Power System Protection Manual Note: This manual is in the formative stage. Not all the experiments have been covered here though they are operational in the laboratory. When the full manual is ready, we will make it available here. Electrical Engineering Department, Birla Vishvakarma Mahavidyalaya (BVM) Engineering College, Vallabh Vidyanagar, Gujarat, India Pin:388120.
Experiment No-1 Radial Feeder Protection Theoretical Background: Whole of the power system can be subdivided in to number of radial feeders fed from one end. Generally such radial feeders are protected by over current and earth fault relays used as primary relays for 11 kv and 66 kv lines. For lines of voltage rating beyond 66 kv, distance protection is applied as a primary protection whereas over current and earth fault relays are used as back up relays. A simplified radial feeder network without transformers (in actual practice transformers do exist at substations) is shown in single line diagram of fig. 1.1 below. Section I Section II Section III S So urce A B C D Transmission Line R1 R2 R3 Fuse Fig. 1.1. A Typical Radial Transmission Line To Load If the fault occurs in distribution network, fuse should isolate the faulty section. Should the fuse fail, relay R 3 shall give back-up protection. Relays R 1, R 2, and R 3 act as primary relays for faults in section I, section I, and section III respectively. If fault in section III is not cleared by relaying scheme at relaying point R 3, relay R 2 will act as a back-up. Similarly back-up protection is provided by relay R 1 for faults in section II. A,B, C and D are substations in fig. 1.1. Generally Inverse time overcurrent relays with Definite Minimum Time feature (IDMT relays) are used in practice. There are many types of such relays available in relaymarket, viz. normal inverse relays, very inverse relays and extremely inverse relays. The characteristics of these relays are shown in fig. 1.2. The other types of o/c relays are 3 second relay and 1.3 second relay. This means the time of operation of the relay is either 3 or 1.3 second at Plug Setting Multiplier (PSM) equal to 10. Long time inverse relays are used for o/c cum overload application. Voltage restrains o/c relays have their own application. Very inverse relays are less prone to the ratio Z S /Z L. Extremely inverse relays are yet better. Very inverse relays are faster in operation for close-in faults yet maintaining the discrimination with fuse and other relays. Extremely inverse relays are more meritorious 1
in this aspect too. Instantaneous o/c relays are not immune to Z S /Z L ratio. Definite time o/c relays are 100 % immune to this ratio. Very inverse relays can be used with an additional advantage while protecting a machine or a transformer as they match with the heating characteristic of equipment better than their normal inverse equivalent. Extremely inverse relays can best co-ordinate with the fuse characteristic. The aim of this experiment is to reveal these facts experimentally. Fig. 1.2 Normal, Very and Extremely Inverse Characteristics L aboratory Simulations: Referring to a.c. circuit of fig. 1.3. a live model of a radial feeder fed from one end can be self-understood. Section - I Section - II Section - III A C1-1 C2-1 C3-1 10/5 9Ω 10/5 9Ω 10/5 230 V, 50 Hz 1-phase, ac supply V R1 R2 R3 S1 S2 S3 MCB 550Ω (Load) 18Ω (Fault resistance) Fig. 1.3 Main AC Circuit in the Experiment. 2
This is only a single phase version of a radial feeder. Transmission lines are simulated by 9 ohms resistors as we are studying only the steady state behavior of the relays and the network. Circuit breakers are simulated by contactors. Distributor is protected by a 0.5 Amp M.C.B. Semaphore indicators on the panel show the status of the contactor(whether ON or OFF). Visual neon lamp indictors are also used. Faults in different sections can be created by switches S1, S2 and S3. Fault limiting resistance of 18 ohms is used for practical purposes only, as otherwise the source would get shorted for a fault at start of the first section. For fault in distributor, Ohmic value of load rheostats can be decreased. MCB simulates fuses or MCCB. In actual practice C.T. secondary rated current (1 Amp or 5 Amp) and relay rating should be same. Here C.T. secondary rating is 5 Amp and relay rating is 1 Amp. This is contradicting the practice for for practical purpose. A1-1 A2-1 A3-1 PB-3 PB-3 T2 PB-3 T3 11 0 V DC Supply L1 L2 L3 C1 C2 C1-2 C2-2 PB-1 PB-1 PB-2 PB-2 C3 A1-2 C3-2 PB-1 PB-2 A1 R1-1 A2-2 A2 R2-1 A3-2 A3 R3-1 (a) C1-3 C2-3 C3-3 110 V DC Supply S-1 S-2 S-3 bulb Buzzer A1-3 A2-3 A3-3 S-1, S-2, S-3 are semaphore indicators (b) Fig. 1.4 Control Circuit in the Experiment. 3
Referring to control circuit of figure 1.4(a) and (b), any section can be manually charged or made off using start (PB 1 ) and stop (PB 2 ) push buttons, which are spring loaded. On occurrence of fault, the corresponding section relay will operate and the concerned auxiliary relay A 1, A 2 or A 3 will energize giving signal to the concerned contactor and making it off. This will also activate the buzzer and bulb which can be reset using Accept pushbutton PB 3. Back up can be shown by using switches T 1 and T 2 on the panel. Time of operation of relays can be measured by a time interval counter connected as shown in fig. 1.4 (c). S1 A1-4 S2 S3 A2-4 A3-4 To timer To timer Start Stop terminals terminals (c) Fig. 1.4 Control Circuit in the Experiment. O bservations and calculations: 1. Measure the fault currents for extreme faults in sections I, II and III by adjusting the corresponding rheostat in minimum (zero resistance) and maximum (full resistance) positions and using the corresponding fault-switch S 1, S 2, or S 3 (refer fig. 1.3). Maximum fault currents in sections I, II and III are denoted by F 1, F 2, F 3 respectively and the minimum fault currents by F 1, F 2, F 3 respectively. Deactivate the relays for this purpose. Record the readings in table 1: Fault Location F 1 F 1 F 2 F 2 F 3 F 3 Fault Current (A) Table:1.1 Measured fault currents for extreme faults in each section 2. Calculate the plug settings (or Tap Value) of relays R 1, R 2 and R 3. (Plug settings will be same irrespective of type of relays ). For this purpose, assume the number of distributors each of 0.5 Amp rating from the following possibilities (The experiment 4
3. is a simulation and hence the distributor current is 0.5 Amp. In actual practice it may be 500 Amp or more or less): Number of distributors: 1, 2, 3 or 4. (Calculations of the plug setting shall be done w.r.t. following considerations.) (I) The plug setting shall be more than or equal to the maximum full-load current passing through the relay. (II) The pick up of the relay varies from 1.05 to 1.3 times the plug setting of the relay. (III) For back up, relay R 1 shall reach for the fault F 2 and R 2 for the fault F 3. Time Settings. (Normal Inverse Relays) For deciding time-settings, co-ordinate the characteristic of relay R 3 with that of an MCB.( Refer figures 1.5 and 1.6.) Use discriminating time-interval of 1.0 second between two characteristics. This will decide TMS of R 3. Why discrimination time of 1.0 second? (Try answer to this question.) For coordinating R 2 with R 3, use the worst possible current to decide TMS of R 2 (F 3 in table:1.1). Similarly decide TMS of R 1 by using F 2 and setting of R 2. Use discriminating time interval between two successive relays as 0.4 seconds (why 0.4 seconds?) for these calculations. Tabulate the results as follows: Normal inverse Relays. Relay R 1 R 2 R 3 P.S. Table:1.2 Calculated relay settings. TMS 4. Time-settings (Very Inverse Relays) Similar exercise as at Sr. No. 3 above can be carried out for very inverse relays also. The results are to be tabulated as in the case of normal inverse relays as per calculated settings. Use figure 1.5 and 1.7 for deciding TMS of very inverse relays. 5. Set the normal inverse relays as per the settings in table 1.2. 6. Calculate the time of operation of the main and back-up relays for extreme faults in each section using the relay settings in table:1.2 and the fault current readings recorded in table:1.1. Enter these in table:1.3 as Calculated Time of Operation. 7. Vary the 550 ohms load resistance such that the current in the radial feeder varies from 0.5 Amp to about 4 Amp. See that MCB trips and relay R 3 does not trip. 8. Now create extreme faults (one by one) in each section starting from section III. For each fault, measure the time of operation of the main and the back-up relay (to measure time of operation of back-up relay, the main relay has to be deactivated using switches T 2 or T 3 (as the case may be) in fig. 1.4). Record these in table 1.3 as Measured Time of Operation. Calculate the error between the calculated and the measured time of operation for each fault and record it in table: 1.3. 9. Derive the table 1.4 from table 1.3 as follows: 5
Relays Table 1.4 Difference in time of operation of relays for extreme faults. Section Observed Calculated R 1 R 2 R 3 I II III 10. Replace the normal inverse relays by very inverse relays. 11. Repeat steps at Sr. No. 5 to 9 for very inverse relays. 12. Draw your own conclusion. Questions: 1. Explain the circuit of the experiment. 2. What is the function of semaphore indicator? 3. Why are the settings of the earth-fault relays lower than the settings of the overcurrent relays? 4. Draw an a.c. circuit and d.c. control circuit for two overcurrent and one earth-fault scheme of protection of a feeder used in practice. How does our experimental scheme differ from that? Why? 5. What do you understand by time discrimination? 6. What do you understand by overshoot of a relay? 7. What is the significance of resetting time of a relays? 8. What do you understand by back-up protection? Explain remote back-up protection. 9. How does the source impedance affect the choice of relay to be used in radial feeder protection? 10. Why are the IDMT relays popular in practice? 11. What are the factors to be considered for deciding settings of phase relays and ground relays? 6
7
8
EXPERIMENT NO. 2 PARELLEL FEEDER PROTECTION Theoretical Background: In power systems, power is also fed by parallel feeders. A simplified single-line diagram of parallel feeder network (without transformers) is shown in fig. 1. Referring to the fig. 1 let the fault occur at F 1. There are two paths for the fault current. (I) AF 1 and (II) DCBF 1 1 F A 1 2 B S Source D R 1 3 F 2 4 R 3 C To Load R 2 R 4 Fig. 1 Parallel feeder network It is required that for fault at F 1, CB 1 and 2 should trip. Similarly CB 3, 4 should trip for fault at F 2. In this scheme if we use all overcurrent relays, the following problem arises: If fault occurs at F 1, R 3 should operate and R 4 should give back-up to relay R 3. This means, time of operation of R 4 should be more than that of R 3, current being same. Let the settings be done according to this criterion. Now, for the fault at F 2, it is required that relay R 4 operate and relay R 3 operates to provide backup. This fact requires that relay R 3 should operate later than R 4. But the relay settings are not satisfying this condition. Hence by using simple IDMT o/c relays it is impossible to fulfill both the criteria. To solve this problem we have to introduce an additional directional feature in the relays R 3 and R 4. These relays will operate only if the fault current is flowing away from the bus. Relays R 1 and R 2 are kept as non-directional IDMT o/c relays. Now, for the fault at F 2 since the current through R 3 is in the non-operative direction, the relay doest not operate. Relay R 4 operates as the directional criterion is satisfied. Hence for fault at F 2, R 2 and R 4 operate and if R 4 fails to operate R 1 provides the backup. Relays operate on similar lines for fault at F 1. 1
L aboratory Simulations: Referring fig. 2, a model for parallel feeders fed from one end is self explanatory. This is a single-phase version of the parallel feeders. One feeder is simulated by two 9 Ω, 12 A rheostats in series. Load is simulated using a 185 Ω, 1.1 A rheostat. Faults are simulated using switches S 1 and S 2 and a limiting resistance of 9Ω, 12 A is used to limit the fault current. The control circuit is similar to that in the radial feeder simulation (Refer to fig. 3, 4 and 5) except for the additional PT to feed directional overcurrent relay. Remote backup can be seen by closing a CT secondary shorting switch at the rear of the panel (thus, making the corresponding relay non-operational). O bservations and Calculations: Calculate the Plug Setting P.S (or Tap Value T.V.) and T.M.S. of relays theoretically such that if one line trips the other healthy line should be able to feed the full power till the load is shed. Remember relay R 1 should provide backup to R 4 and R 2 to R 3. Select some arbitrary T.M.S. for R 3 and R 4 and find the T.M.S. of R 1 and R 2 by keeping proper discrimination time arrival. Tabulate the settings as follows: SR NO Relay P.S. T.M.S. 1 R 1 2 R 2 3 R 3 4 R 4 Procedure: 1. Do the connections as per the circuit diagram. 2. Set the relays according to your calculations. 1. 3 Create the faults and observe whether relays are operating in desired manner. 3. Observe the backup protection using C.T. shorting switch at rear of the panel. Q uestions: 1. Draw at least three single line diagrams of power system network where Directional feature is required. 2. What are the problems of Directional relays. 3. What do you understand by Zero Torque Line, Maximum Torque Line, MTA characteristic, Dead Zone, Characteristic Angle? 4. Out of 30 0, 60 0 and 90 0 connection, which would you prefer? 5. Draw figures for phase advancing techniques in phase and ground directional relays. 2
EXPERIMENT NO.3 Principle of Reverse Power Protection: Motivation: Reversal of power takes place when turbine does not get steam or water (in thermal/hydro plants). When alternator is connected to infinite bus and its turbine fails, machine works as synchronous motor driving the turbine as a pump; thus direction of current reverses. This experiment is designed to understand and demonstrate all possible conditions of prime-mover failure. Theoretical Background: When prime mover fails (steam input to turbine gets lost), generator works as a synchronous motor driving the turbine as the pump but obviously drawing a very small current w.r.t. its rated value. This condition is not dangerous for the alternator. The damage occurs to the prime mover. In modern power plants there are low forward power relays rather than reverse power relays. This protection is very important as not only it protects the prime mover, but also all class-b protections are routed through this protection (low forward power protection). Experimental Model: The main circuit is shown in fig. 1. The normal operating condition is achieved by keeping the switches S 1 and S 2 open. In this case, there are two parallel paths, each with 9 Ω resistance feeding a load simulated by a 185 Ω rheostat. The two paths model two 240V generators in parallel feeding a load of 185 Ω. The generator currents are measured through CT 1 and CT 2 and fed to relays D 1 and D 2 respectively. D 1 and D 2 are reverse power relays (in fact, low forward power relays to be precise), and hence are also provided with a voltage input through a Voltage Transformer (VT) shown in fig. 1. It is important to note here that D 1 and D 2 remain energized during normal condition. Closing of switch S 1 will make the current through CT 1 equal zero. Closing of S 2 will make the current through CT 2 equal zero. Either of these conditions simulates primemover failure and relay will de-energize in response. S 1 and S 2 are Double Pole Double Throw (DPDT) switches used simultaneously in the main (fig. 1) and control (fig. 2) circuits. The control circuit is shown in fig. 2. It shows that the contactors can be energized by push buttons PB1 and de-energized by PB2, same as for other experiments. The switch S 3 simulates the steam valve interlock. This interlock does not allow the generator to start if the steam valve feeding steam to the turbine is closed. S 3 should be kept open to bypass the lock, i.e., to let contactors close. Procedure: 1. Connect the required circuit. Keep S 3 open. This means A 1 and A 2 can never energize. So, A 1-1 and A 2-1 are always closed. The steam valve interlock is bypassed. 2. Make the contactors ON with pushbuttons. Circuit will energize and relays D 1 and D 2 will operate. Now close S 3. This means that in the control circuit of fig. 2, D 1-1 and D 2-1 will close energizing D 1 X/D 2 X. Hence contacts D 1 X-1 or D 2 X-1 will open out, not allowing auxiliary relays A 1 and A 2 to operate. Thus, contactors remain ON. 1
3. Now to simulate prime-mover failure, make switch S 1 on. Relay D 1 will de-energize, hence D 1 X, deenrgizes. D 1 x-1 will close and the auxiliary relay A 1 will operate. A 1-1 will isolate the faulty turbo alternator from bus. Buzzer, bulb and semaphore indicators operate the same way as for other experiments. 4. Open S 1. Then open S3. Now make contactor C 1 On with push button. 5. Close S 2 to simulate zero power in alternator-2. On similar lines to what is explained in steps 1-3, contactor C 2 will open. 6. Draw your own conclusion. 2
3
EXPERIMENT NO: 4 DIFFERENTIAL PROTECTION OF TRANSFORMER Theoretical Background: The transformer is one of the most important devices and a very expensive equipment of a power system. The internal fault current due to short-circuit of winding within a transformer results in enormous damage to the winding and the core. Hence we use the most reliable, fast and the selective protective scheme, the differential protection scheme, which operates only for the internal faults. Some problems envisaged with differential protection scheme of transformer are listed below with their remedies: 1. The voltage ratings of both sides of transformer are different; hence the CTs used are not identical. So there may be some spill current flowing through the relay that may cause mal-operation. Remedy: Use biased differential relay. 2. The full load current of transformer primary and secondary are different. CT should be selected such that the pilot wire current is the same on both sides otherwise maloperation of relay would occur. To achieve this, the CT ratios required on primary and secondary side are different from the standard ratio. Remedy: CTs of standard ratios are employed in conjunction with interposing CTs. 3. At no load, the transformer secondary current is zero but the magnetizing current flows in the primary. Hence mal-operation of relay may occur due to the presence of spill current. Remedy: As the magnetizing current is only 1-2% of the rated current the basic setting of relay is done such that it does not operate at no-load. 4. During heavy external faults, CT ratio error will result in significant spill current flowing through the operating coil of the relay. Hence the relay can mal-operate. Remedy: Use biased differential relay. 5. 5.The power transformers are always provided with a tap-changing arrangement to regulate output voltage as demanded by the load. But CTs are chosen with respect to the normal tap. Hence, tap changing results in a spill current through the relay causing mal-operation. Remedy: Use biased differential relay. 6. When the transformer is switched on there is a large magnetizing inrush which depends on the instant of switching. This current is 8-10 times the normal rated current. Hence relay will mal-operate and will not allow the switching of transformer. Remedy: Use of DTIM-11 relay, or any comparable relay that provides second harmonic restraint. 7. The modern transformers are designed to operate near the knee point voltage of the B-H curve of their core. Hence 20% voltage rise would cause the rise in magnetizing current by 400%. This may cause mal-operation of relay. As the wave shape of the magnetizing current under this condition is rich of 5 th harmonic, the 5 th harmonic bypass circuit of the relay avoids the mal-operation. 1
In early days, a time delay was used to deal with the magnetizing inrush. This compromised relay performance when the transformer energizes with a fault already present in the system. The present remedies are: 1. Even harmonic cancellation. 2. Harmonic restraint. 3. Harmonic blocking. 4. De-biasing. Laboratory Simulation Fig. 1 shows the main circuit for demonstrating the biased differential protection. The function of PB1, PB2 and PB3 is to start, stop and to reset the circuit respectively. The switches S1, S2, and S3 are provided to make the fault at different locations. S1 is for an external fault, S2 is for an internal fault, and S3 is for a severe internal fault. The transformer has primary taps for 230V, 199.8 V, 115 V, 53.3 V and secondary taps for 110 V, 95.48 V, 55 V and 25.5 V. Procedure: PART-A Application of DTTM-11 for protection of single phase transformer 1. From the rating of the transformer (230/115 V, 1 kva), the CT ratings (10/5 A on primary, 20/5A on secondary), calculate the basic and bias settings of the relay, as well as the setting of the high-set unit. 2. Set the DTTM-11 relay according to calculations. Connect the circuit as shown. 3. Keep the secondary circuit open-circuited. Push PB1 button to energize the transformer with no load. Observe that the transformer can be energized without the relay operating. Repeat this process several times and observe the relay never operates while energizing the transformer. This is because DTTM-11 has the second harmonic restraint feature. This obviates the necessity of time delay during starting. 4. After switching the circuit off, complete the secondary circuit and make the circuit ON. Observe that relay does not operate for normal condition. 5. Make S1 ON, and observe that the relay does not operate for external faults as proved in the calculations. Switch S1 OFF. 6. Energize the circuit. Make S2 ON (internal fault on secondary side), and observe that relay 87 operates as proved in the calculations. Switch S2 OFF. 7. Energize the circuit. Make S3 ON (internal severe fault) and observe that unit 87 and the high set unit both operate as proved in the calculations. Switch S3 OFF. 8. Introduce a variable resistance across the secondary of CT on the secondary as shown in fig. 1. This simulates CT saturation or CT mismatch. Energize the circuit. Create an external fault using S1. Observe if the relay mal-operates. Decrease the CT secondary resistance till the relay mal-operates. Now switch off the supply, increase the bias setting and switch on the circuit again. Observe that for the same external fault, the relay does not operate. PART-B Effect of tap-changing and effect of inrush current: 1. Shift the transformer to another table that has similar set-up but has ABB-make type T relay. This relay does not have second harmonic restraint capability. Set this relay 2
with the same settings as for DTTM-11 relay. However, simulate the load resistance with an 18 Ω rheostat. 2. Repeat step 3 of part-a above. Observe that the relay mal-operates several times. The intensity of inrush depends on the instantaneous value of the input voltage when the transformer is energized, hence the relay does not mal-operate all the times. Now repeat the same process with increased time delay setting (this setting is provided with this relay). Observe the mal-operation ceases. However, this remedy is dangerous if the transformer is energized with an existing fault, especially in a part of system that has high fault levels. Keep the time delay in effect for the rest of the experiment. 3. Switch off the circuit and connect the secondary circuit. Switch on the transformer and observe the relay not operating for normal operation. 4. Change the primary tap to 199.8 V and repeat step-3. Observe the relay mal-operates. 5. Increase the bias setting and switch on the circuit. Observe that the relay does not operate. 3
4
5