PROBLEMSETEE443.docx

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Akshay Nerurkar
nerurkar@usc.edu
Abstract
Studyof differentparametersof electrical engineeringusingpowerworldsimulator
PROBLEM SET
EE 443 PROF. CASTRO

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Table of Contents PG
NO.
1. The study of Power flow analysis………………………………………………………… 2
2. Fault analysis………………………………………………………………………………………. 19
3. Unsymmetricalfault in a circuit…………………………………………………………… 36
4. Power factor correction……………………………………………………………………….54
5. Power circle diagram for a transmission line……………………………………….. 61
6. Phaseshifting transformer…………………………………………………………………… 66
7. Choice of conductor for a transmission line…………………………………………. 81
8. Transmission line evaluation………………………………………………………………… 86
9. Load angle estimation for a power line…………………………………………………. 96
Shortcircuit duty…………………………………………………………………………………. 101

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10.
PROBLEM#01 THE STUDY OF POWER FLOWANALYSIS
In this problemwe are going to study about real and reactive power flow and its
relationship with voltage and angle δ.
VS/δo
VR/0O
SS=PS + jQS TRANSMISSIONLINE SR=PR + jQR
LOAD
GENERATING STATION
#A TWO-BUS POWER SYSTEM
A power flow study (load-flow study) is a steady-stateanalysis whosetargetis to
determine the voltages, currents, and real and reactive power flows in a system
under a given load conditions. The purposeof power flow studies is to plan ahead
and accountfor various hypotheticalsituations. For example, if a transmission line
is be taken off line for maintenance, can the remaining lines in the systemhandle
the required loads withoutexceeding their rated values. In the given diagramVR is
the receiving end voltage and VS is the sending end voltage and SS and SR are the
sending end and receiving end MVA’s. And we shall see an relation between real
and reactive power flow with voltage and angle δ.
APPROACH
1. The motive here is to analyzethe flow of real and reactive power. I have
decided to study the flow of power between buses NICOL69 and DAVIS69.
G

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2. Then, I am going to measurethe voltage levels and load angles at each of
the buses. And then calculate the direction of real and reactive power flow
direction in the circuit.
3. And I am going to establish the relation between real power and load angle
and voltage and reactive power.
4. And then I am going to simulate for the same in the power world analysis.
Figure1.1 shows the Power World example of Metropolis Light and Power
Problem 1

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The following figure is a zoomed view of the buses NICOL69 and DAVIS69. (Figure
1.2)
I am going to consider two cases. Onewith base caseand the other one with the
load increased one of the buses.
CASEI
Let us consider the basecase for the buses DAVIS69 and NICOL69.
DAVIS69
We can look at the Real and Reactive power levels on the bus DAVIS69by right
clicking on the bus and selecting the Quick Power Flow List option. This is what
appeared when I selected the Quick Power Flow Option for the bus, DAVIS69.
(Figure1.3)
FIG 1.3

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Q= 12.72MVAR
As can be seen from the Quick Power Flow List for the bus, DAVIS69, theReal
Power Flow at the bus is 38 MW and the reactive power flow is 12.72 MVAR.
P = 38 MW, Q = 12.72 MVAR
Figure1.4 gives the power triangle for the bus, DAVIS69.
P= 38 MW
Figure1.4
There is a lagging power factor at the bus.
Power factor = pf = cosθ
𝑡𝑡𝑡 𝑡=
38
12.72
This gives,
Θ1 = 71.44⁰(lagging)
NICOL69
We can look at the Real and Reactive power levels on the bus NICOL69 by right
appeared when I selected the Quick Power Flow Option for the bus, NICOL69.
(Figure1.5)

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FIG 1.5
As can be seen from the Quick Power Flow List for the bus, NICOL69, theReal
Power Flow at the bus is 28 MW and the reactive power flow is 6 MVAR.
P = 28 MW, Q = 6 MVAR
Figure1.4 gives the power triangle for the bus, NICOL69.
Q= 6MVAR
P= 28 MW
Figure1.4
There is a lagging power factor at the bus.
tanθ =28/6
This gives,

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Θ2 = 77.90⁰(lagging)
δ = θ1 – θ2
δ =-71.44 ⁰+77.90⁰
δ = 6.46⁰
Thus, theoretically the Real Power should flow from NICOL69 to DAVIS69.
Now,
VNICOL = 1.01 pu. (Base= 69 KV)
VDAVIS =1.02 pu. (Base = 69 KV)
Therefore, the Reactive Power mustflow from the bus DAVIS69 towards NICOL69.
We know that the value of the real power, P is given by the formula,
P = (VA+VB/X)*𝑡𝑡𝑡 𝑡
We know the value of VA, VB and δ. We need to find the value of X to find the
value of P.
The base voltage at the buses is 69 KV. The base MVA can be obtained by right
clicking on the generator on the bus DAVIS69 and selecting the generator
information dialog

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P=VNICOL*VDAVIS 𝑡𝑡𝑡 𝑡
X

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VNICOL=69*1.01 KV
VNICOL=69.69 KV
VDAVIS=69*1.02 KV
VDAVIS=70.38 KV
P= 69.69*70.38 𝑡𝑡𝑡6.46
47.61
PTHEOROTICAL=11.61 MW.
Now that we have theorotical figures and predictions, I can simulate the model
and verify my calculations in the power world model.
FIG. 1.6
In the given figure the green arrow representthe active power flowing in the
systemand he blue arrow indicate the reactive power flowing in the system.
Fromthe FIG.1.6itis evident that the real power flows fromNICOL69 toDAVIS69
and the reactive power flows fromDAVIS69 to NICOL69.To calculatethe total Real
Power flow, right click on the transmission lines between the two buses and
select the Line Information Dialog. Add the Real Power flows for both the
transmission lines and get the total flow.

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FIG 1.7
As can be seen from figure1.7, the total real power flow between the bus
NICOL69 and DAVIS69comes outto be 9.51 MW. Thus the total power flow is
PActual = 9.51 MW fromthe NICOL69 to DAVIS69 bus.
There is a loss of 19.5% in the calculation of the theorotical and actual power
measurement sincethe theorotical power and the actual power have a diffrence
of 2.24 MW due to the losses in the other branches.
CASE2
In this caseim adding a 30 MVARcapacitor to the systemand weare going to see
its effect on the real and reactive power hence on the load angle and voltage.

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FIG 1.8
DAVIS69
We can look at the Real and Reactive power levels on the bus DAVIS69by right
appeared when I selected the Quick Power Flow Option for the bus, DAVIS69.
(Figure1.9)
FIG 1.9

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P=38 MW
Q= 4.2 MVAR
As can be seen from the Quick Power Flow List for the bus, DAVIS69, theReal
Power Flow at the bus is 38 MW and the reactive power flow is -4.2 MVAR.
P = 38 MW, Q = -4.2 MVAR
FIG 1.10
There is a leading power factor at the bus.
𝑡𝑡𝑡 𝑡=
4.2
38
This gives,
Θ1 = 6.30⁰(leading)
NICOL69
We can look at the Real and Reactive power levels on the bus NICOL69 by right
appeared when I selected the Quick Power Flow Option for the bus, NICOL69.
(Figure1.11)

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P=28 MWQ=25.78 MVAR
P=28 MW
FIG 1.11
As can be seen from the Quick Power Flow List for the bus, NICOL69, theReal
Power Flow at the bus is 28 MW and the reactive power flow is 25.78 MVARas
there is shuntcapacitor.
P = 28 MW, Q = 31.78-6.00 MVAR
Q=25.72 MVAR
Figure1.11 gives the power triangle for the bus, NICOL69.
FIG 1.11

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There is lagging power factor at NICOLA69 bus.
tanθ =28/25.72
This gives,
Θ2 = -42.56⁰(lagging)
δ = θ1 – θ2
δ =6.30 ⁰ + 47.43⁰
δ = 53.73⁰
Thus, theoretically the Real Power should flow from NICOL69 to DAVIS69.
Now,
VNICOL = 1.03 pu (Base= 69 KV)
VDAVIS =1.02 pu (Base = 69 KV)
Therefore, the Reactive Power mustflow from the bus NICOL69 towards DAVIS69.
We know that the value of the real power, P is given by the formula,
P = (VA+VB/X)*𝑡𝑡𝑡 𝑡
We know the value of VA, VB and δ. We need to find the value of X to find the
value of P.
The base voltage at the buses is 69 KV. The base MVA can be obtained by right
clicking on the generator on the bus DAVIS69 and selecting the generator
information dialogue box.

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P=VNICOL*VDAVIS 𝑡𝑡𝑡 𝑡
X
VNICOL=69*1.03 KV
VNICOL=71.07 KV
VDAVIS=69*1.02 KV
VDAVIS=70.38 KV
P= 71.07*70.38 𝑡𝑡𝑡53.46
47.61
PTHEOROTICAL=80.36MW.
Now that we have theorotical figures and predictions, I can simulate the model
and verify my calculations in the power world model.

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FIG 1.13
As can be seen from figure1.7, the total real power flow between the bus
NICOL69 and DAVIS69comes outto be 10.73 MW. Thus the total power flow is
PActual = 10.73 MW fromthe NICOL69 to DAVIS69 bus.
There is a loss of 80.5% in the calculation of the theorotical and actual power
measurement sincethe theorotical power and the actual power have a diffrence
of 70 MW due to the losses in the other branches. Bt stil due to the changein the
load angle there is changein the real power for the system.
Hence fromthe given example the relationship between real power and load
angle and reactive power and voltage is proved.

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Real powerACTUAL
MW
Load angle δο
Reactive
powerBRANCH
MVAR
Bus Voltage
KV(NICOL69&DAVIS69
)
9.51 6.46 12.40 69.69&70.31
10.73 53.43 4.04 71.07&70.38
I would also like to put forth my results via the matlab analysis for the equation.
DISCUSSION:
Fromthe abovecalculations and simulations we can safely correlate the real
power flows with the load angle and the reactive power flows with the voltage
levels. Real power flows fromhigher to lower load angle bus and the reactive
power flows fromhigher voltage to lower voltage level bus. We also were able to
predict the amountof amount of real power that might flow from say bus A to
bus B. Fromthe 2 cases I can also make a statement that when the reactive power
flow flow between two buses is almost nil, it becomes easier to predict accurate
values of real power flows.
PROBLEM#02 FAULTANAlYSIS

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Itis not practical to design and build electrical equipment or networks so as to
completely eliminate the possibility of failure in service. Itis therefore an
everyday factof life that different types of faults occur on electrical systems,
however infrequently, and at randomlocations. Faults can be broadly classified
into two main areas which have been designated “Active” and “Passive”.
➢ Active Faults
The “Active” fault is when actual current flows fromone phaseconductor to
another (phase-to-phase) or alternatively fromone phase conductor to earth
(phase-to-earth). This type of fault can also be further classified into two areas,
namely the “solid” fault and the “incipient” fault. The solid fault occurs as a result
of an immediate complete breakdown of insulation as would happen if, say, a pick
struck an underground cable, bridging conductors etc. or the cable was dug up by
a bulldozer. In mining, a rockfallcould crush a cable as would a shuttle car. In
these circumstances the fault current would be very high, resulting in an electrical
explosion. This type of fault must be cleared as quickly as possible, otherwise
there will be: Greatly increased damage at the fault location. (Fault energy = 1² x
Rf x t wheret is time). Danger to operating personnel(Flash products). Danger of
igniting combustiblegas such as methane in hazardous areas giving riseto a
disaster of horrendous proportions. Increased probability of earth faults
spreading to other phases. Higher mechanical and thermal stressing of all items of
plant carrying the current fault. (Particularly transformerswhosewindings suffer
progressiveand cumulative deterioration becauseof the enormous
electromechanical forces caused by multi-phase faults proportionalto the current
squared). Sustained voltage dips resulting in motor (and generator) instability
leading to extensive shut-down atthe plant concerned and possibly other nearby
plants. The “incipient” fault, on the other hand, is a fault that starts fromvery
small beginnings, fromsay some partial discharge(excessiveelectronic activity
often referred to as Corona) in a void in the insulation, increasing and developing
over an extended period, until such time as it burns away adjacentinsulation,
eventually running away and developing into a “solid” fault.
➢ Passive Faults

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Passivefaults are not real faults in the true senseof the word but are rather
conditions that are stressing the systembeyond its design capacity, so that
ultimately active faults will occur. Overloading - leading to overheating of
insulation (deteriorating quality, reduced life and ultimate failure). Overvoltage -
stressing the insulation beyond its limits. Under frequency - causing plant to
behave incorrectly. Power swings - generators going out-of-step or synchronism
with each other.
● Types of Faults on a Three PhaseSystem
The types of faults that can occur on a three phaseA.C. systemare as follows:
Types of Faults on a Three PhaseSystem.
(A) Phase-to-earth fault
(B) Phase-to-phasefault
(C) Phase-to-phase-to-earth fault
(D) Three phase fault
(E) Three phase-to-earth fault
(F) Phase-to-pilotfault *
(G) Pilot-to-earth fault *
* In underground mining applications only

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Itwill be noted that for a phase-to-phasefault, the currents will be high, because
the fault currentis only limited by the inherent (natural) series impedance of the
power systemup to the point of faulty (refer Ohms law). By design, this inherent
series impedance in a power systemis purposely chosen to be as low as possible
in order to get maximum power transfer to the consumer and limit unnecessary
losses in the network itself in the interests of efficiency. On the other hand, the
magnitude of earth faults currents will be determined by the manner in which the
systemneutral is earthed. Solid neutral earthing means high earth fault currents
as this is only limited by the inherent earth fault (zero sequence) impedance of
the system. Itis worth noting at this juncturethat it is possibleto control the
level of earth fault current that can flow by the judicious choice of earthing
arrangements for the neutral. In other words, by the useof Resistance or
Impedancein the neutral of the system, earth fault currents can be engineered to
be at whatever level is desired and are therefore controllable. This cannot be
achieved for phase faults.
● Transient & PermanentFaults
Transient faults are faults which do not damage the insulation permanently and
allow the circuit to be safely re-energised after a shortperiod of time. A typical
example would be an insulator flashover following a lightning strike, which would
be successfully cleared on opening of the circuit breaker, which could then be
automatically reclosed. Transientfaults occur mainly on outdoor equipment
whereair is the main insulating medium. Permanent faults, as the name implies,
are the resultof permanent damage to the insulation. In this case, the equipment
has to be repaired and reclosing mustnot be entertained.
● Symmetrical & AsymmetricalFaults
A symmetricalfault is a balanced fault with the sinusoidalwaves being equal
about their axes, and represents a steady state condition. An asymmetricalfault
displays a d.c. offset, transient in nature and decaying to the steady state of the
symmetricalfault after a period of time:
Amongstthese faults assosiated with power systemsystemim gonna discuss in
shortabout symmetricsland asymmetricslfaults and discuss its effects with help
of the power world simulator:
1. Symmetrical Faults

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2. AsymmetricalFaults
Symmetrical Faults
When there is a 120o
equalphase shiftbetween the conductor due to thid
symmetry it is refered to a symmetricla fault.When there is shortcircuit condition
on a 3 phaseline such kind of fault occurs. This fault condtion rarely occurs most
of the time asymmetricla fault condition occuring but symmetricalcondition
impose moredamage on the breakers and are more severe.
Asymmetrical Faults
This is the frequently occuring fault on a power line an asymmetric or unbalanced
fault does not affect each of the three phases equally. Common types of
asymmetric faults, and their causes:
● line-to-line - a shortcircuit between lines, caused by ionization of air, or
when lines come into physicalcontact, for example due to a broken
insulator.
● line-to-ground - a shortcircuit between one line and ground, very often
caused by physicalcontact, for example due to lightning or other storm
damage
● double line-to-ground - two lines come into contact with the ground (and
each other), also commonly due to stormdamage.
The following is the curverepresenting the symmetricaland asymmetrical
faults in a system.

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In this problemim going to show with power world how a symmetricalas well as
a symmetricalfault might occour and affect the characteristic impedence of a
power line. For this im going to assumea fault on the bus and line fromSLACK345
to TIM345.
APPROACH:
1. Firstim going to find the sequence impedencevalue for a given line in our
case it is the line between buses SLACK345 and TIM345fromthebranch
information option by double clicking on the branch.
2. Then im going go to the Run option and then choose the Fault analysis
option and Choosethe single fault options
3. Then im going to choosethe fault location and fault types and its effect on
the line impeadance.
4. Then we are going to find the effects of the differenttypes of fault using
ETAP softwareas a tool.
Firstwe are going to choosea case Design case_6.1

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Design case_6.1
FIG. 2.1
The branch between buses SLACK345 and TIM345is the branch weare going to
analyzefor the effects of the different kinds of faults.
Firstdouble click on the branch SLACK345 and TIM345 and record the
characteristic impedance of the branch

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FIG 2.2
Fromthe FIG_xx
Z=R+jX=0.02+j0.03 pu;
Y=G+
𝑡
𝑡
=0+
𝑡
0.18
pu;
Now we go to the tool ribbon and click on the fault analysis and choosethe Fault
location in our case we will choosethe in-line fault now we choosethe single line
to ground fault and calculate the fault currentvalue and the characteristic
impedance value

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FIG 2.3
Fromthe FIG 2.3
Z=R+jX=0.015+j0.022;
Y=G+
𝑡
𝑡
=0+
𝑡
0.13
;
If
”
=44 p.u.=7833A.
WE CAN ALSO SET THE LOATIONOF THEFAULTON THE LINE.

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FROM THE FIG 2.4 LOCTIONOF THEFAULTFROM THE FARBUS.
Now we go to the tool ribbon and click on the fault analysis and give the location
% for the fault on the line we can vary it from1% to 100%.
NOW WE CAN CAN CALCULATE THE CHANGE OF CHARACTERISTICIMPEDANCE
AND ITS AFFECTONTHE LONG TRANSMISSIONLINEPARAMETER
WE CAN CALCULATE THE INITIAL LOSSLESS AND FAULTLESS PARAMETERS
FIRSTWEKNOW THAT FORA TRANSMISSIONLINETHESENDING END AND
RECEIVING END VOLTAGES AREGIVENAS-
● VS=AVR+BIR
● IS=CVR+DIR
THE EXACT MODEL GIVENFORTHE LONG LINEIS GIVENAS
● VS=AVR+BIR
● IS=CVR+DIR
BUT A=D=𝑡𝑡𝑡𝑡 𝑡𝑡 & C=
1
𝑡 𝑡
𝑡𝑡𝑡𝑡 𝑡𝑡 & B=ZC 𝑡𝑡𝑡𝑡 𝑡𝑡
Firstwe have to find the receiving end current

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P=√3VI𝑡𝑡𝑡 𝑡
INITIALLYWEASSUMETHATIT’S A LOSSLESS YSTEMHENCE WE ASSUME A UNITY
POWERFACTOR
HENCE FROMTHE FIG 2.4
FIG 2.5
TIM345 BUS THEPOWER IS 122 MW AND REACTIVEPOWERIS 18 MVAR
AND V=345 KV WE HAVETO CONVERTITINTO PHASEVOLTAGEV=
345
√3
=200KV

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FIG 2.6
IR=
122𝑡𝑡
√3∗200∗1
=0.325̷ 0O
Fromthe fig 2.6 wecan see that the impedance and admittance for the line in
ohms/mile and mhos/miles which we requirefor our calculation.
Z=0.2+J3.5=3.6̷ 86O
ohms/mile
Y=
𝑡
15.28
=0.065̷ 90O
mhos/miles
ϒL=𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡 𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 𝑡𝑡 𝑡𝑡𝑡𝑡𝑡√ZY=10 ∗ √3.600̷ 86O
*0.065/ 90O
ϒL=0.4805̷ 88O
ϒL=0.013+j0.485
Cosh(αl+βl)=
1
2
(eαl
/βl+e-αl
/-βl)

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/βl=0.48 radians=/27.06o
e0.013
/ 27.06o
=1.01/ 27.06o
=0.89+j0.46
e-0.13
/ -27.06o
=0.98/-27.06=0.87-j0.445
Hence,
cosh ϒL=0.88+j0.02=0.88/1.30o
similarly, sinh ϒL=0.01+j0.455=0.455/88.72o
Zc=√
𝑡
𝑡
=7.44/-2.0o
A=D=cosh ϒL=0.88/1.30o
B=Zcsinh ϒL=3.236/86.72o
C=sinh ϒL/Zc=0.061/91o
Now,
● VS=AVR+BIR
● IS=CVR+DIR
Vs=0.88/1.30 *200/0 + 3.236/86.72*0.325/0
=176/1.30+1.051/86.72
=176.61+j5.097
=176/1.67KV
IS=0.061/91*200/0+0.88/1.30*0.325/0
=12.2/91+0.3/1.30
=0.087+j12.20
=12.20/90kA
The power factor of the line is=cos(90-4)=0.07 leading which is way of the unity
power factor we assumed therefore there is a error in the systemdue to other
interconnection in the system.

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FIG 2.7
FIG 2.8
You can see fromthe FIG 2.8 that receiving end of the transmission line receives
122 MW real power and 33 MVARreactive power fromthe fault pint instead of
the sending end bus because of fault on the line there is no power flow between
the buses SLACK345 and TIM345 henceweare going to assumethat the fault

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point is the sending end bus instead of the SLACK35 bus which is are actual
sending end bus.
FIG 2.8
Fromthese two figures we can calculate the sending end voltage and sending end
currentfor the system. And we know fromFIG_xxthe reactance,resistanceand
capacitance values.So calculating the following we get,
TIM345 BUS THEPOWER IS 122 MW AND REACTIVE POWERIS 22 MVAR
AND V=345 KV WE HAVETO CONVERTITINTO PHASEVOLTAGEV=
345
√3
=200KV
IR=
122𝑡𝑡
√3∗200∗1
=0.325̷ 0O

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FIG 2.9
Z=1.7815+J2.67=3.2026̷ 56O
pu.
Y=
𝑡
11.45
=0.085̷ 90O
pu.
ϒL=𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡 𝑡𝑡𝑡 𝑡𝑡𝑡𝑡√ZY=10 ∗ √3.2026/ 56O
*0.085̷ 90O
ϒL=0.52/ 72O
ϒL=0.16+j0.49
Cosh(αl+βl)=
1
2
(eαl
/βl+e-αl
/-βl)
/βl=0.49 radians=/28.07o
e0.16
/ 28.07o
=1.17/ 28.07o
=1.04+j0.55
e-0.16
/ -28.07o
=0.85/-28.07=0.75-j0.39

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Hence,
cosh ϒL=0.93+j0.04=0.93/2.46o
Zc=√
𝑡
𝑡
=6.15/-17o
& C=sinh ϒL/Zc=0.16/91o
Now,
● VS=AVR+BIR
● IS=CVR+DIR
Vs=0.93/2.46*200/0+0.019/55.72*0.325/0
=186/2.46+0.019/55.72
=185.83+j7.9
=185/2.44KV
IS=0.16/91*200/0+0.93/2.46*0.325/0
=32/91+0.3/2.46
=-0.26+j32.19
=32.20/90kA
The power factor of the line is=cos(90-4)=0.07 leading which is 7% of the unity
power factor we assumed at the receiving end bus therefore there is a error in
the system dueto other interconnection in the systemand we cannot assumea
power factor for large systemto be one as a systemcannot be lossless.And also
the sending end currenthas been affected by the change in the characteristic
impedance of the circuit.
DISCUSSION:

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● Fromthe firstpart of the problem we simulated the value of the impedance
and calculated the sending end voltage and current for a stedy state
lossless mediumi.e. the long transmission line.
● For the second part still assuming the line to be lossless wesimulated a
fault at 25% fromthe bus SLACK345 and calculated the impedance anf
hence the values of the sending end voltage and current
● Fromthe calculation and simulation we can clearly see that the
characteristic impedance of the systemhas reduced hence the sending end
currentfor the systemhas increased .
● Therefore for the given systemthe fault alters the characteristic values of
impedance and therefore the circuit parameters.
● And hence for a systemif the characteristic impedance is reduces the
currentin the circuit increases and vise-versa.
● In this way the characteristic impedance influences the currentflowing in
the circuit. We can also compare the characteristic impedances to the line
impedance. Itappears to me that characteristic impedance Z0 is a
transmission line parameter, depending only on the transmission line
values R, G, L, and C. Whereas line impedance is Z depends on the
magnitude and phaseof the two propagating wavesV ( ) z + and V (z ) − --
values that depend not only on the transmission line, but also on the two
things attached to either end of the transmission line.
PROBLEM#03 UNSYMMETRICAL FAULTS IN POWER SYSTEM

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Intent of the problem: This problem we deduces the zero sequence network of the
system and hence calculate the fault in the sequence network.
What is a sequence network why is it necessary?
In symmetricalcomponent analysis (e.g. for unbalanced faults), a balanced three-
phaseelectrical network can be broken down into three sequence networks,
which are independent, de-coupled sub-networks comprising only quantities in
the samesequence, i.e. the positive sequence network contains only positive
sequence quantities, the negative sequencenetwork contains only negative
sequence quantities and the zero sequencenetwork contains only zero sequence
quantities.
The phase faults are unique since they are balanced i.e. symmetrical in three
phase, and can be calculated fromthe single phase positive sequenceimpedance
diagram. Therefore three phase fault currentis obtained by,
Where IF is the total three phasefault current, v is the phase to neutral voltage z1
is the total positive sequence impedance of the system; assuming that in the
calculation, impedance are represented in ohms on a voltage base.
Symmetrical ComponentAnalysis
The above fault calculation is made on assumption of three phase balanced
system. The calculation is made for one phaseonly as the currentand voltage
conditions are same in all three phases. When actual faults occur in electrical
power system, such as phaseto earth fault, phaseto phase fault and double
phaseto earth fault, the systembecomes unbalanced means, the conditions of
voltages and currents in all phases are no longer symmetrical. Such faults are
solved by symmetricalcomponent analysis. Generally three phase vector diagram
may be replaced by three sets of balanced vectors. Onehas opposite or negative
phaserotation, second has positivephase rotation and last one is co-phasal. That
means these vectors sets are described as negative, positive and zero sequence,
respectively.

39 | Page
For any location in the system, the sequence networks can be reduced to
Thevenin equivalent circuits for illustration of the application of thevenin theorem
for determing the equivalent sequencenetworks, consider a simple power
systemshown in the FIG 3.1
Fig 3.1
1. The impedances are all constantand independent of currents.
2. The synchronous generator is a salient pole type which may generate under
faiult condition, negative and zero-sequenceemfs which are small and are
negligible. Thus for all purposes the machine is assumed togenerate only
positive sequenceemfs.
The network under such conditions can be represented by 3 independent
single phasesequence network.
Approach:
1. I will firstdo the calculation by hand to find the fault currentat 50% line using
the selected circuit shown in Figure _xx and draw the sequence network for the
circuit.

40 | Page
2. Using the Power World software, I willthen find the fault currentat line at 50%
given by Power World.
3. With the fault current calculated by hand and through Power World, I will find
the error and hypothesizepossiblereasons for such an error.
Figure
FIG 3.2
Solution:
FirstI need to select the base MVA value. I can do this by right clicking on the
generator information dialog and obtaining the baseMVA value. (Figure 3.2)

41 | Page
FIG 3.3
Itcan be seen from the Generator Information Dialog that the base MVA value for
the generator is 20 MVA. Thus, I am going to use this MVA value as the base MVA
value for the fault calculations.
The following circuit is obtained for the selected buses in figure3.4. (Figure3.4)
BASEMVA =20 MVA BASEKV =16 KV

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ZONE 1
BASEMVA =20 MVA VB2= 16*(
138
16
) = 130𝑡 VB3= 138*(
16
138
) = 16𝑡
BASEKV =16 KV IB2=
𝑡𝑡𝑡
√3∗𝑡𝑡
=
20000
√3∗138
=84A IB3=
𝑡𝑡𝑡
√3∗𝑡𝑡
=
20000
√3∗16
=722A
IB1=
𝑡𝑡𝑡
√3∗𝑡𝑡
=
20000
√3∗16
=722A ZB2=
𝑡𝑡2
𝑡𝑡𝑡
=
1382
20
= 952 Ω ZB3=
𝑡𝑡2
𝑡𝑡𝑡
=
162
20
= 13 Ω
ZB1=
𝑡𝑡2
𝑡𝑡𝑡
=
162
20
= 13 Ω
To effectively calculate all these values for the generators, transformers and the
transmission lines, I need to obtain these values using Power World simulator.
The data for the generator can be obtained using the Generator Information
Dialog. The data for the transmission lines and the transformer can be obtained
using the Branch Information Dialog. Thus I obtained the reactance and MVA
value for all the generators, transformersand transmission lines using the
information dialogues.

43 | Page
FIG 3.5
XG, NEW= 0.1*(
16
16
)2
*(
20
18720
)=0.1 p.u.

44 | Page
FIG 3.6
XT, NEW= 0.12*(
16
16
)2
*(
20
20
)=0.12p.u.

45 | Page
FIG 3.7
We have the transmission line reactance as X=0.10 p.u. fromthe FIG 3.7 as we
know the value in p.u. we don’thave to calculate the value manually.

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FIG 3.8
XT, NEW= 0.12*(
138
138
)2
*(
20
20
) = 0.12 𝑡.𝑡.
Now we draw the sequence network for the systemand evaluate the fault current
manually.
WE ASSUMETHE POSITIVE, NEGATIVEAND ZERO SEQUENCEREACTANCES FOR
THE GENERATOR TO BE SAME X2=X1=X0=0.1 P.U.

47 | Page
FIG 3.9
THE FAULT CURRENT FOR THE SYSTEMAT MIDPOINTOF THETRANSMISSIONLINE
IS
IF=
3∗𝑡 𝑡
𝑡2+𝑡1+𝑡0
=
3
0.10+0.10+0.155
=8.51 P.U.
AS THE FAULTCURRENT IS OCCURING INZONE2,
IF=8.51*ZB2=8.51*84=720A
NOW WE EVALUATE FORPOWERWORLD ANALYSIS TO FIND THEFAULT
CURRENT.
WE WILL TRY WITH ALL FOURFAULT TYPES AND FIND THEP.U AS WELL AS
AMPERE FAULT CURRENT FOR THE SYSTEM.FIRSTSETTHELOCATION AT50% OF
THE TRANSMISSIONLINE.

48 | Page
FIG 3.10
NOW WE CAN ASSUME THE DIFFERENTTYPES OF FAULTS ONE BY ONE
1. SINGLELINETO GROUND FAULT

49 | Page
FIG 3.11
AS YOUCAN SEE FROM THE FIG 3.11 THATTHIS NOTTHETYPE OF FAULTTHAT IS
OCCURING INOURGIVENCIRCUIT.
2. 3-PHASEBALANCED FAULT
I. INPER UNIT
II. INMAGNITUDEORAMPERES

50 | Page
3. LINETO LINEFAULT
I. INPER UNIT
FIG 3.13
II. INAMPERES

51 | Page
FIG 3.14
4. DOUBLELINETO GROUND FAULT
FIG 3.15

52 | Page
FROMTHE POWERWORLD ANALYSIS WESEE THAT THERE IS A ERROR OF 50% IN
THE SYSTEM THAT DUETO OUR ASSUMPTIONAND POSSIBLEERRORINTHE
CIRCUITITSELF
FROMTHE ANALYSIS INTHEPOWERWORLD WE ALSO CAN CLERLY SEE THAT
SINGLELINETO GROUND AND DOUBLELINET GROUND DOESN’TOCCURINOUR
CIRCUITONLYTHE3-PHASEBALANCED AND LILETO LINEFAULT CAN OCCURIN
OUR CIRCUIT
DISCUSSION
● We have obtained a substantial error in our calculation and simulated
results due to our assumptions and possiblesmall error in the circuit. Thus
Power World has emerged as an effective tool in helping engineers to
simulate complicated models and get accurate results. Wewere also
successfulin determining that the 3 phase fault currentis the highest. Thus
all protection schemes are designed with this 3 phasefault currents in
mind.

53 | Page
PROBLEM#04 Power Factor Correction
Intent of the problem: This problem calculates the Power factor of a system and
the power factor improvement techniques so as to improve the total power quality
of the system.
WHAT IS POWER FACTOR AND WHY DO WE NEED TO IMPROVE IT?
Firstof all whatis power factor and how do we measure it.
➢ In an AC circuit, power is used mostefficiently when the currentis aligned
with the voltage.Efficient AC Current
➢ However, mostequipment tend to draw currentwith a delay, misaligning it
with the voltage. What this means is more currentis being drawn to deliver
the necessary amountof power to run the equipment. And the more an
equipment draws currentwith a delay, the less efficient the equipment
is.InefficientAC Current

54 | Page
➢ Power factor is a way of measuring how efficiently electrical power is being
used within a facility's electrical system, by taking a look at the relationship
of the components of electric power in an AC circuit. These components are
referred to as Real Power, Reactive Power and ApparentPower:
➢ Real power (kW) — the work-producing power thatis used to actually run
the equipment
➢ Reactive power (kVAr) —the non-work producing power thatis required to
magnetize and startup equipment
➢ Apparentpower (kVA) —the combination of real power and reactive
power
The purposeof power factor improvement is simply to reduce the load current
drawn fromthe supply. This allows conductors of smaller cross-sectionalarea to
be utilised, reducing the amount(and cost) of copper used in those conductors
and other supply plant. In the caseof larger commercial and industrialloads,
power factor is part of the tariff, and loads with low power factors may be
financially penalised, so higher power factors aredesirable as a means of reducing
utility bills.
There are many practical ways to do so The following devices and equipments are
used for Power Factor Improvement.
➢ Static Capacitor
➢ Synchronous Condenser
➢ PhaseAdvancer
In this problemwe are going to show the effect of a power factor improvement
on a systemusing power world analysis.
Approach:

55 | Page
Step 1: Firstwe are going manually calculate the power factor of circuit for buses
SANDER69 to BOB69 and note down its effect on the circuit parameters.
Step 2: Then we are going to add a shuntcapacitor to the circuit and calculate the
power factor of the circuit for the samebuses and note down its effect on the
circuit parameters
Step 3: We are going to comparethe results and using power world we are going
to simulate the results for the design case.
Solution:
Firstright click on the bus DAVIS69 and then choosethe quick power flow option
FIG 4.1
Now we note down the real power and reactive power at the bus and calculate
the power factor for the given bus.
P = 75 MW, Q = 50 MVAR

56 | Page
The power factor at DAVIS69bus.
tanθ =50/75
This gives,
Θ = 34⁰ (leading)
Pf=cos Θ=0.82
Now we to improvethe power factor to 0.95 so we are going to calculate the
shuntcapacitor required for the system
Cos Θ=0.95
Θ=𝑡𝑡𝑡−1
𝑡=18.19o
P = 75 MW, Q = X MVAR
FIG 1.11
Assuming the real power to be constantthe MVARrequired for the systemis
tan18.19 =X/75
This gives,

57 | Page
X=24.64 MVAR
Therefore the new shuntreactance to be added should be
XNEW=25 MVAR
Adding this new shuntreactance we get new MVAR value for the system
therefore consideing the values and calculating the new power factor we get,
FIG 4.2
As can be seen from the Quick Power Flow List for the bus, SANDER69, the Real
Power Flow at the bus is 28 MW and the reactive power flow is 6 MVAR.
P = 75 MW, Q = 33 MVAR
Figure1.4 gives the power triangle for the bus, SANDER69.
Q= 33MVAR
P= 75 MW

58 | Page
The power factor at SANDER69 bus.
tanθ =33/75
This gives,
Pf=cos Θ=0.92
Now we to improvethe power factor to 0.92 so we have an error in the
calculation due to the presenceof other interconnection to the buses wedidn’t
consider.
Now we connect an load to the systemand see the changes in the system
FIG 4.3
P = 75 MW, Q = 43 MVAR

59 | Page
Q= 43MVAR
P= 75 MW
tanθ =43/75
This gives,
Pf=cos Θ=0.86
Due to the additional load to the systemthere is a MVAR addition to the system
and the power factor decreases to improveit to 95% i.e. 0.95 weincrease the
shuntreactance value the value required we first are going to calculate manually
Cos Θ=0.95
Θ=𝑡𝑡𝑡−1
𝑡=18.19o
P = 75 MW, Q = X MVAR
FIG 1.11
Assuming the real power to be constantthe

60 | Page
MVAR required for the systemis
tan18.19 =X/75
This gives,
X=24.64 MVAR
Therefore the new shuntreactance to be added should be
XNEW=25 MVAR
Adding this new shuntreactance or increasing the value of the existing shunt we
get new MVAR value for the systemthereforeconsidering the values and
calculating the new power factor we get,
P = 75 MW, Q = 17 MVAR
tanθ =17/75
This gives,
Pf=cos Θ=0.97
Q= 33MVAR
P=75 MW

61 | Page
Now we to improvethe power factor to 0.97 i.e. 97% so we havean error in the
calculation due to the presenceof other interconnection to the buses that we
didn’t consider.
DISCUSSION:
I was unable to incorporate the effect of the other parts of the circuit in
improving the power factor at the selected bus. Thus, Power World has once
again proved its worth as a simulator which can provide moreaccurate results
than manual calculations. When adding the load, additional real and reactive
power came from the rest of the circuit since the values of the other generators
changed. Since some reactive power and real power were used in the
transmission lines, I had to depend on the real and reactive power values given in
Power World to calculate accurate values.

62 | Page
PROBLEM#05 Power Circle Diagramfor Transmissionline
Intent of the problem: This problem calculates the Sending end and Receiving end
power and we draw and explain these parameters via the power circle diagram.
What is a Circle Diagram & What is it significance?
Electrical lines circle diagram is a graphical representation of its equivalent circuit.
This means that whatever information can be obtained from the equivalent circuit,
the same can also be obtained from the circle diagram. The advantages of a circle
diagram are its simplicity and quick estimation of the machines operating
characteristics.
APPROACH
1. Firstwe are going to calculate the transmission line parameters and then
we are going to calculate the sending end and receiving end power.
2. Then we are going to plot the circle diagramusing these values as refrences
and other assumptions if necessary.
ASSUMPTIONS
For these problem we are going to assumea 300 kmline having R= 0.08
𝑡
𝑡𝑡
; X=0.4
𝑡
𝑡𝑡
and Y=5.15*10-6 𝑡
𝑡𝑡
. Fromthe diagram we will find the sending end voltage,
currentand power factor angle when the line is delivering a load of 192 MW at
0.8 pf (lagging) and 275 kV. Assuming a π-configuration.
Solution:
Resistance of the line per phase, R= 300*0.08=24 Ω.
Reactance of the line per phase, X=300*0.4=120 Ω.
Impedanceof line per phase, Z=24 +j120=122.38/78.69o
Ω.
Admittance of line per phase, Y=300*5.15*10-6
/90o
S=1.545*10-3
/90o
S.
Line constants A=D=1+
𝑡𝑡
2
=1+0.5*122.38 /78.69o*
1.545*10-3
/90o

63 | Page
=1+0.094/168.69=0.9073 +j 0.01854=0.9075/1.17
B=Z=122.38/78.69 Ω
Load power factor angle, ΦR=𝑡𝑡𝑡−1
0.8 = /36.9o
Load current, IR= 𝑡𝑡𝑡𝑡 𝑡𝑡 𝑡𝑡∗10
6
√3∗𝑡 𝑡𝑡 ∗𝑡𝑡𝑡 𝑡 𝑡
=
192∗106
√3∗275000∗0.8
=504 A
For given line A=0.9075; α=1.17o
; B=122.38 Ω and β=78.69o
For receiving-end power circle diagram
Horizontalcoordinate=-
𝑡
𝑡
VRL 𝑡𝑡𝑡(𝑡 − 𝑡)=-
0.9075
122.38
𝑡𝑡𝑡(78.69 − 1.17) *2752
=-121
MW
Vertical coordinate=-
𝑡
𝑡
VRL 𝑡𝑡𝑡 (𝑡− 𝑡) =-
0.9075
122.38
𝑡𝑡𝑡(78.69 − 1.17)*2752
=-548
MW
Take scale 1 cm= 100 MW horizontally and 1 cm=100 MVARvertically.
FIG 5.1

64 | Page
Locate point N having coordinates-1.21 cmand -5.48 cmand draw load line OP at
an angle 𝑡𝑡𝑡−1
0.8 i.e. 36.9o
inclined to the horizontaland to represent
√3∗𝑡 𝑡𝑡∗𝑡 𝑡
1000
MVAR i.e.
√3∗275∗504
1000∗100
cm=2.4 cm
Draw circle diagramwith N (-1.21 cm, - 5.48 cm) as center and NP as radius.
Fromcircle diagram NP= 7.6 cm………..(fromMeasurement)
∴ NP=
𝑡 𝑡𝑡∗𝑡 𝑡𝑡
𝑡
=100*7.6=760MVA.
∴ VSL=
760∗𝑡
𝑡 𝑡𝑡
=
760∗122.38
275
=338 kV
Β-α=64o
Δ=β-64o
=78.69o
-64=14.69o
For sending end power circle diagram
Horizontalcoordinate=-
𝑡
𝑡
VSL 𝑡𝑡𝑡(𝑡 − 𝑡)=-
0.9075
122.38
𝑡𝑡𝑡(78.69 − 1.17) *3382
=183
MW
Vertical coordinate=-
𝑡
𝑡
VSL 𝑡𝑡𝑡 (𝑡− 𝑡) =-
0.9075
122.38
𝑡𝑡𝑡(78.69 − 1.17)*3382
=827
MW
Radius of circle=
𝑡 𝑡𝑡∗𝑡 𝑡𝑡
𝑡
=7.6 cm
Draw a circle diagram with N (1.83 cm, 8.27 cm) as center and 7.6 cm as radius.
Draw NP inclined at angle β+δ i.e. 78.68 + 14.69 =93.38o
to the horizontalcutting
the arc of sending-end power circle at P. Join OP
Sending-end power factor angle=ΦS=15o
(frommeasurement)
Sending-end power factor=𝑡𝑡𝑡 𝑡 𝑡 =𝑡𝑡𝑡15=0.966.
OP=2.4 cm= 2.4*100=240 MVA
∴OP=
√3∗𝑡 𝑡𝑡∗𝑡 𝑡
1000
MVAR
∴IS=
𝑡𝑡∗1000
√3∗𝑡 𝑡𝑡
=410 A

65 | Page
Discussion
● Hence we have drawn the circle diagram for the sending end and receiving
end power and we can see that a circle diagramgives the results which are
sufficient accuratefor practical purpose, despitethe fact that an
approximate equivalent circuit is used in a circle diagramand provides a
panoramic view of how operating characteristics areaffected by changes in
the machines parameters, voltage, frequency etc.
● We also can see that for a long transmission linewe have assumed a π-
configuration
● And this can be done becauseit generates very low discrepancies in the
solution but a shorttransmission line solution might fail in this case and
hence cannot be assumed for the case.
PROBLEM#06 Phase shifting transformer

66 | Page
Intent of the problem: This problem calculates the Sending end and Receiving end
power and we draw and explain these parameters via the power circle diagram.
What is a PST & Why is it required?
A Phase-Shifting Transformer is a device for controlling the power flow through
specific lines in a complex power transmission network.
The basic function of a Phase-Shifting Transformer is to change the effective
phasedisplacement between the input voltage and the output voltage of a
transmission line, thus controlling the amount of active power that can flow in the
line.
APPROACH
1.Firstweare going to simulate a circuit in a power world simulator for a
transmission line and evaluate the power flowing through a line and the total
losses in a line.
2. Then we aregoing to add a phaseshifting transformer to the same
transmission line.
3. We are going to controlthe phasebetween the transmission line with the
help of the PSTand thereby evaluate the power flow.
SOLUTION:
Firstwe are going to consider the case3.60 for the question we evaluating,

67 | Page
FIG 6.1
In the given FIGURE6.1 weare assuming a two bus systemfor evaluation, the
values of the generator parameters are:
FIG 6.2

68 | Page
We set the limiting Max value for the generator and for the given generator
parameter we set the value of the load to 500 MW and 100 MVAR. And with help
of the load field we can controlthe value of the generator load.
FIG 6.3
The load field is given set the value of the delta or cahnge in the input power with
respect to the output power to 50 MW per click.

69 | Page
FIG 6.4
We can see fromthe simulator for that for the output of 500 W and 100 MVAR
the input power being generated is 534 MW and 236 MVAR fromFIG 6.1. And the
voltage angle set at the generating bus is 0O
for refrence hence the voltage angle
at the receiving end bus is

70 | Page
Vreceving end=297/-14o
kV
The phase angle at the receiving end is -14o
, now weare going to add a phase
shifting transformer to the circuit and set the phasevalue for the transformer to
0o
.

71 | Page
Now we can see from the FIG_xxthat the total power being transmitted to the
output has an alternate path and the reactive and active power are divided into
the total seprate paths to the load without the load being affected. We have set
the the degree field to th phaseshift of the load bus.

72 | Page
We are going to set the limit for thr phase shiftwe are applying to the circuit we
are considering.
FIG 6.8
For the circuit we havegiven a max and min phaseangle value to +/- 30o
for the
phaseshifting transformer. Now weare going to phaseshift a transformer by 10o
and evalute the change in the power transfer value.

73 | Page
FIG 6.9
Fromthe figureyou can see that the bulk of the power is flowing fromthe phase
shifting transformer and hence by changing the output power phasewe have
managed to reduce the power flow fromthe transmission line.
Now we change the phasefrom10o
to 20 o
and take a look at the change,
FIG 6.10
As the phaseis changed to 20o
the power flow fromthe transmission line is
further reduced and the power transmitted fromthe transformer is further

74 | Page
enhanced. We can also see that the transformer produces power morethan what
is required at the load and hence the direction of this residual power is changed
and now it flows fromthe load to the generator. You can see the change in the
phaseangle at the receiving end too.
FIG 6.11
Vreceving end=326/2o
kV
You can see that there is positivephase angle at the receiving end and now its
leads the sending end power line voltage.
Now we change the phaseto the maximum value of the phase shifting
ttansformer i.e.

75 | Page
FIG 6.12
Fromthe figureyou can see that the a significantamount of the power is being
transmitted fromthe load to the generator. Hence with the useof the phase
shifting transformer wehave managed to change the flow of power through a
circuit.

76 | Page
Vreceving end=319/6o
kV
You can see that there is increased value of positive phaseangle at the receiving
end and now its leads the sending end power line voltage.
Now we make a negative phasechange and evalute the circuit, now that the
phaseis reversed

77 | Page
FIG 6.14
Since the phaseis reversed the excess power of the load flows fromthe
transformer instead of the power line and reactive power is reversed to and it
flows fromthe power line as the MVAR at the sendinf end and receiving end have
little diffrencethe magnitude is small and there is no significantreactive power
flowing in the power line. Now we reversethe phasefurther and evalute the
circuit.
Now we reversethe phaseto the minimum limit of the transformer and evaluate
the changein the circuit

78 | Page
Since the phaseis reversed further the a significant excess power of the load
flows fromthe transformer instead of the power line and the reactive power
flows frombus 2 to bus 1 but as there is still no significant diffrencein the sending
end and receiving end voltage angle.
FIG 6.15
Vreceving end=319/-20o
kV
Hence we conclude that through a useof a phase shifttransformer wemanaged
to control the power flow in the circuit either through the transformer or the
power line.

79 | Page
DISCUSSION:
● A PST is a usefulmeans of control of active power flow, as is proved by
hands-on experience obtained from the varied applications. A simulation in
a power world simulator illustrates the ability to regulate the active power
transmitted over a line.
FIG 6.16

80 | Page
FIG 6.18
● But a draw back of a PSTis the reactive power losses which are significantly
high for positive as well as negative phasechanges. So its high power
applictaions are limited. Or they require really expensivecompensators.
PROBLEM#07 Choice of TransmissionLine Conductor

81 | Page
Intent of the problem: To calculate the impedance and admittance for a
transmission line using two different conductors and make a choice depending on
the losses for same input parameters.
Why do we need to differentiate between conductors?
WE need to because of the costof the conductors fora long transmission line to be
laid this might help reduce the infrastructure costand also minimization of losses
as one of the important factors as different conductorshave different conductivity
and losses margin for the different type of transmission lines.
Approach:
Step 1: First we are going to choose two conductors which we are going to
differentiate for our case study
Step 2: Then we are going to manually calculate the impedances and admittances
for the conductors.
Step 3: Then we are going to insert these values into the power world simulator
and for the same circuit we are going to evaluate the total loss in the system and
make a choice for the conductor to be best suited for the circuit.
For our case study we are going to choose BLUEBIRD and OSTRICH
Technical considerations
I will first assumethat the conductors arearranged in an couple fashion with a
distance of 15 feet between each of the conductors.
Firstwe consider BLUEBIRD conductor.
I will also assumethe GMR for thr circuit i.e. GMR=√(𝑡 𝑡 ∗
6")=3.33”/Φ=0.2775’/Φ
0.788*radius of theconductor

82 | Page
GMD=√ 𝑡3
∗ 𝑡 ∗ 2𝑡
=√11
3
∗ 11 ∗ 22
=13.86’
L=0.7411 log
𝑡𝑡𝑡
𝑡𝑡𝑡
=1.25mH/Miles/Φ
C=
0.0388
𝑡𝑡𝑡
𝑡𝑡𝑡
𝑡𝑡𝑡
=0.0228uF/Miles/Φ
Frequency we assumeto be=60 H
Hence,
XL=2πfL=2*π*60*1.25=0.471Ω/miles/Φ
XC=
1
2𝑡𝑡𝑡
=0.116MΩ-miles/Φ
Z=R+jXL;
Z=0.0476+j0.471…………………………………………………….(RfromtheA3 Table for a
BLUEBIRD conducter)
Y=B+j/XC;
Y=0+j/0.116=8.6 uʊ/mile…………………………………………………………Assuming B=0
Now we insert these values into the power world simulator circuit we asssume
Design case 1
_2010 Caseand asssumeany line for the case study.Wechoosethe line between
buses SANDER69 and DEMAR69
We go to the edit mode and double click the line between the buses and click on
calculate the impedances option there we isert our actual values and it is then
converted into the specific per unit values.

83 | Page
FIG 7.2
We can see that on the right side we havethe per unit values for the given actual
values.
Now we go to the run mode and run the simulation and now we note the loss
value for the line. first we go to the line information box option on right clicking
the line we are using for us line between the buses DEMAR69 and SANDER69.

84 | Page
FIG 7.3
We can see that the losseare 0.385 MW and 0.025 MVARfor the BLUEBIRD
conducter.
Now we choose the OSTRICH conductor and calculate the the parameters.
Firstwe consider OSTRICH conductor.
I will also assumethe GMR for thr circuit i.e. GMR=√(𝑡 𝑡 ∗
6")=3.33”/Φ=0.2775’/Φ
0.788*radius of theconductor
GMD=√ 𝑡3
∗ 𝑡 ∗ 2𝑡

85 | Page
=√11
3
∗ 11 ∗ 22
=13.86’
L=0.7411 log
𝑡𝑡𝑡
𝑡𝑡𝑡
=1.25mH/Miles/Φ
C=
0.0388
𝑡𝑡𝑡
𝑡𝑡𝑡
𝑡𝑡𝑡
=0.0228uF/Miles/Φ
Frequency we assumeto be=60 H
Hence,
XL=2πfL=2*π*60*1.25=0.471Ω/miles/Φ
XC=
1
2𝑡𝑡𝑡
=0.116MΩ-miles/Φ
Z=R+jXL;
Z=0.3070+j0.471…………………………………………………….(RfromtheA3 Table for a
OSTRICH conductor)
Y=B+j/XC;
Y=0+j/0.116=8.6 uʊ/mile…………………………………………………………Assuming B=0
Now we repeat the same procedureas before and find the total loss in the
system.

86 | Page
We can see that the losseare 0.4 MW and 0.023 MVARfor the OSTRICH
conducter.
DISCUSSION:
We can say fromour calculations and assumption that for the same line between
the buses SANDER69 and DEMAR69 for the OSTRICH conductor wehavehigher
losses due to are assumption thses losses arenot high enough but we still
consider them with the error that the BUEBIRD conductor is better suited for the
given line.

87 | Page
PROBLEM#08 TransmissionLine Evaluation
Intent of the problem: This problem calculates the parameters of a short, medium
and long transmission line and compare the result to see whether their models are
compatible with a different transmission line for e.g. short can be used on a long
line.
CASE: Power World Design Case 9
Approach:
Step 1: First we are going to assume the appropriate values needed or our
calculation
Step 2: Then we are going to calculate manually the different parameters for the
short, medium and long transmission lines.
Step 3: Then we are going to insert these values into the power world case we are
assuming and simulate it
Step 4: Then we are going to compare the results and for accurate answers.
SOLUTION:
Firstwe are going to assumea case for our case study
FIG 8.1

88 | Page
We are assuming a 300 mile line with receiving end power to be 50 MW with a
voltage of 220 kV and a power factor of 0.9 and the impedance of the line is
40+j125 Ω and shuntadmittance of 10-3
ʊ
Z=40+j125 Ω=131.2/72.3o
; Y=10-3
ʊ=10-3
/90o
IR=
50
√3∗220∗0.9
=0.1664/-36.7o
kA;
VR=
220
√3
=127/0o
(a) Shorttransmission line approaximation:
VS=131.2/72.3o
*0.1664/-36.7o
+127
VS=145/4.9o
VS=251.2kV
IS=IR=0.1664/20okA;
Sending end power factor=36.7+4.9=41.6o
=0.746 lagging
Sending end power =√3 ∗ 251 ∗ 0.164 ∗ 0.746=52.16.
(c) Exact transmission lineapproximation

89 | Page
ϒL=√ZY=√131.2 /72.3O
*10-3
/ 90O
ϒL=0.362/ 73O
ϒL=0.0554+j0.3577
Cosh(αl+βl)=
1
2
(eαl
/βl+e-αl
/-βl)
/βl=0.3577 radians=/20.49o
e0.53
/ 20.49o
=1.057/ 20.49o
=1.03+j0.46
e-0.53
/ 20.49o
=0.946/-24.06=0.886-j0.331
Hence,
cosh ϒL=0.91+j0.06=0.91/1.2o
Zc=√
𝑡
𝑡
=362.2/-8.85o
& C=sinh ϒL/Zc=128.2/72.65o
Now,
● VS=AVR+BIR
● IS=CVR+DIR
Vs=0.91/1.2 *127/0+128.2/72.65*0.164/-37.70
=182/3.77+0.011/56.72
=181.61+j11.97
=136.97/6.2o
KV=237.23kV
IS=9.77*10-4
/90.4*127/0+0.91/1.2o
*0.164/-20
=1050/91+0.3/3.77
=-18+j1049
=0.1286/15.23kA

90 | Page
Sending end power factor=15.3-6.2=9.1o
=0.987 lagging
Sending end power =√3 ∗ 237.23 ∗ 0.164 ∗ 0.980=52.15 MW.
Now that we have calculated the value of the parameters manually we will use
power world to do so as well and compre the values.
FIG 8.2
Fromthe figurewe can see that the we have set the value according to our
assumptions

91 | Page
Z=40+j125 Ω=131.2/72.3o
; Y=10-3
ʊ=10-3
/90o
We have to convertit to Ω/miles s we divide the impedance and admittance
values by the length of the line i.e. 300 miles.
Hence we get Z=0.11+j0.41 Ω/miles; Y=10-3
ʊ=0.3 ʊ/miles
Fron the fig wecan see that for our assumptions thereis a diffrencein the current
value. Now we will see the receiving end voltage angle to compare.

93 | Page
FIG 8.5
We can also see that the MW value for the systemis different fromthe value we
calculated i.e. 58 MW for simulation and 52 MW for calculated value.
Now we insert the the values of medium and long transmission line and observe
the changes in the circuit
FIG 8.6

94 | Page
We can see that fromassumption and the values we have inserted that the
currentvalue we calculated and the current value wesimulated arevery close
and the our assumptions arerightbut wealso see that the power in MW i.e. 57
MW is quite away fromour calculated values due to the discrepensies in the
circuit. Hence for a long line our sjortline assumptions produces errors and hence
it cannot be applied for a long transmission lineassumptions buta medium
transmission line assumptions havethe same currentvalue at the sending end for
our assumptions.so wecan usemedium transmission lineassumption for a long
line.

95 | Page
DISCUSSION:
We have calculated the transmission line parameter for assumptions wehave
made manually and then we have simulated the results with the help power
simulator through which we haveconcluded that for a long transmission line we
can apply medium line mdel but we cannot apply the shortline model because of
its discrepensies in the circuit.
PROBLEM#09 Load Angle EstimationFor a Power Line

96 | Page
Intent of the problem: This problem we calculate the load angle for system and
introduces a concept called equal area criterion to do so.
What is EQUAL AREA CRITERION?
Firstwe will consider a Power Transfer Equation Single Machine-InfiniteBus
System. For a simple lossless transmission lineconnecting a generator and infinite
bus as shown in Figure
Figure9.1: One machine against infinite bus diagram.
If V1 = U1, V2 = U2cos δ + j U2 sin δ,
Z = R + j X;
Itis well known that the active power P transferred between two generators for
a lossless line can be expressed as:
P= | U2 | | U1| *sinδ
X
Where,
V1 is the voltage of the infinite bus (reference voltage), volt
V2 is the voltage of the generator bus δ is the angle difference between the
generator and infinite bus, rad
X is the total reactance of the transmission lineand generator, ohm
The impedance is reduced to the reactance of the line becausethe resistanceis
often small and gives little contribution to the
The Power Angle Curve

97 | Page
FIG 9.2
The generator in Figure is in stable operation at a phaseangle of δ compared to
the infinite bus, i.e. the voltage at the generator bus U2 is leading the voltage at
the infinite bus U1 by an angle δ. The mechanical power input Pm and the
electrical power output Pe drawn in Figure describes the power balance of the
generator. The curves intersectat two points, the stable equilibrium point and the
unstable equilibrium point.
Approach:
1. We will consider a casefor the of a generator connected to an infinite bus
via interconnected two line system. Then we will find the maximum delta
angle for the systemin a fault condition
2. We are going to make necessary assumption to provethe same. This
maximum delta angle will dtermine the stability of the system.
SOLUTION:

98 | Page
● For a systemE=1.2 pu.;V=1 pu;X’d=0.2 pu,X1=X2=0.4 pu. Initialpower
supply of the systemis 1.5 pu.
Determine Max delta angle for a fault condition on one of the two
interconnnected lines.
FIG 9.3
For the power angle curve;
FIG 9.4
For the curveA i.e. when oth the lines are in operation,
Transfer reactances X=X’d+
𝑡1
2

99 | Page
=0.4 pu.
Steady state power limit, PMAX=
𝑡𝑡
𝑡
=3 pu.
For initial condition Po=1.5 pu.
Initial load ange=𝑡𝑡𝑡−1 𝑡0
𝑡 𝑡𝑡𝑡
=30o
For curveB i.e. when one of the line trips due to fault conditions
Transfer reactances X=X’d+X1(or X2)
=0.6 pu.
Steady state power limit, PMAX=
𝑡𝑡
𝑡
=2 pu.
Electrical power Developed P’E=P’max=2𝑡𝑡𝑡 𝑡
Initial load ange=𝑡𝑡𝑡−1 𝑡 𝑡
𝑡′ 𝑡𝑡𝑡
=48.6o
And load angle δm=180-48.6=131.4
The area A is given as;
A=∫
𝑡 𝑡
𝑡 𝑡
( 𝑡 𝑡 − 𝑡
′
𝑡) 𝑡𝑡=∫
0.848
0.524
(1.5 − 2𝑡𝑡𝑡 𝑡) 𝑡𝑡=0.0769
The systemstability depends on whether there is enough negative area B in the
interval δc<δ<δm to match the area A
B=∫
𝑡 𝑡
𝑡 𝑡
( 𝑡 𝑡 − 𝑡
′
𝑡) 𝑡𝑡=∫
2.293
0.848
(1.5 − 2𝑡𝑡𝑡 𝑡 ) 𝑡𝑡=-0.4777
Since B is greater in magnitude then in A,the systemis stable
The maximum angle Delta for the systemis given by the conditon A=B
∫
𝑡 𝑡
𝑡 𝑡
( 𝑡 𝑡 − 𝑡
′
𝑡) 𝑡𝑡==0.0769
or∫
𝑡 𝑡
0.848
(1.5 − 2𝑡𝑡𝑡 𝑡 ) 𝑡𝑡 = −0.0769
or 1.5 − 2𝑡𝑡𝑡 𝑡=-0.0769+1.5*2cos0.848=2.158
The equation is non-linear and by solving this equation weget;

100 | Page
ΔM=69.8o
Discussion:
We conclude that for a stable system we have managed to find the Maximum
delta angle for a system using our own assumptions. And we know that fo stable
system the maximum delta value is 90o
but for a fault condition an delta
maximum value can never be 90o
.
PROBLEM#10 SHORT CIRCUIT DUTY
Intent of the problem: This problem calculates the Short Circuit Duty (SCD) of
generator connected in SERIES with a transformer. It also uses the concept of fault
analysis shown in previous problems. This SCD can be extended to be used in fault
analysis when calculating fault current from and for different areas/zones.

101 | Page
What is SHORT CIRCUIT DUTY?
Short circuit duty is a description of the severity of overcurrent that a device can
reasonably be expected to withstand transformer windings must be stout enough to
absorb the heat and withstand the magnetic forces during such an event, circuit
breaker contacts must be robust enough to not melt, and in big ones there must be
nonconductive vanes surrounding the contacts to break up the arc cut open a big
round industrial fuse and you'll probably find it stuffed with sand to quench the arc.
CASE: Power World Design Case 6
Approach:
Step 1: First we are going to divide the area we are considering for the given case
to be divided into zone 1 and zone 2. Given the MVA rating of the generators and
KV rating of the generators and transformer, we are able to find out the zone
currents and zone impedances. Assuming the rated MVA value, we will find out
the per unit equivalent reactance on base of 100 MVA in terms of individual
reactance of the generators and transformer. The equivalent reactance and the
system reactance we will add to the system and the fault current will give us the
short circuit duty of generator.
Step 2: Using power world simulator, the fault current at bus SANDER69 can be
calculated.
Step 3: Another generator of the same rating will be added to the bus and 3-phase
fault current will be calculated using power world. Since we assume the values of
reactance of all the generators to be same, the fault current can be verified by
manual calculation.
Now we select a power world design case 6

102 | Page
Design power world case_6 _FIG 10.1
Firstwe have to find the generators and transformer value. We can find this from
the power world example.
➢ To effectively calculate all these values for the generators, transformers
and the transmission lines, I need to obtain these values using Power World
simulator. The data for the generator can be obtained using the Generator

103 | Page
Information Dialog. The data for the transmission lines and the transformer
can be obtained using the Branch Information Dialog. Thus I obtained the
reactance and MVA value for all the generators, transformers and
transmission lines using the information dialogues.
FIG 10.3

104 | Page
FIG. 10.4
Fromthe two figures FIG 10. 2. And FIG 10. 3. To effectively calculate all these
values for the generators, transformers and thetransmission lines, I need to
obtain the MVA, KV and reactance values using Power World simulator. The data
for the generator can be obtained using the Generator Information Dialog. The
data for the transmission lines and the transformer can be obtained using the
Branch Information Dialog. Thus I obtained the reactance and MVA value for all
the generators, transformersand transmission lines using the information
dialogues.
Hence I have obtain the data for all the elements.

105 | Page
Generator G1 and G2 – 100 MVA and 69 KV and 138 KV
Transformer T- 187 MVA and 69KV/138KV
The one line diagramfor the circuit we are considering is given as
BASEMVA =100 MVA BASEKV =69 KV GENERATOR 2 SCD=100 MVA
T
ZONE 1 ZONE 2
BASEMVA =100 MVA BASE
MVA =100 MVA
VB1= 69 KV VB2= 69*(
138
69
)
IB1=
𝑡𝑡𝑡
√3∗𝑡𝑡
=
100000
√3∗69
= 837 𝑡 IB2=
𝑡𝑡𝑡
√3∗𝑡𝑡
=
100000
√3∗138
= 419 𝑡
ZB1=
𝑡𝑡2
𝑡𝑡𝑡
=
692
100
= 48 Ω ZB2 =
1382
100
= 191 Ω
XT, NEW= 0.05*(
69
69
)2
*(
100
187
)
XT, NEW=0.02 p.u
XG2, NEW= 0.05*(
138
138
)2
*(
100
100
)

106 | Page
XG2, NEW=0.05 p.u
Now using power world simulator we calculate the fault at the bus SANDER69.
We must firstgo to the run mode then select fault analysis then choosesingle
fault and then bus fault fromfault location then we can choosethe bus wherewe
want to calculate the fault then in our casewhich is SANDER69 then we choose
fault type i.e. phase fault in our case. Then we click on calculate fromthe top left
then we get the fault at the given bus in p.u and in amps
FIG .10.6

107 | Page
FIG. 10.7
The bus currents are-
IP.U=47 p.u
IAMP=391 KA
Ibus=ibus*Ib1;
FIG 10.8

108 | Page
391KA=ibus*837A;
ibus=391/837;
ibus=468 p.u
ibus=
𝑡 𝑡
𝑡 𝑡 𝑡𝑡𝑡𝑡𝑡
=
1
𝑡 𝑡2+𝑡 𝑡𝑡𝑡
=
1
0.1+𝑡 𝑡𝑡𝑡
;
(0.1+Xsys)*(468) =1;
46.8+468Xsys=1;
Xsys=0.097⁄90p.u;
Xsys*ZB1=0.097*0.09=4.8Ω;
Xsys=KV/√3/Isc;
Isc=KV/√3/Xsys=8299A;
MVA=√3V*I=√3*69*8.2=97MVA.
Hence the SCD for the generator can be found out.
If weadd a generator to the given bus SANDER69 wecan calculate and analyze
the circuit and its effect on the Fault currentand SCD.

109 | Page
FIG. 10.9
FIG.10.10

110 | Page
We can clearly see the risein the fault current.
Ip.u=54 p.u;
Imagnitude=440KA
DISCUSSION:
Hence the solution demonstrates the following:
1. The fault currentdue to a new generator increases. This means that the
generator will add to the fault currentand the added current depends on the
MVA rating of the generator as well as the reactance. Itcan be seen that the fault
currentis directly proportionalto the SCD and the p.u. reactance of the
generator. Hence the systemreactance can be modeled for the generator fault
duty for a specific value of SCD or shortcircuit current.
2. Short circuit duty is an important parameter for modeling the shortcircuit
currentthat a generator can kick-off at a given voltage. Oncewe determine SCD,
we can calculate the shortcircuit current at any voltage- base, rated or operating.
Itgives a tie link between voltage, shortcircuit currentand SCD. Ittells us the
maximum currentthe generator can produce in case of a fault, hence helping in
determining the ratings of fuseand circuit breakers.
3. Practical application of SCD is when power systemengineers need to model
large and heavily interconnected systems. Itis usefulwhen the network is divided
into certain areas with separatecharacteristics, but since they are
interconnected, they cannot be ignored in analysis problems. This is wherethe
ShortCircuit Duty comes into play. When the SCD for an area is known, it can be
easily modeled for large systems and complicated analysis doneby power
engineers.

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