQP +−

( )( )*k2V

21Y

23Y

( )1k1V +

( )k3V

( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQP

Y

1V


45




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°)

= 0.9816 ∠-1.0570°

33Y31Y

32Y

33 jQP −

( )*k3V

( )1k1V +

( )1k2V +

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQP

Y

1V


46




ΔV2 = V2(1) - V2

(0)

= 1.0071∠1.1705° - 1.0∠0°

⏐ΔV2⏐ = 0.0217

ΔV3 = V3(1) - V3

(0)

= 0.9816∠-1.0570° - 1.0∠0°

⏐ΔV3⏐ = 0.0259


47




Iteration 2 (k = 1): V1(2) = 1.0∠0°

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.]

Let, V2(1) = 1.0∠1.1705°


48




V2(2) =

1___________________9.2233 ∠ -71.5650

0.2 - j0.0073

1.0 ∠ -1.1705°______________

- (-1.25 +j3.75) (1.0 ∠ 0°)

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

= 0.9966 ∠ 0.5819°

( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQP

Y

1V


49




V3(2) =

1_____________________21.0819 ∠ -71.5650

-0.6 + j0.25

0.9816 ∠ 1.0570°___________________

- (-5 +j15) (1.0 ∠ 0°)

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

= 0.9783 ∠ -1.2166°

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQP

Y

1V


50




= 0.9966 ∠ 0.5819° - 1.0071 ∠ 1.1705°

⏐ΔV2⏐ = 0.0125

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

⏐ΔV3⏐ = 0.004

ΔV2 = V2(2) - V2

(1)

ΔV3 = V3(2) - V3

(1)


51




Iteration 3 (k = 2):

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.]

Let, V22 = 1.0 ∠ 0.5819°

V1(2) = 1.0∠0°


52




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°

( )( )

( )( )( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +

++ k

3231k

121*k2

1k22

22

1k2 VYVY

V

jQP

Y

1V


53




V33 =

1_____________________21.0819 ∠-71.5650

-0.6 + j0.25

0.9783 ∠ 1.2166°

___________________

- (-5 +j15)(1.0∠0°)

- (5.2705∠108.4349°)(0.9990∠0.4129°)

= 0.9788∠-1.2560°

( )( )( )

( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

−= +++ 1k

2321k

131*k3

33

33

1k3 VYVY

V

jQP

Y

1V


54




= 0.9990∠0.4129° - 1.0∠0.5819°

⏐ΔV2⏐ = 0.003 < 0.005

= 0.9788∠-1.2560° - 0.9783∠-1.2166°

⏐ΔV3⏐ = 0.0008 < 0.005

The solution has converged.

ΔV2 = V2(3) - V2

(2)

ΔV3 = V3(3) - V3

(2)


55




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 ]


56




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 ]


57




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]


58




p qypo yqo

ypqIpq IqpIlineVp 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


59




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


60




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

pqppqpqpq IVQj PS *=−=

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

Power FLOWS

qpqqpqpqp IVQj PS *=−=


61




The line losses are:

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




62




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

qppqlosslosslossSSQj PS ++++====++++====

Line Losses


( ) *

pqqp

*

pqq

*

pqp

IVV

IVIV

+=

−=

63




SUMMARY OF BASIC INFORMATION

Voltage Profile

Injected Power (Pp and Qp)

Line Currents (Ipq and Ipq)

Power Flows (Ppq and Qpq)

Line Losses (I2R and I2X)


64




OTHER INFORMATION

Overvoltage and Undervoltage Buses

Critical and Overloaded Transformers and Lines

Total System Losses


65




� Non-Linear Formulation of Load Flow Equations

� Newton-Raphson Load Flow Solution

� Numerical Example

Newton-Raphson Load Flow

66





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

n

p pq qq 1

I Y E=

= ∑Substituting (2) into (1)

(1)

(2)

(3)q

n

1qpq

*

ppp EYEjQP ∑=

•=−

67




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

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=

= − + −∑

(4)

(5)

(6)

Separating the real and imaginary components

68




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

69




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=

(7)M M

M M

70




Equation (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 ...

Δδ Δδ Δδ Δ Δ Δδ δ δ


Δδ Δδδ 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

Δδ Δ Δ Δδ


Δδ Δδ Δδ Δ Δδ 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


∂ ∂ ∂ ∂ ∂ ∂= + + + + + + + +

∂ ∂ ∂ ∂ ∂ ∂

M M M M M M M M M

M M M M M M M M M


71




1 1 1 1 1 1s p e c c a lc1 1

1 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 P

V V

P P P P P P

V V V

Q Q Q Q Q Q

V V V

Q Q Q Q Q Q

V V V

Q Q Q Q Q Q

V 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 δ

Δ δ

Δ

Δ

Δ

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

M

M

72




Newton-Raphson Load Flowor simply

⎥⎦

⎤⎢⎣

⎡ΔΔ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡ΔΔ

VV

QQV

PP

Q

P δ

δ

δ

⎥⎥⎦

⎤

⎢⎢⎣

⎡ΔΔ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡ΔΔ

VV

V

QV

QV

PV

P

Q

P δ

δ

δ

73




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

= ≠

∂⎧= + −⎪ ∂⎪

⎨ ∂⎪ = − + −⎪ ∂⎩

∑

1 2

3 4

J JP

VJ JQ V

ΔδΔ

ΔΔ

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

Newton-Raphson Load Flow Solution


74




p 2p p p p p p p

p

2p

q p q p q p q q pq

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

= ≠

∂⎧= +⎪ ∂⎪

⎨ ∂⎪ = + −⎪ ∂⎩∂⎧

= + −⎪ ∂⎪⎨ ∂⎪ = − + −⎪ ∂⎩

∑


75




p 2p p p p p p q

p

4p

q p q p q p q q pq

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.


76




At the kth iteration,

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

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

δ δ Δδ

Δ

+

+

= +

= +

The process is terminated once convergence is achieved whrein

( k ) ( k )p qMAX P and MAX QΔ ε Δ ε≤ ≤


77




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


78




Branch Data


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



79




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 71.5651

Y 1,25 j3.75 3.9528 108.4349

Y 5 j15 15.8114 108.4349

Y 1.25 j .375 3.9528 108.4349

Y 2.9167 j8.75 9.2233 71.5649

Y 1.6667 j5 5.2705 108.4349

Y 5 j15 15.811

= − = ∠ −= − + = ∠= − + = ∠= − + = ∠= − = ∠ −= − + = ∠= − + =

32

33

4 108.4349

Y 1.6667 j5 5.2705 108.4349

Y 6.6667 j 20 21.0819 71.5650

∠= − + = ∠= − = ∠ −

80




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 V

V

P P PP V

V

Q Q Q VQ V

V V

Δ Δδδ δ

Δ Δδδ δ

ΔΔδ δ

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

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

Bus 1: Swing Bus (Not included)

Bus 2: Generator Bus (Compute for P2)

Bus 3: Load Bus (Compute for P2 and Q2)


81




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


82




Compute 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

= + −+ + + −

= + ++ −+

=


83




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=


84




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=


85




Compute Power Mismatch

02P 0 .2 0 .0 0 .2Δ = − =

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

Evaluate elements of Jacobian Matrix

P PV

JQ Q

VV

δ δ

δ

∂ ∂⎡ ⎤⎢ ⎥∂ ∂⎢ ⎥=⎢ ⎥∂ ∂⎢ ⎥⎢ ⎥∂ ∂⎣ ⎦


86




Elements of J1:

22 1 21 21 1 2

2

2 3 23 23 3 2

PV 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

∂= + −

∂+ + −

= + −+ + −

=


87




Elements 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

∂= − + −

∂= − + −= −


88




Elements of J1:

33 1 31 31 1 3

3

3 2 23 32 2 3

PV 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

∂= + −

∂+ + −

= + −+ + −

=


89




Elements of J2:

233 3 3 33 33

3

2

PV P V Y cos

V

0.0 ( 1.0 ) ( 8.2233 )cos( 71.5649 )

2.9167

θ∂= +

∂

= + −=

( )232323323

23 cos δδθ ++=

∂∂

YVVV

PV

( )( )( ) ( )6667.1

0.00.04349.108cos2705.50.10.1

−=−+=


90




Elements of J3:3

3 2 32 32 2 32

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

QV 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

∂= + −

∂+ + −

= + −+ + −

= −


91




Elements of J4:

8.75 5 1.6667

5 20 2.9167

1.6667 6.6667 20

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦

In Matrix Form,

33332

333

33 sinθYVQ

V

QV −=

∂∂

( ) ( ) ( )20

5649.71sin0819.210.10.0 2

=−−=


92




Solving for Gradients,

2

3

3 3

0.2 8.75 5 1.6667

0.6 5 20 2.9167

0.25 1.6667 6.6667 20 V /V

ΔδΔδ

Δ

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

02 0.003984rad . 0.2283degΔδ = =

03

3

V0.02145

V

Δ= − 0

3V 0.02145Δ = −

deg4822.1.rad02587.00

3 −=−=δΔ


93




Update Initial Estimates

11

11

12

V 1.0

0.0

V 1.0

δ

=

=

=

1 0 02 2 2δ δ Δδ= +12

1 0 03 3 3

13

1 0 03 3 3

13

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 Δδ

δ

Δ

= + =

= +

= − = −

= +

= − =

Specified Variables


94




Update Estimates of Injected Power

( )( )( )

YVV YVV

YVVP

23232332

222222

313121121

2

coscos

cos

δδθθ

δδθ

−+++

−+=

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

0.1975

=−−+

−+++=

2283.04822.14349.108cos2705.597855.00.15649.71cos2233.90.10.1

4822.10.04349.108cos9528.30.10.1


95





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

−= −


96





[

][

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

−

= −


97





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

Δ

Δ

Δ

= −

= −

=

= −

= − +

=

= −

= − +

= Proceed to Iteration 2


98




Evaluate Elements of Jacobian Matrix

Elements of J1:

22 1 21 21 1 2

2

2 3 23 23 3 2

PVVY 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

∂= + −

∂+ + −

= + −+ − −

=


99




Elements of J1:

22 3 23 23 3 2

3

PVVY 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

PVV Y sin( )

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

4.8419

θ δ δδ

∂= − + −

∂= − + +

= −


100




Elements of J1:

33 1 31 31 1 3

3

3 2 23 32 2 3

PV 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

∂= + −

∂+ + −

= + ++ + +

=


101




Elements of J2:

233 3 3 33 33

3

2

PV P V Y cos

V

0.6633 (0.97855 ) ( 21.0819 )cos( 71.5650 )

5.7205

θ∂= +

∂

= − + −=

( )232323323

23 cos δδθ ++=

∂∂

YVVV

PV

( )( )( ) ( )4842.1

2283.04822.14349.108cos2705.5097855.00.1

−=−+=


102




Elements of J3:3

3 2 32 32 2 32

QVV 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 3

3

3 2 32 32 2 3

QV 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

∂= + −

∂+ + −

= + ++ + +

= −


103




Elements of J4:

In Matrix Form,

233 3 3 33 33

3

2

PV Q V Y sin

0.2375 (0.97855 ) ( 21.0819 )sin( 71.565 )

18.9137

θδ

∂= −

∂

= − − −=

8.6942 4.9393 1.4842

4.8419 19.3887 5.7205

1.7762 7.0470 18.9137

− −⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−⎣ ⎦


104




Solving for Gradients,

2

3

3 3

0.0025 8.6942 4.9393 1.4842

0.0633 4.8419 19.3887 5.7205

0.0125 1.7762 7.0470 18.9137 V / V

ΔδΔδ

Δ

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

001

2 1458.0rad/180 x rad0025.0 == πδΔ001

3 2150.0rad/180 x rad0038.0 == πδΔ

0005.0V

V

3

0

3 =Δ ( )( ) 0005.097855.00005.0V 1

3 ==Δ


105




Update Previous Estimates

11

11

12

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 1.2672

V V V

0.97855 0.0005 0.9791

δ δ Δδ

δ δ Δδ

Δ

= +

= + =

= +

= − + = −

= += + =


106




Update 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

− −=


107





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

−= −


108





[]

[

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

−

= −


109





The solution has converged

2 2 ,sp 2 ,c a lc

3 3 ,sp 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

Δ

Δ

Δ

= −

= −

=

= −

= − +

=

= −

= − +

=


110




The solution of the Load Flow Problem is

0

3

0

2

0

1

2672.19791.0V

3741.00.1V

00.1V

−∠=

∠=

∠=


111




� 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

Backward/Forward Sweep Load Flow

112





Bus1

Utility Grid

Bus2Bus3

Bus4V1 = 67 kV



V2 = ?

V4 = ?

V3 = ?

I23 , Loss23 = ?

I24 , Loss24 = ?

I12 , Loss12 = ?

P1 , Q1 = ?P2 , Q2 = ?

P3 , Q3 = ?

P4 , Q4 = ?

Load Flow for Radial Distribution System

0.635 + j1.970 ΩΩΩΩ

0.4223 + j0.7980 ΩΩΩΩ

0.131 + j1.595 ΩΩΩΩ

113





Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Equivalent Circuit

~

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

114





Iterative Solution

Continue iteration by Backward-Forward Sweep until convergence is achieved

1. 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 Mismatch

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

115




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Initialization

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

V1(0) = 1/0

V2(0) = 1/0

V3(0) = 1/0

V4(0) = 1/0

Initialize,


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

116




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Solving for Injected Currents

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

I1(0) = 0

I2(0) = 0

I3(0) = S3* /[V3

(0)]* = __________

I4(0) = S4* /[V4

(0)]* = __________

Solve Injected Currents by Loads

I1(0) = 0

I2(0) = 0

I3(0) = S3* /[V3

(0)]* = __________

I4(0) = S4* /[V4

(0)]* = __________


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

117




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Backward Sweep

~

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) = _______

Solve Line Currents

(Backward Sweep)


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

118




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Forward Sweep

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 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] = ________


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

119




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Solving for Injected Power

~

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

120




Bus1

Utility Grid

Bus2Bus3

Bus4

V1 = 67 kV

V2

V4

V3

Solving for Power Mismatch

~

V1

1 + j0 pu 0.085 + j0.05267 pu

0.17 + j0.10536 pu

Solve Power Mismatch

ΔS1(1) = S1

(sp) - S1(calc) = ____________

ΔS2(1) = S2

(sp) – S2(calc) = ____________

ΔS3(1) = S3

(sp) – S3(calc) = ____________

ΔS4(1) = S4

(sp) – S4(calc) = ____________


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu

0.0242+j0.0458 pu

121




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)


122




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)]* = ___________


123




Iterative Solution


ΔS1(2) = S1

(sp) - S1(calc) = ____________

ΔS2(2) = S2

(sp) – S2(calc) = ____________

ΔS3(2) = S3

(sp) – S3(calc) = ____________

ΔS4(2) = S4

(sp) – S4(calc) = ____________

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


124




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)


125




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)]* = ___________


126




Iterative Solution


ΔS1(3) = S1

(3) - S1(2)

ΔS2(3) = S2

(3) – S2(2) = ____________

ΔS3(3) = S3

(3) – S3(2) = ____________

ΔS4(3) = S4

(3) – S4(2) = ____________

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


127




Bus1

Utility Grid

Bus2Bus3

Bus4



VOLTAGE PROFILE

V1 = ________

V2 = ________

V3 = ________

V4 = ________

INJECTED POWER

P1 + jQ1 = ________ + j ________

P2 + jQ2 = ________ + j ________

P3 + jQ3 = ________ + j ________

P4 + jQ4 = ________ + j ________


0.0364 +j 0.1131 pu

0.0075+j 0.0915 pu0.0242+j0.0458

128




Bus1

Utility Grid

Bus2Bus3

Bus4



POWER 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 ________


0.0364 +j 0.1131 pu

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

129




Bus1

Utility Grid

Bus2Bus3

Bus4



Branch Currents

I12 = ________

I23 = ________

I24 = ________ POWER LOSSES

I2R12 + jI2X12 = ________ + j ________

I2R23 + jI2X24 = ________ + j ________

I2R24 + jI2X24 = ________ + j ________


0.0364 +j 0.1131 pu

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

130




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 1516

1718

1920

21 22 2324 25 26

3130292827

32 33 34

35

Layer 1

Layer 2

Layer 3

Layer 4

Layer 5

Layer 6

Layer 7

Layer 8


8

131




Three-Phase Forward/ Backward Sweep Method

( )( )( )

)1k(

ic

ib

ia

*ic

*ib

*ia

)1k(icic

)1k(ibib

)1k(iaia

)k(

ic

ib

ia

V

V

V

Y

Y

Y

V/S

V/S

V/S

I

I

I−

∗−

∗−

∗−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

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

Step 1: Nodal current calculation

icibia

icibia

icibia

icibia

Y,Y,Y

V,V,V

S,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

132





∑∈ ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

Mm

)k(

mc

mb

ma

)k(

jc

jb

ja

)k(

lc

lb

la

J

J

J

I

I

I

J

J

J

Step 2: Backward Sweep to sum up line section currentStarting 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

Is the set of line sections connected to node jMand l

133





)k(

lc

lb

la

l,ccl,bcl,ac

l,bcl,bbl,ab

l,acl,abl,aa

)k(

ic

ib

ia

)k(

jc

jb

ja

J

J

J

zzz

zzz

zzz

V

V

V

V

V

V

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

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

134





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

−−=

−−=

−−=

∗

∗

∗

Δ

Δ

Δ

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.

135




� Prime mover and excitation control of generators

� Reactive Var Compensation (e.g., Capacitors)

� Control of tap-changing and voltage regulating transformers

Principles of Load Flow Control

136




II

jXjX

EEii∠δ∠δ VVtt∠∠00

Generator Voltage & Power Control

~

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

[ ] [ ] ⎥⎦

⎤⎢⎣

⎡ ∠−∠∠=∠=+

jX

VEVIVjQP ti

tttt

000 * δ

⎥⎦⎤

⎢⎣⎡= δsin

X

VEP ti

t ⎥⎦

⎤⎢⎣

⎡−=

X

V

X

VEQ tti

t

2

cosδ


137





⎥⎦⎤

⎢⎣⎡= δsin

X

VEP ti

t ⎥⎦

⎤⎢⎣

⎡−=

X

V

X

VEQ 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


138





⎥⎦⎤

⎢⎣⎡= δsin

X

VEP ti

t ⎥⎦

⎤⎢⎣

⎡−=

X

V

X

VEQ 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


139




Capacitor Compensation

Ipqp

q

PL - jQL

+ jQc

~

p

qpq

p

qpqpq V

PXj

V

QXVE −−=

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


140




Tap-Changing Transformera:1

q r

s p

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

pqya

a2

1−pqy

a

a 1−

pqya

1

Observation:

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


141




� Sensitivity Analysis with Load Flow Study

� Analysis of Existing Conditions

� Analysis for Correcting PQ Problems

� Expansion Planning

� Contingency Analysis

� System Loss Analysis

Uses of Load Flow Study

142




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

143




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 Overloads� Line Overloads

• Check for system losses� Caps on Segregated DSL


144




2) ANALYSIS FOR CORRECTING PQ PROBLEMS

• Voltage adjustment by utility at delivery point� Request TransCo to improve voltage at

connection point� TransCo as System Operator will determine

feasibility based on Economic Dispatch and other adjustments such as transformer tap changing and reactive power compensation


145




• Transformer tap changing� Available Taps

� At Primary Side� At Secondary Side� Both Sides

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



146




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

• System configuration improvement



147




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


148




4) CONTINGENCY ANALYSIS

Reliability analysis of the Transmission (Grid) and Subtransmission System

5) SYSTEM LOSS ANALYSIS

Segregation of System Losses


149




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Load Flow Analysis

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