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Gardner
How does earth fault loop impedance and disconnect time requirements change for 3phase 3 wire 230 volt systems where the phase to earth voltage is 133 volts instead of 230 volts?
Discuss Earth fault loop impedance in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net
R= V/5In for a B type MCB. so you'd use 133V for the value of V.
Disconnection times only double to 0.8 s once the Uo decreases below 120V rms, so at 133 V rms the disconnection times will remain the same as for a Uo of 230 V.
However using Tel's formula above, decreasing Uo from 230 V to 133 V will reduce the impedance ( by a factor of 1.73).
So say for a 32 A type B BSEN60898 with a Zs at Uo of 230 V = 230 / 160 = 1.44 Ω
but for a 32 A type B BSEN60898 with a Uo of 133 V Zs = 133 / 160 = 0.83 Ω.
But only the Ze is known, the circuit isn't installed yet.
if you know the type and size of cable, work out R1+R2 from the milliohms/m x length.But only the Ze is known, the circuit isn't installed yet.
if you know the type and size of cable, work out R1+R2 from the milliohms/m x length.
You have a Ze value (lets say this is 0.3Ω). You are running a 30m long circuit from a 20A MCB type B to BSEN60898 at origin.
Your design voltage is Uo of 133 V.
PEFC(max) = Uo/Ze = 133 / 0.3 = 443.3 A
Adiabatic equation
View attachment 29333
From the graphs in appendix 3, 443 A is off the graph so take the instantaneous trip value of 100A and a time of 0.1s, k = 115
minimum csa for the protective conductor is 0.27 mm².
However the minimum size csa for a conductor in a power circuit from table 52.3 is 1.5mm² therefore 1.5mm² is fine.
Then you need to calculate volt drop on 20A design current (max) on 30m of 2.5mm² line conductor and neutral conductor = 18mV/A/m *20A*30m = 10.8 V this is less than 11.5V max for power so you are OK.
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