Avaliable fault current higher for lower voltages
- Thread starter philly
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Electric-Light
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It all depends on system impedance.
O.K.
Lets say that the utility has a given voltage and an avaliable fault current. If we then connect this through two identical size and impedance transformers but have two different voltage levels on the secondary of the transformers, I know that the smaller voltage on the secondary will have a higher fault current value.
I also know that the fault current downstream of a transformer can be higher than the fault current at the higher voltage on the primary of the transformer depending on the impedance ration between the transormer and the utility source impedance.
My question is for a given utility voltage and avaliable fault current will this current be higher for higher voltage or be lower? If you said that this depends on the source impedance then lets assume that the source impedance is 1 pu. So therefore for any given voltage using 1pu our fault current will be 1pu. Therefore in this case the fault current will be higher for lower voltages. Is this correct?
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- Lockport, IL
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- Retired Electrical Engineer
You are making the problem too hard. This is a simple Ohm's Law application. The current at any location will be the voltage at the source divided by the total impedance along the path from the source, to the fault point, and back to the source. So if you start with two identical utility connections, and go through two separate transformers that have identical impedances by different turns rations, and from there you have identical secondary feeder sizes and lengths and identical panels and identical everything else, so that the impedance seen from the point of view of the source looking out at the system, then the fault current will be higher on the system that starts with the higher voltage. 480 divided by a value of impedance will be higher than 240 divided by that same impedance. Ohm's Law shall prevail.
But in a practical sense, if you have different secondary voltages on two transformers, then the stuff downstream of the two will not be identical. The higher voltage transformer will have a lower value of 100% rated current, so its secondary conductors will be smaller. That means they will have a higher impedance for the same length. Therefore, if you postulate a fault at the same distance from two transformers, one at 480 volts and the other at 240 volts, you will get one fault current of 480 divided by an impedance value, and another fault current of 240 divided by a lower impedance value. Which one has the larger result? Without knowing the exact values of secondary conductor sizes, the answer cannot be guessed.
But in a practical sense, if you have different secondary voltages on two transformers, then the stuff downstream of the two will not be identical. The higher voltage transformer will have a lower value of 100% rated current, so its secondary conductors will be smaller. That means they will have a higher impedance for the same length. Therefore, if you postulate a fault at the same distance from two transformers, one at 480 volts and the other at 240 volts, you will get one fault current of 480 divided by an impedance value, and another fault current of 240 divided by a lower impedance value. Which one has the larger result? Without knowing the exact values of secondary conductor sizes, the answer cannot be guessed.
I'm not sure I agree with you here in regards to the higer voltage resulting in a higer fault current for the same impedance.
Lets say that in our example the utility source is 4160V and connected to our two identical size, and impedance transformers with different turns ratios to provide 480V and 240V at the secondaries of the two transformers. With this, both of the P.U. impedances would be the same between the utility and the secondary of each transformer. With the same P.U. impedance's we will have equal P.U. currents for a fault on the secondary of each transformer. However since the 240V secondary has a higher base current (2x) that of the 480V base current the same P.U. current would result in a higher fault current value at the secondary 240V terminals. Do you agree with this?
For this exact reason I have seen where fault currents can be higher downstream at a lower voltage level than the value at the upstream higher voltage level.
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- Lockport, IL
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- Retired Electrical Engineer
The PU system sometimes complicates, rather than simplifies, the situation. Recall that the base voltages also change from one side of the transformer to the other. So in your example there are three base voltages at play (4160V primary, 480V secondary, and 240V secondary), as well as three base currents and three base impedances. Thus, if you say that you have the same PU impedance, then you have different actual impedances, and you the transformers are not identical in all other respects than the turns ratio.
You can't get around the basic principle that a higher voltage will push more current through the same impedance. So I conclude that if you have a practical example in which a 240V secondary has a higher fault current than a 480V secondary, then you are not comparing "otherwise identical" installations.
You can't get around the basic principle that a higher voltage will push more current through the same impedance. So I conclude that if you have a practical example in which a 240V secondary has a higher fault current than a 480V secondary, then you are not comparing "otherwise identical" installations.
That is true for the following example. If you look at the fault in kva it is simpler to see. Take 2 1000 kva transformers with 5% impedance. The FLa of the 208 volt transformer is 2780 amps. The FLa amps for the 480 volt transformer is 1204 amps. Calculating the approximate Fault 208 volt = 2780/0.05 = 55600 480 volt = 1204/0.05 = 24084
The kva fault is about the same.
gar
Senior Member
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- EE
110422-2039 EDT
philly:
Consider two transformers that are identical and connected to the same primary supply, except for the secondary voltage. This means the secondary coil occupies the same physical space in both transformers.
There are three different impedances to consider.
The input impedance of the transformer when the secondary is shorted. For both transformers this will be the same. Thus, primary fault current for each of these transformers with the secondary shorted is the same.
The output impedance of each transformer is different. If the secondary voltages are in a 2 to 1 ratio, then the higher voltage will have an output impedance 4 times that of the lower. Thus, the fault current on the shorted secondary of the lower voltage output will be 2 times that of the higher voltage secondary.
So you need to determine whether it is the primary or secondary fault current that is of interest.
.
philly:
Consider two transformers that are identical and connected to the same primary supply, except for the secondary voltage. This means the secondary coil occupies the same physical space in both transformers.
There are three different impedances to consider.
The input impedance of the transformer when the secondary is shorted. For both transformers this will be the same. Thus, primary fault current for each of these transformers with the secondary shorted is the same.
The output impedance of each transformer is different. If the secondary voltages are in a 2 to 1 ratio, then the higher voltage will have an output impedance 4 times that of the lower. Thus, the fault current on the shorted secondary of the lower voltage output will be 2 times that of the higher voltage secondary.
So you need to determine whether it is the primary or secondary fault current that is of interest.
.
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