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Volume 3 nr 1 / 2011 - Academia Oamenilor de Stiinta din Romania

Volume 3 nr 1 / 2011 - Academia Oamenilor de Stiinta din Romania

Volume 3 nr 1 / 2011 - Academia Oamenilor de Stiinta din Romania

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140 Alexandru Ionuț Chiuță, Liviu Mihai Sima, Nicoleta Doriana SecăreanuUNUAYAUBYBUCY VY Y Yand if the neutral circuit has Y admittance:UAYAUBYBU VY Y Yand so, there is no neutral displacement, then V = 0.5.3. Capacitive unbalanceAABBCCUTo <strong>de</strong>termine the currents passing through capacities from the ground andreturning through the ground (if the homopolar circuit admittance is infinite), theequivalent circuit can be <strong>de</strong>termined either in the form of a voltage equivalentequal to the neutral displacement voltage in series with the capacitive admittanceseither as an injected current.Equation (5.2) can be written in the form:IN UNC YYA Y B Y C U Y KKYCCKkA,B,CFrom this relationship, the capacitive unbalance current I in the normal regime is<strong>de</strong>termined, calculating U, I. This has the same value like the current <strong>de</strong>terminedby neutral displacement voltage, applied to the admittance equal to the sum of thecapacitive admittances.In a three-phase power line, the capacities between conductors and betweenconductors and earth cause a capacitive consumption.~IRR RCFig. 4. Equivalent circuits for <strong>de</strong>termining the capacitive currents returning through the earth.Copyright © Editura Aca<strong>de</strong>miei <strong>Oamenilor</strong> <strong>de</strong> Știință <strong>din</strong> România, <strong>2011</strong>Watermark Protected

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