سیستم پریونیت
روابط ریاضی

| Equation |  |  |
| --- | --- | --- |
|  {\text{Base number selection}} |  {\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{Per-unit volts}}={\frac {\text{volts}}{\text{base volts}}} |  |
: 
|  4 |  {\text{Per-unit amperes}}={\frac {\text{amperes}}{\text{base amperes}}} |  |
|  5 |  {\text{Per-unit ohms}}={\frac {\text{ohms}}{\text{base ohms}}} |  |
|  |  {\text{Alternatively, choosing base volts and base kva values, we have,}} |  |
|  |  {\text{in single-phase systems:}} |  |
|  6 |  {\text{Base amperes }}={\frac {\text{base kva * 1000}}{\text{base volts}}} |  |
|  7 |  {\text{Base amperes }}={\frac {\text{base kva}}{{\text{base kv}}_{L-L}}} |  |
|  8 |  {\text{Base ohms }}={\frac {\text{base volts}}{\text{base amperes}}} |  |
|  |  {\text{and in three-phase 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}}_{L-L}}} |  |
|  11 |  {\text{Base ohms }}={\frac {\text{base volts}}{{\sqrt {3}}*{\text{base amperes}}}} |  |
|  |  {\text{Working out for convenience per-unit ohms directly, we have}} |  |
|  |  {\text{for single-phase and three-phase systems:}} |  |
|  12 |  {\text{Base ohms }}={\frac {\text{ohms * base kva}}{kv_{L-L}^{2}*1000}} |  |
|  {\text{Short-Circuit Calculation Formulas}} |  {\text{Short-Circuit Calculation Formulas}} |  |
:  {\text{Ohms conversions:}}
|  13 |  {\text{Per-unit ohms reactance}}={\frac {{\text{ohms reactance * }}{\text{kva base}}}{kv_{L-L}^{2}*1000}} |  |
|  14 |  {\text{Ohms reactance}}={\frac {\%{\text{ reactance}}*kv_{L-L}^{2}*10}{\text{kva base}}} |  |
|  15 |  {\text{Per-unit 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 short-circuit symmetrical rms kva or current, convert to per-unit as follows:}} |  |
|  18 |  {\text{0/1 reactance}}={\frac {\text{kva base used in reactance in studied calculation}}{\text{system short-circuit kva}}} |  |
|  19 |  {\text{0/1 reactance}}={\frac {\text{kva base used in reactance in studied calculation}}{{\text{system short-circuit current * }}{\sqrt {3}}{\text{ * system kv}}_{L-L}}} |  |
:  {\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{Short-circuit 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}}_{L-N}^{2}}{{\text{ohms reactance}}{\text{ * 1000}}}} |  |
|  26 |  ={\frac {{\text{kv}}_{L-L}^{2}{\text{ * 1000}}}{\text{ohms reactance}}} |  |
|  |  {\text{Symmetrical short circuit current:}} |  |
|  27 |  ={\frac {\text{100 * kva base}}{\%{\text{ X}}*{\sqrt {3}}*{\text{kv}}_{L-L}}} |  |
|  28 |  ={\frac {\text{kva base}}{{\text{0/1 X}}*{\sqrt {3}}*{\text{kv}}_{L-L}}} |  |
|  29 |  ={\frac {{\text{kv}}_{L-L}{\text{ * 1000}}}{{\sqrt {3}}*{\text{ohms reactance}}}} |  |
|  |  {\text{Asymmetrical short-circuit current and kva:}} |  |
|  30 |  {\text{Asymmetrical short-circuit current = symmetrical current *  X/R factor }} |  |
|  31 |  {\text{Asymmetrical short-circuit kva = symmetrical kva *  X/R factor }} |  |