Available L-G fault current higher than 3-phase fault current

[mull982]

I have seen cases where the unbalanced L-G fault current was greater than the 3-phase balanced bolted fault current at a particular location. What are the system paramaters that would cause the L-G fault current to be higher than the 3 phase fault current?

I've been told that this typically happens close to a transformer where the zero sequence impedane is low however as you move further from the transformer the pos and neg impedances become much greater than the zero sequence and therfore reduce the level of the L-G fault current?

Can anyone explain a cause for the L-G fault being higher?

 

[mivey]

I have seen cases where the unbalanced L-G fault current was greater than the 3-phase balanced bolted fault current at a particular location. What are the system paramaters that would cause the L-G fault current to be higher than the 3 phase fault current?

I've been told that this typically happens close to a transformer where the zero sequence impedane is low however as you move further from the transformer the pos and neg impedances become much greater than the zero sequence and therfore reduce the level of the L-G fault current?

Can anyone explain a cause for the L-G fault being higher?

You just did.

 

[G._S._Ohm]

this typically happens close to a transformer where the zero sequence impedane is low however as you move further from the transformer the pos and neg impedances become much greater than the zero sequence and therfore reduce the level of the L-G fault current?

If this is true you should be able to figure out how much impedance gives you a toss-up situation. It doesn't hurt to know these benchmarks ahead of time.

". . .when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely in your thoughts advanced to the state of Science, whatever the matter may be."

 

[mivey]

Here is some actual fault data from one of our 12 kV substations showing the line-ground & 3-phase resistance, reactance, impedance and kilo-amps fault as you get further from the substation transformer:

1) R_LG = 0.15 , X_LG = 0.78 , Z_LG = 0.79 , kA_LG = 9.1 , R_3P = 0.18 , X_3P = 0.93 , Z_3P = 0.94 , kA_3P = 7.6

2) R_LG = 0.19 , X_LG = 0.88 , Z_LG = 0.90 , kA_LG = 8.0 , R_3P = 0.20 , X_3P = 0.99 , Z_3P = 1.01 , kA_3P = 7.1

3) R_LG = 0.24 , X_LG = 0.98 , Z_LG = 1.01 , kA_LG = 7.1 , R_3P = 0.23 , X_3P = 1.05 , Z_3P = 1.08 , kA_3P = 6.7

4) R_LG = 0.28 , X_LG = 1.08 , Z_LG = 1.12 , kA_LG = 6.5 , R_3P = 0.26 , X_3P = 1.12 , Z_3P = 1.15 , kA_3P = 6.3

5) R_LG = 0.32 , X_LG = 1.18 , Z_LG = 1.22 , kA_LG = 5.9 , R_3P = 0.29 , X_3P = 1.18 , Z_3P = 1.21 , kA_3P = 5.9

6) R_LG = 0.36 , X_LG = 1.28 , Z_LG = 1.33 , kA_LG = 5.4 , R_3P = 0.31 , X_3P = 1.24 , Z_3P = 1.28 , kA_3P = 5.6

You can see the 3-phase impedance starts out as the highest but as you move down the line, the line-ground impedance eventually becomes the highest.

 

[mivey]

I've been told that this typically happens close to a transformer where the zero sequence impedane is low however as you move further from the transformer the pos and neg impedances become much greater than the zero sequence and therfore reduce the level of the L-G fault current?

I just noticed you had this backwards. The zero sequence impedance grows higher than the positive sequence as you move away from the transformer.

The fault currents are calculated as:
I_LG = 3V /(Z1 + Z2 + Z0)
I_3PH = V / Z1

You can see when the zero sequence is lower than the positive sequence impedance Z1 (assume the negative sequence impedance Z2 to be about the same as Z1 for simplicity), that I_LG is greater than I_3PH.

As you move down the line, Z0 gets bigger than Z1 and then I_3PH becomes greater than I_LG.

 

[mull982]

I just noticed you had this backwards. The zero sequence impedance grows higher than the positive sequence as you move away from the transformer.

The fault currents are calculated as:
I_LG = 3V /(Z1 + Z2 + Z0)
I_3PH = V / Z1

You can see when the zero sequence is lower than the positive sequence impedance Z1 (assume the negative sequence impedance Z2 to be about the same as Z1 for simplicity), that I_LG is greater than I_3PH.

As you move down the line, Z0 gets bigger than Z1 and then I_3PH becomes greater than I_LG.

Thanks! The equations that you posted makes sense and make it clear why the L-G fault can be higher than 3-phase if zero sequence impedance is lower than the positive and negative sequence impedances.

Do you know what physical aspect of the transformer makes the zero sequence less than the positive or negative sequences?

Do you know what physical nature of the cables make the zero sequence increase at a larger rate than the positive or negative sequences as you move further from the transformer?

 

[mivey]

Thanks! The equations that you posted makes sense and make it clear why the L-G fault can be higher than 3-phase if zero sequence impedance is lower than the positive and negative sequence impedances.

Do you know what physical aspect of the transformer makes the zero sequence less than the positive or negative sequences?

Do you know what physical nature of the cables make the zero sequence increase at a larger rate than the positive or negative sequences as you move further from the transformer?

The zero sequence current do not travel the same path as the positive sequence currents and the impedances are for a complete loop. The impedance of this path might be higher.

For a distribution line, the impedance is not only a function of the line resistance but of the return path as well. Depending on the configuration, this may include the ground wire, earth path, cable shield, etc. These not only include the resistance but the mutual inductance between the wires, shields, and earth path. Here is a good reference source:
http://www.basler.com/downloads/ReviewOfSysGrounding.pdf

The transformer impedance will also depend on the paths in the core for the zero sequence current. For a wye winding that is earthed, an impedance (Zn) is sometimes added to reduce the level of ground fault current (like the grounding resistor used for a generator). For a wye-delta bank, the zero sequence looking into the delta will be infinity. Here is a very good paper on transformer zero sequence impedances:
http://www.basler.com/downloads/ZeroSequenceCircuit.pdf

Here's another good paper:
http://www.basler.com/downloads/ZeroSeqImped.pdf

 

[mull982]

The zero sequence current do not travel the same path as the positive sequence currents and the impedances are for a complete loop. The impedance of this path might be higher.

For a distribution line, the impedance is not only a function of the line resistance but of the return path as well. Depending on the configuration, this may include the ground wire, earth path, cable shield, etc. These not only include the resistance but the mutual inductance between the wires, shields, and earth path. Here is a good reference source:
http://www.basler.com/downloads/ReviewOfSysGrounding.pdf

The transformer impedance will also depend on the paths in the core for the zero sequence current. For a wye winding that is earthed, an impedance (Zn) is sometimes added to reduce the level of ground fault current (like the grounding resistor used for a generator). For a wye-delta bank, the zero sequence looking into the delta will be infinity. Here is a very good paper on transformer zero sequence impedances:
http://www.basler.com/downloads/ZeroSequenceCircuit.pdf

Here's another good paper:
http://www.basler.com/downloads/ZeroSeqImped.pdf

Thanks Mivey! These papers look like the perfect explanations to my question. I'll read them and digest them and let you know my comments.

 

[powerqualityworld]

LG fault values are usually larger than 3P at the secondary bus of a transformer that has an ungrounded primary and solidly grounded secondary.

As a simple illustration:

3P = 1/(Z1)
SLG = 3/(Z1+Z2+Z0)

where Z1=Z2=(Zu+Zt) and
Z0 = Zt (assuming Zt is equal in both positive and zero sequence)

If the transformer has an ungrounded primary and solidly grounded secondary, you will only consider Zt as equal to Z0 in a secondary bus fault. In this case, Z0 < Z1. Consequently, the SLG fault becomes larger than 3P.

 

[mull982]

LG fault values are usually larger than 3P at the secondary bus of a transformer that has an ungrounded primary and solidly grounded secondary.

As a simple illustration:

3P = 1/(Z1)
SLG = 3/(Z1+Z2+Z0)

where Z1=Z2=(Zu+Zt) and
Z0 = Zt (assuming Zt is equal in both positive and zero sequence)

If the transformer has an ungrounded primary and solidly grounded secondary, you will only consider Zt as equal to Z0 in a secondary bus fault. In this case, Z0 < Z1. Consequently, the SLG fault becomes larger than 3P.

What are you calling Zu and Zt in your example?

 

[kwired]

I get a little lost in the discussion here mostly because I don't have to deal with this much.

But I want to question something even more basic. Assuming a center tapped or wye neutral (both being the mid point of the windings making up the system) don't you have half the length of winding conductor to one of the phase conductors than you have from phase to phase which will have less impedance and allow more current to flow?

You will have less voltage but more current, which only makes sense with common basic formulas P=IxE. I understand that during a fault that current and voltage will not be directly proportional in a line to line vs line to neutral but these concepts are a start at understanding why the lower voltage results in higher current.

Correct me if I am totally wrong.

 

[powerqualityworld]

What are you calling Zu and Zt in your example?

Zu = utility positive seq impedance
Zt = transformer %Z

For transformer with ungrounded primary and solidly grounded secondary (i.e.Delta-wye G), Z0 = Zt.

Thanks...