CHAPTER 8: APPLICATION OF SETTINGS SETTINGS EXAMPLEC70 CAPACITOR BANK PROTECTION AND CONTROL SYSTEM – INSTRUCTION MANUAL 8-158The maximum rated voltage across the healthy bottom section:Eq. 8-13The voltage distribution factor for the healthy perfectly balanced bank for the 87V-2 and 87V-3 elements:Eq. 8-14The impedance of the string for the failed capacitors elements (for one failed element, n f = 1:Eq. 8-15The voltage at the tap in the section with one failed capacitor element in the string:Eq. 8-16The voltage of the top section with one capacitor element failed:Eq. 8-17The overvoltage at the affected capacitors with one capacitor element failed:Eq. 8-18The differential voltage between two taps with one failed capacitor element in the upper string, assuming K = 1 for 87V-1,in per-unit values is:Eq. 8-19The differential voltage between the bus voltage and the tap with one failed capacitor element in the upper string,assuming K = 1, for 87V-3 in per-unit values is:Eq. 8-20The unbalance current in the window type CT between two strings with one failed capacitor element in the upper string:Eq. 8-21The following table summarizes the calculated values for the number of failed capacitors used to set the protectionelements.V bottomV max Z Cb×Z Ct---------------------- 81515.37 V 4.08 Ω×1894.92Ω------------------------------------------------ 175.37 V= = =K tapV topV bottom----------------- 81340.40 V175.37 V-------------------------- 463.8194= = =Z Csf Z Ce N n× n f–( ) Z Cb+ 45.01Ω 7 6× 1–( ) 4.08 Ω+× 1849.41Ω= = =V tap2V max Z Cb×Z Csf---------------------- 81515.37 V 4.08 Ω×1849.91 Ω------------------------------------------------ 179.64 V= = =V top V max V tap2–=OV V f top( ) N n××V top N n× n f–( )×------------------------------------------ 81335.74 V 7× 6×81340.40 V 7 6× 1–( )×-------------------------------------------------------- 1.024 pu= = =V diff tap( )V tap1 1 V tap2×–VT nom tap( )--------------------------------------- 175.374 V 1 179.64 V×–171.59 V---------------------------------------------------------- 0.0249 pu= = =V diff string( )V f top( ) K tap V tap2×–VT nom bus( )------------------------------------------------ 175.374 V 463.8194 179.64 V×–79674.34 V----------------------------------------------------------------------------- 0.0249 pu= = =I unbalV maxZ Csf CT tap×------------------------ V maxZ Cs CT tap×-----------------------– 81515.37 V1845.33Ω 50×----------------------------------- 81515.37 V1890.34Ω 50×-----------------------------------– 0.0209 pu= = =
