# Single Phase Transformer (Experiment) Report.

Electrotechnology Coursework

Single Phase Transformer (Experiment) Report.

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.

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 = V(I0a)2 + (I0r)2 and cos ?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 ?1 and cos ?2 are the primary and secondary power factors respectively.

* V2 = V'2(N2/N1)

Because the no-load current is relatively ...