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### Single Phase Transformer Experiment

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Single Phase Transformer Experiment

Aims.

The aims of this experiment are to calculate the turn's ratio of the
transformer (in the open and short circuit test), calculate the
inductive reactance and rm from the values in the open circuit test.
In the short circuit test calculate R1 and X1. Then finally in the
load test calculate voltage regulation and the efficiency compare
these results to the voltage regulation and efficiency from the
equivalent circuit. I will then predict the voltage regulation and
efficiency if the load used above had a power factor of 0.8 lagging.

Objectives.

To determine the approximate equivalent circuit of a single-phase
transformer. This will enable me to calculate all the different
parameters in the open-and short- circuit tests. Enabling me to
predict results for an actual circuit and also compare values between
actual and equivalent circuits to see how accurate the estimation or
prediction is.

Equipment.

TecQuipment electrical machines teaching unit NE8010 or NE8013, with
the B-phase transformer (EMTU-TT01) on the bench. One feedback
electronic wattmeter, one Multi-range moving-iron ammeter and one
instrument voltage transformer. Electrical wires where used for the
connections between the components of the circuits.

Theory and Introduction.

A Transformer is a device that transfers electric energy from one
alternating-current circuit to one or more other circuits, either
increasing (stepping up) or reducing (stepping down) the voltage.
Transformers are employed for widely varying purposes; e.g. to reduce
the voltage of conventional power circuits to operate low-voltage
devices, such as doorbells and toy electric trains, and to raise the
voltage from electric generators so that electric power can be
transmitted over long distances. Transformers are widely used in power
systems for the stepping up the generated voltage from about 20kV to
400kV for efficient transmission and then down again in stages
(typically 132kV, 33kV and 11kV) for distribution to industrial
consumers and finally to 230V for domestic purposes. Power system
transformers are usually 3-phase devices, but their basic concepts are
most easily demonstrated using a single-phase transformer.

Transformers change voltage through electromagnetic induction; i.e. as
the magnetic lines of force (flux lines) build up and collapse with
the changes in current passing through the primary coil, current is
induced in another coil, called the secondary. The secondary voltage
is calculated by multiplying the primary voltage by the ratio of the
number of turns in the secondary coil to the number of turns in the
primary coil, a quantity called the turns ratio. So transformers
operate by mutual induction, with energy being transferred between two
(or more) separate windings via a coupling magnetic field. Their
performance can be modelled, predicted and analysed using equivalent
circuits.

When a transformer is not connected to a load, the secondary coil does
nothing. The primary coil nevertheless takes a small current, I0, made
up of two components: the magnetising current which provides the
useful flux linking the coils, Im, and the current supplying the power
lost in the core, Il. I0 remains constant whether the transformer is
loaded or not. Im lags Il by 90o, so the transformer's equivalent
circuit must contain a core loss resistance, R0, in parallel with an
inductive reactance, Xm, both in parallel with the primary winding.
The primary and secondary coils each have small resistance, R1 and R2,
in series with the windings, representing the coil resistance.

[IMAGE]

the single impedance Z1 comprises

* A primary circuit resistance R1, which is the sum of the primary
and referred secondary winding resistances.

* A primary circuit reactance X1, which is the sum of the primary
and referred secondary leakage reactances.

I0 is the no-load primary current, it comprises a reactive or
magnetising component I0r, which produces the flux and an active or
power component I0a, which supplies the losses and is in phase with V1.

The no-load current and power-factor are respectively

I0 = âˆš(I0a)2 + (I0r)2 and cos Î¸0 = I0a

I0

The power factor is usually low on no-load because I0>> I0a.

* I'2 is the component of the primary current which compensates for
the secondary load current. I'2 = I2(N2/N1)

* I1 is the total primary current, the phasor sum I0 and I'2.

* I'2R1 is the voltage drop associated with the winding resistance
and is in phase with I1.

* I'2X1 is the voltage drop associated with the total leakage
reactance and is in phase quadrature with I1.

* Cos Î¸1 and cos Î¸2 are the primary and secondary power factors
respectively.

* V2 = V'2(N2/N1)

Because the no-load current is relatively small rm and xm are often
neglected when considering the behaviour near full-load.

Expressions for both the voltage regulation and the efficiency can be
derived from the approximate equivalent circuit.

Voltage regulation is the change in the secondary voltage between
no-load and full-load. It is expressed as a fraction (either per-unit
or per-cent) of the no-load voltage with the primary voltage with the
primary voltage assumed constant.

Efficiency is the ratio of output power and input power. It can be
expressed either in per-unit or per-centage terms.

Measurements.

Open-circuit test.

The actual circuit is shown in Appendix 2.

V1 /Volts

I0 /Amps

Poc /Watts

V2 /Volts

Deflection

AmpRange

VoltRange

80

0.042

2.4

40

0.24

0.1

100

100

0.049

3.5

50

0.35

0.1

100

120

0.056

4.6

59

0.23

0.1

200

140

0.064

6.2

69

0.31

0.1

200

160

0.073

7.6

79

0.38

0.1

200

180

0.084

9.4

89

0.47

0.1

200

200

0.098

11.0

99

0.22

0.1

500

220

0.117

13.0

109.5

0.13

0.2

500

240

0.141

16.0

119

0.16

0.2

500

260

0.173

18.0

129

0.18

0.2

500

Here is an equivalent circuit diagram for this test: -

[IMAGE]

Short-circuit test.

The actual circuit is shown in Appendix 3.

I2 = 6A

I1 = 2.97A

Vsc = 123/5 = 24.6V

Amp range = 5

Volt range = 50

Deflection = 0.12

Psc = 30 W

Here is an equivalent circuit diagram for this test: -

[IMAGE]

Load Test.

The actual circuit is shown in Appendix 4.

V1 = 230V

I1 = 0.131A

Switch

V1 /Volts

I1 /Amps

V2 /Volts

I2 /Amps

Deflection

AmpRange

VoltRange

Pin/ Watts

Off

231

0.131

115

0

0.14

0.2

500

14

Pos 1

231

2.74

110

5.5

0.255

5

500

637.5

Output Power = V2I2 = 110 x 5.5 = 605 W

Here is an equivalent circuit for the load test: -

[IMAGE]

Calculation of Equivalent Circuit Parameters.

Open-circuit test.

The graph of Primary current vs. Primary voltage is shown in Appendix
5, the second graph with Input power and Power lost against Primary
Voltage is shown in Appendix 6, the results below are read directly
from the graph and table above.

The rated primary voltage = 240 Volts.

The values for the following at this rated voltage are;

* Primary current = 0.141 Amperes

* Secondary voltage =

If the primary voltage is 240 Volts (the rated voltage) then I will
use the turn's ratio formula to find the secondary voltage.

N1 = V1 = 2.02 = 240 Therefore V2 = 240 = 118.8 Volts.

N2 V2 V2 2.02

* Input power = 16.0 Watts

Turns-Ratio = N1 = V1 = 160 = 2.02

N2 V2 79

rm = V12 = 1602 = 3368.42

Poc 7.6

V1 160

[IMAGE][IMAGE]Xm = = = 2886.40

âˆš (Io)2 - (V1/rm)2 âˆš 0.0732 - (160/3368.42)2

The lines on the graph involving the power loss and input power
against the primary voltage are the same to one decimal place, which
is not realistic. If power loss = power input then there would be no
power output. If I look closely at the results worked out for the
graph, say for 220V (the rated voltage), gives 13.00000 Watts for the
input power, where for power loss the value is 12.99999. This shows
that there is a small amount of power which is not lost; this power is
then amplified at the secondary coil.

Input Power /W

Power loss /W

13

12.99999

Short-circuit test.

R1 = Psc = 30 = 3.40â„¦

[IMAGE][IMAGE] I12 2.972

[IMAGE][IMAGE]

[IMAGE][IMAGE]X1 = (Vsc)2 - R12 = 13.86

âˆš ( I1 )

[IMAGE]

Load Test

Voltage regulation = V2(no load) - V2(loaded) [x 100%] =

V2(no load)

115-110 [x100%] = 4.35%

115

Efficiency = output power = V2 I2 [x 100%] = 110 x 5.5 [x 100%]

Input power Pin 637.5

Efficiency = 94.9%

Calculation of the Voltage regulation and Efficiency for the load test
using the Equivalent circuit parameters.

(In these calculations the equivalent circuit parameters calculated
and listed above are used, Î¸ in these calculations will be considered
as being zero)

Voltage regulation = I'2 (R1 cos Î¸2 + X1 sin Î¸2) [x 100%]

V1

= 2.72 (3.4) [x 100%] = 4.00%

231

Efficiency = V2 I2 cos Î¸2 [x 100%] =

V2 I2 cos Î¸2 + V12/rm + I'22 R1

Efficiency = 110 x 5.5 [x 100%] =

110 x 5.5 + (2312/ 3368.42) + 2.72 x 3.4

Efficiency = 96.0%

Comparison between the load test and the voltage regulation
calculations.

I worked out the voltage regulation and the efficiency of the load
test in 2 different ways, firstly I used the method where I use to
values from the actual experiment and use them in the formulae above.
The second method was where I used the equivalent circuit parameters
calculated to using the results from the open and short circuit tests,
and then used these in the formulae to calculate the voltage
regulation and the efficiency.

My results show that when I calculate the voltage regulation with the
experiment values (actual circuit), the change in the secondary
voltage between no-load and full-load is 4.35%. This value is 0.35%
greater than the value calculated using the equivalent circuit
parameters. So therefore the voltage regulation calculated with the
equivalent circuit formula is 4.00%.

The Efficiency values are the other way around. For the experiment
values used in the efficiency equation, the efficiency value is 94.9%.
This value is less than the efficiency value produced when the
equivalent circuit parameters are used in the equation. This value is
96.0%.

My values are very close to the correct trend from my knowledge. With
the circuit not being exactly perfect due to losses, such as power
losses due to heating in the wires around the circuit. There are also
eddy currents and hysteresis within the core of the transformer. All
these losses add up to alter the efficiency of the circuit. With the
calculations from the actual circuit the efficiency was calculated and
it was lower then the efficiency calculated using the equivalent
circuit formula. I can see that the actual circuit formula takes all
the losses into account, due to the value of the voltage recorded is
used in the formula. With the equivalent circuit parameters being only
estimation, this leaves with the opinion of it being less accurate,
with also the knowledge of the formula not taking into account the
heat losses and power losses in the wire, which is bound to affect the
efficiency of the circuit performance.

The voltage regulation difference shows that in the equivalent circuit
the change from voltage with no load to voltage with a load has a
smaller change (smaller percentage) therefore the circuit is more
efficient. So with the actual having a larger change (larger
percentage) then the efficiency will be less.

Areas of Application of Transformers.

The control of transformer ratio under load is a desirable means of
regulating the voltage of high-voltage feeders and of primary
networks. It may be used for the control of the bus voltage in large
distributing substations. It finds a wide field of application in
controlling the ratio on step-up transformers operating from power
stations whose bus voltage must be varied to suit local distribution.

In industrial work, it is used for the control of current in a variety
of furnace operations and electrolytic processes. It all furnishes a
convenient means for voltage regulation of concentrated industrial
loads.

Power transformers are for use in connection with a.c. power supplies
are chiefly characterized by a constant primary voltage and a single
frequency. These are used for power distribution, particularly in
aircraft.

A wide-band transformer is one which has to work well over a wide
range of frequencies. The most common example is the audio-frequency
transformer for transmitting speech and music; there are also
applications in the supersonic range.

Television scanning transformers have a close analogy with the current
transformer; they are used to handle the special current-waveforms
which produce the 'raster' or rectangle of parallel lines on the
screen of a television receiver. Each line is a result of a rapid
horizontal movement of the cathode-ray spot. A slower vertical
movement of the spot causes the repeated horizontal movements to build
up the raster. The deflections of the cathode-ray beam are usually
brought about by two pairs of magnetic deflecting coils carrying
currents of saw-tooth waveform. These currents may be produced in a
number of different ways, but a common method is to pass the anode
current of a pentode through the primary winding of the scanning
transformer with its secondary closed by a pair of deflecting coils. A
saw-tooth voltage applied to the grid, of the pentode produces the
desired deflections of the spot.

The widespread use of radar in World war two involved the development
of transformers capable of supplying electrical energy in brief,
intense pulses to the radio transmitting valves. The pulse technique
finds numerous applications outside the field of radar. In electrical
measurements of all kinds when the continuous application of voltage
is not allowable, because of over-heating or for other reasons, the
use of pulses offers a solution.

Discussion with ECP and experiment.

Equivalent Circuits are used throughout transformer businesses to
predict the voltage regulation and efficiency incase of problems in
circuit, consistency in a circuit or maybe boosts at predicted times
like in T.V viewing when there is something very popular being viewed,
a lot more people watch T.V and watch a specific channel. In this
experiment I have used the equivalent circuit formulae to check and
compare with the actual values. In a simplified equivalent circuit
shown in the theory section would give a voltage regulation around 3-5
per cent. The values I calculated were 4.00% and 4.35% which are
within the ideal values. Due to the ability of equivalent circuit
prediction your able to predict the value of voltage and current say
for houses or factories, because if you don't have enough voltage then
nothing will work, but if you have too much then you will blow the
accessories. So it is very helpful being able to predict and estimate
fairly accurately.

The efficiency calculations where reasonable in the way that the
transformers where efficient and that they followed the trend of the
voltage regulation, where the higher the voltage regulation the less
efficient the circuit was.

The calculations have proved to me that the equivalent circuit method
has worked, enabling me to predict the results of a transformer and
effects with certain loads.

Â· Now to predict the voltage regulation and efficiency if the load
used above had a power factor of 0.8 lagging.

(So now instead of the Î¸ being a value of zero, the cos Î¸ now becomes
the value 0.8 therefore the value for sin Î¸ is 0.6,and due to it
lagging then in the voltage regulation equation the sign in the middle
must be an add and not a minus.)

Firstly I had to calculate the primary current which compensates for
the secondary load current (I'2).

Using the formula I'2 = I2 (N2/N1) = 2.72

Voltage regulation = I'2 (R1 cos Î¸2 + X1 sin Î¸2) [x 100%]

V1

= 2.72 (3.4 x 0.8 + 13.86 x 0.6) [x 100%] = 13.0%

231

Efficiency = V2 I2 cos Î¸2 [x 100%] =

V2 I2 cos Î¸2 + V12/rm + I'22 R1

Efficiency = 110 x 5.5 x 0.8 [x 100%] =

110 x 5.5 x 0.8 + (2312/ 3368.42) + 2.72 x 3.4

Efficiency = 95.1%

Voltage regulation goes up a lot due to the load being 0.8 lagging.
Now it is 0.8 lagging, and therefore cos Î¸ = 0.8, so therefore sin Î¸ =
0.6. This then involves X1 into the equation, which is the leakage
reactance. The leakage flux largely determines how good the
transformer is at maintaining its voltage load, so it is essential to
minimise it. With the load being added, the transformers effectiveness
decreases and produces more leakage flux, therefore producing a bigger
difference between the voltage without load and the voltage with load.
To reduce the voltage regulation value the following can be done: -

* Wind primary and secondary over each other - interleaving the
layers helps further.

* Sandwich the coils.

* Make the coils longer and thinner by making the limbs longer and
the yoke shorter.

* Use a core which completely encloses the winding.

With the primary wound around one limb and the secondary coil wound on
another limb, as its excessive leakage flux would lead to very poor
regulation.

Looking at the results, I can say that they look realistic, with the
efficiency obviously decreasing due to the losses in the transformer
with the leakage flux and also around the load. One of the differences
between ideal and practical transformers are the losses in real
transformers which entail a power input that is always slightly
greater than the power output. Though small, these losses are of great
economic and practical importance; all the electricity generated in a
power station is transformed several times before it is put to use. If
the losses at each transformer were 1%, it would result in 4 or 5% of
all electricity production being wasted as heat in transformers. The
losses emanate from two sources: core losses (WFe) and coil losses (Wcu):

Pin = Pout + WFe + Wcu

Maximum efficiency: - The efficiency of a transformer is maximised if
the core losses are equal to the copper losses.

WFe = Wcu

With the circuit having a power of 0.8 lagging, makes the circuit out
of phase this is shown in Appendix 1, the equivalent circuit is on the
left and the phasor diagram is shown on the right, displaying the
phase difference. This will have an effect on the efficiency of the
circuit aswell.

Transformers can achieve maximum power transfer between the source and
the load, by a method called resistance matching, using the formula
below: -

R1 = N1 2 RL

N2

This formula allows the resistance of R1 to equal the resistance RL by
altering the turns ratio depending on these two values, this can be
very productive when needing maximum power output for a system which
is very large. This method could possibly used for the transfer of
electricity across the country via the power lines.

Conclusion

The experiment went very well for me, the results worked out fine
especially when they were compared with the results given from the
equivalent circuits, and the voltage regulation values were between 3
and 5 per cent. The circuit was not supplying the most efficient or
maximum efficiency, due to the core losses not equalling the copper
losses. The load used, which produced a power factor of 0.8 lagging
allowed the equivalent circuit method to be tested, it gave normal
results, or results that would have predicted before the calculations
giving a smaller efficiency and the voltage regulation went up high
due to the effect on the transformer.

Now I have done this experiment I now realise how important it is to
use the equivalent circuit method for prediction and estimation. Using
this method will enable large businesses, and especially the electric
board to estimate costs of use and to enable them to estimate the
amount of voltage required for input and the amount of power output.
This will save the businesses a lot of money and a lot of trouble, it
they were not able to predict or estimate then they could be a risk of
having a too high voltage output and therefore causing injury, and
damaging appliances. They could be a problem of having too little
voltage and houses been abundant of electrical use.

To improve the experiment I would have taken twice as many values for
the Open-circuit test, I would have improved the transformer itself,
as said in the discussion; we could overlap the layers, or lengthen
the coils e.t.c. The connections in the circuit could be improved by
using gold connections and better quality copper wire. The accuracy of
my results were not the most accurate to get the best results. So
therefore the ammeter and the voltmeter readings could have had a more
accurate reading, e.g. digital, with the dials being a manual reading
would have led to the inaccuracy of our own eyes. The

References.

* Transformer Engineering (2nd Edition):- Blime, Camili, Lennox,
Minneci, Montsinger and Boyajian

* Transformers and generators for power systems: - Berthelot.R.

* Transformers and Inductors: - K.A. Macfadyen.

* Electronic and Electrical engineering (2nd Edition): - Lionel
Warnes.

Appendices.

Appendix 1.

[IMAGE]

Appendix 4.

Appendix 2.

Appendix 3.

[IMAGE]

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