Perunit system
In the power systems analysis field of electrical engineering, a perunit system is the expression of system quantities as fractions of a defined base unit quantity. Calculations are simplified because quantities expressed as perunit do not change when they are referred from one side of a transformer to the other. This can be a pronounced advantage in power system analysis where large numbers of transformers may be encountered. Moreover, similar types of apparatus will have the impedances lying within a narrow numerical range when expressed as a perunit fraction of the equipment rating, even if the unit size varies widely. Conversion of perunit quantities to volts, ohms, or amperes requires a knowledge of the base that the perunit quantities were referenced to.
The main idea of a per unit system is to absorb large difference in absolute values into base relationships. Thus, representations of elements in the system with per unit values become more uniform.
A perunit system provides units for; power, voltage, current, impedance, and admittance. Except impedance and admittance, any two of these are independent and can be arbitrarily selected as base values, usually power and voltage. All quantities are specified as multiples of selected base values. For example, the base power might be the rated power of a transformer, or perhaps an arbitrarily selected power which makes power quantities in the system more convenient. The base voltage might be the nominal voltage of a bus. Different types of quantities are labeled with the same symbol (pu); it should be clear from context whether the quantity is a voltage, current, etc.
Perunit being used in power flow, short circuit evaluation and motor starting studies, it is important for all power engineers to be familiar with the concept.
Contents
 Purpose 1
 Base quantities 2

Relationship between units 3
 Singlephase 3.1
 Threephase 3.2
 Example of perunit 4
 Perunit system formulas 5
 In transformers 6
 References 7
Purpose
There are several reasons for using a perunit system:
 Similar apparatus (generators, transformers, lines) will have similar perunit impedances and losses expressed on their own rating, regardless of their absolute size. Because of this, perunit data can be checked rapidly for gross errors. A per unit value out of normal range is worth looking into for potential errors.
 Manufacturers usually specify the impedance of apparati in per unit values.
 Use of the constant \scriptstyle \sqrt{3} is reduced in threephase calculations.
 Perunit quantities are the same on either side of a transformer, independent of voltage level
 By normalizing quantities to a common base, both hand and automatic calculations are simplified.
 It improves numerical stability of automatic calculation methods
 Per unit data representation yields important information about relative magnitudes.
The perunit system was developed to make manual analysis of power systems easier. Although powersystem analysis is now done by computer, results are often expressed as perunit values on a convenient systemwide base.
Base quantities
Generally base values of power and voltage are chosen. The base power may be the rating of a single piece of apparatus such as a motor or generator. If a system is being studied, the base power is usually chosen as a convenient round number such as 10 MVA or 100 MVA. The base voltage is chosen as the nominal rated voltage of the system. All other base quantities are derived from these two base quantities. Once the base power and the base voltage are chosen, the base current and the base impedance are determined by the natural laws of electrical circuits. Note the base value should only be magnitudes, while the perunit values are phasors. The phase angles of complex power, voltage, current, impedance etc. are not affected by the conversion to per unit values.
By convention, we adopt the following two rules for base quantities:
 The value of base power is the same for the entire power system of concern.
 The ratio of the voltage bases on either side of a transformer is selected to be the same as the ratio of the transformer voltage ratings.
With these two rules, a perunit impedance remains unchanged when referred from one side of a transformer to the other. This allows us to eliminate ideal transformer from a transformer model.
Relationship between units
The relationship between units in a perunit system depends on whether the system is singlephase or threephase.
Singlephase
Assuming that the independent base values are power and voltage, we have:
 P_{base} = 1 \mathrm{pu}
 V_{base} = 1 \mathrm{pu}
Alternatively, the base value for power may be given in terms of reactive or apparent power, in which case we have, respectively,
 Q_{base} = 1 \mathrm{pu}
or
 S_{base} = 1 \mathrm{pu}
The rest of the units can be derived from power and voltage using the equations S = IV, P = S\cos(\phi), Q = S\sin(\phi) and \underline{V} = \underline{I} \underline{Z} (Ohm's law), Z being represented by \underline{Z} = R + j X = Z\cos(\phi) + j Z\sin(\phi). We have:
 I_{\mathrm{base}} = \frac{S_{\mathrm{base}}}{V_{base}} = 1 \mathrm{pu}
 Z_{\mathrm{base}} = \frac{V_{\mathrm{base}}}{I_{base}} = \frac{V_{\mathrm{base}}^{2}}{I_{\mathrm{base}}V_{\mathrm{base}}} = \frac{V_{\mathrm{base}}^{2}}{S_{\mathrm{base}}} = 1 \mathrm{pu}
 Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 1 \mathrm{pu}
Threephase
Power and voltage are specified in the same way as singlephase systems. However, due to differences in what these terms usually represent in threephase systems, the relationships for the derived units are different. Specifically, power is given as total (not perphase) power, and voltage is linetoline voltage. In threephase systems the equations P = S\cos(\phi) and Q = S\sin(\phi) also hold. The apparent power S now equals S_{\mathrm{base}}= \sqrt{3}V_{\mathrm{base}} I_{\mathrm{base}}
 I_{\mathrm{base}} = \frac{S_{\mathrm{base}}}{V_{\mathrm{base}} \times \sqrt{3}} = 1 \mathrm{pu}
 Z_{\mathrm{base}} = \frac{V_{\mathrm{base}}}{I_{\mathrm{base}} \times \sqrt{3}} = \frac^2}}{S_{\mathrm{base}}} = 1 \mathrm{pu}
 Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 1 \mathrm{pu}
Example of perunit
As an example of how perunit is used, consider a threephase power transmission system that deals with powers of the order of 500 MW and uses a nominal voltage of 138 kV for transmission. We arbitrarily select S_{\mathrm{base}} = 500\, \mathrm{MVA}, and use the nominal voltage 138 kV as the base voltage V_{\mathrm{base}}. We then have:
 I_{\mathrm{base}} = \frac{S_{\mathrm{base}}}{V_{base} \times \sqrt{3}} = 2.09 \, \mathrm{kA}
 Z_{\mathrm{base}} = \frac{V_{\mathrm{base}}}{I_{base} \times \sqrt{3}} = \frac{V_{\mathrm{base}}^{2}}{S_{\mathrm{base}}} = 38.1 \, \Omega
 Y_{\mathrm{base}} = \frac{1}{Z_{\mathrm{base}}} = 26.3 \, \mathrm{mS}
If, for example, the actual voltage at one of the buses is measured to be 136 kV, we have:
 V_{\mathrm{pu}} = \frac{V}{V_{\mathrm{base}}} = \frac{136 \, \mathrm{kV}}{138 \, \mathrm{kV}} = 0.9855 \, \mathrm{pu}
Perunit system formulas
The following tabulation of perunit system formulas is adapted from Beeman's Industrial Power Systems Handbook.
Equation  

\text{Base number selection}  
\text{Arbitrarily selecting from ohm's law the two base numbers: base voltage and base current}  
1  \text{We have, Z}=\frac{E}{I}  
2  \text{Base ohms}=\frac{\text{base volts}}{\text{base amperes}}  
3  \text{Perunit volts}=\frac{\text{volts}}{\text{base volts}}  
4  \text{Perunit amperes}=\frac{\text{amperes}}{\text{base amperes}}  
5  \text{Perunit ohms}=\frac{\text{ohms}}{\text{base ohms}}  
\text{Alternatively, choosing base volts and base kva values, we have,}  
\text{in singlephase systems:}  
6  \text{Base amperes }=\frac{\text{base kva * 1000}}{\text{base volts}}  
7  \text{Base amperes }=\frac{\text{base kva}}{\text{base kv}_{LL}}  
8  \text{Base ohms }=\frac{\text{base volts}}{\text{base amperes}}  
\text{and in threephase systems:}  
9  \text{Base amperes }=\frac{\text{base kva * 1000}}{\sqrt{3 } * \text{base volts}}  
10  \text{Base amperes }=\frac{\text{base kva}}{\sqrt{3 } * \text{base kv}_{LL}}  
11  \text{Base ohms }=\frac{\text{base volts}}{\sqrt{3 } * \text{base amperes}}  
\text{Working out for convenience perunit ohms directly, we have}  
\text{for singlephase and threephase systems:}  
12  \text{Base ohms }=\frac{\text{ohms * base kva}}{kv_{LL}^2 * 1000}  
\text{ShortCircuit Calculation Formulas}  
\text{Ohms conversions:}  
13  \text{Perunit ohms reactance} =\frac{\text{ohms reactance * }\text{kva base}}{kv_{LL}^2*1000}  
14  \text{Ohms reactance} =\frac{%\text{ reactance}*kv_{LL}^2 * 10}{\text{kva base}}  
15  \text{Perunit ohms reactance} = \frac{\text{per cent ohms reactance}}{100}  
\text{Changing ohms from one kva base to another:}  
16  \%\text{ ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1} * \%\text{ ohms reactance on base}_1  
17  \text{0/1 ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1}\text{ * 0/1 ohms reactance on base}_1  
\text{Changing incoming system reactance:}  
\text{a. If system reactance is given in percent, use Eq. 16 to change from one kva base to another.}  
\text{b. If system reactance is given in shortcircuit symmetrical rms kva or current, convert to perunit as follows:}  
18  \text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system shortcircuit kva}}  
19  \text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system shortcircuit current * }\sqrt{3} \text{ * system kv}_{LL}}  
\text{Calculating approximate motor kva base:}  
\text{a. For induction motors and 0.8 power factor synchronous motors}  
20  \text{kva base} \approx\text{ horsepower rating}  
\text{b. For unity power factor synchronous motors}  
21  \text{kva base} \approx\text{ 0.8 * horsepower rating}  
\text{Converting ohms from one voltage to another:}  
22  \text{Ohms on basis of voltage}_1 =(\frac{\text{voltage}_1}{\text{voltage}_2})^2 \text{ * ohms on basis of voltage}_2  
\text{Shortcircuit kva and current calculations}  
\text{Symmetrical short circuit kva:}  
23  =\frac{\text{100 * kva base}}{%\text{ X}}  
24  =\frac{\text{kva base}}{\text{0/1 X}}  
25  =3 * \frac{\text{Voltage}_{LN}^2}{\text{ohms reactance}\text{ * 1000}}  
26  =\frac{\text{kv}_{LL}^2 \text{ * 1000}}{\text{ohms reactance}}  
\text{Symmetrical short circuit current:}  
27  =\frac{\text{100 * kva base}}{%\text{ X}* \sqrt{3} *\text{kv}_{LL}}  
28  =\frac{\text{kva base}}{\text{0/1 X}* \sqrt{3} *\text{kv}_{LL}}  
29  =\frac{\text{kv}_{LL} \text{ * 1000}}{\sqrt{3} * \text{ohms reactance}}  
\text{Asymmetrical shortcircuit current and kva:}  
30  \text{Asymmetrical shortcircuit current = symmetrical current * X/R factor }  
31  \text{Asymmetrical shortcircuit kva = symmetrical kva * X/R factor } 
In transformers
It can be shown that voltages, currents, and impedances in a perunit system will have the same values whether they are referred to primary or secondary of a transformer.^{[1]}
The full load copper loss of a transformer in perunit form is equal to the perunit value of its resistance:
\begin{align} P_{cu,FL}&=\text{fullload copper loss}\\ &= I_{R1}^2R_{eq1}\\ \end{align}
\begin{align} P_{cu,FL,pu}&=\frac{P_{cu,FL}}{P_{base}}\\ &= \frac {I_{R1}^2R_{eq1}} {V_{R1}I_{R1}}\\ &= \frac {R_{eq1}} {V_{R1}/I_{R1}}\\ &= \frac {R_{eq1}} {Z_{B1}}\\ &= R_{eq1,pu}\\ \end{align}
Therefore, it may be more useful to express the resistance in perunit form as it also represents the fullload copper loss.^{[2]}
References
 ^ Sen, P. C. (1997). Principles of electric machines and power electronics. New York: John Wiley & Sons. p. 85.
 ^ Sen, P. C. (1997). Principles of electric machines and power electronics. New York: John Wiley & Sons. p. 86.
 Beeman, Donald (1955). "ShortCircuitCurrent Calculating Procedures". In Beeman, Donald (ed.). Industrial Power Systems Handbook. McGrawHill. pp. see esp. 38–41, 52–55.
 Elgerd, Olle I. (2007). "§2.5 PerUnit Representation of Impedances, Currents, Voltages and Powers". Electric Energy Systems Theory: An Introduction (1971 1st ed.). Tata McGrawHill. pp. 35–39.
 Yuen, Moon H. (Mar–Apr 1974). "Short Circuit ABCLearn It in an Hour, Use It Anywhere, Memorize No Formula". IEEE Trans. on Industry Applications. IA10 (2): 261–272.
 William D., Jr, Stevenson, (1975). Elements of power system analysis (3rd ed.). New York: McGrawHill.
 Weedy, B.M. (1972). Electric power systems (2nd ed.). London ; Toronto: J. Wiley.
 Glover, J. Duncan; Sarma, Mulukutla; Overbye, Thomas J. (2011). Power System Analysis and Design. Cengage Learning. pp. 108–116.