EXPERIMENTAL SECTION L- G Fault Analysis

1.1 EXPERLMENTAL OBJECTIVE To analyze and calculate different fault currents that occur due to the introduction of faults (L-G. ) in transmission line using three phase fault analysis trainer. 1. Line Ground fault (L-G) 1.2 RELATED THEORY 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 Thevenin's 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
• Loads currents are negligible compared to fault currents
• Line charging currents can be completely neglected

1.2.1 Basic Voltage Current Network equations in Sequence Components The generated voltages in the transmission system are assumed balanced prior to the fault, so that they consist only of the positive sequence component Vf(pre-fault voltage). This is in fact the Thevenin's equivalent at the point of the fault prior to the occurrence of the fault. 1.2.2 Analysis of Asymmetrical Faults The common types of asymmetrical faults occurring in a Power System are single line to ground faults and line to line faults, with and \\ impedance. These will he analysed in the hollowing sections. 1.2.3 Single Line to Ground faults (1. - C faults) The single line to ground fault can occur in any of the three phases. I lowever, it is sufficient to analyse only one of the cases. Looking at the symmetry ()I' the symmetrical component matrix, it is seen that the simplest to analyse would be the phase a. Consider an L-G fault with zero fault impedance as shown in figure 2.15. Since the fault impedance is 0, at the fault Vaz-^-- 0, lb= 0, lc = 0 since load currents are neglected. These can be converted to equivalent conditions in symmetrical components as follows. Va= V?o+ Vai Vat = 0 Fault Suppl) side 1 irwie 2 1'3 - L-G fault on phase ii

	a 0 ja2	- oivina I 0 = (47,
and

Mathematical analysis using the network equation in symmetrical components would yield the desired result for the fault current 1r= Ia. L-G Fault with Fault Impedance In this case, la corresponds to the fault current If, which in turn corresponds to 3 times any one of the components (Iao = Iai = Ia2= Ia/3). Thus the network would also yield the same fault current as in the mathematical analysis. In this example, the connection of sequence components is more convenient to apply than the mathematical analysis. Thus for a single line to ground fault (L-G fault) with no fault impedance, the sequence networks must be connected in series and short circuited. Consider now an L-G fault with fault impedance Zf as shown in figure 2.17. at the fault 1.5 RESOURCES 1 . Auto Transformer (30, 415V) - 1 no

1. VPL 87 Trainer - 1 no
2. Multimeter -I no
3. Connecting wires - As Required

1.6 PRECAUTIONS 1. Connections should be made as per the connection diagram. 2. After creating the fault, the current should not be increased above 3 A. 3. All connections should be tight. 4. Do not touch live conductors. 5. Readings should be taken carefully. 1.7 PROCEDURE 1.7.1 Connections

1. Connect the (R, Y, B, N) terminals of 30 supply to the input (R, Y, B, N) terminals of 30 auto transformer respectively.
2. Connect the output (R, Y, B, N) terminals of 31) auto transformer to the (R, Y, B, N) terminals of Variac Out respectively.
3. Connect the terminal 'R' and 'N' (Variac Out) to the terminals of voltmeter respectively. S. In order to connect the transmission line terminals with the fault section, connect terminals of VPL-87 Trainer to the FAULT terminals (R,Y,B) of VPL-

87 Trainer respectively 1.7.2 Testing 6. Connections should be made as per the connection diagram. Initially keep all the switches in OFF position 1.7.3 Line Ground Fault

1. For creating L-G fault. connect the circuit as for 1.-C Fault
2. Connect earth terminal to the (L) terminal of 1.-(r I Al I. T section.
3. Switch ON the I(.13 power supply
4. increase the soltage upto 15 V (Vm) and put rotary switch Si is changed from Of f positxxi to ON position.

I1. Vary the auto transformer until' rated current (3A) is reached

1. Note down -- readings of the soltagc and currents.
2. Turn the rotary sw itch SI from ON position to OFF position.
3. Switch OFF the \1CB power supply .

1.8 OlisFAVATIONS AND CALCULATIONS Now calculate the fault current using formula as above it should be approximately equal to what ?e are getting in meter Ir 3E. Z. +Z, -Z here se have not used and fault impedance 1.9 INTERPRETATION RESULT NNe halve verified for L-G fault the %alue of fault current s ithout fault impedance is dependent of sequence impedances of the line and can be calculated pre fault for designing the line EXPERIMENTAL SECTION L- L Fault Analysis 1.1 EXPERIMENTAL OBJECTIVE To _analyze _and _calculate different fault currents that occur due to the introduction of faults (L-L,) in transmission line using three phase fault analysis trainer. 1. Line Line fault (L-L) 1.2 RELATED THEORY 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 Thevenin's 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
• Loads currents are negligible compared to fault currents
• Line charging currents can be completely neglected

1.2.1 Basic Voltage Current Network equations in Sequence Components The generated voltages in the transmission system are assumed balanced prior to the fault, so that they consist only of the positive sequence component Vf (pre-fault voltage). This is in fact the Thevenin's equivalent at the point of the fault prior to the occurrence of the fault. Va0= 0 ZO la Vat Ef ZI lai Vat= 0 Z2 1a2 This may be written in matrix form as

a0		0		Z0 0 0	.1.0
Val	=	E		0 Z1 0	Ict
Va2				0 0 Z2	1,2

These may be expressed in Network form as shown in figure Ef Z-'

_____________________________ > V I Positive Sequence Network

______________ > V) > Vo Negative Sequence Network Zero Sequence Network

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power-sys-lab....pdf