LAB REPORT: 01 OPEN CIRCUIT AND SHORT

CIRCUIT TESTS

OF A SINGLE-PHASE TRANSFORMER

ECX 3232 ELECTRICAL POWER

NAME : M. S. D. PERERA.

REGNO : 311089590

CENTER : COLOMBO.

DATE OF SUBMISSION: 07/11/2012

Q. NO MARKS

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Experiment 1: Open circuit and short circuit tests of a single-phase

transformer.

Apparatus:

500VA, 230V/230V 1:1 transformer.

(0-250V,AC) voltmeter.

(0-150V,AC) voltmeter.

(0-1A,AC) Ammeter.

Wattmeter.

230/(0-250V) variac.

Leads.

Theory

Working with a ideal transformer is easy. But life get complicated when it

comes to theoretical electrical engineering to real world electrical

engineering. Ideal model no longer useful in industrial electrical

engineering applications. So we have to come up with a model that represent

an normal non-ideal industry transformer.

The real transformer have following things to be included and modeled when in

when we drawing it’s equivalent circuit.

Losses,

There are two main losses related to a power transformer. And they are,

* Core Losses

* Hysteresis loss.

* Eddy current loss.

* Copper loss.

Leakage flux.

No load flux(also known as magnetizing flux).

In an equivalent circuit we represent core losses as a parallel resistor

because it’s proportional to the number of turns in the winding. And

magnetizing flux could also represent as a parallel component as well as it’s

also proportional to the number of windings.

We represent copper loss as a serial resistive component , because it’s just

equal to a pure resistor passing current through it and disparaging energy.

And also we represent leakage flux as also a serial inductive component, we

could imagine it as a series inductor outside the transformer which is

blocking some potential difference across it so it will reduce the gross

potential difference among ideal transformer terminals.

Bellow figure depicts this model diagrammatically.

Since these two windings are magnetically coupled, we could get it’s thevean

equivalent circuit as we seen from the primary. (This could be done to the

secondary too).

Bellow figure depicts how we see it from primary side.

In the case of transforming secondary side to primary side we have to

multiply each inductive/resistive component by square of turns ration. Which

means,

When you transforming primary into the secondary side, you have to divide it

by square of turns ratio,

EXPREMENT:

PROCEDURE:

Part A: Open Circuit Test.

(a) The voltage ratings of the transformer is,

500VA, 230/230V 1:1 transformer.

So KVA rating is ½ KVA.

(b) Rated voltages,

(c) :

This is open circuit test. We log no load current and iron while changing the

voltage through variac device.

This is the data we have collected.

Impressed

Voltage(V)

No Load Current

(I/A)

Iron Loss(W)

230 0.6115 15

180 0.224 12

160 0.195 10

140 0.116 8

120 0.142 6

100 0.121 5

80 0.101 4

60 0.081 3

40 0.060 2

Calculations:

Since there are no power desperation on the secondary side we have only

power desperations on the primary side. They are sum of copper loss+ core

loss. But in here, since we have very little current flowing through primary

winding, we could ignore copper loss and assuming that reading in the

wattmeter is equal to core losses. So through that we could find two

variables.

Graphs and Characteristics:

Part B: Short Circuit Test

Now we are going to short circuit the secondary side. We need to take caution

here, because there is a potential to burn the fuses in the learning panel if

we won’t be careful. So we keep the variac device at it’s lower position and

powering up the switches. Here we are getting a one reading only. It’s at

wattmeter and ammeter readings while variac kept at 9V.

We use such a very small (9V) potential thus because this is a short circuit

test and we are not supposing to burn that expensive learning panels.

Theory:

In here there are no power desperation on the secondary side. And just

because Rc and Xm are very large values, we could ignore them. So it’s safe

to assume that all the power desperation is now equal to the copper loss.

So,

Observations,

9Ammeter Voltmeter Wattmeter

2.17 9 32.5

So we get,

Discussion:

Why HV side is open circuited and LV side is short-circuited when performing

the practical?

Well as it term derives it’s meaning that HV side will generate high

voltages. So that will ramp up the short circuit current to a very large

value just because there are no any resistance to limit current flow on

secondary side. So that’s why we need to use LV side to be short circuited as

well as we should use very low voltage ( like 9V in our experiment ) to avoid

damaging or frying transformers, fuses or breakers.

Experiment 2 : Load Test Of Transformers

Apparatus

1. 500VA, 230V/230V single phase transformer 2. 0-250V,AC voltmeter 3. 0-5A,AC ammeter 4. Wattmeter 5. 230/(0-250V) variac 6. Resistor bank 7. Capacitor bank 8. Leads

Theory

Voltage regulation is a principle to keep voltage value independent of the

load. When it comes to voltage regulation we have to consider bellow facts

into consideration.

Different loads will take different currents at same voltage.

Different loads will have different leading/lagging reactive

components.

A load may vary how much it will draw dynamically, take a washing

machine for a example, When it washing clothes it will have drive

motors and there will be a lagging current component and when it

switched to drying clothes it will turn it’s motors off and turn on

it’s heaters which will dynamically change gross load inductive load to

a resistive load.

Above facts are making voltage regulation a difficult subject. So it’s not

possible to get a ideal constant voltage, it will vary at least by a fraction

of a million when it’s load current changes.

By the way, we should have some standard index to measure how much bad or

good a particular device could regulate against varying load currents.

In transformers we use ,

And phasor diagram of a transformer when loaded with power factor load.

And the efficacy of the transformer is given by,

The first experiment is about resistive loads, so we could use as zero.

Observations And Calculations:

Graphs:

For a capacitive load

Here we can’t assume that power factor is 1, we have to calculate it.

Since ,

Graphs: