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DC short circuit calculation formula for copper busbar
- Thread starter hojat
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How about I=V/R?
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???
Where is the problem? The equation I gave is perfect. What else do you need to know?
Where is the problem? The equation I gave is perfect. What else do you need to know?
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There's noting much to interpret. It is Ohm's law - straightforward.
R = resistance
V = voltage
I = current
Knowing any two of these variables you can calculate the third. To arrive at the short circuit currrent of the busbar yolu need to know its resistance (should be a very, very small Ohm value) and the voltage.
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Sorry, I can't follow you. From Ohm's law you get I=V/R. So where is the issue? This equation is valid regardless of the type of voltage or current, be it DC, AC or mixed.
Your busbar has a finite resistance, although a very low one. Probably a few milliOhms. You'll have to know this resistance to calculate short circuit current from the applied voltage.
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Which transformer? You've been asking about a DC busbar...
For any circuit, be it a transformer, busbar or whatever you get I=V/R. Know the open circuit voltage V and the resistance R )e.g. winding resistance, wire resistance) and calculate I.
Maybe this helps?
For any circuit, be it a transformer, busbar or whatever you get I=V/R. Know the open circuit voltage V and the resistance R )e.g. winding resistance, wire resistance) and calculate I.
Maybe this helps?
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The resistance of a length of copper is calculated from the specific resistance rho and the geometry (length l and cross sectional area A). R = rho*l/A
For copper rho = 1.68×10−8 Ω*m
Therefore R = 1.68×10−8 Ω*m * 10m / (390 mm * 20 mm) = 2.15×10−5 Ω per busbar.
Putting 5 bars in parallel gives R = 1/5*2.15×10−5 Ω = 4.3×10−6 Ω
This is assuming the short circuit happens at the end (10 m) of the busbar. For a short circuit in between, the shorter length (<10 m) has to be taken into account.
For copper rho = 1.68×10−8 Ω*m
Therefore R = 1.68×10−8 Ω*m * 10m / (390 mm * 20 mm) = 2.15×10−5 Ω per busbar.
Putting 5 bars in parallel gives R = 1/5*2.15×10−5 Ω = 4.3×10−6 Ω
This is assuming the short circuit happens at the end (10 m) of the busbar. For a short circuit in between, the shorter length (<10 m) has to be taken into account.
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This is for the busbar only. The cables feeding the busbar have a resistance, too, which needs to be taken into account. You need to look at the whole system, not the busbar as an isolated item.
Not when your AC values are RMS. RMS values are defined such that the AC RMS value is equivalent to the same DC value.
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@dorke : I'm sorry, too, but this amperage is really common in the electrolysis of metals.
@hojat : What irks me is how you come to deal with such massive amounts of energy while at the same time lacking some real fundamental knowledge. I only hope you are under the supervision of someone with expert knowledge. Otherwise whatever you try to accomplish with the information you're getting here can result in a disaster.
@hojat : What irks me is how you come to deal with such massive amounts of energy while at the same time lacking some real fundamental knowledge. I only hope you are under the supervision of someone with expert knowledge. Otherwise whatever you try to accomplish with the information you're getting here can result in a disaster.
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You have been given some good advice and a reference by @Harald Kapp but you have provided no information on why you need to calculate the short-circuit current in the bus bar. You know the copper bus bars are capable of providing a 38000 ampere current to your process. They are certainly capable of more current than that, but how much short-circuit current can your power supply provide? How do you interrupt a fault current that could possibly produce a 100 kA arc?
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