Ground wire in distribution board

[KundaliniZero]

Hi
I have a doubt. I have seen an electrical diagram of a future instalation. I saw that the wire of ground/earth has a seccion that is lower than the phases. The three phases and neutral (95 mm2) are conected to a transformer DY with neutral conected to earth, but the ground wire of distribution board has been conected to the ground bar mounted in the electrical room and has a seccion of 50 mm2.

¿Could i reduce the earth wire in this situation?, for me the answer is negative.

 

[Julius Right]

In BS7671 it is a formula how to calculate the minimum cross section area of a protecting conductor: S=SQRT(I^2*t)/k where I is the short-circuit current and t is the fault clearing time.
The maximum safety duration for 230 V phase-to-ground has to be 0.4 [usually less].
k is a constant which depends on insulation rated temperature , ambient temperature and starting
temperature. For instance if the conductor is a part of PVC insulated cable the starting short-circuit
temperature is 70oC and if the ambient temperature is 30oC then k=115[copper conductor].
The short circuit current of the 400 kVA transformer 4% short-circuit impedance will be 14400 A.
then S=79.4 mm^2 [minimum].
However it is a very approximate calculation.
You may calculate the short-circuit current considering the actual supply system short-circuit and the actual short-circuit impedance of the transformer, and you may consider an other rated insulation [even bare conductor] outside the cable then to reduce the initial temperature to ambient[30-40oC] and then you may reduce the protection fault clearing time to 0.3 sec for instance and then 50 mm^2 could be enough.

 

[topgone]

In BS7671 it is a formula how to calculate the minimum cross section area of a protecting conductor: S=SQRT(I^2*t)/k where I is the short-circuit current and t is the fault clearing time.
The maximum safety duration for 230 V phase-to-ground has to be 0.4 [usually less].
k is a constant which depends on insulation rated temperature , ambient temperature and starting
temperature. For instance if the conductor is a part of PVC insulated cable the starting short-circuit
temperature is 70oC and if the ambient temperature is 30oC then k=115[copper conductor].
The short circuit current of the 400 kVA transformer 4% short-circuit impedance will be 14400 A.
then S=79.4 mm^2 [minimum].
However it is a very approximate calculation.
You may calculate the short-circuit current considering the actual supply system short-circuit and the actual short-circuit impedance of the transformer, and you may consider an other rated insulation [even bare conductor] outside the cable then to reduce the initial temperature to ambient[30-40oC] and then you may reduce the protection fault clearing time to 0.3 sec for instance and then 50 mm^2 could be enough.

You are using the adiabatic method as provided in IEC 60364-5-54 there.
If you use the simplified method, you size according to = phase cond. size /2, for phase conductors greater than 50 mm2. In this case = 95mm2/2 = 47.5 mm2 ~ 50 mm2!

 

[Julius Right]

You are right, it use the adiabatic calculation-entire heat loss for conductor temperature rise.
No heat evacuation to the ambient.
However, up to 5 sec. fault clearing time the difference is negligible.
Calculated according to BS7454 the permissible current for 1 second will be 1.7% more but for 0.4 second will be 1% more, only. So instead of 79.4 will be 78.6 mm^2 required.

 

[KundaliniZero]

I have seen a device that can trip between 0,1 and 1 seg, so i think the answer only depend of shortcircuit current.

 

[Tony S]

Are you working to BS7671 (IEC) or NFPA-NIC?