CPT3 - Load Flow Analysis-6th Batch

Competency Training and Certification Program in Electric Power Distribution System Engineering

Certificate in

Power System Modeling and Analysis T r a i n i n g Co Co u r s e i n

L o a d Fl Fl o w A n a l y s i s

U. P. NATIONAL ENGINEERING CENTER NATIONAL ELECTRIFICATION ADMINISTRATION

Training Course in Load Flow Analysis

2

Co u r s e O u t l i n e 1. The Lo Load Flo low w Pro Prob blem 2. Powe Powerr Sys Syste tem m Mod Model els s for for Lo Load ad Fl Flow ow Analysis 3. Gau auss ss-S -Sei eide dell Lo Load ad Flo low w 4. Newton-Raphson Load Flow 5. Ba Back ckwa ward rd/F /For orwa ward rd Sw Swee eep p Load Load Flo Flow w 6. Pr Prin inci cipl ples es of Lo Load ad Fl Flow ow Co Cont ntro roll 7. Use ses s of of Loa Load d Flo Flow w St Stud udie ies s U. P. National Engineering Center

Competency Training & Certification Program in


3

T h e Lo L o a d Fl Fl o w Pr o b l e m 

Basic Electrical Engineering Solution



Load Flow of Distribution System



Load Flow of Transmission and Subt Su btra rans nsmi miss ssio ion n Sy Syst stem em



Load Flow of a Contemplated System



Load Flow of a Single Line

U. P. National Engineering Center



4

T h e L o a d Fl Fl o w Pr o b l e m B a si s i c El El e ct c t r i ca c a l En En g i n e er e r i n g So l u t i o n H ow d o y o u d e t e r m i n e t h e v ol t a g e, cu r r e n t , p o w er , a n d p o w e r f a ct c t o r a t v a r i o u s p o i n t s i n a p o w e r s y st st e m ? Sending End

Line

1.1034 + j2.0856 ohms/phase

Receiving End

Solve for: VS = ?

ISR = ?

VR = 13.2 kVLL

VOLTAGE DROP D ROP = V S - V R

Load 2 MVA, 3Ph 85%PF

1) ISR = (SR /VR )* 2) VD = ISRZL 3) VS = VR + VD 4) SS = VSx(ISR)*




5

T h e L o a d Fl Fl o w Pr o b l e m Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = ?

S1φ

V R

ISR = ?

Receiving End

1) ISR = (SR /VR )*

VR = 13.2 kVLL

= ( 2 ,000 ,000 / 3 )∠ cos−1 ( 0.85 ) = 666 ,666.67 ∠31.79 VA VA

= ( 13, 200 /

3 )∠0

= 7621.02∠0

Solve for:


2) VD = ISRZL 3) VS = VR + VD 4) SS = VSx(ISR)*

V

∗

I

⎛ 666 ,666.67 ∠31.79 ⎞ 8=7.48 ∠3−1.79 A = SR ⎜ ⎟ 7621.02∠0 ⎝ ⎠ VD = ( 87.48∠ − 31.79 ) ( 1.1034 + j 2.0856 ) = 178.15 + j104.23 VS = (7621.02 + j0 ) + ( 178.15 + j104.23 ) = 7,799.87∠0.77 V VS

= 7,799.87 ∠0.77 /1000*

3

= 13.51


V

k V Competency Training & Certification Program in


6

T h e L o a d Fl Fl o w Pr o b l e m L o a d Fl F l o w F r o m t h e Re Re a l W o r l d Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

How do you solve sol ve for: f or: 1 ) I SR = ?

Receiving End

VR = ? Load 2 MVA, 3Ph 85%PF

2) VD = ? 3) VR = ? 4 ) SS = ? U. P. National Engineering Center



7

T h e L o a d Fl Fl o w Pr o b l e m L o a d Fl F l o w o f D i st s t r i b u t i o n Sy Sy s t e m Bus2

Utility Grid

I24 , Loss24 = ?

V4 = ? P4 , Q4 = ? Bus4

V1 = 67 kV P 1 , Q1 = ?

Bus3 V3 = ?

I23 , Loss23 = ?

Bus1 I12 , Loss12 = ?

V2 = ?

Lumped Load A 2 MVA 85%PF

P 2 , Q2 = ?

How do you y ou solve solv e for the Voltages, Currents, P ow er and Losses? U. P. National Engineering Center

P 3 , Q3 = ?

Lumped Load B 1 MVA 85%PF



8

T h e L o a d Fl Fl o w Pr o b l e m L o a d Fl F l o w o f Tr T r a n s m i ss ssi o n a n d S u b t r a n sm s m i ss s s i o n S y st st em G

G

Line 1

2

1

How do you y ou solve solv e for the Voltages, Curre urrents nts and a nd P ow er o f a LOOP power system?


Line 2

Line 3 3



9

T h e L o a d Fl Fl o w Pr o b l e m L o a d Fl F l o w o f a Co Co n t e m p l a t e d Sy Sy s t e m How about if there are contemplated changes in the System? How will will you you determine determine in advance advance the effects effects of: of: • Growth or addition of new loads • Addition of generating plants • Upgrading of Substation • Expansion of distribution lines before the proposed changes are implemented?

F L O W A N A L YS YSI S Answer: L O A D FL U. P. National Engineering Center



10

T h e L o a d Fl Fl o w Pr o b l e m 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

T h e L o a d Fl Fl o w Pr o b l e m L o a d Fl F l o w o f a Si S i n g l e Li Li n e Sending End

Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?

Injected Power at Receiving End SR = VR x (ISR)*

Solving for the Current ISR = (SR / VR)* U. P. National Engineering Center


Voltage at Sending End VS = VR + Z x ISR

Voltage at Receiving End VR = VS - Z x SR*/VR* Competency Training & Certification Program in


12


Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?


Converting Quantities in Per Unit Base Power = 1 MVA

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

Base Voltage = 13.2 kV

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

Base Impedance = [13.2]2 /1

Zpu = (1.1034 + j2.0856)/174.24

= 174.24 ohms U. P. National Engineering Center

= 0.00633 + j0.01197 Competency Training & Certification Program in


13


Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?


VR(k) = VS - Z x [SR]* / [VR(k-1) ]* Let

VR(0) = 1/0

For k = 1

For k = 2

VR(1) = __________

VR(2) = __________

∆V(1) = __________

∆V(2) = __________




14


Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?


VR(k) = VS - Z x [SR]* / [VR(k-1) ]* VR(2) = __________ For k = 3

For k = 4

VR(3) = __________

VR(4) = __________

∆V(3) = __________

∆V(4) = __________




15


Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?


VS = __________

ISR = __________

VR = __________

SR = __________

VD = VS – VR

SS = VS x [ISR]*

VD = __________

SS = __________




16


Line

1.1034 + j2.0856 ohms/phase

VS = 13.2 kVLL

ISR = ?

Receiving End

VR = ?


PFR = PR / SR

SLoss = PLoss + QLoss

PFR = _________

SLoss = SS - SR

PFS = PS / SS

PLoss = _________

PFS = _________

QLoss = _________




17

P o w e r Sy s t e m M o d e l s f o r Lo a d f l o w A n a ly s i s 

Bus Admittance Matrix, Ybus



Network Models



Generator Models



Bus Types for Load Flow Analysis




18

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s 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

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s N et e t w o r k M o d el el s The static components (transformers and lines) are represented by the bus admittance matrix, Y bus

[YBUS] =

Y 11

Y 12

Y 13

L Y 1 n

Y 21

Y 22

Y 23

L Y 2 n

Y 31

Y 32

Y 33

L Y 3 n

M

M

Y n 1

M

Y n 2

M

Y n 3 L Y nn

The number of buses (excluding the neutral bus) determines the dimension of the bus admittance, Ybus. U. P. National Engineering Center



20

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s N et e t w o r k M o d el el s Lin Li n e N o . B us Co Cod d e Imp mpe e d a n c e Z p q (p.u.) 1 2 3

1 -2 1 -3 2 -3

0.08 + j0.24 0.02 + j0.06 0.06 + j0.18

Line 1 2

1 Line 2

Line 3

Set-up the Ybus

3 U. P. National Engineering Center



21

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s N et e t w o r k M o d el el s Compute the branch admittances to set up Y b u s : 1 1 y12 = ____ = ______________ = 1.25 - j3.75 z12 0.08 + j0.24 1 1 y13 = ____ = ______________ = 5 - j15 z13 0.02 + j0.06 1 1 y23 = ____ = ______________ = 1.667 - j5 z23 0.06 + j0.18 U. P. National Engineering Center



22

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s Set-up the bus admittance matrix: Y11 = y12 + y13 = (1.25 - j3.75) + (5 - j15) = 6.2 6.25 5 - j1 j18. 8.75 75 = 19. 19.76 7642 42 ∠ -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° U. P. National Engineering Center



23

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s Y22 = y12 + y23 = (1 (1..25 - j3.7 .75 5) + (1 (1.6 .66 667 - j5) = 2.9 2.916 167 7 - j8 j8.7 .75 5 = 9. 9.22 2233 33 ∠ -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)

= 6.6 6.666 667 7 - j2 j20 0 = 21 21.0 .081 819 9 ∠ -71.5650° U. P. National Engineering Center



24

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s Ge n e r a t o r M o d e l s 



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 S wing generator U. P. National Engineering Center



25

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s B u s Ty T y p e s f o r L o a d Fl o w 

Generators and loads are connected from bus to neutral.

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

Bus Bus Bus Bus

voltage magnitu de, V p voltage phase angle, p in jected active ac tive pow er, P p in jected reactive react ive pow er, Q p




26

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s Swing Bus or Slack Bus 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. P,Q + Type 1: ∠ V δ G Sw ing Bus -


Specify: V, δ Unknown: P, Q



27

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s 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 P,Q power injection. + Type 2: ∠ V δ G G enerator Bu s U. P. National Engineering Center

Specify: P, V Unknown: Q, δ



28

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s 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,Q

Type 3: Load Bus

+

V∠δ -


Specify: P, Q Unknown: V, δ Competency Training & Certification Program in


29

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s SUM M ARY OF BUS TY P E S Bus Typ e

K n o w n Q u a n titie s

U n k n o w n Q u a n titie s

Type1: S w in g

V p , δ p

P p, Q p

Type 2: G e n e ra to r

P p , V Vp

Q p, δ p

Type 3: Load

P p, Q p

V p , δ p




30

P o w e r Sy s t e m M o d e l s f o r L o a d Fl Fl o w A n a l y s i s G

G

Line 1

2

1

B u s Ty Ty p e s

Line 2

Line 3 3

Bus Voltage Generation Load Remarks No. V (p.u.) δ P Q P Q 1 1.0 0.0 * * 0 0 Swing Bus 2 1.0 * 0.20 * 0 0 Gen Ge n Bus 3 * * 0 0 0.60 0.25 Load Bus U. P. National Engineering Center



31

G a u s ss - Se i d e l L o a d Fl o w 

Linear Formulation of Load Flow Equations



Gauss-Seidel Load Flow Solution



Numerical Example




32

G a u s ss - Se i d e l L o a d Fl o w L i n e a r Fo r m u l a t i o n o f L o a d Fl o w Equations The real and reactive power into any bus P is: P p + jQ p = V p I p*

or

(1)

P p - jQ p = V p* I p

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

G a u s ss - Se i d e l L o a d Fl o w Equation (1) may be rewritten as: I p =

P p - jQ p _________

(2)

V p*

From the Bus Admittance Matrix equation, the current injected into the bus are: I p = Y p1V 1 + Y p2V 2 + … + Y ppV p + … + Y pnV n

(3)

I 1 = Y 11V 1 + Y 12V 2 + Y 13V 3 I 2 = Y 21V 1 + Y 22V 2 + Y 23V 3 I 3 = Y 31V 1 + Y 32V 2 + Y 33V 3 U. P. National Engineering Center



34

G a u s ss - Se i d e l L o a d Fl o w Substituting (3) into (2) P p - jQ p _________ = Y V + Y V + … + Y V + … + Y V p1 1 p2 2 pp p pn n * V p P 1 – jQ1 _________ = Y V + Y V + Y V 11 1 12 2 13 3 * V 1

(4)

P 2 – jQ2 _________ = Y V + Y V + Y V 21 1 22 2 23 3 * V 2 P 3 – jQ3 _________ = Y V + Y V + Y V 31 1 32 2 33 3 * V 3 U. P. National Engineering Center



35

G a u s ss - Se i d e l L o a d Fl o w Solving for V p in (4) Y 11V 1 =

V 1

=

=

- (___ + Y 12V 2 + Y 13V 3 )

V 1*

1 Y 11

Y 22V 2 =

V 2

P 1 – jQ1 _______

⎡ P − jQ ⎤ ⎢ V − Y V − Y V ⎥ ⎣ ⎦ 1

1

13

3

- (Y 12V 2 + ___ + Y 13V 3 )

V 2*

Y 22

2

1

P 2 – jQ2 _______

1

12

*

⎡ P − jQ ⎤ ⎢ V − Y V − Y V ⎥ ⎣ ⎦ 2


2

*

21

1

13

3

2



36

G a u s ss - Se i d e l L o a d Fl o w Y 33V 3 =

V 3

=

P 3 – jQ3 _______

- (Y 13V 1 + Y 23V 2 + ___)

V 3*

1 Y 33

⎡ P − jQ ⎤ ⎢ V − Y V − Y V ⎥ ⎣ ⎦ 3

3

31

*

1

32

2

3

n P p - jQ p 1 _______ V p = ___ - Σ Y pqV q Y pp V p* q=1

(5)

q p




37

G a u s ss - Se i d e l L o a d Fl o w Ga u s s - Se i d e l L o a d Fl o w So l u t i o n Generalizing the Gauss-Seidel Load Flow, the estimate for the voltage V p at bus p at the kth iteration is: n P p - jQ p 1 _______ ___ k+1 - Σ Y pqV q V p = Y pp (V pk )* q=1

(6)

q p

where,

α=k α=k+1


if p < q if p > q Competency Training & Certification Program in


38

G a u s ss - Se i d e l L o a d Fl o w 

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 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. U. P. National Engineering Center



39

G a u s ss - Se i d e l L o a d Fl o w N u m e r i ca c a l Ex a m p l e 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 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.

G

G

Line 1

2

1 Line 2

Line 3 3




40

G a u s ss - Se i d e l L o a d Fl o w Branch Data Lin Li n e N o . B u s Co Cod d e I mp mpe e d a nc nce e Zp q (p.u.) 1 2 3

1 -2 1 -3 2 -3

0.08 + j0.24 0.02 + j0.06 0.06 + j0.18

Bus Data Bus Voltage Generation Load Remarks No. V (p.u.) δ P Q P Q 1 1.0 0.0 * * 0 0 Swing Bus 2 1.0 * 0.20 * 0 0 Gen Ge n Bus 3 * * 0 0 0.60 0.25 Load Bus U. P. National Engineering Center



41

G a u s ss - Se i d e l L o a d Fl o w Specified Variables: V1 = 1.0

δ1 = 0.0

V2 = 1.0

P2 = 0.2

P3 = -0.6

Q3 = -0.25

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

Initial Estimates of Unknown Variables:

δ20 = 0.0 V30 = 1.0

δ30 = 0.0 U. P. National Engineering Center



42

G a u s ss - Se i d e l L o a d Fl o w The Bus Admittance Matrix elements are: Y11 = 6.2 6.25 5 - j1 j18. 8.75 75 = 19. 19.76 7642 42 ∠ -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 = 2.9 2.916 167 7 - j8 j8.7 .75 5 = 9.2 9.223 233 3 ∠ -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 6.6 .666 667 7 - j2 j20 0 = 21 21.0 .081 819 9 ∠ -71.5650° U. P. National Engineering Center



43

G a u s ss - Se i d e l L o a d Fl o w G auss-S auss-Seidel eidel Equation s Bus 1: Swing Bus V 1

( k + 1 )

= 1∠ 0

for all iterations

Bus 2: Generator Bus Q2 must first be determined from: P 2 - jQ2(k+1) = (V 2(k))* [Y 21V 1(k+1) + Y 22V 2(k) + Y 23V 3(k)]

then substitute it to: V 2

( k + 1 )

=

1 Y 22

⎡ P − jQ (k + 1 ) ⎤ (k + 1 ) ( k ) 2 2 ⎢ Y 21 V 1 Y 23 V 3 ⎥ − − * ⎢⎣ (V 2 (k ) ) ⎥⎦




44

G a u s ss - Se i d e l L o a d Fl o w Bus 3: Load Bus V 3

( k + 1 )

=

1 Y 33

⎡ P − jQ ⎤ ( + ) ( + ) k 1 k 1 3 ⎢ 3 ⎥ Y 31 V 1 Y 32 V 2 − − * ⎢⎣ (V 3 (k ) ) ⎥⎦

I t e r a t i o n 1 ( k = 0 ) : V1 (1) = 1.0∠0° P2 - jQ2(1) = (1.0∠0°) [(-1.25 + j3.75)(1.0∠0°)

+ (2. (2.91 9167 67 - j8 j8.7 .75) 5)(1 (1.0 .0∠0°) + (-1.6667 + j5)(1.0∠0°) = 0.0 + j0.0 Q2(1) = 0.0 [This value is within the limits.] U. P. National Engineering Center



45

G a u s ss - Se i d e l L o a d Fl o w V 2

( k + 1 )

V2

(1)

=

1 Y 22

⎡ P − jQ (k + 1 ) ⎤ + ( ) ( ) k 1 k 2 ⎢ 2 − − Y 21V 1 Y 23 V 3 ⎥ * ⎢⎣ (V 2 (k ) ) ⎥⎦

1 0. 2 - j 0. 0 ___________________ ___________

=

9.2233∠-71.5650

− jQ2 (k +1)

1.0∠0°

Y 22 Y 21

P2

V 1

( k +1 )

(V ) ( k )

*

2

- ((-1. 1.25 25 +j +j3. 3.75 75)) (1. (1.0 0∠0°) Y 23

V 3

( k )

- ((-1. 1.66 6667 67 + j5) j5) (1 (1.0 .0∠0°) = 1.0071∠1.1705° U. P. National Engineering Center



46


( k + 1 )

=

1 Y 33

⎡ P − jQ ⎤ + + ( ) ( ) k 1 k 1 3 ⎢ 3 ⎥ − − Y 31V 1 Y 32 V 2 * ⎢⎣ (V 3 (k ) ) ⎥⎦ P − jQ 3

V31

= Y 33

1 _____________________ 21.0819∠-71.5650 Y 31

V 1

3

-0.6 + j0.25 ____________ 1.0∠0°

(V )

k *

3

( k +1 )

- (-5 +j1 j15 5) (1 (1..0∠0°) Y 32

V 2

( k +1 )

- (5.2705∠108.4349°)(1.0071∠1.1705°) = 0.9816 ∠-1.0570° U. P. National Engineering Center



47

G a u s ss - Se i d e l L o a d Fl o w ∆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 U. P. National Engineering Center



48

G a u s ss - Se i d e l L o a d Fl o w I t e r a t i o n 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. 0.29 299 95 - j0 j0.0 .007 073 3 Q2 (2) = 0.0073

[This value is within the limits.]




49


( k + 1 )

V2

(2)

=

( k + 1 ) 1 ⎡ P2 − jQ 2

⎢ Y 22 ⎢ ⎣

(V ( ) ) k

*

− Y 21V 1

( k + 1 )

2

− Y 23 V 3

( k )

⎤ ⎥ ⎥⎦

1 ___________________

0.2 - j0.0073 ______________

9.2233 ∠ -71.5650

1.0 ∠ -1.1705°

=

- ((-1. 1.25 25 +j +j3. 3.75 75)) (1. (1.0 0 ∠ 0°) - (5.2705 ∠ 108.4349° ) (0.9816 ∠ -1.0570°) = 0.9966 ∠ 0.5819° U. P. National Engineering Center



50


( k + 1 )

=

1 Y 33

⎡ P − jQ ⎤ ( ) ( ) + + k 1 k 1 3 ⎢ 3 ⎥ − − Y 31 V 1 Y 32 V 2 * ⎢⎣ (V 3 (k ) ) ⎥⎦

1 _____________________

-0.6 + j0.25 ___________________

21.0819 ∠ -71.5650

0.9816 ∠ 1.0570°

V3(2) =

- (-5 +j15) (1 (1..0 ∠ 0°) - (5.2705 ∠ 108.4349°) (0.9966 ∠ 0.5819° ) = 0.9783 ∠ -1.2166° U. P. National Engineering Center



51


= 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 U. P. National Engineering Center



52

G a u s ss - Se i d e l L o a d Fl o w I t e r a t i o n 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. 0.22 2287 87 - j0 j0.0 .047 472 2 Q22 = 0.0472

[This value is within the limits.]




53


( k + 1 )

V23

=

=

( k + 1 ) 1 ⎡ P2 − jQ 2

⎢ Y 22 ⎢ ⎣

(V ( ) ) k

*

− Y 21V 1

( k + 1 )

2

− Y 23 V 3

( k )

⎤ ⎥ ⎥⎦

1 ___________________

0.2 - j0.0472 ______________

9.2233 ∠ -71.5650

1.0 ∠ -0.5819°

- ((-1. 1.25 25 +j +j3. 3.75 75)) (1 (1.0 .0 ∠ 0°) - (5.2705 ∠ 108.4349° ) (0.9783 ∠ -1.2166° ) = 0.9990 ∠ 0.4129°




54


( k + 1 )

V33 =

=

1 Y 33

⎡ P − jQ ⎤ ( ) ( ) + + k 1 k 1 3 ⎢ 3 ⎥ − − Y 31 V 1 Y 32 V 2 * ⎢⎣ (V 3 (k ) ) ⎥⎦

1 _____________________

-0.6 + j0.25 ___________________

21.0819 ∠-71.5650

0.9783 ∠ 1.2166°

- (-5 +j15)(1.0∠0°) - (5.2705∠108.4349°)(0.9990∠0.4129°) = 0.9788∠-1.2560° U. P. National Engineering Center



55


= 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. U. P. National Engineering Center



56

N e w t o n - Ra R a p h s o n L o a d Fl Fl o w 

Non-Linear Formulation of Load Flow Equations



Newton New ton-Ra -Raphs phson on Loa Load d Flo Flow w Sol Soluti ution on



Numerical Example




57

N e w t o n - Ra R a p h s o n L o ad a d Fl o w N o n - L i n e a r Fo F o r m u l a t i o n o f L o a d Fl Fl o w Equations The complex power injected into Bus p is

Pp

−

=E

jQ p

* p

Ip

(1)

and the current equation may be written as Ip

=

n

∑Y q =1

pq

Eq

(2)

Substituting (2) into (1) P p

−

jQ

n

= E • ∑ *

p

p

q


=1

Y pq E q

(3)



58

N e w t o n - Ra R a p h s o n L o ad a d Fl o w Let E p

Ypq

= V p ∠ δ p = Y p q ∠ θ p q

E q

= V q ∠ δ q

Substituting into equation (3),

Pp

−

n

jQ p

=∑

q =1

V pV q Y p q

∠ ( θ p q +δ q − δ p

)

(4)

Separating the real and imaginary components n

Pp

=∑

q =1

V pV q Y p q c o s ( θ p q +δ q n

Qp

= −∑

q =1

V pV q Y p q s i n ( θ p q +δ q


− δ p

)

− δ p

)

(5) (6)



59

N e w t o n - Ra R a p h s o n L o ad a d Fl o w The 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. Equatio Equ ations ns Qp Qp are writt written en for for Load Load Buse Buses s only 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

N e w t o n - Ra R a p h s o n L o ad a d Fl o w The system of equations may be written as

P1 P2

= P1 ( δ 1 ,δ 2 , . . . ,δ i , v 1 , v 2 . . . . , v j ) = P2 ( δ 1 ,δ 2 , . . . ,δ i , v1 , v 2 . . . . , v j )

M

Pi

M

=

Pi ( δ 1 ,δ 2 , . . . ,δ i , v 1 , v 2 . . . . , v j )

(7)

= Q1 ( δ 1 ,δ 2 ,...,δ i ,v1 ,v2 ....,v j ) Q2 = Q2 ( δ 1 ,δ 2 , ...,δ i , v1 , v2 ...., v j ) Q1 M

Qj

M

= Q ( δ 1 ,δ 2 ,...,δ i ,v1 ,v2 ....,v j ) j




61

N e w t o n - Ra R a p h s o n L o ad a d Fl o w Equation Equatio n (7) may may be linear linearized ized usi using ng a First-O First-Orde rderr Taylor-Series Expansion P1spec P2

spec

= P2calc

M M M

M

Pi spec

= Pi calc

Q1spec

= Q1calc

spec

= Q2calc

Q2

∂P1 ∂P1 ∆δ 1 + ∆δ +... + ∂δ1 ∂δ 2 2 ∂P2 ∂P2 ∆δ 1 + ∆δ + +... + ∂δ 1 ∂δ 2 2

= P1calc +

M M M Q

spec j

= Q

calc j

M

M

∂Pi ∂Pi ∆δ 1 + ∆δ +... + + ∂δ1 ∂δ 2 2 ∂Q1 ∂Q1 ∆δ 1 + ∆δ + +... + ∂δ 1 ∂δ 2 2 ∂Q2 ∂Q2 ∆δ 1 + ∆δ + +... + ∂δ1 ∂δ 2 2 +

M∂Q

j

∂δ 1

∆δ1

M ∂Q

+

∂P1 ∂P1 ∂P1 ∆δ i + ∆V1 + ∆V +... + ∂δ i ∂V1 ∂V2 2 ∂P2 ∂P2 ∂P2 ∆δ i + ∆V1 + ∆V +... + ∂δ i ∂V1 ∂V2 2

M

M

∂P1 ∆V ∂V j j ∂P2 ∆V ∂V j j

M

∂Pi ∂Pi ∂Pi ∂Pi ∆δ i + ∆V1 + ∆V2 +... + ∆V ∂δ 2 ∂V1 ∂V2 ∂V j j ∂Q1 ∂Q1 ∂Q1 ∂Q1 ∆δ i + ∆V1 + ∆V2 +... + ∆V ∂δ i ∂V1 ∂V2 ∂V j j ∂Q2 ∂Q2 ∂Q2 ∂Q2 ∆δ i + ∆V1 + ∆V2 +... + ∆V ∂δ i ∂V1 ∂V 2 ∂V j j

M

j

∂δ 2

∆δ 2

∂Q j M ∂Q j M ∂Q j M ∂Qj ∆δ i + ∆V1 + ∆V2 +... + ∆V +... + ∂δ i ∂V1 ∂V2 ∂V j j




⎡ P1 s p e c ⎢ ⎢ ⎢ sp ec ⎢ P2 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ sp ec ⎢ Pi ⎢ ⎢ ⎢ sp ec ⎢Q1 ⎢ ⎢ ⎢ Q 2s p e c ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ sp ec ⎢⎣ Q j

−

P1c a l c

−

c a lc 2

P

M

−

Pi c a l c

−

Q

c a lc 1

−

Q 2c a l c

M

−

Q

c a lc j

⎡ ∂ P1 ⎤ ⎢ ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ∂ P2 ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂ Pi ⎥ ⎢ ⎥ ⎢ ∂δ 1 ⎥ = ⎢ ⎥ ⎢ ∂Q1 ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ∂Q 2 ⎥ ⎢ ⎥ ⎢ ∂δ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∂Q j ⎥⎦ ⎢ ∂ δ 1 ⎣

62

∂ P1 ∂δ 2

L

∂ P1 ∂ δ i

∂ P1 ∂V1

∂ P1 ∂V2

L

∂ P2 ∂δ 2

L

∂ P2 ∂ δ i

∂ P2 ∂ V 1

∂ P2 ∂V2

L

M

M

M

M

∂ Pi ∂δ 2

L

∂ Pi ∂ δ 2

∂ Pi ∂V1

∂ Pi ∂V2

L

∂Q1 ∂δ 2

L

∂Q1 ∂ δ i

∂Q1 ∂V1

∂Q1 ∂V2

L

∂Q2 ∂δ 2

L

∂Q2 ∂ δ i

∂Q 2 ∂V1

∂Q2 ∂V2

L

M

M

M

M

M

L

∂Q j ∂ δ i

∂Q j ∂V1

∂Q j ∂V2

∂Q j ∂δ 2


L

∂ P1 ⎤ ⎡ ∆ δ 1 ∂ V j ⎥⎥ ⎢ ⎢ ∂ P2 ⎥⎥ ⎢ ∆ δ 2 ∂ V j ⎥ ⎢ ⎥⎢ ⎥⎢ M ⎥ ⎢ M ⎥⎢ ⎥⎢ ∂ Pi ⎥ ⎢ ⎢ ∆ δ i ∂ V j ⎥ ⎢ ⎥ ⎢ ∂ Q 1 ⎥⎥ ⎢ ∆ V 1 ∂ V j ⎥ ⎢ ⎥⎢ ∂ Q 2 ⎥ ⎢⎢ ⎥ ∆ V 2 ∂ V j ⎥ ⎢ ⎥⎢ ⎥ ⎢ M M ⎥⎢ ⎥⎢ ⎢ ∂Q j ⎥ ⎢ ⎥ ∆ V j ∂ V j ⎥⎦ ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦



63

N e w t o n - Ra R a p h s o n L o ad a d Fl o w or simply

⎡ ∂P ∂P ⎤ ⎡ ∆P ⎤ ⎢ ∂δ ∂V ⎥ ⎡ ∆δ ⎤ ⎢∆Q ⎥ = ⎢ ∂Q ∂Q ⎥ ⎢∆V ⎥ ⎣ ⎦ ⎢ ⎥⎣ ⎦ ⎣ ∂δ ∂V ⎦ ⎡ ∂P ⎡ ∆P ⎤ ⎢ ∂δ ⎢∆Q ⎥ = ⎢ ∂Q ⎣ ⎦ ⎢ ⎣ ∂δ

∂P ⎤ V ∂V ⎥ ⎡⎢ ∆δ ⎤⎥ ∂Q ⎥ ⎢⎣∆V V ⎥⎦ ⎥ V ∂V ⎦




64

N e w t o n - Ra R a p h s o n L o ad a d Fl o w N ew e w t o n - Ra R a p h s o n L o ad a d Fl Fl o w So l u t i o n

⎡ ∆ P ⎤ ⎡ J 1 ⎢ ⎥ ⎢ ⎢ ⎥=⎢ ⎢ ∆ Q ⎥ ⎢ J ⎣ ⎦ ⎣ 3

J 1

⎡ ∆ δ ⎤ ⎥ ⎥⎢ ⎥ ⎥⎢ ⎢ ∆V ⎥ ⎥ J 4 ⎦ ⎢ ⎣ V ⎥⎦ J 2 ⎤

n ⎧ ∂ P p ⎪ ∂ δ = ∑ V p V q Y p q s i n ( θ p q + δ q − δ p ⎪ p q = 1 ,q ≠ p ⎨ ⎪ ∂ P p = − V V Y s i n ( θ + δ − δ ) p q pq pq q p ⎪ ∂ δ q ⎩ U. P. National Engineering Center

)



65

N e w t o n - Ra R a p h s o n L o ad a d Fl o w ∂ P p ⎧ 2 = + V P V ⎪ p p pY p pc o s θ p p ∂ V p ⎪ J 2 ⎨ ⎪ V ∂ P p = V V Y c o s ( θ q p q pq p q + δ q − δ p ) ⎪ ∂ V q ⎩ n ⎧ ∂ Q p = ∑ V pV qY p q c o s ( θ p q + δ q − δ ⎪ q = 1 ,q ≠ p ⎪ ∂ δ p J 3 ⎨ ⎪ ∂ Q p = − V V Y c o s ( θ ) p q pq p q + δ q − δ p ⎪ ∂ δ q ⎩ U. P. National Engineering Center

p

)



66

N e w t o n - Ra R a p h s o n L o ad a d Fl o w

J 4

⎧ ⎪V p ⎪ ⎨ ⎪V ⎪ q ⎩

∂ Q p = Q p − V p2 Y p p s i n θ p q ∂ V p ∂ Q p = −V pV q Y pq s in ( θ pq + ∂ V q

δq

− δ p

)

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




67

N e w t o n - Ra R a p h s o n L o ad a d Fl o w At th the e kth kth it iter erat atio ion, n,

δ

( k +1 ) p

=δ

( k +1 )

= Vp( k ) + ∆V p( k )

Vp

(k ) p

+ ∆δ

(k ) p

The process is terminated once convergence is achieved whrein

MAX∆ (Pk )

≤ ε p

and


(k) MAX∆ Q

≤ εq



68

N e w t o n - Ra R a p h s o n L o ad a d Fl o w N u m e r i ca c a l Ex a m p l e 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 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.

G

G

Line 1

2

1 Line 2

Line 3 3




69

N e w t o n - Ra R a p h s o n L o ad a d Fl o w Branch Data Lin Li n e N o . B u s Co Cod d e I mp mpe e d a nc nce e Zp q (p.u.) 1 2 3

1 -2 1 -3 2 -3

0.08 + j0.24 0.02 + j0.06 0.06 + j0.18

Bus Data Bus Voltage Generation Load Remarks No. V (p.u.) δ P Q P Q 1 1.0 0.0 * * 0 0 Swing Bus 2 1.0 * 0.20 * 0 0 Gen Ge n Bus 3 * * 0 0 0.60 0.25 Load Bus U. P. National Engineering Center



70

N e w t o n - Ra R a p h s o n L o ad a d Fl o w The elements of the Bus Admittance Matrix are: Y1 1 Y1 2 Y1 3 Y2 1 Y2 2 Y2 3 Y3 1 Y3 2 Y3 3

= 6 .2 5 − j 1 8 .7 5 = 1 9 .7 6 4 2 ∠ − 7 1 .5 6 5 1 = − 1 , 2 5 + j 3 .7 5 = 3 .9 5 2 8 ∠ 1 0 8 .4 3 4 9 = − 5 + j 1 5 = 1 5 .8 1 1 4 ∠ 1 0 8 .4 3 4 9 = − 1 .2 5 + j .3 7 5 = 3 .9 5 2 8 ∠ 1 0 8 .4 3 4 9 = 2 .9 1 6 7 − j 8 .7 5 = 9 .2 2 3 3 ∠ − 7 1 .5 6 4 9 = − 1 .6 6 6 7 + j 5 = 5 .2 7 0 5 ∠ 1 0 8 .4 3 4 9 = − 5 + j 1 5 = 1 5 .8 1 1 4 ∠ 1 0 8 .4 3 4 9 = − 1 .6 6 6 7 + j 5 = 5 .2 7 0 5 ∠ 1 0 8 .4 3 4 9 = 6 .6 6 6 7 − j 2 0 = 2 1 .0 8 1 9 ∠ − 7 1 .5 6 5 0



CPT3 - Load Flow Analysis-6th Batch

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