Transcript short circuit test

,
TOPIC : O.C. , S.C. AND
SUMPNER TEST
Prepared
by :
(1) Patel Sandip A.
(130450109046)
(2) Patel Tarun R.
(130450109047)
(3) Patel Yash P.
(130450109048)
(4) Prajapati Jigar V.
(130450109049)
(5) Prajapati Urvashi H. (130450109050)
1. OPEN CIRCUIT TEST (O.C TEST)
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Figure shows the O.C test on transformer.
The primary voltage of the transformer is connected to
a rated ac voltage by using a variac.
A voltmeter is connected across the primary winding to
measures a primary winding.
The ammeter is used to measure a primary current.
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The secondary is open circuited because it is
an open circuit test.
Note that the ac supply voltage is applied at
the low voltage side which is primary side.
At the secondary side higher voltage is used.
The wattmeter is used to measure the input
power.
Procedure
1.
2.
3.
4.
5.
:
Connect the circuit as shown in
figure.
Keep the variac at its minimum
voltage position.
Switch on the power supply and
adjust the variac to the rated voltage.
Now measured a primary current,
voltage and power using voltmeter,
ammeter and wattmeter.
The ammeter gets primary current Io
and wattmeter gets power Wo.
6.
The observation table for the O.C test is as follows.
Voltmeter
reading
V1
(rated voltage)
7.
8.
Ammeter
reading
Wattmeter
reading
Io
(ampere)
Wo
(watt)
The no load current Io is very small as compared to
the full load primary current.
As I2 is zero, the secondary copper loss is zero.
9.
The primary copper loss will be negligible because of
Io is small.
10.
There for the total copper loss is very small and can
be assumed to equal to zero.
11.
Hence the wattmeter reading Wo represents the iron
losses
Wo = Pi = Iron losses
SHORT CIRCUIT TEST (S.C TEST)
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Variac is used to adjust the input voltage to
the rated voltage.
The voltmeter is connected to measure the
primary voltage.
The ammeter measures the short circuit rated
primary current.
The wattmeter measures the short circuit
input power.
The secondary is short circuited.
Procedure
:
1.
Connect the circuit as shown in figure.
2.
Short circuit the secondary which is low
voltage high current, low resistance
winding.
3.
Keep the variac at its minimum voltage and
switch on the ac supply.
4.
By increase the primary voltage gradually
note down the wattmeter, voltmeter and
ammeter reading.
5.
The observation table is shown
Voltmeter
Reading
Ammeter
Reading
Wattmeter
Reading
Vsc
Volts
Isc
Ampere
Wsc
Power
Parameter
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calculation :
The primary and secondary currents are the rated
currents.
Therefore the total loss is the full load copper loss.
The iron losses are a function of applied voltage. As the
applied voltage in S.C the iron loss will be negligibly
small.
Hence the wattmeter reading
Wsc = Full load copper loss = Pcu
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We can calculate the parameters R1, X1 and Z1of the
equivalent circuit from the short circuit.
We know that Wsc = Vsc Isc Osc
Hence the short circuit power factor is given by
cos Osc = Wsc / Vsc Isc
But wattmeter reading Wsc indicated the full load
copper loss
Wsc = copper loss = I’sc X R1
R1 = Wsc / I’sc
Efficiency
calculation from
O.C and S.C test :

We can obtained the value of efficiency at any power
factor cos O, or at any load which is a fraction of full
load.
efficiency(n) = x V2 I2 cos O / (x V2 I2 cos O) +
Wo + x’ Wsc
Calculation
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of regulation :
The percentage regulation (%R) is given by,
%R = I1 R1 cos O + I1 X1 sin O / V1
X100
%R = I2 R2 cos O + I2 X2 sin O / V2
X100
We can the obtain the parameters such as R1, X1, R2,
X2 from the S.C test of the transformer.
Where as the voltage V1, V2, and current I1, I2 in the
above equation known from the given transformer.
Polarity
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of transformer :
The transformer works on the principle of mutual
inductance.
To indicate the winding direction easily the dot
convention is used.
If two coils are mutually coupled then the terminals
bearing the dots will be having in phase induced
voltage.
The dots are marked on the primary and secondary
winding to indicate the similar polarity of the two
winding.
SUMPNER OR BACK TO BACK TEST :
The open circuit test and short circuit test carried
out over a transformer are useful to obtain its
equivalent circuit.
 But the o.c. and s.c. tests are not useful for the
heat run test in which the temperature rise in
the transformer is to be monitored by fully
loading it continuously.
 This is because in o.c. test the transformer is
subjected only to the core loss will in s.c. test it is
subjected only to the copper loss but the
transformer is not subjected simultaneously to
both this losses.

The solution to this problem is to carry out the
sumpner test.
 The sumpner test can be carried out
simultaneously on to identical transformers.
 The set up for sumpner test is shown in fig. this
test is also called as back to back test .

SET UP :
As shown in fig. the primary windings of the two
transformers are connected in parallel with each
other across the rated voltage supply V1.
 The secondary windings of the two transformers
are connected in series opposition as shown. The
indication of correct connection is that the
voltage across the terminals T2 and T4 is zero.
 A low voltage supply (V2 ) is connected across the
series connection of the two secondaries to inject
a current Ifl at low voltage into the secondary
voltage windings.
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OPERATION :
 With
V2 assumed to be zero :
According to superposition theorem ,V2 source is
assumed to be short circuited , then the two
transformers will appear in open circuit to source
V1.
This is due to the fact that the secondary
windings are in phase opposition and so no
current can flow through them.
So the total current drown from source V1 is the
sum of no load currents of the two transformers
i.e. 2Io.
So the reading of wattmeter W1 is given by ,
W1 = 2Po = 2Pi
= 2 × core loss of each transformers
Table shows the reading of various meters on the
primary side to V2 = 0.
V1
Rated
voltage
I1
2Io
W1
2Po
 With

V1 assumed to be zero :
Now imagine that V1 is short circuited. So V1= 0.
this is equivalent to two transformers in series
connected across V2 and are short circuited on
their primary side.
So the impedance seen by the source V2 will be
2 ZΩ.
the source voltage V2 is adjusted such that a full
load current Ifl is circulated on the secondary
side. So I2 = Ifl.

And the second wattmeter W2 reading will be
2 Pc i.e. twice the full load copper loss of each
transformer.
W2 = 2 Pc
= 2 × full load copper loss of each transformer.
table shows the reading of various meters on the
secondary side for V1 = 0.
V2
Adjusted to
get I2 = Ifl
I2
Ifl
W2
2 Pc
CONCLUSION :
In the sumpner test , even though the
transformers are not supplying any load , full
iron loss takes place in their cores (2 Po) and full
copper loss takes place in that windings (2 Pc).
 So the net power input is given by ,
Pт = W1 + W2 = 2 (Po + Pc)
 Hence the heat run test can be conducted on the
two transformers when no actual load is supplied
but only the losses are being supplied.
It is possible to conduct this test on the three
phase transformers as well.

THANK YOU