1. What you mean by accuracy limit factor?
The ratio between the accuracy limited primary current to rated primary current is
called the accuracy limit factor.
2. What is the characteristic of inverse time over current relay?
If the fault current increases the time of the operation of the relay will be decreases.
3. What are the two errors in instrument transformer?
a. Ratio error.
b. Phase angle error.
4. Where core balance CT is used?
Core balance CT is used in earth fault protection.
5. Define knee point voltage of a CT.
When the primary of a CT is open circuited and supply (variable) of system
frequency is given to secondary, then a 10% increase in voltage constitutes 50%
increase in current. That voltage is the knee point voltage.
At this point the core is saturated and a little increase in voltage constitutes a great
increase in current. kpv decides the opening range of the CT. Above kpv the ratio of
transformer will not be applicable.
kpv = RCT + RLEADS + RRELAY
6. What do you mean by the term 5P10?
This indicates the type of relay, Its % error and accuracy limit factor.
5 – composite error (Phase angle error + ratio error) 5%.
P – Protection CT.
10 – Accuracy limit factor.
7. Mention the important properties of relay contacts.
a. Should be robust in construction.
b. Self-cleaning (oxides easily breakdown).
c. Corrosion resistant.
d. Bounces free and striction free (low contact resistance).
e. Able to carry rated continuous current and short time rated current.
8. What is a composite error and write down the formula for composite error?
Basically composite error = Ratio error + Phase angle error. It is the ratio error
integrated over one cycle at steady state of operation.
Composite error =100 * 1 oςT (kn * Is – Ip)2 dt
T Ip
9. Define pickup value and reset value.
Pickup value: It is the smallest value of actuating quantity when its value is increased
from zero to pickup value, the relay will energise.
Drop out value: It is the largest value of the actuating quantity when its value is
decreased from pickup value, the relay will reset or de-energize.
10. Draw the circuit diagram for finding out the knee point voltage and explain the
procedure.
0 – 5 A
A CT
V 0 – 300V Sec Primary
240 V AC
Variac V Saturation
kpv = RCT + RLEADS + RRELAY
Knee point
Ankle point A
Connect the circuit as shown. O/P of variac should be zero. Increase it to 5 Volts and
take down the value of current from the ammeter. Now increase the voltage by 10%
(5 + 10% = 5.5 V) and take the current reading. Now increase the voltage by 10%
(5.5 V +0.55 V =6.05 V) and note down the current. Now keep on increasing voltage
by 10% and note down current reading. At some value there will be 50% increase in
current for 10% increase in voltage.
Example 40 V 0.2 A
40.4 V 0.3 A (0.2 + 50% = 0.3 A).
That point is the knee point voltage of that particular CT. From this point onwards a
little increase in voltage will lead to a large increase in current, because the core is
saturated fully. When we plot all the values on a graph taking current as X-axis and
voltage as Y-axis, we will get the above graph. Protective relays operate between
ankle point and knee point. Above this they cannot detect the fault correctly.
Measuring CT operate in the ankle region.
11. Explain the procedure for finding out the polarity and ratio test in a CT with circuit
diagram.
Polarity test:
Connect the circuit as shown in figure with a battery, switch and
AVOmeter.
Now momentarily close the switch S and see the deflection in the
AVOmeter.
If it is in the direction as shown in the figure, then the polarity of the CT
is correct. If it is in opposite direction the polarity of CT is not correct. Polarity test is
very important because if polarity is not correct in differential protection the relay
will fail to act when fault occurs.
Ratio test: Connect circuit as shown in figure (2). Slowly increase the current. Take
down the readings of A1 and A2. Then see whether it confirms to reading of
nameplate. Ratio = A1/A2.
SECONDARY INJECTION KIT
+AVO - A A1
0 – 30 A
+ - 240 V AC
S1 S2
P1 P2
+ -
S B
POLARITY TEST (FIG 1) Fig – 2 Ratio test
Ratio – A1 : A2 A2
A 0 – 15A
12. Explain the principle of operation of attracted armature relay with equation and
characteristics curve.
Principle: It works on the principle that when a current is passed through a coil
magnetic lines of force develop and the coil behaves like a magnet. When we place a
magnetic material inside the coil it is attracted.
In attracted armature type of relays there is a spring that keeps the contact open, a
plunger that tends to close the contact and a coil through which current is passed.
The spring force and magnetic force oppose each other. When these both are equal
the relay will pickups.
At verge (time) of pickup Instantaneous select
k1 I2 = k2
f = k1 I2 = k2 Time Time delay select
I = k2 / k1
Where f – force.
k1 – magnetic force constant.
k2 – spring tension constant. current
I – current in the coil.
We can see the inverse characteristics from the above formulae. Usually attracted
armature relays are instantaneous. That is there is no intentional (fixed) time delay. If
we want a time delay we can add a slug in the armature core.
13. Mention the initial commissioning checks on CT’s, PT’s and relays.
Commissioning checks on
CT
a) Terminal marking correctness.
b) Polarity of terminals.
c) Insulation resistance between primary and secondary.
d) Insulation resistance between primary to earth and secondary to earth.
e) Magnetization characteristic and knee point voltage test.
f) Ratio test.
PT
a) Terminal markings.
b) Polarity checks of terminals.
c) Insulation resistance between primary and secondary.
d) Insulation resistance between primary to earth and secondary to earth.
e) Ratio test.
f) Whether PT can supply as per the burden of load check.
RELAYS
a) Pickup and dropout value check.
b) Insulation resistance of contacts and relay coil.
c) Time delay (if relay is not instantaneous), operating time value check of relay.
d) See that the correct circuit breaker trips on energisation of the particular relay.
e) Continuity checks of contacts after energisation of relay.
f) See if plug-shorting contacts are correct.
g) See if CT’s and PT’s are corrected in correct polarity.
h) Burden check of relay.
i) Primary injection test.
j) Secondary injection test.
14. Explain with simple diagram the core balance CT.
In core balance CT all the three phases go through the core and the resultant
magnetic flux is zero. Because the flux of three phases cancel each other. So the
secondary output of CT is zero and the relay will not energise.
When there is a earth fault in one of the phase the fluxes cannot balance each other
and there is a voltage induced in secondary of the CT and the relay is energised to
trip the circuit. Saturation is no problem because the core size is very big.
+R
Ground fault
R R Y
Relay dropped Relay pickup
B R
R Y B R Y B - R
Normal operation. During earth fault. Resultant diagram.
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