Appendix D

The Per-Unit System

D.1 Introduction

Because of various advantages involved, it is customary in power system analysis calculations to use impedances, currents, voltages, and powers in per-unit values (which are scaled or normalized values) rather than in physical values of ohms, amperes, kilovolts, and megavoltamperes (or megavars, or megawatts).

A per-unit system is a means of expressing quantities for ease in comparing them. The per-unit value of any quantity is defined as the ratio of the quantity to an “arbitrarily” chosen base (i.e., reference) value having the same dimensions. Therefore, the per-unit value of any quantity can be defined as physical quantity

$\begin{array}{cc}\text{Quantity in per unit} =\frac{\text{Physical quantity}}{\text{Base value of quantity}}\hfill & (\text{D}.1)\hfill \end{array}$

where “physical quantity” refers to the given value in ohms, amperes, volts, etc. The base value is also called unit value since in the per-unit system it has a value of 1, or unity. Therefore, a base current is also referred to as a unit current.

Since both the physical quantity and base quantity have the same dimensions, the resulting perunit value expressed as a decimal has no dimension and, therefore, is simply indicated by a subscript pu. The base quantity is indicated by a subscript B. The symbol for per unit is pu, or 0/1. The percent system is obtained by multiplying the per-unit value by 100. Hence,

$\begin{array}{cc}\text{Quantity in percent}=\frac{\text{Physical quantity}}{\text{Base value of quantity}}\times 100\hfill & (\text{D}.2)\hfill \end{array}$

However, the percent system is somewhat more difficult to work with and more subject to possible error since it must always be remembered that the quantities have been multiplied by 100.

Thus, the factor 100 has to be continually inserted or removed for reasons that may not be obvious at the time. For example, 40% reactance times 100% current is equal to 4000% voltage, which, of course, must be corrected to 40% voltage. Hence, the per-unit system is preferred in power system calculations. The advantages of using the per-unit system include the following:

1. Network analysis is greatly simplified since all impedances of a given equivalent circuit can directly be added together regardless of the system voltages.
2. It eliminates the $\sqrt{3}$ multiplications and divisions that are required when balanced three-phase systems are represented by per-phase systems. Therefore, the factors $\sqrt{3}$ and 3 associated with delta and wye quantities in a balanced three-phase system are directly taken into account by the base quantities.
3. Usually, the impedance of an electrical apparatus is given in percent or per unit by its manufacturer based on its nameplate ratings (e.g., its rated voltamperes and rated voltage).
4. Differences in operating characteristics of many electrical apparatus can be estimated by a comparison of their constants expressed in per units.
5. Average machine constants can easily be obtained since the parameters of similar equipment tend to fall in a relatively narrow range and, therefore, arc comparable when expressed as per units based on rated capacity.
6. The use of per-unit quantities is more convenient in calculations involving digital computers.

D.2 Single-Phase System

In the event that any two of the four base quantities (i.e., base voltage, base current, base voltamperes, and base impedance) are “arbitrarily” specified, the other two can be determined immediately.

Here, the term arbitrarily is slightly misleading since in practice the base values are selected so as to force the results to fall into specified ranges. For example, the base voltage is selected such that the system voltage is normally close to unity.

Similarly, the base voltampere is usually selected as the kilovoltampere or megavoltampere rating of one of the machines or transformers in the system, or a convenient round number such as 1, 10, 100, or 1000 MVA, depending on system size. As aforementioned, on determining the base voltamperes and base voltages, the other base values are fixed. For example, current base can be determined as

$\begin{array}{cc}{I}_B=\frac{S_B}{V_B}=\frac{V{A}_B}{V_B}\hfill & \left(\text{D}.3\right)\hfill \end{array}$

where

• IB is the current base in amperes
• SB is the selected voltampere base in voltamperes
• VB is the selected voltage base in volts

Note that

$\begin{array}{cc}{S}_B=V{A}_B=P_B=Q_B=V_B{I}_B\hfill & \left(\text{D}.4\right)\hfill \end{array}$

Similarly, the impedance base* can be determined as

$\begin{array}{l}\begin{array}{cc}{Z}_B=\frac{V_B}{I_B}\hfill & \left(D.5\right)\hfill \end{array}\hfill \end{array}$

where

$\begin{array}{cc}{Z}_B=X_B=R_B\hfill & \left(\text{D}.6\right)\hfill \end{array}$

Similarly,

$\begin{array}{cc}{Y}_B=B_B=G_B=\frac{I_B}{V_B}\hfill & \left(\text{D}.7\right)\hfill \end{array}$

Note that by substituting Equation D.3 into Equation D.5, the impedance base can be expressed as

$\begin{array}{cc}{Z}_B=\frac{VB}{V{A}_B/V_B}=\frac{V_B^2}{V{A}_B}\hfill & \left(\text{D}.8\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_B=\frac{{\left(k{V}_B\right)}^2}{MV{A}_B}\hfill & \left(\text{D}.9\right)\hfill \end{array}$

where

• kVB is the voltage base in kilovolts
• MVAB is the voltampere base in megavoltamperes

The per-unit value of any quantity can be found by the normalization process, that is, by dividing the physical quantity by the base quantity of the same dimension. For example, the per-unit impedance can be expressed as

$\begin{array}{l}\begin{array}{cc}{Z}_{\text{pu}}=\frac{Z_{\text{physical}}}{Z_B}\hfill & \left(\text{D}.10\right)\hfill \end{array}\hfill \\ \hfill \end{array}$

or

$\begin{array}{cc}{Z}_{\text{pu}}=\frac{Z_{\text{physical}}}{V_B^2/\left(kV{A}_B\times 1000\right)Z_B}\hfill & \left(\text{D}.11\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_{\text{pu}}=\frac{\left(Z_{\text{physical}}\right)\left(kV{A}_B\right)\left(1000\right)}{V_B^2}\hfill & \left(\text{D}.12\right)\hfill \end{array}$

or

$\begin{array}{l}\begin{array}{cc}{Z}_{\text{pu}}=\frac{\left(Z_{\text{physical}}\right)\left(kV{A}_B\right)}{{\left(k{V}_B\right)}^2\left(1000\right)}\hfill & \left(\text{D}.13\right)\hfill \end{array}\hfill \end{array}$

or

$\begin{array}{cc}{Z}_{\text{pu}}=\frac{\left(Z_{\text{physical}}\right)}{{\left(k{V}_B\right)}^2/MV{A}_B}\hfill & \left(\text{D}.14\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_{\text{pu}}=\frac{\left(Z_{\text{physical}}\right)\left(MV{A}_B\right)}{{\left(k{V}_B\right)}^2}\hfill & \left(\text{D}.15\right)\hfill \end{array}$

Similarly, the others can be expressed as

$\begin{array}{cc}{I}_{\text{pu}}=\frac{I_{\text{physical}}}{I_B}\hfill & \left(\text{D}.16\right)\hfill \end{array}$

or

$\begin{array}{cc}{V}_{\text{pu}}=\frac{V_{Physical}}{V_B}\hfill & \left(\text{D}.17\right)\hfill \end{array}$

or

$\begin{array}{l}\begin{array}{cc}k{V}_{\text{pu}}=\frac{k{V}_{\text{physical}}}{k{V}_{\text{B}}}\hfill & \left(\text{D}.18\right)\hfill \end{array}\hfill \\ \hfill \end{array}$

or

$\begin{array}{l}\begin{array}{cc}V{A}_{\text{pu}}=\frac{V{A}_{\text{physical}}}{V{A}_B}\hfill & \left(\text{D}.19\right)\hfill \end{array}\hfill \end{array}$

or

$\begin{array}{l}\begin{array}{cc}kV{A}_{\text{pu}}=\frac{kV{A}_{\text{physical}}}{kV{A}_B}\hfill & \left(\text{D}.20\right)\hfill \end{array}\hfill \end{array}$

or

$\begin{array}{cc}MV{A}_{pu}=\frac{MV{A}_{\text{physical}}}{MV{A}_B}\hfill & \left(\text{D}.21\right)\hfill \end{array}$

Note that the base quantity is always a real number, whereas the physical quantity can be a complex number. For example, if the actual impedance quantity is given as Z∠θ Ω, it can be expressed in the per-unit system as

$\begin{array}{cc}{\text{Z}}_{\text{pu}}=\frac{Z\angle \theta }{Z_B}=Z_{\text{pu}}\angle \theta \hfill & \left(\text{D}.22\right)\hfill \end{array}$

that is, it is the magnitude expressed in per-unit terms.

Alternatively, if the impedance has been given in rectangular form as

$\begin{array}{cc}\text{Z}=R+jX\hfill & \left(\text{D}.23\right)\hfill \end{array}$

then

$\begin{array}{cc}{\text{Z}}_{\text{pu}}=R_{\text{pu}}+jX\text{pu}\hfill & \left(\text{D}.24\right)\hfill \end{array}$

where

$\begin{array}{cc}{R}_{\text{pu}}=\frac{R_{\text{physical}}}{Z_B}\hfill & \left(\text{D}.25\right)\hfill \end{array}$

and

$\begin{array}{cc}{X}_{\text{pu}}=\frac{X_{\text{physical}}}{Z_B}\hfill & \left(\text{D}.26\right)\hfill \end{array}$

Similarly, if the complex power has been given as

$\begin{array}{cc}\text{S}=P+jQ\hfill & \left(\text{D}.27\right)\hfill \end{array}$

then

$\begin{array}{cc}{\text{S}}_{\text{pu}}=P_{\text{pu}}+j{Q}_{\text{pu}}\hfill & \left(\text{D}.28\right)\hfill \end{array}$

where

$\begin{array}{cc}{P}_{\text{pu}}=\frac{P_{\text{physical}}}{S_B}\hfill & \left(\text{D}.29\right)\hfill \end{array}$

and

$\begin{array}{cc}{Q}_{\text{pu}}=\frac{Q_{\text{physical}}}{S_B}\hfill & \left(\text{D}.30\right)\hfill \end{array}$

If the actual voltage and current values are given as

$\begin{array}{cc}\text{V}=\text{V}\angle {\theta }_{\text{v}}\hfill & \left(\text{D}.31\right)\hfill \end{array}$

and

$\begin{array}{cc}\text{I}=I\angle {\theta }_I\hfill & \left(\text{D}.32\right)\hfill \end{array}$

the complex power can be expressed as

$\begin{array}{cc}\text{S}=\text{VI*}\hfill & \left(\text{D}.33\right)\hfill \end{array}$

or

$\begin{array}{cc}S\angle \theta =\left(V\angle \theta v\right)\left(I\angle -{\theta }_I\right)\hfill & \left(\text{D}.34\right)\hfill \end{array}$

Therefore, dividing through by SB,

$\begin{array}{cc}\frac{S\angle \varphi }{S_B}=\frac{\left(V\angle {\theta }_{\text{v}}\right)\left(I\angle {\theta }_I\right)}{S_B}\hfill & \left(\text{D}.35\right)\hfill \end{array}$

However,

$\begin{array}{cc}{S}_B=V_B{I}_B\hfill & \left(\text{D}.36\right)\hfill \end{array}$

Thus,

$\begin{array}{cc}\frac{S\angle \theta }{S_B}=\frac{\left(V\angle {\theta }_{\text{v}}\right)\left(I\angle -{\theta }_I\right)}{V_B{I}_B}\hfill & \left(\text{D}.37\right)\hfill \end{array}$

or

$S_{\text{pu}}\angle \theta =(v_{\text{pu}}\angle {\theta }_{\text{V}})(I_{\text{pu}}\angle -{\theta }_{\text{I}})(\text{D}.38)$

or

$\begin{array}{cc}{S}_{\text{pu}}=V_{\text{pu}}{I}_{\text{pu}}^{*}\hfill & \left(\text{D}.39\right)\hfill \end{array}$

Example D.1

A 240/120 V single-phase transformer rated 5 kVA has a high-voltage winding impedance of 0.3603 Ω. Use 240 V and 5 kVA as the base quantities and determine the following:

1. The high-voltage side base current.
2. The high-voltage side base impedance in ohms.
3. The transformer impedance referred to the high-voltage side in per unit.
4. The transformer impedance referred to the high-voltage side in percent.
5. The turns ratio of the transformer windings.
6. The low-voltage side base current.
7. The low-voltage side base impedance.
8. The transformer impedance referred to the low-voltage side in per unit.

Solution:

1. The high-voltage side base current is

   $\begin{array}{lll}IB(HV)\hfill & =\hfill & \frac{S_B}{V_B{}_{(HV)}}\hfill \\ \hfill & =\hfill & \frac{5000\text{VA}}{240\text{V}}\hfill \\ \hfill & =\hfill & 20.8333\text{A}\hfill \end{array}$

2. The high-voltage side base impedance is

   $\begin{array}{lll}{Z}_{B\left(\text{HV}\right)}\hfill & =\hfill & \frac{{\text{V}}_{B\left(\text{HV}\right)}}{I_{B\left(\text{HV}\right)}}\hfill \\ \hfill & =\hfill & \frac{240\text{V}}{20.8333\text{A}}\hfill \\ \hfill & =\hfill & 11.52\mathrm{\Omega }\hfill \end{array}$

3. The transformer impedance referred to the high-voltage side is

   $\begin{array}{lll}{Z}_{\text{pu}\left(\text{HV}\right)}\hfill & =\hfill & \frac{Z_{\text{HV}}}{Z_{B\left(\text{HV}\right)}}\hfill \\ \hfill & =\hfill & \frac{0.3603\mathrm{\Omega }}{11.51\mathrm{\Omega }}\hfill \\ \hfill & =\hfill & 0.0313\text{pu}\hfill \end{array}$

4. The transformer impedance referred to the high-voltage side is %

   $\begin{array}{lll}\%Z_{\text{HV}}\hfill & =\hfill & Z_{\text{pu}\left(\text{HV}\right)}\times 100\hfill \\ \hfill & =\hfill & \left(0.0313\text{pu}\right)100\hfill \\ \hfill & =\hfill & 3.13\%\hfill \end{array}$

5. The turns ratio of the transformer windings is

   $\begin{array}{lll}n\hfill & =\hfill & \frac{{\text{V}}_{\text{HV}}}{{\text{V}}_{\text{LV}}}\hfill \\ \hfill & =\hfill & \frac{240\text{V}}{120\text{V}}\hfill \\ \hfill & =\hfill & 2\hfill \end{array}$

6. The low-voltage side base current is

   $\begin{array}{lll}{I}_{B\left(\text{LV}\right)}\hfill & =\hfill & \frac{S_B}{{\text{V}}_{\text{B}\left(\text{LV}\right)}}\hfill \\ \hfill & =\hfill & \frac{5000\text{VA}}{120\text{V}}\hfill \\ \hfill & =\hfill & 41.6667\text{A}\hfill \end{array}$

   or

   $\begin{array}{lll}{I}_{B\left(\text{LV}\right)}\hfill & =\hfill & n{I}_B{}_{\left(\text{LV}\right)}\hfill \\ \hfill & =\hfill & 2\left(20.8333\text{A}\right)\hfill \\ \hfill & =\hfill & 41.6667\hfill \end{array}$

7. The low-voltage side base impedance is

   $\begin{array}{lll}{Z}_{B\left(\text{LV}\right)}\hfill & =\hfill & \frac{V_{B\left(\text{LV}\right)}}{I_{B\left(\text{LV}\right)}}\hfill \\ \hfill & =\hfill & \frac{120\text{V}}{41.667\text{A}}\hfill \\ \hfill & =\hfill & 2.88\mathrm{\Omega }\hfill \end{array}$

   or

   $\begin{array}{lll}{Z}_{B\left(\text{HV}\right)}\hfill & =\hfill & \frac{Z_{B\left(\text{LV}\right)}}{n^2}\hfill \\ \hfill & =\hfill & 2.88\mathrm{\Omega }\hfill \end{array}$

8. The transformer impedance referred to the low-voltage side is

   $\begin{array}{lll}{Z}_{\text{LV}}\hfill & =\hfill & \frac{Z_{\text{HV}}}{n^2}\hfill \\ \hfill & =\hfill & \frac{0.3603\mathrm{\Omega }}{2^2}\hfill \\ \hfill & =\hfill & 0.0901\mathrm{\Omega }\hfill \end{array}$

   Therefore,

   $\begin{array}{lll}{Z}_{\text{pu}\left(\text{LV}\right)}\hfill & =\hfill & \frac{Z_{\text{LV}}}{Z_{\text{B}\left(\text{LV}\right)}}\hfill \\ \hfill & =\hfill & \frac{0.0901\mathrm{\Omega }}{2.88\mathrm{\Omega }}\hfill \\ \hfill & =\hfill & 0.0313\mathrm{\Omega }\hfill \end{array}$

   or

   $\begin{array}{lll}{Z}_{\text{pu}\left(\text{LV}\right)}\hfill & =\hfill & Z_{\text{pu}\left(\text{HV}\right)}\hfill \\ \hfill & =\hfill & 0.0313\text{pu}\hfill \end{array}$

   Notice that in terms of per units the impedance of the transformer is the same whether it is referred to the high-voltage side or the low-voltage side.

Example D.2

Redo the Example D.1 by using MATLAB®.

1. Write the MATLAB program script.
2. Give the MATLAB program output.

Solution:

1. Here is the MATLAB program script:

   % MATLAB SCRIPT for Example D.1

   clear

   clc

   %System Parameters

    VBhv = 240;

    VBlv = 120;

    SB = 5e3;

    Zhv = 0.3603;

   %Solution for part (a)

    IBhv = SB/VBhv

   %Solution for part (b)

    ZBhv = VBhv/IBhv

   %Solution for part (c)

    Zpu_hv = Zhv/ZBhv

   %Solution for part (d)

    percent_Zhv = Zpu_hv*100

   %Solution for part (e)

    n = VBhv/VBlv

   %Solution for part (f)

    IBlv = SB/VBlv

    IBlv = n*IBhv

   %Solution for part (g)

    ZBlv = VBlv/IBlv

    ZBlv = ZBhv/n^2

   %Solution for part (h)

    Zlv = Zhv/n^2

    Zpu_lv = Zlv/ZBlv

    Zpu_lv = Zpu_hv

2. Here is the MATLAB program output:

    IBhv =

    20.8333

    ZBhv =

    11.5200

    Zpu_hv =

    0.0313

    percent_Zhv =

    3.1276

    n =

    2

    IBlv =

    41.6667

    IBlv =

    41.6667

    ZBlv =

    2.8800

    ZBlv =

    2.8800

    Zlv =

    0.0901

    Zpu_lv =

    0.0313

    Zpu_lv =

    0.0313

    >>

D.3 Converting From Per-Unit Values to Physical Values

The physical values (or system values) and per-unit values are related by the following relationships:

$\begin{array}{cc}\text{I}={\text{I}}_{\text{pu}}\times I_{\text{B}}\hfill & \left(\text{D}.40\right)\hfill \end{array}$

$\begin{array}{cc}\text{V}={\text{V}}_{\text{pu}}\times V_B\hfill & \left(\text{D}.41\right)\hfill \end{array}$

$\begin{array}{cc}\text{Z}={\text{Z}}_{\text{pu}}\times Z_B\hfill & \left(\text{D}.42\right)\hfill \end{array}$

$\begin{array}{cc}R=R_{\text{pu}}\times Z_B\hfill & \left(\text{D}.43\right)\hfill \end{array}$

$\begin{array}{cc}X=X_{\text{pu}}\times Z_B\hfill & \left(\text{D}.44\right)\hfill \end{array}$

$\begin{array}{cc}VA=V{A}_{\text{pu}}\times V{A}_B\hfill & \left(\text{D}.45\right)\hfill \end{array}$

$\begin{array}{cc}P=P_{\text{pu}}\times V{A}_B\hfill & \left(\text{D}.46\right)\hfill \end{array}$

$\begin{array}{cc}Q=Q_{\text{pu}}\times V{A}_B\hfill & \left(\text{D}.47\right)\hfill \end{array}$

D.4 Change Of Base

In general, the per-unit impedance of a power apparatus is given based on its own voltampere and voltage ratings and consequently based on its own impedance base. When such an apparatus is used in a system that has its own bases, it becomes necessary to refer all the given per-unit values to the system base values. Assume that the per-unit impedance of the apparatus is given based on its nameplate ratings as

$\begin{array}{cc}\begin{array}{ccc}{Z}_{\text{pu(given)}}\hfill & =\hfill & \left(Z_{\text{physical}}\right)\frac{MV{A}_B\left(\text{given}\right)}{{\left[k{V}_{B\left(\text{given}\right)}\right]}^2}\hfill \end{array}\hfill & \left(\text{D}.48\right)\hfill \end{array}$

and that it is necessary to refer the very same physical impedance to a new set of voltage and voltampere bases such that

$\begin{array}{cc}\begin{array}{ccc}{Z}_{\text{pu(new)}}\hfill & =\hfill & \left(Z_{\text{physical}}\right)\frac{MV{A}_B\left(\text{new}\right)}{{\left[k{V}_{B\left(\text{new}\right)}\right]}^2}\hfill \end{array}\hfill & \left(\text{D}.49\right)\hfill \end{array}$

By dividing Equation D.48 by Equation D.49 side by side,

$\begin{array}{cc}{Z}_{\text{pu}\left(\text{new}\right)}=Z_{\text{pu}\left(\text{old}\right)}\left[\frac{MV{A}_{B\left(\text{old}\right)}}{MV{A}_{B\left(\text{given}\right)}}\right]{\left[\frac{k{V}_{B\left(\text{given}\right)}}{k{V}_{B\left(\text{old}\right)}}\right]}^2\hfill & \left(\text{D}.50\right)\hfill \end{array}$

In certain situations, it is more convenient to use subscripts 1 and 2 instead of subscripts “given” and “new,” respectively. Then Equation D.50 can be expressed as

$\begin{array}{cc}{Z}_{\text{pu}\left(2\right)}=Z_{\text{pu}\left(1\right)}\left[\frac{MV{A}_{B\left(2\right)}}{MV{A}_{B\left(1\right)}}\right]{\left[\frac{k{V}_{B\left(1\right)}}{k{V}_{B\left(2\right)}}\right]}^2\hfill & \left(\text{D}.51\right)\hfill \end{array}$

In the event that the kV bases are the same but the MVA bases are different, from Equation D.50,

$\begin{array}{cc}{Z}_{\text{pu}\left(\text{new}\right)}=Z_{\text{pu}\left(\text{given}\right)}\frac{MV{A}_{B\left(\text{new}\right)}}{MV{A}_{B\left(\text{given}\right)}}\hfill & \left(\text{D}.52\right)\hfill \end{array}$

Similarly, if the megavoltampere bases are the same but the kilovolt bases are different, from Equation D.50,

$\begin{array}{cc}{Z}_{\text{pu}\left(\text{new}\right)}=Z_{\text{pu}\left(\text{given}\right)}{\left[\frac{k{V}_{B\left(\text{given}\right)}}{k{V}_{B\left(\text{new}\right)}}\right]}^2\hfill & \left(\text{D}.53\right)\hfill \end{array}$

Equations D.49 through D.52 must only be used to convert the given per-unit impedance from the base to another but not for referring the physical value of an impedance from one side of the transformer to another [3].

Example D.3

Consider Example B.1 and select 300/150 V as the base voltages for the high-voltage and the low-voltage windings, respectively. Use a new base power of 10 kVA and determine the new per-unit, base, and physical impedances of the transformer referred to the high-voltage side.

Solution:

By using Equation D.50, the new per-unit impedance can be found as

$\begin{array}{lll}{Z}_{\text{pu}\left(\text{new}\right)}\hfill & =\hfill & Z_{\text{pu}\left(\text{old}\right)}\left[\frac{MV{A}_{B\left(\text{old}\right)}}{MV{A}_{B\left(\text{given}\right)}}\right]{\left[\frac{k{V}_{B\left(\text{given}\right)}}{k{V}_{B\left(\text{old}\right)}}\right]}^2\hfill \\ \hfill & =\hfill & \left(0.0313\text{pu}\right)\left(\frac{10,000\text{VA}}{300\text{V}}\right){\left(\frac{240\text{V}}{300\text{V}}\right)}^2\hfill \\ \hfill & =\hfill & 33.334\text{ A}\hfill \end{array}$

The new current base is

$\begin{array}{lll}{I}_{B\left(HV\right)\text{new}}\hfill & =\hfill & \frac{S_B}{V_{B\left(\text{HV}\right)\text{new}}}\hfill \\ \hfill & =\hfill & \frac{10,000\text{VA}}{300\text{V}}\hfill \\ \hfill & =\hfill & 33,334\text{ A}\hfill \end{array}$

Thus,

$\begin{array}{lll}{Z}_{B\left(\text{HV}\right)\text{new}}\hfill & =\hfill & \frac{V_{B\left(\text{HV}\right)\text{new}}}{I_{B\left(\text{HV}\right)\text{new}}}\hfill \\ \hfill & =\hfill & \frac{300\text{V}}{33.334\text{ A}}\hfill \\ \hfill & =\hfill & 9\mathrm{\Omega }\hfill \end{array}$

Therefore, the physical impedance of the transformer is still

$\begin{array}{lll}{Z}_{\text{HV}}\hfill & =\hfill & Z_{\text{pu},\text{new}}\times Z_{B\left(\text{HV}\right)\text{new}}\hfill \\ \hfill & =\hfill & \left(0.0401\text{pu}\right)\left(9\mathrm{\Omega }\right)\hfill \\ \hfill & =\hfill & 0.3609\mathrm{\Omega }\hfill \end{array}$

D.5 Three-Phase Systems

The three-phase problems involving balanced systems can be solved on a per-phase basis. In that case, the equations that are developed for single-phase systems can be used for three-phase systems as long as per-phase values are used consistently. Therefore,

$\begin{array}{cc}{I}_B=\frac{S_{B\left(1\varphi \right)}}{V_{B\left(L-N\right)}}\hfill & \left(\text{D}.54\right)\hfill \end{array}$

or

$\begin{array}{cc}{I}_B=\frac{V{A}_{B\left(1\varphi \right)}}{V_{B\left(L-N\right)}}\hfill & \left(\text{D}.55\right)\hfill \end{array}$

and

$\begin{array}{cc}{Z}_B=\frac{V_{B\left(L-N\right)}}{I_B}\hfill & \left(\text{D}.56\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_B=\frac{{\left[k{V}_{B\left(L-N\right)}\right]}^2\left(1000\right)}{kV{A}_{B\left(1\varphi \right)}}\hfill & \left(\text{D}.57\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_B=\frac{{\left[k{V}_{B\left(L-N\right)}\right]}^2}{MV{A}_{B\left(1\varphi \right)}}\hfill & \left(\text{D}.58\right)\hfill \end{array}$

where the subscripts 1ϕ and L–N denote per phase and line to neutral, respectively. Note that, for a balanced system,

$\begin{array}{cc}{V}_{B\left(L-N\right)}=\frac{V_{B\left(L-L\right)}}{\sqrt{3}}\hfill & \left(\text{D}.59\right)\hfill \end{array}$

and

$\begin{array}{cc}{S}_{B\left(3\varphi \right)}=\frac{S_{B\left(3\varphi \right)}}{3}\hfill & \left(\text{D}.60\right)\hfill \end{array}$

However, it has been customary in three-phase system analysis to use line-to-line voltage and three-phase voltamperes as the base values. Therefore,

$\begin{array}{cc}{I}_B=\frac{S_{B\left(3\varphi \right)}}{\sqrt{3}{V}_{B\left(L-L\right)}}\hfill & \left(\text{D}.61\right)\hfill \end{array}$

or

$\begin{array}{cc}{I}_B=\frac{kV{A}_{B\left(3\varphi \right)}}{\sqrt{3}k{V}_{B\left(L-L\right)}}\hfill & \left(\text{D}.62\right)\hfill \end{array}$

and

$\begin{array}{cc}{Z}_B=\frac{V_{B\left(L-L\right)}}{\sqrt{3}{I}_B}\hfill & \left(\text{D}.63\right)\hfill \end{array}$

$\begin{array}{cc}{Z}_B=\frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2\left(1000\right)}{kV{A}_{B\left(3\varphi \right)}}\hfill & \left(\text{D}.64\right)\hfill \end{array}$

or

$\begin{array}{cc}{Z}_B=\frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2}{MV{A}_{B\left(3\varphi \right)}}\hfill & \left(\text{D}.65\right)\hfill \end{array}$

where the subscripts 3ϕ and L–L denote per three phase and line, respectively. Furthermore, base admittance can be expressed as

$\begin{array}{cc}{Y}_B=\frac{1}{Z_B}\hfill & \left(\text{D}.66\right)\hfill \end{array}$

or

$\begin{array}{cc}\hfill & \left(\text{D}.67\right)\hfill \end{array}$

where

$\begin{array}{cc}{Y}_B=B_B=G_B\hfill & \left(\text{D}.68\right)\hfill \end{array}$

The data for transmission lines are usually given in terms of the line resistance R in ohms per mile at a given temperature, the line inductive reactance XL in ohms per mile at 60 Hz, and the line shunt capacitive reactance Xc in megohms per mile at 60 Hz. Therefore, the line impedance and shunt susceptance in per units for 1 mi of line can be expressed as

$\begin{array}{cc}{Z}_{\text{pu}}=\left(\text{Z,}\mathrm{\Omega }\text{/mi}\right)\frac{MV{A}_{B\left(3\varphi \right)}}{{\left[kVB\left(L-L\right)\right]}^2}\text{pu}\hfill & \left(\text{D}.69\right)\hfill \end{array}$

where

$Z=R+j{X}_L=Z\angle \theta \mathrm{\Omega }/\text{mi}$

and

$\begin{array}{cc}{B}_{\text{pu}}=\frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2\times {10}^{-6}}{\left[MV{A}_{B\left(3\varphi \right)}\right]\left[X_c,M\mathrm{\Omega }/\text{mi}\right]}\hfill & \left(\text{D}.70\right)\hfill \end{array}$

In the event that the admittance for a transmission line is given in microsiemens per mile, the perunit admittance can be expressed as

$\begin{array}{cc}{Y}_{\text{pu}}=\frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2\left(Y,\text{μ}S\right)}{\left[MV{A}_{B\left(3\varphi \right)}\right]\times {10}^6}\hfill & \left(\text{D}.71\right)\hfill \end{array}$

Similarly, if it is given as reciprocal admittance in megohms per mile, the per-unit admittance can be found as

$\begin{array}{cc}{Y}_{\text{pu}}=\frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2\times {10}^{-6}}{\left[MV{A}_{B\left(3\varphi \right)}\right]\left[Z,M\mathrm{\Omega }/\text{mi}\right]}\hfill & \left(\text{D}.72\right)\hfill \end{array}$

Figure 4.29 shows conventional three-phase transformer connections and associated relationships between the high-voltage and low-voltage side voltages and currents. The given relationships are correct for a three-phase transformer as well as for a three-phase bank of single-phase transformers. Note that in the figure, n is the turns ratio, that is,

$\begin{array}{cc}n=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{I_2}{I_1}\hfill & \left(\text{D}.73\right)\hfill \end{array}$

where the subscripts 1 and 2 are used for the primary and secondary sides. Therefore, an impedance Z2 in the secondary circuit can be referred to the primary circuit provided that

$\begin{array}{cc}{Z}_1=n^2{Z}_2\hfill & \left(\text{D}.74\right)\hfill \end{array}$

Thus, it can be observed from Figure 4.29 that in an ideal transformer, voltages are transformed in the direct ratio of turns, currents in the inverse ratio, and impedances in the direct ratio squared, and power and voltamperes are, of course, unchanged. Note that a balanced delta-connected circuit of ZΔ Ω/phase is equivalent to a balanced wye-connected circuit of ZY Ω/phase as long as

$\begin{array}{cc}{Z}_Y=\frac{1}{3}{\mathrm{Z}}_{\mathrm{\Delta }}\hfill & \left(\text{D}.75\right)\hfill \end{array}$

The per-unit impedance of a transformer remains the same without taking into account whether it is converted from physical impedance values that are found by referring to the high-voltage side or low-voltage side of the transformer. This can be accomplished by choosing separate appropriate bases for each side of the transformer (whether or not the transformer is connected in wye–wye, delta–delta, delta–wye, or wye–delta since the transformation of voltages is the same as that made by wye–wye transformers as long as the same line-to-line voltage ratings are used). In other words, the designated per-unit impedance values of transformers are based on the coil ratings.

Since the ratings of coils cannot be altered by a simple change in connection (e.g., from wye–wye to delta–wye), the per-unit impedance remains the same regardless of the three-phase connection. The line-to-line voltage for the transformer will differ. Because of the method of choosing the base in various sections of the three-phase system, the per-unit impedances calculated in various sections can be put together on one impedance diagram without paying any attention to whether the transformers are connected in wye–wye or delta–wye.

Example D.4

Assume that a 19.5 kV 120 MVA three-phase generator has a synchronous reactance of 1.5/Ω and is connected to a 150 MVA 18/230 kV delta–wye connected three-phase transformer with a 0.1/Ω reactance. The transformer is connected to a transmission line at the 230 kV side. Use the new MVA base of 100 MVA and 240 kV base for the line and determine the following:

1. The new reactance value for the generator in per unit ohms.
2. The new reactance value for the transformer in per unit ohms.

Solution:

1. Using Equation B.49, the new per unit impedance of the generator is

   $Z_{\text{pu}\left(\text{new}\right)}=Z_{\text{pu}\left(\text{old}\right)}\left[\frac{MV{A}_{B\left(\text{new}\right)}}{MV{A}_{B\left(\text{old}\right)}}\right]{\left[\frac{k{V}_{B\left(\text{old}\right)}}{k{V}_{B\left(\text{old}\right)}}\right]}^2$

   But, first determining the new kV base for the generator,

   $\mathrm{k}{\text{V}}_{B\left(\text{new}\right)}^{\text{gen}}=\left(240\mathrm{k}\text{V}\right)\left(\frac{18\text{kV}}{230\text{kV}}\right)=18.783\text{kV}$

   Thus, the new and adjusted synchronous reactance of the generator is

   $\begin{array}{lll}{X}_{B\left(\text{new}\right)}^{\text{gen}}\hfill & =\hfill & X_{B\left(\text{new}\right)}^{\text{gen}}\left[\frac{MV{A}_{B\left(\text{new}\right)}}{MV{A}_{B\left(\text{old}\right)}}\right]{\left[\frac{k{V}_{B\left(\text{old}\right)}}{k_{B\left(\text{new}\right)}}\right]}^2\hfill \\ \hfill & =\hfill & \left(1.5\text{pu}\right)\left[\frac{100\text{MVA}}{120\text{MVA}}\right]{\left[\frac{19.5\text{kV}}{18.783\text{kV}}\right]}^2\hfill \\ \hfill & =\hfill & 1.347\text{pu}\hfill \end{array}$

2. The new reactance value for the transformer in per unit ohms, referred to high-voltage side is

   $\begin{array}{lll}{X}_{\text{pu}\left(\text{new}\right)}^{\text{trf}}\hfill & =\hfill & \left(0.1\right)\left[\frac{100\text{MVA}}{150\text{MVA}}\right]{\left[\frac{230\text{kV}}{240\text{kV}}\right]}^2\hfill \\ \hfill & =\hfill & 0.061\text{pu}\hfill \end{array}$

   And referred to the low-voltage side is

   $\begin{array}{ccc}{X}_{\text{pu(new)}}^{\text{trf}}\hfill & =\hfill & (0.1\text{pu})\left[\frac{100\text{MVA}}{150\text{MVA}}\right]{\left[\frac{18\text{KV}}{18.783\text{KV}}\right]}^2\hfill \\ \hfill & =\hfill & 0.061\text{pu}\hfill \end{array}$

   Note that the transformer reactance referred to the high-voltage side or the low-voltage side is the same, as it should be!

Example D.5

A three-phase transformer has a nameplate ratings of 20 MVA, 345Y/34.5Y kV with a leakage reactance of 12% and the transformer connection is wye–wye. Select a base of 20 MVA and 345 kV on the high-voltage side and determine the following:

1. Reactance of transformer in per units.
2. High-voltage side base impedance.
3. Low-voltage side base impedance.
4. Transformer reactance referred to high-voltage side in ohms.
5. Transformer reactance referred to low-voltage side in ohms.

Solution:

1. The reactance of the transformer in per units is 12/100, or 0.12 pu. Note that it is the same whether it is referred to the high-voltage or the low-voltage sides.
2. The high-voltage side base impedance is

   $\begin{array}{lll}{Z}_{B\left(\text{HV}\right)}\hfill & =\hfill & \frac{{\left[k{V}_{B\left(\text{HV}\right)}\right]}^2}{MV{A}_{B\left(3\varphi \right)}}\hfill \\ \hfill & =\hfill & \frac{{345}^2}{20}=5951.25\mathrm{\Omega }\hfill \end{array}$

3. The low-voltage side base impedance is

   $\begin{array}{lll}{Z}_{B\left(\text{LV}\right)}\hfill & =\hfill & \frac{{\left[k{V}_{B\left(\text{LV}\right)}\right]}^2}{MV{A}_{B\left(3\varphi \right)}}\hfill \\ \hfill & =\hfill & \frac{{34.5}^2}{20}=59.5125\mathrm{\Omega }\hfill \end{array}$

4. The reactance referred to the high-voltage side is

   $\begin{array}{lll}{X}_{\left(\text{HV}\right)}\hfill & =\hfill & X_{\text{pu}}\times X_{B\left(\text{HV}\right)}\hfill \\ \hfill & =\hfill & \left(0.12\right)\left(5951.25\right)=714.15\mathrm{\Omega }\hfill \end{array}$

5. The reactance referred to the low-voltage side is

   $\begin{array}{lll}{X}_{\left(\text{LV}\right)}\hfill & =\hfill & X_{\text{pu}}\times X_{B\left(\text{LV}\right)}\hfill \\ \hfill & =\hfill & \left(0.12\right)\left(59.5125\right)=7.1415\mathrm{\Omega }\hfill \end{array}$

   or from

   $\begin{array}{lll}{X}_{\left(\text{LV}\right)}\hfill & =\hfill & \frac{X_{\left(\text{HV}\right)}}{n^2}\hfill \\ \hfill & =\hfill & \frac{714.15\mathrm{\Omega }}{{\left(\frac{345/\sqrt{3}}{34.5/\sqrt{3}}\right)}^2}=7.1415\mathrm{\Omega }\hfill \end{array}$

   where n is defined as the turns ratio of the windings.

Example D.6

A three-phase transformer has a nameplate ratings of 20 MVA, and the voltage ratings of 345Y/34.5Δ kV with a leakage reactance of 12% and the transformer connection is wye–delta. Select a base of 20 MVA and 345 kV on the high-voltage side and determine the following:

1. Turns ratio of windings.
2. Transformer reactance referred to low-voltage side in ohms.
3. Transformer reactance referred to low-voltage side in per units.

Solution:

1. The turns ratio of the windings is

   $n=\frac{345/\sqrt{3}}{34.5}=5.7735$

2. Since the high-voltage side impedance base is

   $\begin{array}{lll}{Z}_{B\left(\text{HV}\right)}\hfill & =\hfill & \frac{{\left[k{V}_{B\left(\text{HV}\right)}\right]}^2}{MV{A}_{B\left(3\varphi \right)}}\hfill \\ \hfill & =\hfill & \frac{{345}^2}{20}=5951.25\mathrm{\Omega }\hfill \end{array}$

   and

   $\begin{array}{ccc}{X}_{(\text{HV})}\hfill & =\hfill & X_{\text{pu}}\times X_{B(\text{HV})}\hfill \\ \hfill & =\hfill & (0.12)(5951.25)=714.15\mathrm{\Omega }\hfill \end{array}$

   Thus, the transformer reactance referred to the delta-connected low-voltage side is

   $\begin{array}{lll}{X}_{\left(\text{HV}\right)}\hfill & =\hfill & X_{\text{pu}}\times X_{B\left(\text{HV}\right)}\hfill \\ \hfill & =\hfill & \frac{714.14\mathrm{\Omega }}{{5.7735}^2}=21.4245\mathrm{\Omega }\hfill \end{array}$

3. The reactance of the equivalent wye connection is

   $\begin{array}{lll}{Z}_Y\hfill & =\hfill & \frac{Z_{\Delta }}{3}\hfill \\ \hfill & =\hfill & \frac{21.4245\mathrm{\Omega }}{3}=7.1415\mathrm{\Omega }\hfill \end{array}$

   Similarly,

   $\begin{array}{lll}{Z}_{B\left(\text{LV}\right)}\hfill & =\hfill & \frac{{\left[k{V}_{B\left(LV\right)}\right]}^2}{MV{A}_{B\left(3\varphi \right)}}\hfill \\ \hfill & =\hfill & \frac{{34.5}^2}{20}=59.5125\mathrm{\Omega }\hfill \end{array}$

   Thus,

   $\begin{array}{lll}{X}_{\text{pu}}\hfill & =\hfill & \frac{7.1415\mathrm{\Omega }}{Z_{B\left(\text{LV}\right)}}\hfill \\ \hfill & =\hfill & \frac{7.1415\mathrm{\Omega }}{59.5125\mathrm{\Omega }}=0.12\text{pu}\hfill \end{array}$

   Alternatively, if the line-to-line voltages are used,

   $\begin{array}{lll}{X}_{\left(\text{LV}\right)}\hfill & =\hfill & \frac{X_{\left(\text{HV}\right)}}{n^2}\hfill \\ \hfill & =\hfill & \frac{714.14\mathrm{\Omega }}{{\left(345/34.5\right)}^2}=7.1415\mathrm{\Omega }\hfill \end{array}$

   and therefore,

   $\begin{array}{lll}{X}_{\text{pu}}\hfill & =\hfill & \frac{X_{\left(\text{LV}\right)}}{Z{B}_{\left(\text{LV}\right)}}\hfill \\ \hfill & =\hfill & \frac{7.1415\mathrm{\Omega }}{59.5125\mathrm{\Omega }}=0.12\text{pu}\hfill \end{array}$

   as before.

Example D.7

Consider a three-phase system that has a generator connected to a 2.4/24 kV, wye–wye connected, three-phase step-up transformer T1. Suppose that the transformer is connected to three-phase power line. The receiving end of the line is connected to a second, wye–wye connected, three-phase 24/12 kV step-down transformer T2. Assume that the line length between the two transformers is negligible and the three-phase generator is rated 4160 kVA, 2.4 kV, and 1000 A and that it supplies a purely inductive load of Ipu = 2.08∠–90° pu. The three-phase transformer T1 is rated 6000 kVA, 2.4Y–24Y kV, with leakage reactance of 0.04 pu. Transformer T2 is made up of three single-phase transformers and is rated 4000 kVA, 24Y–12Y kV, with leakage reactance of 0.04 pu. Determine the following for all three circuits, 2.4, 24, and 12 kV circuits:

1. Base kilovoltampere values.
2. Base line-to-line kilovolt values.
3. Base impedance values.
4. Base current values.
5. Physical current values (neglect magnetizing currents in transformers and charging currents in lines).
6. Per-unit current values.
7. New transformer reactances based on their new bases.
8. Per-unit voltage values at buses 1, 2, and 4.
9. Per-unit apparent power values at buses 1, 2, and 4.
10. Summarize results in a table.

Solution:

1. The kilovoltampere base for all three circuits is arbitrarily selected as 2080 kVA
2. The base voltage for the 2.4 kV circuit is arbitrarily selected as 2.5 kV. Since the turns ratios for transformers T1 and T2 are

   $\begin{array}{ccc}\frac{N_1}{N_2}=10\hfill & \text{or}\hfill & \frac{N_2}{N_1}=0.10\hfill \end{array}$

   and

   $\frac{N_1^{\text{'}}}{N_2^{\text{'}}}=2$

   the base voltages for the 24 and 12 kV circuits are determined to be 25 and 12.5 kV, respectively.

3. The base impedance values can be found as

   $\begin{array}{lll}{Z}_B\hfill & =\hfill & \frac{{\left[k{V}_{B\left(L-L\right)}\right]}^2\left(1000\right)}{kV{A}_{B\left(3\varphi \right)}}\hfill \\ \hfill & =\hfill & \frac{{\left[2.5kV\right]}^{2}1000}{2080\text{kVA}}=3.005\mathrm{\Omega }\hfill \end{array}$

   and

   $Z_B=\frac{{\left[25kV\right]}^{2}1000}{2080\text{kVA}}=300.5\mathrm{\Omega }$

   and

   $Z_B=\frac{{\left[12.5kV\right]}^{2}1000}{2080\text{kVA}}=75.1\mathrm{\Omega }$

4. The base current values can be determined as

   $\begin{array}{lll}{I}_B\hfill & =\hfill & \frac{kV{A}_{B\left(3\varphi \right)}}{\sqrt{3}{\text{kV}}_{B\left(L-L\right)}}\hfill \\ \hfill & =\hfill & \frac{2080\text{kVA}}{\sqrt{3}\left(2.5\text{kV}\right)}=480\text{A}\hfill \end{array}$

   and

   $\begin{array}{ccc}{I}_B\hfill & =\hfill & \frac{2080\text{kVA}}{\sqrt{3}\left(25\text{kV}\right)}=48\text{A}\hfill \end{array}$

   and

   $\begin{array}{ccc}{I}_B\hfill & =\hfill & \frac{2080\text{kVA}}{\sqrt{3}\left(12.5\text{kV}\right)}=96\text{A}\hfill \end{array}$

5. The physical current values can be found based on the turns ratios as

   $\begin{array}{lll}I\hfill & =\hfill & 1000\text{A}\hfill \\ I\hfill & =\hfill & \left(\frac{N_2}{N_1}\right)\left(1000\text{A}\right)=100\text{A}\hfill \\ I\hfill & =\hfill & \left(\frac{N{\cancel{c}}_1}{N{\cancel{c}}_2}\right)\left(100\text{A}\right)=200\text{A}\hfill \end{array}$

6. The per-unit current values are the same, 2.08 pu, for all three circuits.
7. The given transformer reactances can be converted based on their new bases using

   $Z_{\text{pu(new)}}{\text{ = Z}}_{\text{pu(given)}}\left[\frac{kV{A}_{B\left(\text{new}\right)}}{kV{A}_{B\left(\text{given}\right)}}\right]{\left[\frac{k{V}_{B\left(\text{given}\right)}}{k{V}_{B\left(\text{new}\right)}}\right]}^2$

   Therefore, the new reactances of the two transformers can be found as

   $Z_{\text{pu(}{T}_1\text{)}}\text{ = }j0.04\left[\frac{2080\text{kVA}}{6000\text{kVA}}\right]{\left[\frac{2.4\text{kV}}{2.5\text{kV}}\right]}^2=j0.0128\text{pu}$

   and

   $Z_{\text{pu(}{T}_2\text{)}}\text{ = }j0.04\left[\frac{2080\text{kVA}}{4000\text{kVA}}\right]{\left[\frac{12\text{kV}}{12.5\text{kV}}\right]}^2=j0.0192\text{pu}$

8. Therefore, the per-unit voltage values at buses 1, 2, and 4 can be calculated as

   $\begin{array}{lll}{\text{V}}_1\hfill & =\hfill & \frac{2.4k\text{V}\angle 0^{\circ }}{2.5\text{kV}}=0.96\angle 0^{\circ }\text{pu}\hfill \\ {\text{V}}_2\hfill & =\hfill & {\text{V}}_1-{\text{I}}_{\text{pu}}{\text{Z}}_{\text{pu}\left(T_1\right)}\hfill \\ \hfill & =\hfill & 0.96\angle 0^{\circ }-\left(2.08\angle -{90}^{\circ }\right)\left(0.0128\angle {90}^{\circ }\right)=0.9334\angle 0^{\circ }\text{pu}\hfill \\ {\text{V}}_4\hfill & =\hfill & {\text{V}}_2-{\text{I}}_{\text{pu}}{\text{Z}}_{\text{pu}\left(T_2\right)}\hfill \\ \hfill & =\hfill & 0.9334\angle 0^{\circ }-\left(2.08\angle -{90}^{\circ }\right)\left(0.0192\angle {90}^{\circ }\right)=0.8935\angle 0^{\circ }\text{pu}\hfill \end{array}$

9. Thus, the per-unit apparent power values at buses 1, 2, and 4 are

   $\begin{array}{ll}{S}_1\hfill & =2.00\text{pu}\hfill \\ S_2\hfill & =V_2{I}_{\text{pu}}=\left(0.9334\right)\left(2.08\right)=1.9415\text{pu}\hfill \\ S_4\hfill & =V_4{I}_{\text{pu}}=\left(0.8935\right)\left(2.08\right)=1.8585\text{pu}\hfill \end{array}$

10. The results are summarized in Table D.1.

    Table D.1

    Results of Example D.7

    Quantity	2.4 kV Circuit	24 kV Circuit	12 kV Circuit
    kVAB(3ϕ)	2080 kVA	2080 kVA	2080 kVA
    kVB(L-L)	2.5 kV	25 kV	12.5 kV
    ZB	3005 Ω	300.5 Ω	75.1 Ω
    IB	480 A	48 A	96 A
    Iphysical	1000 A	100 A	200 A
    Ipu	2.08 pu	2.08 pu	2.08 pu
    Vpu	0.96 pu	0.9334 pu	0.8935 pu
    Spu	2.00 pu	1.9415 pu	1.8585 pu

Problems

1. D.1 Solve Example D.1 for a transformer rated 100 kVA and 2400/240 V that has a high-voltage winding impedance of 0.911.
2. D.2 Consider the results of Problem D.1 and use 3000/300 V as new base voltages for the high-voltage and low-voltage windings, respectively. Use a new base power of 200 kVA and determine the new per-unit, base, and physical impedances of the transformer referred to the high-voltage side.
3. D.3 A 240/120 V single-phase transformer rated 25 kVA has a high-voltage winding impedance of 0.65 Ω. If 240 V and 25 kVA are used as the base quantities, determine the following:
   1. The high-voltage side base current.
   2. The high-voltage side base impedance in Q.
   3. The transformer impedance referred to the high-voltage side in per unit.
   4. The transformer impedance referred to the high-voltage side in percent.
   5. The turns ratio of the transformer windings.
   6. The low-voltage side base current.
   7. The low-voltage side base impedance.
   8. The transformer impedance referred to the low-voltage side in per unit.
4. D.4 A 240/120 V single-phase transformer is rated 25 kVA and has a high-voltage winding impedance referred to its high-voltage side that is 0.2821 pu based on 240 V and 25 kVA. Select 230/115 V as the base voltages for the high-voltage and low-voltage windings, respectively. Use a new base power of 50 kVA and determine the new per-unit base, and physical impedances of the transformer referred to the high-voltage side.
5. D.5 After changing the S base from 5 to 10 MVA, redo the Example D.1 by using MATLAB.
   1. Write the MATLAB program script.
   2. Give the MATLAB program output.

* It is defined as that impedance across which there is a voltage drop that is equal to the base voltage if the current through it is equal to the base current.