A Lab Manual of Power System Protectionfinal

A Lab Manual Of Power System Protection

3ee -203A Lab Manual Of Power System Protection U.V. PATEL COLLEGE OF ENGINEERINGDEPARTMENT OF ELECTRICAL ENGINEERINGGANPAT VIDHYANAGARKHERVA - 382 711

U.V.Patel College of EngineeringLABORATORYJOURNAL

M.TECH. SEMESTER II (E.E.)3EE 203 power system protection

NAME :

ROLL NO: BATCH:

U.V.PATEL COLLEGE OF ENGINEERING

GANPAT VIDHYANAGAR KHERVA 382 711DIST. :- MEHSANAPHONE NO. :- (02762) 286805

U.V.PATEL COLLEGE OF ENGINEERING

GANPAT VIDHYANAGAR KHERVA 382 711DIST. :- MEHSANAPHONE NO. :- (02762) 286805

C E R T I F I C A T E

This is to certify that Shree / Kum.

Roll No.& Exam No.Of class

has satisfactorily Completed the course in

Within four walls

Of U.V.Patel College of Engineering, Kherva.

Date of submission Staff in charge

HEAD OF DEPARTMENT INDEXSR.NO.

TITLE OF PRACTICALPAGENO.

ST.DATECOMP.DATEMARKS/ GRADSSIGN.OFSTAFF

1To study current transformer models and fault simulation

2To study the inrush phenomena when energizing a transformer

3To perform parallel operation of two single-phase transformers

4To perform the Sumpners Test on Single Phase Transformer & obtain efficiency and regulation

5To obtain Two phase supply from Three phase supply.

6To perform load test on D.C. compound generator

7To perform load test on D.C. Series Generator.

8To perform the speed control of a D.C. Shunt Motor.

9To perform magnetization characteristics of shunt generator.

10To perform the internal and external characteristics of a DC shunt motor.

11To find the performance of 3 phase induction motor on load condition.

12To study the D.C. Motor Starter.

EXPERIMENT NO: 1AIM: To study current transformer models and fault simulation.

Objectives:

Understanding models related to fault simulation. Comparing different CT models.

Procedure:T1.1 Create a new case by using either the Menu or Toolbar. A new case should appear in the Workspace settings entitled noname [psc]. Right-click on this Workspace settings entry and select Save As and give the case a name. NOTE: Do not use any spaces in the name!Create a folder called c:/PscadTraining/Faults. Save the case as case01.psc

T1.2 Open the main page of your new case. The single line diagram shown below is a part of a substation feeding a shunt reactor. The reactor is modeled in two parts to enable a fault at point B, inside the turns. The component data is as shown. (make the transformer losses zero to limit the number of nodes if using the student version)

SIMULATION CIRCUIT:

RESULT:

Figure:1 dc offset present in load voltage

T1.3 Build the case in PSCAD and enter the component data.

T1.4 Plot the current IL and the voltage EL.T1.5 Use the fault component to simulate a phase A to ground fault at location A at 0.1s.

T1.6 Observe the fault curent, IL. What is the reason for the presence of the initial DC exponential component?

T1.7 What affects the rate of decay of the DC components. Change the resistance of the short line to 1 Ohm and observe the results.

Figure 2: line current during faultT1.8 Does the instant of the fault inception have an effect on the DC offset.?

T1.9 What negative impacts can the DC offset have on the system protection.?

T1.10 Connect the phase A line current at point A to the CT model as shown below. The CT ratio is 5:400. The CT burden is 0.15 Ohms in series with 0.8mH. Plot the secondary current and the flux density.

T1.11 Increase the burden resistance to 4 Ohms and observe the results. Note the half cycle saturation effects due to the dc offset in the primary current.

Figure 3: current IL1 of C.T. 1T1.12 The reactor is protected by a differential relay scheme. Use the 2-CT model in PSCAD to connect one phase of the reactor protection scheme.

T1.13 Verify the burden current in the differential CT connection for faults at A and B.

Figure 3: current Ir1 and Ir2 of 2-C.T. model

T1.14 Does the impedance of the connection leads have an effect on the results.? How is this impedance accounted for.?T1.15 Open the case ftdiff.psc. Check the performance of the differential relay during transformer energization.

Conclusion:

----------------------------------------------------------------------------------------------------------------Date of Completion:-Signature of Faculty:-

EXPERIMENT: 2AIM: To study the inrush phenomena when energizing a transformer.Objective: Learning about the transformer and breaker operation Creating inrush phenomena when energinzing a transformer Understanding plotting and controlProcedure: Create a new case by using either the Menu or Toolbar. A new case should appear in the Workspace settings entitled noname [psc]. Right-click on this Workspace settings entry and select Save As and give the case a name. NOTE: Do not use any spaces in the name! Create a folder called c:/PscadTraining/Tutorial_01. Save the case as case01.psc T1.2 Open the main page of your new case. Build a case to study the inrush phenomena when energizing a transformer. The component data is as shown. The transformer is rated 66/12.47 kV.

Fig.1 Transformer Energizing Circuit

T1.3 Plot the currents (Ia) and voltages (E_66) on the HV side of the transformer. Note: Ia and Ea contains the three waveforms of the three phases.

Fig.2 Basic steps to create a graph with a selected signal.RESULT:

Figure:3 graph of Ia

Figure :4 graph of 66-kv Voltage

T1.4 The LV side of the transformer is not connected to a load or any other system equipment. The breaker is closed at 0.5 s to energize the transformer 66 kV side. Inrush is related to core saturation. Verify that saturation is included in the model used for this simulation.Ask your instructor to explain the large resistance connected to the HV side.Inrush current magnitude depends on the point on wave switching conditions. Use a manual switch to operate the breaker. Note the point on wave dependency of the inrush peak.

Fig.5 Two state switch attached to a control panel

.

Figure 6: current after operation of C.B.

Conclusion:


EXPERIMENT No. 4AIM: To Study Ferranti effect of Transmission Line for 400km.SPECIFICATIONS: Sending End Transformer:Pri: 0-110-220V, Sec:220V, VA:1250 Receiving End Transformer:No. Of section: 5 each section: 80 kmsEach Section : L=35 mH(5A max.)Resistance : R=2.6Capacitance : C=1.6F Line Parameters:R=0.0325/km, L=0.4375mH/km, C=0.015F/km, 0.02F and 0.035F(Selectable)For 400kms:R=13, L=175mH, C=6F, 8F & 1.68F(Selectable)

External Potential Transformers:

Pri:440V, Sec:220V, VA:25VA-3 Nos.-external for meters


Pri:10A, Sec:5A, VA:5, CL.0.5- 3 Nos.-external for meters

For OC Relay:

Pri:5A, Sec:1A, 5VA, CL 0.5

FOR SENDING END POWER ANALYZER: SE1PT: NILCT: Pri 10A, Sec: 5A, VA: 5, CL: 0.5

MCBS

Mains MCB: 20A, 2 PoleControl MCB: 2AVariac- Sending End: 6AVariac- Receiving End: 6A

METERS:

Voltmeter- 1 No.0-750VAC, Meco-SMP45S

Ammeter- 1 No.0-20A, Meco-SMP35S+

Time Interval Meter- 1 No.0.01 sec to 9999 sec, TI103C- SelectronPower Analyzer- 4 Nos.Circutor- model: CVM-NRG-96-ITF-RS485C, AUX: 230V ACPush Button: Schneider1. CB-1 ON/OFF- SENDING END2. CB-1 ON/OFF- RECEIVING END3. CB-3 ON/OFF- LOAD4. ACCEPT OR ACKNOWLEDGE FOR FAULT5. FAULT/RESET

INTRODUCTION:Ferranti Effect:Ferranti Effect occurs in transmission lines which carry electricity from one place to another. The receiving end voltage will be more than the sending end voltage. This is called ferranti effect.It occurs in longer transmission lines due to the effect of Inductance and Capacitance. The longer transmission lines draw a sufficient amount of charging current during transmission. When this line is open circuited or lightly loaded, the reactive power supplied by the capacitors is combined with the voltage at the receiving end. Hence the voltage at the receiving end is higher than the sending end.This phenomenon can be explained with following reasonings:(1) Assume no load conditions:

Reduces to

When x=l and Ir=0At l=0

As l increases, the incident component of sending end voltage increases exponentially and turns the vector anti-clockwise through and angle I whereas the reflected part of sending end voltage decreases by the same amount and is rotated clockwise through the same angle I. The sum of these two components of sending end voltage gives a voltage which is smaller than Vr.

PROCEDURE:1. Select initial length of line(say 400kms). 2. Start the supply.3. Note down the sending end voltage Vs, sending end current Is and receiving end voltage Vr.4. Reduce the voltage and disconnect the supply.5. Decrease the length of line. 6. Repeat the steps 2-5.

RESULTS:The experiment was carried out for line lengths of 400 Kmtransmission lines. In each case the voltage of sending end and receiving end for three phases were observed. The sending end and receiving end were tabulated.In every case it is observed that the value of voltage increases from sending endside to receiving end side at no load condition.

Sending End ParametersReceiving End Parameters

CVsIsCos VrIrCos

0.6220.00.5100.0391233.50090

0.8219.70.6860.090237.90090

1.4221.80.1800.090253.90090

CONCLUSION:

The over voltage can be confirmed due to ferranti effect along the transmission lineLength. Itoccurs when the line is energized but there is a very light load or the load is disconnected. The effect is due to the voltage drop across the line inductance being in phase with the sending end voltages. Therefore inductance is responsible for producing this phenomenon. The ferranti effect will be more pronounced the longer the line and the higher the voltage applied. From the knowledge of ferranti effect and by compensating this effect the temporary overvoltage in the transmission line can be reduced and thus the line can be protected.

EXPERIMENT No. 5AIM: To study about the series compensation of a transmission line.SPECIFICATION:1. Sending End Transformer:Pri: 0-110-220V, Sec:220V, VA:12501. Receiving End Transformer:No. Of section: 5 each section: 80 kmsEach Section : L=35 mH(5A max.)Resistance : R=2.6Capacitance : C=1.6F1. Line Parameters:R=0.0325/km, L=0.4375mH/km, C=0.015F/km, 0.02F and 0.035F(Selectable)For 400kms:R=13, L=175mH, C=6F, 8F & 1.68F(Selectable)Capacitor Selector Switch- 5 nos. 1 for each sectionPOS1: C=0.6FPOS2: C=0.8FPOS3: C=1.68F1. Loading Units:RLC Unit-1:Variable Potentiometer: 1k, 100WResistance Bank: @ 220V SW1: 5k/10W & 10k/5W SW2: 2k4/20W SW3: 1k2/40W SW4: 605E/80W SW5: 302E/160WInductor Bank: @ 220V, 7.5VA to 112.5VA SW1: 7.5VA SW2: 15VA SW3: 15VA SW4: 30VA SW5: 45VACapacitor Bank: SW1: 0.8F SW2: 1.2F SW3: 3.15F SW4: 6F SW5: 8F

RLC Unit-2:Variable Potentiometer: 1k, 100W

Resistance Bank: SW1: 2k/20W SW2: 969E/50W SW3: 484E/100W SW4: 242E/200W SW5: 141E/400WInductor Bank: SW1: 15VA SW2: 30VA SW3: 60VA SW4: 60VA SW5: 10mH-unity pf

Capacitor Bank: SW1: 1.2F SW2: 3.15F SW3: 6F SW4: 8F SW5: 12.5F





Internal CT for Impedance Relay:

Pri:7A, Sec:3.5A, VA:5, CL.0.5- (10A primary max.) 1 No.

For OC Relay:



MCBS


Output Voltage Selector Switch: 1 Pole 12 Way, 6A

All Toggle Switch: 15AAll Connectors: BTI-15, Elcom Make

Impedance Relay: High Voltage Make, Aux: 230V AC (-15%,+10%)

Over Current Relay: L&T, CT Sec. 1A, Aux. 90-240V AC/DC

METERS:



Time Interval Meter- 1 No.1. sec to 9999 sec, TI103C- Selectron.Power Analyzer- 4 Nos.Circutor- model: CVM-NRG-96-ITF-RS485C, AUX: 230V AC(4.2VA, -15%, +10%), I/P: 240Fault Switch- 2 Pole, 2Way, WO/O, 32A- SalzerPush Button: Schneider1. CB-1 ON/OFF- SENDING END1. CB-1 ON/OFF- RECEIVING END1. CB-3 ON/OFF- LOAD1. ACCEPT OR ACKNOWLEDGE FOR FAULT1. FAULT/RESETINTRODUCTION:The line impedance, which consists of resistance and inductive reactance, limits line performance in several important ways:1. The resistance causes real power (I2R) losses, and IR voltage drop.1. The inductive reactance causes reactive (I2X) losses and IX voltage drop.1. The inductive reactance limits the amount of power that can be transferred:The series impedance, which consists of resistance and inductive reactance, limits line performance in several important ways:1. The resistance causes real power (I2R) losses, and IR voltage drop.1. The inductive reactance causes reactive (I2X) losses and IX voltage drop.1. The inductive reactance limits the amount of power that can be transferred:

Since X is in the denominator, reducing the X increases the Pmax. Alternatively, reducing the X reduces the amount of voltage phase angle displacement required to achieve a certain power transfer. This tends to help improve the power system dynamic stability.Series compensation is most effective on the higher-voltage transmission lines (230 kV and above) because they have relatively high ratios of series X to series R; typically 7 to 20 or so. For example, the typical 345 kV line has an X/R ratio of about 10 and a 500 kV line has an X/R of about 18.Series capacitors are an important and valuable technique for improving the electrical performance of a line, and the power system of which it is a part. The reduction achieved in net series line impedance improves power system performance from the standpoint of both dynamic stability and voltage stability. Unlike shunt capacitors, series capacitors do not ordinarily need to be switched in or out in response to changes in power system conditions, and their reactive output is not dependent on bus voltage.

PROCEDURE:

1. Open circuit the receiving end.1. Adjust the sending-end voltage to 220V. Note the receiving-end voltage.1. Introduce the compensation equipment and adjust its L and C till the receiving end voltage also reached 220V. (Vs=Vr)1. Note the current taken by the compensator (which includes L and C in parallel ) and determine its Var rating and its ohmic valueNOTE: METER DISPLAY WILL READ LAG = +SIGN, LEAD IS VE SIGN.LINE COMPENSATION AT NO LOAD USING L AND C, C=0.015MFD / KMSENDING END PARAMETERS.

EXPERIMENT No. 6AIM: To determine the shunt compensation to counteract the voltage rises on no-load.SPECIFICATION: Sending End Transformer:Pri: 0-110-220V, Sec:220V, VA:1250 Receiving End Transformer:No. Of section: 5 each section: 80 kmsEach Section : L=35 mH(5A max.)Resistance : R=2.6Capacitance : C=1.6F Line Parameters:R=0.0325/km, L=0.4375mH/km, C=0.015F/km, 0.02F and 0.035F(Selectable)For 400kms:R=13, L=175mH, C=6F, 8F & 1.68F(Selectable)Capacitor Selector Switch- 5 nos. 1 for each sectionPOS1: C=0.6FPOS2: C=0.8FPOS3: C=1.68F Loading Units:RLC Unit-1:Variable Potentiometer: 1k, 100WResistance Bank: @ 220V SW1: 5k/10W & 10k/5W SW2: 2k4/20W SW3: 1k2/40W SW4: 605E/80W SW5: 302E/160WInductor Bank: @ 220V, 7.5VA to 112.5VA SW1: 7.5VA SW2: 15VA SW3: 15VA SW4: 30VA SW5: 45VACapacitor Bank: SW1: 0.8F SW2: 1.2F SW3: 3.15F SW4: 6F SW5: 8F










For OC Relay:



MCBS






METERS:



Time Interval Meter- 1 No.0.01.1 sec to 9999 sec, TI103C- Selectron.Power Analyzer- 4 Nos.Circutor- model: CVM-NRG-96-ITF-RS485C, AUX: 230V AC(4.2VA, -15%, +10%), I/P: 240Fault Switch- 2 Pole, 2Way, WO/O, 32A- SalzerPush Button: Schneider11. CB-1 ON/OFF- SENDING END12. CB-1 ON/OFF- RECEIVING END13. CB-3 ON/OFF- LOAD14. ACCEPT OR ACKNOWLEDGE FOR FAULT15. FAULT/RESET

INTRODUCTION:Shunt compensation:Shunt compensation is used by utilities at both transmission and distribution levels.

Fig 1: shunt compensation In the shunt compensation applications, passive shunt capacitors are extensivelyused for power factor correction and to improve the system voltage profile by economicalmeans. The principal advantages of shunt capacitors could be listed as their low cost, flexibility on installation and practical operations. To show the effect of shunt compensation on the system voltage profile, a simple two-bus system given in Fig. 1 is used and a shunt capacitor is added.. The effect of shunt compensation can be introduced using Thevenins theorem by keeping the load phase angle at a constant value. For this case, the Thevenin equivalent voltage (Vth) and impedance (Zth) seen from the end of the line (point of r) can be written for each line model, given in Fig. 1, by neglecting the active losses of the shunt capacitor.

PROCEDURE:

Open circuit the receiving end. Adjust the sending-end voltage to 220V. Note the receiving-end voltage. Introduce the compensation equipment and adjust its L and C till the receiving end voltage also reached 220V. (Vs=Vr) Note the current taken by the compensator (which includes L and C in parallel ) and determine its Varrating and its ohmic valueNOTE: METER DISPLAY WILL READ LAG = +SIGN, LEAD IS VE SIGN.LINE COMPENSATION AT NO LOAD USING L AND C, C=0.015MFD / KMSENDING END PARAMETERS

RESULTS:CBefore CompensationBefore CompensationVAR Load

SE-VsRE-VrSE-VsRE-VrVIPFKW

220.0 V232.3220.0 V232.3232.300

R213.6213.6213.6213.60000

L00002200.2760.8

C0000000

LINE COMPENSATION AT NO LOAD, C=0.015 MFD/KM.CBefore CompensationBefore CompensationVAR Load


0.6220.0 V232.3220.0 V232.32200.17200.02

LINE COMPENSATION AT NO LOAD USING L AND CC=0.015 MFD/KMCBefore CompensationBefore CompensationVAR Load


220.0 V232.3220.0 V232.3232.300

R213.6213.6213.6213.60000

L00002200.2760.8

C0000000

INPUT SE: INPUT: 221.3 V, I=0.316 A, W=46, VARC: 48, VA:69, PF:0.67, COS :0.70RE: 221.4 V, 0.277 A,49 W, 37 VARL, 61 VA, PF: 0.8, =0.8INDUCTOR 15 VA + 30VA + 0.8 (V RESISTANCE ADDED)ITS OHMIC VALUE IS: 638.611

CONCLUSION:After performing this experiment we can conclude that voltage sensitivities of load have significant effects on receiving end voltage magnitude of the line, transferred power and power losses for different sizes of series and shunt capacitors. Effects of shunt and series capacitors on the load voltage increase when voltage sensitivities of loads decrease. It is concluded that for a fixed load voltage level, the required shunt compensation levels decrease with increasing voltage sensitivity of the loads.

EXPERIMENT No. 7AIM: To study about the fault occurs in Transmission Line.SPECIFICATION: Sending End Transformer:Pri: 0-110-220V, Sec:220V, VA:1250 Receiving End Transformer:Pri: 0-220V, Sec:110,130,160,170,190,210,220,240V, VA:1000No. Of section: 5 each section: 80 kmsEach Section : L=35 mH(5A max.)Resistance : R=2.6Capacitance : C=1.6F Line Parameters:R=0.0325/km, L=0.4375mH/km, C=0.015F/km, 0.02F and 0.035F(Selectable)For 400kms:R=13, L=175mH, C=6F, 8F & 1.68F(Selectable)Capacitor Selector Switch- 5 nos. 1 for each sectionPOS1: C=0.6FPOS2: C=0.8FPOS3: C=1.68F Loading Units:RLC Unit-1:Variable Potentiometer: 1k, 100WResistance Bank: @ 220V SW1: 5k/10W & 10k/5W SW2: 2k4/20W SW3: 1k2/40W SW4: 605E/80W SW5: 302E/160WInductor Bank: @ 220V, 7.5VA to 112.5VA SW1: 7.5VA SW2: 15VA SW3: 15VA SW4: 30VA SW5: 45VACapacitor Bank: SW1: 0.8F SW2: 1.2F SW3: 3.15F SW4: 6F SW5: 8F










For OC Relay:



MCBS






METERS:



Time Interval Meter- 1 No.0.01.2 sec to 9999 sec, TI103C- Selectron.Power Analyzer- 4 Nos.Circutor- model: CVM-NRG-96-ITF-RS485C, AUX: 230V AC(4.2VA, -15%, +10%), I/P: 240Fault Switch- 2 Pole, 2Way, WO/O, 32A- SalzerPush Button: Schneider16. CB-1 ON/OFF- SENDING END17. CB-1 ON/OFF- RECEIVING END18. CB-3 ON/OFF- LOAD19. ACCEPT OR ACKNOWLEDGE FOR FAULT20. FAULT/RESET

INTRODUCTION:High-voltage systems have an effectively grounded neutral. When a ground faultoccurs on an overhead transmission line, in a power network with grounded neutral, the fault current returns to the grounded neutral through the tower structures, ground return paths and ground wires. It is based on Kirchhoffs theorems, an analytical method in order to determine the ground fault current distribution in effectively grounded power network is presented. A phase-to-ground fault that appears on a phase of a transmission line, dividesthe line into two sections, each extending from the fault towards one end of the line. PROCEDURE:2. Select initial length of line (say 400kms). 2. Start the motor-generator set.2. Note down the sending end voltage Vs, sending end current Is and receiving end voltage Vr.2. Disconnect the supply of motor-generator set.2. Measure the distances when fault will occur and note down the trip time.2. Decrease the length of line. 2. Repeat the steps 2-5.RESULTS:Sr. No.Distance when fault occur in kmsTrip time in scc.Trip current

1.80

2.160

3.240

4.320

5.400

CONCLUSION:

EXPERIMENT-1Aim:-To study the operation of Electro mechanical type an inverse time over current relay.Apparatus required:-1. VPL-102A module1. Power chord -1 no1. Patch chord -2 nosConnection procedure:-1. Connection the patch chord from electro mechanical type relay contact that may be NO or NC depending upon provisions of contacts in the relay. Some manufactures may provide both NO and NC contact.1. Keep the MCB ON/ OFF switch in OFF position.1. Keep the current adjustment knob in minimum position.1. Connect the power chord at the back side of VPL-102A module.Experiment procedure:-1. Select the desired plug setting of electro mechanical relay by using plug. Now switch ON power ON/OFF switch and keep MCB ON/OFF switch in ON position.1. Reset the digital stop clock.1. Press the start switch S1. Now the start LED glows to indicate the relay in START mode.Procedure to check the relay in normal conditions (Without occurrence of fault):-1. By using current adjustment knob of VPL-102A module set the actual current below the plug setting value. Here we say plug setting value is 8A and actual current is 3A.1. Here the actual current (3A) is below the plug setting value (8A) i.e. normal condition which means there is no fault in the system. So the relay does not operate.Procedure to check the relay in abnormal conditions (With occurrence of fault):-1. Repeat the steps 5 to 7.1. By using current adjustment knob of VPL-102A module set the actual current above the setting value. Here for example plug setting value is 8A and actual current is 15A.1. Press the STOP button and reset the digital stop clock. After relay disc operating is stopped press the START button. 1. Here the actual current (15A) is above plug setting value (8A) i.e. abnormal condition which means there is fault in the system. So the disc of the relay rotates and the moving contact touching the fixed contact and relay proceeding the tripping of circuit.1. After relay is tripped the relay time is stopped in digital stop clock.1. By setting the various plug setting values, perform the relay in abnormal condition and tabulated the readings.

TABULATION:-[1] plug setting = 2.5 amp.S.NO.PLUG SETTING CURRENT(A)ACTUAL RELAY CURRENT(A)TIME TAKEN TO TRIP THE RELAY(S)

13.979.936.24

25.3813.455.05

32.125.3010.60

[2] plug setting = 3 amp.S.NO.PLUG SETTING CURRENT(A)ACTUAL RELAY CURRENT(A)TIME TAKEN TO TRIP THE RELAY(S)

13.1715.866.58

22.010.0010.13

31.2086.0428.50

[3] plug setting = 7.5 amp.

S.NO.PLUG SETTING CURRENT(A)ACTUAL RELAY CURRENT(A)TIME TAKEN TO TRIP THE RELAY(S)

11.331019.48

22.0215.039.73

32.6719.967.38


EXPERIMENT 2AIM :-To study the operation of micro controller based earth fault relay using VPL -04 Module. APPARATUS REQUIRED :-1. VPL-04 MODULE2. VPL-01 ACC3. Power Chords -2 nos4. Patch Chords 3 nos5. PC interface cable 1 no

CONNECTION DIAGRAM :-

CONNECTION PROCEDURE:-1. Connections should be given as shown in Figure 4.1 2. Connect output terminal (1) of variable AC current sources to NC contact (3) of earth fault relay.3. Connect output terminal (2) of variable AC current sources to current coil input (L2) of earth fault relay.4. Connect NC contact (4) of earth fault relay to current coil input (L1) of the earth fault relay.5. Connect the power chords at the back side of unit of variable AC current source and microcontroller based earth fault relay.6. Connect serial communication port of earth fault relay to PC by using PC interface cable .7. Keep the current source variac at zero position .8. Now switch on the power supply of variable AC current source and earth fault relay.

TABULATION:-

FOR DMT MODE:-

S.NOSet Current (A)Fault Current (A)Set Time(Sec.)Actual relay tripping Time (Sec)

10.51.0855.05

20.52.025.55.5

30.51.086.56.5

FOR IDMT MODE :-

[1] Take set current and fault current constant.S.NOSet Current (A)Fault Current (A)Set Time(Sec.)Actual relay tripping Time (Sec)

10.51.090.54.57

20.51.090.76.48

30.51.090.98.33

[2] Take set current and set time constant.S.NOSet Current (A)Fault Current (A)Set Time(Sec.)Actual relay tripping Time (Sec)

112.020.54.37

212.560.53.72

313.040.53.14

412.00.76.64

512.450.75.48

612.970.74.5


A Lab Manual of Power System Protectionfinal

Documents

fault simulation

fault component

load voltage t1

current transformer

dc shunt motor

line current

fault inception

ground fault