Editing Per-unit system (section)

==Per-unit system formulas==
The following tabulation of per-unit system formulas is adapted from Beeman's ''Industrial Power Systems Handbook''.
{| 
! Equation
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| style="background:##ffdead;" colspan="2"| Base number selection
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| || Arbitrarily selecting from ohm's law the two base numbers: base voltage and base current
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|style="text-align:center;" | 1||<math>\text{We have, Z}=\frac{E}{I}</math>
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|style="text-align:center;" | 2||<math>\text{Base ohms}=\frac{\text{base volts}}{\text{base amperes}}</math>
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|style="text-align:center;" | 3||<math>\text{Per-unit volts}=\frac{\text{volts}}{\text{base volts}}</math>
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|style="text-align:center;" | 4||<math>\text{Per-unit amperes}=\frac{\text{amperes}}{\text{base amperes}}</math>
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|style="text-align:center;" | 5||<math>\text{Per-unit ohms}=\frac{\text{ohms}}{\text{base ohms}}</math>
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| || Alternatively, choosing base volts and base kva values, we have,
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| || in single-phase systems:
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|style="text-align:center;" | 6||<math>\text{Base amperes }=\frac{\text{base kva}\cdot 1000}{\text{base volts}}</math>
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|style="text-align:center;" | 7||<math>\text{Base amperes }=\frac{\text{base kva}}{\text{base kv}_{L-L}}</math>
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|style="text-align:center;" | 8||<math>\text{Base ohms }=\frac{\text{base volts}}{\text{base amperes}}</math>
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| || and in three-phase systems:
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|style="text-align:center;" | 9||<math>\text{Base amperes }=\frac{\text{base kva}\cdot 1000}{\sqrt{3 } \cdot \text{base volts}}</math>
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|style="text-align:center;" | 10||<math>\text{Base amperes }=\frac{\text{base kva}}{\sqrt{3 } \cdot \text{base kv}_{L-L}}</math>
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|style="text-align:center;" | 11||<math>\text{Base ohms }=\frac{\text{base volts}}{\sqrt{3 } \cdot \text{base amperes}}</math>
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| || Working out for convenience per-unit ohms directly, we have
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| || for single-phase and three-phase systems:
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|style="text-align:center;" | 12||<math>\text{Base ohms }=\frac{\text{ohms} \cdot \text{base kva}}{kv_{L-L}^2 \cdot 1000}</math>
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| style="background:##ffdead;" colspan="2"| Short-Circuit Calculation Formulas
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| || style="background:##ffdead;" colspan="2"| Ohms conversions:

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|style="text-align:center;" | 13||<math>\text{Per-unit ohms reactance} =\frac{\text{ohms reactance}\cdot\text{kva base}}{kv_{L-L}^2 \cdot 1000}</math>
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|style="text-align:center;" | 14||<math>\text{Ohms reactance} =\frac{\%\text{ reactance}\cdot kv_{L-L}^2 \cdot 1000}{\text{kva base}}</math>
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|style="text-align:center;" | 15||<math>\text{Per-unit ohms reactance} = \frac{\text{per cent ohms reactance}}{100}</math>
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| || style="background:##ffdead;" colspan="2"| Changing ohms from one kva base to another:
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|style="text-align:center;" | 16||<math>\%\text{ ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1} \cdot \%\text{ ohms reactance on base}_1</math>
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|style="text-align:center;" | 17||<math>\text{0/1 ohms reactance on kva base}_2=\frac{\text{kva base}_2}{\text{kva base}_1}\cdot 0/1 \text{ ohms reactance on base}_1</math>
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| || style="background:##ffdead;" colspan="2"| Changing incoming system reactance:
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| || a. If system reactance is given in percent, use Eq. 16 to change from one kva base to another.
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| || b. If system reactance is given in short-circuit symmetrical rms kva or current, convert to per-unit as follows:
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|style="text-align:center;" | 18||<math>\text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system short-circuit kva}}</math>
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|style="text-align:center;" | 19||<math>\text{0/1 reactance} = \frac{\text{kva base used in reactance in studied calculation}}{\text{system short-circuit current}\cdot\sqrt{3} \cdot\text{system kv}_{L-L}}</math>
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| || style="background:##ffdead;" colspan="2"| Calculating approximate motor kva base:
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| || a. For induction motors and 0.8 power factor synchronous motors
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|style="text-align:center;" | 20||<math>\text{kva base} \approx\text{ horsepower rating}</math>
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| || b. For unity power factor synchronous motors
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|style="text-align:center;" | 21||<math>\text{kva base} 0.8 \cdot\approx\text{horsepower rating}</math>
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| || style="background:##ffdead;" colspan="2"| Converting ohms from one voltage to another:
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|style="text-align:center;" | 22||<math>\text{Ohms on basis of voltage}_1 =\left(\frac{\text{voltage}_1}{\text{voltage}_2}\right)^2 \cdot\text{ohms on basis of voltage}_2</math>
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| ||align="left" style="background:##ffdead;" colspan="2"| Short-circuit kva and current calculations
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| || Symmetrical short circuit kva:
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|style="text-align:center;" | 23||<math>=100 \cdot \frac{\text{kva base}}{\%\text{ X}}</math>
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|style="text-align:center;" | 24||<math>=\frac{\text{kva base}}{\text{0/1 X}}</math>
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|style="text-align:center;" | 25||<math>=3 \cdot \frac{\text{Voltage}_{L-N}^2}{\text{ohms reactance}\cdot 1000}</math>
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|style="text-align:center;" | 26||<math>=\frac{\text{kv}_{L-L}^2 \cdot 1000}{\text{ohms reactance}}</math>
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| || Symmetrical short circuit current:
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|style="text-align:center;" | 27||<math>=100 \cdot\frac{\text{kva base}}{\%\text{ X}\cdot \sqrt{3} \cdot\text{kv}_{L-L}}</math>
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|style="text-align:center;" | 28||<math>=\frac{\text{kva base}}{\text{0/1 X}\cdot \sqrt{3} \cdot\text{kv}_{L-L}}</math>
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|style="text-align:center;" | 29||<math>=\frac{\text{kv}_{L-L} \cdot 1000}{\sqrt{3} \cdot \text{ohms reactance}}</math>
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| || Asymmetrical short-circuit current and kva:
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|style="text-align:center;" | 30||<math>\text{Asymmetrical short-circuit current = symmetrical current}\cdot \text{X/R factor}</math>
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|style="text-align:center;" | 31||<math>\text{Asymmetrical short-circuit kva = symmetrical kva}\cdot  \text{X/R factor}</math>
|}