LESSON 6. Construction of single phase transformer

Transformers types

Transformers are manufactured in many types the most widely used in power systems are classified with their core types as seen below.  Core type (Fig. 3.3 A) where each of the windings are wound on one leg of the core, while the shell core type (Fig. 3.3 B), in which both windings are wound on the same leg. Each type has its own advantages and disadvantages. Core type is very reliable and easy to maintain, but take larger space, however, shell type is smaller but not reliable.

Fig 3.3a                                             Fig 3.3b

A) Core type                                                               B) Shell type

Figure 3.3 Single phase transformer construction

Ideal Transformer:

This is the most known relationships that relates the induction between the two coils (primary/Secondary) of an ideal transformer to turns ratio and their currents

e1

Ideal Unloaded Transformers

  • Winding resistances are zero, no leakage inductance and iron loss

  • Magnetization current generates a flux that induces voltage in both windings

    Fig 3.4

Fig. 3.4 Current, voltages and flux in an unloaded ideal transformer

Ideal loaded transformer

  • Loaded transformer , means current will be also feeding the load and that secondary side voltage, current (load Side)  will be calculated using the previous relations

Fig 3.5

Fig. 3.5 Currents and fluxes in a loaded ideal transformer

 e2

Ideal transformers equivalent circuit

fig-3.6

Fig. 3.6 Equivalent circuit of an ideal transformer

Transferring impedances through a transformer

Fig 3.7

Fig. 3.7 Thevenin equivalent of transformer circuit

Transferring impedances through a transformer

To transfer impedances of either side then the square of the turns ratio is used. Fig 3.8

Fig. 3.8 Equivalent circuit when secondary impedance is transferred to primary side and ideal transformer eliminated

Fig 3.9

Fig. 3.9 Equivalent circuit when primary source is transferred to secondary side and ideal transformer eliminated

Example:  A 200 kV A, 6600 V/400 V, 50 Hz single-phase transformer has 80 turns on the secondary. Calculate:

(i) the appropriate values of the primary and secondary currents;

(ii) the approximate number of primary turns

e3

6600/400 (80)= 1220 turns

Transformer rating = Primary Voltage x Primary current

    = Secondary Voltage x Secondary Current

 For Primary 200 KVA = (6600) (I1)

 I1  = 200KVA/6600v = 30.3 A

 For Secondary 200 KVA= 400 I2

I2  = 200 KVA/400v = 500 A

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no-load current is 3 A at a power factor of 0.2 lagging.

Calculate the primary current and power factor when the secondary current is 280 A at a power factor of 0.8 lagging. Assume the voltage drop in the windings to be negligible.

Answer: 58.3 A, 0.78 pf lagging.

Adapted from EE360 course on Electric Energy Engineering, KFUPM Open Courseware

Last modified: Tuesday, 1 October 2013, 7:07 AM