LOAD FLOW ANALYSIS - PART - 18 – CALCULATIONS OF LINE FLOWS AND TRANSMISSION LOSSES

LOAD FLOW ANALYSIS - PART - 17 – CALCULATIONS OF REAL AND REACTIVE POWER USING G-S METHOD

LOAD FLOW ANALYSIS - PART - 16 – GAUSS-SEIDEL METHOD – ONE PROBLEM

LOAD FLOW ANALYSIS - PART - 15 – COMPUTATION OF LINE FLOWS ANS LINE LOSSES

LOAD FLOW ANALYSIS - PART - 14 – LINEARIZING STATIC LOAD FLOW OR POWER FLOW EQUATIONS

LOAD FLOW ANALYSIS - PART - 13 – STATIC LOAD FLOW OR POWER FLOW EQUATIONS

LOAD FLOW ANALYSIS - PART - 12 – CLASSIFICATION OF BUSES & STEPS FOR LOAD FLOW STUDIES

LOAD FLOW ANALYSIS - PART - 11 – POWER FLOW IN A SIMPLE TWO TERMINAL POWER LINE

LOAD FLOW ANALYSIS - PART - 10 - EFFECT OF VOLTAGE AND FREQUENCY ON LOAD

LOAD FLOW ANALYSIS - PART - 09 - RELATION BETWEEN PER UNIT CHANGE IN VOLTAGE, REACTIVE POWER AND SCC OF THE LOAD BUS

LOAD FLOW ANALYSIS - PART - 08 - THE ROLE OF TORQUE ANGLE IN POWER SYSTEM

LOAD FLOW ANALYSIS - PART - 07 - ROLE OF REACTIVE POWER ON VOLTAGE REGULATION

LOAD FLOW ANALYSIS - PART - 06 - RELATION BETWEEN REACTIVE POWER AND BUS VOLTAGE

LOAD FLOW ANALYSIS - PART - 05 – GENERATION AND ABSORPTION OF REACTIVE POWER

LOAD FLOW ANALYSIS - PART - 04 – NATURAL LOAD OR SURGE IMPEDANCE LOAD

LOAD FLOW ANALYSIS - PART - 03 - POWER TRANSFER THROUGH IMPEDANCE – X = 1.732R

LOAD FLOW ANALYSIS - PART - 02 - POWER FLOW EQUATION

LOAD FLOW ANALYSIS - PART - 01 - COMPLEX POWER S = P + j Q & S = P - j Q

UNSYMMETRICAL FAULT CALCULATIONS - PART – 50 - CONCLUSION SPEECH

UNSYMMETRICAL FAULT CALCULATIONS – PART – 49 – PROBLEM – 09 – DOUBLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 48 – PROBLEM – 08 – DOUBLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 47 – PROBLEM – 07 – LINE TO LINE FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 46 – PROBLEM – 06 – LINE TO LINE FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 45 – PROBLEM – 05 – SINGLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 44 – PROBLEM – 04 – SINGLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 43 – PROBLEM – 03 – SINGLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 42 – PROBLEM – 02 – SINGLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 41 – PROBLEM – 01 – SINGLE LINE TO GROUND FAULT

UNSYMMETRICAL FAULT CALCULATIONS – PART – 40 – PROBLEMS - POWER IN TERMS OF SYMMETRICAL COMPONENTS

UNSYMMETRICAL FAULT CALCULATIONS – PART – 39 – POWER IN TERMS OF SYMMETRICAL COMPONENTS

UNSYMMETRICAL FAULT CALCULATIONS – PART – 38 – FIVE PROBLEMS ON SEQUENCE NETWORKS

UNSYMMETRICAL FAULT CALCULATIONS – PART – 37 – SEQUENCE NETWORKS OF A LOAD

UNSYMMETRICAL FAULT CALCULATIONS – PART – 36 – SEQUENCE NETWORK OF A TRANSMISSION LINE

UNSYMMETRICAL FAULT CALCULATIONS – PART – 35 – SEQUENCE NETWORK OF TRANSFORMER

The presence of a fault at any point in a system means generally that the system is put into an unbalanced state of operation.

In unsymmetrical fault calculations, each piece of equipment will have three value of impedance offered by each sequence current.

The zero sequence network is a single-phase component and its existence is dependent upon a closed path that must be completed through reference ground.

1. For positive or negative sequence networks, the neutral of the generator is taken as the reference bus. The reason is the positive and negative sequence components represent balanced sets and hence all the neutral point must be at the same potential for positive or negative sequence currents.

2. If there is no ground return, zero sequence currents cannot flow and in the corresponding zero-sequence networks are considered an open circuit.

3. In star connected transformer if the neutral is grounded zero-sequence current can flow in the winding.

4. If the star point is ungrounded zero sequence currents cannot flow in the winding.

5. In a delta connected, there is no neutral, hence zero-sequence currents cannot flow in the line because there is no return path. However, zero-sequence currents can circulate in the legs of the delta connected winding.

6. Due to the presence of zero sequence voltages the zero-sequence currents flowing around a delta winding.

7. When the magnetizing current in a transformer is neglected, the primary ampere-turns balance the secondary ampere-turns and current can flow in the primary only if the there is a current in secondary.

8. If the neutral point of the star connection is grounded through an impedance Zn and impedance 3Zn appears in series with Z0 in the zero-sequence network.

9. The zero sequence impedance depends upon earth connection. If there is a through the circuit for earth current, zero-sequence impedance is equal to positive sequence impedance otherwise it will be infinite.

10. Positive sequence impedance = Negative sequence impedance 

zero sequence impedance = Positive sequence impedance if there is a circuit for earth current (infinite, if there is no circuit for earth current)

UNSYMMETRICAL FAULT CALCULATIONS – PART – 34 – SEQUENCE NETWORK OF LOADED THREE-PHASE GENERATOR SEQUENCE NETWORKS

A single-phase equivalent circuit of power system consisting of impedances to the current of any one sequence only is called sequence network. [OR] An equivalent network for the balanced power system under an imagined operating condition such that only one sequence component of voltages and currents is present in the system.

Sequence network is composed of impedances offered to that sequence current in the system. Three sequence networks for a given power system namely

(i) Positive sequence network (ZO) – The positive sequence network for a given power system shows all the paths for the flow of positive sequence currents in the system.

(ii) Negative sequence network (Z1) – The negative sequence network for a given power system shows all the paths for the flow of negative sequence currents in the system

(iii) Zero sequence network (Z2) - The zero sequence network for a given power system shows all the paths for the flow of zero sequence currents in the system.

Each sequence network is replaced by a Thevenin’s equivalent circuit between two points i.e. each sequence network can be reduced to a single voltage and single impedance.

One point is the fault point ‘F’ and the other point is the zero potential of reference bus N.

Va1 = Vf - Z1 IR1, Va2 = -Z2 Ia2 and Va0 = -Z0 Ia0

Ia is the current flowing from the system into the fault, its components Ia1, Ia2 and Ia0.

Z1, Z2 and Z0 are the total impedance of the positive, negative and zero sequence networks up to the fault point.

For a fault on the unloaded generator with excitation voltage Eg the following will be the sequence voltage drops.

Va1 = Eg - Z1 IR1, Va2 = -Z2 Ia2 and Va0 = -Z0 Ia0

ELECTRICAL DIAGRAMS - PART - 05 DIFFERENCE BETWEEN IMPEDANCE AND REACTANCE DIAGRAM

ELECTRICAL DIAGRAMS – PART – 04 - SINGLE LINE DIAGRAM (SLD) IMPORTANCE AND APPLICATIONS

SINGLE LINE DIAGRAM

The single-line diagram is the blueprint for electrical power system analysis. It is the first step in preparing a critical response plan and to understand thoroughly familiar with the electrical distribution system layout and design.

IMPORTANCE OF SLD

1. The SLD is the roadmap for all future testing, service and maintenance activities.

2. It provides a picture of the facility at a moment in time.

3. It needs to change as your facility changes to ensure that your systems are adequately protected.

4. The SLD is making the electrical system easily understandable for any technical person inside/outside of the factory.

APPLICATIONS OF SINGLE LINE DIAGRAM

1. Short circuit calculations             2. Coordination studies

3. Load flow studies                        4. Safety evaluation studies

5. Electrical safety procedures         6. Efficient maintenance

ELECTRICAL DIAGRAMS - PART – 03 - DIFFERENCE BETWEEN BLOCK AND LAYOUT DIAGRAM

BLOCK DIAGRAM - It is a functional drawing which shows and describes the main operating principles of the equipment or devices. It consists of principle functions or parts represented by blocks and is connected through lines that show the relationship between the blocks.

LAYOUT DIAGRAM - A drawing meant to show the physical arrangement of the wires and the components they connect is called layout design. 

ELECTRICAL DIAGRAMS PART – 02 - DIFFERENCE BETWEEN PICTORIAL AND SCHEMATIC DIAGRAM

ELECTRIC CIRCUIT – An electrical circuit is a path through which an electrical current flows. A closed circuit makes electrical current flow possible. An open circuit does not allow electrical current to flow.

An electric circuit made up of the basic elements resistance, inductance, and capacitance in a simple arrangement such that its performance would duplicate that of a more complicated circuit or network.

PICTORIAL CIRCUIT DIAGRAM - A pictorial circuit diagram uses simple images of components.

SCHEMATIC CIRCUIT DIAGRAM - A schematic circuit diagram is used to give a visual representation of an electrical circuit to an electrician.

ELECTRICAL DIAGRAMS PART – 01 - DIFFERENCE BETWEEN VECTOR AND PHASOR DIAGRAM

VECTOR - A quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.

VECTOR DIAGRAM - Vector diagram shows the direction and relative magnitude of a vector quantity by a vector arrow. Vector diagrams can be used to describe the velocity of a moving object during its motion. Vector diagrams can be used to represent any vector quantity. For example, acceleration, force, and momentum.

PHASOR - A phasor is a line whose direction represents the phase angle in electrical degrees and whose length represents, the magnitude and of the electrical quantity.

PHASOR DIAGRAM - The graphic representation of the phasors of sinusoidal quantities taken all at the same frequency and with proper phase relationships with respect to each other is called a phasor diagrams. The relative position of the phasors or phase difference which is important in a.c. calculations. 

UNSYMMETRICAL FAULT CALCULATIONS – PART – 32 – DOUBLE LINE TO GROUND AND LINE TO LINE FAULT - PROBLEM

A 33 kV bus bar has a three phase fault level of 1000 MVA. The negative and zero sequence source reactances are 2/3 and 1/3 of positive sequence reactance. The zero sequence resistance of the source is 50 ohms. A 30 MVA, 33 / 132 kV solidly grounded delta/star transformer having a reactance of 0.2 per unit. Is fed from 33 kV bus. Find fault current and fault MVA at 132 kV bus for the following faults. (a) Double line-to-ground fault and (b) Line to Line fault.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 31 – THREE PHASE FAULT AND SINGLE LINE TO GROUND FAULT - PROBLEM

A 33 kV bus bar has a three-phase fault level of 1000 MVA. The negative and zero sequence source reactances are 2/3 and 1/3 of positive sequence reactance. The zero sequence resistance of the source is 50 ohms. A 30 MVA, 33 / 132 kV solidly grounded delta/star transformer having a reactance of 0.2 per unit. Is fed from 33 kV bus. Find fault current and fault MVA at 132 kV bus for the following faults. (a) Three phase fault and (b) Single line to ground fault.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 30 – LINE-TO-LINE FAULT - PROBLEM

A 25 MVA, 11 kV, three-phase, 50Hz generator has Z1 = j0.25 p,u., Z2 = j0.35 p,u., Z0 = j0.1 p.u. and Zn = 0.3 ohms. If a line to line fault occurs on Y and B phases. Calculate the fault current and per phase voltages.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 29 – LINE-TO-LINE FAULT (Zn & Zf)

UNSYMMETRICAL FAULT CALCULATIONS – PART – 29 – LINE-TO-LINE FAULT (Zn & Zf)

Consider a three-phase circuit diagram of an unloaded generator, the neutral is connected through impedance Zn and fault impedance is Zf. If a line-to-line fault between Y and B phases derive the fault current and per phase voltages.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 28 – LINE-TO-LINE FAULT

Consider a three-phase circuit diagram of an unloaded generator, if a line-to-line fault between Y and B phases derive the fault current and per phase voltages.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 27 – DOUBLE LINE-TO-GROUND FAULT - PROBLEM

Two generators A and B are of rating 50 MVA, 11kV, three-phase and 50 Hz are connected in parallel and supply a substation feeder.

Generator A and B positive, negative and zero sequence impedances are Z1= j0.6 ohms, Z2 = j0.4 ohms, Z0 = 0.2 ohms and neutral impedance Zn = j0.1 ohms.

Feeder positive, negative and zero sequence impedances are

Z1 = Z2 = (0.4 + j0.7) ohms, Z0 = (0.7 + j3.0) ohms.
If fault involves in Y and B phases, calculate the fault current.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 26 – DOUBLE LINE-TO-GROUND FAULT - PROBLEM

A 25 MVA, 11 kV, 50 Hz and three-phase generator has Z1 = j0.25 p,u., Z2 = j0.35 p,u., Z0 = j0.1 p.u. and Zn = 0.3 ohms. If a double line to ground fault occurs on Y and B phases. Calculate the line current, fault current, line to neutral voltages and line to line voltages during the occurrence of fault conditions.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 25 – DOUBLE LINE-TO-GROUND FAULT - PROBLEM

A 25 MVA, 11 kV, 50 Hz and three-phase generator has Z1 = j0.25 p,u., Z2 = j0.35 p,u. and Z0 = j0.1 p.u. If a double line to ground fault occurs on Y and B phases. Calculate the line current, fault current, line to neutral voltages and line to line voltages during the occurrence of fault conditions.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 24 – DOUBLE LINE-TO-GROUND FAULT – PROBLEM

PROBLEM

A 25 MVA, 11 kV, 50 Hz and three-phase generator has Z1 = j0.2 p,u., Z2 = j0.2 p,u. and Z0 = j0.05 p.u. If a double line to ground fault occurs on Y and B phases. Calculate the line current, fault current and line to neutral voltages during the occurrence of fault conditions

UNSYMMETRICAL FAULT CALCULATIONS – PART – 23 – DOUBLE LINE-TO-GROUND FAULT (Zs & Zf)

An unloaded generator with a fault on phases Y and B through impedance Zf to ground. The neutral of the generator is grounded through an impedance. Assume the generator is initially under no load.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 22 – DOUBLE LINE-TO-GROUND FAULT

Consider a double line-to-ground fault, calculate the fault current and phase voltages.

UNSYMMETRICAL FAULT CALCULATIONS - PART - 21 - SINGLE LINE TO GROUND FAULT - PROBLEM

PROBLEM

Two generators G1 and G2 are connected in parallel the rating of each generator is 25 MVA, 11kV, three-phase, 50 Hz generator has positive sequence reactance = X1 = 20%, negative sequence reactance = 10% and zero sequence reactance = 15%.

CASE – 01 – G1 is grounded through 5% reactance and G2 is ungrounded

CASE – 02 – G1 is grounded through 5% reactance G2 is grounded

CASE – 03 – G1 and G2 are grounded through 5% reactance

Calculate the fault current.

UNSYMMETRICAL FAULT CALCULATIONS - PART - 20 - SINGLE LINE TO GROUND FAULT - PROBLEM

PROBLEM

Two generators G1 and G2 are connected in parallel the rating of each generator is 25 MVA, 11kV, three-phase, 50 Hz generator has positive sequence reactance = X1 = 20%, negative sequence reactance = 10% and zero sequence reactance = 15%.

CASE – 01 – ungrounded neutral

CASE – 02 – grounded neutral

CASE – 03 – G1 is grounded and G2 is ungrounded. Calculate the fault current.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 19 - SINGLE LINE TO GROUND FAULT – TWO PROBLEMS

PROBLEM – 01

A star point of a 3 kV, 3 MVA three phase synchronous generator is solidly grounded. It's positive, negative and zero sequence impedances are 2.4, 0.45 and 0.3 ohms. The generator operating unloaded sustains a resistive fault between the R phase and ground. This fault has a resistance of 1.2 ohms. Calculate the fault current and the voltage to ground of the R phase.

PROBLEM – 02

A three phase 30 MVA, 33 kV alternator having X1 = 0.18 pu and X2 = 0.12 pu and X0 = 0.10 pu, based on its rating, is connected to a 33 kV overhead line having X1 = 6.3 ohms and X2 = 6.3 ohms and X0 = 12.5 ohms per phase. A single line to ground fault occurs at the remote end of the line. The alternator neutral is solidly grounded. Calculate the fault current.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 18 – SINGLE LINE TO GROUND FAULT – TWO PROBLEMS

PROBLEM – 01

A star connected alternator with earthed neutral. A single line to ground fault occurs near the generator terminal of phase R. The fault current is 300 A. Calculate the sequence currents in the other phases and also calculate the if a reactance of 0.3 ohms is connected to the neutral calculate the fault current.
PROBLEM – 02

A three phase 10 MVA, 13.2 kV generator with earthed neutral supplied a feeder. The sequence impedances of generator and feeder are mentioned below. If a single line to ground fault occurs at the far end of the feeder, calculate the fault current.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 17 – SINGLE LINE TO GROUND FAULT – PROBLEM

PROBLEM

A three-phase generator rated 25 MVA, 13.2 kV has a solidly grounded neutral. It's positive, negative and zero sequence impedances are 30%, 40%, and 5% respectively. Resistances are negligible.

(i) What value of reactance must be placed in the generator neutral so that the fault current for a line to a ground fault of zero fault impedance shall not exceed the rated line current?

(ii) What value of resistance in the neutral will serve the same purpose?

UNSYMMETRICAL FAULT CALCULATIONS – PART – 16 – SINGLE LINE TO GROUND FAULT – PROBLEM

PROBLEM

Calculate the shunt fault current and also determine the fault current when Xn = j0.15 p.u, Xn = j0.3 p.u and Xn = j0.6 p.u and also compare the results.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 15 – SINGLE LINE TO GROUND FAULT – PROBLEM

PROBLEM

A 25 MVA, 11 kV three-phase generator has a direct sub-transient reactance of 0.25 pu, negative sequence reactance on 0.35 pu and zero sequence reactance of 0.1 pu. The neutral is grounded solidly grounded. Determine the fault current and the line to line voltages for sub-transient conditions when the single line to ground fault occurs on phase R. The generator operating unloaded at rated voltage.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 14 – SINGLE LINE TO GROUND FAULT – PROBLEM

PROBLEM

A 25 MVA, 11 kV synchronous generator has Z1= 0.2, Z2 = 0.2 and Z0 = 0.05 p.u. A single line to ground fault occurs on the phase R of the generator. Find the fault current and line to line voltages during the fault condition. Assume that the generator neutral is solidly grounded and the generator is operating at no load and at rated voltage at the occurrence of a fault.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 13 – SINGLE LINE TO GROUND FAULT CURRENT USING MATRIX MANIPULATION

POSITIVE SEQUENCE IMPEDANCE (Z1)

The impedance offered by an equipment or circuit to positive sequence current is called positive sequence impedance.

NEGATIVE SEQUENCE IMPEDANCE (Z2)

The impedance offered by an equipment or circuit to negative sequence current is called negative sequence impedance.

ZERO SEQUENCE IMPEDANCE (Z0)

The impedance offered by an equipment or circuit to zero sequence current is called zero sequence impedance.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 12 PART – 12 – SINGLE LINE TO GROUND FAULT

In a balanced three phase system negative and zero sequence currents are zero.

In a three phase, four wire system the magnitude sequence current is one-third of the current in the neutral.

The neutral is not earthed the zero sequence impedance is infinity, hence the fault current is zero.

The most common type of three phase fault is the single line to ground fault (70%).

ASSUMPTIONS

1. The impedance of the fault is zero.

2. Load currents are neglected.

3. The generated e.m.f. the system is of positive sequence only.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 11

DELTA-CONNECTED WINDING

In delta connected winding no zero sequence currents flow in the lines because there is no return path for these currents.

STAR-CONNECTED WITHOUT NEUTRAL WINDING

In a three-phase, a three-wire system without a neutral network, the zero sequence line currents are zero.

STAR-CONNECTED WITH NEUTRAL GROUNDED

Three phase with neutral return zero sequence currents can flow both in the phase and windings as well as in the lines.

If the neutral is solidly grounded Zn = 0.

Total zero sequence impedance from point Ground to P is equal to the Total zero sequence impedance from Neutral to P.

The voltage to neutral or voltage to ground is the same in the case of positive and negative sequence systems.

For solidly grounded system zero sequence impedance is infinity.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 10 – PROBLEM

PROBLEM

A star connected load consists of three equal resistance of 1 ohms. The load is assumed to be connected to an Unsymmetrical three-phase supply, the line voltages are 220V, 385V and 400 V. Find the symmetrical three-phase current in the R phase by the symmetrical component method.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 09 – PROBLEMS

PROBLEM

An unbalanced three phase network of pure resistance draws currents of 10. 10 and 14.14 amperes respectively from the R, Y and B lines of a three phase balanced supply. Find the symmetrical components of the line currents, if the load is delta connected.

PROBLEM

Prove that (i) positive sequence of the line component is equal to [(a – 1) / a ] x positive sequence of the phase voltage.

Prove that (i) negative sequence of the line component is equal to [(a^2 – 1) / a^2 ] x positive sequence of the phase voltage.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 08 – PROBLEMS

PROBLEM

The line currents in R, Y and B phases are given referred to the same reference vector. Find the symmetrical components of currents.

PROBLEM

The current in the neutral to ground connection is 18 A. Calculate the zero phase sequence components in the phases.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 07 - PROBLEMS

PROBLEM

A delta connected balanced resistive load is connected across an unbalanced three-phase supply shown in the figure. Current in R and Y phases are specified. Determine the symmetrical components of the current.

PROBLEM

A delta connected load is supplied from a three-phase supply. The fuse in the blue is removed and the current in the other two lines are specified. Find the symmetrical components of currents.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 06 - PROBLEM - 04

PROBLEM -04

A balanced star connected load takes 45 A from a balanced three-phase. Four-wire supply. If the fuses in two of the lines are removed, find the symmetrical components of the currents before and after the fuses are removed.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 05 - PROBLEM - 03

PROBLEM

(a) A system of unbalanced three-phase voltages are given. Determine the three symmetrical components of the system.

(b) The positive, negative and zero sequence voltages are given. Determine the unbalanced phase voltages Va, Vb and Vc in a circuit. 

UNSYMMETRICAL FAULT CALCULATIONS – PART – 04 – PROBLEM – 02

PROBLEM – 02 

In a balanced three phase four wire system on R, Y and B line under abnormal conditions are given below. Calculate the positive, negative and zero sequence currents respectively of the line currents.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 03 – PROBLEM – 01

PROBLEM -01

A balanced star connected load takes 9 amps from a balanced three-phase, four wire supply. Calculate the positive, negative and zero sequence current in the system.

UNSYMMETRICAL FAULT CALCULATIONS - PART – 02 - FACTS ABOUT SEQUENCE CURRENTS

1. A balanced three-phase system consists of positive consequence components only.

2. In a balanced three-phase system the negative and zero sequence components being zero.

3. The presence of negative or zero sequence component introduces asymmetry in the three phase system.

4. The presence of negative and zero sequence currents are the indication of abnormality in the system.

5. The vector sum of the positive and negative sequence currents of an unbalanced three phase system is zero.

6. The vector sum of all sequence currents in three phase system unbalanced is equal to IRo + IYo + IBo.

7. Three phase four wire system zero sequence current is equal to one-third of the sum of current in the neutral wire.

8. The current of a single phase load drawn from a three phase system comprises equal positive, negative and zero sequence components.

9. In the unbalanced three phase system, the magnitude of negative sequence component cannot exceed that of the positive sequence component. The phase sequence should be reversed if the negative sequence components were the greater.

10. A delta-connected load has no neutral path and the line currents

to delta-connected load can contain no zero-sequence components.

UNSYMMETRICAL FAULT CALCULATIONS – PART – 01 - INTRODUCTION TO SYMMETRICAL COMPONENTS

When the system is unbalanced the voltages, currents and the phase impedances are in general unequal.

Such a system can be resolved by a symmetrical per phase technique known as the method of symmetrical components or three-component method. 

The analysis of unbalanced cases is greatly simplified by the use of the technique of symmetrical components.

PROFILE OF FORTESCUE

Charles LeGeyt Fortescue (1876–1936) was an electrical engineer born in York Factory (Manitoba, Canada).

Fortescue demonstrated that any set of N unbalanced phasors — that is, any such "polyphase" signal — could be expressed as the sum of N symmetrical sets of balanced phasors known as symmetrical components.

This method was proposed in the year 1918.

The paper was judged to be the most important power engineering paper in the twentieth century.

He was awarded the Franklin Institute's 1932 Elliott Cresson Medal for his contributions to the field of electrical engineering.

A fellowship awarded every year by the IEEE in his name commemorates his contributions to electrical engineering. 

He was born in York Factory (Manitoba, Canada).

FORTESCUE’S THEOREM

An unbalanced set of n phasors may be resolved into (n-1) balanced n-phase systems of different phase sequence on one zero-phase sequence system. A zero-phase sequence system is one in which all phasors are of equal magnitude and angle or they are all identical.

PHASE SEQUENCE

A phase sequence is of the phasors is the order in which they pass through a positive maximum.

R Y B represents positive sequence

R B Y represents negative sequence

The direction of phasor rotation is anticlockwise.

SYMMETRICAL COMPONENT METHOD

According to Fortescue’s theorem, three unsymmetrical and unbalanced phasor voltages or currents of a three-phase system can be resolved into the following components.

1. POSITIVE-SEQUENCE COMPONENTS

Positive phase sequence component – A balanced system of three-phase currents/voltages having positive or normal phase sequence.

Three balanced phasors, equal in magnitude and displaced from each other by 120° same phase sequence as the original phasors (for example R-Y-B)

2. NEGATIVE-SEQUENCE COMPONENTS

Negative phase sequence component – A balanced system of three-phase currents/voltages having negative or opposite phase sequence.

Three balanced phasors, equal in magnitude and displaced from each other by 120° opposite phase sequence to the original phasors (for example R-B-Y)

3. ZERO-SEQUENCE COMPONENTS

Zero phase sequence component – A system of three-phase currents/voltages having equal magnitude and having zero phase displacement.

Three unbalanced phasors of a 3-phase system can be resolved into 3 balanced systems of phasors.

Three equal phasors, equal in magnitude and zero phase displacement from each other. 
The subscripts 1, 2 and 0 are generally used to indicate positive, negative and zero phase sequence.