eee department power systems lab manual · 2019-04-15 · eee department power systems lab manual...

EEE Department Power systems lab manual

Anurag college of Engineering, Aushapur

Experiment no:1

Characteristics of IDMT over current relay

Aim: To study the Operation of a Non- Directional (I D M T relay) and plot the inverse time

current characteristics.

Apparatus:

Non-Directional Over current relay --1 no

Time Totalizer --1 no

Fault creation Panel --1 no

Digital Ammeter --1 no

Theory:

IDMT relay is inverse definite minimum time relay. It is one in which Time of operation is

inversely proportional to magnitude of fault current near pickup value and becomes substantially

constant slightly above the pickup value of the Relay. Fault current and measure relay operation

time is used to conduct the experiment. Values recorded for various TSMs and PSMs.

Characteristics studied with the help of a graph and correlated with theory.



Circuit Diagram:



Procedure:-

1. Switch ON the MCB.

2. Initially rotor switch should be in OFF position.

3. Now set the described fault current by using the current source. For that switch

ON the rotor switch and move the current till the described fault current is

indicated in the ammeter.

4. Now move the rotor switch is OFF position and press the green button. Note down

the time in seconds after relay operated.

5. Repeat the same procedure for various T.S.M and P.S.M

6. Plot the graph between time take for relay to operate Vs P.S.M for various T.S.M.

Precautions:-

1. Disc must be stationary before applying fault current.

2. TSM setting must be changed with due care.

Readings and Tabular forms:-

PSM = TSM =

SNO. Fault

Current

Time of

operation

.

Expected graphs:



Result:

Conclusion:

Discussion questions:

1. Why CT is required in this experiment?

2. Can we design the experiment without Current Injection Unit?

3. What is TSM & PSM and why different TSM & PSM?

4. Identify different terminals of the relay and explain their use. Write them in your record.



Experiment no:2

Differential Protection on Single Phase Transformer

Aim: To study the differential protection scheme for a single phase transformer with unequal

turn’s ratio

Apparatus:

Single phase transformer, Current transformer, Single phase variac, Suitable ammeters and

over current relay.

Theory: A Differential relay responds to vector difference between two or more similar

electrical quantities. From this definition the Differential relay has at least two actuating

quantities say 1-1 and 2-1. The two or more actuating quantities should be same.

Ex: Current/Current.

The Relay responds to vector difference between 1-1 &2-1which includes magnitude and /or

phase angle difference. Differential protection is generally unit protection. The protection zone

is exactly determined by location of CTs. The vector difference is actuated by suitable

connection of CTs or PTs secondaries. Most differential relays are current differential relays

in which vector difference between current entering the winding & current leaving the winding

is used for relay operation. Differential protection is used for protection of Generators,

Transformers etc. Internal fault is created using switch and relay operation observed for various

TSMs. Relay operations for external faults can also be studied.



Circuit Diagram:

Procedure:

1. Make the connections as shown in fig.1

2. Select the transformation ratio 2:1 and the C.T. ratios of 2:1 and 4:1 Set PSM of the

relay equal to 0.5.

3. Apply rated voltage 230V to primary by varying the variac.

4. Without applying fault, note down different meter readings.

5. By applying load observe whether the relay is operating of not .

6. Now close the switch so as to create an internal fault.

7. Note the various ammeter readings when relay operates.

8. Create internal fault at different loads and note the various meter readings.

9. Now create an external fault and observe whether the relay operates or not, note the

various meter readings.



Readings and Tabular form:

Observation:- For Internal Fault

S.No I Primary I Secondary I relay I fault Relay

operates/doesn’t

operate

For External Fault

S.No I Primary I Secondary I relay I fault Relay

operates/doesn’t

operate

Result:

Conclusion:

Discussion Questions:

1. Why identical CTs are required in this scheme.

2. How would you take into account CT imbalances?

3. What do you understand by internal fault?



GHBH

Experiment no:3

Characteristics of Microprocessor Based Over Voltage and Under Voltage Relay

Aim:

Apparatus:

Theory:

Panel Layout

CIRCUIT DIAGRAM

MV12



Procedure:

(1) Switch on the MCB.

(2) Initially Rotary switch should be in OFF Position.

3) Now to set the desired fault voltage we will be using voltage source. For that switch ON

the Rotary switch marked as voltage set and move the voltage source till the desired fault

voltage is indicated on the voltmeter, it is quit possible that while adjusting the fault voltage

the trip LED will start blinking at a frequency of one second and will continue till relay trips It

will stop blinking and will remain permanently ON as soon as relay trips. Then you have to

reset the relay by pressing reset button. The Rotary switch must be brought in OFF position.

(4) Now the desire Fault voltage is SET. Now move the Rotary Switch on OFF position 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 voltage may be note down.

(6) Now plot the graph between time take for the relay to operate Vs Plug Setting Multiplier

at various T.M.S.

Tabular Column

Plug Setting = Voltage 121V

TMS= 1.0

TMS=

0.8 TMS= 0.6

S NO

Applied

Voltage Volts

Operating

Time Sec

Applied

Voltage

volts

Operating

Time Sec.

Applied

Voltage

volts

Operating

Time Sec.

1

2

3

4



5

Graph:

A graph of operation time v/s applied voltage for any one plug setting.

Result:

Conclusion:

Discussion questions:



Experiment no:4

Testing of CT, PT’s and Insulator strings

Aim: Testing of Polarity, Ratio and magnetisation characteristics of current Transformer and

Potential Transformer.

Theory:

Current Transformer and potential Transformer are known as instrument transformer. In a high

voltage and high current circuit direct measurement of voltage, current and other electrical

parameter instrument transformer are used. PT is connected to the main bus of the associated

power system, stepped down secondary voltage is measured by the instrument displayed actual

high voltage after proper calibration. Primary of CT connected in series with the equipment

whose current is to be measured and the secondary current measured by the instrument

displayed actual high primary current after proper calibration. Using CT and PT we can

electrically isolate the measuring devices from the main circuit and measuring device can be

placed any far distance in control room.

In case of a PT primary voltage rating according to the high voltage circuit but secondary

voltage rating is fixed to 110 V. In case of CT primary rating will be according to the high

current in primary circuit but secondary current rating is fixed to either 1A or 5A.

VA burden of CT and PT is being calculated from their voltage and current loading to the

maximum polarities of transformer define the phase shift between two sides, which is either 00

or 1800. There are two type of CT one is measuring CT and other is protection CT.

When a single CT output is fed to an ammeter or a PT output fed to a voltmeter there is no

need to know the polarity of transformer. But when CT and PT connected in star or delta in

case of three phase circuit they must be connected according to their polarity. Performing

polarity test we can find out the symmetrical terminals between primary and secondary

wingding of a CT and PT.

Using ratio test we can find out the voltage ratio for PT and current ratio for CT between

primary and secondary at different point throughout its full operating range. Through this test

we can also find out whether it is constant at all point or not.

Transformer core must not be saturated within its operating zone. If it saturated then due to

change of input voltage or current in primary make no change in secondary that cause the error

in measurement. We can determine its property from the data of ratio test. If the transformer

ratio never changes between its full operating ranges means that core of the transformer is

healthy condition.



Circuit diagram for Polarity test of PT using AC supply:

Circuit diagram of ratio test of PT using AC supply:

Circuit diagram for Polarity test of CT using DC supply:



Circuit diagram of ratio test of CT using AC supply:

Instrument And Equipment Used:

Procedure:

Polarity test of PT using AC supply:

Connect the circuit as per circuit diagram and marked the each terminals of primary and

secondary wingding input voltage at suitable value. Read the input voltage (V1) and output

voltage (V2).

If the voltmeter connected between primary and secondary reads the sum

of voltages of (V1 + V2). Then P1 and S1 are opposite polarity.

If the voltmeter connected between primary and secondary reads the difference of voltages of

(V1 – V2). Then P1 and S1 are same polarity.

Ratio test of PT:

Connect the circuit as per circuit diagram.

Connect the PT primary to the output of variable voltage source.

Sl. No Name of the

Apparatus

Quantity Type Marker’s Name

01

02

03



Increase the input voltage and take input and output voltage reading at least 10 point over its

operating range.

Find out the transformation ratio. Also find error from the given ratio of the transformer.

Polarity test of CT using DC supply:

Connect the circuit as per circuit diagram and marked the each terminals of primary and

secondary wingding.

Be sure the input voltage polarity to the input terminals and polarity of centre galvanometer

connection in the output terminals.

Operate the push button for a fraction of time. Find the direction of deflection of the zero centre

galvanometer.

If deflection is right hand side that means supply positive terminal and galvanometer positive

terminal is same polarity. If deflection is left hand side that means supply positive terminal and

galvanometer positive terminal is opposite polarity.

Repeat the whole experiments after reversing the source.

Ratio test of CT:

Connect the circuit as per circuit diagram.

Connect the CT primary to the output of variable current source.

Increase the input current and take input and output current reading at least 10 point over its

operating range.

Find out the transformation ratio. Also find error from the given ratio of the transformer.

Be sure that at any condition CT secondary will not open during primary energies.

Observation And Result:

Polarity test of PT:

Input voltage (V1) Output voltage(V2) Voltmeter between primary

and secondary

Polarity according to

marking

Ratio test of PT:

No of

observation

Input voltage (V1) Output voltage (V2) Transformation Ratio (V1 / V2)

01



Polarity test of CT:

Ratio test of CT:

Result:

Conclusion:

Questions:

Why secondary of CT is not kept open?

What do you know about ratio error & phase angle error?

02

03

Supply DC polarity to

the primary terminal

according to marking

Connected

galvanometer to the

secondary terminal

marking

Direction of deflection Polarity of terminals

according to marking

No of

observation

Input current (I1) Output current (I2) Transformation Ratio (I1/ I2)

01

02

03



Experiment no:5

Determination of Sequence Impedances of a Synchronous machines

Aim: To determine the Positive, Negative and Zero sequence of impedances or sequence

impedances of the given three phase alternator.

Apparatus:-

Three phase alternator DC motor set

Ammeter (0 –2.5 A) -----MC

(0—10 A) -----MI

Wattmeter

Auto-Transformer ( 0—230V)

Theory:

The positive, Negative and Zero sequence impedances of rotating machines are generally

different. The +ve sequence impedance of Synchronous generator is equal to the Synchronous

impedance of the machine. Experimental set up to conduct OCC and SCC is made available.

With the help of observations Synchronous impedance can be calculated. The –ve sequence

impedance is much less than +ve Sequence impedance. The zero sequence impedance is a

variable item and if its value is not given, it may be assumed to be equal to the +ve sequence

impedance. For Zero sequence impedance a separate model is used to conduct of experiment.



CIRCUIT DIAGRAMS: (A) POSITIVE SEQUENCE IMPEDANCE

(B) NEGATIVE SEQUENCE IMPEDANCE:



(C) DETERMINATION OF ZERO-SEQUENCE IMPEDANCE:

Procedure:-

I. POSITIVE SEQUENCE IMPEDANCE

1. Obtain the O.C.C of the Alternator by connecting the Alternator As in fig .1.

2. Obtain the S.C.C of the Alternator by connecting the Alternator as in fig .2.

3. Determine the stator resistance / phase by connecting as shown in fig.3.

Zs = O.C voltage / S.C current (at a same value of field current)

II NEGATIVE SEQUENCE IMPEDANCE

1. Connect the Circuit as shown in the fig.4.

2. Run the machine at rated speed with low excitation to the field of the Alternator.

3. The lines B and C shorted and the meters are connected as Shown

Cos Ø = W/ (VI)

Z 2 = V/ ( √(3 I)



III ZERO SEQUENCY IMPEDANCE

1. Connect the circuit as shown in the fig. 5.

2. Three phase winding are connected in series.

3. Apply low voltage to the Armature, so that rated full load current flow in the series field

winding.

Z 0 = V ao / Iao = 3E/I

Readings and Tabular forms:

Si.No. E I P Zo Xo

Result:

Conclusion:


1. Define +ve, -ve, Zero sequence impedances.

2. Why are they different for Alternators?

3. Can we analyze an unbalanced system otherwise?

4. What is the effect on the value of zero sequence reactance if the synchronous machine

is rotated at synchronous speed during the above experiment?

5. Out of all the reactance of synchronous machine which one is lowest?

6. What is the typical value of zero sequence reactance in per unit per large rating salient

pole alternator?



Experiment no:6

Finding the Sequence Impedance of Three Phase Transformer

Aim: -

To determine the Positive, Negative and Zero sequence (sequence impedance) of the

given three phase transformer

Apparatus: -

Ammeter -------- 0 – 5 A MI ----- 01

Voltmeter --------- 0 – 60 MI ----- 01

Theory:

Since Transformers have the same impedance with reversed phase rotation, their +ve and

–ve sequence impedances are equal. This value being equal to the impedance of the

Transformer. However, Zero sequence impedance depends upon the Earth connection. If

there is a through Circuit for the earth current, zero sequence impedance will be equal to

the +ve sequence impedance otherwise it will be infinite. Lab experiment is planned to

find out sequence impedances by creation of faults at secondary suitably and measure

impedances. Proper care is taken to ensure readings would not damage the equipment.

Circuit Diagram:



Procedure: -

1. Connect the given three phase transformer as shown in the circuit

Diagram 1

2. After short-circuiting the low voltage side adjusts the voltage on high

voltage side with the help of the autotransformer such that the rated

current flows in the windings.

3. Note down the voltage and current.

4. From these readings determine the transformer positive sequence

Impedance which is also equal to negative sequence impedance.

5. Determine the zero sequence impedance of the transformer by making

connections as shown in the 4. Circuit Diagram 2.

6. Note down the voltage and current.

I

VZ

31

12 ZZ

I

VZ

30

Readings and Tabular form:

Result:

Conclusion:




1. Define Sequence Impedances.

2. Explain symmetrical components method.

3. Why +ve, -ve sequence impedances same for a transformer?



Experiment no:7

LG, LL and 3 phase fault Analysis of 3 phase Synchronous Machine

Aim: To determine the fault currents on an unloaded synchronous generator for

(a) Line to ground fault (L-G Fault)

(b) Line to Line fault (L-L fault)

(c) Double Line to ground fault (LL-G Fault)

Apparatus Required:

Ammeter (0-10A) MI -- 1 No

Ammeter (0-2A) MC -- 1 No

Voltmeter (0-500V) -- 1 No

Procedure:

(a) L-G Fault:

1) Connect the circuit as per the circuit diagram for a line to ground fault on phase A

.

2) Calculate the determinate value of the fault current from impedances

(+,-, Zero sequences).

3) Run the generator rated speed.

4) Increase the field current of excitation so that terminal voltage is constant value.

5) Close the switch to create the L-G fault on Phase A.

6) Note the current and voltage in the ammeter and voltmeter.

7) Open the switch and remove the L-G fault on phase A.

8) Reduce the excitation and open the field circuit switch and switch of the prime

mover.

(b) L-L fault:

(1) Repeat the steps 1 to 6 for L-L and L-G faults.

(2) Connect the respective circuit in step 1.

(3) Generator is excited its a certain voltage as mention in step 2 of the procedure.

Note: This voltage must be such that it does not cause the rated current of the machine to

be exceeded.



(a) Circuit diagram for L-G fault:

Tabular column:

S.No. I in Amps Ef in Volts

Ia =3Ef/Z1+Z2+Z0 ------------------------------ (1)

Where Ia is fault current

Ef is the voltage to which the machine is excited

Z1+Z2+Z0 are the positive, negative and zero sequence impedances of the machine.



Verify the theoretical value calculated by using equation 1 with the actual value noted by

the ammeter.

(b) Circuit diagram for Line to line fault (L-L fault):

Tabular column:


Ia1 = Ef/Z1+Z2

Ib = a2Ia1+aIa2; Ia2 = -Ia1 Where a2 = (-0.5-j0.866)

a= (-0.5+j0.866)

Fault current calculated which must be verified with the actual value.



(c) Circuit Diagram for Double line to ground fault (LL-G fault):

Tabular column:


Calculations:

Va1 = Va2=Va0=Ef-Ia1Z1

Ia1 = Ef/Z1+(Z2xZ0/Z2+Z0)

Ia2 = -Va2/Z2 ; Ia0= -Va0/Z0

Ib = a2Ia1+a Ia2+Ia0

Where a2 =(-0.5+j0.866) ; a=(-0.5+l0.866)



In = 3 Ia0 = Ib+IC

Ic = - Ib=aIa1-a2Ia1

Result:

Conclusion:

eee department power systems lab manual · 2019-04-15 · eee department power systems lab manual...

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