\[ \text{KVAR} = \frac{C \cdot V^2\cdot 2\cdot 3.1415 \cdot F }{1000000000} \]
For 60Hz:
\[ I_{60Hz} = 1330 \cdot \sqrt{\frac{\text{KVAR}}{L}} \]
For 50Hz:
\[ I_{50Hz} = 1456.7 \cdot \sqrt{\frac{\text{KVAR}}{L}} \]
This program calculates the current based on the capacitance (KVAR) and inductance (L) at 50Hz and 60Hz frequencies. It first calculates the KVAR from the given capacitance and voltage, and then uses this value along with the inductance to calculate the current at both frequencies.
Let's assume the following values:
Step 1: Calculate KVAR:
\[ \text{KVAR} = \frac{100 \cdot 230^2\cdot 2\cdot 3.1415\cdot 50}{1000000000} = 1.662 \, \text{kVAR} \]
\[ \text{KVAR} = \frac{100 \cdot 230^2\cdot 2\cdot 3.1415\cdot 60}{1000000000} = 1.994 \, \text{kVAR} \]
Step 2: Calculate Current:
For 60Hz:
\[ I_{60Hz} = 1330 \cdot \sqrt{\frac{1.661}{50}} = 265.62 \, \text{A} \]
For 50Hz:
\[ I_{50Hz} = 1456.7 \cdot \sqrt{\frac{1.994}{50}} = 265.57 \, \text{A} \]
This shows the derivation of the inrush calculations from IEEEC37.99.