Transformers HNCD Engineering Science Transformers No not them

Transformers HNC/D Engineering Science

Transformers No not them

Transformers – the electrical ones Transformers come in many shapes and sizes but basically operate the same way. Potential-transformer Three-phase pole-mounted step-down transformer. Laminated core transformer showing edge of laminations at top of unit

Transformers A transformer is a device that converts alternating voltage at one level to an alternating voltage at a higher or lower level. The above diagram shows a simple double wound transformer. The coil on the left is called the primary and the coil or winding on the right is called the secondary. The bit in the middle is called the core.

Double Wound Transformers Most double wound transformers look slightly different in that the two windings primary and secondary will be on the same leg of the core. This helps reduce losses.

Three-phase Transformers In a three-phase transformer, the transformer is slightly different. There a series of three windings on the primary and secondary. The windings on a transformer are sandwiched or wound on top of each other to help the distribution of magnetic flux.

Three Phase Supply In a three phase supply there are effectively three supplies all rising and falling at the same time. The secondary windings can differentiate between the varying magnetic fields because each produces a particular phase and these are at maximum and minimum at different times.

Instrument Transformers In large power circuits it is not possible to measure either current or voltage directly. This is because the meter used would have to be very large for the levels of voltage and current. When measuring instruments placed in circuits is not appropriate, because of the large voltage and current, then voltage transformers and current transformers are used.

Air Cooled Transformers The simplest kind of cooling available is air. There are two types of air cooling, these are natural air (AN) and forced air (AF) The air cooled transformer will be labelled with one (AN) or the other (AF) or possibly both. It is important to check for damage before installing the transformer. This type of transformer must be installed indoors and surrounded by an earthed metal case or enclosure. If the transformer has been stood for some time it is necessary to carry out an insulation resistance test.

Oil Cooled Transformer With oil cooled transformers we have to consider the cooling of the transformer and the cooling of the oil itself. The labelling on oil-cooled transformers has four letters, possibly ONAN (oil-natural air-natural) The first two letters describe the removal of the heat from the transformer and the second two denote the removal of the heat from the oil. When installing oil-cooled transformers there needs to be a means of containing the oil and a means of containing the oil should it leak.

Transformer on no-load ¡ An ideal transformer is one where all the magnetic flux is confined to the core and links both the primary and secondary coils

Calculations for transformers ¡ ¡ U = voltage N = number of turns I = current p = primary, s = secondary

Calculations Example 1 An idea transformer has a primary with 900 turns. How many turns will there need to be on the secondary coil if a 240 V a. c input is to give rise to a no-load output of 80 V ? Example 2 An idea transformer has 800 primary turns and 200 secondary turns. What will be the secondary voltage when the primary is supplied from a 240 V a. c supply ?

Transformer – resistive load ¡ ¡ If there is a resistive load connected to the secondary. A secondary current flows and hence power is dissipated. The secondary coil current produces its own alternating flux in the core. If the power losses is negligible the power supplied to the primary coil must equal the secondary coil when there is a current in the primary coil l I 1 V 1 = I 2 V 2

Calculations Example 1 An ideal transformer is used to light a 12 V, 15 W lamp from the 240 V mains supply. Determine the turns ratio required and the current taken from the mains supply. Example 2 An ideal transformer has 20 primary turns and 80 secondary turns and supplies a secondary current of 1. 5 A. What is the value of the ampere-turns of the transformer and the primary current ?

Losses ¡ Losses are inherent to the nature of the equipment that is used and fall into two key types in transformers: ¡ Copper losses are created by the current flowing through the cables. All conductors have resistance and when current flows then heat is dissipated. This is a loss. Copper losses are never affected by frequency. ¡ Solution = cooling

Losses ¡ Eddy current, and hysteresis losses are called ‘iron losses’. This type of loss is a magnetic loss. ¡ The eddy current, particularly within a material such as iron, can cause quite a large increase in temperature, and consequent power loss. ¡ Solution = laminated cores (these reduce the losses) ¡ Hysteresis is a magnetic phenomenon which causes residual magnetism to hang around after the field has collapsed, again causing heat, this happens more with changes in frequency ¡ Solution = silicon steel as the core material is better

Power Rating ¡ Power rating in VA = V 2 I 2 FL V 2 = the secondary voltage ¡ I 2 FL = the secondary current on full load ¡ ¡ Full Load = Maximum power output l (determined by the heat generated by internal losses dissipated)

Power Rating ¡ ¡ ¡ Power Loss = I 2 R The loss is termed the copper loss This is the current which can be obtained before the transformer overheats When a transformer is said to be operating at half load, it means half its rated power output Therefore the secondary current is half the full load value

Calculations Example 1 A 2. 75 KVA single phase transformer has a primary with 600 turns and a secondary with 100 turns. If the alternating primary voltage is 3300 V what will be the minimum resistance which can be connected as the load ?

Efficiency ¡ ¡ Power loss that occurs with a transformer are : Copper Loss – generated by current through primary and secondary winding l l l ¡ Power Loss = I 12 R 1 + I 22 R 2 I 1 = current through primary of resistance R 1 I 2 = current through primary of resistance R 2 Iron Loss – is the loss resulting from hysteresis and eddy currents.

Efficiency ¡ ¡ Efficiency = output power x 100 input power Output Power = V 2 I 2 cos. Ф (this equals the input power minus losses)

Calculations Example 1 A transformer delivers a power of 5 k. W when on-load. What will be its efficiency if the copper loss is then 100 W and the iron loss 75 W ? Example 2 A 40 k. VA single phase transformer has an iron loss of 400 W and a full load copper loss of 600 W. What is the efficiency of the transformer for a power factor of 0. 8 at (a) full load, (b) half load ?