laboratory 5 psa

5/24/2018 Laboratory 5 psa

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LABORATORY 5FAULT LEVEL CALCULATIONS USING EMTC/PSCAD

INTRODUCTION

This experiment investigated the analysis of Symmetrical and Unsymmetrical Three-Phase

faults in power systems.

Theory Background

Under normal conditions, a power system operates under balanced conditions with allequipment carrying normal load currents and the bus voltages within the prescribed limits.This condition can be disrupted due to a fault in the system. A fault in a circuit is a failure

that interferes with the normal flow of current. A short circuit fault occurs when theinsulation of the system fails resulting in low impedance path either between phases or

phase(s) to ground. This causes excessively high currents to flow in the circuit, requiring theoperation of protective devices to prevent damage to equipment. The short circuit faults can

be classified as:

Symmetrical faults Unsymmetrical faults

Symmetrical faults:

A three phase symmetrical fault is caused by application of three equal fault impedances to the three phases, as shown in Fig. 1. If = 0 the fault is called a solid or a bolted fault.These faults can be of two types: (a) line to line to line to ground fault (LLLG fault) or (b)line to line to line fault (LLL fault). Since the three phases are equally affected, the systemremains balanced. That is why, this fault is called a symmetrical or a balanced fault and thefault analysis is done on per phase basis. The behaviour of LLLG fault and LLL fault isidentical due to the balanced nature of the fault. This is a very severe fault that can occur in a

system and if , this is usually the most severe fault that can occur in a system.Fortunately, such faults occur infrequently and only about 5% of the system faults are three

phase faults.

Figure 1Symmetrical Faults


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Unsymmetrical faults:

Faults in which the balanced state of the network is disturbed are called unsymmetrical orunbalanced faults. The most common type of unbalanced fault in a system is a single line to

ground fault (LG fault). Almost 60 to 75% of faults in a system are LG faults. The othertypes of unbalanced faults are line to line faults (LL faults) and double line to ground faults(LLG faults). About 15 to 25% faults are LLG faults and 5 to 15% are LL faults. Thesefaults are shown in Fig. 4.40.

Figure 2 - Unsymmetrical Fault

Majority of the faults occur on transmission lines as they are exposed to external elements.Lightening strokes may cause line insulators to flashover, high velocity winds may causetower failure, ice loading and wind may result in mechanical failure of line or insulator andtree branches may cause short circuit. Much less common are the faults on cables, circuit

breakers, generators, motors and transformers.

Fault analysis is necessary for selecting proper circuit breaker rating and for relay settingsand coordination. The symmetrical faults are analysed on per phase basis while the

unsymmetrical faults are analysed using symmetrical components. Further, the Busmatrixis very useful for short circuit studies.

Symmetrical or Balanced three phase fault analysis:

In this type of fault all three phases are simultaneously short circuited. Since the networkremains balanced, it is analysed on per phase basis. The other two phases carry identical

currents but with a phase shift of 120. A fault in the network is simulated by connectingimpedances in the network at the fault location. The faulted network is then solved usingThevenins equivalent network as seen from the fault point. The bus impedance matrix isconvenient to use for fault studies as its diagonal elements are the Thevenins impedance ofthe network as seen from different buses. Prior to the occurrence of fault, the system isassumed to be in a balanced steady state and hence per phase network model is used. Thegenerators are represented by a constant voltage source behind a suitable reactance whichmay be sub-transient, transient or normal d-axis reactance. The transmission lines arerepresented by their -modelswith all impedances referred to a common base. A typical busof an n- buspower system network is shown in Fig. 3. Further, a balanced three phase fault,

through a fault impedance is assumed to occur at kth bus as shown in the figure. A pre-fault load flow provides the information about the pre-fault bus voltage.


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Let [Bus (0)] be the pre-fault bus voltage vector = (0) . . . Vk(0) . . . Vn(0)p.u. Thefault at kth bus through an impedance will cause a change in the voltage of all the buses

[BUS] due to the flow of heavy currents through the transmission lines. This change canbe calculated by applying a voltage (0) at kth bus and short circuiting all other voltagesources. The sources and loads are replaced by their equivalent impedances.

Figure 3 -Network representation for calculating BUS

In Figure 3 above, and are the equivalent load impedances as bus i and k respectively, is the impedance of line between ith and kth buses. is the appropriate generatorreactance, is the fault impedance, is the fault current and (0) is the pre-faultvoltage at kth bus. From the superposition theorem, the bus voltages due to a fault can be

obtained as the sum of pre-fault bus voltages and the change in bus voltages due to fault, i.e.,

Figure 4Fault at Kth Bus of a Power System

(1)


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Where,

[Bus (F)] = Vector of bus voltages during fault = (F) . . . (F) . . . n(F)

[Bus (0)] = Vector of pre-fault bus voltages = (0) . . . (0) . . . n(0)

[BUS] = Vector of change in bus voltages due to fault = [

Also the bus injected current Bus]can be expressed as,

(2)

where, [BUS] is the bus voltage vector and [BUS] is the bus admittance matrix. With allthe bus currents, except of the faulted bus k, equal to zero, the node equation for the networkof Fig. 3 can be written as

(3)

As the fault current (F) is leaving the bus it is taken as a negative current entering the bus.Hence,

(4)

[BUS] can be calculated as:

(5)

where, [BUS] is the bus impedance matrix = [BUS. Substituting the expression of[BUS] from equation (4) in equation (1) one can write,


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(6)

Expanding the above equation one can write,

(7)

The bus voltage of kth bus can be expressed as:

(8)

Also from Fig. 4,

(9)

For a bolted fault = 0 and hence, (F) = 0. Thus the fault current k(F) for bolted faultcan be expressed using equation (8) as,

(10)

For faulty with non-zero fault impedance , the fault current can be calculated as:


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(11)

The quantity in equation (10) and equation (11) is the Thevenins impedance oropencircuit impedance of the network as seen from the faulted bus k. From equation (4.70),the bus voltage after fault for the un-faulted or healthy buses can be written as:

(12)

Substituting (F) from equation (10) , (F) can be expressed as:

(13)

The fault current (F) flowing in the line connecting ith andjthbus can be calculated as

(14)

where zij is the impedance of line connecting buses i and j.


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Unsymmetrical or Unbalanced fault analysis:

For the analysis of unsymmetrical or unbalanced faults, symmetrical component method isused. The use of symmetrical components simplifies the analysis procedure of unbalancedsystem and also helps in improving the understanding of the system behavior during fault

conditions.

Symmetrical components:

Any unbalanced set of three phase voltage or current phasors can be replaced by threebalanced sets of three phase voltage or current phasors. These three balanced set of voltage orcurrent phasors are called symmetrical components of voltages or currents. Let the vectors Ia,

Ib, andIcbe an arbitrary set of three current phasors representing phase currents. Then usingsymmetrical components they can be expressed as:

(15)

Or,

(15)

Where,


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The graphical representation of the sequence components is shown in Fig. (5). Let anoperator a be defined such that a = .120 . Any phasor multiplied by a undergoes acounter clockwise rotation of 120without any change in the magnitude. Further,

Or,

(16)

where, is the angle of phase a positive sequence current.

Figure 5 - Representation of Symmetrical Components

(17)


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The phase sequence of the positive component set is abc.

Similarly the negative sequence set can be written as:

(18)

where, is the angle of phase a negative sequence current.

(19)

The phase sequence of the negative component set is acb.

The zero-sequence component set can be written as:

(20)

where, ._0 is the angle of phase a zero sequence current.Hence, equation (15) can besimplified as:

(21)


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It can also expressed in a compact form as:

(22)

(23)

To summarize:

For Voltage:

(24)

(25)


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where, abcis the set of phase voltages, and is the set of sequence voltages.

For Current:

(26)

(27)

where,is the set of phase voltages, and is the set of sequence voltages.

Assumptions Commonly Made in Three Phase Fault Studies

The following assumptions are usually made in fault analysis in three phase transmission

lines.

All sources are balanced and equal in magnitude & phase Sources represented by the Thevenins voltage prior to fault at the fault point Large systems may be represented by an infinite bus-bars Transformers are on nominal tap position Resistances are negligible compared to reactances Transmission lines are assumed fully transposed and all 3 phases have same Z Loadscurrents are negligible compared to fault currents Line charging currents can be completely neglected

AIM AND OBJECTIVES

The objective of this experiment is to become familiar with the use of the Software

EMTDC/PSCAD in the analyse of Symmetrical and Unsymmetrical Faults in Power

Systems.


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PROCEDURE

Figure 6 shows the single-line diagram of a 4-bus power system. The system parameters aregiven in Table 1.

Figure 6Bus Power System Schematic

Table 1The System Data

Figure 7Positive sequence impedance Figure 8: Negative sequence impedance


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Figure 9 - Zero sequence impedance diagram

1. Using the system data given in Table 1 and the impedance diagrams shown in Figs 7-9,compute the positive, negative and zero sequence Y-bus matrices.

2. Invert the Y-bus matrices to obtain the Z-bus matrices and the Theveninimpedances for each bus.

3. Run the PSCAD program.

4. Choose the fault location, type, start time and duration as per the instructor

given values.

5. Calculate the rms fault current sequence componentsFigure

Apparatus

PSCAD softwarePrinter


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RESULTS

Y and Z Bus Matrices

Positive = Negative Y-Bus Matrix

j

Zero Y-Bus Matrix

j

Positive = Negative Z-Bus Matrix

j

Zero Z-Bus Matrix

j

-14.16 2.5 1.66 0

2.5 -14.88 1.38 1

1.66 1.38 -13.13 2.5

0 1 2.5 -11.83

0. 0744 0.0138 0. 0115 3.608

0.0138 0.071 0.0108 8.29

0.0115 0.0108 0.0822 0.0182

3.608 8.29 0.0182 0.089

-11.660 5.000 3.330 0.000

5.000 -13.110 2.770 2.000

3.330 2.770 -14.600 5.000

0.000 2.000 5.000 -8.080

0.136 0.075 0.066 0.059

0.075 0.130 0.067 0.074

0.066 0.067 0.129 0.097

0.059 0.074 0.097 0.202


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Graph 1 No-Fault

As can be seen in Graph 1, under normal

operating conditions, the voltages and currents

in each phase are equal in magnitude and 120

apart from each other. Hence the system isstable and balanced in all 3 phases.

Graph 2Phase a to Ground Fault

In graph 2 a fault Line-to-Ground hasoccurred between phase a and ground. In

this type of fault there will be heavy currentflowing on phase ato ground. Currents of

phases b and c remain the same. Thismeans that phases b and c are notexperiencing heavy current because they arenot faulted or shorted (connected to ground).This fault introduces unbalance between thethree phases where voltage of phase a

becomes zero, whereas voltage of phase b

and cremain almost the same as before thefault has occurred. Note that L-G fault is themost common and severe one in powersystems. It accounts for about 75% to 80% ofall faults related to power systems. ThereforeSystem protection schemes are to be evolvedand implemented for the reliability and safety of

power systems..


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Graph 3Phase ab toGround Fault

In graph 3, a fault Line-to-Line Ground has

occurred on phases a and b. As a result

voltages on phase a and b are zero because

they are grounded. On the other hand phasesa and b carry a heavy current.

Graph 4Phase abc to Ground Fault Power System

In this type of fault where all three phases are

shorted to ground, as is shown in Graph 4, all

the three phases experience the same

conditions which are:


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Graph 5Phase ab Shorted Fault

Graph 6Phase abc Shorted Fault


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DISCUSSION


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APPENDIX

Reactances in pu:

(|)


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Currents in Actual unit:


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laboratory 5 psa

Documents

line fault lll fault

system faults

line faults ll faults

ground fault lllg fault

type of fault

severe fault

bolted fault

phase symmetrical fault