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EIEN15 Electric Power Systems Lab 1

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Electr ic Power Systems Laboratory Exercise 1

Voltage and Fault Currents in Power System Networks

Two different issues are studied in this laboratory exercise. First, the voltage profile overtransmission and distribution lines is investigated. The attention is focused on the dependency of

voltage on active and reactive power transmission as a function of the X/R ratio of a line. The

impact of distributed generation and reactive compensation on the voltage profile is alsoinvestigated. Second, single-line-to-ground faults in distribution networks are investigated.

Common grounding practices for distribution networks in Europe and some basic principles of

ground fault protection are considered.

Section I in this manual reviews the necessary theory to solve this lab exercise, Section II

includes several preparatory exercises, which must be made before you come to the lab. The

experimental part is described in Section III.

Section I: TheoryI.1. Active and Reactive Power Transfer on a Transmission Line

A simplified model of a transmission line is represented by a series impedance. For atransmission line, the series resistance is small compared to the series reactance and it can be

neglected. Consider the system in Figure 1, where an ideal generator supplies a load through a

line. The sending end is considered to be a stiff point in the network with a fixed voltage and

angle. The receiving end voltage and angle depend on the active and reactive power transmittedthrough the line. The active and reactive (Pand Q) power received at the load end can be written

as in (1) and (2).

Figure 1: Single line diagram of the examined network

sinX

EVPP LG (1)

X

V

X

EVQQ LG

2

cos (2)

By eliminating we obtain (3). Solving for V2yields (4):

22

22

X

EVP

X

VQ (3)

X

EQP

X

EQX

EV

22

2

422

42 (4)

Thus, the problem has real positive solutions if (and only if):

G

E

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2

422

4X

E

X

EQP (5)

The expression (5) can be normalized by using the short-circuit power Ssc= E2/X, and theinequality (6) is then obtained.

2

2

2

SCSC

SQSP (6)

By setting Q = 0in (6), we get the maximum active power that can be received at the load end:

2max

SCSP

In a similar way, we setP = 0in (6) and get the maximum active power, notation Pmax, that can

be received at the load end. We find that

4max

SCSQ

Figure 2 shows the relation between load voltage and load power for different reactive loadings.

We see that the reactive power strongly influences the line drop as well as the maximumcapability of the line. This representation of the PV characteristic is often called PV-curves.

These curves are characterized by a parabolic shape, which describes how a specific power can

be transmitted at two different voltage levels, high and low voltage. The desired working pointsare those at high voltage, corresponding to low current thus minimizing power transmission

losses voltages. The vertex of the parabola determines the maximum power that can be

transmitted by the system and it is often called the point of maximum loadability or point ofcollapse. If the point of collapse is reached, thermostats and tap changing controllers may cause

the voltage to start decreasing quickly to very low voltages, which is termed voltage collapse.

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4

LoadVoltage

(V)

Active Load (P)

tan p hi=0.50

tan phi=0.25

tan phi=0

tan phi=-0.25

tan phi=-0.50

Figure 2: PV-characteristic (nose curve), illustrates the dependency of receiving end voltage to load power for a transmissionline. tan phi=Q/P.

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I.2. Reactive Power Compensation

Consider the standard pi-link (e.g. Glover-Sarma-Overbye fig. 5.3) representation of a lossless

transmission line terminated by a resistive load (R). The reactive power losses absorbed and the

reactive power generated by the line are given by equations (7) and (8) respectively.

2LIQloss (7)2CVQgen (8)

Note that Equation 8 is exactly true only if the line has a flat voltage profile, i.e. the voltage is V

all along the line. This is true if generation and absorption of line reactive power are in balance:

C

L

I

VRCVLI 0

22 (9)

R=R0 is called the characteristic impedance, and the corresponding active power transportedP0=V

2/R0, is called natural loading or surge impedance loading (SIL). In a line operating at

natural loading there is no reactive power transport whatsoever, and this is therefore its most

economical operating point. In reality, power system lines are however frequently operated quitea bit above their natural loading, whereas cables (with their high capacitance) are operated below

their natural loading.

The purpose of reactive power compensation is to reduce the amount of reactive power

transported over the lines, and thereby also reducing active losses. There are several alternatives:

Series capacitance, lowers the line series reactance and increases the natural loading;

Shunt reactance, lowers the line shunt capacitance and lowers the natural loading; Shunt capacitance, increases the shunt line capacitance and increases the natural loading.

Series reactance is not used for line compensation. It is used instead to limit short-circuit currents

if they are excessive.

The approximate amount of shunt compensation needed to adjust the voltage at a bus is

calculated from (10), the sensitivity of the bus voltage to injection of (positive or negative)

reactive power: Describe the system behind the bus using a Thvenin equivalent as in Figure 1.

112

cos

X

V

X

E

V

Q

Q

V

(10)

The equivalent system is unloaded and therefore =0 and E=V=1. Since the p.u. short-circuit

capacity is SSC=1/X, the expression can be simplified:

.).(

1.).(

211

upSupX

E

X

X

V

X

E

V

Q

Q

V

SC

(11)

Increasing the voltage by 0.01 p.u, requires connecting a reactive load of 0.01SSC p.u., i.e. acapacitor bank rated 1% of the short-circuit capacity.

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I.3. Voltage Profile in Distribution Lines

In section I.1 the series resistance of the transmission line has been neglected. This is reasonable,

since for a transmission line the series resistance per km is typically an order of magnitude

smaller than the series inductive reactance per km. This fact no longer holds for a distributionline, for which the series resistance and the series inductive reactance are of the same order of

magnitude. The series resistance must be included in the model of a distribution line, as shown inFigure 3.

Figure 3: Single line diagram of the examined network with a distribution line

Equations (1) and (2) derived for a transmission line are not valid in the case of a distribution

line. An exact expression for the dependence of the voltage drop over the line in Figure 3 on theactive and reactive power transfer can be obtained in case of an impedance load. However, only

an approximate expression valid for a lightly loaded line is used here.

Suppose the load current, active and reactive power are I,Pand Qrespectively and thatRX. InFigure 4, Vphis the phase-neutral voltage. Neglecting the vertical component of the sending end

phase-neutral voltageEph, it can be expressed as:

Figure 4: Phasor diagram for circuit in Figure 3

sincos XIRIVE phph (12)

The load active and reactive power are given as:

cos3 IVP phsin3 IVQ ph

Substituting into (12) and expressing in terms of the line-line voltagesE and Vleads to

V

XQ

V

RPVE (13)

And finally, the line-line voltage difference between the receiving and sending ends is given as

V

XQRPEVV (14)

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Equation (14) is a second order equation in V,which can be easily solved with known P, Q and

E. However, even without solving it, equation (14) shows that in the case of a distribution line

with RX, the active and reactive power affect equally the voltage drop. Consuming (drawing)

active and reactive power (with the chosen load convention, power is positive) causes thereceiving end voltage to drop. Producing (injecting) active and reactive power (with the chosen

load convention, power is negative) increases the receiving end voltage. Notice that for R=0 and0, equation (14) reduces to equation (2), valid for a transmission line.

Distribution lines are operated radially and traditionally the power flow has been unidirectionalalong the line from generation to load. Consequently, the voltage profile was consistently

decreasing from the substation to the line end. The voltage was controlled within an acceptable

band of, for example, 10% around the nominal value by means of tap-changing transformers

and/or shunt compensation. The growing amount of installed distributed generation (windpowerand solar electricity) in the distribution network poses new challenges to voltage control. With

distributed generation, the voltage profile over a line is no longer monotonic.

I.4. The p.u. System

The p.u. system has been already introduced during the lectures. A brief description is included in

this laboratory report. Further information can be found in the course textbook, in Chapter 3,Section 3. The p.u system is often used to present the voltage, current, power and impedance as a

function of selected base quantities, in order to simplify calculations and avoid numerical errors.

The method also avoids the manipulation of the component values, which are located on one sideor another of the transformers. The per-unit quantity is given by equation (15):

valuebasequantityactualquantityunitper (15)

Both the actual quantity and the base value have the same units; the actual quantity is the value

of a quantity in the real system. The base value is a real number, which means that thetransformation does not change the angle of the actual quantity. By selecting the voltage base,

Vbase(line-to-line voltage) and the power base, Sbase, at one point, other base quantities likeIbase(line

current) andZbasecan be easily obtained from equations (16)-(19). The value of Sbaseis the same forthe entire power system, while the Vbase is different for each voltage level, and therefore it is

selected so that the ratio of the voltage bases on either side of a transformer is the same as the

ratio of the transformer voltage ratings.

basebasebase SQP (16)

base

basebase

V

SI

3 (17)

base

base

base

basebasebasebase

S

V

I

VXRZ

2

3 (18)

basebasebasebase

ZBGY

1 (19)

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I.5. Symmetrical and Unsymmetrical FaultsPower systems are designed to be symmetric or balanced, i.e. in a three-phase system, the three

line-to-neutral voltages have the same magnitude and differ in phase by 120o, and the line

currents have the same magnitude and differ in phase by 120o. Therefore, if a system is balancedthe sum of the 3-phase currents will be equal to zero. Sometimes asymmetry can occur in currents

and voltages due to unbalanced loads connected to the system. In Sweden, residential loads aresingle-phase and since these are connected to different phases, three-phase currents and voltages

at the lowest distribution voltage level are not balanced, i.e. the sum of the 3-phase currents will

not be zero. This sum will be equal to the current that circulates through the neutral of the system(if it exists). Loads at higher voltage levels are balanced.

At higher voltage levels the asymmetries are mainly due to system faults. The main two types of

faults areshunt faults, where a (low) impedance is (shunt) connected between line and ground,and series faults, where a (high) impedance is connected in series with the line. The most

extreme, but also most common, series fault is the open-circuit. This occurs for example when acircuit breaker or disconnector is open only in one or two phases or when a phase conductor of a

line is broken (but does not hit ground). Series faults will not be treated in this course. The most

extreme, and also most common, shunt fault is the short-circuit. Short-circuits faults ordered byoccurrence are classified into:

Single-line-to-ground (SLG); Unsymmetrical fault between one phase and ground. Thephase current magnitudes will be no longer identical. The Swedish power system can stilldeliver power to the load through the other two phases.

Line-to-line (LL); Unsymmetrical fault between two phases.

Double-line-to-ground or Line-to-line-to-ground (LLG); Unsymmetrical fault between

two phases and ground.

Three-phase short-circuit (3); It is a symmetrical fault that affects the 3-phases of thepower system. The most severe short-circuit.

For all the above faults, the path of the fault current may be limited by nonzero impedance. If the

impedance is equal to zero, the short-circuit is called bolted. Short- circuit faults often occur as a

consequence of damage to cables and lightning strikes or trees falling on overhead lines. One ofthe most common reasons for three-phase short-circuit to occur is when a line or busbar is

energized with grounding equipment left connected. This equipment is used for safety reasons

while repairmen work on the power equipment, and it should be removed after the work is

completed, and before the equipment is energized.

I.6. Single-Line-to-Ground (SLG) Faults in Distribution NetworksSLG faults represent the vast majority of all faults in public distribution networks and a correct

handling of SLG faults is of primary concern in order to achieve continuity of service to

customers. The magnitude of the fault currents and of the overvoltages during such faults isdetermined by the grounding method of the system. System grounding is important because itdirectly concerns safety. System grounding methods also determine the setup of the protection

system to detect and disconnect (clear) SLG faults.

Swedish regulations prescribe fast and automatic disconnection of permanent ground faults in

distribution networks with a fault resistance up to 5 k. In Sweden, medium voltage distributionnetworks are primarily resonant-grounded, through the use of a Petersen coil. Impedance

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grounded and ungrounded distribution networks are most common in Europe, while distribution

networks are directly grounded in the United States. Directly grounding is used also in Sweden in

the low voltage network and in transmission (high and extra high voltage) networks.

This section briefly presents the theory on SLG faults with regard to ungrounded and resonant-

grounded systems. The fault current at the fault point, the zero sequence voltage and the zerosequence current flowing at the beginning of each line will be calculated. The zero sequence

current flowing at the beginning of each line is important because this is the current measured by

the protection system. For further details about theoretical and practical aspect of networkgrounding methods, refer to the chapter on Neutral earthing in the Protection Guide,

Schneider Electric.

I.6.1 Ungrounded Distribution Networks

Consider the distribution network withNradial lines (feeders) shown in Figure 5. A ground faultwith fault resistanceRfis applied on phase aon theNthline. Each line k, which may represent anoverhead line or a cable, is represented by its zero sequence capacitance to ground Cok. All series

impedances can be neglected, since normally they have much lower magnitude compared to theshunt impedance to ground, 1/(jCok).The feeding transformer neutral point is ungrounded on

the Medium Voltage (MV) side.

Figure 5: MV ungrounded distribution network withNlines. A SLG fault is applied on phase a of theNthline

The SLG fault current at the fault point can be calculated using the equation in the formula sheet

or by inspection of Figure 6.

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Figure 6. Equivalent circuit with sequence diagrams for SLG fault on phase a, ungrounded network

The positive and negative sequence series impedances of the lines are small compared to the zero

sequence impedance and can be neglected. The zero sequence impedance is due to thecapacitances Cok, which are connected in parallel. The total capacitance to ground of the system

is:

N

k

kCC

1

00 (20)

Given the positive sequence phase to ground voltageE, the positive, negative and zero sequencecurrent at the fault point are given as

ff

f

npRCj

ECj

CjR

EI

III0

0

0

0311

33

(21)

The SLG fault current at the fault point is given as:

ff

RCj

ECjI

0

0

31

3

(22)

From Figure 6, the zero sequence voltageE0can be expressed as:

fRCjEI

CjE

00

00

311

(23)

These two equations show that the fault current and the zero sequence quantities decrease with

increasing fault resistance. A small amount of zero sequence voltage may exist in a distribution

network also during healthy conditions because of asymmetries in the capacitances to ground.For very high fault resistance, the zero sequence quantities during a fault may be of the same

order of magnitude as during healthy conditions, rendering difficult the detection of the fault. The

fault current during a fault withRf=0is proportional to the total capacitance to ground C0. Cablelines have higher capacitance to ground per km than overhead lines. Hence, in networks with

cable lines the SLG fault current has higher magnitude than in networks with only overhead lines.

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The zero sequence current flowing at the beginning of each healthy line is easily calculated,

given the zero sequence voltage in (23):

0

000,0

31 CRj

ECjECjI

f

kkhealthyk

(24)

Because of Kirchoffs law, the zero sequence current at the beginning of the faulted Nthline must

be equal to the opposite of the sum of the zero sequence currents on all other (healthy) lines:

0

00000

1

1

00

1

1

,0,031 CRj

ECCjECCjCEjII

f

NN

N

k

k

N

k

healthykfaultedN

(25)

By comparing equation (24) and (25), it can be seen that the zero sequence currents at the

beginning of a healthy or faulted line differ in phase by 180relative to the zero sequence voltage

E0. As you will learn during the lab, this fact can be used to selectively detect and hencedisconnect only the faulted line thus minimizing the number of customers experiencing a serviceinterruption during permanent SLG faults. Also, notice that the zero sequence current flowing at

the beginning of the faulted line in equation (25) differs from the zero sequence current at the

fault point by the zero sequence current contribution due to the capacitance to ground of the Nth

line, C0N.

I.6.2 Resonant-grounded Distribution Networks

Figure 7 shows the diagram of a resonant-grounded radial distribution network with N linesduring a SLG fault on phase aon theNthline. The feeding transformer neutral point is grounded

through a Petersen coil, with inductanceL.

Figure 7: MV resonant-grounded distribution network. A SLG fault is applied on phase a of theNthline

The fault current at the fault location can be calculated by inspecting the diagram in Figure 8.

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Figure 8. Equivalent circuit with sequence diagrams for SLG fault on phase a,resonant-grounded network

Since the positive and negative sequence series impedances can be neglected, only the zero-sequence impedance and the fault resistance limit the fault current magnitude. If only a zero-

sequence voltage would be applied to the healthy system, a zero sequence current would flow

through the Cok of all lines and through the Petersen coil. Therefore, the zero-sequenceimpedanceZ0is given by the parallel connection of j3Land 1/(jC0), where C0is given as in

equation (20). The factor 3 is due to the fact that the zero sequence currents in each phase of the

transformer sum up in the neutral (and hence in the Petersen coils 3I0flows).

The positive, negative and zero sequence current at the fault point are given as

LjRLCR

ELC

LC

LjR

E

LjCj

R

EI

IIIff

ff

f

np

393

31

31

333//

13

30

2

02

02

0

0

(26)

The SLG fault current at the fault point is given as:

LjRLCR

ELCI

ff

f

393

313

02

02

(27)

From Figure 8, the zero sequence voltageE0can be expressed as:

LjRLCR

LEjILj

CjE

ff

393

33//

1

020

00

(28)

From the diagram in Figure 8 and inspecting equations (26) and (27), it is evident that, choosing

L as L 1/(32C0), the fault current at the fault location can be reduced possibly to zero.Reducing the fault current increases the probability of self-extinguishing for temporary ground

faults in overhead lines. In practice, the value ofLcan be adjusted to accommodate for different

values of C0due to changes in network topology. Since the value of Lis chosen to compensatethe total network capacitive current, it is common practice in the industry to use an ampere

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value for the Petersen coil. Notice that both the zero sequence current and voltage decrease with

increasing fault resistance.

It is relevant for protection relay considerations to calculate the zero-sequence currentcontribution of the unfaulted lines, i.e. the zero sequence current measured by the protection

system on the healthy lines, and the current into the Petersen coil.

00,0 ECjI khealthyk (29)

000 3 ECjLj

EIL

(30)

In equation (30) it has been assumed that perfect compensation exists. Because of Kirchoffs law,

the zero-sequence current flowing into the faulted line (this is the zero sequence current measured

by the protection system on the faulted line) must be opposite to the sum of the zero-sequencecurrents on the healthy lines and of the Petersen coil current divided by 3.

0000

000

1

1

,0,03

3

3ECj

ECjECCj

III NN

LN

k

healthykfaultedN

(31)

The phase relationship between the zero sequence currents flowing at the beginning of healthy

and faulted lines and the zero sequence voltage does not allow in this case the discrimination of

the faulted line. The zero sequence current in equations (29) and (31) are both 90 degrees aheadof the zero sequence voltage E0.Compare these results with those obtained in equation (24) and

(25) for an ungrounded network.

To allow for a selective detection of the faulted line, a resistor RLis normally added in parallel

with the Petersen coil. The zero sequence current due to RL is in phase with the zero sequence

voltage and (its opposite) flows only into the faulted line. As you will learn during the lab, with aproper choice of the resistanceRL, the resistive component of the zero sequence current permits

to selectively detect the faulted line in resonant-earthed networks. Figures 1 and 2 on page 10 of

the Schneider Protection Guide help clarifying this concept.

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Section II: Preparatory Exercises

Matlab Exercises

Figure 9: Single line diagram of the examined transmission network (X=L, B=C)

The base quantities are chosen as Sbase=100VA and Vbase=24V. The p.u. bus admittance

matrix for the network in Figure 9, not including the load resistor, is:

8181.18335.1

8335.18181.1,

jj

jjY pubus p.u.

1. Compute the natural loading of the line asP0=E2/R0whereR0=(L/C).

2. Plot the P-V characteristic of the load bus by varying the parameterRLand calculating P

and V.UseRLequal to 0, 50%, 100%, 200% of X and also 10000% (no-load) to properlycapture the shape of the P-V curve. Remember to use per unit for all or for no quantities.

3. Repeat 2. with a 32 F shunt capacitor at the load bus. Represent the capacitor byincluding it in element Y22 of the reduced Ybus matrix as follows:

Y22,New=Y22,Original+jCload. Do not confuse Cloadwith the line capacitance C.

4. Repeat 2. with a 250 mH shunt reactor at the load bus. Include the reactor like you have

done with the capacitor (Y22,New=Y22,Original-j/(Lload)).

PowerWorld Exercises

Figure 10: Single line diagram of the examined distribution network (X=L, B=C). Note: the network is three-phase and E=24Vis the phase-to-ground voltage. Sbase=100VA. Zbase=17.28.

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Open the file lab1_DG_2feeders.pwb in PowerWorld, which models the system in Figure

10. Select Run Mode. The wind turbine is initially disconnected. The tap position of the

transformer can be changed with up/down arrows. By varying the tap position, the voltage in

the network can be regulated.

1. Increase the active load at the end of feeder 2 until the voltage somewhere in the networkreaches the 10% of nominal voltagelimits (you can vary PLby clicking on the up-down

arrows close to the MW indication. To run a load flow select Tools Play (green)).

Now adjust the transformer tap changer to bring all voltages into the acceptable range 10%.

2. What is the maximum load that can be supplied while still maintaining the voltage in the

whole network within 10% with the tap changer? What is the voltage at the main bus in

this case? And what is the transformer tap position?

Disconnect the load on line 2 (click on the breaker). Adjust the transformer tap position to get

1 p.u. voltage at the main bus (Tap position = 0).

3. Connect the distributed wind turbine generator to feeder 1, node 4 (click on the breaker).

Find out the maxPDGthat can be injected while the voltage is within a 10% band.(you

can vary PDGby clicking on the up-down arrows close to the MW indication). Keep QDG=0. The voltage on feeder 1 increases withPDG, in accordance with equation (14).

4. WithPDG=0, check how QDGaffects the voltage at node 4.

5.

Set QDG=0. Adjust the transformer tap position to find the maximum PDG that can beinjected without violating the voltage limit in the network. What is the voltage at the main

bus in this case? And what is the transformer tap position?

Set the transformer tap position to zero. Connect the load on line 2 and the distributed wind

turbine on feeder 1.

6. Try to maximize bothPL andPDG, without violating the 10% voltage limits.7.

Can you increase both the maximumPL andPDGby regulating only the tap position?

8. Do the same as in 7) but by also adjusting QDG. In practice, |QDG| is kept below PDG/2.

Figure 11: Distribution network, where different grounding methods of the transformer neutral can be selected. Sbase=100MVA,Vbase=10kV, Zbase=1

3F

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Open the file lab1_Faults with the model in Figure 11. Check the voltage levels at the

various buses. There is one Dy transformer in the system, initially solidly grounded on the

secondary side. You can change the grounding method of the transformer (double-click onthe transformer Fault Info Configuration. Choose Delta-Grounded Wye for

directly and resonant grounded systems. Choose Delta-Wye for an ungrounded system. Tochoose the value of the Petersen coil Ground Impedance X ).

1. Find the three-phase short-circuit current at Bus 3 for the system when this is:

a) solidly grounded, b) ungrounded, c) resonant-grounded with X=100 p.u.

What is the impedance limiting the three-phase fault current in the different cases?

(To calculate short-circuit current, right-click on a bus Fault Single Fault.

Fault Type 3 Phase Balanced. Select Amps under Units. Press Calculate).

2. With the system solidly grounded find the SLG fault current for a zero fault on phase aat bus 3. What is the impedance limiting the fault current in this case?Could an overcurrent relay (or fuse) be used to disconnect the line with a permanent zero

ohm SLG fault in this case? Assume line rated current is 100 A.

3. Do the same as in 2. but with the system ungrounded and resonant-grounded with perfect

compensation.

Choose theFault Info Ground Impedance X = 1/(3C0) with C0 from

Figure 11. Note that p.u.base impedance is 1.

Could an overcurrent relay (or fuse) be used to disconnect the line with a permanent zeroohm SLG fault in these cases?

4. With a resonant-grounded system, choose the value of the Petersen coil which gives 5 A

fault current for a SLG fault. Change the fault resistance from 020 k (In the FaultAnalysis panel Fault Impedance R. Note that p.u.base impedance is 1).

How does this affect the zero sequence voltage at bus 2 and the fault current?

Reset theFault Impedance Rparameter to zero. The line in the test system represents

a 10 km cable, with a capacitance to ground C=0.3 F/km which results in a Fault Info -Zero Sequence Impedance - C = 0.00094 p.u. with Zbase=1 p.u. The SLG fault current

magnitude depends on the total capacitance to ground in the network. Cables and overheadlines have very different capacitance to ground.

5. With an ungrounded networks, determine the SLG fault current if the Fault Info - Zero

Sequence Impedance - C= 0.0000345 p.u., corresponding to a 10 km overhead line with a

capacitance to ground of 11 nF/km. Compare the result with the first answer in 3.

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Section III: Experiments

Connect the system as shown in Figure 12 with a variable transformer, three sections of the line

model in series and a variable resistance as load.

Figure 12: Single line diagram of the examined transmission network (X=L, B=C)

To begin with, the line model represents a transmission line with negligible resistance. Therefore,

short-circuit all series resistances of the line. Let the capacitor bank and shunt reactance beunconnected.

III.1. Transmission line: PV-Curve and Reactive power compensation1. Form the circuit seen in Figure 12. Use a Wattmeter to measure the active power

consumption in the variable resistor. Use an oscilloscope to measure the voltage angle

of the load bus relative to that of the generator bus. Use a variable transformer and themains to generate the fixed voltage equal to 24 V. Note that this voltage has to be

adjusted during the measurements, because of the internal voltage drop in the

transformer!2. Plot the P-V and P-characteristic of the system with a) no reactive compensation.

0 10 20 30 400

5

10

15

20

25

30

35

P [W]

V

[V]

0 10 20 30 400

20

40

60

80

100

P [W]

Angle[o]

Figure 13. P-V and P-characteristic for the proposed system

3. Find the no-load voltage and the maximum transfer limit when b) a 32 F shuntcapacitor is connected at the load bus, and c) when a 250 mH shunt reactor isconnected at the load bus. Draw these points on the same plot as the PV curve. What

do you observe? How do the shunt reactor and the capacitor affect the transfer limits

and the voltage stability limit?4. Determine the voltage sensitivity to reactive power compensation given by equation

(11), by using the reading of the voltage at P = 0, for the three cases a), b) and c).

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Compare the results to those obtained by using the rule of thumb

V /Q 1/SSC(p.u.)

V

QVcompVuncomp

Qcomp

5. Use the measured PV-characteristic for the uncompensated system to determine the

natural loading of the line. Compare it with the calculated value.6. Plot the voltage as a function of the position on the line, when the line is loaded

below, at and above its natural loading.

III.2. Short-circuit impedance and powerMake a short circuit test on the load bus as follows (disconnect any shunt compensation).

Determine the short-circuit-power Sscby measuring the no-load voltage and the short-circuitcurrent and multiply the two. The short-circuit impedance is found by dividing the no-load

voltage with the short-circuit-current.

III.3. Voltage Profile on Distribution LinesDisconnect the voltage supply and connect the second line model (also three sections) in parallel

with the first model, as shown in Figure 10. The line models should now represent distributionoverhead lines with anX/Rratio of approximately one. Hence, insert the series resistances of the

-model in series with the inductive reactances. For short distribution overhead lines, the shuntcapacitance can be neglected. Hence, disconnect the capacitances of the -model from the circuit.In the lab set-up, the distributed generation is not a wind turbine. Instead, a battery in series with

a resistor Rinv (which may represent a solar PV panel) can be connected to the line through aninverter which implements a Maximum Power Point tracking (MPPT) algorithm. The power

delivered from the battery to the grid can be varied by varying the resistorRinv.

Use a variable transformer and the mains to generate a fixed voltage equal to 50 V. Connect a

variable load resistor at the end of one line and use a Wattmeter to measure the active powerconsumption. Do not connect the inverter yet. In the following, always keep the voltage in the

whole network within 10% of nominal voltage.

7. What is the maximum power that can be delivered to the load resistor if the voltage at

the variable transformer bus is regulated to nominal? What are the voltages at allnodes of the two line models in this case? Draw the voltage profile at all nodes.

8.

Can the maximum power delivered to the load be increased without violating thevoltage limits by acting on the variable transformer?

Regulate the variable transformer to get 50V at the transformer bus and maximum power

delivered to the load. Connect the battery with the inverter at the end of the second line model tostudy the impact of distributed generation on the voltage profile.

9. Vary the delivered active power from the inverter by means of the variable resistorRinv. How is this affecting the voltage profile on the second line?

10.While adjusting the voltage at the transformer bus to 50V, find the maximum active

power that can be delivered by the inverter without violating the voltage limit. Drawthe voltage profile at all nodes.

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Keep the inverter power fixed. Acting only on the variable transformer, it is not possible to

increase the power delivered to the load without violating the voltage limit in the network. Try!

11.

Suggest how the maximum power delivered to the load could be increased withoutviolating the voltage limit of 10%. Try your proposal!

III.4. Grounding in Distribution NetworksThe model of the distribution network used in this lab exercise is shown in Fig 13. You will

either work with a model representing two cables or with a model representing two overheadlines. To begin with, you will investigate the fault current in an ungrounded system.

Figure 13: Distribution network model used in the lab.

III.4.1 Ungrounded Distribution Network

Be sure that the neutral point at the secondary side of the distribution transformer is ungrounded.

Connect the three-phase transformer to the mains and set the voltage to 30V.1. Inspect voltages and currents on the LabView measuring panel. Is there any zero-

sequence voltage or current in the system?

2. On the LabView measuring panel, check the value of the potential difference betweenthe neutral point of the transformer and reference ground (VnG).

3. Apply a solid SLG fault (phase a to ground) at the end of feeder 1. Inspect the

following quantities on the measuring panel and report their complex value(magnitude and angle):

I0, eeder 1

I0, eeder 2

Ia, eeder 1= I

V0

VaG

VbG

VcG

4. What happens to the line-ground voltage on the two unfaulted phases? Can you

explain with reference to the diagram for a three-phase symmetrical voltage system?What is a requirement for the isolation in ungrounded networks?

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5. With the SLG fault applied, check again the value of the potential difference between

the transformer neutral point and the reference ground (VnG) on the LabView

measuring panel. This is the zero sequence voltage!

6.

You have seen that the zero-sequence quantities were all zero before the fault. Can thezero-sequence voltage V0 be used for protection to detect a SLG fault? What is the

major disadvantage of using only V0 for fault detection?

In case of a permanent SLG fault in real distribution systems, it is important to selectively

disconnect only the faulted feeder and to continue to deliver power to the customers connected onthe unfaulted part of the system.

7. What is the phase relationship between V0and I0 in the unfaulted and in the faulted

feeder respectively? Propose a principle on which a selective protection system could

be based. Take help also from equations (24) and (25).8. What happens to V0andI0if the SLG fault is applied through a fault resistor, when the

value of the resistance is increased?

III.4.2 Resonant-grounded Distribution Network

Disconnect any fault, bring the voltage down to 0V and disconnect the mains. Connect a variableinductor between the neutral of the distribution transformer secondary side and the reference

ground. The distribution network is now resonant-grounded. Reconnect the mains and set thevoltage to 30V.

9. Apply a solid SLG fault on phase aof feeder 2. Tune the inductor to match the total

capacitance to ground C0,Tot. How can you know when the system is perfectly tuned?10.What is the value of Lin this case? Why the fault current is not zero even though the

system is perfectly compensated?

Ideally, with no resistive losses and in case of perfect compensation, the current magnitude at the

fault location would be equal to zero. However, the zero-sequence currents at the beginning of

each feeder and from the Petersen coil would not be zero.

11.Inspect the following quantities on the measuring panel and report their complexvalue (magnitude and angle):

I0, eeder 1

I0, eedere 2

InG= IL

Ia, eeder 1= I

V0 VaG

VbG

VcG

12.What are the voltages on the unfaulted phases and the zero sequence voltage V0? What

is the relation betweenI0,feeder_1,I0,feeder_2andIL (= InG)?

Perfect tuning of the system is difficult to achieve during all conditions. Mistuning may arise

because of topological changes in the network, for ex. because one line is taken out of service.

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This changes the total capacitance to ground in the network. You will study the effect of

mistuning by changing the value of the coil inductance with both lines connected.

13.Change the value of the variable coil inductance and observe how this affects the fault

currentIfand the zero sequence voltage V014.Vary the coil inductance so to perfectly compensate the system zero sequence

capacitance. What is the relationship between V0, andI0,feeder_1,I0,feeder_2respectively?Could a protection system based on the phase relationship between these quantities

selectively detect the faulted line?

Keep the system perfectly tuned. Disconnect the SLG fault. Bring the voltage down to 0V and

disconnect the mains. Insert a resistor RLin parallel with the Petersen coil. Reconnect the mains

and set the voltage to 30V.

15.Apply a SLG fault on feeder 1. Inspect the following quantities on the measuringpanel and report their complex value (magnitude and angle):

I0, eeder 1

I0, eeder 2

InG= IL

Ia, eeder 1= I

V0

VaG

VbG

VcG

16.What is the major difference compared with the case without resistor RL? What is the

advantage with the resistorRLconcerning selective fault detection?

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