Representation of Transformers | Devices | Electrical Engineering

Transformer is a static machine which is used to transform electric power from one circuit to another without change in frequency. It may raise or lower the voltage in a circuit but with a corresponding decrease or increase in current.

Thus transformer plays a vital role in power systems. Practically every transformer is provided with taps for ratio control, and thus it is also used for controlling secondary voltage level. Most of the transformers are provided with off-load tap-changers while some of the transformers are provided with on-load tap-changers.

A power transformer with turn-ratio K can be represented by its impedance Z_L or admittance Y_L in series with an ideal transformer, as illustrated in Fig. 6.12. However, this model does not serve the purpose here. Though it may prove useful in other cases.

We shall write the injected power equations in terms of the network bus admittance matrix, which is only possible if all branch elements (including transformer) are resolved into equivalent π-network; i.e., transformer model depicted in Fig. 6.12 needs transformation into the model depicted in Fig. 6.13.

The parameter of the π-equivalent model can be derived by equating the terminal currents of the transformer model shown in Fig. 6.12 with the corresponding currents of the π-equivalent model given in Fig. 6.13.

From the transformer model shown in Fig. 6.12, we have –

I₁ = I₂/K                                                                    …(6.28)

and I₂ = [V₂ – (V₁/K)] Y_L

= Y_L V₂ – [Y_L (V₁/K)]                                            …(6.29)

From Eqs. (6.28) and (6.29), we have –

I₁ = [– (Y_L/K)V₂] + [Y_L (V₁/K²)]                        …(6.30)

where Z_L = Leakage impedance of transformer

Y_L = 1/Z_L. Leakage admittance of transformer

K = turn-ratio of transformer = N₁/N₂

From the π-equivalent model of transformer depicted in Fig. 6.12, we have-

I₁ = (Y₁ + Y₃)V₁ – Y₃ V₂                                          …(6.31)

I₂ = – Y₃V₁ + (Y₂ + Y₃)V₂                                       …(6.32)

Equating like coefficients, we have –

Y₁ + Y₃ = Y_L/K²                                                          …(6.33)

Y₃ = Y_L/K                                                                     …(6.34)

Y₂ + Y₃ = Y_L                                                                 …(6.35)

From the above equation, we have –

Y₁ =  [(Y_L/K²) – (Y_L/K)] = [(Y_L/K²)(1 – K)]          …(6.36)

Y₂ = [(Y_L/K) (K – 1)]                                                 …(6.37)

and Y₃ = Y_L/K                                                             …(6.38)

So our mathematical model in matrix form comes out to be –

With the above equation and the equivalent circuit depicted in Fig. 6.13, desired requirement in load flow equation is complete.

Auto-Transformer and Phase-Shifter in Tandem:

Now let us make a model for a phase-shifting transformer connected in series with an auto-transformer, as shown in Fig. 6.14.

Phase shifter is used for regulating power flow through a line element or occasionally through a set of lines and also in the area interchange control.

For the circuit given in Fig. 6.14, we have –

I_K = Y_ik (V_k – V₂)                                     …(6.40)

I₂ = Y_ik (V₂ – V_k)                                     …(6.41)

where Y_ik = 1/Z_ik, equivalent line impedance (line impedance plus leakage impedances of auto- transformer and phase-shifting transformer) between nodes (or buses) i and k. Shunt admittance of the line is neglected.

At the phase- shifter V₂ = (1 ∠ɸ) V₁                              …(6.42)

V₁I₁* = V₂ I₂*                                                                            …(6.43)

At the auto-transformer –

K V₁ = V_i                                                                                    …(6.44)

V₁ I₁* = V_i I_i*                                                                           …(6.45)

from Eqs. (6.40), (6.42) and (6.44), we have –

I_k = Y_ikV_k – Y_ik (1/K ∠ɸ) V_i                                                …(6.46)

From Eqs. (6.41) and (6.43), and (6.45) we have –

 

 

Using Eqs. (6.42) and (6.44), equation (6.47) may be reduced to –

I_i = (1/K ∠– ɸ) (Y_ik V₂ – Y_ik V_k)                                      …(6.48)

Again using Eqs. (6.42) and (6.44), equation (6.48) is further reduced to –

I_i = – Y_ik [1/K∠– ɸV_k] + Y_ik V_i/K²                                 …(6.49)

Now Eqs. (6.46) and (6.49) may be written in matrix form as below:

 

 

 

 

 

where ɸ is phase shift angle between V₂, and V₁.

For the use of above Eq. (6.50) it is necessary to consider the reference bus for the phase shifter and the tap-side bus for the tap-changer. It is noteworthy that in Fig. 6.14 either could be the tapped side or phase-shifter reference.

We may define here that –  

ϵ = + 1 if the p bus is untapped side

= 0 if the p bus is tapped side

ϵ’ = + 1 if the p bus is phase-shifter reference

= 0, if no phase shifter

= – 1 if the q bus is phase-shifter reference

Thus Fig. 6.15 represents complete model expressed in terms of equivalent admittances looking from the p bus or written in equation form as –

 

 

K = 1 in case there is no auto-transformer and only phase-shifter is present.

ϵ ‘ = 0 or ɸ = 0 if there is no phase-shifter and only auto-transformer is present.

It is to be noted that Eq. (6.50) in matrix form is not the π-equivalent model of our power system as is obvious from the admittance matrix elements. Mathematical model represented by Eq. (6.51) can be easily used in the construction of bus admittance matrix of the power system under power flow studies.

Let us derive Eqs. (6.46) and (6.49) from our general mathematical Eq. (6.51) as an illustration.

For determining I_i, we substitute in Eq. (6.51) –

This is same equation as Eq. (6.49).

For determining I_k, we substitute in Eq. (6.51) –

This is the same equation as Eq. (6.46).