GPR calculation for Low Voltage system 2

[kssschsekhar]

Dear Folks,
If the fault current is 50kA (Ph-E) in a low voltage system-400V, the calculated Grid potential rise (GPR)= 50kA * 1 ohm= 50kV....It is against Ohms law and the potential rise should be less than 400V.... I would like to know why system voltage is not considered in calculating GPR.....

 

[cuky2000]

The reason why Ground potential rise (GPR) is independent of the system voltage may be found in how the GPR voltage is generated: GPR is the voltage with respect a reference potential (zero earth voltage) produced by the fault current injected into the earth (I_e) through out the earth impedance (~R_e). GPR=I_e.R<sub>e

Beware that the current injected into the ground is usually a fraction of the fault current (Ie= S.Ipe, S<1: Split factor). Also in LV system the fault current is reduced significantly along the cable because of its high impedance.

For additional details in the subject, search for "Zone of Influence (ZOI)" applicable GPR and transfer potentials.</sub>

 

[7anoter4]

I agree with Cuki2000 the GPR does not depends on voltage. But the short-circuit current does.
Let's say the Ztrf=0.00462 ohm per phase impedance of a 2000 kVA 5.8% uk% transformer
In a 3 phase short-circuit [metallic] I3k=0.4/sqrt(3)/0.00462=50 kA. But in the of one phase to ground case the short-circuit current will be:
Io=sqrt(3)*kVrated/sqrt((X1+X2+Xo)^2+(3*Rsc+Rearth)^2)
X1=X2=Xtrf Xo=0.95*Xtrf for DY connection transformer Rsc=Xtrf/10. Xtrf=Ztrf/sqrt(1+1/100)=0.004617 ohm
Rtrf=0.0004617 ohm.
Then Io=sqrt(3)*.4/sqrt((2.95*0.004617)^2+(3*0.0004617+1)^2)=0.692 kA
and GPR=0.6924*1=0.6924 kV.
But GPR is only a theoretical data as it is the difference between the grid potential and the infinite.
What does matter is Emash and this could be less.[see IEEE-80].

 

[kssschsekhar]

Fine... Let us assume transformer (Dyn11) is located remotely and neutral is solid grounded....If there is a ground fault at another remote location where load is connected, obviously GPR should not be more than Ph-E voltage for load side ground fault, where earth is referring to neutral point connected to ground...

 

[7anoter4]

First of all I have to apologize in my above calculation is a mistake: the Rearth has to be 3 time =3 ohm
So the Io=231A and GPR=231 V [for the record!]
If Rearth=0.33 ohm [it could be!] then the above calculation is correct.
Second, I cannot catch the point in your last post. Could be, please, more specific?

 

[pwrtran]

kssschsekhar

In your example, you cannot have 50kA and 1 ohm in the same circuit!
Who drives the fault current? The 400V source! What is your 1 ohm consist of? Total of Z1, Z2 and Z0? or ground resistance? If it is the sum of total sequential components then your L-G fault @ 400V system is only 692.8A. If it is the grounding resistance, that is nothing to do with GPR as the fault current will return from the neutral. If it is ground fault resistance then you will multiply it by 3 then add to the sum of Z1, Z2, and Z0, the fault current will even less than 692.8A

 

[cuky2000]

Hi all,

Bellow are a few comments associated with this post:

7anoter4 :

But GPR is only a theoretical data as it is the difference between the grid potential and the infinite.

The GPR is not a theoretical data. The isue about infinite refer to a location far from the fault to avoid voltage rise.

kssschsekhar:

For remote transformer with neutral solidly grounded..GPR should not be more than Ph-E voltage for load side ground fault, where earth is referring to neutral point connected to ground...

The GPR could be any value since depend of the earth fault current and the earth/grid resistance. [small]Some code such as the CA Code limit the GPR up to 3000 V. IEEE 80 do not limit directly the GPR but some how the grid resistance is limited by many utilities below 5 for distribution applications and less than 1 Ohm for larger substations. [/small].

pwrtran:

If it is ground fault resistance then you will multiply it by 3 then add to the sum of Z1, Z2, and Z0, the fault current will even less than 692.8A

Because of the asymmetry of the fault current path (neutral, ground, earth, etc) the symmetrical component method is not recommended for grounding analysis.

 

[Marmite]

Cuky, The GPR caused by a fault on the LV side can't be more than 230V. Go back to basics and think it through. Ohm's law. It can be any value upto 230V.

"The GPR could be any value since depend of the earth fault current and the earth/grid resistance." would be correct if you had said any value upto 230V.

If you have a 230V source and an earth resistance of 1 Ohm the maximum fault current is going to be 230A, like it or not.
You can't do it the other way around and say my calculated fault current is 1000A and the earth resistance is 1 Ohm therefore my GPR is 1000V.

Regards
Marmite

 

[7anoter4]

Thank you,Marmite.You are right of course!

 

[pwrtran]

cuky2000

I think you messed up time domain with phasor & network domain. Symmetrical sequential component method is the proper way to do all types of fault analysis in power system.

The "symmetrical" and the "asymmetrical" you are talking about is in the time domain. You are referring symmetrical fault current and the asymmetrical fault current I believe. The difference between them is the dc component (all together 1.6 times of the symmetrical current as per IEEE) that will normally decay after several cycles. The calculation involves L/R network time constant in the exponential expression.

On the other hand, in the phasor domain, any fault can be split by the sequential components. In a L-G fault, the positive, negative and zero sequence impedance plus the positive sequence source compose the network. The L-G fault current is 3 times of zero sequence current.

 

[pwrtran]

As I said earlier, you can NOT have 400V, 50kA and 1ohm in the same circuit!

I assume the OP is talking about a 3-phase 4-wire system, otherwise 50kA Ph-E fault will NOT hold stand and we are all off topic.

So, question for Marmite:

If you have a 230V source and an earth resistance of 1 Ohm the maximum fault current is going to be 230A, like it or not...

I don't think your statement is true. What do you mean by "earth resistance of 1 ohm"? Is it grounding resistance? If yes then it is nothing to do with the GPR because the fault current will not flow through it and go back to the remote source. If it is the fault resistance, then you have to multiply by 3 and add to your (2*Z1+Z0). Therefore, you need to ask what is the positive and zero sequence impedance of the 400V system.

Even if someone saying the 1 ohm is all summed of total impedance, then your calculation is still not correct. 230A you got is just the zero sequence current, it is only 1/3 of the L-G fault current!

Regards

 

[cuky2000]

I did not realize that my statements will create this level of controversial. We could take this synergy in benefit of all in this forum revisiting the concept associated with fault current return path to the source and the impact on the GPR.

This also may be a good opportunity to explore practical engineering methods including symmetrical component, direct phase analysis, etc. Here is some reference to start (see the enclosed file below).

Marmita & Pwrtran, do you would like taking the lead to show us why the GPR should not be greater than the source voltage?
[small]SUGGESTIONS: Consider in your analysis simplify scenarios such as: 1) grounded both ends via metallic return conductor. 2) System grounded at one end only with ground fault predicated at the receiving end. 4) Practical case with hybrid grounding return path (current split between earth and neutral/ground conductor).[/small]

 

[7anoter4]

Hi pwrtran
You are right to ask what represent 50 kA and 1 ohm since kssschsekhar in the o.p. did not say.
But from experience only a metallic 3 phases -or one phase to neutral [not to Ground!]- short-circuit at low voltage terminals of a transformer [or close] may produce such short-circuit.
Second 1 ohm it seems to be what IEEE 80 call Rg grounding grid resistance since kssschsekhar in his o.p. said: GPR=Ig*Rg.
See IEEE-80:
"Ground potential rise (GPR): The maximum electrical potential that a substation grounding grid may
attain relative to a distant grounding point assumed to be at the potential of remote earth. This voltage, GPR, is equal to the maximum grid current times the grid resistance".
The confusion was that 50 kA is three phase short-circuit I3k and Ig has to be 3*Io the homopolar symmetric component of a phase to ground fault.
Then Ig [neglecting decrement factor and fault current division factor-if there are other metallic means for returning part of the fault current through as medium voltage cable shield or static wire or railway or other.]
Ig=sqrt(3)*VL-L/sqrt((x1+x2+xo)^2+(r1+r2+ro+3*Rg)^2)
If x1,x2,xo,r1,r2,ro are small and we could neglect these all then Ig[maximum]=sqrt(3)*VL-L/(3*Rg)=VL-0/Rg and GPR[maximum]=VL-0.It is clear if we would take into consideration all was neglected above the GPR could be less than VL-0.

 

[pwrtran]

I did not realize that my statements will create this level of controversial. We could take this synergy in benefit of all in this forum revisiting the concept associated with fault current return path to the source and the impact on the GPR....

cuky2000 - don't you think it is the nature of an electrical engineer, isn't it?

I cannot say taking the lead "to show us why the GPR should not be greater than the source voltage", however, I am very happy to discuss with you.

Please think about these:

1) when a fault occurs, a fault loop circuitry is formed as such it consists of the source, the line impedance, fault resistance and the return path that could combined with earth, neutral and the overhead ground wire.
2) the KVL Law applies, which means within a complete loop, the sum of voltage rise (the source) and the voltage drop on each circuit element is ZERO - yes, that is the key word "zero"
3) let us further assume that there is NO L/C resonance takes place in a L_G fault circuit as the fault impedance is much inductive than capacitive (e.g. some 80 degree lagging as X>>R)
4) how do you come across that at any point within the loop the potential is higher than the source? It only happens by error calculation where you use the current inside the loop to apply to a passive impedance element elsewhere outside the circuit! For example, you have 10kA fault current and you take it for grant that the 10kA is the grid current (be aware of that If does NOT equal Ig), and then multiply it to a Rg that If does NOT flow through! There are some details in IEEE 80

I will be surprised to see if you could give me an example that against the KVL Law!

Regards

 

[jghrist]

This also may be a good opportunity to explore practical engineering methods including symmetrical component, direct phase analysis, etc. Here is some reference to start (see the enclosed file below).

cuky,
Could you please re-attach the linked paper? I get a blank.

 

[cuky2000]

Thanks. I am convinced that GPR can not be larger than the source voltage.