Power Power System Lab Manual
INDEX
List of Experiments 1.
To the study time vs.voltage characteristcs characteristcs of over voltage induction relay
2.
To the study time vs. current current characteristics characteristics of over current induction induction relay
3.
To the study time vs. differential differential current characteristics characteristics of differential relay
4.
To study the time vs. current current characteristics characteristics of earth fault relay
5.
To study the time vs. temperature of thermal relay characteristics characteristics
6.
To study and verify the operation of buchholz relay.
7.
To study and performance of the long T/M line under no load test and under light load (R,L,C) condition and Ferranti effect.
8.
To find the location of of a fault in a UG Cable Cable using Murray loop test method method
9.
Study the single line to ground fault.
10. To study the three phase faults as practical application in transmission line.
EXPERIMENT NO. 1
OVER VOLTAGE RELAY AIM: 1. Study the operating current & de-operating voltage of disc. 2. Study the characteristics between voltage and time.
MATERIAL REQUIRED: 1. Voltmeter (0-300 V) 96 x 96 mm 2. Auto Transformer 0-270V 3. Indicating Light 4. Over Voltage Relay (Electromechanical Type) 5. Timer with Start & Stop facility 6. Push Button for Timer START & STOP 7. Rotary Switch 8. MCB/Isolator 9. Insulating terminals INTRODUCTION There are several instruments in power system which may burn at over voltage like motor, transformer, capacitors etc. To protect these instruments from over voltage, the over voltage relay is necessary. The several type of over voltage relays is available. One of the type from these relays is studying here. It is electromechanical type. It is very much rough & tough in use, its failure almost negligible. As the accuracy required to the electric instrument, the sufficient accuracy is available in this type of relay. THEORY Here the relay have an operating coil which core developed a rotating flux in disk due to this the disk rotates at certain voltage. Before the certain voltage a helical spring makes force in opposite direction so the disk is not in the condition to rotate. A socket is given to different setting of tripping voltage. Time setting multiplier is given to setting the tripping time. A flag relay is given to extending DC voltage to the circuit breaker. The resistances and diode is providing to operating of flag relay. We apply the tripping voltage to pin No 9 and 10 to this relay as given in Fig –‘1’. The pin No. 9 will extend to a transformer, disk operating coil and through moving this contact to a bridge. Pin No 10 will extend to the common pole of plug setting socket. On the plug setting socket all the primary tapings are connected and secondary will fed to operating disk coil and bridge rectifier. The output of bridge is filtered by capacitor C1 and resistance R1. The filtered DC voltage is fed to flag relay through resistance R2. Resistance R2 and R3 is meant for increasing the operating voltage capacity of flag relay. Resistance R4 and R5 is meant for increasing the operating voltage capacity of disk operating coil.
PROCEDURE: Study.1. Study the operating current & de-operating current of disc. (i) Keep the voltage source at minimum. (ii) The amp adj / relay test rotary switch is kept at VOLT ADJ. (iii) Switch ON the test set. (iv) Push the ON button. (v) Increase the voltage slowly and pay pa y attention at disc of relay. (vi) At certain voltage, it just moves in forward for ward direction, this voltage is operating voltage and note it. (vii) Now decrease the voltage through voltage source and pay hard attention at disc. (viii) The disc will stop at certain voltage and moves in reverse direction just after reducing the voltage. This current is de-operating voltage and note its value. Observation Table:
2. Study the characteristics between PSM & time (i) We have to study the graph between PSM and time at TMS-1 etc. (ii) Keep the voltage source at minimum. (iii) The volt adj / relay test rotary switch is kept at VOLT ADJ. (iv) Switch ON the MCB/Isolator. (v) Push the ON button.
(vi) Set the voltage at which the time has to observe and note down this voltage in table. (vii) Push the OFF button. (viii) The volt adj / relay test rotary switch is kept at Relay Test. (ix) Push the ON button. (x) The relay will trip the system then note down the time in given table. (xi) Repeat the above procedure to observe the time for different PSM and note these
PRECAUTIONS 1. The relay should be used very carefully and fast. 2. Suppose the 1 amp relay can withstand upto 1 amp continuously and if more than 1 amp the relay should trip the system within few seconds
Result:
EXPERIMENT NO.2
I.D.M.T. OVER CURRENT RELAY AIM: 1. Study the operating current & de-operating current of disc. 2. Study the time current characteristics at various multiples of plug setting current.
MATERIALS REQUIREDS: 1. Voltmeter (0-300 V) Digital 2. Ammeter (0-10 A) Digital 3. Loading C.T. 4. Auto Transformer 0-270V 5. Indicating Light 6. I.D.M.T. Relay Type CDG 7. Timer with Start & Stop facility 8. Push Button for Timer START & STOP 9. Rotary Switch 10. DP Switch 11. Insulating terminals
Theory:
There are several over current protection such as fuse, thermal relay & IDMT Relay. IDMT (Inverse Definite Minimum Time) Relay is a high accuracy over current relay. If we does not want to flow the current in lines more than 1 Amp, we will set the tripping current in our relay 1 Amp. As the current will become 1.10 or 1.20, the relay disc will start forward and trip the breaker after certain time. It is widely used to prevent over current on transmission lines, power transformers etc, because the error & tripping time of the relay is tolerable by the lines and transformer.
As the requirement of system is that t he faulted line should be o pen instantaneously. If the faulted line breaker fails to open the faulted line, the next supply breaker have to be open to for making dead the faulty line. The next breaker may be at higher voltage line or the same voltage. The next breaker should open only after the first breaker failure. So we will allow approx 0.4 sec time to operate first breaker. If first breaker dose not become open within 0.4 sec than it will be assume failure and the next breaker will become functional. T hese time and current distinguish is made by IDMT relay.
1
The time multiplier setting for an inverse time relay is defined as :T TMS = Tm Where T
= the required time of operation.
Tm = The time obtained from the relay characteristics curve at TMS = 1.0 and using PSM equivalent to max fault current. Current setting is adjusted by means of a tapped plug bridge hence known as PSM.
Fault Current PSM = Plug Setting
The operating time of an over current relay tends to become asymptotic to a definite minimum value with increase in the value of c urrent. This is inherent in electromagnetic relays due to saturation of magnetic circuit. So, by varying the saturation, different characteristics are obtained. If the core is made to saturate at large stage, it is called IDMT relay. The time characteristics in inverse over the same range and then after saturation occurs and definite min. time is reached. At low value of operating current the shape of current is determined by effect of restraining force of control spring while at higher values the effect of saturation predominates.
Different time multiplier setting (TMS) are obtained by varying the travel of disc or cup required to close the contracts. Higher is the TMS, greater is the spring restraining force. In order to make the relay operate at constant value o f minimum trip current for any TMS greater holes are cut in the disc.
TYPE CDG RELAY Type CDG Relay consists of an induction disc unit with an operation indicator and in some cases, an instantaneous high set unit all assembled on a standard frames.
The disc shaft carried a silver rod moving contact which completes the aux-unit circuit through the fixed contact. Permanent magnet is used to control the disc speed.
The setting is adjusted by the movement of the back stop which is controlled by rotating a knurled moulded disc at the base of the graduated time multiplier. 2
The self powered relay has an auxiliary unit which is powered by a secondary winding on the electro-magnet through a bridge rectifier. When the current through the relay exceeds its setting, the disc unit operated and closes its contacts to complete the path for secondary winding. The auxiliary unit connected across the rectifier now picks up and one of its own N/O contacts reinforces the disc contact. The other contacts of this auxiliary unit are available for tripping and alarm purposes.
Relay modes fitted with instantaneous element are also available. The instantaneous high set unit, type CAG13/CAG17 attracted armature relay, if powered, is fitted at the top left hand corner of the relay, with an operation indicator reset by a separate push rod protruding through the relay case. The pick up current of CAG 13 unit is adjusted by means of means of knurled knob which rotates a calibrated scale. The pick up current of C AG 17 unit is adjusted by rotating calibrated white dial fitted above the CAG 17 unit. A separate publication covers the operation and commissioning instruction for these relays.
For exact connections, references should be made to the panel schematics and relay wiring diagram. The self powered auxiliary lelement is fitted with indicator. The single disc contacts is rated to make and carry for 0.5 Sec. 2500 VA with maximum of 10 Amps and 660 Volts AC or DC. The two pairs of electrically separate, self or and rest contacts provoded in the auxiliary unit are rated to make and carry of 0.5 Sec. 7500 VA with maximum of 30 amps and 660 volts AC or DC.
SELECTION OF TIME SETTING For selective operation when there are a no of relays connected in series, the relay farthest from the source should be set to operate in the min possible time. For succeeding relays towards source a time delay step is given. For over current relays (inverse time) the time setting should be done at the max fault current. If the relay has proper selectivity at max. fault current it will automatically have a higher selectivity at the min fault current as the curve is more inverse on lo wer current region.
SELECTION OF CURRENT SETTING It is necessary to calculate the max fault current which can occur at each relay position. On a radial system the lowest setting must be at the farthest end, the setting being in creased for the subsequent relay towards the source. As per Indian standards the operating value should not exceed 130% of the setting. 3
Minimum Sht. Ckt Current I setting = 1.3
Procedure: 1.Study the operating current & de-operating current of disc. (i)
Keep the current source at minimum.
(ii)
The amp adj / relay test rotary switch is kept at AMP ADJ.
(iii)
Switch ON the test set.
(iv)
Increase the current source slowly and pay attention at disc of relay.
(v)
At certain current, it just moves in forward direction, this current is operating current and note the current.
(vi)
Now decrease the current through current source and pay hard attention at disc.
(vii)
The disc will stop at certain current and moves in reverse direction just after reducing the current. This current is de-operating current and note its value.
Observations: S NO. 1. 2. 3. 4. 5. 6. 7.
PLUG SETTING 1A
OPERATING CURRENT 1.06 A
DE-OPERATING CURRENT 0.90 A
5
3.
Study the time current characteristics at various multiples of plug setting current.
(i)
We have to study the graph between PSM and time at TMS-1 etc.
(ii)
Keep the current source at minimum.
(iii)
The amp adj / relay test rotary switch is kept at AMP ADJ.
(iv)
Switch ON the MCB/Isolator.
(v)
Increase the fault current upto required PSM (refer table). it is quit possible that while adjusting the fault current the FLAG of the Relay might trip for that you have to RESET the FLAG by moving the marked shaft UPWARD denoted by (RELAY FLAG RESET) for resetting the FLAG the Toggle switch must be brought in OFF position and the marked shaft move UPWARD.
(vi)
Now the desire Fault Current is SET and relay FLAG RESET - Only when the disk has move fully anti clockwise. Now move the Toggle Switch in Relay Test and press the green push button and timer counting will START and counting will STOP once the relay is operated. Note down the time in seconds.
(vii)
Now for various T.M.S. (Time Multiplier Setting) and P.S.M. (Plug Setting Multiplier), the time taken by the relay to operate at various fault current may be noted down.
(viii)
Now plot the graph between time take for the relay to operate Vs Plug Setting Multiplier at various T.M.S.
Observations: S NO. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
PLUG SETTING (PS) 1A
PSM 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 10.0
FAULT CURRENT = PS x PSM 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 10
TRIPPING TIME
6
4.
Study to find out the plug setting and TMS (Time Multiplier Setting).
(i)
When we want to use this relay, we know only the fault current and the delay time, the system can withstand the fault.
(ii)
The fault current is given 1000 amp, the delay time is 0.9 secs, and the CT used 100/1. Normal running current less than 100 amps and our system can tolerate upto 100 amps.
(iii)
Plug setting
= Tolerable current / Primary current of CT. = 100 / 100 = 1 Amp
(iv)
PSM
= Fault Current in secondary of CT / PS = 10 Amp / 1 = 10 Amp
(v)
From the already discussed graph at PSM 10, TMS at 1, the delay time is shown 3.5 sec approximate.
(vi)
TMS
= T / Tm = 0.9 / 3.5 = 0.25
(vii) (viii) 5.
Verify the practically delay time by setting the TMS and PS as calculated. Practically delay time = 0.8686
Study of front plate numerical values.
(i)
At front plate plug setting (PS) is given to set the relay tripping current.
(ii)
Some numerical values given between plug setting multiplier (PSM) and time.
(iii)
We will calculate the PSM from the formula as discussed.
(iv)
The seconds shows the delay in tripping at given PSM.
(v)
Auxiliary volt is used in some relays to operating the flag relay.
(vi)
The amp shows approx tolerance of relay i.e. 1 amp relay should not in use more than 1 amp. 7
MERITS OF THE RELAY 1.
The relay is electromechanical type can tolerate rough-tough use.
2.
High accuracy in respect of thermal relays.
3.
Normally no false tripping .
DE-MERITS OF THE RELAY 1.
The relay is electromechanical type so it should be used in dust free environment.
2.
The frequency variations can change the parameters.
PRECAUTIONS 1.
2.
The relay should be used very carefully and fast, beyond its rating.
Suppose the 1 amp relay can withstand upto 1 amp continuously and if more than 1 amp the relay should trip the system within few seconds.
RESULT:
EXPERIMENT NO. 3 PERCENTAGE DIFFERENTIAL RELAY STATIC TYPE
AIM: Study of operation of relay on different operating currents. Find operational characteristics of the relay.
MATERIAL REQUIREDS: 1.
Relay Single Pole Version 1 A (Numerical Type) 'AREVA' make MBCH-12
2.
Timer
3.
Auto Transformer 0-270V, 10 A
4.
Ammeter (10 A, AC) - 2 Nos.
5.
Neon lamp 1A, AC, 230 V
6.
Rheostat 5 A, 45 Ohms
7.
Rheostat 10 A, 20 Ohms - 1 Nos.
8.
Isolation Transformer 9.
- 1 Nos.
Auxiliary DC Supply Unit with T ransformer
2 THEORY It is a very important protection of the transformer. It is based on the ratio of H.T. current and L.T. current should be constant. Consider the Fig No '1', here we considering the single pole of 132/33 KV Transformer. It's H.T. current and L.T. current ratio will be 1:4. If the CT of H.T. side is considered 100/1 Amp, so the CT of L.T. side will be 400/1 Amp. The secondary current of L.T. side CT and H.T. side CT will always equal in normal condition. Both the secondary of CTs will enter in Numerical type % Differential Relay. The secondary of CT connection is make in such a way that the CT current will flow only through coil circuit and no extra current is to flow from Differential coil. As soon as the fault occurs in transformer, the H.T. current will high. The ratio of H.T. current and L.T. current will change. The secondary of H.T. side CT current will become high with respect to secondary of L.T. side CT current. So the difference of current will flow through differential winding. The secondary of differential winding transformer will go to an electronic circuit that will operate a tripping relay to trip the breaker of main transformer. The through windings are used to restraining the differential relay. It will more clearly by drawing the curve between through current and differential current.
PRINCIPAL OF OPERATION Considering the drawing given below, that two t ypes of proportionate DC is available. One is proportional to through current and another is proportional to differential current.
The through current always restraining the tripping and differential current always operating the tripping relay. With a very simple way, we are understanding the circuit that in an amplifier we applies differential current DC to the base of transistor to operate the tripping coil and through current DC applied to the emitter of transistor to restrain/block the tripping. By the resistance we set the blocking and tripping of relay as per our requirement.
PROCEDURE:
Experiment No 1. Study of operation of relay on different operating currents. 1.
Turn the current source to minimum.
2.
Connect the Rheostat 5 Amp, 45 Ohms to differential point.
3.
Select the moving part of Rheostat at center.
4.
Select the current by DIP switch at the relay.
5.
Switch ON the MCB.
6.
Push the green start button.
7.
Increase the differential current gradually by the current source.
8.
Observe that the differential current will increase the set value by DIP switch,
9.
The relay will issue the tripping command and trip LED will become glow.
10. We can reset the trip relay b y resetting button. Experiment No 2. Find operational characteristics of the relay 1.
Turn the current source to minimum.
2.
Connect the Rheostat 5 Amp, 45 Ohms to differential points.
3.
Connect the Rheostat 10 Amp, 20 Ohms to through points.
4.
Select the moving part of Rheostats 5 Amp, 45 Ohms at maximum ohms.
5.
Select the moving part of Rheostats 10 Amp, 20 Ohms at center.
6.
Select the current by DIP switch at the relay.
7.
Switch ON the MCB.
8.
Push the green start button.
9.
Increase the through current gradually as given in table by the current source.
10. Increase the differential current gradually upto the tripping of the relay and record in the table. 11. Repeat the same process to get the differential values at different through current.
Observation Table: S.No
Restraining Current (A)
Operating Current (A)
Precautions: 1.
The relay is designed to handle 1 Amp current. The study of relay more than 1 Amp, required fast operation.
2.
Do not touch any electronic part with hand, otherwise the relay may damaged b y static current.
RESULT:
EXPERIMENT NO. 4
EARTH FAULT RELAY AIM: Find operational characteristics of the relay for time and current setting.
MATERIAL REQUIREDS: 1. 2. 3. 4. 5. 6. 7. 8. 9.
Voltmeter (0-300 V) Ammeter (0-10 A) Loading C.T. Auto Transformer 0-270V Neon lamp 230 V Earth Fault Relay Electromechanical type CDG Timer with Start & Stop facility Push Button for Timer START & STOP DP Switch
THEORY: This manual covers the commissioning and maintenance instruction for nondirectional inverse time over current relays belonging to the CDG 11 family, which are self powered. There is no need for any separate auxiliary DC or AC supply for these type of relays. These relays are available in the standard current setting ranges of 50-200% for present case, 20-90% and 10-40% of 1 A or 5 A. Triple Pole version designate as CDG 31 consists of identical 3 units. Type CDG 11 is an inverse time over current relay with a definite minimum time. It is used for protection against phase and earth faults. The 50 Hz version has a time current characteristics which confirms to IS 3231 (1965).
The time multiplier setting for an inverse time relay is defined as :T TMS = Where T
Tm = the required time of operation.
Tm = The time obtained from the relay characteristics curve at TMS = 1.0 and using PSM equivalent to max fault current.
PROCEDURE: 1.
Switch ON the MCB, the Voltmeter will show the line voltage.
2.
Initially Rotary Switch should be in Amp Adjust Position.
3. Now to set the desired fault current we will be using current source. For that switch ON the Rotary switch and move the current source till the desired fault current is indicated on the Ammeter, it is quit possible that while adjusting the fault current the FLAG of the Relay might trip for that you have to RESET the FLAG by moving the marked shaft UPWARD denoted by (RELAY FLAG RESET) for resetting the FLAG the Toggle switch must be brought in OFF position and the marked shaft move UPWARD.
4. Now the desire Fault Current is SET and relay FLAG RESET - Only when the disk has move fully anti clockwise. Now move the Rotary Switch in Relay Test and press the green push button and timer counting will START and counting will STOP once the relay is operated. Note down the time in seconds. 5. Now for various T.M.S. (Time Multiplier Setting) and P.S.M. (Plug Setting Multiplier), the time taken by the relay to operate at various fault current may be noted down. 6. Now plot the graph between time take for the relay to operate Vs Plug Setting Multiplier at various T.M.S.
OBSERVATION: (i)
TMS = 1.00, 0.9, 0.7, 0.5
Current
RESULT:
PS = 1.0 Amp
Time
PSM
EXPERIMENT NO.5
THERMAL RELAY
EXPERIMENT NO.5
THERMAL RELAY
AIM: 1. 2.
Find operational characteristics of the relay Determine time current characteristics of given fuse.
MATERIAL REQUIREDS : 1.
Auto Transformer
2.
Thermal Relay
3.
Fuse 3A
4.
Ammeter 10 Amp
5.
Loading C.T.
6.
Timer
THEORY: Thermal relays are current sensitive devices. They operate on thermal effect of electric current. This relay senser current by temperature rise, which is produced by the higher current. These relays can also respond to understand 3 currents, which cause rise in temperature due ti their negative sequence component. Fuse: Fuse is a current interrupting device which opens the circuit in which it is inserted, when the current in circuit exceeds a certain value. The part of the fuse which is designed to melt, when the fuse melts it is called fuse link
PROCEDURE: For Thermal relay: 1.
Switch ON the DP Isolator Switch.
2.
Select both the rotary switches to Thermal relay side.
3.
Now put the current source at minimum setting.
4.
Push the Green Push Button marked on right hand side of the panel.
5.
Now set desired current by current source and note the corresponding tripping time of the relay.
For Fuse: 1.
Switch ON the DP Isolator Switch.
2.
Select both the rotary switches to Fuse T est Side.
3.
Now put the current source at minimum setting.
4.
Push the Green Push Button marked on right hand side of the panel.
5.
Now set desired current-by-current source and note the corresponding tripping time of the r elay.
6. Press the push button for fuse test till the fuse block and timer stops thus giving the blowing time of the fuse.
RESULT:
EXPERIMENT NO. 6
BUCHHOLZ'S RELAY AIM: To study the Buchholz's Relay
MATERIAL REQUIREDS : 1. 2. 3. 4.
Experimental setup consisting of Buchholz Relay, Transformer Tank, Air Cylinder, Conservator, 5. Breather and Air Pump THEORY:
Buchholz's Relay is a gas actuated relay used for protecting oil immersed transformer against all types of internal faults and marks use of the fact that faults decompose oil thus generating gases.
Fig-'1' shows such as a relay. It is connected by means of pipes between the top of the transformer tank and the conservator as shown in Fig-'2' and is, therefore, under normal condition of operation, full of oil.
It consists of a cast housing containing a hinged hollow float, inside of which is mercury switch, the float being located in the upper part of the housing, and a hinged flat value to which is attached a similar mercury switch, located directly in the path of the oil between the bank and the conservator. When an incipient fault (such as insulation fault between the turns, breakdown
or core insulation, core heating, bad switch contact, fault joints) occurs in the transformer bubbles of gas are evolved by the heat generated and rise upto the tank surface and thence into housing of the devices where their through passage is prevented by the flap valve. The bubble pass to the top of the housing, causing the oil level to fall, where upon the hollow float tilts and when the mercury switch contacts are short-circuited by the mercury, an alarm device is operated. If however a serious fault such as internal short-circuited between phases or turns, earth fault puncture of bushing insulator inside tank occurs the volume of gas generated is considerable which in moving through the relay causes the gas surge flap value to be deflected, thereby closing the mercury contact switch and energising the trip coils of the circuit breakers. The connecting pipe between the tank and conservator should be as straight possible and should slope upwards towards the conservator at a small angle from the horizontal.
PROCEDURE: 1. Connect the pump to air cylinder and close the outlet valve outlet valve of air cylinder. 2. Pressurize the air cylinder upto the pressure of the 30 lbs. 3. Release the air in the transformer tank by opening outlet valve of air cylinder. 4. NC contact of relay will change to open position and NO contact will change to close position and will comeback to open position. 5. Change in contact position is indicated by glowing of lamp on the panel board.
6. Release the air from the relay by slowly opening the air release valve of the relay. 7. Properly close the air release valve.
RESULT:
EXPERIMENT NO.7 PERFORMANCE OF LONG TRANSMISSION LINE UNDER NO & LIGHT LOAD
AIM: To study the performance of a long transmission line under no load and under light load conditions.
MATERIAL REQUIREDS: 1. 2. 3. 4.
Transmission line load model of Pie network Two Digital voltmeter and Ttwo digital ammeter Resistive, Inductive, Capacitive loa
LPF wattmeter
5. Patch cards
THEORY: Ferranti Effect: A simple explanation of Ferranti Effect can be given by approximating the distributed parameters of the line by lumped impedance as shown in Fig -'1'.
Line representation (Lumped) under No Load Condition Fig 1
Phasor Diagram for Line representation (Lumped) under No Load Condition Fig 2
Since usually the capacitor reactance of the line is quite large as compared to the inductive reactance under no load or lightly loaded conditions the line current is of leading P.F. The phasor diagram is given as fig -'2' for thus operating conditions.
The charging current produces drop in the reactance if the line which is in phase opposition to the receiving end voltage and hence the sending end voltage becomes smaller than the receiving end voltage.
For finding out inductance KVAR required to obtain zero% voltage regulation variable shunt inductor is connected at the receiving end a nd its value is changed till we get voltage at receiving end equal to that of voltage at the sending end.
PROCEDURE: 1.
Apply the voltage (200 V max.) to the sending end and connect power factor
meter. Also connect 1 ammeter and voltmeter to each end (receiving and sending). 2.
Connect the load comprising of R, L and C at the receiving end and note down
the value of receiving end voltage.
3.
Now remove the load from the receiving end and note down the voltage on
receiving end. This voltage at the receiving end is quite large as co mpared to sending end voltage.
OBSERVATION: Load Resistive Inductive Capacitance No Load
RESULT:
Vs
Is
Vr
Ir
EXPERIMENT NO. 8
FAULT LOCATION IN CABLE BY MURRAY LOOP TEST AIM: To locate fault in a cable by Murray Loop Test. MATERIAL REQUIREDS : 1. Rheostat 1.1 A, 800 Ohms - 2 Nos
2. Digital Multimeter 3. Galvanometer - 1 No. 4. Measuring Tape (5M) - 1 No 5. 3Core Cable (25M) - 1 No 6. DC Power Source - 1 No
THEORY: MURRAY LOOP TEST This test is carried out for locating a ground or a short circuit fault, provided that a cable runs along with the grounded cable or with two cables (or with two cores of a multi-core cable) which are short circuited. The advantage of loop test is that t he resistance of the fault does not affect the r esults obtained. Provided this resistance is not very high. Otherwise it may adversely affect the sensitivity. The connection diagram for conducting Murray Loop Test is shown in Fig '1'
.
Connections of Fig -'1' are used for locating a ground fault and those of Fig -'2' for a short circuit fault. As is evident from these figures, both circuits are essential
Connection diagram for locating short circuit fault by Murray Loop Test – Fig – 2
PROCEDURE 1. Take a multicore cable (say 3 core) of known length (say 25M). 2.Measure the resistance of each core. 3. Make connections as shown in Fig - '1'. Short circuit the two cores of the cable at the other end. Adjust P and Q such that balance is obtained. Note P, Q and calculate distance of fault x. Take three-four observations and take the mean of calculated value of length of the fault from each set of readings. This length should be equal to the distance of fault from the lower end of resistance Q. 4.Note down the actual distance of fault by measuring the actual distance of fault and calculate the % error. 5.Make connections as shown in Fig -'2'. Short circuit any two cores of cable to create short circuit fault. Adjust P and Q such that balance is obtained. Note P and Q in the observation table. Calculate x with. 6.Take three-four observations and find average of x. calculate the distance of short circuit fault from the measuring end of the cable. 7.Note down the actual distance of fault by measuring of actual fault distance and calculate the %age error.
OBSERVATION Total length of the cable = 25 Meter r = resistance per meters = 0.9/25 = 0.036 Localization of Earth Fault :S P Q No. (Ohm) (Ohm)
x = 2l [Q/(P+Q)]
1. 2.
699 609
172.7 148.3
9.91 9.79
3.
799
194
9.77
Distance of fault from measuring end = x (m) (Average of x1, x2, x3 ) 9.82
Actual location of fault
10.00
% Error
1.8
Localization of Short Circuit Fault :S P Q No. (Ohm) (Ohm)
1. 2. 3.
Result:
161 185 165
670 761 678
X = 2l [P/(P+Q)]
10.32 9.78 9.77
Distance of fault from measuring end = x (m) (Average of x1, x2, x3 ) 9.96
Actual location of fault
10.00
% Error
0.4
EXPERIMENT NO. 9 SINGLE LINE TO GROUND FAULT
AIM: To Study Single Line to Ground Fault as practical application in transmission lines.
MATERIAL REQUIREDS : 1. Instantaneous Earth Fault Relay Type CAG 2. 'Areva' make Electro Mechanical Type 3. Digital AC Voltmeter 0-300 V, 4. Push Button 5. Rotary Switch 6. Indicating Light 7 .Fuse 8. DP Switch 9. Insulating Terminals 10. Transformers 11. Line Impedances 12.1 Ph Variac with loading CT
Theory: The Single line to Ground fault comes under Shunt type of fault (Short Circuit) this fault occur due to one broken conductor on transmission line. The shunt faults are characterise by increase in voltage & frequency and fall in current in the faulted phase. The need for line fault study is because these fault mostly occur in transmission lines. These are broadly classified as 3 Phase fault & 1 Ph to ground fault. Fault point which may occur in transmission line should be isolated from rest supply system after the fault occurs. The rest transmission line should remain normal as usual without hindrance. Here to study the above fault in the lab we will be creating mock faults at various points and first calculating the theoretical values of current at the point where the fault occurs and then practically verifying the calculated values. For this purpose we have considered a single phase system (generally used in railway Traction). Operating voltage for MAIN TRANSFORMER referred to as (T) is 230/125 V and the transmission line impedance to be 2+2 Ohms each and the secondary transformer referred to as (t) as 125/74 V. Also we have provided an Instantaneous Earth Fault Relay (Electro Mechanical Type) showing the tripping of line at the ti me when fault occurs.
WORKING OF RELAY 1. Whenever breakdown or fault occurs on a transmission line some kind of protection is essential in order to isolate the faulty point from the rest of the transmission line. The Instantaneous Earth Fault Relay is one of the protective relays. 2. This relay is generally used in the distributing lines such as 33 KV, 11 KV and 0.4 KV etc. 3. As soon as a fault occurs on any line tripping command issues from this relay to breakers, 24 V DC is extended through this Relay to tripping coils of the 33 KV or 11 KV Breakers. In the Breakers of 0.4 KV, AC/DC 220V/110V is given as per design and the line is again charged. In our lab experiment we are using a Contactor as a Breaker. T his tripping Relay shorts the coils of an other Auxiliary relay thereby coil denergizes and the supply to contactor coil is disconnected. Locking and also over current is recovered the Instantaneous Earth Fault Relay becomes normal and Auxiliary Relay whose coil was short circuited by the Instantaneous Earth fault Relay is again operated, therefore our system again comes in the normal manually start position. If it happens practically on site, the patrolling is making in Fault area, fault locates and removes.
WORKING OF PANEL
At first we have to see that any fault has not been framed on the p anel. If it is so then we have to make it free from any such fault by disconnecting all the leads.
1.
VOLTAGE RATIO TEST
(i)
Turn the knob of the variac to bring it to zero indication.
(ii)
Variac O/P is connected to the Transformer (T) I/P.
(iii)
Connect the transformer (T) Input & Output to the Voltmeter.
(iv)
Press the 'ON' push button.
(v)
Increase the Variac voltage upto 230 volts & the secondary voltage should be 125 volts.
(vi)
Feed the Output from Transformer (T) to the Input of Transformer (t).
(vii)
The Output of Transformer (t) shall be 74 volts.
st
Voltage Ratio Test 1 Transformer (T) Input - 230 Volts Output - 125 Volts
-
115 x 2 = 230 62 x 2 = 124
Voltage Ratio Test 2 nd Transformer (t) Input - 125 Volts Output - 74 Volts
2.
-
62 x 2 = 124 36 x 2 = 72
PERCENTAGE IMPEDANCE OF TRANSFORMER (%Z)
Full load short ckt voltage Formula =
x 100% Applied Voltage
(a)
for Transformer (T)
(i)
Full Load current of Transformer (T) 124 Volts side 1000 = = 8.06 A 124
(ii)
Zero the Variac knob.
(iii)
Connect the Variac Output to 230 Volts side of Transformer (T).
(iv)
Connect the Voltmeter to Variac output.
(v)
Short the 125 Volts side.
(vi)
Connect the Tong tester at 125 volts side lead.
(vii)
Increase the voltage by Variac gradually so that the current in tong tester reaches 8.00 A.
(viii)
Read the voltage at output of Variac it should be 14.0.
% impedance of Transformer (T) =
11.62 x 100 230
= 5.05 %
(b)
for Transformer (t) (i)
Full Load current of transformer (t) 74 Volts side 500 = = 6.94 A 72
(ii)
Zero the Variac knob.
(iii)
Connect the Variac Output to 230 Volts side of Transformer (T).
(iv)
Connect the transformer (T) 125 V side to 125 V side of transformer (t).
(v)
Connect the voltmeter to 125 volts of transformer (T).
(vi)
Short the 74 Volts side.
(vii)
Connect the Tong tester at 74 volts side lead.
(viii)
Increase the voltage by Variac gradually so that the current in tong tester reaches 6.76 A.
(ix)
Read the voltage at output of Transformer (T) it should be 12 Volts.
14.14 % impedance of Transformer (t) =
x 100 124
= 11.40 %
% IMPEDANCE OF TRANSFORMER 'Z' ON 100 MVA BASE
% Z x 100 Formula : %Z on base MVA = MVA (Transformer Capacity) For Transformer (T) %Z on base MVA 5.05 x 100 =
= 505217.39 %
0.001 For Transformer (t) %Z on base MVA 11.40 x 100 =
= 2280645.16 % 0.0005
LINE IMPEDANCE ON 100 MVA BASE Line Impedance x 100 x 100 Line Impedance On 100 MVA Base
= 2
(KV) Line Impedance = 2 + 2 = 4 KV = 0.125
=
4 x 100 x 100 (.125)
= 2560000%
2
Framing of Network for faults Red Jumpers are to be connected as network design or samples as given below. 1. 2. 3.
Only one transformer & Line. One transformer line & other transformer. Line & other transformer etc.
SAMPLE - I As a sample-I we are considering the given design & fault at point F. Considering zero line impedance.
Equivalent N/W design on 100 MVA Base
I2
I1 505217.39
I3 2280645.16
Fault Impedance = 609000 + 1920000 = 2785862.55 % 100 x 100 Formula
:
Fault MVA = Fault Impedance 100 x 100 Fault MVA = 2560000 = 0.003589552
Fault MVA x 1000 Formula
I = Volts in KVA 0.003589552 x 1000 = 49.85 Amp
I3 = .072
0.003589552 x 1000 I2 =
= 28.94 Amp .124 0.003589552 x 1000
I1 =
= 15.60 Amp .230
Practical Value (i)
Zero the Variac knob.
(ii)
Design the N/W as per design on panel.
(iii)
Apply the voltage 23 Volts at port 230 Volts side of Transformer (T).
The measured current values are at 23V :I3 = 5.03 A
So for 23 x 10 = 230 V
Theoretical Value :-
I3 = 5.03 x 10 = 50.3 A
I3 = 49.85 A
I2 = 3.10 A
I2 = 3.10 x 10 = 31.0 A
I2 = 28.94 A
I1 = 1.78 A
I1 = 1.78 x 10 = 17.8 A
I1 = 15.60 A
Hence the practical value & theoretical values are nearly same and little difference due to lead impedance only.
SAMPLE - II As a sample II we are considering the given design fault at point F & line impedance 4 Ohms.
Equivalent N/W design on 100 MVA Base I2
I1 505217.39
I2 2280645.16
I3 2560000
Fault Impedance = 609000 + 1920000 + 2560000 = 5345862.55 100 x 100 Fault MVA =
= 0.001870605 5345862.55
0.00187 x 1000 I3 =
= 25.90 Amp .072 0.00187 x 1000
I2 =
= 14.96 Amp .124 0.00187 x 1000
I1 =
= 8.13 Amp .230
Practical Values Procedure as take before. The measured current values are at 23V :I3 = 2.52 A
So for 23 x 10 = 230 V
Theoretical Value :-
I3 = 2.52 x 10 = 25.2 A
I3 = 25.90 A
I2 = 1.53 A
I2 = 1.53 x 10 = 15.3 A
I2 = 14.96 A
I1 = 0.91 A
I1 = 0.91 x 10 = 9.1 A
I1 = 8.13 A
Hence the practical value & theoretical values are nearly same and verified.
PRECAUTIONS
1. time.
The voltage at 230 volts side of transformer (T) should not exceed 250 volts at any
2. time.
The voltage at 125 volts side of transformer (t) should not exceed 140 volts at any
Result:
EXPERIMENT NO. 10
THREE PHASE FAULT Aim : To Study Three Phase Fault as practical application in transmission li nes. INSTRUMENTS REQUIRED 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
Three Phase Earth Fault Relay, Static T ype Digital AC Voltmeter 0-600 V, Push Button Rotary Switch Indicating Light TP Isolator DP Switch Insulating Terminals Transformers Line Impedances Digital Clamp on Meter
-
1 No. 2 No.
-
1 No
-
2 No.
-
1 No.
WHY THREE PHASE OCCUR ON TRANSMISSION LINES? Three Phase occurs in 11 KV and 33 KV due to :(a) (b) (c)
Tree falling By Birds Oftenly occurred due to hotline wire broken.
NEED OF FAULT STUDY CALCULATIONS 1. The point of state electricity supply advice the fault occur (either three phase or single line to ground fault) must be isolated from the transmission line supply so that the rest of the transmission supply line should be normal without any hindrance. For this purpose 11 KV fault point should be isolated immediately and 33 KV system at 11 KV fault should be tripped after 0.5 second. 2. First the chance is to be given at 11 KV protection and if that fails 33 KV protection will operate and isolate the fault zone. Similarly 132 KV, 220 KV and 400 KV protection will operate after giving the chance at lower voltage side. 3. A definite time relay is used to give the proper time of lower voltage side protection. For operation of protections the fault current at various points should be known, hence after knowing the minimum & maximum fault current various protections are used hence this study is required. THEORY The three phase fault comes under Shunt t ype of fault (Short Circuit) this fault occur due to one broken conductor on transmission line. The shunt faults are characterize by increase in voltage & frequency and fall in current in the faulted phase. The need for line fault study is because these fault mostly occur in transmission lines. These are broadly classified as 3 Phase fault & 1 Ph to ground fault. Fault point which may occur
in transmission line should be isolated from rest supply system after the fault occurs. The rest transmission line should remain normal as usual without hindrance. Here to study the above fault in the lab we will be creating mock faults at various points and first calculating the theoretical values of current at the point where the fault occurs and then practically verifying the calculated values. For this purpose we have considered a three-phase (four wire) system. Operating voltage for MAIN TRANSFORMER referred to as (T) is 400/220 V and the transmission line impedance to be 2+2 Ohms each phase and the secondary transformer referred to as (t) as 220/133 V. Also we have provided an Instantaneous Earth Fault Relay (Static Type) showing the tripping of line at the time when fault occurs. BASIC BLOCK DIAGRAM
WORKING OF RELAY 1.Whenever breakdown or fault occurs on a transmission line some kind of protection is essential in order to isolate the faulty point from the rest of the transmission line. The Instantaneous Earth Fault Relay is one of the protective relays. 2.
This relay is generally used in the distributing lines such as 33 KV, 11 KV and 0.4 KV etc.
3. As soon as a fault occurs on any line tripping command issues from this relay to breakers, 24 V DC is extended through this Relay to tripping coils of the 33 KV or 11 KV Breakers. In the Breakers of 0.4 KV, AC/DC 220V/110V is given as per design and the line is again charged. In our lab experiment we are using a Contactor as a Breaker. This tripping Relay shorts the coils of an other Auxiliary relay thereby coil denergises and the supply to contactor coil is disconnected. Locking and also over current is recovered the Instantaneous Earth Fault Relay becomes normal and Auxiliary Relay whose coil was short circuited by the Instantaneous Earth fault Relay is again operated, therefore our system again comes in the normal manually start position. If it happens practically on site, the patrolling is making in Fault area, fault locates and removes. 4. It is a static definite time relay whose definite time is almost 1.5 to 2.0 seconds as soon as out of two lines, any line current will increase the preset amp values at relay and remains for almost 2 seconds, the relay will issue the trip command and contactor will open in case the failure of auxiliary supply or low voltage less then 300 Volt the protection will fail so the relay will also issue the trip command to open the contact point. T he relay is manual reset type and tripping indication will appear at the relay. After resetting the relay can be reused. 5. First we will create the fault which is having high fault impedance so that our system can tolerate in our lab i.e. the fault current should be in permissible limit so that our equipments and supply should able to bear the fault current values.
6. During fault condition our source should be able to supply fault current so that fault condition occurs by short circuiting 133 V side of transformer (t). Both the line resistance should be in fault. In this condition the current at 400 side is approx 12 Amp and our relay can tolerate upto 5 amp and relay will issue a trip command thus the relay is operated.
WORKING OF PANEL At first we have to see that any fault has not been framed on the panel. If it is so then we have to make it free from any such fault by disconnecting all the leads. 1.
VOLTAGE RATIO TEST (i)
Turn the knob of the three phase variac to bring it to zero indication.
(ii)
Variac O/P is connected to the Transformer (T) I/P.
(iii)
Connect the transformer (T) Input & Output to the Voltmeter.
(iv)
Press the 'ON' push button.
(v)
Increase the Variac voltage upto 400 volts & the secondary voltage should be 220 volts.
(vi)
Feed the Output from Transformer (T) to the Input of Transformer (t).
(vii)
The Output of Transformer (t) shall be 133 volts.
Voltage Ratio Test 1 st Transformer (T) Star/Star Input - 440 Volts Output - 220 Volts
nd
Voltage Ratio Test 2
Transformer (t) Delta/Star
Input - 220 Volts Output - 132 Volts
2.
PERCENTAGE IMPEDANCE OF TRANSFORMER (%Z) Full load short ckt voltage Formula =
x 100% Applied Voltage
(a)
for Transformer (T) MVA - 0.003
Voltage Ratio - 400/226 V Phase to Neutral voltage secondary side is - 127 V Full load secondary current = 7.87 A (T) (ii)
Zero the Variac knob.
(iii)
Connect the Variac Output to 400 Volts side of Transformer (T).
(iv)
Connect the Voltmeter to Variac output.
(v)
Short the 220 Volts side.
(vi)
Connect the Tong tester at 220 volts side lead.
(vii)
Increase the voltage by Variac gradually so that t he current in tong tester reaches 7.87 A.
(viii)
Read the voltage at output of Variac it should be approx 13 V.
Primary voltage at secondary full load short ckt current - 13 V 13 % ZT =
x 100 = 3.25 400
(b)
for Transformer (t) MVA - 0.0015 Voltage Ratio - 220/132 V Phase to Neutral voltage secondary side is - 76.21 V Full load secondary current - 6.56 A (T) (ii)
Zero the Variac knob.
(iii)
Connect the Variac Output to 440 Volts side of Transformer (T).
(iv)
Connect the transformer (T) 220 Volts side of transformer (t).
(v)
Connect the voltmeter to 220 volts of transformer (T).
(vi) (vii)
Short the 132 Volts side. Connect the Tong tester at 132 volts side lead.
(viii)
Increase the voltage by Variac gradually so that the current in tong tester reaches 6.56 A.
(ix)
Read the voltage at output of Transformer (T) it should be 9 Volts.
Primary voltage at secondary full load short ckt current - 9 V 9 %Z
t
=
x 100 = 4.09 220
% IMPEDANCE OF TRANSFORMER 'Z' ON 100 MVA BASE % Z x 100 Formula : %Z on base MVA = MVA (Transformer Capacity) For Transformer (T) %Z on base MVA 3.25 x 100 =
= 108333.33 % 0.003
For Transformer (t) %Z on base MVA 4.09 x 100 =
= 272726.66 % 0.0015
LINE IMPEDANCE ON 100 MVA BASE Line Impedance x 100 x 100 Line Impedance On 100 MVA Base
= (KV)
2
Line Impedance = 2 Ohms KV = 0.220 2 x 100 x 100 =
= 413223.14 (.220)
2
Framing of Network for faults Jumpers are to be connected as network design or samples as given below. 1. 2. 3.
Only one transformer & Line. One transformer line & other transformer. Line & other transformer etc.
SAMPLE - I As a sample-I we are considering the given design & fault at point F. Considering zero line impedance.
Equivalent N/W design on 100 MVA Base I2
I1 108333.33
I3 272726.66
Fault Impedance = 108333.33 + 272726.66 = 381059.99 100 x 100 Formula
:
Fault MVA = Fault Impedance at 100 MVA Base 100 x 100 Fault MVA = 381059.99 = 0.02624
•
Fault current at transformer (t) output Fault MVA x 1000
Formula
I = 1.732 x KV 0.02624 x 1000
I3 =
= 114.00 Amp 1.732 x 0.132
•
Fault current at transformer (T) output
0.02624 x 1000 I2 =
= 68.8 Amp 1.732 x 0.220
•
Fault current at transformer (T) input 0.02624 x 1000 = 37.8 Amp
I1 = 1.732 x 0.4 Practical Value (i) (ii) (iii)
Zero the Variac knob. Design the N/W as per design on panel. Apply the voltage 20 Volts at port 400 Volts side of Transformer (T).
The measured current values are at 20V :I3 = 5.44 A
These current values at 400 V Theoretical Value :shall be :I3 = 114.0 A I3 = 5.44 x 20 = 108.8 A
I2 = 3.58 A
I2 = 3.58 x 20 = 71.6 A
I2 = 68.8 A
I1 = 1.91 A
I1 = 1.91 x 20 = 38.2 A
I1 = 37.8 A
Hence the practical value & theoretical values are nearly same and little difference due to lead impedance only. SAMPLE - II As a sample II we are considering the given design fault at point F & line impedance 2 Ohms.
Line Impedance = 2 KV = 0.220 2 x 100 x 100 =
= 413223 (.220)
2
Equivalent N/W design on 100 MVA Base
I2
I1 108333.33
I2 413223.00+413223.00
I3 272726
Fault Impedance = 1207505 100 x 100 Fault MVA =
= 0.0082815 1207505
•
Fault current at transformer (t) output Fault MVA x 1000
Formula I = 1.732 x KV 0.0082815 x 1000 = 36.22 Amp
I3 = 1.732 x 0.132 •
Fault current at transformer (T) output 0.0082815 x 1000 = 21.7 Amp
I2 = 1.732 x 0.220 •
Fault current at transformer (T) input 0.0082815 x 1000 I1 =
= 11.95 Amp 1.732 x 0.4
Practical Values Procedure as take before. The measured current values are at 400 Volt Applying 40 V :I3 = 1.31 A
These current values at 400 V Theoretical Value :shall be :I3 = 1.31 x 10 = 13.1 A I3 = 11.95 A
I2 = 2.45 A
I2 = 2.45 x 10 = 24.5 A
I2 = 21.7 A
I1 = 3.64 A
I1 = 3.64 x 10 = 36.4 A
I1 = 36.22 A
Hence the practical value & theoretical values are nearly same and verified. PRECAUTIONS 1.
The voltage at 400 volts side of transformer (T) should not exceed 440 volts at any time.
2.
The voltage at 220 volts side of transformer (t) should not exceed 230 volts at any time.
3.
Neutral is to be connected on input side.
MPORTANT FORMULAS Transformer %Z x 100 %Z on 100 MVA Base = MVA
Full Load Short Ckt x 100 %Z = Applied Voltage Line Impedance Z x 100 x 100 Line Impedance on 100 MVA Base = (KV)
2
100 x 100 Fault MVA = Fault Impedance Fault MVA x 1000 Fault Current I
=
f
1.732 x KV 220 KV line Imp/KM = 0.418 Ohms
79.1 O
132 Line Imp/KM = 0.16 + 0.408 or Magnitude Value = 0.44 Ohms 33 KV Line Imp/KM = 0.484 Ohms 100 x 100 Fault MVA
= Zf
Fault MVA x 1000 Fault MVA
= KV (Ph to N)
After T/F Delta/Star Ph to P h fault feet then fault current Fault MVA x 100 = KV (Ph to Ph)
