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.                                                  

CALCULATION OF SUB-TRANSIENT, TRANSIENT, STEADY STATE REACTANCE – PART – 05

PROBLEM
A synchronous generator is connected to an infinite bus through a 138 kV transmission line, as shown in the figure. A three phase fault occurs on the line near the CB1. 
Before the short circuit, the receiving-end voltage was 1.0 per unit, unity power factor and the generator was 80 percent loaded, on the basis of its MVA rating. Determine the sub-transient, transient, and synchronous short-circuit currents.

CALCULATION OF SUB-TRANSIENT, TRANSIENT, STEADY STATE REACTANCE – PART – 04

PROBLEM

A 10000 kVA, 13.8 kV three phase alternator connected through a 5  cycle oil circuit breaker has sub-transient reactance, transient reactance and synchronous reactance of 0.09 p.u, 0.15 p.u and 1 p.u respectively. It is running on load with rated voltage develop across the terminals when three phase symmetrical short circuit occurs just beyond the CB. Determine: (1) the sustained short circuit current[M1]  in the CB (2) the initial symmetrical rms current (3) the maximum possible d.c. component of the short circuit current (4) the momentary current rating of the CB (5) the current to be interrupted by the breaker and (6) interrupting capacity in MVA.

CALCULATION OF SUB-TRANSIENT, TRANSIENT, STEADY STATE REACTANCE – PART – 03

PROBLEM

A 20 MVA, 6.6 kV three phase alternator connected through a 3 cycle oil circuit breaker has sub-transient reactance, transient reactance and synchronous reactance of 0.1 p.u, 0.15 p.u and 0.8 p.u respectively. It is running on load with rated voltage develop across the terminals when three-phase symmetrical short circuit occurs just beyond the CB. Determine: (1) the steady short circuit current (2) the initial symmetrical current (3) the maximum possible d.c. component of the short circuit current (4) the making capacity of the CB in kA (5) the rms value of the symmetrical breaking current (6) asymmetrical breaking current and (7) the interrupting capacity in MVA.

CALCULATION OF SUB-TRANSIENT, TRANSIENT, STEADY STATE REACTANCE – PART – 02

CIRCUIT BREAKER CAPACITIES

CB capacities are quoted under specified conditions of severity e.g., power factor, recovery voltage and rate rise of recovery voltage.

1. VOLTAGE RATING

Standard voltage ratings of power circuit breakers are in terms of three phase line-to-line voltage. These ratings are based on operation at an altitude of 1,000 meters or less. For higher altitudes, the rating should be de-rated.

2. NORMAL CURRENT RATING

It is the rms value of current which the CB is capable of carrying current continuously as its rated frequency under specified conditions.

3. SHORT TIME RATING

It is the period for which the CB is able to carry fault current while remaining closed.

4. ARC VOLTAGE

It may be defined as the voltage that appears across the contact during the arcing period.

5. RECOVERY VOLTAGE

It is approximately equal to the system voltage. When contacts of the circuit breaker are opened, current drops to zero after every half cycle.

6. RESTRIKING VOLTAGE

It may be defined as the voltage that appears across the breaking contact at the instant of arc extinction.

7. RATE OF RAISE OF RE-STRIKING VOLTAGE (R.R.R.V)

It is defined as the ratio of the peak value of the restriking voltage to the time taken to reach to peak value. It is usually in kV and time in microseconds so that R.R.R.V is in kV/micro-sec.

It is one of the most important parameters as if the rate at which the dielectric strength developed between the contacts is greater than RRRV, and then the arc will be extinguished.

8. DC OFFSET

A periodic waveform has a DC offset if the average value of the waveform over one period is not zero.

9. BREAKING CURRENT

It is the current (rms) that a CB is capable of breaking at given recovery voltage and under specified conditions. (p.f, RRRV)

10. MAKING CURRENT

The making current of a CB is the peak value of the maximum current wave (including the d.c. component) in the first cycle of the current after the circuit is closed by the CB.

11. SYMMETRICAL BREAKING CURRENT

Symmetric breaking current is the rms value of the a.c component of the short circuit current at the instant of contact separation.

In India breaking current is equal to the rms value of ac component.

12. ASYMMETRICAL BREAKING CURRENT

Asymmetrical breaking current is the rms value of the total current comprising a.c and d.c components of the current, at the instant of contact separation. It is equal to the rms value of total current.

In America breaking current is equal to the asymmetrical breaking current.

13. BREAKING CAPACITY

The breaking capacity of CB is generally expressed in terms of MVA. It is the product of rated breaking current expressed in kA and the corresponding rated voltage expressed in kV.

The factor which is depending on the type of circuit for which the apparatus is intended.

1 for single phase circuit and 1.732 for three phase circuit.

[OR]

It is current (r.m.s) that a circuit breaker is capable of breaking at given recovery voltage and under specified conditions (e.g. power factor and RRRV)

14. SYMMETRICAL BREAKING CAPACITY

It is equal to 1.732 x V x I x 10^-6 MVA (Three-phase)

I = rated breaking current, V = rated service line voltage

15. MAKING CAPACITY

The peak value of current (including d.c component) during the first cycle of current wave after the closure of CB is known as making capacity. (Doubling effect value = 1.8)

Making capacity = 1.414 x 1.8 x symmetrical breaking capacity

Making capacity = 2.55 x symmetrical breaking capacity