EXPERIMENT 3

LOAD FLOW ANALYSIS - I : SOLUTION OF LOAD FLOW AND RELATED PROBLEMS USING GAUSS-SEIDEL METHOD 3.1 AIM

(i) To understand, the basic aspects of steady state analysis of power systems that are required for effective planning and operation of power systems. (ii) 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 3.2 OBJECTIVES

i. 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. ii. To investigate the convergence characteristics of GSLF algorithm for normally loaded small system for different acceleration factors. iii. 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.

3.3 SOFTWARE REQUIRED

GAUSS – SEIDEL METHOD module of AU Powerlab or equivalent 3.4 THEORETICAL BACKGROUND

3.4.1 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.

3-1

3.4.2 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 3.1. 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 transmission line is ���������������� ����������� ��� � ������ "!$#�%&��'�(*)� �+�)��� �(�,.-0/��1(2+3#�'5476�89+�:;��+=<> �(�/?8@636 -nominal tap ratio is A�B�C�D�C�AE�F�D�G�H�F�IKJMLNEOB�F�G"DQP"R SUT�V�W�X�Y$Z�[2T�\�]_^�X�`�^�W�X�]�acb�dMW@\�]e^�f9g;h$T�\�ijZ�]kX"\�l>^�Z�h@Z�^�X�]�f�`minf9`0`3T�Z�^]kf9`jioZ�`�Trepresented 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.

The steady state of the system is given by the state vector X defined as X = p q 1 q 2 rsrut N V1 V2 …….VN)T = ( v 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. 3.4.3 Development of Load Flow Model

The Load Flow model in complex form is obtained by writing one complex power matching equation at each bus.

3-2

. Vt

PIk +jQIk = (PGk - PDk) + j (QGk - QDk) PGk+ jQGk G PDk+ jQDk k Vk k Vk (Pk + jQk) Ik (Pk+ jQk) Ik (a) (b) Referring to Fig 3.1 (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 (3.1) In expanded form (PGk - PDk) + j (QGk - QDk) = VkIk

* (3.2) The network equation relating bus voltage vector V with bus current vector I is YV = I (3.3) Taking the kth component of I from (3.3) and substituting for Ik* in (3.2) we get the power flow model in complex form as N

PIk + jQIk = Vk w x km* Vm*; k=1,2,………… N (3.4) m=1

In (3.4) 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 (3.4) 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 y{z�|9}5~��@���2z��M�e��~�����}5~��_���9���O���������c~2���7~O�9�c�2~"�$z����$�>�O�����5�7�������cz���~��n�@~��@z���~��Fig 3.2. 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 3.2) Vk a:1 Pmk + jQmk Vm Ik Pkm + jQkm It ykm Im t k jbc jbc m

Fig 3.1 Complex Power Balancing at a Bus

Fig 3.2 PI Equivalent Circuit of a Transmission Line / Transformer

3-3

Pkm + jQkm = Vk Ik* = Vt It * (3.5) Noting that Vk / Vt = a (3.6) It = (Vt -Vm) ykm + Vt (jbc) (3.7) Substituting equations (3.6) and (3.7) in equation (3.5) we get Pkm + jQkm = (Vk/a) [(Vk * /a) – V m *)] y km * + (Vk/a)2 (jbc)* (3.8) 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 * + V2

m (jbc)* (3.9) The complex power loss in line / transformer k-m, PLkm + jQLkm, is given by the sum of the two expressions (3.8) and (3.9) 3.4.4 Classification of Buses

From the Load Flow model in equation (3.4) and from the definition of complex bus voltage, Vk as Vk = |Vk| � � k one can observe that there ���������9�@���@�����&���� 2��¡�¢e£U¤�¢0¥¦¤3¢�§ ¨�©�ª¬« ­;«$¨¯®�®7°M±�²2¨³�´�ª1µK²�³�¶·´�¨�±�¶¹¸Mº5®¼»Any two of these four may be treated as independent variables (that is specified) while the other two may be computed by solving power flow equations. The buses are classified based on the variables specified. Three types of buses classified based on practical requirements are given below: Slack bus: While specifying a generation schedule for a given system demand, one can fix up the generation setting of all the generation buses except one bus because of the limitation of not knowing the transmission loss in advance. This leaves us with the only ¨�½"³�´�¾�©�¨�³k²"¿@´À°MÁ®�Ã$´�±�²&ÁkÄ�²"©�Åc³OµK°Æ¿@¨�¾�²&¨�¸�½&´�®*Ç 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 ®�Ä�®3³�´�Èɨ�©�ª=³O¶$´¦®�Ã@´�±�²2Á�²2±�¨�³k²&°9©{°@Á0Ç s as zero, makes Vs as reference phasor. Thus for the slack ¸Mº�®7ÊM¸$°9³�¶{Ç Ë�Ì$ÍÀÎ Ï;ÎË�Ð3Ñ=ÒmÓ@Ñ�Ô�Õ2Ö3Õ"Ñ�Í×Ë�Ì$ÍNØUÙÚË�Ì�ÍNÛ*ÙÜË�Ð3ÑÞÝ�ßKà�Ñ�Ô�ß9á{ÓMâ�Ý�Ñ�ÍNß�Ì�ã�äÀË�Ö�Ý�Ñ�Ð_ÝOå�Ñ*Õ�Ý�Ñ�Ð3ËÝ�Õ�æ@Ñsolution 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. çÞè5é�ê;ë�è$ì>ímî�ïî�ì�ð@ï�ñ�ò&ï�ó�é&ì�ô�õ>ò�í�îkëNó$ì*ö�ë9÷{øMùMî�ì�ú{ï�îûî�ü�òOíýóMù�í�þ9ÿ�ü�ìÞñ�ì�ï�öî�ò�ð@ì¦ø$ë��=ì�ñ���ì�è$ì�ñ3ïî�ò2ë9è����

3-4

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 ��� ������������������������������� ���!��#" $%"&�'(���)�*���,+���-%./�/���0�21 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 = ( � 1�

2 343 � NP V1V2 ……..V NQ) 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. 3.4.5 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. They are treated in the experiment 4. Taking the complex conjugate of equation (3.4) and transferring Vk to the left hand side, we obtain N

Vk = [(PIk – jQIk) / Vk* - 576 km Vm] / Ykk;

m=1

8�9 k = 1,2, …..(N -1) (Slack bus excluded) (3.10)

Define Ak = (PIk – jQIk) / Ykk (3.11)

Bkm = Ykm / Ykk (3.12)

3-5

The voltage equation to be solved during the hth iteration of G.S method is obtained from

(3.10), (3.11) and (3.12) as

k-1 N Vk

(h+1) = (Ak / Vk(h)* ) – : ; kmVm

(h+1) – < ; kmVm(h) (3.13)

m= 1 m= k+1 3.4.6 GSLF Algorithm

The algorithm for GSLF is given in the flow chart Fig 3.3 Convergence Check: Referring to Flow chart Fig 3.3, during every iteration h, the maximum change in bus =�>�?A@CB0D/E�@�F�B&@GF�BIHJ>LK!K!M/NON(E0P Q�HRH(@)>�NSETP QAU V W�XZY*[]\�^�_/`ba�cd e�cTfbg�h i jlknmporqtstu�v wyx{z

k(h+1)

w&|}w&x{~k(h+1)| ; k = 1,2,……N; �t� ���}�)�&�/�

�*����O�,�J� k

(h+1) � ��� k(h+1) ���R���

k(h+1) = Vk

(h+1) - Vk(h)

�������0�L�/���!�S�L�!���0�l�� �����T�¡��0¢p£/¤¥�0��¦#§�¨!�(�A���ª©J«l¬®­p¯r° �²±��p±����� �§��!���b³S�´�T¢l±C��µb�!�S¨����� ¶T·

Additional Computation for P-V Bus

The flow chart in Fig.3.3 does not have provision for voltage – controlled buses. However , if the link between X and Y in Fig.3.3 is removed and the P-V bus module in Fig.3.4 is introduced, then P-V buses can be handled. Referring to Fig.3.3 and Fig.3.4, for each P-V bus during the hth iteration, before updating bus voltage, the following computations are made: Step 1:

Adjusting the complex voltage Vk(h) ¸º¹

k (h) +j fk

(h) to correct the voltage magnitude to the scheduled value,|Vk|sch as follows: »

k(h) = arc tan(fk

(h) / ek(h)) (3.15)

V k(new) (h) = | Vk|sch e¼ ½ k(h) (3.16) Step 2: Compute the reactive power generation using the V k(new)

(h) as QG k

(h) = QDk + Q k (h) (V(h)) (3.17)

k-1 N Q k

(h) (V(h)) = -Imag V k(new) (h)* ¾ Ykm Vm

(h+1) + Ykk V k(new) (h) + ¾ ¿ km Vm

(h)

m=1 m=k+1 If the inequality QGk

min ÀÂÁ�à k(h) ÀtÁ*à k max is satisfied, then Vk

(h) is set as V(h) k(new).

3-6

Go to step 3.

If QGk(h) > QGk

max , then set QGk(h) = QGk

max, go to step 3.

If QGk(h) < QGk

min , then set QGk(h) = QGk

min, go to step 3. Step 3: Recompute Ak using (3.11) and the updated QG(h)

k

Acceleration Factor

Experience has shown that the number of iterations required for convergence can be considerably reduced if the correction in bus voltage computed at each iteration is multiplied by a factor greater than unity (termed as acceleration factor) to bring the voltage closer to the value to which it is converging. For example, during the hth iteration the accelerated value of the voltage at kth bus is calculated using V k, acc (h+1) = V k (h) Ä Å%Æ)Ç

k (h+1) – V k

(h)) (3.18) ÈpÉ�Ê!ËOÊ Å ÌÎÍ0Ï!Ï ÊTÐbÊ!Ë ÍÑÓÒbÔ�Õ¥ÖOÍ!Ï�ÑÓÔ Ë V k

(h) = accelerated value obtained in the (h-1)th iteration

V k (h+1) = value computed during hth iteration using equation (3.13)

Then set V k (h+1) = V k,acc

(h+1)

3-7

Fig.3.3 Flow chart for GSLF ( P-V buses – not present)

3-8

No

Read data

Form Y

Compute Ak and Bkm; k = 1,2,…….N; ×ÙØ m = 1,2,…….N using equations (3.11) & (3.12)

Set iteration count h = 0

Set Bus Count k = 1

Is k = s?

Start

X

Stop

Y

Update Voltage Vk(h+1)

using equation (3.13)

k = k+1

Is k ÚJÛÝÜ

Check convergence ÞÓß à á{âtã{äæå ç è

Compute line flows, line loss, slack bus power and print results

Yes

h=h+1

Yes

No No

Yes

Fig.3.4 P-V Bus Module for GSLF Algorithm.

3.5 EXERCISES

3.5.1 Write a program in C language for iteratively solving non-linear Load 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, transformers, shunt elements, bus generation and loads (Refer Fig in Annexure 3.1). 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.

A P-V bus ?

éGê&ëRìí&îbï ðk

(h) using (3.15)

Compute V(h) k(new) using (3.16)

Compute QGk (h) using (3.17)

Is QGk (h) > QGk

max ?

Is QGk(h) < QGk

min ? QGk(h) = QGk

max

Vk(h) = V(h) k(new) QGk

(h) = QGkmin

Recompute Ak using (3.11) and QGk

(h)

Yes No

No

Yes

Yes

X

Y

3-9

(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 load) 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.

(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 2.5.1 under Experiment 2.

3-10

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 magnitude is satisfied. Typical value of tolerance for voltage magnitude convergence is 0.001 p.u

Using the values of converged bus voltages compute:

(i) Forward and reverse real and reactive power flows in lines and transformers (ii) Real and reactive power losses in lines and transformer (iii) Reactive power absorbed / generated by shunt elements (iv) Reactive power generated at the P-V buses (v) Real and reactive power generations at the slack bus Note:

i. Real power to be printed in MW ii. Reactive power to be printed in MVAR

III Output section

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

(i) Student information: As specified in exercise 2.5.1. under experiment 2. (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:

Generation Demand Bus Voltage Compensation Bus Id.No MW MVAR MW MVAR Magnitude

p.u. Angle

degrees MVAR

3-11

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. System Performance

In this section following details must be provided:

(i) Total active power generation in MW (ii) Total active power load in MW

(iii) Total system active power loss as a percentage of generation

Repeat (i) (ii) and (iii) for reactive power and in addition include total reactive power generated / absorbed by shunt elements 3.5.2. Using a text editor create, an input data file in the sequence given in Exercise 3.5.1

for Load Flow solution of the 6-bus system given in Annexure 3.1 run the program developed in Exercise 3.5.1 using an acceleration factor of 1.0 and print the output file. Check the results obtained using the available software.

3.5.3. Solve the base case in exercise 3.5.2 for different values of acceleration factor, draw

the convergence characteristic “Iterations taken for convergence versus acceleration factor” and determine the “best” acceleration factor for this 6 -bus system.

3.5.4. Solve the base case in exercise 3.5.2. after removing the two shunt capacitors and

comment on the results obtained especially on voltage magnitude of load buses and transmission system losses.

3.5.5. Solve the base case in exercise 3.5.2. for different voltage specified for generator

G1&G2 (1.0, 1.01, 1.02, 1.03 and 1.04 per unit) and comment on the voltage magnitude of the load buses and transmission system losses.

3.5.6. Solve the base case in exercise 3.5.2. for different tap settings (0.85, 0.90, 0.95,

1.05 and 1.1) of the transformer T1 and T2 and comment on the voltage magnitude of load buses and transmission system losses.

3.5.7. Solve the base case in exercise 3.5.2. by shifting bulk generation from slack bus to

bus 2 and comment on the line/transformer loadings obtained.

3-12

ANNEXURE 3.1

6-BUS, 7-LINES / TRANSFORMER POWER SYSTEM

Single-Line Diagram

S1 L2 T2 L3 a:1 L1 L5 T1 L4 a:1 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:

Generation, MW Demand Gen. Limit MVAR

Bus ID No.

Schedule Max Min MW MVAR Max Min

Scheduled Volt (p.u)

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

Demand Bus ID No MW MVAR

Volt. Mag. Assumed (p.u)

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

G

1

6

4 3

2

G

5 S2

3-13

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 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

3-14