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	Subject - Available Fault Current YukonRay Can anyone tell me the available fault current "let-through fault" of the secondary side of this transformer? Here are some givens: Delta/Wye 480:208/120 volts KVA = 150 Operating @ 100% capacity Z = 3% Have fun frenchelectrician well let me try this but i have a listing here but it dont show the Z rating on my chart so i use the worst case sisuation for 208 volts secondary at 150 kva transformer the fault curret will run at 35,000 amps that at the transformer source. but if fault is located away from transformer the fault curret do drop some. but i try to figure out the Z formila but i dont have it with me . merci, marc YukonRay Hint: Using the Infinite Buss Method or "Quick Method" All the information needed is stated in the post. Remember, "as impeadance goes down - fault current goes up" KSsparky Ray: You might find the calculation you are looking for at www.circuitprotection.ca/arcflash.html Ryan_J I'm coming up with 6,000. Scratch that! 13,900. KSsparky I agree with Ryan. My calculation shows 13,879 Amps of bolted fault current. If you know the type of short circuit protective device, breaker, current limiting fuses, etc., you can determine the incident energy in cal/cm squared and thus the proper level of PPE. Ryan_J KS: I think using the infinite buss primary method will give you an inacuratley high amount of fault current. That is fine for a design...certainly better than the contrary. When it comes to PPE selection, however, the true short circuit available should be used because you may get a lower level of PPE than would be required. KSsparky Ryan: How does one determine the true short circuit current available ? I really don't want to overkill our PPE levels. That can get expensive. Thanks for your help. Ryan_J Please don't take my opinion as gospel, but I beleive you must get the real values from the Power Company. They don't typically gic=ve you these values, but if you call them and tell them that it is for PPE selection, I think it would be irresponsible of them to not tell you. Again, short circuit calc's ARE NOT my strong point, so take this with a grain of salt. :) YukonRay Gentlemen, Thanks for support and participation. You have the correct answer. At least for our purposes. Available fault-current on the load side of a transformer (any transformer) with an impeadance of Z is equal to: I/%Z In the example: 150KVA/(208*1.732)= 416A 416A/.03Z = 13,866A If no other factors were taken into consideration you would have to use equipment that was rated at 22,000 AIC. With the demand for more energy efficiency we will see transformers with less impedance. This is just a practice to make sure that we install the proper switches, switchgear, and other devices. Thanks Ryan_J KS: I think you may have misunderstood my 1st post. After re-reading it, I probably should have worded it different. Using the true fault current will result in a higher requirement of PPE, not a lower one. Using the Z of the transformer may make your PPE inadaquete. KSsparky Ryan: After doing some further investigation, it appears the only way to get a truly accurate cal/cm2 value is to have an engineered point to point study done. Talking the powers that be out of the funding for that should be fun. Thanks again for your help. wareagle YukconRay The %impedance has little to do with the efficiency of the transformer. The customer will order the transformer with a specific impedance to reduce the fault current. Utilities get the larger transformers with 5% impedance. Using the infinite buss method of caculating the fault current will give you the max fault but when adding the utility system impedance, the fault may be cut in half. This is a problem when trying to caculate the arc fault information. KSsparky you must do a point to point to get the correct info for the cal/cm2. It will change at every point on the circuit. YukonRay Wareagle, I can't vouche for your statement one way or the other. I have just had two different EE's tell me that the newer more efficeint transformers have a lower impedance. The lower the internal resitance the greater the efficiency. It makes sense to me - but I don't know for sure. Although, several different code experts on the 2005 NEC have been relating that fault currents are becoming a real issue because of energy demands and lower Z factors. We all know that utility's initial available fault current is used to establish base line fault current calcs. The question was, what is the maximum available fault current from the secondary side of the stated transformer. A simple problem in electric theory. It is also a question on the Minnesota Masters Exam. It was just for fun. But what I have found out is that we have alot of experts in out mist. It will be fun to learn more from you guys. wareagle YukonRay You are correct when you say that the new transformers have a higher efficiency. That is due to a lower resistance of the coil. Less watts lost. The impedance is made up of inductance and resistance. The major component is the inductance which does not contribute to the watts lost. As I said many times the utility will specify an impedance of 5% or something in that range for the larger transformers to reduce the fault current. If you use the infinite buss method(no utility system impedance) then the fault current caculated is the theortical maximum. In reality the utility system is part of the faulted circuit and that impedance will result in a lower actual fault. When sizing equipment for fault current most designers use the infinite buss method because it provides a higher fault current. When cacuclating information for arc flash this can be a problem because the actual fault will be less. Safety with regards to arc flash depends on the arc being eliminated in a few cycles. If the fault current is less the arc will last longer becasue the OD device will take longer to take it out. LIVEWIRE This topic is getting to be fun. I don't know what "watts lost" is. There is such a thing as increased VA required to supply a given Wattage load. I was taught that Impedance is made up of Resistance and "Inductive or Capacitive" reactance. And that reactance is measured in ohms. Therefore, inductance is a/the contributor to increased VA since without Inductive or capacitive reactance, Watts=VA ,i.e. unity(1.0 PF). Remember those vector diagrams? wareagle Livewire Posted - 05/25/2004 : 10:38:34 Watts Loss is caculated by the formula I^2 x Resistance. The loss in the transformer due to the resistance of the windings. You are correct in that the impedance is made up of the reactance and the resistance. However there is no power loss due to reactance only resistance.