GPR rise in star grounded generator 1

[Kani77]

Hi,

I am designing a ground grid for 1000kW, 3Phase+Neutral 277/480VAc 50Hz Generator. The Gen Neutral is star solidly grounded. What happen If the L-G fault at the generator. Do I need to calculate GPR rise as per IEEE 80 or only the touch and step voltage. The station is 400m x 400m. and the all loads are with in the area.

Please share your knowledge with me.

Thanks,

Kani

 

[7anoter4]

In my opinion, if all the supplied “premises”-installation, facilities etc.- are entirely above a grounding grid well connected the fault current will flow through the grid conductors-or grounded conductors- and a very small part through earth. Then no dangerous GPR –touch or step potential –will involved and following NEC art.250 will be enough.

 

[Kani77]

Thanks!

Is there any way to find the earth current %. I followed the IEEE 80-2000, but there was no formulas when the fault at the station.

The soil is 2800 Ohm.m and they need 1 Ohm. It is impossible to achieve this with a grid unless they spend $$$ on coppers. Any experience with encased electrodes.

 

[7anoter4]

If the generator is not connected to the grid usually you have only cables [feeders] and the impedance of the neutral or grounding wire .Let’s take a very simple example and make an approximate calculation.
A cable of 4*240 sqr.mm [Copper] presents 0.08 ohm/km reactance and a grounding wire closed to live conductor will present a similar reactance. The generator x”d=10% [approx.] then the X”d=x”d%*Vrated^2/S
S=1000 kVA [1 MVA] X”d=0.1*0.48^2/1=0.02304 ohm [That is X1~X2 but Xo~0.3*X”d=0.0069 ohm.]
Then if the short-circuit will be at 0.4 km from generator the phase-to-ground [it is not actually “phase-to-ground” but “phase-to-grounded wire” short-circuit] current will be:
3*Io=3*277/[2*(0.02304)+0.00069+3*(0.032)]=5820 A [neglected the cable resistance].
The equivalent impedance of the grounding wire and grounding electrode it is [60 ohm will be a grounding grid of 20*20m resistance-2800 ohm.m resistivity]:
Xocable*Rg/( Xocable*Rg)=0.032*60/(0.032+60)=0.032[approx.]
Then the current flowing through 60 ohm grounding grid will be: 3*Io*Req/Rg=5820*.032/60=3.1 A
GPR=60*3.1=186.2 V and the Mesh voltage 12.75 V [10 m distance between horizontal laid conductors]. [Permissible for 1 sec fault clearance time 810 V].

 

[7anoter4]

Sorry, 2 [minor?]corrections:
1)Touch potential it is not 12.75 V but 12.75% of permissible [actually 97 V].
2)Permissible touch potential of 810 V it is for 70 kg body. For 50 kg will be 623 V and the Emash 17.7%.

 

[Kani77]

Thank you very much!

This is very nicely explain. Just Ohm low and short circuit calculation.

I think I got the detail what I was looking for.

Thanks once again!

 

[Kiribanda]

7anoter4,
The equivalent impedance of the grounding wire and grounding electrode it is [60 ohm will be a grounding grid of 20*20m resistance-2800 ohm.m resistivity]:
Just for curiosity, did you do a calculation using IEEE 80 equations to obtain 6Ohms, or otherwise?

 

[7anoter4]

Yes, I did. Now I see for 20*20 m 10 m distance between horizontal conductors and 8 vertical electrodes of 20 ft. length at the periphery and 2800 ohm.m earth resistivity is about 58.7 ohms, actually. But for this approximate calculation it is not so important. Thank for your remark, any way. It could be my old excel template double crossed me?

 

[karls2]

7anoter4
Can we say in that kanishka68 case : as the Generator and the loads are on the same site and directly wire connected (following the Neutral system choice), the earth is only use for fixing the 0V Reference, all the homopolar fault current will not use earth, so no GPR and so the value of the earth is not really important.
Thanks for your answer