Power System Analysis in Grid

TET 4115 POWER SYSTEM ANALYSIS 2011

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NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY

FACULTY OF INFORMATION TECHNOLOGY, MATHEMATICS AND ELECTRICAL

ENGINEERING

TRONDHEIM, NORWAY

A

PROJECT REPORT

ON

TET4115

POWER SYSTEM ANALYSIS

SUBMITTED BY: SUBMITTED TO:

MAMTA MAHARJAN (729299) PROFESSOR KJETIL UHLEN

NABARAJ POKHAREL (729305) Department of Electrical Power Engineering.

NAMRATA TUSUJU SHRESTHA (729295)

NITU MANDAL (729288)

(Group 2)


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1. Introduction:

1.1 Background:

The goal of a power flow study is to obtain complete voltage (angle and magnitude), power

(active and reactive) information for each bus in a power system. Once this information is

known, symmetrical and unsymmetrical fault calculation can be analytically determined. Due to

the nonlinear nature of this problem, numerical methods are employed to obtain a solution that is

within an acceptable tolerance.

The solution to the power flow problem begins with identifying the known and unknown

variables in the system. The known and unknown variables are dependent on the type of bus. A

bus without any generators connected to it is called a Load Bus. With one exception, a bus with

at least one generator connected to it is called a Generator Bus. The exception is one arbitrarily-

selected bus that has a generator. This bus is referred to as the Slack Bus.

In power engineering, a symmetrical fault is a fault which affects each of the three-phases

equally. Asymmetric fault is a fault which does not affect each of the three phases equally. For

these reasons thus symmetrical faults are more severe than the former. In a practical circuit the

experience has shown that between 70-80% of transmission line faults are single line to ground

faults (unsymmetrical fault), 5% of all faults involve all three phases so called symmetrical fault

in the project data given we mainly deal with the fault calculations of these fault types.

1.2 Objective:

This Project gives us the brief insight in the understanding of various network modeling

concepts. The project also deals with the analysis and simulation of a practical case where load

flow analysis along with fault calculation was realized. The constraints given at practical case are

investigated to be in given range with the use of simulation tools Matlab and Matpower.

Impedance matrix of positive, negative and zero sequences are used to analyze symmetrical and

unsymmetrical fault analysis.


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1.3 Organization of the report:

The report contains the load flow result of the network provided. The result of the load flow did

not comply with the constrains provided. The “mpoption” in Matlab are adopted to get the

desired voltages at different bus is also included. Equivalent diagram for the positive, negative

and zero sequence along with its per-unit calculated values for the system is included. Here the

impedance matrix is determined for each of the positive, negative and zero sequence which is

also included. The report also contains the m.file for the symmetrical and unsymmetrical faults in

the appendix. The results obtained for the symmetrical and unsymmetrical faults (i.e. fault

current, bus voltages and line currents) are included in the report.

2.0 Problem description

Part 1-Power flow analysis:

1.1Power Flow Solution

The grid consists of 8 bus system. Generally highest power generator is selected as slack bus so

Bus 1 Tokke is considered to be slack Bus. And rest three generator Bus namely Vinje, Songa,

Vemork; Bus 2, Bus 3 and Bus 4 respectively is assumed to be PV Bus and remaining four

Buses; Bus 5,Bus 6, Bus 7 and Bus 8 namely Rjukan, Flesaker, Tveiten and rød are considered

as PQ Bus.

In slack bus, voltage and reference angle is fixed, in PV bus real power and voltage is fixed

while in PQ BUS real and reactive power is fixed. Out of the four variables P, Q, V and angle

remaining two variables must be found out. The parameters must be within the constraint. Given

in the Appendix 1 is the power flow problem Figure 1.1and the table indicating the constraint of

the power system Table 1.1

Given Sref=900MVA and Vref=300kV. Implies Zref=100ohm Iref=1.732KA and line data is

calculated and tabulated in appendix 1, showing impedance in Table1.2

All these data are fed into the case file of MATPOWER. Then, the result that was obtained is

shown by Table 1.3 in appendix 1.From there all of the parameters are under the limited value


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but Q generated at bus 3 and 4 exceeded the limit. Now changing the type of bus 3 and bus4 as

PQ bus limiting the Q as 0 (lower limit) to bus 3and 180 (upper limit) to bus 4 and again run the

file in MATLAB and the result obtained is shown as in Table1.4 in appendix 1. Now, the total

power loss is 42.548 MW and 490.31MVAr.

When the type of bus 3 is changed from PV to PQ, Q is limited to its lower value (0 MVar). But

for minimising the loss the upper value (135Mvar) is selected. The loss is now decreased to

41.858 MW and 483.09 MVAr and all the parameters are also under the constraint.

The data which is obtained below is used for further use.

Table 1: Bus and branch data of given grid


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Part 2–Analysis of symmetrical faults:

2.1 Positive Sequence

This part of the problem deals with the analysis of symmetrical fault. The power grid system can

be simulated into electrical circuits. In order to analyze the symmetrical fault, the electrical grid

is represented by positive sequence system. Generator is represented by current source in parallel

with positive sequence synchronous reactance. Given: Xd1 = 1.1 pu Nominal generator voltage:

Ugn = 12 kV. This is true for all the voltage level.

The transformer which is connected from bus 1 to bus 4 is delta to star transformer (Δ-Y, phase

shift δ=30°) so the transformer is represented by ideal transformer with short circuit impedence

x1 = 0.07.The Y-side is connected to the 300 kV grids and the Y-connection is directly grounded

from the neutral point. All transformers have transformer ratio 12:300 kV.

Load is represented by impedance. There is transformer of star to star type in the load side which

connected in bus 5, 7 and 8. But load is referred to 300kv side in the load flow calculation, so the

load impedance can be assumed to include the contribution from the transformer short circuit.

The neutral point on the 300 kV side is isolated from ground, and the neutral point on the load

side is grounded through a Peterson coil. Magnetizing reactance is not given in both the

transformer which will be very high so it can be assume to be infinite.

The interconnected power system can be represented by current source whose short circuit

reactance is 18ohm. The equivalent diagram of positive sequence is attached in appendix 4.

2.2 Global Per Unit Values

All the figures given are in local per unit system. So, the global per unit system is calculated.


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Table 2: Impedance calculation

Description S=P+iQ Vref. Z* Z Zpu

Line 5 100+60i 300 661,764-397.058i 661,764+397.058i 6,617+3.970i

Line 6 1080+225i 300 79.866-16.638i 79.866+16.638i

0.798+0.166i

Line 7 180+45i 300 470,588-117.647i

470.588+117.647i

4.705+1.176i

Line 8 225+45i 300 384.615-76.923i 384.615+76.923i

3.846+0.769i

Table 3: Calculation of PU values.

Impedence LOCAL Pu

S (MVA)

LOCAL

Global Pu

( Positive) (Negative) ( Positive) (Negative)

XG1 1.1 0.25 1200 0.825 0.1875

XG2 1.1 0.25 500 1.98 0.45

XG3 1.1 0.25 500 1.98 0.45

XG4 1.1 0.25 300 3.3 0.75

XT1 0.07 0.07 1200 0.0525 0.0525

XT2 0.07 0.07 500 0.126 0.126

XT3 0.07 0.07 500 0.126 0.126

XT4 0.07 0.07 300 0.21 0.21


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2.3 Node Impedance Matrix

The node impedance matrix is then generated using “zbuild”. Here “linedata” is fed for the

zbuild which consists of information regarding impedence (resistance and reactance) between the

various nodes. The result of zbuild for the positive matrix is shown below.

Table 4: Line data for Zbuild

From To R X

0 1 0 0.9040

0 2 0 2.1315

0 3 0 2.1471

0 4 0 3.6202

0 5 7.3945 3.4455

0 6 0 0.1793

0 7 4.8624 0.8146

0 8 3.9474 0.4548

1 2 0.0092 0.0984

2 3 0.0075 0.0808

3 4 0.0191 0.2055

4 5 0.0036 0.0399

5 6 0.0142 0.2278

6 7 0.0109 0.1741

7 8 0.0198 0.2124

1 8 0.0423 0.4549

1 6 0.0467 0.5018


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Table 5: Zbus for positive sequence reactance

2.4 Short Circuit Current and Voltage at all Buses

There is three phase short circuit fault at bus 5. Neglecting arc resistance, short circuit current

and voltage at all the bus is calculated by Matlab. The Matlab code is attached in the appendix 2(

Table 2.1) .After fault voltage is calculated and tabulated.

Table 6: After fault voltage

Bus Magnitude Angle (Ө)

1 0.5342 6.957

2 0.4809 10.22

3 0.3753 12.166

4 0.08162 11.4475

5 0 0

6 0.48166 -25.65

7 0.47456 -23.074

8 0.472862 -14.97


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2.5 Fault Current Between Bus 4 and Bus 5

The fault current at bus 5 is calculated as

-0.2140 - 3.8979i= 3.904<-93.140 pu

Iref=300/( *100)kA=1.732kA

Therefore fault current If5=6.7619<-93.140KA

Now current flowing through bus 4 and bus 5

=2.042‹-73.1840pu

=3.53‹-73.184 KA

The Matlab coding is attached in appendix 2, table 2.1.

Part 3-Analysis of unsymmetrical faults

3.1 Per Phase Equivalent for Negative and Zero Sequence System

This part of the problem is based on analysis of unsymmetrical fault. Negative sequence

reactance of all generator is Xd2=0.25 pu. Negative sequence short circuit impedance of

transformer is equal to positive sequence reactance. And line impedance is also calculated for

negative and zero sequence. The equivalent diagram of negative and zero sequence is shown in

appendix 4.

The load side transformer is Y-Y connected and its primary side is isolated and the secondary

side is connected to ground using the Petersons coil. It is done so because whenever there is fault

in the load side then it is not transferred to the grid. Here in zero sequence equivalent circuit

capacitive reactance is not considered. Capacitive reactance is very large compared to the line

parameters. So, even if that value is neglected the equivalent impedance does not have much

significant change. In ∆-Y connection of generating unit there is isolation in primary side of

transformer and no current flow in generator in zero sequence.


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3.2 Node Impedance Matrices for Negative and Zero Sequence System

Using Zbuild node impedance matrices is calculated for the grid. The negative sequence and zero

sequence node impedance matrices are as follows:

Table 7: Zbus for negative sequence

Table 8:Zbus for zero sequence

3.3 Short Circuit Current and Voltage at All Buses

There is single phase to ground fault at bus 5. Arc resistance is assumed to be 0. Here is single

phase to ground fault. It is assumed that there is fault on phase A to ground. In this condition

Va=0, Ifb=0, Ifc=0 at bus 5. Using the flowchart and Matlab code given in Appendix 3, The

result is as below:


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Fault current in bus 5: If5a=-0.1865 - 4.7720i= 4.7756‹-87.760pu=8.2713‹-87.76

0 kA. Fault

current in other phase is zero as this is single phase to ground. And fault current in all other

nodes are also zero because there is fault only in bus 5. The voltage at all bus in all phase is

tabulated as below.

Table 9: Voltage for phase A, Phase B and Phase C

PHASE A PHASE B PHASE C

Magnitude Angle Magnitude Angle Magnitude Angle

1 0.739 2.07805 0.8721 -114.71 0.8585 117.5

2 0.6967 3.5383 0.87 -112.8 0.8496 117.5

3 0.5879 3.285 0.8366 -112.9 0.8135 115.755

4 0.1869 -5.4025 0.8274 -122.15 0.8306 110.09

5 0 0 0.839 -129.56 0.8793 112.12

6 0.54163 -25.022 0.83 -136.58 0.8331 100.94

7 0.5661 -22.66 0.8194 -136.58 0.8184 100.5

8 0.5998 15.6882 0.8159 -135.56 0.8119 104.04

After fault the current flowing from 4 to 5 in all the phase respectively are tabulated below:

Table 10: current flowing from bus 4 to 5 in all the phase

From-To Phase Magnitude Angle

4 to 5

A 3.119 -79.74

B 1.051 -121.26

C 0.39 118.50

Part 4: Additional Problems

4.1 Power Flow Static Var Compensator

Here a Static Var Compensator is installed in a step of 50 MVAr from 0 to 250 MVAr at bus 7.

When the SVC reached its limit then voltage at bus 3 exceeds limit hence installing 250 MVAr


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SVC is not permissible. From the Table 3.2 shown in appendix 3, when 200Mvar is selected, bus

3 exceed its limit so the VAr compensation should not be increased more than this. So 150Mvar

SVC is selected which could build voltage at bus7 from 0.912 to 0.961pu.

SVC of 250 MVAr capacity is available but installing 250 MVAr in any load bus leads to exceed

limit of voltage at least one bus. Therefore trying to get better voltage profile at all buses and less

active and reactive power loss a point is reached where installing 200 MVAr of SVC at bus 8

gives better voltage profile at all buses and reduced active and reactive losses compared to using

SVC in all other load bus. This is referred in Table3.2 of Appendix 3.

4.3 Symmetrical Fault

Putting the subtransient reactance of the generators which is equal to the negative sequence

reactance and the effect of the fault at bus no. 5 is to be analysed in this section.

The fault current at bus 5 is calculated as

-0.3851-5.4974i = 5.510‹-940

pu

Iref=300/ ( *100) kA=1.732kA

Therefore fault current If5=9.54‹-940 kA.

There is a increase in the fault current as the generator reactance is decreased in this case

compared to the problem 2.

If fault occur in bus no. 6 which is further away from the generators, the fault current is:

-2.0161-8.0771i= 8.32‹-1040

pu

Iref=300/ ( *100) kA=1.732kA

Therefore fault current If5=14.394‹-1040 KA.


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Load no. 6 is the interconnected system which is consuming maximum load. So when the fault

occurs on that node then maximum fault current occurs.

If fault occur in bus no. 4 which is near to the generators, the fault current is:

-0.2664-5.7478i= 5.75‹-92.650

pu

Iref=300/( *100) kA=1.732kA

Therefore fault current If5=9.948‹-92.65°KA.

Fault current is decreased compared to the fault occurring at bus 6.

Conclusions:

This project governs with the power flow calculation of the grid depicted in the practical cases of

main grid in Telemark, Buskerud and Vestfold. From this problem analysis, it can be seen that to

limit the reactive power generation when it is beyond the limit we can change the type of bus.

While talking about fault condition, one can see that whenever there is fault at a bus then the

fault current will be relatively higher if the fault is unsymmetrical with compared to that of

symmetrical fault.

SVC increases voltage level of buses and reduces active and reactive losses but there should be

trade-off between increased voltage level and reduced loss to choose appropriate value of SVC.

Whenever subtransient reactance is used in fault calculation, the generator reactance is reduced

and hence the fault current is increased. The other important thing to be noticed is whenever

there is fault near a bus having higher load it will have higher fault current compared to in the

case when fault occurs at bus having lower load.


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LITERATURE:

1. Book: Hadi Saadat: “Power System Analysis”, Second or Third edition,

http://www.psapublishing.com/ (selected chapters)

2. Lecture notes: “TET 4115 Power System Analysis”, (selected chapters)

3. Lecture notes: “TET 4115 Network design”

4. G.B. Hasmon, L.H.C.C. Lee, "Distribution Network Reduction for Voltage Stability Analysis

and Load Flow Calculations", Electric Power Energy Systems 1991, 13(1), pp. 9-13.


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APPENDIX 1

Figure 1.1: Power Flow Problem


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Table 1.1: Constraints of the power system

Table 1.2: Impedance and Capacitance Values.


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Table 1.3: Initial Result


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Table 1.4: Power Flow solution after “Enforcing Q”


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APPENDIX 2

Table 2.1: Matlab Coding for Symmetrical fault

n=8 %no. of bus

f=5; %fault occur at f bus

%FOR SYMMETRICAL FAULT

linedata=[0 1 0 0.904;0 2 0 2.1315;0 3 0 2.1471;0 4 0 3.6202;0 5 7.3945 3.4455;0 6 0

0.1793;0 7 4.8624 0.8146;0 8 3.9474 0.45475;1 2 0.00916 0.0984;2 3 0.00752 0.0808;3 4

0.01912 0.2055;4 5 0.00363 0.0399;5 6 0.01424 0.2278;6 7 0.01088 0.17408;7 8 0.01976

0.2124;1 8 0.0423 0.4549;1 6 0.04668 0.5018];

zbus=zbuild(linedata);

up=[1.02 1.0299+0.0126i 1.0026-0.0047i 0.9769-0.1093i 0.9574-0.1358i 0.8689-0.2671i

0.8728-0.2671i 0.9021-0.2135i];

for x=1:n

zb(x)= zbus(x,f); %zb =z15, z25 etc

end

fc=up(f)*inv(zbus(f,f));%fault current

for y=1:n

uaf(y)=up(y)-fc*zb(y);

end

ybus=inv(zbus);

c45=(uaf(4)-uaf(5))*inv((0.00363+0.03993i)); %fault current between 4 and 5


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APPENDIX 3

Table 3.1: Matlab Coding for Unsymmetrical fault

% FOR ASYMETRICAL FAULT

%zbus for negative sequence

%zbus for zero sequence ignoring the capactive reactace.

pzbus=zbus;

pybus=inv(pzbus);

linedatap=linedata;

linedata=[0 1 0 0.2419;0 2 0 0.5779;0 3 0 0.579;0 4 0 0.9681;0 5 7.3945 3.4455;0 6 0

0.1793;0 7 4.8624 0.8146;0 8 3.9474 0.45475;1 2 0.00916 0.0984;2 3 0.00752 0.0808;3 4

0.01912 0.2055;4 5 0.00363 0.0399;5 6 0.01424 0.2278;6 7 0.01088 0.17408;7 8 0.01976

0.2124;1 8 0.0423 0.4549;1 6 0.04668 0.5018];


nzbus=zbus;

nybus=inv(nzbus);

linedatan=linedata;

linedata=[0 1 0 0.0525;0 2 0 0.126;0 3 0 0.126;0 4 0 0.21;0 6 0 0.36;1 2 0.04809 0.1947;2

3 0.0395 0.1598;3 4 0.1004 0.4063;4 5 0.0242 0.0908;5 6 0.1353 0.5268;6 7 0.1034

0.4026;7 8 0.1037 0.4199;1 8 0.2222 0.8993;1 6 0.24507 0.9919];


ybus=inv(zbus);

if012=up(f)*inv((pzbus(f,f)+nzbus(f,f)+zbus(f,f)));

if3a=3*if012; %if3b=if3c=0 for 0 for single phase to ground


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%positive sequence after fault

for x=1:n

zbp(x)= pzbus(x,f); %zb =z15 z25 etc

end

%fc=up(f)*inv(zbus(f,f));%fault current

for y=1:n

uafp(y)=up(y)-if012*zbp(y);

end

c45p=(uafp(4)-uafp(5))*inv((0.00363+0.03993i));

%negative sequence after fault

for x=1:n

zbn(x)= nzbus(x,f); %zb =z15 z25 etc

end


for y=1:n

uafn(y)=-if012*zbn(y);

end

c45n=(uafn(4)-uafn(5))*inv((0.00363+0.03993i));

%zero sequence after fault

for x=1:n

zbz(x)= zbus(x,f); %zb =z15 z25 etc

end


for y=1:n

uafz(y)=-if012*zbz(y);

end

c45z=(uafz(4)-uafz(5))*inv((0.0242+0.09075i));


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%current flowing from 4 to 5 in each phase

A=[1 1 1; 1 -0.5-0.866i -0.5+0.866i;1 -0.5+0.866i -0.5-0.866i];

Izpn=[c45z; c45p; c45n];

I45abc=A*Izpn; %Iabc is current flowing through 4 5 in all the phase

%voltage at all bus at all the phase after fault

for(x=1:8)

Uafa(x)=uafz(x)+uafp(x)+uafn(x); %Uafa is after fault A phase voltage

end

for (x=1:8)

Uafb(x)=uafz(x)+(-0.5-0.866i)*uafp(x)+(-0.5+0.866i)*uafn(x);%Uafa is after fault B

phase voltage

end

for (x=1:8)

Uafc(x)=uafz(x)+(-0.5+0.866i)*uafp(x)+(-0.5-0.866i)*uafn(x);%Uafa is after fault C

phase voltage

End


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Figure 3.1: Flow chart

Finding Z bus for

positive/negative/

zero sequence

If 50=If 51=If 52= Up (Prefault Voltage)

Z550+Z55+ +Z55-

If 5a =If0 +If1+If2

Positive Sequence Voltage in all Bus

= - If5+

Negative Sequence Voltage in all Bus

= - If5-


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Zero Sequence Voltage in all Bus

= - If50

=

=

=


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Table 3.2: Static Var Compensator

Bus Power Ploss Qloss V3 V4 V5 V6 V7 V8

7

0 41.858 483.09 1.029 0.985 0.969 0.911 0.912 0.928

50 40.559 468.17 1.03 0.991 0.976 0.921 0.928 0.939

100 39.392 454.8 1.032 0.996 0.982 0.932 0.944 0.95

150 38.369 443.14 1.033 1.002 0.989 0.943 0.961 0.962

200 37.503 433.35 1.035 1.008 0.995 0.954 0.978 0.974

250 36.811 425.6 1.037 1.014 1.002 0.9666 0.996 0.987

6

50 40.517 467.52 1.03 0.992 0.977 0.923 0.922 0.935

100 39.293 453.28 1.032 0.999 0.985 0.936 0.933 0.943

150 38.193 440.45 1.034 1.006 0.993 0.949 0.944 0.95

5 50 40.868 472.6 1.032 0.996 0.982 0.919 0.919 0.933

100 39.995 463.44 1.035 1.007 0.994 0.927 0.926 0.938

8

50 40.717 470.74 1.03 0.989 0.974 0.918 0.923 0.944

100 39.714 458.54 1.031 0.994 0.979 0.926 0.935 0.96

150 38.862 448.73 1.032 0.998 0.984 0.934 0.947 0.977

200 38.176 440.8 1.033 1.002 0.989 0.942 0.959 0.994

250 37.672 434.91 1.035 1.007 0.994 0.951 0.972 1.012

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Power System Analysis in Grid

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