Department of Electrical Engineering
Course Name: Electric Machines Lab
Instructor Name: Omar Tamimi
Course #:63392
Experiment Name: OCT & SCT Test for
single phase Transformer
Experiment #: 3
Academic Year: 2023-202
Students:
1. Omar khalil
2.
3.
Report's Outcomes
ILO A, B= (50) %
ILO D = (20) %
ILOA= (20) %
Evaluation Criterion
Introduction
Sufficient, Clear and complete statement of Objectives
Apparatus/Procedure
Apparatus sufficiently described to enable other experimenters
to identify the equipment needed to conduct the experiment.
Procedure sufficiently described.
Experimental Results and Calculations
Results analyzed correctly. Experimental findings adequately
and specifically summarized, in graphical, tabular and or written
form
Conclusions
Conclusions summarize the major findings from the
experimental results with adequate specificity.
References
Complete and consistent bibliographic information that would
enable the reader to find the reference of interest
Total
ILO K = (10) %
Grades
Points
Objectives:
This experiment demonstrates the concept of open-circuit and short circuit
testing will be performed on single-phase transformer. From these tests, we
can determine the transformer parameters, voltage regulation, To determine
the efficiency and regulation of a given single phase transformer by conducting
the OC and SC test and also to draw its equivalent circuit. In addition, we will
study the excitation and magnetization currents.
Apparatus Required:
-
Power supply unit PS189
2x Resistor banks LU178
Transformer trainer TT179
Standalone transformer
External wattmeter
Theory:
Electrical transformer: is a static electrical machine which transforms
electrical power from one circuit to another circuit, without changing the
frequency. Transformer can increase or decrease the voltage with
corresponding decrease or increase in current
Working Principle of Transformer The basic principle behind working of a
transformer is the phenomenon of mutual induction between two windings
linked by common magnetic flux. The figure at right shows the simplest form of
a transformer. Basically, a transformer consists of two inductive coils; primary
winding and secondary winding. The
coils are electrically separated but
magnetically linked to each other.
When, primary winding is connected
to a source of alternating voltage,
alternating magnetic flux is produced
around the winding. The core
provides magnetic path for the flux,
to get linked with the secondary
winding. Most of the flux gets linked
with the secondary winding which is
P a g e 2 | 15
called as 'useful flux' or main 'flux', and the flux which does not get linked with
secondary winding is called as 'leakage flux'. As the flux produced is
alternating (the direction of it is continuously changing), EMF gets induced in
the secondary winding according to Faraday's law of electromagnetic
induction. This emf is called 'mutually induced emf', and the frequency of
mutually induced emf is same as that of supplied emf. If the secondary winding
is closed circuit, then mutually induced current flows through it, and hence the
electrical energy is transferred from one circuit (primary) to another circuit
(secondary).
Basic Construction of Transformer Basically,
a transformer consists of two inductive windings
and a laminated steel core. The coils are insulated
from each other as well as from the steel core. A
transformer may also consist of a container for
winding and core assembly (called as tank),
suitable bushings to take the terminals, oil
conservator to provide oil in the transformer tank
for cooling purposes etc. The figure at left
illustrates the basic construction of a transformer.
In all types of transformers, core is constructed by assembling
(stacking) laminated sheets of steel, with
minimum air-gap between them (to
achieve continuous magnetic path). The
steel used is having high silicon content
and sometimes heat treated, to provide
high permeability and low hysteresis loss.
Laminated sheets of steel are used to
reduce eddy current loss. The sheets are
cut in the shape as E, I and L. To avoid
high reluctance at joints, laminations are
stacked by alternating the sides of joint.
That is, if joints of first sheet assembly are
at front face, the joints of following assemble are kept at back face.
P a g e 3 | 15
Open Circuit and Short Circuit Test on Transformer:
It is possible to predict the performance of a transformer at various levels of
load by knowing all the equivalent circuit parameters. These circuit parameters
are supplied in terms Open Circuit (OC) and Short Circuit (SC) test data of a
transformer. Without actually loading the transformer, these two assessed
tests give the test results, which are used to determine the equivalent circuit
parameters.
By these parameters, we can easily predetermine the efficiency and
regulation of the transformer at any power factor condition as well as at any
load condition. This method of finding the parameters of a transformer is called
as an Indirect Loading Method.
RemarkingThese two transformer tests are performed to find the
parameters of equivalent circuit of transformer and losses of the
transformer. Open circuit test and short circuit test on transformer are
very economical and convenient because they are performed without
actually loading of the transformer.
Open Circuit or No-Load Test on Transformer:
This test is performed to find out the shunt or no-load branch
parameters of equivalent circuit of a transformer. This test results the
iron losses and no-load current values, thereby we can determine the
no load branch parameters with simple calculations.
As the name itself indicates, secondary side load terminals of the
transformer are kept open and the input voltage is applied on the
primary side. Since this test is carried out without placing any load,
this test is also named as No Load Test.
How to Perform Open Circuit Test?
The open circuit (OC) test is carried out by connecting LV side (as
primary) of the transformer to the AC supply through variac,
P a g e 4 | 15
ammeter, voltmeter and wattmeter instruments. The secondary side
or HV side terminals are left open and in some cases a voltmeter is
connected across it to measure the secondary voltage.
The primary side voltmeter reads the applied voltage to the
transformer, ammeter reads the no load current, wattmeter gives the
input power and the variac used to vary the voltage applied to
transformer so that rated voltage is applied at rated frequency. The
OC test arrangement of a transformer is shown in below figure:
- 𝑅𝑒𝑞 𝑝 = 𝑅𝑝 + 𝑎2 𝑅𝑠
𝑎𝑛𝑑
𝑋𝑒𝑞 𝑝 = 𝑋𝑃 + 𝑎2 𝑋𝑠
When a single-phase supply is given to the transformer, the rated value of the
primary voltage is adjusted by varying the variac. At this rated voltage, the
ammeter and wattmeter readings are to be taken. From this test, we get rated
voltage VO = ES, input or no-load current IO=IS and input power Po=Ps.
P a g e 5 | 15
The no load power factor= Ɵ, cosƟ =
𝒀𝒐𝒄 =
𝑷𝑶
𝑽𝑶 𝑰𝑶
𝑰𝒐𝒄
𝑷𝒐𝒄
∠ − 𝐜𝐨𝐬 −𝟏
𝑽𝒐𝒄
𝑽𝑶𝑪 × 𝑰𝒐𝒄
𝒀=
𝟏
𝟏
+ 𝒋
,
𝑹𝑪
𝑿𝑴
We know that, when the transformer is on no load, the no load current or
primary current is very small, typically 3 to 5 percent of the rated current value.
Thus, the copper loss in the primary winding is negligible.
In OC test, transformer is operated at rated voltage at rated frequency so the
maximum loses will be the flux in the core. Since the iron or core losses are at
rated voltage, the power input is drawn to supply the iron losses by the
transformer under no load.
Short Circuit Test on Transformer:
The connection diagram for short circuit test or impedance test on transformer is as
shown in the figure below. The LV side of transformer is short circuited and wattmeter (W),
voltmeter (V) and ammeter (A) are connected on the HV side of the transformer. Voltage is
applied to the HV side and increased from the zero until the ammeter reading equals the
rated current. All the readings are taken at this rated current.
The ammeter reading gives primary equivalent of full load current (Isc).
The voltage applied for full load current is very small as compared to rated voltage.
Hence, core loss due to small applied voltage can be neglected. Thus, the wattmeter
reading can be taken as copper loss in the transformer.
P a g e 6 | 15
Therefore, P = Isc2Req....... (where Req is the equivalent resistance of transformer)
𝐙𝐬𝐜 =
𝐕𝐬𝐜
𝐏𝐬𝐜
∠ 𝐜𝐨𝐬 −𝟏
𝐈𝐬𝐜
𝐕𝐬𝐜 × 𝐈𝐬𝐜
Therefore, equivalent reactance of transformer can be calculated from the
formula Zeq2 = Req2 + Xeq2.
These, values are referred to the HV side of the transformer.
Hence, it is seen that the short circuit test gives copper losses of transformer and
approximate equivalent resistance and reactance of the transformer.
Why Transformers Are Rated In KVA?
From the above transformer tests, it can be seen that Cu loss of a
transformer depends on current, and iron loss depends on voltage. Thus, total
transformer loss depends on volt-ampere (VA). It does not depend on the phase
angle between voltage and current, i.e. transformer loss is independent of load
power factor. This is the reason that transformers are rated in kVA.
Transformer losses and efficiency:
Iron Losses
Eddy Currents
The magnetic core of a transformer consists of many laminations of a high-grade
silicon steel of a definite thickness. The power absorbed by the core of a transformer
is due to eddy currents and hysteresis and is called iron losses.
When the alternating flux cuts the steel core, an EMF is induced in each lamination,
causing a current (called an eddy current) to flow in the closed electrical circuit of the
lamination. This eddy current flows through the resistance in each lamination,
causing heat to be generated in the laminations and therefore in the core as a whole.
Although eddy-current losses are effectively reduced by using laminations for the
core, they are never entirely eliminated.
Hysteresis
The alternating flux also causes changes in the
alignment of the magnetic domains in the magnetic
core, with the magnetic polarity reversing 100 times a
second. This change is energy consuming and heat
is produced within the core. The energy loss is
referred to as hysteresis loss, the degree of loss
being dependent on the nature of the material used
for the laminations.
Silicon steel has low hysteresis losses, making it
suitable for electrical laminations. Figure 1 shows a
comparison of two hysteresis curves for different
P a g e 7 | 15
materials. It can be seen that the silicon steel curve has a smaller area, representing
a lower energy loss and reduced heat production.
The total iron losses represent the power
absorbed by the iron core and so are proportional
to Ie, the energy component of I0 in Figure 2. The
mutual flux Φ remains fairly constant from no load
to full load, therefore it follows that the excitation
current I0 producing that flux, and so Ie , will also
be constant.
The iron losses will be constant irrespective of the
load applied to the transformer. These iron losses
can be obtained by measuring the power
consumed on no load in what is known as a ‘noload or open circuit test’.
Figure 3 No-load or open-circuit test
The transformer is connected as in Figure 3 to a
supply at the rated voltage and frequency. The
primary current on no load is usually less than 3 per
cent of the full-load current, so the primary I2 R loss
on no load is negligible compared with the iron loss.
The wattmeter reading can then be taken as being
the total iron loss of the transformer.
Copper Losses
Another form of loss that occurs in a transformer is
copper loss, which is the energy lost in the windings
when the transformer is loaded. The resistance of each
winding is relatively low, but since the power dissipated
in each winding is proportional to the square of the
current flowing through that winding, it follows that the
copper loss is significant when the load current is high.
The total copper loss is PCu = I12 R1 + I22 R2,
where R1 and R2 are the resistance values of the
primary and secondary windings respectively. The
copper losses are not constant, but change according
to the square of the load current. The value of the
losses can be obtained by performing the short-circuit
test, as shown in Figure 4. The typically shaped curve
of copper losses can be seen in Figure 5.
P a g e 8 | 15
As Figure 4 shows, the secondary winding of the transformer under test is shorted
through the ammeter A2. An adjustable autotransformer is used to provide a lowvoltage supply to the primary winding of the transformer on test. The output of the
autotransformer is increased until the full rated current flows in the primary and
secondary circuits.
The supply voltage to the transformer is low, and the flux in the iron core is also
low, and so the iron losses are negligible. The power registered on the wattmeter W
can be taken as the total copper losses in the transformer on full load.
The efficiency varies with output current
𝜂=
𝑃𝑜𝑢𝑡
× 100%
𝑃𝑖𝑛
𝜂=
𝑃𝑜𝑢𝑡
× 100%
𝑃𝑜𝑢𝑡 + 𝑃𝑐𝑜𝑟𝑒 + 𝑃𝑐𝑢
𝜂=
𝑉𝑠 𝐼𝑠 𝑐𝑜𝑠𝜃
× 100%
𝑉𝑠 𝐼𝑠 𝑐𝑜𝑠𝜃 + 𝑃𝑜𝑐 + 𝑃𝑠𝑐
Why the SCT and OCT are performed?
Open circuit test The OCT is performed to estimate the core loss and
magnetizing reactance. In this test, one side is left in open circuit condition
while the rated voltage is applied to other side
P a g e 9 | 15
Short circuit testThe SCT is performed to estimate the series resistance and
reactance of the two windings. In this test, the secondary side is short
circuited, and the rated current is allowed to flow into the transformer from
primary side
Procedures:
- Open circuit test:
P a g e 10 | 15
- We connected the circuit as shown in figure7.
- The primary voltage adjusted at 220v
- The following table shows our readings:
𝑽𝟏 (𝑶𝑪)
volt
220
𝑰𝟏 (𝑶𝑪)
Amp
0.155
𝑷𝟏 (𝑶𝑪)
watt
12
𝑽𝟐
volt
232
- Short circuit test
P a g e 11 | 15
- We connected the circuit as shown in figure 8
- The secondary current adjusted at 1.36A
𝑽𝟏 (𝑺𝑪)
volt
9.6
𝑰𝟏 (𝑺𝑪)
Amp
1.46
𝑷𝟏 (𝑺𝑪)
watt
13.9
𝑰𝟐
Amp
1.36
- Efficiency test:
- The open and short circuit tests are performed to determine the
transformer’s efficiency.
- The core losses in a given transformer remains always unchanged
- In this part we will vary the supply voltage to change the secondary
current, then we will observe how the efficiency changes with the
current
- The efficiency can be found from the equation:
𝑃𝑜𝑢𝑡
𝜂=
× 100%
𝑃𝑜𝑢𝑡 + 𝑃𝑐𝑜𝑟𝑒 + 𝑃𝑐𝑢
%of full
load
𝐼2 (𝐴𝑚𝑝)
𝑃𝑜𝑢𝑡 (𝑤𝑎𝑡𝑡)
𝑃𝑐𝑜𝑟𝑒 (𝑤𝑎𝑡𝑡)
𝑃𝑐𝑢 (𝑤𝑎𝑡𝑡)
𝜂%
25%
50%
75%
100%
125%
0.34
0.68
1.02
1.36
1.7
75
150
225
300
375
12
12
12
12
12
0.98
3.4
8.2
14.5
22.0
85.24
90.69
91.76
92.88
91.68
P a g e 12 | 15
Analysis:
- Based on OCT results we can calculate the core loss and the magnetizing
reactance:
𝒀𝒐𝒄 =
𝑌𝑜𝑐 =
𝑅𝑐 =
0.155
220
1
2.48×10−4
∠ − 𝑐𝑜𝑠 −1
𝑰𝒐𝒄
𝑷𝒐𝒄
∠ − 𝐜𝐨𝐬 −𝟏
𝑽𝒐𝒄
𝑽𝑶𝑪 × 𝑰𝒐𝒄
12
0.155×220
= 4.033 𝑘. 𝑜ℎ𝑚 ,
= 2.48 × 10−4 − 𝑗6.59 × 10−4 mho
𝑋𝑚 =
1
6.59×10−4
= 1517.5 𝑜ℎ𝑚
- Based on SCT results we can calculate the equivalent resistance and the
equivalent reactance:
𝑍𝑠𝑐 =
𝑍𝑠𝑐 =
𝑉𝑠𝑐
𝑃𝑠𝑐
∠ cos −1
𝐼𝑠𝑐
𝑉𝑠𝑐 × 𝐼𝑠𝑐
9.6
13.9
∠ cos −1
= 6.33 + 𝑗1.77
1.46
9.6 × 1.46
𝑅𝑒𝑞 = 6.33 , 𝑋𝑒𝑞 = 1.77 𝑜ℎ𝑚
Equivalent circuit
- To find 𝐼𝑀 𝑎𝑛𝑑 𝐼ℎ+𝑒 , we know that the voltage across Rc the same as
Vp.
𝑰𝒆+𝒉 =
𝟐𝟐𝟎∠𝟎
𝟐𝟐𝟎∠𝟎
= 𝟎. 𝟎𝟓𝟒𝟓∠𝟎 𝑨 , 𝑰𝑴 =
= 𝟎. 𝟏𝟒𝟓∠ − 𝟗𝟎 𝑨
𝟑
𝟒. 𝟎𝟑𝟑 × 𝟏𝟎
𝒋𝟏𝟓𝟏𝟕. 𝟓
P a g e 13 | 15
𝑰𝒐 = 𝑰𝒉+𝒆 + 𝑰𝑴 = 𝟎. 𝟏𝟓𝟓∠ − 𝟔𝟗. 𝟒 𝑨
- The following graph shows the phasor diagram for currents:
eff.
100
90
80
70
60
50
40
30
20
10
0
0
0,5
1
1,5
2
2,5
P a g e 14 | 15
- We can see from the efficiency graph that as we increase the current,
the efficiency increases until we reach more than 100% of full load, at
which point it starts to decrease.
- At maximum efficiency Poc and Psc values are very close “approx. Equal”
Conclusion:
The open circuit test on transformer is used to determine core losses in
transformer and parameters of shunt branch of the equivalent circuit of
transformer. the Short Circuit test on transformer is used to determine copper
loss in transformer at full load and parameters of approximate equivalent
circuit of transformer, we observed how efficiency changes as the current is
changed. Reading the values with analog mustimeters was the main source of
inaccuracies in this experiment.
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