Earth fault loop impedance

[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?

 

[telectrix]

R= V/5In for a B type MCB. so you'd use 133V for the value of V.

 

[Gardner]

R= V/5In for a B type MCB. so you'd use 133V for the value of V.

This would mean a larger PE since the impedance is higher?

I could be wrong, but I read somewhere in the IEC that the disconnect times can be longer for 133 volts.

 

[Richard Burns]

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.

 

[Gardner]

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.

Makes sense. Looks like I will have to increase the PE then. Would a full size 2.5mm2 PE sound right for a 20amp MCB at 30 meters?

 

[telectrix]

no. you work out your actual Zs first. Ze+ ( R1+R2), then calculate the PEFC from I = V/Zs.

from a design viewpoint you could use the max. Zs for the breaker but be prepared for it to be higher if the actual Zs is lower.

 

[Gardner]

But only the Ze is known, the circuit isn't installed yet.

 

[davesparks]

But only the Ze is known, the circuit isn't installed yet.

You have the predicted Zs from the circuit design though presumably?

 

[telectrix]

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.

 

[Gardner]

if you know the type and size of cable, work out R1+R2 from the milliohms/m x length.

I can do that, but no easy equation to help me choose a minimum PE?

 

[Richard Burns]

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

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.

 

[telectrix]

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.

but not in OP's case as his line voltage is 133V so a max. VD of 5% is 6.65V.

 

[Richard Burns]

Oops... think before you post RB!