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03-22-2008, 05:54 PM   #16
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The "bang" produced is, of course, related to the applied voltage but not totally.

Eg 1.
A transformer with a secondary voltage of 120 volts is connected to a circuit, the impedance of which is X ohms. The transformer has the capability of supplying 50 kA (Prospective Fault Current).

Eg 2.
A transformer with a secondary voltage of 240 volts is connected to a circuit, the impedance of which is X ohms (same as Eg 1). The transformer has the capability of supplying 25 kA (Prospective Fault Current).

Which of the above examples will give the bigger "bang" under "bolted" short conditions? Both circuits have the same impedance. Both transformers have the same impedance.

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Last edited by elkangorito; 03-22-2008 at 06:35 PM.

 03-22-2008, 08:15 PM #17 Idiot Emeritus   Join Date: Mar 2008 Location: Fernley, Nevada (near Reno) Posts: 1,849 Rewards Points: 1,492 This is actually a valid question, I think. What you'd have in each scenario is called a 'bolted fault'. This is different from an arcing fault, because here the breaker will have to open the entire short circuit current. In an arcing fault, the current is reduced by the arc itself. All circuit breakers and fuses have an AIC rating, and a voltage rating. It stands for 'Amp Interrupting Current'. This is the absolute maximun current the breaker can safely open at its rated voltage. If this amount is exceeded (either current or voltage), there's no gaurantee that the breaker will not harm the operator. As in blow up! As in high speed pieces becoming embedded in your flesh. A typical residential breaker has an AIC rating of 10,000 amps at 120/240 volts. The first voltage rating is phase to ground, (120) the second is phase to phase (240). This means it can safely open 10,000 amps to either neutral or ground, and a 2 pole unit can open 10,000 amps at 240 across both poles. A typical house panel can really only come up with 1,000 to 6,000 amps, so no problem. A 3,000 amp industrial service might be capable of 80,000 amps, so those breakers have higher ratings. And are a bit more expensive, several thousand \$. Length and size of wire have a HUGE influence on fault current. A bolted fault at the breaker terminals will have way more fault current that a bolted fault 20' away with #12's. Even big wire (service entrance) will cut it down considerably. If you've ever wondered why experienced electricians (especially industrial ones) always stand to the side of a breaker when turning it on....... Rob
03-22-2008, 09:37 PM   #18
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Quote:
 Originally Posted by elkangorito The "bang" produced is, of course, related to the applied voltage but not totally. Eg 1. A transformer with a secondary voltage of 120 volts is connected to a circuit, the impedance of which is X ohms. The transformer has the capability of supplying 50 kA (Prospective Fault Current). Eg 2. A transformer with a secondary voltage of 240 volts is connected to a circuit, the impedance of which is X ohms (same as Eg 1). The transformer has the capability of supplying 25 kA (Prospective Fault Current). Which of the above examples will give the bigger "bang" under "bolted" short conditions? Both circuits have the same impedance. Both transformers have the same impedance.
Well, I'm no design engineer, so I would only be guessing, but my two years as a physics major tell me that the two scenarios are mathematically equivalent, as far as available energy. But, my electrical experience tells me that most likely the 120 V fault, having a larger number of amperes, will operate an overcurrent device faster, and thus limit the total "bang" energy developed.

The 240 V fault will take longer to blow a fuse or circuit breaker, and therefore will reach a greater total peak number of joules released. Am I close or totally off base?

InPhase277

 03-23-2008, 04:18 AM #19 UAW SKILLED TRADES     Join Date: Jan 2007 Location: Kansas Posts: 5,341 Rewards Points: 2,652 Assuming an infinite utility and if your talking center tapped transformers in the USA these would be 120/240 volts... and I am assuming your talking the fault at the secondary terminals of the transformer and not load side of the Overcurrent protective device at some distance. In #1 the transformer is around 160 kva (660 amps fla) to give a short circuit fault current of 50 KVA. L-N at the secondary terminals is approx. 1.5 times 50 KVA or 75 KVA bolted faulted fault current available at the transformer. In #2 your transformer is around 50 KVA (208 amps fla) and 25 KVA fault current available at the transformer secondary L-L. L-N in #2 is 37.5 KVA short circuit current available for bolted fault. If your talking strictly a line to neutral transformer at 120 volts or 240 volts no line to line then you answered your own questions given the examples. #1 is obviously greater than #2. I may not be following what your eluding to but you give the fault currents the fault currents then elude to the transformer size. If we consider 120/240 center tapped transformer (160KVA) size the example in #1 (120 volts) for a fault 25 feet after the ocpd using copper 500 kcmil in the conduit we would have a fault current of 31,225 amps. In example two (240 volts) with a 50KVA transformer with a fault current of 25 KVA we would have 20,247 amps fault current possible at the location of the fault. So I suppose what we are after here is that the lower voltage produces the larger fault current given the same size transformer and impedances. Had you reversed the fault currents in 1 and 2 this same calculation would have resulted in the 240 volt short circuit being 23,476 Amps and the 120 volt short circuit being 22,005 amps. Last edited by Stubbie; 03-23-2008 at 06:24 AM.
 03-23-2008, 10:35 AM #20 Newbie   Join Date: Mar 2008 Posts: 21 Rewards Points: 10 I don't think I really understood the question. If you took a 120V 20 gauge wire that wasn't solid, split the ends and hooked it to the outlet for the stove, it would work. Would probably light up the clock. You may even be able to run one of the small eyes on the stove. Past that it would throw the breaker. You could only draw 10 amps per (240V) wire on the 20 amp circuit. Plus if you didn't have an equal number of wires it could possibly burn one of the wires in half before the threw the breaker. Last edited by wesc; 03-23-2008 at 10:42 AM.
 03-23-2008, 12:40 PM #21 Once fried, twice shy.     Join Date: Sep 2007 Location: Thailand Posts: 251 Rewards Points: 250 InPhase277 is spot on (correct)! Each of the 2 examples have the same amount of available fault current (energy). The method is as follows; Fault levels are normally expressed in MVA (Mega volt Amps). The following equations apply where PFC is the Prospective Fault Current in Amps & E is the supply voltage. Eg 1. VA = PFC x 1.732 x E VA = 50 000 x 1.732 x 120v MVA = 10.392 Eg 2. VA = PFC x 1,732 x E VA = 25 000 x 1.732 x 240v MVA = 10.392 Whilst voltage seems to increase the amount of work done (Joules) by the power of 2 (squared), this does not equate to Fault Current, which causes the bang & destruction. The Work done as a result of a "bolted" short will be difficult to measure because the supply voltage normally falls to a much lower value. In most cases, the supply voltage is known under short circuit conditions with a transformer. Example, a transformer has an impedance of 5%. This means that the Primary voltage can be at 5% of the supply voltage in order to supply a Full Load Current at the shorted Secondary terminals. This can change somewhat under conditions whereby the supply current is unlimited (which doesn't exist). Fault Current generally equates to the square of the current multiplied by time or (I squared t). Fault Current is not as simple as this. X/R relationships are very important & so are the "downstream" contributing loads like motors etc (symmetric vs asymmetric loads). __________________ Switchboard design engineer & Licensed Electrician (Australia).
 03-25-2008, 07:42 AM #22 When is fishing season?     Join Date: Feb 2008 Posts: 613 Rewards Points: 500 Okay, I have a new "what if" question... This is strictly hypothetical. I have not done this, not do I plan on it, nor do I suggest anyone else do it. I just want to know to satisfy my own curiousity. What would happen if you took a straght 240v circuit - just 2 hots - and hooked them to a 120v device, hooking one of the hots to the neutral? This is strictly hypothetical. I have not done this, not do I plan on it, nor do I suggest anyone else do it. I just want to know to satisfy my own curiousity. __________________ I DON'T OWN MY HOUSE...MY HOUSE OWNS ME!
 03-25-2008, 07:59 AM #23 Once fried, twice shy.     Join Date: Sep 2007 Location: Thailand Posts: 251 Rewards Points: 250 Just to clarify, is the neutral that you speak of, the neutral terminal on the load? If it is then what you would be doing is connecting 240 volts across a 120 volt device. In most cases, this would result in the failure of the equipment. __________________ Switchboard design engineer & Licensed Electrician (Australia).
 03-25-2008, 08:24 AM #24 Once fried, twice shy.     Join Date: Sep 2007 Location: Thailand Posts: 251 Rewards Points: 250 Here's another "what if" question. A 2 phase, 3 wire supply (2 hots & a neutral - both phases are 90 electrical degrees apart) has a voltage of 340 volts when measured phase to phase. What would happen if you connected one phase & neutral to a 240 volt load? __________________ Switchboard design engineer & Licensed Electrician (Australia). Last edited by elkangorito; 03-25-2008 at 08:35 AM. Reason: Changed phase displacement from 120 to 90 degrees.
03-25-2008, 08:59 AM   #25
When is fishing season?

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Quote:
 Originally Posted by elkangorito Just to clarify, is the neutral that you speak of, the neutral terminal on the load? If it is then what you would be doing is connecting 240 volts across a 120 volt device. In most cases, this would result in the failure of the equipment.
Yes, I mean to the neutral of the load.
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03-25-2008, 11:15 AM   #26
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Quote:
 Originally Posted by elkangorito Here's another "what if" question. A 2 phase, 3 wire supply (2 hots & a neutral - both phases are 90 electrical degrees apart) has a voltage of 340 volts when measured phase to phase. What would happen if you connected one phase & neutral to a 240 volt load?
Good luck finding a 2 phase supply around here! I think there may still be some small isolated pockets of 2 phase in Pittsburgh or Philidelphia, but major 2 phase distribution in the U.S. went out in the 40s.

But anyhow, I'm gonna say that the since the line voltage in your question is 240 V, that the equipment would function properly, assuming the supply could deliver the necessary current.

InPhase277

03-25-2008, 11:15 AM   #27
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Quote:
 Originally Posted by CowboyAndy Okay, I have a new "what if" question... This is strictly hypothetical. I have not done this, not do I plan on it, nor do I suggest anyone else do it. I just want to know to satisfy my own curiousity. What would happen if you took a straght 240v circuit - just 2 hots - and hooked them to a 120v device, hooking one of the hots to the neutral? This is strictly hypothetical. I have not done this, not do I plan on it, nor do I suggest anyone else do it. I just want to know to satisfy my own curiousity.
Poof!

InPhase277

03-25-2008, 11:43 AM   #28
When is fishing season?

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Quote:
 Originally Posted by InPhase277 Poof! InPhase277
Well, I figured that part. I was hoping for a bit more of a technical explination!
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03-25-2008, 11:50 AM   #29
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Quote:
 Originally Posted by CowboyAndy Well, I figured that part. I was hoping for a bit more of a technical explination!
Your 100 watt light bulb would become a 200 watt bulb. For a little while. Then poof.

03-25-2008, 11:53 AM   #30
When is fishing season?

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Quote:
 Originally Posted by jrclen Your 100 watt light bulb would become a 200 watt bulb. For a little while. Then poof.
So what if you hooked something like a saw to it...

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