Transformers of various sizes, ratings, voltage ratios can seen being used in a power system

The parameters of the equivalent circuits of these machines vary over a large range

The comparison of the machines can be made simpler if all the parameters are normalized

If simple scaling of the parameters is done then one has to carry forward the scaling factors in the calculation

Expressing in percent basis is one example of scaling

However if the scaling is done on a logical basis one can have a simple representation of the parameters without the bother of the scaling factors

Also different units of measurement are in use in the different countries (FPS, CGS, MKS, etc.), which also undergo several revisions over the years

If the parameters can be freed from the units then the system becomes very simple

The ’per-unit’ system is developed keeping these aspects in mind

The parameters of the transformer are referred to some base values and thus get scaled

A common base value is adopted in view of the different ratings of the equipment

In case of individual equipments, its own nominal parameters are used as base values

Choose \(V_{base}\) and \(I_{base}\) as nominal voltage and current on the primary side of the transformer, and compute \(S_{base}\) and \(Z_{base}\)

\[\begin{aligned} P_{base},Q_{base},S_{base} & =V_{base}*I_{base}\\ R_{base},X_{base},Z_{base} & =\dfrac{V_{base}}{I_{base}}\\ G_{base},B_{base},Y_{base} & =\dfrac{I_{base}}{V_{base}} \end{aligned}\]

Normally \(S_{base}\) and \(V_{base}\) are known from name plate details. Other base values can be derived from them \[\begin{aligned} V_{p.u} & =\dfrac{V(volt)}{V_{base}(volt)}\\ I_{p.u} & =\dfrac{I\left(Amps\right)}{I_{base}\left(Amps\right)}=\dfrac{I}{\left(\dfrac{S_{base}}{V_{base}}\right)}\\ Z_{p.u} & =\dfrac{Z\left(ohm\right)}{Z_{base}\left(ohm\right)}=Z(ohm)*\dfrac{I_{base}}{V_{base}}=Z(ohm)*\dfrac{S_{base}}{V_{base}^{2}} \end{aligned}\]

Many times, when more transformers are involved in a circuit one is required to choose a common base value for all of them

Parameters of all the machines are expressed on this common base

The conversion of the base values naturally lead to change in the per unit values of their parameters

An impedance \(Z_{p.u.,old}\) on the old base of \(S_{base,old}\) and \(V_{base,old}\) shall get modified on new base \(S_{base,new}\), \(V_{base,new}\) as

\[Z_{p.u.,new}=Z_{p.u.,old}*\dfrac{V_{base,old}^{2}}{S_{base,old}}*\dfrac{S_{base,new}}{V_{base,new}^{2}}\]

If all the equivalent circuit parameters are referred to the secondary side and per unit values of the new equivalent circuit parameters are computed with secondary voltage and current as the base values, there is no change in the per unit values. This can be easily seen by

\[Z_{p.u.}^{'} = Z_{ohm}^{'}\cdot \dfrac{S_{base}^{'}}{V_{base}^{'2}}~\mbox{but}~~Z_{ohm}^{'}=\dfrac{1}{a^2}\cdot Z_{ohm}\] where \[\begin{aligned} a & =\mbox{turns ratio of primary to secondary}\\ Z & =\mbox{impedance as seen by primary}\\ Z^{'} & =\mbox{impedance as seen by secondary}\\ S_{base}^{'} & =S_{base}~\mbox{as transformer rating is unaltered}\\ V_{base}^{'} & =V_{base}\cdot\dfrac{1}{a} \end{aligned}\] Thus, \(Z_{p.u.}^{'}=Z_{p.u.}\)

Parallel operation means two or more transformers are connected to the same supply bus bars on the primary side and to a common bus bar/load on the secondary side

Reasons that necessitate parallel operation:

Non-availability of a single large transformer to meet the total load requirement

Increase in power demand leading to augmentation of the capacity

Improved reliability : Even if there is fault in one transformer or taken out for maintenance the load can continued to be serviced

Transportation problems limit installation of large transformers at site, it is easier to transport smaller ones to site and work them in parallel

\(a_1, a_2 :\) turns ratio of Transformers

\(Z_A, Z_B :\) equivalent impedance referred to secondary

\(Z_L, V_L, I_L :\) secondary load impedance, voltage and current

\(I_A, I_B :\) current supplied to the load by secondary of the transformersConditions for parallel operation of transformers sharing a common load:

The voltage ratio must be the same

The per unit impedance of each machine on its own base must be the same

The polarity must be the same, so that there is no circulating current between the transformers

The phase sequence must be the same and no phase difference must exist between the voltages of the two transformers

\[\begin{aligned} I_{A}+I_{B} & =I_{L}\\ V_{L} & =\dfrac{V_{1}}{a_{1}}-I_{A}Z_{A}\\ V_{L} & =\dfrac{V_{1}}{a_{2}}-I_{B}Z_{B}=\dfrac{V_{1}}{a_{2}}-\left(I_{L}-I_{A}\right)Z_{B} \end{aligned}\] Solving (2) and (3), we get \[\begin{aligned} I_{A} & =\dfrac{Z_{B}I_{L}}{Z_{A}+Z_{B}}+\dfrac{V_{1}\left(a_{2}-a_{1}\right)}{a_{1}a_{2}\left(Z_{A}+Z_{B}\right)}\\ I_{B} & =\dfrac{Z_{A}I_{L}}{Z_{A}+Z_{B}}+\dfrac{V_{1}\left(a_{2}-a_{1}\right)}{a_{1}a_{2}\left(Z_{A}+Z_{B}\right)} \end{aligned}\]

There are two components in current \(I_A\) and \(I_B\). The first component represents the transformer share of the load current and the second component is the circulating current in the secondary winding \[I_c = \dfrac{E_A-E_B}{Z_A+Z_B} = \dfrac{V_1/a_1-V_1/a_2}{Z_A+Z_B} = \dfrac{V_{1}\left(a_{2}-a_{1}\right)}{a_{1}a_{2}\left(Z_{A}+Z_{B}\right)}\]

Undesirable effects of the circulating current:

increase in copper loss

overload one transformer and reduce the permissible load KVA

Equal Voltage Ratio:

To eliminate circulating currents, the voltage ratio must be identical, i.e. \(a_1=a_2\)

Under this condition, \[\begin{aligned} I_A & = \dfrac{Z_BI_L}{Z_A+Z_B}\\ I_B & = \dfrac{Z_AI_L}{Z_A+Z_B} \end{aligned}\] From Equations, we get \[\dfrac{I_A}{I_B} = \dfrac{Z_B}{Z_A}\]

Thus, transformer currents are inversely proportional to impedance

Thus for efficient paralleling, the potential differences at full load across the internal impedance should be equal

The condition ensures that the load sharing between the transformers are according to their rating

If per unit equivalent impedance are not equal transformers will not share the load in proportion to their KVA ratings, and as a result the overall rating of the transformer bank will be reduced \[I_{A}Z_{A} = I_{B}Z_{B}\]

The Volt-Amperes of the Transformers \[\begin{aligned} S_{A} & =V_{L}I_{A}=V_{L}I_{L}\left(\dfrac{Z_{B}}{Z_{A}+Z_{B}}\right)=\left(\dfrac{Z_{B}}{Z_{A}+Z_{B}}\right)S_{L}\\ S_{B} & =V_{L}I_{B}=V_{L}I_{L}\left(\dfrac{Z_{A}}{Z_{A}+Z_{B}}\right)=\left(\dfrac{Z_{A}}{Z_{A}+Z_{B}}\right)S_{L}\\ \Rightarrow\dfrac{S_{A}}{S_{B}} & =\dfrac{Z_{B}}{Z_{A}} \end{aligned}\]

VA load on each transformer is inversely proportional to its impedance

Hence, to share the load in proportion to their ratings, the transformers should have the impedance which is inversely proportional to their ratings