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Tests of Single-Phase Transformer
Experiment Findings · October 2019
DOI: 10.13140/RG.2.2.19891.58401
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Ahmed M. T. Ibraheem Al-Naib
Northern Technical University
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Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Experiment (1)
Open Circuit Test (O.C. T.) of Single-Phase Transformer
(No Load Test)
Objective:
The main purpose of this test is to find the iron core losses ( ) and no load
current ( ) which are useful in calculating core loss resistance ( ) and
magnetizing reactance ( ) of the transformer (equivalent circuit parameters of
the core).
Theory:
The equivalent circuit of the transformer is shown in figures (1 - 2).
Fig.1: Elementary transformer
Fig.2: Development of transformer equivalent circuit
As secondary is open (no load),
= 0, hence secondary copper losses (in
) are zero. And the no load current which equal to
( = ) is very
low ( ˂ 6% of load current), hence copper losses on primary (
) are
also very low. Thus the total copper losses (primary and secondary) in this test are
negligibly small, hence neglected. Therefore the wattmeter reading ( ) in this test
gives core losses (in
&
) which remain constant for all the loads (i.e.
will
only flow through core circuit, for this reason, it is possible to calculate the core
circuit parameter (
)) as shown in figure 3.
1
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Fig. 3: Equivalent Circuit on No−Load
The no load current
has two components as shown in phasor diagram as
shown in figure below: these component are:
, and
Fig. 4: phasor diagram
Power input can be written as:
Then :
Once
and
are known we can determine core circuit parameters as:
&
Procedure:
1- Connect the circuit shown in the Fig. 5; the primary winding is connected to AC
supply through ammeter and wattmeter, keeping secondary open. Usually low
voltage side is used as primary and high voltage side as secondary to conduct O.C.
test.
2
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Fig. 5: Circuit diagram of O.C. test
2- Switch on the supply after checking connection by concerned teacher.
3- The primary is excited by (80 - 90 - 100 - 110V (rated voltage)), which is adjusted
precisely.
4- When primary voltage is adjusted to its rated value with the help of variac (autotransformer), readings of ammeter and wattmeter are to be recorded (Take the
readings of all meters at each voltage step; record it in table (1) below).
Table (1): Open circuit test readings
(V)
80
90
100
110
(W)
(A)
(V)
5- Reduce the voltage slowly, and then Switch off the supply and remove
connections.
Report:
1- Calculate the value of equivalent circuit parameters of core (
,
) at each
value of applied voltage (80 – 90 – 100 – 110V) according the above equations
depending on the table (1).
2- What are the approximations are made in this experiment?
3- Why the transformer core made from lamination plates?
4- Why core losses remain almost constant at any load?
5- Why O.C.T. is generally performed on L.V. side (not at H.V. side)?
3
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Experiment (2)
Short Circuit Test (S.C.T.) of Single-Phase Transformer
Objective:
The main purpose of this test is to find:
- Full load copper loss, this loss is used in calculating the efficiency of the
transformer.
- Winding parameters (
&
or
&
).
Theory:
In this test usually LV side is shorted with the help of ammeter and meters are
connected on HV side. A variable low voltage is applied (usually 5 to 10% of
normal primary voltage) to the HV winding with the help of an auto-transformer.
This voltage is varied till the full load currents flowing both in the HV and LV
side.
The wattmeter indicates the full load copper losses and iron losses. As the
voltage applied is low which a small fraction of the rated voltage and iron losses
are function of applied voltage, hence iron losses are negligibly small. Now the
current flowing through the windings are rated current hence the total copper loss
is full load copper loss. Hence the wattmeter reading represent the full load copper
losses (
) for the whole transformer. The equivalent circuit of the transformer
under short cicuit condtions is shown in fig. 1.
Fig. (1): Approximate circuit of transformer on short circuit
If
is the voltage required to circulate rated load currents, then
Also
= Full load copper loss
√
4
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Thus we get the equivalent circuit parameters
,
and
. Knowing the
transformation ratio K, the equivalent circuit parameters referred to secondary also
can be obtained.
Procedure:
1) Connect the circuit as shown in fig. (2).
Fig. 2: Circuit diagram of short circuit test
2) Switch on the supply after checking connection by concerned teacher.
3) Increase the input voltage very carefully and slowly so that the current in
secondary winding reaches rated value (from 2A to 5A rated value).
4) Record the readings of meters in table (1).
Table (1): Short circuit test readings
(V)
(A)
(W)
(A)
2
3
4
5) Reduce the voltage slowly, and then switch off the supply and remove
connections.
Report:
1. Draw the copper losses ( ) against the input current ( ).
2. Calculate the rated value of
,
, and
.
3. What are the approximations are made in this experiment?
4. Why S.C. test is generally performed with L.V. side short circuited?
5. Why the wattmeter readings are considered the copper losses?
6. Discuss all your results clearly and briefly.
5
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Experiment (3)
Load Test of Single-Phase Transformer
(Resistive Load)
Objective: to study the characteristics of the transformer under resistive load
condition.
Theory:
When secondary is loaded, the secondary current
and phase of
with respect to terminal voltage
- If load is resistive then
will be in phase with
is set up. The magnitude
depends on the type of load:
, thus the power factor = 1, as
shown in fig. 1-a.
- For inductive load
will lag behind
, as shown in fig. 1-b.
- For capacitive load it will lead the voltage
*(a)
), as shown in fig. 1-c.
(b)
Fig. 1: Vector diagram for various load conditions
Because of this secondary current
(c)
, there is a drop in terminal voltage
.
Drop in voltage depends on the impedance of load & power factor. For leading
power factor voltage drop may be negative and for lagging power factor it is
always positive.
Procedure:
1) Connect the circuit as shown in fig. 2.
6
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Fig. 2: Circuit diagram of load test
2) Apply the rated voltage (220V) to the primary side.
3) Connect the resistive load across the secondary winding of the transformer;
(switch on Sw).
4) Record reading of all measurement instruments in table (1) before and after
connected the load with different values.
Note: Read the name plate of the transformer:
Rated Apparent Power (S) of the transformer =
⁄
Calculate the rated load current ( ) =
=
KVA, secondary voltage =
V
A.
Table (1): Load test readings
%
PF
(V)
(A)
(W)
(A)
(V)
(W(
220
0
110
220
1
1
220
2
1
220
3
1
220
4
1
1
0
0
0
5) Reduce the load current slowly, and then Switch off the supply and remove
connections.
6) Complete the calculations according the following equations:
Report:
1- Plot the efficiency against secondary current.
2- Plot voltage regulation against secondary current.
3- List the different losses which take place in a transformer.
4- What are the factors that affect the efficiency?
5- What are the various causes of voltage drop in a transformer?
6- Draw the vector diagram at unity power factor (resistive load).
7- Why the transformer rating in KVA?
8- What is the condition for maximum efficiency? Derive it.
9- List the different losses which take place in a transformer.
7
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Experiment (4)
Parallel Operation of Single-Phase Transformer
Objective:
Two transformers may work together (in parallel) to share the load which is
higher than the rated load of any them.
Theory:
For supplying a load in excess of the rating of an existing transformer, a second
transformer may be connected in parallel.
Let
&
be the total impedance referred to the secondary side of Tr. 1 and
Tr. 2 respectively. The equivalent circuit referred the secondary side is shown in
fig. (1) below.
⁄
⁄
⁄
⁄
Since
then
⁄
⁄
⁄
&
Thus the relative sharing can be varied by inserting additional impedance in
either of the secondaries of the two transformers.
8
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
Procedure:
1- Connect the circuit as shown in fig. (2).
Fig. 2: Circuit diagram of parallel operation test of single phase transformer
2- Be certain that both switch ( w1 & w2) are open.
3- Apply a small percentage of the rated primary voltage across the two primaries,
then note:
a- If the reading of
voltmeter equal to zero (or value approximate to zero),
this mean that two transformers are ready to operate in parallel.
b- If the reading of
voltmeter equal to summation of voltages across
secondary side for two transformers, this mean be twice the secondary
voltage and one of the windings has to be reversed to obtain (
When
= 0 then switch on w1
4- Apply a rated primary voltage 220 V across the two primaries.
9
= 0).
Ahmed M. T. Ibraheem Alnaib, Lecturer.
Dep. of Electrical Power Technology Eng., Technical college / Mosul, Northern Technical University.
5- Switch on w2, then apply load in steps up to 120% of the rated secondary
current and take the reading of all reading of all meters as in table (1).
Table (1): Reading of parallel operation test
V1 (V) Iin(A)
I1(A)
I2(A)
IL (A) VL (V)
P1(W)
P2 (W)
Report:
1- Draw the current I1 and I2 against load current.
2- Draw the power P1 and P2 against load current.
3- Draw the current phasor diagram for two transformers connected in parallel.
4- What are the necessary conditions for successful operation of two single phase
transformers?
5- How the load can be divided between the two transformers?
……………………………………………………………………………
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