problemsetee443.docx

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Akshay [email protected]

AbstractStudy of different parameters of electrical engineering using power world simulator

PROBLEM SETEE 443 PROF. CASTRO

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Table of Contents PG NO.

1. The study of Power flow analysis………………………………………………………… 2 2. Fault analysis………………………………………………………………………………………. 193. Unsymmetrical fault in a circuit…………………………………………………………… 364. Power factor correction………………………………………………………………………. 545. Power circle diagram for a transmission line……………………………………….. 616. Phase shifting transformer…………………………………………………………………… 667. Choice of conductor for a transmission line…………………………………………. 818. Transmission line evaluation………………………………………………………………… 869. Load angle estimation for a power line…………………………………………………. 96

Short circuit duty…………………………………………………………………………………. 101

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10.PROBLEM #01 THE STUDY OF POWER FLOW ANALYSIS

In this problem we 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 TRANSMISSION LINE SR=PR + jQR

LOAD

GENERATING STATION

#A TWO-BUS POWER SYSTEM

A power flow study (load-flow study) is a steady-state analysis whose target is to determine the voltages, currents, and real and reactive power flows in a system under a given load conditions. The purpose of power flow studies is to plan ahead and account for various hypothetical situations. For example, if a transmission line is be taken off line for maintenance, can the remaining lines in the system handle the required loads without exceeding their rated values. In the given diagram VR 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 analyze the flow of real and reactive power. I have decided to study the flow of power between buses NICOL69 and DAVIS69.

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G

2. Then, I am going to measure the 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.

Figure 1.1 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. One with base case and the other one with the load increased one of the buses.

CASE I

Let us consider the base case for the buses DAVIS69 and NICOL69.

DAVIS69

We can look at the Real and Reactive power levels on the bus DAVIS69 by 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. (Figure 1.3)

FIG 1.3

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Q= 12.72MVAR

As can be seen from the Quick Power Flow List for the bus, DAVIS69, the Real Power Flow at the bus is 38 MW and the reactive power flow is 12.72 MVAR.

P = 38 MW, Q = 12.72 MVAR

Figure 1.4 gives the power triangle for the bus, DAVIS69.

P= 38 MW

Figure 1.4

There is a lagging power factor at the bus.

Power factor = pf = cosθ

tanθ= 3812.72

This gives,

Θ1 = 71.44⁰ (lagging)

NICOL69

We can look at the Real and Reactive power levels on the bus NICOL69 by 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, NICOL69. (Figure 1.5)

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FIG 1.5

As can be seen from the Quick Power Flow List for the bus, NICOL69, the Real Power Flow at the bus is 28 MW and the reactive power flow is 6 MVAR.

P = 28 MW, Q = 6 MVAR

Figure 1.4 gives the power triangle for the bus, NICOL69.

Q= 6MVAR

P= 28 MW

Figure 1.4

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 must flow 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)*sin δ

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 sin δ

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 sin 6.46

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 represent the active power flowing in the system and he blue arrow indicate the reactive power flowing in the system.

From the FIG.1.6 it is evident that the real power flows from NICOL69 toDAVIS69 and the reactive power flows from DAVIS69 to NICOL69.To calculate the 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 figure 1.7, the total real power flow between the bus NICOL69 and DAVIS69 comes out to be 9.51 MW. Thus the total power flow is

PActual = 9.51 MW from the NICOL69 to DAVIS69 bus.

There is a loss of 19.5% in the calculation of the theorotical and actual power measurement since the theorotical power and the actual power have a diffrence of 2.24 MW due to the losses in the other branches.

CASE 2

In this case im adding a 30 MVAR capacitor to the system and we are 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 DAVIS69 by 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. (Figure 1.9)

FIG 1.9

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Q= 4.2 MVAR

P=38 MW

As can be seen from the Quick Power Flow List for the bus, DAVIS69, the Real 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.


tanθ=4.238

This gives,

Θ1 = 6.30⁰ (leading)

NICOL69

We can look at the Real and Reactive power levels on the bus NICOL69 by 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, NICOL69. (Figure 1.11)

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P=28 MW

Q=25.78 MVARP=28 MW

FIG 1.11

As can be seen from the Quick Power Flow List for the bus, NICOL69, the Real Power Flow at the bus is 28 MW and the reactive power flow is 25.78 MVAR as there is shunt capacitor.

P = 28 MW, Q = 31.78-6.00 MVAR

Q=25.72 MVAR

Figure 1.11 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 must flow 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)*sin δ

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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FIG 1.12

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P=VNICOL*VDAVIS sin δ

X

VNICOL=69*1.03 KV

VNICOL=71.07 KV

VDAVIS=69*1.02 KV

VDAVIS= 70.38 KV

P= 71.07*70.38 sin 53.46

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 figure 1.7, the total real power flow between the bus NICOL69 and DAVIS69 comes out to be 10.73 MW. Thus the total power flow is

PActual = 10.73 MW from the NICOL69 to DAVIS69 bus.

There is a loss of 80.5% in the calculation of the theorotical and actual power measurement since the 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 change in the load angle there is change in the real power for the system.

Hence from the 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:

From the above calculations 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 from higher to lower load angle bus and the reactive power flows from higher voltage to lower voltage level bus. We also were able to predict the amount of amount of real power that might flow from say bus A to bus B. From the 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 FAULT ANAlYSIS

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It is not practical to design and build electrical equipment or networks so as to completely eliminate the possibility of failure in service. It is therefore an everyday fact of life that different types of faults occur on electrical systems, however infrequently, and at random locations. 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 from one phase conductor to another (phase-to-phase) or alternatively from one 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 rockfall could 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 where t is time). Danger to operating personnel (Flash products). Danger of igniting combustible gas such as methane in hazardous areas giving rise to 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 transformers whose windings suffer progressive and cumulative deterioration because of the enormous electromechanical forces caused by multi-phase faults proportional to the current squared). Sustained voltage dips resulting in motor (and generator) instability leading to extensive shut-down at the plant concerned and possibly other nearby plants. The “incipient” fault, on the other hand, is a fault that starts from very small beginnings, from say some partial discharge (excessive electronic 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 adjacent insulation, eventually running away and developing into a “solid” fault.

➢ Passive Faults

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Passive faults are not real faults in the true sense of the word but are rather conditions that are stressing the system beyond 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 Phase System The types of faults that can occur on a three phase A.C. system are as follows:

Types of Faults on a Three Phase System.

(A) Phase-to-earth fault

(B) Phase-to-phase fault

(C) Phase-to-phase-to-earth fault

(D) Three phase fault

(E) Three phase-to-earth fault

(F) Phase-to-pilot fault *

(G) Pilot-to-earth fault *

* In underground mining applications only

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It will be noted that for a phase-to-phase fault, the currents will be high, because the fault current is only limited by the inherent (natural) series impedance of the power system up to the point of faulty (refer Ohms law). By design, this inherent series impedance in a power system is 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 system neutral 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. It is worth noting at this juncture that it is possible to 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 use of Resistance or Impedance in 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 & Permanent Faults Transient faults are faults which do not damage the insulation permanently and allow the circuit to be safely re-energised after a short period 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. Transient faults occur mainly on outdoor equipment where air is the main insulating medium. Permanent faults, as the name implies, are the result of permanent damage to the insulation. In this case, the equipment has to be repaired and reclosing must not be entertained.

● Symmetrical & Asymmetrical Faults A symmetrical fault is a balanced fault with the sinusoidal waves being equal about their axes, and represents a steady state condition. An asymmetrical fault displays a d.c. offset, transient in nature and decaying to the steady state of the symmetrical fault after a period of time:

Amongst these faults assosiated with power system system im gonna discuss in short about symmetricsl and asymmetricsl faults and discuss its effects with help of the power world simulator:

1. Symmetrical Faults

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2. Asymmetrical FaultsSymmetrical Faults

When there is a 120o equal phase shift between the conductor due to thid symmetry it is refered to a symmetricla fault.When there is short circuit condition on a 3 phase line such kind of fault occurs. This fault condtion rarely occurs most of the time asymmetricla fault condition occuring but symmetrical condition impose more damage 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 short circuit between lines, caused by ionization of air, or when lines come into physical contact, for example due to a broken insulator.

● line-to-ground - a short circuit between one line and ground, very often caused by physical contact, 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 storm damage.

The following is the curve representing the symmetrical and asymmetrical faults in a system.

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In this problem im going to show with power world how a symmetrical as well as a symmetrical fault might occour and affect the characteristic impedence of a power line. For this im going to assume a fault on the bus and line from SLACK345 to TIM345.

APPROACH:

1. First im going to find the sequence impedencevalue for a given line in our case it is the line between buses SLACK345 and TIM345 from the branch 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 Choose the single fault options

3. Then im going to choose the fault location and fault types and its effect on the line impeadance.

4. Then we are going to find the effects of the different types of fault using ETAP software as a tool.

First we are going to choose a case Design case_6.1

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Design case_6.1

FIG. 2.1

The branch between buses SLACK345 and TIM345 is the branch we are going to analyze for the effects of the different kinds of faults.

First double click on the branch SLACK345 and TIM345 and record the characteristic impedance of the branch

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FIG 2.2

From the FIG_xx

Z=R+jX=0.02+j0.03 pu;

Y=G+ jX =0+ j

0.18 pu;

Now we go to the tool ribbon and click on the fault analysis and choose the Fault location in our case we will choose the in-line fault now we choose the single line to ground fault and calculate the fault current value and the characteristic impedance value

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FIG 2.3

From the FIG 2.3

Z=R+jX=0.015+j0.022;

Y=G+ jX =0+ j

0.13 ;

If”=44 p.u.=7833A.

WE CAN ALSO SET THE LOATION OF THE FAULT ON THE LINE.

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FROM THE FIG 2.4 LOCTION OF THE FAULT FROM THE FAR BUS.

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 from 1% to 100%.

NOW WE CAN CAN CALCULATE THE CHANGE OF CHARACTERISTIC IMPEDANCE AND ITS AFFECT ON THE LONG TRANSMISSION LINE PARAMETER

WE CAN CALCULATE THE INITIAL LOSSLESS AND FAULTLESS PARAMETERS

FIRST WE KNOW THAT FOR A TRANSMISSION LINE THE SENDING END AND RECEIVING END VOLTAGES ARE GIVEN AS-

● VS=AVR+BIR● IS=CVR+DIR

THE EXACT MODEL GIVEN FOR THE LONG LINE IS GIVEN AS

● VS=AVR+BIR

● IS=CVR+DIR

BUT A=D=cosh γL & C=1

ZCsinh γL & B=ZCsinh γL

First we have to find the receiving end current

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P=√3VIcosθ

INITIALLY WE ASSUME THAT IT’S A LOSSLESS YSTEM HENCE WE ASSUME A UNITY POWER FACTOR

HENCE FROM THE FIG 2.4

FIG 2.5

TIM345 BUS THE POWER IS 122 MW AND REACTIVE POWER IS 18 MVAR

AND V=345 KV WE HAVE TO CONVERT IT INTO PHASE VOLTAGE V= 345√ 3 =200KV

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FIG 2.6

IR=122 MW

√❑=0.32 05 5̷ O

From the fig 2.6 we can see that the impedance and admittance for the line in ohms/mile and mhos/miles which we require for our calculation.

Z=0.2+J3.5=3. 866 6̷ O ohms/mile

Y= J15.28=0.06 905 5̷ O mhos/miles

ϒL=LENGTH OF THE LINE∈MILES √ZY=10∗√3.60 860 0̷ O*0.065/ 90O

ϒL=0.480 885 5̷ O

ϒL=0.013+j0.485

Cosh(αl+βl)= 12(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=√ ZY =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 system due to other interconnection in the system.

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FIG 2.7

FIG 2.8

You can see from the FIG 2.8 that receiving end of the transmission line receives 122 MW real power and 33 MVAR reactive power from the 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 hence we are going to assume that 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

From these two figures we can calculate the sending end voltage and sending end current for the system. And we know from FIG_xx the reactance,resistance and capacitance values.So calculating the following we get,

TIM345 BUS THE POWER IS 122 MW AND REACTIVE POWER IS 22 MVAR

AND V=345 KV WE HAVE TO CONVERT IT INTO PHASE VOLTAGE V= 345√ 3 =200KV

IR=122 MW

√❑=0.32 05 5̷ O

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FIG 2.9

Z=1.7815+J2.67=3.202 566 6̷ O pu.

Y= J11.45=0.08 905 5̷ O pu.

ϒL=LENGTH OF THE LINE √ZY=10∗√3.2026/ 56O*0.08 905 5̷ O

ϒL=0.52/ 72O

ϒL=0.16+j0.49


/βl=0.49 radians=/28.07o

e0.16/ 28.07o =1.17/ 28.07o=1.04+j0.55

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e-0.16/ -28.07o=0.85/-28.07=0.75-j0.39

Hence,

cosh ϒL=0.93+j0.04=0.93/2.46o


Zc=√ ZY =6.15/-17o


B=Zcsinh ϒL=0.019/55.72o & 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 due to other interconnection in the system and we cannot assume a power factor for large system to be one as a system cannot be lossless.And also the sending end current has been affected by the change in the characteristic impedance of the circuit.

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DISCUSSION:

● From the first part of the problem we simulated the value of the impedance and calculated the sending end voltage and current for a stedy state lossless medium i.e. the long transmission line.

● For the second part still assuming the line to be lossless we simulated a fault at 25% from the bus SLACK345 and calculated the impedance anf hence the values of the sending end voltage and current

● From the calculation and simulation we can clearly see that the characteristic impedance of the system has reduced hence the sending end current for the system has increased .

● Therefore for the given system the fault alters the characteristic values of impedance and therefore the circuit parameters.

● And hence for a system if the characteristic impedance is reduces the current in the circuit increases and vise-versa.

● In this way the characteristic impedance influences the current flowing in the circuit. We can also compare the characteristic impedances to the line impedance. It appears 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 phase of 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 symmetrical component analysis (e.g. for unbalanced faults), a balanced three-phase electrical network can be broken down into three sequence networks, which are independent, de-coupled sub-networks comprising only quantities in the same sequence, i.e. the positive sequence network contains only positive sequence quantities, the negative sequence network contains only negative sequence quantities and the zero sequence network contains only zero sequence quantities.

The phase faults are unique since they are balanced i.e. symmetrical in three phase, and can be calculated from the single phase positive sequence impedance diagram. Therefore three phase fault current is obtained by,

Where IF is the total three phase fault 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 Component Analysis

The above fault calculation is made on assumption of three phase balanced system. The calculation is made for one phase only as the current and voltage conditions are same in all three phases. When actual faults occur in electrical power system, such as phase to earth fault, phase to phase fault and double phase to earth fault, the system becomes unbalanced means, the conditions of voltages and currents in all phases are no longer symmetrical. Such faults are solved by symmetrical component analysis. Generally three phase vector diagram may be replaced by three sets of balanced vectors. One has opposite or negative phase rotation, second has positive phase rotation and last one is co-phasal. That

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means these vectors sets are described as negative, positive and zero sequence, respectively.

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 sequence networks, consider a simple power system shown in the FIG 3.1

Fig 3.1

1. The impedances are all constant and independent of currents.2. The synchronous generator is a salient pole type which may generate under

faiult condition, negative and zero-sequence emfs which are small and are negligible. Thus for all purposes the machine is assumed togenerate only positive sequence emfs.

The network under such conditions can be represented by 3 independent single phase sequence network.

Approach:

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1. I will first do the calculation by hand to find the fault current at 50% line using the selected circuit shown in Figure _xx and draw the sequence network for the circuit.

2. Using the Power World software, I will then find the fault current at 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 hypothesize possible reasons for such an error.

Figure

FIG 3.2

Solution:

First I need to select the base MVA value. I can do this by right clicking on the generator information dialog and obtaining the base MVA value. (Figure 3.2)

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FIG 3.3

It can 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 figure 3.4. (Figure 3.4)

BASE MVA =20 MVA BASE KV =16 KV

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ZONE 1

BASE MVA =20 MVA VB2= 16*( 13816

)=130V VB3= 138*( 16138

)=16 V

BASE KV =16 KV IB2= KVA√❑

=84A IB3= KVA√❑

=722A

IB1= KVA√❑

=722A ZB2= KV 2

MVA=1382

20=952Ω ZB3= KV 2

MVA=162

20=13Ω

ZB1= KV 2

MVA=162

20=13Ω

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, transformers and transmission lines using the information dialogues.

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FIG 3.5

XG, NEW= 0.1*¿*( 2018720

)=0.1 p.u.

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FIG 3.6

XT, NEW= 0.12*¿*( 2020

)=0.12p.u.

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FIG 3.7

We have the transmission line reactance as X=0.10 p.u. from the FIG 3.7 as we know the value in p.u. we don’t have to calculate the value manually.

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FIG 3.8

XT, NEW= 0.12*¿*( 2020

)=0.12 p .u .

Now we draw the sequence network for the system and evaluate the fault current manually.

WE ASSUME THE POSITIVE, NEGATIVE AND ZERO SEQUENCE REACTANCES FOR THE GENERATOR TO BE SAME X2=X1=X0=0.1 P.U.

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FIG 3.9

THE FAULT CURRENT FOR THE SYSTEM AT MIDPOINT OF THE TRANSMISSION LINE IS

IF=3∗V F

Z2+Z1+Z0= 3

0.10+0.10+0.155 =8.51 P.U.

AS THE FAULT CURRENT IS OCCURING IN ZONE 2,

IF=8.51*ZB2=8.51*84=720A

NOW WE EVALUATE FOR POWER WORLD ANALYSIS TO FIND THE FAULT CURRENT.

WE WILL TRY WITH ALL FOUR FAULT TYPES AND FIND THE P.U AS WELL AS AMPERE FAULT CURRENT FOR THE SYSTEM.FIRST SET THE LOCATION AT 50% OF THE TRANSMISSION LINE.

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FIG 3.10

NOW WE CAN ASSUME THE DIFFERENT TYPES OF FAULTS ONE BY ONE

1. SINGLE LINE TO GROUND FAULT

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FIG 3.11

AS YOU CAN SEE FROM THE FIG 3.11 THAT THIS NOT THE TYPE OF FAULT THAT IS OCCURING IN OUR GIVEN CIRCUIT.

2. 3-PHASE BALANCED FAULTI. IN PER UNIT

II. IN MAGNITUDE OR AMPERES

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3. LINE TO LINE FAULTI. IN PER UNIT

FIG 3.13

II. IN AMPERES

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FIG 3.14

4. DOUBLE LINE TO GROUND FAULT

FIG 3.15

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FROM THE POWER WORLD ANALYSIS WE SEE THAT THERE IS A ERROR OF 50% IN THE SYSTEM THAT DUE TO OUR ASSUMPTION AND POSSIBLE ERROR IN THE CIRCUIT ITSELF

FROM THE ANALYSIS IN THE POWER WORLD WE ALSO CAN CLERLY SEE THAT SINGLE LINE TO GROUND AND DOUBLE LINE T GROUND DOESN’T OCCUR IN OUR CIRCUIT ONLY THE 3-PHASE BALANCED AND LILE TO LINE FAULT CAN OCCUR IN OUR CIRCUIT

DISCUSSION

● We have obtained a substantial error in our calculation and simulated results due to our assumptions and possible small error in the circuit. Thus Power World has emerged as an effective tool in helping engineers to simulate complicated models and get accurate results. We were also successful in determining that the 3 phase fault current is the highest. Thus all protection schemes are designed with this 3 phase fault currents in mind.

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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?

First of all what is power factor and how do we measure it.

➢ In an AC circuit, power is used most efficiently when the current is aligned with the voltage.Efficient AC Current

➢ However, most equipment tend to draw current with a delay, misaligning it with the voltage. What this means is more current is being drawn to deliver the necessary amount of power to run the equipment. And the more an equipment draws current with a delay, the less efficient the equipment is.Inefficient AC Current

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➢ 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 Apparent Power:

➢ Real power (kW) — the work-producing power that is used to actually run the equipment

➢ Reactive power (kVAr) — the non-work producing power that is required to magnetize and start up equipment

➢ Apparent power (kVA) — the combination of real power and reactive power

The purpose of power factor improvement is simply to reduce the load current drawn from the supply. This allows conductors of smaller cross-sectional area to be utilised, reducing the amount (and cost) of copper used in those conductors and other supply plant. In the case of larger commercial and industrial loads, power factor is part of the tariff, and loads with low power factors may be financially penalised, so higher power factors are desirable 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➢ Phase Advancer

In this problem we are going to show the effect of a power factor improvement on a system using power world analysis.

Approach:

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Step 1: First we 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 shunt capacitor to the circuit and calculate the power factor of the circuit for the same buses and note down its effect on the circuit parameters

Step 3: We are going to compare the results and using power world we are going to simulate the results for the design case.

Solution:

First right click on the bus DAVIS69 and then choose the 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


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The power factor at DAVIS69 bus.


tanθ =50/75

This gives,

Θ = 34⁰ (leading)

Pf=cos Θ=0.82

Now we to improve the power factor to 0.95 so we are going to calculate the shunt capacitor required for the system

Cos Θ=0.95

Θ=cos−1Θ=18.19o

P = 75 MW, Q = X MVAR


FIG 1.11

Assuming the real power to be constant the MVAR required for the system is

tan18.19 =X/75

This gives,

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X=24.64 MVAR

Therefore the new shunt reactance to be added should be

XNEW=25 MVAR

Adding this new shunt reactance 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

Figure 1.4 gives the power triangle for the bus, SANDER69.

Q= 33MVAR

P= 75 MW

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The power factor at SANDER69 bus.


tanθ =33/75

This gives,


Pf=cos Θ=0.92

Now we to improve the power factor to 0.92 so we have an error in the calculation due to the presence of other interconnection to the buses we didn’t consider.

Now we connect an load to the system and see the changes in the system

FIG 4.3

P = 75 MW, Q = 43 MVAR


Q= 43MVAR

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P= 75 MW



tanθ =43/75

This gives,


Pf=cos Θ=0.86

Due to the additional load to the system there is a MVAR addition to the system and the power factor decreases to improve it to 95% i.e. 0.95 we increase the shunt reactance value the value required we first are going to calculate manually

Cos Θ=0.95

Θ=cos−1Θ=18.19o

P = 75 MW, Q = X MVAR


FIG 1.11

Assuming the real power to be constant the MVAR required for the system is

tan18.19 =X/75

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This gives,

X=24.64 MVAR

Therefore the new shunt reactance to be added should be

XNEW=25 MVAR

Adding this new shunt reactance or increasing the value of the existing shunt we get new MVAR value for the system therefore considering 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

Now we to improve the power factor to 0.97 i.e. 97% so we have an error in the calculation due to the presence of other interconnection to the buses that we didn’t consider.

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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 more accurate 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.

PROBLEM#05 Power Circle Diagram for Transmission line

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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. First we 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 diagram using these values as refrences and other assumptions if necessary.

ASSUMPTIONS

For these problem we are going to assume a 300 km line having R= 0.08 Ωkm ; X=0.4

Ωkm and Y=5.15*10-6 S

km . From the diagram we will find the sending end voltage,

current and 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.

Impedance of 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+ ZY2

=1+0.5*122.38 /78.69o*1.545*10-3/90o

=1+0.094/168.69=0.9073 +j 0.01854=0.9075/1.17

B=Z=122.38/78.69 Ω

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Load power factor angle, ΦR=cos−1 0.8=¿/36.9o

Load current, IR= Load∈MW∗106

√❑= 192∗106

√❑=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

Horizontal coordinate=- AB VRLcos (β−α )=-0.9075

122.38cos (78.69−1.17)*2752=-121 MW

Vertical coordinate=- AB VRLsin( β−α )=-0.9075

122.38cos (78.69−1.17)*2752=-548 MW

Take scale 1 cm= 100 MW horizontally and 1 cm=100 MVAR vertically.

FIG 5.1

Locate point N having coordinates-1.21 cm and -5.48 cm and draw load line OP at an angle cos−1 0.8 i.e. 36.9o inclined to the horizontal and to represent √❑

❑ MVAR

i.e. √❑❑ cm=2.4 cm

Draw circle diagram with N (-1.21 cm, - 5.48 cm) as center and NP as radius.

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From circle diagram NP= 7.6 cm………..(from Measurement)

∴ NP= V SL∗V RL

B=100*7.6=760 MVA.

∴ VSL= 760∗B

V RL=760∗122.38

275 =338 kV

Β-α=64o

Δ=β-64o=78.69o-64=14.69o

For sending end power circle diagram

Horizontal coordinate=- DB VSLcos (β−δ)=-0.9075

122.38cos (78.69−1.17)*3382=183 MW

Vertical coordinate=- DB VSLsin( β−δ)=-0.9075

122.38sin(78.69−1.17)*3382=827 MW

Radius of circle=V SL∗V RL

B =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 horizontal cutting the arc of sending-end power circle at P. Join OP

Sending-end power factor angle=ΦS=15o (from measurement)

Sending-end power factor=cos∅ s=cos15=0.966.

OP=2.4 cm = 2.4*100=240 MVA

∴OP=√❑❑ MVAR

∴IS=OP∗1000√❑

=410 A

Discussion

● Hence we have drawn the circle diagram for the sending end and receiving end power and we can see that a circle diagram gives the results which are sufficient accurate for practical purpose, despite the fact that an

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approximate equivalent circuit is used in a circle diagram and provides a panoramic view of how operating characteristics are affected by changes in the machines parameters, voltage, frequency etc.

● We also can see that for a long transmission line we have assumed a π-configuration

● And this can be done because it generates very low discrepancies in the solution but a short transmission line solution might fail in this case and hence cannot be assumed for the case.

PROBLEM#06 Phase shifting transformer

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.

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The basic function of a Phase-Shifting Transformer is to change the effective phase displacement 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.First we are 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 are going to add a phase shifting transformer to the same transmission line.

3. We are going to control the phase between the transmission line with the help of the PST and thereby evaluate the power flow.

SOLUTION:

First we are going to consider the case 3.60 for the question we evaluating,

FIG 6.1

In the given FIGURE 6.1 we are assuming a two bus system for evaluation, the values of the generator parameters are:

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FIG 6.2

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 control the value of the generator load.

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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.

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FIG 6.4

We can see from the simulator for that for the output of 500 W and 100 MVAR the input power being generated is 534 MW and 236 MVAR from FIG 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

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Vreceving end=297/-14okV

The phase angle at the receiving end is -14o, now we are going to add a phase shifting transformer to the circuit and set the phase value for the transformer to 0o.

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Now we can see from the FIG_xx that 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 phase shift of the load bus.

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We are going to set the limit for thr phase shift we are applying to the circuit we are considering.

FIG 6.8

For the circuit we have given a max and min phase angle value to +/- 30o for the phase shifting transformer. Now we are going to phase shift a transformer by 10o and evalute the change in the power transfer value.

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FIG 6.9

From the figure you can see that the bulk of the power is flowing from the phase shifting transformer and hence by changing the output power phase we have managed to reduce the power flow from the transmission line.

Now we change the phase from 10o to 20 o and take a look at the change,

FIG 6.10

As the phase is changed to 20o the power flow from the transmission line is further reduced and the power transmitted from the transformer is further

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enhanced. We can also see that the transformer produces power more than what is required at the load and hence the direction of this residual power is changed and now it flows from the load to the generator. You can see the change in the phase angle at the receiving end too.

FIG 6.11

Vreceving end=326/2okV

You can see that there is positive phase angle at the receiving end and now its leads the sending end power line voltage.

Now we change the phase to the maximum value of the phase shifting ttansformer i.e.

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FIG 6.12

From the figure you can see that the a significant amount of the power is being transmitted from the load to the generator. Hence with the use of the phase shifting transformer we have managed to change the flow of power through a circuit.

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Vreceving end=319/6okV

You can see that there is increased value of positive phase angle at the receiving end and now its leads the sending end power line voltage.

Now we make a negative phase change and evalute the circuit, now that the phase is reversed

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FIG 6.14

Since the phase is reversed the excess power of the load flows from the transformer instead of the power line and reactive power is reversed to and it flows from the power line as the MVAR at the sendinf end and receiving end have little diffrence the magnitude is small and there is no significant reactive power flowing in the power line. Now we reverse the phase further and evalute the circuit.

Now we reverse the phase to the minimum limit of the transformer and evaluate the change in the circuit

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Since the phase is reversed further the a significant excess power of the load flows from the transformer instead of the power line and the reactive power flows from bus 2 to bus 1 but as there is still no significant diffrence in the sending end and receiving end voltage angle.

FIG 6.15

Vreceving end=319/-20okV

Hence we conclude that through a use of a phase shift transformer we managed to control the power flow in the circuit either through the transformer or the power line.

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DISCUSSION:

● A PST is a useful means 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

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FIG 6.18

● But a draw back of a PST is the reactive power losses which are significantly high for positive as well as negative phase changes. So its high power applictaions are limited. Or they require really expensive compensators.

PROBLEM#07 Choice of Transmission Line Conductor

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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 cost of the conductors for a long transmission line to be laid this might help reduce the infrastructure cost and also minimization of losses as one of the important factors as different conductors have 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 assume that the conductors are arranged in an couple fashion with a distance of 15 feet between each of the conductors.

First we consider BLUEBIRD conductor.

I will also assume the GMR for thr circuit i.e. GMR=√ ¿=3.33”/Φ=0.2775’/Φ

0.788*radius of the conductor

GMD=3√d∗d∗2d

=3√11∗11∗22

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=13.86’

L=0.7411 logGMDGMR =1.25mH/Miles/Φ

C=0.0388

log GMDGMR

=0.0228uF/Miles/Φ

Frequency we assume to be=60 H

Hence,

XL=2πfL=2*π*60*1.25=0.471Ω/miles/Φ

XC=1

2 πfC =0.116MΩ-miles/Φ

Z=R+jXL;

Z=0.0476+j0.471…………………………………………………….(R from the A3 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 Case and asssume any line for the case study.We choose the 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.

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FIG 7.2

We can see that on the right side we have the 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.

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FIG 7.3

We can see that the losse are 0.385 MW and 0.025 MVAR for the BLUEBIRD conducter.

Now we choose the OSTRICH conductor and calculate the the parameters.

First we consider OSTRICH conductor.

I will also assume the GMR for thr circuit i.e. GMR=√ ¿=3.33”/Φ=0.2775’/Φ

0.788*radius of the conductor

GMD=3√d∗d∗2d

=3√11∗11∗22

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=13.86’

L=0.7411 logGMDGMR =1.25mH/Miles/Φ

C=0.0388

log GMDGMR

=0.0228uF/Miles/Φ

Frequency we assume to be=60 H

Hence,

XL=2πfL=2*π*60*1.25=0.471Ω/miles/Φ

XC=1

2 πfC =0.116MΩ-miles/Φ

Z=R+jXL;

Z=0.3070+j0.471…………………………………………………….(R from the A3 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 procedure as before and find the total loss in the system.

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We can see that the losse are 0.4 MW and 0.023 MVAR for the OSTRICH conducter.

DISCUSSION:

We can say from our calculations and assumption that for the same line between the buses SANDER69 and DEMAR69 for the OSTRICH conductor we have higher losses due to are assumption thses losses are not high enough but we still consider them with the error that the BUEBIRD conductor is better suited for the given line.

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PROBLEM#08 Transmission Line 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:

First we are going to assume a case for our case study

FIG 8.1

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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 shunt admittance of 10-3ʊ

Z=40+j125 Ω=131.2/72.3o; Y=10-3ʊ=10-3/90o

IR=50√❑

=0.1664/-36.7okA;

VR=220√ 3 =127/0o

(a) Short transmission 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 =√❑=52.16.

(c) Exact transmission line approximation

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ϒL=√ZY=√131.2 /72.3O*10-3/ 90O

ϒL=0.362/ 73O

ϒL=0.0554+j0.3577


/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=√ ZY =362.2/-8.85o


B=Zcsinh ϒL=0.0336/56.72o & 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

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Sending end power factor=15.3-6.2=9.1o=0.987 lagging

Sending end power =√❑=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

From the figure we can see that the we have set the value according to our assumptions

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Z=40+j125 Ω=131.2/72.3o; Y=10-3ʊ=10-3/90o

We have to convert it 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=10-3ʊ=0.3 ʊ/miles

Fron the fig we can see that for our assumptions there is a diffrence in the current value. Now we will see the receiving end voltage angle to compare.

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FIG 8.4

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FIG 8.5

We can also see that the MW value for the system is different from the 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

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We can see that from assumption and the values we have inserted that the current value we calculated and the current value we simulated are very close and the our assumptions are right but we also see that the power in MW i.e. 57 MW is quite away from our calculated values due to the discrepensies in the circuit. Hence for a long line our sjort line assumptions produces errors and hence it cannot be applied for a long transmission line assumptions but a medium transmission line assumptions have the same current value at the sending end for our assumptions.so we can use medium transmission line assumption for a long line.

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DISCUSSION:

We have calculated the transmission line parameter for assumptions we have made manually and then we have simulated the results with the help power simulator through which we have concluded that for a long transmission line we can apply medium line mdel but we cannot apply the short line model because of its discrepensies in the circuit.

PROBLEM#09 Load Angle Estimation For a Power Line

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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?First we will consider a Power Transfer Equation Single Machine-Infinite Bus System. For a simple lossless transmission line connecting a generator and infinite bus as shown in Figure

Figure 9.1: One machine against infinite bus diagram.

If V1 = U1, V2 = U2cos δ + j U2 sin δ,

Z = R + j X;

It is 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 line and generator, ohm

The impedance is reduced to the reactance of the line because the resistance is often small and gives little contribution to the

The Power Angle Curve

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FIG 9.2

The generator in Figure is in stable operation at a phase angle 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 intersect at two points, the stable equilibrium point and the unstable equilibrium point.

Approach:

1. We will consider a case for the of a generator connected to an infinite bus via interconnected two line system. Then we will find the maximum delta angle for the system in a fault condition

2. We are going to make necessary assumption to prove the same. This maximum delta angle will dtermine the stability of the system.

SOLUTION:

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● For a system E=1.2 pu.;V=1 pu;X’d=0.2 pu,X1=X2=0.4 pu. Initial power supply of the system is 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

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For the curve A i.e. when oth the lines are in operation,

Transfer reactances X=X’d+X1

2

=0.4 pu.

Steady state power limit, PMAX=EVX =3 pu.

For initial condition Po=1.5 pu.

Initial load ange=sin−1 P0

Pmax=30o

For curve B 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=EVX =2 pu.

Electrical power Developed P’E=P’max=2sin δ

Initial load ange=sin−1 PS

P 'max=48.6o

And load angle δm=180-48.6=131.4

The area A is given as;

A=∫δo

δc

(PS−P'E)dδ= ∫

0.524

0.848

(1.5−2sin δ )dδ=0.0769

The system stability depends on whether there is enough negative area B in the interval δc<δ<δm to match the area A

B=∫δc

δm

(PS−P'E)dδ= ∫

0.848

2.293

(1.5−2sin δ)dδ=-0.4777

Since B is greater in magnitude then in A,the system is stable

The maximum angle Delta for the system is given by the conditon A=B

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∫δC

δ M

(PS−P'E)dδ==0.0769

or ∫0.848

δM

(1.5−2 sin δ)dδ=−0.0769

or 1.5−2sin δ=-0.0769+1.5*2cos0.848=2.158

The equation is non-linear and by solving this equation we get;

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.

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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.

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

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Design power world case_6 _FIG 10.1

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First we 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 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.

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FIG 10.3

FIG. 10.4

From the two figures FIG 10. 2. And FIG 10. 3. To effectively calculate all these values for the generators, transformers and the transmission 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

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Branch Information Dialog. Thus I obtained the reactance and MVA value for all the generators, transformers and transmission lines using the information dialogues.

Hence I have obtain the data for all the elements.

Generator G1 and G2 – 100 MVA and 69 KV and 138 KV

Transformer T- 187 MVA and 69KV/138KV

The one line diagram for the circuit we are considering is given as

BASE MVA =100 MVA BASE KV =69 KV GENERATOR 2 SCD=100 MVA

T

ZONE 1 ZONE 2

BASE MVA =100 MVA BASE MVA =100 MVA

VB1= 69 KV VB2= 69*( 13869

)

IB1= KVA√❑

IB2= KVA√❑

ZB1= KV 2

MVA= 692

100=48Ω ZB2 ¿ 1382

100=191Ω

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XT, NEW= 0.05*¿*( 100187

)

XT, NEW=0.02 p.u

XG2, NEW= 0.05*¿*( 100100

)

XG2, NEW=0.05 p.u

Now using power world simulator we calculate the fault at the bus SANDER69. We must first go to the run mode then select fault analysis then choose single fault and then bus fault from fault location then we can choose the bus where we want to calculate the fault then in our case which is SANDER69 then we choose fault type i.e. phase fault in our case. Then we click on calculate from the top left then we get the fault at the given bus in p.u and in amps

FIG .10.6

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FIG. 10.7

The bus currents are-

IP.U=47 p.u

IAMP=391 KA

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FIG 10.8

Ibus=ibus*Ib1;

391KA=ibus*837A;

ibus=391/837;

ibus=468 p.u

ibus=Eg

ε X total=

1Xg 2+X sys

=1

0.1+X sys;

(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/√❑/Xsys=8299A;

MVA=√ 3V*I=√ 3*69*8.2=97MVA.

Hence the SCD for the generator can be found out.

If we add a generator to the given bus SANDER69 we can calculate and analyze the circuit and its effect on the Fault current and SCD.

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FIG. 10.9

FIG.10.10

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We can clearly see the rise in the fault current.

Ip.u=54 p.u;

Imagnitude=440KA

DISCUSSION:

Hence the solution demonstrates the following:

1. The fault current due to a new generator increases. This means that the generator will add to the fault current and the added current depends on the MVA rating of the generator as well as the reactance. It can be seen that the fault current is directly proportional to the SCD and the p.u. reactance of the generator. Hence the system reactance can be modeled for the generator fault duty for a specific value of SCD or short circuit current.

2. Short circuit duty is an important parameter for modeling the short circuit current that a generator can kick-off at a given voltage. Once we determine SCD, we can calculate the short circuit current at any voltage- base, rated or operating. It gives a tie link between voltage, short circuit current and SCD. It tells us the maximum current the generator can produce in case of a fault, hence helping in determining the ratings of fuse and circuit breakers.

3. Practical application of SCD is when power system engineers need to model large and heavily interconnected systems. It is useful when the network is divided into certain areas with separate characteristics, but since they are interconnected, they cannot be ignored in analysis problems. This is where the Short Circuit Duty comes into play. When the SCD for an area is known, it can be easily modeled for large systems and complicated analysis done by power engineers.

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