POWER SYSTEM SIMULATION LAB-1 MANUAL (ELECTRICAL - POWER SYSTEM ENGINEERING )

Power System Simulation Lab - 1

M.E (Power Systems Engineering) MATHANKUMAR.S, AP/EEE

EXPERIMENT: 1 Date:

COMPUTATION OF PARAMETERS OF

TRANSMISSION LINES – SINGLE CIRCUIT

AIM:-

To determine the positive sequence line parameters L and C per phase per kilometer of a three phase single circuit transmission lines for different conductor arrangements.

OBJECTIVES:-

To become familiar with different arrangements of conductors of a three phase single circuit transmission lines and to compute the GMD and GMR for different arrangements.

SOFTWARE REQUIRED:-

LINE CONSTANTS module of AU Power lab & MATLAB or other equivalent.

THEORETICAL BACK GROUND:-

Line Parameters

Transmission line has four electrical parameters - resistance, inductance, capacitance and conductance. The inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. The shunt conductance characterizes the leakage current through insulators, which is very small and can be neglected. The parameters R, L and C are essential for the development of the transmission line models to be used in power system analysis both during planning and operation stages.

While the resistance of the conductor is best determined from manufactures data, the inductances and capacitances can be evaluated using formula. The student is advised to read any other text book before taking up the experiment.

INDUCTANCE

The inductance is computed from flux linkage per ampere. In the case of the three phase lines, the inductance of each phase is not the same if conductors are not spaced equilaterally. A different inductance in each phase results in unbalanced circuit. Conductors are transposed in order to balance the inductance of the phases and the average inductance per phase is given by simple formulas, which depends on conductor configuration and



conductor radius. General Formula:-

The general formula for computing inductance per phase in mH per km of a transmission is given by

L = 0.2 ln GMD/GMRL mH / km

Where,

GMD = Geometric Mean Distance

GMRL = Geometric Mean Radius Value of Inductance

The expression for GMR and GMD for different arrangement of conductors of the transmission lines are given in the following section.

I. Single Phase - 2 Wire Systems:

Fig. Conductor arrangement

GMD = D

GMR = re-1/4 = r’

Where, r = radius of conductor

II. Three Phase - Symmetrical Spacing:

D D

D Fig. Conductor Arrangement

Where, GMD = D

GMR = re-1/4 = r’

r = radius of conductor



III. Three Phase - Asymmetrical Transposed:

Fig. Conductor Arrangement

GMD = Geometric mean of the three distances of the unsymmetrically placed conductors

3GMD = D * D * DAB BC CA

GMR = re-1/4 = r’ (or) D/2

Where, r = radius of conductor = D/2 D = Diameter of conductor

Capacitance

A general formula for evaluating capacitance per phase in micro farad per km

of a transmission line is given by

C = 0.0556 / ln (GMD / GMRC) µF / km

Where,

GMD is the “Geometric Mean Distance” which is the same as that

defined for inductance under various cases.

GMR is the Geometric Mean Radius and is defined case by case below:

(i) Single phase two wires system (for diagram see inductance):

GMD = D

GMRL = r (as against r‟ in the case of L)



(ii) Three phase - symmetrical spacing (for diagram see inductance):

GMD = D

GMR = r in the case of solid conductor

GMR = Ds in the case of stranded conductor to be obtained from manufacturer‟s data.

(iii) Three-phase – Asymmetrical - transposed (for diagram see Inductance):

3GMD = D * D * DAB BC CA (Or) GMD = [DAB DBC DCA]1/3

GMRsC = r (or) D/2; for solid conductor

GMRsC = Ds for stranded conductor

GMRsC = rb for bundled conductor

Where,

SC sp

DGMR = *d

2 for 2 conductor bundle

3SC sp sp

DGMR = *d *d

2 for 3 conductor bundle

4SC sp sp sp

DGMR = 1.09 *d *d *d

2for 4 conductor

bundle

(or) Where,

rb = [r*dsp]1/2 for 2 conductor bundle

rb = [r*dsp2]1/3 for 3 conductor bundle

rb = 1.09 [r*dsp3]1/4 for 4 conductor bundle

Where,

r = radius of each subconductor

D = Diameter of each subconductor

dsp = bundle spacing



EXERCISES:

1.a) :- A three-phase transposed line composed of one ACSR, 1,43,000 cmil, 47/7

Bobolink conductor per phase with flat horizontal spacing of m

between phases a and b and between phases b and c. The conductors have

a diameter of 3.625 cm and a GMR of 1.439 cm. The spacing between the

conductors in the bundle is 45 cm.

1.b) :-

The line is to be replaced by a three conductor bundle of ACSR 477,000-cmil,

26/7 Hawk conductors having the same cross sectional area of aluminum as

the single- conductor line. The conductors have a diameter of 2.1793 cm and a

GMR of 0.8839 cm. The new line will also have flat horizontal configurations,

but it is to be operated at a higher voltage and therefore the phase spacing is

increased to m as measured from the centre of the bundles. The spacing

between the conductors in the bundle is 45 cm.

(i) Determine the inductance and capacitance per phase per kilometer

of the above lines.

(ii) Verify the results using the available program.



Viva Questions

1. What are the line parameters?

2. What is conductor?

3. What are the difference between conductors and insulators?

4. Differentiate between GMR and GMD of a conductor.

5. What are the types of conductor?

6. What are the primary conductor materials for the overhead system?

7. What are the advantages of standard conductor over solid conductor?

8. List out the different arrangements of conductors of the transmission lines.

9. What is meant by stranded conductors? How is a stranded conductor made?

10. Define corona loss. Mention the effect of Corona Losses in various surface of

the conductor.

11. What is meant by ACSR?

12. What is the purpose of transmission lines?

13. Write a general formula for inductance & Capacitance of a transmission line.

14. How power system studies are carried out?

15. What happens if the capacitance of a transmission line is high?

16. Why does a transmission line possess inductance and capacitance?

17. What is meant by cmil?

18. Define bundle spacing.

19. What is transposition?

20. How to reduce or eliminate the corona losses in EHV line?



Result:

Marks split-up Marks

Secured

Marks

Awarded

Basic understanding 15

Theoretical Calculation 20

Conducting 15

Software output with graph

(Aupower & MATLAB) 20

Comparison Results 10

Record 10

Viva - voce 10

Total Marks 100



EXPERIMENT: 2 Date:

COMPUTATION OF PARAMETERS OF

TRANSMISSION LINES – DOUBLE CIRCUIT

AIM

To determine the positive sequence line parameters L and C per phase per kilometer of a three phase double circuit transmission lines for different conductor arrangements.

OBJECTIVES

To become familiar with different arrangements of conductors of a three phase double circuit transmission lines and to compute the GMD and GMR for different arrangements.

SOFTWARE REQUIRED

LINE CONSTANTS module of AU Power lab & MATLAB or equivalent. THEORETICAL BACK GROUND:-

Line Parameters

Transmission line has four electrical parameters - resistance, inductance, capacitance and conductance. The inductance and capacitance are due to the effect of magnetic and electric fields around the conductor. The shunt conductance characterizes the leakage current through insulators, which is very small and can be neglected. The parameters R, L and C are essential for the development of the transmission line models to be used in power system analysis both during planning and operation stages.

While the resistance of the conductor is best determined from manufactures data, the inductances and capacitances can be evaluated using formula. The student is advised to read any other text book before taking up the experiment.

INDUCTANCE

The inductance is computed from flux linkage per ampere. In the case of the three phase lines, the inductance of each phase is not the same if conductors are not spaced equilaterally. A different inductance in each phase results in unbalanced circuit. Conductors are transposed in order to balance the inductance of the phases and the average inductance per phase is given by simple formulas, which depends on conductor configuration and conductor radius.



General Formula:-

The general formula for computing inductance per phase in mH per km of a transmission is given by

L = 0.2 ln GMD/GMRL mH / km

Where,

GMD = Geometric Mean Distance

GMRL = Geometric Mean Radius Value of Inductance

The expression for GMR and GMD for different arrangement of conductors of the transmission lines are given in the following section.

Capacitance

A general formula for evaluating capacitance per phase in micro farad per km

of a transmission line is given by

C = 0.0556 / ln (GMD / GMRC) µF / km

Where,

GMD is the “Geometric Mean Distance” which is the same as that

defined for inductance under various cases.

Three phase - Double circuit transposed:

A three-phase double circuit line consists of two identical three-phase circuits. The phases a, b and c are operated with a1-a2, b1-b2 and c1-c2 in parallel respectively. The GMD and GMR are computed considering that identical phase forms a composite conductor, for example, phase a conductors a1 and a2 form a composite conductor and similarly for other phases.

Fig. Conductor Arrangement

b1

a1

c1

a1

S33

a1

S22

a1

a2

a1

c2 S11 a1

b2

H12

H23



Relative phase position a1b1c1 –c2b2a2. It can also be a1b1c1 – a2b2c2.

Three phase - Double circuit - transposed (for diagrams see inductance):

The inductance per phase in milli henries per km is

L = 0.2 ln (GMD/GMRL) mH/km

Where,

GMRL is equivalent geometric mean radius and is given by

GMRL = (GMRSA*GMRSB *GMRSC)1/3

(Or)

3L SA SB SCGMR = GMR * GMR * GMR

Where,

GMRSA ,GMRSB and GMRSC are GMR of each phase group and given by

Phase A, GMRSA = [GMRESL* Da1a2] 1/2

Phase B, GMRSB = [GMRESL* Db1b2 ] 1/2

Phase C, GMRSC = [GMRESL* Dc1c2 ] 1/2

Where,

ESL act spGMR = GMR *d for 2 conductor bundle

GMRESL= GMR of equivalent conductor of two sub conductor bundle.

GMD is the “equivalent GMD per phase” & is given by

3GMD = D * D * DAB BC CA (or) GMD = [DAB *DBC *DCA]1/3

Where,

DAB, DBC, & DCA are GMD between each phase group A-B, B-C, C-A

which are given by

DAB = [Da1b1 Da1b2 Da2b1 Da2b2]1/4

DBC = [Db1c1 Db1c2 Db2c1 Db2c2]1/4

DCA = [Dc1a1 Dc2a1 Dc2a1 Dc2a2]1/4



Three phase - Double circuit - transposed (for diagrams see capacitance):

C = 0.0556 / ln (GMD/GMRc) µF / km

Where, GMRC is equivalent geometric mean radius and is given by

GMRC = (GMRCSA*GMRCSB *GMRCSC)1/3

(Or)

3C CSA CSB CSCGMR = GMR * GMR * GMR

Where,

GMRCSA ,GMRCSB and GMRCSC are GMR of each phase group and given by

Phase A, GMRCSA = [GMRESC* Da1a2] 1/2

Phase B, GMRCSB = [GMRESC* Db1b2 ] 1/2

Phase C, GMRCSC = [GMRESC* Dc1c2 ] 1/2

Where,

D

2ESC spGMR = *d for 2 conductor bundle

(Or)

ESC spGMR = r*d for 2 conductor bundle

GMRESC= GMR of equivalent conductor of two sub conductor bundle.

Where,

r = radius of each subconductor

D = Diameter of each subconductor

dsp = bundle spacing

*GMD is the same as for Inductance and Capacitance calculations*



EXERCISES:

A 345 kV double circuit three phase transposed line is composed of two ACSR,

1,431, 000cmil, 45/7 bobolink conductors per phase with vertical conductor

configuration as shown in Figure. The conductors have a diameter of

1.427 inch and a GMR of 1.439 cm. The spacing between the conductors in the

bundle is 18 inch.

i) Find the inductance and capacitance per phase per kilometer of the above

line.

ii) Verify the results using the available program.

a1 S11 = m c2

H12= m 18”

b1 S22 = m b2

H23= m

c1 S33 = m a2



Viva Questions

1. State two advantages of bundled conductors. 2. What is composite conductor? 3. Which type of conductor is used in high voltage transmission? 4. What is the major advantage of a double circuit tower? 5. How are the stranded conductors manufactured for three phase system? 6. What does a conductor size 45/7 signify? 7. What are the factors depend on the skin effect. 8. What happens if the capacitance of a transmission line is high? 9. What are the factors governing the inductance & capacitance of a transmission line?

10. How will you compare copper with aluminium as a conductor? 11. What is Lumped and Distributed Transmission Lines? 12. Which conductors are used for neutral circuit? Why? 13. Write a general formula for inductance and capacitance of a transmission line. 14. What are the different arrangements of conductors of the transmission lines? 15. Compare delay in transmission line and overhead transmission line. 16. What are the differences between vertical and horizontal conductor configuration? 17. Define bundle spacing. 18. What are the differences between ACSR, AAAC & ACAR? 19. List the various types of conductor. 20. What is meant by counterpoise?



Result:


Secured

Marks

Awarded



Conducting 15




Record 10

Viva - voce 10

Total Marks 100



EXPERIMENT: 3 Date:

FORMATION OF BUS ADMITTANCE AND IMPEDANCE

MATRICES AND SOLUTION OF NETWORKS AIM

To understand the formation of network matrices, the bus admittance matrix Y and the bus impedance matrix Z of a power network, to effect certain required changes on these matrices and to obtain network solution using these matrices.

OBJECTIVES

To determine the bus admittance and impedance matrices for the given power system network.

To obtain network solution using these matrices.

To obtain certain specified columns of the bus impedance matrix Z

or the full matrix Z using the factors of Y or the inverse of Y.

SOFTWARE REQUIRED

FORMATION OF NETWORK MATRICES module of AU Power lab & MATLAB.

THEORETICAL BACKGROUND

Network Description of a Multimode Power System

The bus admittance matrix Y and bus impedance matrix Z are the two important network descriptions of interconnected power system. The injected bus currents and bus voltages of a power system under steady state condition can be related through these matrices as

Y V = I (3.1)

Z I = V (3.2)



Formation of Bus Admittance Matrix These matrices are important building blocks of power system modelling and analysis. The Z is mainly used in fault analysis while Y is mainly used in power flow and stability analysis.

Two-Rule Method (Based on Node - Voltage Analysis)

Consider a three-bus power system shown in Fig.3.1 .The equivalent power network for the system is shown in Fig.3.2 in which the generator is replaced by Norton equivalent, the loads by equivalent admittance and line by π - equivalent circuits.

Generation G

Load

2

I1 yg1

yd2

1

Short line

Short line Long line

1 2

y12

y13 y23 y23‟

Shunt

Compensation

3

Load

„ Y23

3 y30 yd3

Fig.3.1 A Sample Power System Fig.3.2 Equivalent Power Network

In Fig.3.2, the admittances of the generator, loads and transmission lines are given in per unit to system MVA base. The ground is taken as reference node.

Applying Kirchhoff‟s current law (KCL) to nodes 1,2 and 3.

yg1 V1 + y12 (V1 –V2) + y13 (V1-V3) = I1

yd2 V2 + y12 (V2 –V1 ) + y23 (V2 – V3) + y‟23 V2 = 0 (3.3)

yd3 V3 + y30 V3 + y23 (V3 – V2) + y‟23 V3 + y13 (V3 – V1) = 0

Rearranging these equations

(yg1 + y12 +y13) V1 + (-y12) V2 + (-y13) V3 = I1

(-y12) V1 + (yd2 + y12 + y23 + y‟23) V2 + (-y23) V3 =0 (3.4)

(-y13) V1 + (-y23) V2 + (yd3 + y30 + y23 + y‟23 + y13) V3 =0


In matrix form

Y11 Y12 Y13 V1 I1

Y21 Y22 Y23 V2 = 0 (3.5)

Y31 Y32 Y33 V3 0

Where,

Y11 = (yd1 + y12 +y13)

Y22 = (yd2 + y12 + y23 + y‟23) (3.6)

Y33 = (yd3 + y30 + y23 + y‟ 23 + y13)

Y12 = Y21 = -y12

Y13 = Y31 = -y13

Y23 = Y32 = -y23

The matrix equation (3.5) can be extended to an „n‟ node system. The steps involved in assembling bus admittance matrix may be extracted from equations (3.5) and (3.6) and are given below.

Two-Rule Method for Assembling Y matrix:

1. The diagonal element Yii of the matrix is equal to the sum of the admittances of all elements connected to the ith node.

2. The off-diagonal element Yij of the matrix is equal to the negative of the sum

of the admittances of all elements connected between the nodes i and j.

Equivalent Circuit of A Transformer With Off-Nominal Tap for the Purpose of Formation of Bus Admittance Matrix

A two – winding transformer with off- nominal turns ratio, connected between nodes k and m is shown in Fig 3.3 In this representation, the turns ratio is normalized as a:1 and the non–unity side is called the tap side which is taken as the sending end side. The series admittance of the transformer is connected to the unity side.

Fig.3.3 Transformer with off-nominal tap


Im

K Ik y/a m

Vk y/a {(1/a)-1}

y {1-(1/a)} Vm

Fig.3.4 Equivalent Circuit for Transformer with Off – Nominal tap.

By applying the two-rule method to buses t and m in Fig.3.3 we get

Y -y Vk/a aIk

-y y Vm = Im (3.7)

By elementary matrix operation equation (3.7) can be reduced to

y/a2 -y/a Vk Ik (3.8)

-y/a y Vm = Im

It can be checked by applying the Two-Rule method to buses k and m in Fig.3.4 that the bus admittance matrix for the circuit in fig.3.4 and that in eqn.3.8 are the same.

Hence the π-equivalent circuit for transformer with off – nominal tap is that in fig.3.4 and the contribution of the transformer to the bus admittance matrix Y of the power system to which it is connected is (refer equation (3.8))

Ykk = y/a2 ; Ymm = y ; Ykm = Ymk = -y/a (3.9)

Algorithm for Formation of Bus Admittance Matrix

The algorithm initializes the matrix Y with all the elements set to zero. Then read one element of the network at a time and update the matrix Y by adding the contribution of that element. The contribution of a transformer connected between nodes k and m is given in equation (3.9).


The contribution of transmission line connected between nodes k and m to Y is

Ykk = Ymm = ykm + y‟km (3.10)

Ykm = Ymk = -ykm

Where, ykm and y‟km are respectively the series admittance in p.u and half line changing admittance in p.u of the line.

The contribution of a shunt element connected to node k to Y is

Ykk = y (3.11)

Where y is the admittance in p.u of the shunt element. If S and So are respectively

the MVA rating of the shunt element (capacitor) and base MVA chosen for the

system; then the shunt admittance is given by

y= 0+j(S/So) p.u (3.12)

Algorithm

Step 1: Initialize Y with all elements set to zero

Step 2: Read the line list, one line at-a-time and update Y by adding the respective contribution, equation (3.10)

Step 3: Read the transformer list, one transformer at-a-time and update

Y by adding the respective contribution, equation (3.9)

Step 4: Read the shunt element list, one element at-a-time and update Y by adding the respective contribution, equation (3.11)

Building Algorithm for Bus Impedance Matrix

A building algorithm for bus impedance matrix can be developed by first studying the rules required for modifying an existing Z matrix for addition of new elements. Let us start with a given partial power network with r nodes whose bus impedance matrix Z is known. It is proposed to add new elements, one at a time, to this network and get the modified matrix Zm. Any one of the following four rules can be used depending upon the type of modification.

Modification 1: Add an element with impedance z, connected between the

reference node of the partial network and a new node (r+1).

Rule 1: The modified matrix Zm of dimension (r+1) x (r+1) is given by

Zm = Z 0 (3.13)

0 z


Where, Z is the bus impedance matrix of the partial network.

Modification 2: Add an element with impedance z, connected between an existing node i and a new node (r+1).

Rule 2: The modified matrix Zm of dimension (r+1) x (r+1) is given by

Zm = Z Zi (3.14)

ZiT (Zii + z)

Where, Zi is the ith column of Z

ZiT is the transpose of Zi

Zii is the iith element of Z

Modification 3: Add an element with impedance z, connected between an existing node i and the reference node of the partial network.

Rule 3: The modified matrix Zm of dimension r x r is obtained through a two step process. In the first step, assume that the added element is between the existing node i and a fictitious node (r+1) (instead of the reference node) and obtain the modified matrix Z‟ of dimension (r+1) x (r+1) by augmenting Z with an extra row and column as in (2.14). The second step is to connect the fictitious node (r+1) by zero impedance link to the reference node whose voltage is zero and to obtain the final matrix Zm of dimension r x r by applying Kron‟s -reduction to the last row and column to obtain

Z m jk = Z‟jk - Z‟j, (r+1) Z‟(r+1), k ; j,k = 1,2,….,r (3.15)

Z‟ii + z

Modification 4 : Add an element with impedance z, connected between

existing nodes i and j.

Rule 4 : The modified matrix Zm of dimension r x r is given by

Zm = Z – c b bT (3.16)

Where, b = Zi - Zj (3.17)

c= (z + Zii +Zjj – 2Zij)-1 (3.18)

Zi, Zj : ith and jth columns of Z

Zii, Zjj, Zij: iith , jjthand ijth elements of Z


Note:

(3.19)

Building Algorithm for Z:

The above rules are built into the following step wise procedure to build Z

matrix:

Step 1:

Start with a partial network composed only of those elements connected directly

to reference node. Let the number of these elements be r. The corresponding bus

impedance matrix Z(1) is of dimension r x r and is diagonal with the impedance

values of the elements appearing on the diagonal. This process is equivalent to the

repeated use of rule 1.

Step 2:

Add a new element which brings a new node and modify Z(1) using rule 2.

Continue until all the nodes of the complete network are brought in.

Step 3:

Add a new element connected between existing nodes i and j using rule 4.

Continue until all the elements are connected.

Network Solution Using Factorization And Repeat Solution Network solution in a power network is concerned with the determination of the bus voltage vector V from the network equation (3.20), given the bus admittance matrix Y and bus current source vector I

Y V = I (3.20)

In power system applications, during most of the studies, the network

configuration and parameters remain the same implying that Y remains fixed.

However the operating conditions change resulting in changed bus current

vector I. In such cases, while finding numerical solution for the bus voltage

vector V repeatedly for different bus current vectors I, the computations can be


performed effectively resulting in considerable savings in time if “triangular

factorization” and “repeat solution” are adopted.

First the matrix Y in equation (3.20) is split into factor matrices L and U

using the “triangular factorization process” [1]. The equation (3.20) may be

written as

Y V = L U V = I (3.21)

~ By defining an intermediate voltage vector V as

~ U V = V (3.22)

Equation (3.18) may be written as

~ L V = I (3.23)

The solution for voltage vector V in Eqn. (3.21) is obtained through the “repeat solution process” which comprises the following two steps:

~ (i) Forward Elimination: This involves solving eqn. (3.23) for vector V through

elementary transformations on vector I using the elements of L. ~

(ii) Back substitution: This involves solving eqn.(3.22) for vector V through elementary transformations on vector V using the elements of U.

Factorization Algorithm:

Let us define the L and U factor matrices for a sample Y matrix of dimension 3 x 3 as

Y11 Y12 Y13 1 0 0 U11 U12 U13

Y = Y21 Y22 Y23 = L U= L21 1 0 0 U22 U23

Y31 Y32 Y33 L31 L32 1 0 0 U33

(3.24) After multiplication of L and U matrices, eqn. (3.24) becomes,

Y11 Y12 Y13 U11 U12 U13

Y = Y21 Y22 Y23 = L21U11 (L21U12+U22) (L21U13+U23) Y31 Y32 Y33 L31U11 (L31U12+L32U22) (L31U13+L32U23+U33)

(3.25)


It is seen from (3.24) and (3.25) that the first row of U is readily obtained from

the first row of Y, whereas the first column of L can be obtained by dividing

the first column of Y by U11 which is the same as Y11. All the elements of L

and U can be obtained from the Y matrix in Eqn. (3.25) by performing two

transformations. The first transformation on Y matrix in eqn. (3.25) is as follows.

Y

At the end of this first transformation Y matrix is given by

F11 F12 F13

Y(1)= F21 F22 F23 = F

F31 F32 F33 (3.27)

Now the Y(2) matrix in (3.27) contains all the elements of L and U. This can be

termed as factor matrix F. The above factorization process can be generalized for

a Y matrix of dimension nxn as follows.

kth transformation : k = 1,2,……(n -1)

Y ij (k) = Y ij

(k-1)– (Y ik (k-1) Ykj

(k-1)) / Y kk (k-1) ; i, j = (k+1) , ….. , n (3.28)

and Y ik (k) = Y ik

(k-1) / Y kk (k-1) ; i = (k+1) , ….. , n (3.29)

After the above (n-1) transformations, the Y matrix is converted into the factor matrix F which contain all the elements of L and U.

Repeat Solution Algorithm:

The forward elimination is performed on the given vector I using the elements of L stored in F to obtain the intermediate vector V as shown below.


~ V1 = I1

~

~ 1 0 0 V1 I1

~ F21 1 0 V2 = I2 (3.30)

~ F31 F32 1 V3 I3

V2 = I2(1) ; I2

(1) = I2 – F21 I1 (3.31) ~ V3 = I3

(2) ; I3(1) = I3 – F31 I1 ; I3

(2) = I3 (1)– F32 I2

(1)

The above steps can be generalized as : ~ Vi = Ii

(i-1) ; i = 1,2,………..n

Transformation on Ii ; i = 2,3 …..n is performed using

Ii(j) = Ii

(j-1) - Fij Ij(j-1) ; j= 1,2, ….. (i -1) (3.32)

~ The back substitution is performed on the intermediate vector V using the elements of U stored in F to obtain the solution vector V as shown below:

~

F11 F12 F13 V1 V1

~ 0 F22 F23 V2 = V2 (3.33)

~ 0 0 F33 V3 V3

~ ~ ~

V3 = V3(1) ; V3

(1) = V3 / F33

~ ~ ~ ~ ~ ~ V2 = V2

(2) ; V2 (1) = V2 – F23 V3

(1) ; V2 (2) = V2

(1) / F22 (3.34) ~ ~ ~ ~ ~ ~ ~ ~ ~

V1 = V1 (3) ; V1

(1) = V1 – F12 V2 (2) ; V1

(2) = V1 (1) – F13 V3

(1) ; V1 (3) = V1

(2) / F11


The above steps can be generalized as

i = n, ….., 2, 1

j = n, ….., (i+1), i

No If (j = i)?

Yes

~ ~ ~

Vi = Vi – Fij Vj

~ ~ Vi = Vi / Fij

Fig 3.5 Flow Chart for Repeat Solution

Z matrix through factorization of Y and repeat solution:

The relation between Y and Z is given by eqn. (3.35)

Y Z = I (3.35)

Where, I is the identity matrix.

Eqn. (3.35) can be written as

Y (Z1 Z2 …. Zn) = ( I1 I2 ….. In ) (3.36)

Where, Zi = ith column of Z matrix

and Ii = ith column of identity matrix.

Eqn. (3.36) can be split into n equations

Y Zi = Ii ; i = 1, 2, …. ,n (3.37)

From (3.37) it is clear that the ith column Zi of Z matrix can be obtained

through factorization of Y and repeat solution on Ii.


Instruction: The program should have three sections: Input Section, Compute Section and

Output Section. I. Input Section: Pre-requisite:

Before creating the input data file, draw a single-line diagram showing the buses, lines, transformers, shunt elements, bus generations and demands. (Refer Question) Bus Id numbers are serially given from 1 to NB where NB is the total number of buses in the system.

The data to be read from an input file should contain the general data,

transmission line data, transformer data and shunt element (capacitor / reactor) data in the following sequence:

i. General Data: Number of buses, number of transmission lines, number of transformers, number

of shunt elements (capacitor / reactor) and the base MVA of the network.

ii. T r a n sm i s s i o n Line Data: The following data to be read for all lines, (one line for each transmission line):

Identification number (serial number) of line

Identification number of sending end and receiving end buses of line

Series impedance (R,X)of the line in per unit.

Half-line-charging susceptance, Bc, in per unit. Maximum loadability limit (rating) of line in MVA

iii. Transformer Data: The following data to be read for all transformers (one line for each transformer):

Identification number (serial number) of transformer Identification number of sending end (tap side) bus and receiving end bus of transformer. Impedance (R,X) of transformer in per unit.

Off-nominal tap ratio Maximum loadability limit (rating) of transformer in MVA

iv. Shunt element (capacitor / reactor) Data:

The following data to be read for all shunt elements (one line for each element):

Identification number (serial number) of the shunt element Identification number of the bus Rated capacity in MVA (positive for capacitor and negative for inductor).

II. Compute Section:

To form the Y matrix using the algorithm in section 3.4


III. Output Section:

To create an output file in a report form comprising the following:

(i) Student Information :

(ii) Input Data with proper headings :

(iii) Results with proper headings: Element value (in per unit) of Y printed row wise.

EXERCISES:

BY using AU power software & MATLAB for formation of bus admittance matrix Y

of a power network using the “Two – Rule Method”.

3-BUS, 3-LINES POWER SYSTEM Single-Line Diagram

Base MVA = 100

Transmission Line Data:

Line ID. No

Send Bus No.

Receive Bus No.

Resist P.U

Reactance P.U.

Half Line charging Suscept.

P.U

Rating MVA

1 1 2 0

2 1 3 0

3 2 3 0

1

1

2

2

3

3


Viva Questions

1. What is per-unit? (or) Define the way of representation of power system quantities. (or) Define the per unit value of any electrical quantity.

2. Define bus in a power system network (or) what is called as bus? 3. What is single line diagram (or) one line diagram? 4. List out the components of power system. 5. What is called as driving point and transfer impedances? 6. Mention the advantages of bus admittance matrix Y Bus. 7. What is Triangular factorization? 8. Define primitive network. (or) What is primitive network? 9. What is off nominal transformer ratio? 10. Define unity matrix. 11. What is called as self admittance and mutual admittance matrix? 12. What are the representation of loads? 13. How loop impedance matrices are formed? 14. What is a bus impedance matrix? 15. How bus admittance matrices are formed? 16. What are the methods available for forming bus impedance matrix? 17. How the power system network can be studied using mathematical model? 18. Define bus admittance matrix. 19. How the network equations can be formed? 20. What are the advantages of per unit computations?


Result:


Secured

Marks

Awarded



Conducting 15




Record 10

Viva - voce 10

Total Marks 100


EXPERIMENT: 4 Date:

LOAD FLOW ANALYSIS - I: SOLUTION OF LOAD

FLOW AND RELATED PROBLEMS USING

GAUSS-SEIDEL METHOD

AIM

To understand, the basic aspects of load flow analysis of power systems

that is required for effective planning and operation of power systems.

To understand, in particular, the mathematical formulation of load flow model in complex form and a simple method of solving load flow

problems of small sized system using Gauss-Seidel iterative algorithm.

OBJECTIVES

To write a computer program to solve the set of non-linear load flow equations using Gauss-Seidel Load Flow (GSLF) algorithm and present

the results in the format required for system studies.

To investigate the convergence characteristics of GSLF algorithm for

normally loaded small system for different acceleration factors.

To investigate the effects on the load flow results, load bus voltages and line /transformer loadings, due to the following control actions:

a. Variation of voltage settings of P-V buses

b. Variation of shunt compensation at P-Q buses

c. Variation of tap settings of transformer

d. Generation shifting or rescheduling.

SOFTWARE REQUIRED

GAUSS – SEIDEL METHOD module of AU Power lab or equivalent


Need For Load Flow Analysis

Load Flow analysis, is the most frequently performed system study by

electric utilities. This analysis is performed on a symmetrical steady-state

operating condition of a power system under “normal” mode of operation


and aims at obtaining bus voltages and line / transformer flows for a

given load condition. This information is essential both for long term

planning and next day operational planning. In long term planning, load

flow analysis, helps in investigating the effectiveness of alternative plans

and choosing the “best” plan for system expansion to meet the projected

operating state. In operational planning, it helps in choosing the “best”

unit commitment plan and generation schedules to run the system

efficiently for the next day‟s load condition without violating the bus

voltage and line flow operating limits.

Description of Load Flow Problem

In the load flow analysis, the system is considered to be operating under steady state balanced condition and per phase analysis is used.

With reasonable assumptions and approximations, a power system

under this condition may be represented by a power network as

shown by the single-line diagram in Annexure 4.1.

The network consists of a number of buses (nodes) representing either

generating stations or bulk power substations, switching stations

interconnected by means of transmission lines or power transformers.

The bus generation and demand are characterized by complex powers

flowing into and out of the buses respectively. Each t r a n s m i s s i o n

l i n e i s characterized by its π equivalent circuit. Shunt compensating

capacitor or reactors are represented as shunt susceptance.

Load Flow analysis is essentially concerned with the determination

of complex bus voltages at all buses, given the network configuration

and the bus demands. Let the given system demand (sum of all the bus

demands) be met by a specific generation schedule. A generation

schedule is nothing but a combination of MW generation (chosen within

their ratings) of the various spinning generators the total of which

should match the given system demand plus the transmission losses.

It should be noted that there are many generation schedules

available to match the given system demand and one such schedule is

chosen for load flow analysis.

The “Ideal” Load Flow problem is stated as follows: Given: The network configuration (bus admittance matrix), and all the bus power injections (bus injection refers to bus generation minus bus demand)


To determine: The complex voltages at all the buses.

X = ( δ 1 δ 2 ………… δ n V1 V2 …….VN)T = ( δ T VT)T

Once the „state‟ of the system is known, all the other quantities of interest

in the power network can be computed. The above statement of Load Flow problem will be modified later after

taking into account certain practical constraints.

Development of Load Flow Model

The Load Flow model in complex form is obtained by writing one

complex power matching equation at each bus.

PGk+ jQGk G PDk+ jQDk PIk +jQIk = (PGk - PDk) + j (QGk - QDk)

k Vk

(Pk+ jQk) Ik

k Vk

(Pk + jQk) Ik

(a) (b)

Fig 4.1 Complex Power Balancing at a Bus

Referring to Fig 4.1 (b) the complex power injection (generation minus

demand) at the kth bus is equal to the complex power flowing into the

network at that bus which is given by

PIk + JQIk = Pk + jQk (4.1)

In expanded form, (PGk - PDk) + j (QGk - QDk) = VkIk* (4.2)

The network equation relating bus voltage vector V with bus current

vector I is

YV = I (4.3)

Taking the kth component of I from (4.3) and substituting for Ik* in (4.2) we get the power flow model in complex form as

N

PIk + jQIk = Vk km* Vm*; k=1,2,………… N (4.4)

m=1


Transmission line / Transformer Flow Equation:

In a Load Flow package after solving equation (4.4) for complex bus

voltages using any iterative method, the active and reactive power flows in all the lines/ transformers are to be computed. A common Fig 4.2. For

a transmission line set the variable “a” equal to unity and for a

transformer set variable bc equal to zero. The expression for power flow

in line / transformer k-m from the kth bus to the mth bus, measured at the

kth bus end is given by (refer Fig 4.2)

Fig 4.2 PI Equivalent Circuit of a Transmission Line / Transformer

Pkm + jQkm = Vk Ik* = Vt It * (4.5)

Noting that, Vk / Vt = a (4.6)

It = (Vt -Vm) ykm + Vt (jbc) (4.7)

Substituting equations (4.6) and (4.7) in equation (4.5) we get

Pkm + jQkm = (Vk/a) [(Vk * /a) – V m *)] y km * + (Vk/a)2 (jbc)* (4.8)

Similarly the power flow in line k-m from the mth bus to kth bus measured at the mth bus end is

Pmk + jQmk = VmI m * = Vm[Vm * – (V k * /a)] y km * + V2m (jbc)* (4.9)

The complex power loss in line / transformer k-m, PLkm + jQLkm, is given by the

sum of the two expressions (4.8) and (4.9)

Classification of Buses

From the Load Flow model in equation (4.4) and from the definition of

complex bus voltage, Vk as

Vk = |Vk| └ δ k


one can observe that there Any two of these four may be treated as independent variables (that is specified) while the other two may be computed by solving power flow equations. The buses are classified based on the variables specif ied. Three types of buses classified based on practical requirements are given below:

Slack bus: While specifying a generation schedule for a given system

demand, one can fix up the generation setting of all the generation

buses except one bus because of the limitation of not knowing the

transmission loss in advance. This leaves us with the only s and |Vs|

pertaining to a generator bus (usually a large capacity generation bus is

chosen and this is called as slack bus) and solving for the remaining (N-

1) complex bus voltages from the respective (N-1) complex load

flow equations. Incidentally the specification of |Vs| helps us to fix

the voltage level of the δs as zero, makes Vs as reference phasor. Thus

for the slack both δ and |V| are speci f ied and PG and QG are to

be computed only after a i terat ive solution of bus voltages is

completed.

P-V buses: In order to maintain a good voltage profile over the system,

it is customary to maintain the bus voltage magnitude of each of the

generator buses at a desired level. This can be achieved in practice by

proper Automatic Voltage Regulator (AVR) settings. These generator

buses and other Voltage-controlled buses with controllable reactive power

source such as SVC buses are classified as P-V buses since PG and |V| are

specified at these buses. Only one state variable, δ is to be computed at

this bus. The reactive power generation QG at this bus which is a

dependent variable is also to be computed to check whether it lies

within its operating limits.

P-Q buses: All other buses where both PI and QI are specified are

termed as P-Q buses and at these buses both δ and |V| are to be computed. Hence the “Practical” Load Flow problem may be stated as:

Given: The network configuration (bus admittance matrix), all the

complex bus power demands, MW generation schedules and voltage

magnitudes of all the P-V buses, and voltage magnitude of the slack

bus, To determine: The bus voltage phase angles of all buses except the slack

bus and bus voltage magnitudes of all the P-Q buses. Hence the state

vector to be solved from the Load Flow model is


X = (δ 1 δ 2 δ NP V1V2 …….. VNQ) T Where, NP = N-1

NQ = N-NV – 1

and the NV number of P-V buses and the slack bus are arranged at the end.

Solution to Load Flow Problem A number of methods are available for solving Load Flow problem. In all

these methods, voltage solution is initially assumed and then improved

upon using some iterative process until convergence is reached.

The following three methods will be presented: (i) Gauss-Seidel Load Flow (GSLF) method (ii) Newton-Raphson Load Flow (NRLF) method (iii) Fast Decoupled Load Flow (FDLF) method The first method GSLF is a simple method to program but the voltage

solution is updated only node by node and hence the convergence rate is

poor. The NRLF and FDLF methods update the voltage solution of all the

buses simultaneously in each iteration and hence have faster convergence

rate. Taking the complex conjugate of equation (4.4) and transferring Vk to the left hand side, we obtain

N

Vk = [(PIk – jQIk) / Vk* - Σ Y km Vm] / Ykk m = 1 m ≠ k

k = 1,2, …..(N -1) (Slack bus excluded) (4.10)

Define Ak = (PIk – jQIk) / Ykk (4.11) Bkm = Ykm / Ykk (4.12)


LOAD FLOW SOLUTION (EQUATION) :

The assumptions and approximations made in the load flow equations.

N

P V y V Cos in ni i in inn 1

(1)

N

Q V y V Sin in ni i in inn 1

(2)

Line resistances being small are neglected. Active power loss PL of the

system is zero. Thus in (1) & (2) 90 and 90in ii

ni is small 6

so that Sin n ni i

All buses other than the slack bus (bus (1)) are PV buses.

i.e. Voltage magnitudes at al the buses including the slack bus are

specified & (2) are reduced as

N

P V y V n , i 1, 2......Nni i in in 1

(3)

N 2

Q V y V Cos n V y , i 1, 2......Nni i in i i iin 1

n i

(4)

Since iV ‟S are specified equation (3) represents a set of linear algebraic

equations in i „s which are (N-1) in number as 1 is specified at slack bus

01 .

The Nth equation corresponding to slack bus (N=1) is redundant as the real

power injected at this bus is now fully specified as,

N N

P P P ; P 01 Di LGii 2 i 2

Equation (3) can be solved explicitly for n .............., 32 which is

substituted in (4) yields Qi „s the reactive power bus injections.

FORMULATION OF LOAD FLOW EQUATIONS USING Y Bus MATRIX

The load flow equations can be formed using either the mesh or node basis

equation of a power system. However, from the view point of computer

time and memory, the nod admittance formulation using the nodal voltages

as the independent variables is the mo economic.


The node basis matrix equation of a n-bus system given by

Ybus V=I …………………………….…(1)

Where,

Ybus -- Bus admittance matrix of order (nxn)

V Bus (node) -- Voltage matrix of order (n x 1)

I -- Source current matrix of order (n x 1).

An separating the real and imaginary parts of eqn (1) we get.

..………………. (2)

…………………(3)

..……………….. (4)

The equations (2), (3) and (4) are called load-flow equations of Newton-Raphson method.

LOAD FLOW SOLUTION BY GAUSS-SEIDEL METHOD

The Gauss-Seidel method is an iterative algorithm for solving a set of non-

linear load flow equations. The non-linear load flow equations are given by

equation (2).

When, p = 1,2 n and this equation is presented here for convenience

V

1

1 1*)(

1 p

q

n

pq

qpqqpq

p

pp

pp

p VYVYV

jQP

Y ……………. (5)

Where, P = 1,2,3 ……… n

The variables in the equations obtained from equ (5) for p = 1,2,3 ……. n are

the node voltages V1 ,V2 ,V3 ,……….Vn. In Gauss-Seidel method an initial

value of voltages are assumed and they are denoted as

V1‟ ,V2‟ ,V3‟ ,……….Vn‟. On substituting these initial values in equ (5) and by

taking p = 1, the revised value of bus- 1 voltage V1‟ is computed. The revised

value of bus voltage V1‟ is replaced for initial value V0 and the revised bus-2

voltage V2‟ is computed. Now replace the V1‟ for V1 and V2‟ for V2 and

perform the calculationforbus-3 and soon.


The process of computing all the bus voltages as explained above is called

one iteration. The iterative process is then repeated till the bus voltage

converges within prescribed accuracy. The convergence of bus voltage is

quite sensitive to the initial values assumed. Based on practical experience it

is easier to get a set of initial voltages very close to final solution.

In view of the above discussions the load flow equation [5] can be written in

the modified form as shown below.

V

1

1 1

1

*

1

)(

1 p

q

n

pq

K

qpq

K

qpqK

p

pp

pp

K

p VYVYV

jQP

Y …………………….(6)

where,

Vik = kth iteration value of bus voltage Vi

Vik+1= (k+1)th iteration value of bus voltage Vi

It is important to note that the slack bus is a reference bus and so its voltage

will not change. Therefore in each iteration the slack bus voltage is not

modified

For a generator bus, the reactive power is not specified. Therefore in order to

calculate the phase of bus voltage of a generator bus using equation (6), we

have to estimate the reactive power from the bus voltages and admittances

as shown below.

From equation we get,

……………………….(7)

From equation (7) the equation for complex power in bus-p during (k + i)th

iteration can be obtained as shown in equation (8).

…………………(8)

The reactive power of bus-p during (k + i)th iteration is given by imaginary part of equation (8).


………….. (9) Also, for generator buses a lower and upper limit for reactive powers will

be specified. In each iteration, the reactive power of generator bus is

calculated using equation (9) and then checked with specified limits. If it

violates the specified limits then the reactive power of the bus is equated

to the limited and it is treated as load bus. If it does not violate the limits

then the bus is treated as generator bus.

GAUSS SEIDEL METHOD USED IN POWER FOLW ANALYSIS:

Digital solutions of the power flow problems follow an iterative process

by assigning estimated values to the unknown bus voltages and by

calculating a new value for ach bus voltage from the estimated value at

the other buses and the real and reactive power specified.

A new set of values for the voltage at each bus is thus obtained and used to calculate still another set of bus voltages. Each calculation of a new set of voltages is called ITERATION.

The iterative process is repeated until the changes at each bus are less

than a specified minimum value.

We derive equations for a four bus system with the slack bus designated

as number 1 computations start with bus (2)

If P2sch and Q2sch are the scheduled real and reactive power, entering the

network at bus (2)

From the equation, N*

P jQ V y Vni i i inn 1

(1)

with, i = 2 and N = 4

P sch jQ sch2 2y V y V y V y V21 1 22 2 23 3 24 4*

V2

(2)

Solving for V2 gives

P sch jQ sch1 2 2

V y V y V y V2 21 1 23 3 24 4*y V22 2

(3)


If suppose bus (3) and (4) are also load buses with real and reactive power

specified.

At bus (3),

P sch jQ sch1 3 3

V y V y V y V3 31 1 32 2 34 4*y V33 3

(4)

Similarly at bus (4)

P sch jQ sch1 4 4

V y V y V y V4 41 1 42 2 43 3*y V44 4

(5)

The solution proceeds by iteration based on scheduled real and reactive

power at buses (2), (3) and (4). The scheduled slack bus voltage is

V V1 1 1 and initial voltage estimates 0

4

0

3

0

2 , VandVV at other buses

Solution of equation (3) gives the corrected voltage 1

2V calculated from

P sch Q sch11 0 02, 2,V y V y V y V2 21 1 23 3 24 40 *

y V22 2

(6)

All the quantities in the right hand side expression are either fixed specifications or initial estimates.

The calculated value 1

2V and the estimated value 0

2V will not agree.

Agreement would be reached to a good degree of accuracy after several

iterations.

This value would not be the solution for V2 for the specific power flow

conditions, however because the voltages on which this calculation for V2

depends are the estimated values 0

3V and 0

4V at the other buses, and the

actual voltages are not yet known.

Substituting 1

2V in 4th equation we obtain the first calculated value at

bus (3)

P sch jQ sch11 1 03, 3,V y V y V y V3 31 1 32 2 34 40 *

y V33 3

(7)


The process is repeated at bus (4)

P sch jQ sch11 1 14, 4,V y V y V y V4 41 1 42 2 43 30 *

y V44 4

(8)

This completes the first iteration in which the calculated values are found

for each state variable.

Then, the entire process is carried out again and again until the amount of

correction in voltage at every bus is less than some predetermined precision

index.

This process of solving the power – flow equations is known as the

“GAUSS – SEIDEL ITERATIVE METHOD”

It is common practice to set the initial estimates of the unknown voltages at

all load buses equal to 000.1 per unit.

Such initialization is called a FLAT START because of the uniform

voltage profile assumed.

For a system of N buses the general equation for the calculated voltage at

any bus (i) where P and Q are scheduled is

P sch jQ sch i 1 N1k k k 1i,, i ,

V y V y Vi ij j ij jk 1 * j 1 j i 1y Vii i

(9)

The superscript (K) denotes the number of iteration in which the voltage is currently being calculated and (k-1) indicates the number of the preceding iteration.

Equation (9) applies only at load buses where real and reactive powers are specified.

An additional step is necessary at Voltage controlled buses where voltage magnitude is to remain constant. The number of iterations required may be reduced may be reduced considerably if the connection in voltage at each bus is multiplied by some


constant that increases the amount of correction to bring the voltage closer to the value it is approaching. The multiplier that accomplishes this improved convergence is called an ACCELERATION FACTOR.

The difference between the newly calculated voltage and the best previous

voltage at the bus is multiplied by the appropriate acceleration factor to

obtain a better correction to be added to the previous value. For example,

At bus (2) in the first iteration we have the accelerated value accV .1

2 defined

by

1 0 1

V , acc 1 V V2 2 2

1 0 1 0

V , acc V V V2 2 2 2 (10)

acceleration factor.

Generally, for bus (i) during iteration K, the accelerated value is given by,

k k 1 k

V , acc 1 V Vi i,acc i

k k 1 k k 1

V , acc V V Vi i ,acc i i ,acc

(11)

In power flow studies is generally set at about 1.6 and cannot exceed 2 if

convergence is to occur.

Voltage controlled buses (or) PV – buses:

When voltage magnitude rather than the reactive power is specified at bus

(i), the real and imaginary components of the voltages for each iteration are

found by first computing a value for the reactive power.

From N*

P jQ V y Vni i i inn 1

N*Q I V y Vmi i ij jj 1

(12)

Equivalent algorithmic expression,

i 1 Nk k 1 * k k 1

Q I V y V y Vmi i ij j ij jj 1 j 1

(13)


Im Imaginary par of k

Qi is substituted in equation (9) to find a new value

of k

Vi .The components of the new

kV

i are then multiplied by the ratio of

the specified constant magnitude Vi to the magnitude of k

Vi from (9)th

equation.

In 4 – bus system, if bus (4) is voltage controlled.

Equation (13) becomes,

1 0 * 1 1 0

Q I V y V y V , acc y V , acc y V4 m 4 41 1 42 2 43 3 44 4 (14)

The calculated voltages of buses (2) and (3) are accelerated

values of the first iteration.

Substitute 1

4Q for Q4,sch in (9) for bus (4) yields.

1P , sch jQ11 1 14 4

V y V y V y V acc4 41 1 42 2 43 30 *

y V44 4

(15)

all the quantities on the right hand side are known.

Since 4V is specified,

We correct the magnitude of 1

4V as

1V1 4

V , corr V4 4 1V4

(16)

and proceed to the next step with stored value corrV ,1

4 of bus (4) voltage

having the specified magnitude in the remaining calculations of the

iteration. The reactive power Qg must be within definite limits.

Q Q Qmin g nax

Advantages of Gauss- seidel method:

Calculations are simple and so the programming task is lesser.

The memory requirement is less

Useful for small size system.

Disadvantages:

Requires large number of iterations to reach convergence.

Not suitable for large systems

Convergence time increases with size of the system.


FLOWCHART FOR LOAD FLOW SOLUTION

BY GAUSS-SEIDEL METHOD


Procedure

Write a program f o r iteratively so l v i n g non-linear L o a d Flow

equations using Gauss-Seidel method for small and medium sized

power systems. The program should have three sections i.e. input

section, Compute section and Output section.

I - Input section

Pre-requisite:

Before creating the input data file, draw a single- line diagram showing

the buses, lines, t r a n s f o r m e r s , shunt e l e m e n t s ,

bus g e n e r a t i o n a n d l o a d s ( Refer in Annexure 4 .1). Bus ID

numbers are serially given from 1 to NB where NB is the total

number of buses comprising P-V buses (which includes the slack bus)

and P-Q buses.

The data to be read from an input file should contain general data,

bus data, line data, transformer data and shunt element data in the

following sequence.

(i) General Data

The following data are read in one line

a) Total number of buses

b) Number of P-V buses

This includes all the voltage-controlled buses such as generator buses (including slack bus),synchronous condenser buses and SVC buses for which a specified voltage magnitude is to be maintained.

c) Number of P-Q buses.

This includes all load buses, dummy (zero generation and zero

loads) buses and generator buses in which voltage magnitude

is not controlled.

d) Number of transmission lines. e) Number of transformers. f) ID number of slack bus. g) Number of shunt elements. h) Maximum number of iterations to be performed. i) System MVA base. j) Convergence tolerance in p.u. voltage. k) Acceleration factor to be used.


(ii) Bus Data

The following data are read for each bus in one line.

P-V Bus Data

(a) ID number of bus (b) Active power generation in MW

(c) Active power load in MW

(d) Reactive power load in MVAR

(e) Maximum limit of reactive power generation in MVAR.

(f) Minimum limit of reactive power generation in MVAR.

(g) Scheduled voltage magnitude of the bus in p.u.

P-Q Bus Data:

(a) ID number of bus

(b) Active power load in MW.

(c) Reactive power load in MVAR.

(d) Initial voltage magnitude assumed in p.u.

(iii) Transmission line data (iv) Transformer Data (v) Shunt elements data

The data to be read and the sequence in which it is to be read for

(iii), (iv) and (v) are the same as that given in exercise under

Experiment 4.

II - Compute Section

Starting from the initial bus voltage solution (usually a “flat start” is assumed), update the voltage solution iteratively using Gauss-Seidel method until the convergence criteria on bus voltage m a g n i t u d e is s a t i s f i e d . Typical v a l u e o f t o l e r a n c e f o r v o l t a g e m a g n i t u d e convergence is 0.001 p.u .

III - Output section

Create an output file in a report form comprising the following:

(i) Student information: As specified in exercise under experiment 3.

(ii) Input data: with proper headings.

(iii) Results obtained with proper headings, in the following sequence:


System details

Total No of Buses: No of P-V

Buses: No of P-Q Buses: No of Lines:

No of Transformers: No of Shunt elements: Slack

bus ID number: System base MVA:

Convergence details

Maximum iterations prescribed:

Maximum iterations taken:

Convergence tolerance prescribed:

Convergence limit reached:

Bus Results The results are to be printed under the following headings:

Bus

Id.No

Generation Demand Bus Voltage Compensation MW MVAR MW MVAR Magnitude

p.u.

Angle

degrees

MVAR

For each one of the buses, whether it is a P-V bus or a P-Q bus, one line

covering the above information is to be printed.

Transmission line / Transformer Results

The results are to be printed under the following headings.

Sending Bus No

Receiving Bus No

Flow MW

Flow MVAR

Flow MVA

Rating MVA

P loss MW

Q loss MVAR

For each one of the lines / transformers, two lines are to be printed. The

first line printed should have all the above information pertaining to the

forward direction including Rating, Ploss and Qloss. The second line

to be printed should have the above information, pertaining to the

flow in the reverse direction excluding Rating, Ploss and Qloss.

Exercise:

i. Prepare the data for the 6-bus system described in the Annexure 4.1. Run the GSLF program with an acceleration factor of 1.0 and a convergence for voltage tolerance of 0.001p.u by using available software.



ANNEXURE 4.1

6-BUS, 7-LINES / TRANSFORMER POWER SYSTEM Single-Line Diagram

Buses : 6, numbered serially from 1 to 6 Lines : 5, numbered serially from L1 to L5 Transformers: 2, numbered serially as T1 and T2 Shunt Load : 2, numbered serially as s1 and s2 Base MVA : 100

Bus Data – P-V Buses:

Bus ID No.

Generation, MW Demand Gen. Limit MVAR

Scheduled Volt (p.u)

Schedule Max Min MW MVAR Max Min 1 ? 200 40 0.0 0.0 100.0 -50.0 1.02

2 50.0 100 20 0.0 0.0 50.0 -25.0 1.02

Bus Data – P-Q Buses

Bus ID No Demand

Volt. Mag. Assumed

(p.u) MW MVAR

3 55.0 13.0 1.0

4 0.0 0.0 1.0

5 30.0 18.0 1.0

6 50.0 5.0 1.0



Line ID.

No

Send Bus No.

Receive Bus No.

Resist P.U

Reactance P.U.


P.U

Rating MVA

1 1 6 0.123 0.518 0.0 55

2 1 4 0.080 0.370 0.0 65

3 4 6 0.087 0.407 0.0 30

4 5 2 0.282 0.640 0.0 55

5 2 3 0.723 1.050 0.0 40

Transformer Data:

Transformer ID.No

Send (*) Bus No.

Receive Bus No.

Resist. P.U

Reactance P.U.

Tap Ratio Rating MVA

1 6 (*) 5 0.0 0.300 1.000 30

2 4 (*) 3 0.0 0.133 1.000 55

(*) Note: The sending end bus of a transformer should be the tap side.

Shunt Element Data:

Shunt ID No. Bus ID. No. Rated Capacity MVAR (*)

1 4 2.0

2 6 2.5

(*) Note: Sign for capacitor : +ve Sign for Inductor : -ve

Exercise:

ii. Two voltage sources V 120 5 & V 100 01 2

are connected by a

short line of impedance Z 1 j7 . Determine the real and reactive

power supplied or received by each source and the power loss in the line. Write the program in MATLAB. Compare the results.


Viva Questions

1. What is the need for load flow study? 2. Why power flow analysis is made? 3. Define voltage controlled bus. 4. What are the different types of buses in a power system? 5. What is power flow study or load flow study? 6. States the major steps involved in load flow studies. 7. What are the quantities determined through load flow studies? 8. What is PQ –bus? 9. Define load bus? 10. What is Swing bus? 11. What is the need for slack bus? 12. What are the iterative methods mainly used for the solution of load flow

problems? 13. Write the load flow equation of Gauss-Seidel method. 14. What are the advantages & disadvantages of Gauss-Seidel method? 15. What is an acceleration factor? 16. What do you mean by a flat voltage start? 17. When the generator bus is treated as load bus? 18. What are the information‟s that are necessary to solve the power –flow problem? 19. Why acceleration factor used in the Gauss –Seidel method? 20. What are the operating constraints used in the load flow studies?


Result:


Secured

Marks

Awarded



Conducting 15




Record 10

Viva - voce 10

Total Marks 100


EXPERIMENT: 5 Date:

LOAD FLOW ANALYSIS - I: SOLUTION OF LOAD

FLOW AND RELATED PROBLEMS USING

NEWTON-RAPHSON METHOD

AIM

To understand the following for medium and large scale power systems: Mathematical formulation of the load flow problem in real

variable form. Newton-Raphson method of load flow (NRLF) solution.

To become proficient in the usage of software for practical problem

solving in the areas of power system planning and operation

To become proficient in the usage of the software in solving problems using Newton-Raphson method.

OBJECTIVES

To investigate the convergence characteristics of load flow solutions using NRLF algorithm for different sized systems and compare the same with that of GSLF algorithm.

To investigate the effect of variation of voltage-control parameters such

as generator voltage magnitude setting, off-nominal tap ratio of transformer and MVAR injections of shunt capacitor / inductor on the voltage profile and transmission loss of the system.

To assess the effect of single outage contingencies such as a line

outages and generator outages. SOFTWARE REQUIRED

NEWTON RAPHSON module of AU Powerlab or equivalent


Need For Load Flow Analysis

Load Flow analysis, is the most frequently performed system study by

electric utilities. This analysis is performed on a symmetrical steady-state

operating condition of a power system under “normal” mode of operation

and aims at obtaining bus voltages and line / transformer flows for a

given load condition. This information is essential both for long term

planning and next day operational planning. In long term planning, load


flow analysis, helps in investigating the effectiveness of alternative plans

and choosing the “best” plan for system expansion to meet the projected

operating state. In operational planning, it helps in choosing the “best”

unit commitment plan and generation schedules to run the system

efficiently for the next day‟s load condition without violating the bus

voltage and line flow operating limits.

Description of Load Flow Problem

In the load flow analysis, the system is considered to be operating

under steady state balanced condition and per phase analysis is used.

With reasonable assumptions and approximations, a power system

under this condition may be represented by a power network as

shown by the single-line diagram in Annexure 5.1.

The network consists of a number of buses (nodes) representing either

generating stations or bulk power substations, switching stations

interconnected by means of transmission lines or power transformers.

The bus generation and demand are characterized by complex powers

flowing into and out of the buses respectively. Each t r a n s m i s s i o n

l i n e i s characterized by its π equivalent circuit. Shunt compensating

capacitor or reactors are represented as shunt susceptance.

Load Flow analysis is essentially concerned with the determination

of complex bus voltages at all buses, given the network configuration

and the bus demands. Let the given system demand (sum of all the bus

demands) be met by a specific generation schedule. A generation

schedule is nothing but a combination of MW generation (chosen within

their ratings) of the various spinning generators the total of which

should match the given system demand plus the transmission losses.

It should be noted that there are many generation schedules

available to match the given system demand and one such schedule is

chosen for load flow analysis.

The “Ideal” Load Flow problem is stated as follows: Given: The network configuration (bus admittance matrix), and all the bus power injections (bus injection refers to bus generation minus bus demand)


To determine: The complex voltages at all the buses.

X = ( δ 1 δ 2 ………… δ n V1 V2 …….VN)T = ( δ T VT)T

Once the „state‟ of the system is known, all the other quantities of interest

in the power network can be computed. The above statement of Load Flow problem will be modified later after

taking into account certain practical constraints.

Development of Load Flow Model

The Load Flow model in complex form is obtained by writing one

complex power matching equation at each bus.

PGk+ jQGk G PDk+ jQDk PIk +jQIk = (PGk - PDk) + j (QGk - QDk)

k Vk

(Pk+ jQk) Ik

k Vk

(Pk + jQk) Ik

(a) (b)

Fig 5.1 Complex Power Balancing at a Bus

Referring to Fig 5.1 (b) the complex power injection (generation minus

demand) at the kth bus is equal to the complex power flowing into the

network at that bus which is given by

PIk + JQIk = Pk + jQk (5.1)

In expanded form, (PGk - PDk) + j (QGk - QDk) = VkIk* (5.2)

The network equation relating bus voltage vector V with bus current

vector I is

YV = I (5.3)

Taking the kth component of I from (5.3) and substituting for Ik* in (5.2) we get the power flow model in complex form as

N

PIk + jQIk = Vk km* Vm*; k=1,2,………… N (5.4) m=1

In (5.4) there are N complex variable equations from which the N

unknown complex variables, V1,……V N can be determined.


Transmission line / Transformer Flow Equation:

In a Load Flow package after solving equation (5.4) for complex bus

voltages using any iterative method, the active and reactive power flows

in all the lines/ transformers are to be computed. A common Fig 5.2. For

a transmission line set the variable “a” equal to unity and for a

transformer set variable bc equal to zero. The expression for power flow

in line / transformer k-m from the kth bus to the mth bus, measured at the

kth bus end is given by (refer Fig 5.2)

Fig 5.2 PI Equivalent Circuit of a Transmission Line / Transformer

Pkm + jQkm = Vk Ik* = Vt It * (5.5)

Noting that, Vk / Vt = a (5.6)

It = (Vt -Vm) ykm + Vt (jbc) (5.7)

Substituting equations (5.6) and (5.7) in equation (5.5) we get

Pkm + jQkm = (Vk/a) [(Vk * /a) – V m *)] y km * + (Vk/a)2 (jbc)* (5.8)

Similarly the power flow in line k-m from the mth bus to kth bus measured at the mth bus end is

Pmk + jQmk = VmI m * = Vm[Vm * – (V k * /a)] y km * + V2m (jbc)* (5.9)

The complex power loss in line / transformer k-m, PLkm + jQLkm, is given by the

sum of the two expressions (5.8) and (5.9)

Classification of Buses

From the Load Flow model in equation (5.4) and from the definition of

complex bus voltage, Vk as

Vk = |Vk| └ δ k


one can observe that there Any two of these four may be treated as independent variables (that is specified) while the other two may be computed by solving power flow equations. The buses are classified based on the variables specif ied. Three types of buses classified based on practical requirements are given below:

Slack bus: While specifying a generation schedule for a given system

demand, one can fix up the generation setting of all the generation

buses except one bus because of the limitation of not knowing the

transmission loss in advance. This leaves us with the only s and |Vs|

pertaining to a generator bus (usually a large capacity generation bus is

chosen and this is called as slack bus) and solving for the remaining (N-

1) complex bus voltages from the respective (N-1) complex load

flow equations. Incidentally the specification of |Vs| helps us to fix

the voltage level of the δs as zero, makes Vs as reference phasor. Thus

for the slack both δ and |V| are speci f ied and PG and QG are to

be computed only after a i terat ive solution of bus voltages is

completed.

P-V buses: In order to maintain a good voltage profile over the system,

it is customary to maintain the bus voltage magnitude of each of the

generator buses at a desired level. This can be achieved in practice by

proper Automatic Voltage Regulator (AVR) settings. These generator

buses and other Voltage-controlled buses with controllable reactive power

source such as SVC buses are classified as P-V buses since PG and |V| are

specified at these buses. Only one state variable, δ is to be computed at

this bus. The reactive power generation QG at this bus which is a

dependent variable is also to be computed to check whether it lies

within its operating limits.

P-Q buses: All other buses where both PI and QI are specified are

termed as P-Q buses and at these buses both δ and |V| are to be computed. Hence the “Practical” Load Flow problem may be stated as:

Given: The network configuration (bus admittance matrix), all the

complex bus power demands, MW generation schedules and voltage

magnitudes of all the P-V buses, and voltage magnitude of the slack

bus, To determine: The bus voltage phase angles of all buses except the slack

bus and bus voltage magnitudes of all the P-Q buses. Hence the state

vector to be solved from the Load Flow model is

X = (δ 1 δ 2 δ NP V1V2 …….. VNQ) T


Where, NP = N-1

NQ = N-NV – 1

and the NV number of P-V buses and the slack bus are arranged at the end.

Solution to Load Flow Problem A number of methods are available for solving Load Flow problem. In all

these methods, voltage solution is initially assumed and then improved

upon using some iterative process until convergence is reached.

The following three methods will be presented: (i) Gauss-Seidel Load Flow (GSLF) method (ii) Newton-Raphson Load Flow (NRLF) method (iii) Fast Decoupled Load Flow (FDLF) method

The first method GSLF is a simple method to program but the voltage

solution is updated only node by node and hence the convergence rate is

poor.

The NRLF and FDLF methods update the voltage solution of all the buses

simultaneously in each iteration and hence have faster convergence rate. Taking the complex conjugate of equation (5.4) and transferring Vk to the left hand side, we obtain

N

Vk = [(PIk – jQIk) / Vk* - Σ Y km Vm] / Ykk m = 1 m ≠ k

k = 1,2, …..(N -1) (Slack bus excluded) (5.10)

Define Ak = (PIk – jQIk) / Ykk (5.11) Bkm = Ykm / Ykk (5.12)


NEWTON – RAPHSON METHOD FOR LOAD FLOW PROBLEM To apply the Newton – Raphson method to the solution of the power low equations, we express the bus voltages and line admittances in polar form.

From,

N

P y V V Cosn ni in i in in 1

N

Q y V V Sinn ni in i in in 1

(1)

When, n = i in the above equations and separating the terms by summations,

2 N

P V G V V y Cosi i ii i n in in n in 1

(2)

2 N

Q V B V V y Sini i ii i n in in n in 1n i

(3)

Gii and Bii y y Cos ij j y Sin ijij ij ij ij

G Bij ij

(4)

jBijGijy ijij

Assume all buses, (except the slack bus) as load buses with known

demands Pdi and Qdi. The slack bus has specified values for 1 and V1 ,

For other buses in the network, the two static variable s i and Vi are to be

calculated in the power – flow solution . The power mismatches for the typical load bus (i)

calcPischPiPi ,,

(5)

calcQischQiQi ,,

For real power Pi,

P P P P P Pi i i i i iP V V Vi 2 3 4 2 3 4

V V V2 3 4 2 3 4

(6)

The last three terms can be multiplied and divided by their respective voltage magnitudes without altering their values.

VVP P P P P 32i i i i iP V Vi 2 3 4 2 3

V V V V2 3 4 2 2 3 3

VP 4iV4

V V4 4

(7)

A mismatch equation can be written for reactive power Qi.


VVQ Q Q Q Q 32i i i i iQ V Vi 2 3 4 2 3

V V V V2 3 4 2 2 3 3

VQ 4iV4

V V4 4

(8)

Each non slack bus of the system has two equations like those for pi and

Qi Collecting all the mismatch equations into vector – matrix form yields.

The solution of (9) is found by the following iteration.

Estimate values 0i and

0Vi for static variables.

Use the estimates to calculate 0

P , calci and 0

Q , calci , from (2)

and (3) mismatches 0

Pi and 0

Qi from (5) and the partial

derivative elements of the Jacobian J.

Solve (9) for the initial corrections 0i and

0Vi

0Vi

Add the solved corrections to the initial estimates to obtain. 1 0 0

i i i (10)

1 0 0V V Vi i i

0V0 i

V 1i 0Vi

(11)

Use the new values 1

i and 1

Vi as starting values for

iteration 2 and continue.

The general formulas are

k 1 k k

i i i

(12)

k 1 k k

V V Vi i i

kVk i

V 1i kVi

(13)


For the four bus system sub matrix J11 has the form

p p p2 2 2

2 3 4p p p3 3 3

J112 3 4

p p p4 4 4

2 3 4

(14)

Expressions for the elements of this equation are easily found by differentiating the appropriate number of terms in (2)

The off diagonal element of J11

Pi

V V y Sin ii j ij ij j ij

(15)

The typical diagonal element of J11

NPi

V V y Sin in ni in in in 1

in i

N Pi

n 1 nn i

(16)

By comparing (16) & (3)

P 2iQ V Bi i ii

j

(17)

The formulas for the elements of sub matrix J21 is given by

The off diagonal element of J21 is

Qi

V V y Cos ii j ij ij j ij

(18)

The main diagonal element of J21 is

inininni

N

inni

i iCosyVVQ

1

N Qi

n 1 n

n i

(19)

Comparing (19) with (2) for Pi ] (20)

The off diagonal elements of J12 are simply the negatives of elements in J21.

2QiP V Gi i iii


This is obtained by multiplying

Pi

Vj

by jV

The diagonal elements of J12 are fond by

NPi

V V 2 V G V y Cosn ni i i ii in in in 1Vin i

(21)

Comparing (21) with (19) & (20)

P Q 2 2i iV 2 V G P V Gi i ii i i ii

Vi i

(22)

The off diagonal elements of submatrix J22 of the Jacobian are found using

Q Pi i

V V V j Sinj j i ij ij j iV ij

(23)

The main diagonal elements are given by

2Q P 2i iV 2 V B Q V Bi j ii i i ii

Vi i

(24)

Generally, The off diagonal elements, ji

P Qi iM Vij j

Vj j

(25)

Q Pi iN Vij j

Vj j

(26)

Diagonal elements, j = i

P Q 2i iM V M 2 Vi Bii i ii ii

Vi i

(27)

Q P 2i iN V N 2 Vi Gij i ii ii

Vi i

(28)


Using equation (9)

44

2

4444342444342

3433

2

33332343332

242322

2

222242322

44

2

4444342444342

3433

2

33332343332

242322

2

222242322

2

2

2

2

2

2

BVMNMMNN

MBVMMNNN

MMBVMNNN

GVMNNMMM

NGVNNMMM

NNGVNMMM

4

3

2

4

3

2

4

4

3

3

22

4

3

2

/

Q

Q

Q

P

P

P

V

V

V

V

VV

(29)

When voltage controlled buses are given for example if bus (4) is voltage

controlled, and then 4V has a specified constant value and voltage

correctionV4

0V4

. So 6th column of (29) is multiplied by zero and so it may

be removed.

Q4 is not specified.

4Q Cannot be defined. So sixth row of (29) is removed.

In general,

If there are Ng voltage controlled buses, (except slack bus) will have (2N-Ng-2) rows and columns.

Advantages of Newton – Raphson method

Faster, more reliable and results are accurate

Requires less number of iterations for convergence

Suitable for large systems.

Disadvantages

Programming logic is more complex than Gauss–seidel method. Memory requirement is more. Numbers of calculations per iterations are higher than G.S method.


FLOWCHART FOR NEWTON RAPHSON POWER FLOW METHOD


Procedure

Write a program f o r iteratively so l v i n g non-linear L o a d Flow

equations using Newton - Raphson method for small and medium

sized power systems. The program should have three sections i.e.

input section, Compute section and Output section.

I - Input section

Pre-requisite:

Before creating the input data file, draw a single- line diagram showing

the buses, lines, t r a n s f o r m e r s , shunt e l e m e n t s ,

bus g e n e r a t i o n a n d l o a d s ( Refer in Annexure 5.1). Bus ID

numbers are serially given from 1 to NB where NB is the total

number of buses comprising P-V buses (which includes the slack bus)

and P-Q buses.

The data to be read from an input file should contain general data,

bus data, line data, transformer data and shunt element data in the following sequence.

(i) General Data

The following data are read in one line

l) Total number of buses

a) Number of P-V buses

This includes all the voltage-controlled buses such as generator buses (including slack bus),synchronous condenser buses and SVC buses for which a specified voltage magnitude is to be maintained.

b) Number of P-Q buses.

This includes all load buses, dummy (zero generation and zero

loads) buses and generator buses in which voltage magnitude

is not controlled.

c) Number of transmission lines. d) Number of transformers. e) ID number of slack bus. f) Number of shunt elements. g) Maximum number of iterations to be performed. h) System MVA base. i) Convergence tolerance in p.u. voltage. j) Acceleration factor to be used.


(ii) Bus Data

The following data are read for each bus in one line.

P-V Bus Data

(a) ID number of bus (b) Active power generation in MW

(c) Active power load in MW

(d) Reactive power load in MVAR

(e) Maximum limit of reactive power generation in MVAR.

(f) Minimum limit of reactive power generation in MVAR.

(g) Scheduled voltage magnitude of the bus in p.u.

P-Q Bus Data:

(a) ID number of bus

(b) Active power load in MW.

(c) Reactive power load in MVAR.

(d) Initial voltage magnitude assumed in p.u.

(iii) Transmission line data (iv) Transformer Data (v) Shunt elements data

The data to be read and the sequence in which it is to be read for

(iii), (iv) and (v) are the same as that given in exercise under

Experiment 5.

II - Compute Section

Starting from the initial bus voltage solution (usually a “flat start” is assumed), update the voltage solution iteratively using Newton - Raphson method until the convergence criteria on bus voltage m a g n i t u d e is s a t i s f i e d . Typical v a l u e o f t o l e r a n c e f o r v o l t a g e m a g n i t u d e convergence is 0.001 p.u .

III - Output section

Create an output file in a report form comprising the following:

(iv) Student information: As specified in exercise under experiment 3.

(v) Input data: with proper headings.

(vi) Results obtained with proper headings, in the following sequence:


System details

Total No of Buses: No of P-V

Buses: No of P-Q Buses: No of Lines:

No of Transformers: No of Shunt elements: Slack

bus ID number: System base MVA:

Convergence details

Maximum iterations prescribed:

Maximum iterations taken:

Convergence tolerance prescribed:

Convergence limit reached:

Bus Results The results are to be printed under the following headings:

Bus

Id.No

Generation Demand Bus Voltage Compensation MW MVAR MW MVAR Magnitude

p.u.

Angle

degrees

MVAR

For each one of the buses, whether it is a P-V bus or a P-Q bus, one line

covering the above information is to be printed.

Transmission line / Transformer Results

The results are to be printed under the following headings.

Sending Bus No

Receiving Bus No

Flow MW

Flow MVAR

Flow MVA

Rating MVA

P loss MW

Q loss MVAR

For each one of the lines / transformers, two lines are to be printed. The

first line printed should have all the above information pertaining to the

forward direction including Rating, Ploss and Qloss. The second line

to be printed should have the above information, pertaining to the

flow in the reverse direction excluding Rating, Ploss and Qloss.

Exercise:

i. Prepare the data for the 6-bus system described in the Annexure 5.1. Run the NR program and a convergence for voltage tolerance of 0.001p.u by using available software.



ANNEXURE 5.1

6-BUS, 7-LINES / TRANSFORMER POWER SYSTEM Single-Line Diagram

Buses : 6, numbered serially from 1 to 6 Lines : 5, numbered serially from L1 to L5 Transformers: 2, numbered serially as T1 and T2 Shunt Load : 2, numbered serially as s1 and s2 Base MVA : 100

Bus Data – P-V Buses:

Bus ID No.

Generation, MW Demand Gen. Limit MVAR

Scheduled Volt (p.u)

Schedule Max Min MW MVAR Max Min 1 ? 200 40 0.0 0.0 100.0 -50.0 1.02

2 50.0 100 20 0.0 0.0 50.0 -25.0 1.02

Bus Data – P-Q Buses

Bus ID No Demand

Volt. Mag. Assumed

(p.u) MW MVAR

3 55.0 13.0 1.0

4 0.0 0.0 1.0

5 30.0 18.0 1.0

6 50.0 5.0 1.0



Line ID.

No

Send Bus No.

Receive Bus No.

Resist P.U

Reactance P.U.


P.U

Rating MVA

1 1 6 0.123 0.518 0.0 55

2 1 4 0.080 0.370 0.0 65

3 4 6 0.087 0.407 0.0 30

4 5 2 0.282 0.640 0.0 55

5 2 3 0.723 1.050 0.0 40

Transformer Data:

Transformer ID.No

Send (*) Bus No.

Receive Bus No.

Resist. P.U

Reactance P.U.

Tap Ratio Rating MVA

1 6 (*) 5 0.0 0.300 1.000 30

2 4 (*) 3 0.0 0.133 1.000 55

(*) Note: The sending end bus of a transformer should be the tap side.

Shunt Element Data:

Shunt ID No. Bus ID. No. Rated Capacity MVAR (*)

1 4 2.0

2 6 2.5

(*) Note: Sign for capacitor : +ve Sign for Inductor : -ve

Exercise:

ii. Two voltage sources V _______________& V ________________1 2 are

connected by a short line of impedance Z __________ . Determine the

real and reactive power supplied or received by each source and the

power loss in the line. Write the program in MATLAB. Compare the

results.


Viva Questions

1. What is infinite bus? 2. When the generator bus is treated as load bus? 3. How approximation is performed in Newton- Raphson method. 4. Write the most important mode of operation of power system and mention

the major problems encountered with it. 5. What is the need of load flow solution? 6. List the quantities specified and the quantities to be determined from load

flow study for various types of buses? 7. What are the reasons for changes in bus voltage? 8. What is Jacobian matrix? How the elements of Jacobin matrix are computed? 9. How load flow study is performed? 10. What are the quantities that are associated with each bus in a system? 11. Write the load –flow equation of Newton –Raphson method. 12. Why do we go for iterative methods to solve load flow problems? 13. Compare Gauss –Seidel & Newton –Raphson method of load flow solution. 14. What are the advantages & disadvantages of Newton –Raphson method? 15. How the disadvantages of N-R method are overcome? 16. What will be the reactive power and bus voltage when the generator bus is

treated as load bus? 17. Write the SLFE? 18. What is meant by a flat voltage profile? 19. What is the need for voltage control in a power system? 20. How the convergence of N_R method is speeded up?


Result:


Secured

Marks

Awarded



Conducting 15

Software output



Record 10

Viva - voce 10

Total Marks 100


MATLAB QUESTIONS

1. What is MATLAB?

2. Who is the father of MATLAB?

3. List out the memory management functions in MATLAB.

4. What are the advantages of MATLAB?

5. What is a stem in MATLAB?

6. Is there MATLAB complier in MATLAB?

7. Why MATLAB is so called MATLAB?

8. What are the types of display windows in MATLAB?

9. What are the disadvantages of MATLAB?

10. Function of semicolon (;) in MATLAB.

11. Need for ‘clc’ command in MATLAB

12. What are the arithmetic operations in MATLAB?

13. Mention the various exponential functions of MATLAB.

14. List out the commands for managing variables in MATLAB.

15. Function of percentage symbol (%) in MATLAB.

16. What is the use of ‘disp’ function in MATLAB?

17. What does MATLAB stands for?

18. Mention the main function of command window.

19. What are the types of M-file?

20. Difference between clear all & close all.

21. Function of ‘for’ loop in MATLAB.

22. What are the special characters in MATLAB?

23. Function of ‘who’ in MATLAB.

24. List the built in constant in MATLAB.

25. What is MATLAB Script?

POWER SYSTEM SIMULATION LAB-1 MANUAL (ELECTRICAL - POWER SYSTEM ENGINEERING )

Engineering

Transcript of POWER SYSTEM SIMULATION LAB-1 MANUAL (ELECTRICAL - POWER SYSTEM ENGINEERING )