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| Subject - Parallel conductors of unequal length | |
| LIVEWIRE |
I thought this would be a fun topic. So here it is: Given: A three phase EMT overhead feeder consists of 3 conduits, each with 3#500KCM. No derating issues apply.Assume the lugs in this installation are rated for 75 deg C. Two of the runs have conductors that are 100' long. The third run is 95' long. During an inspection, the AHJ frowned upon this contractors installation. In your opinion: A. What didn't the AHJ approve of? B. In theory, how does the amperage split between these parallel runs? C. At what "theoretical" maximum level of overcurrent protection could you protect this feeder, based on the NEC? D. What would the amperage level be in each of these three runs if the "load" amperage level equaled your answer to Part C? E. Has the feeder been violated based on your answer to Part C and/or D? F. What is the lowest "load" amperage level needed to be to violate the rating of the conductors? G. Per NEC, at what fuse size could you fuse this feeder? H. Assume you are using a 100% rated adjustable-trip circuit breaker w/ Long Time settings of .5-.9. At what maximum dial setting could youu adjust the circuit breaker? I. Assume you are using a 80% rated adjustable-trip circuit breaker w/ Long Time settings of .5-.9. At what maximum dial setting could youu adjust the circuit breaker? J. I know....but I couldn't resist this one. Sould the AHJ approve the installation if the over-current protection was appropriately selected? Yes or No based on ??? Hope this is a fun topic for everyone. If I have left out essential information which leads to assumptions having to be made, let me know and I will respond ASAP. |
| iwire | I think it is a great subject. The straight answer is it is a violation, and no size overcurrent device can be used to 'make it right'. quote: Not any leeway given. This is of course impossible no matter how hard you try they will be different lengths, even if only by a 1/32 of an inch. So any parallel conductors can be failed. Back to the real world, your example is 5% difference in length, most inspectors I know would be good with that. Change the example to one 15' and one 20' still 5' but now a 25% difference in length, I do not know any inspectors that would accept that. I have seen how to figure out the exact current on each conductor and with just two conductors it is pretty close to go with percentages. 5% difference in length will result approximately 5% difference in current. Make it 3, 4, 5, - 8 parallel conductors and it was way over my head to figure the current on each one. I have done isolated phase installations (each phase in it's own raceway) in order to avoid the problem of different length conductors. Bob |
| LIVEWIRE | Let's assume we have found an AHJ with a "soft heart" who says "Boys, how bout you calculate for me how you plan to size the over-current protection for this feeder and I'll give this installation due consideration, prior to layin the old 310.4(1) on ya." |
| iwire | I think the inspector would be putting his rear on the line. By asking for the calculations he has now made himself a party to the code violation. Much easer for the inspector to just not ask, if the @*%* hits the fan they can say the wires appeared to be of the same length. By the way I will tell you that my parallel conductors are never of the same length and have never been failed. Bob |
| LIVEWIRE | Iwire, You have made it perfectly clear where you stand on Item J. Remember, this AHJ has a "soft heart". Care to go for some of the other questions? |
| Dave Nix | I would have to agree with Bob. The AHJ has the ability to "approve" this installation based on his judgement of code intent and conformity. Got any more exaples, bring 'em on! |
| LIVEWIRE | Dave, The issue is that this AHJ is looking at you and saying "Gee Dave, I'd love to approve this installation. But, oh by the the way, what rating of overcurrent protection were you planning to use"? |
| LIVEWIRE | I didn't want this question to be difficult - only fun. Hopefully, I can get us started by rendering my opinion to Part B. In my opinion, the load amperage splits according to the ("paralleled equivalent length of the conductors"/"length of the conductors being analyzed") x "load any given amperage". Paralleled equivalent length =95'//100'//100'=32.76' Any instructors out there? |
| iwire | Livewire all the electrical forums are slow right now. I imagine many people are enjoying the weekend. I am sure people will get in on this, I am not able to the calculations. I guess that is because I have never had too. |
| LIVEWIRE | If you don't have a calculator which simplifies this procedure, paralleled equivalent length of the 3 runs (95', 100', 100') can be found using the "Product over Sum" method as follows: For the first two runs: (95'x100') divided by (95'+100')= 48.72' Calculating for the third run: (48.72'x100') divided by (48.72'+100')=32.76'parallel equivalent length. If you have 150A load current, then the current in the 95' run =32.76'/95'x150A=51.73A The current in each of the 100' runs is 32.76'/100'x150A=49.14A |
| Dave Nix | Livewire, Got a first name? The math is correct, however not for electrical systems. When figuring the parrallel conductor loads, the total load is divided equally between all three (in a perfect installation). When there is a length difference between the conductors, resistance is also a consideration. The shorter length has less resistance than the two longer ones. If we take the total load to be 1200A and divide it equally, each leg will have a load of 400A. Now if one leg is 5% shorter, what would be the amp draw on the other two? It's not the short leg that has a problem, it's the remaining legs! |
| Ryan_J | I'm going to read the rest of this later and crunch some numbers, but first...why would you run 3 500's in parrallel? 3*380A=1140 amps, which can't be rounded up to 1200. 1140 seems like an odd number for a service. But anyway, I'll read the rest and get back. I just wanted to throw that in
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| Dave Nix | Hi Ryan, I just picked 1200A for ease of the math calcs. I think this is a theory question more than a code question! |
| LIVEWIRE | Ryan, I think you answered Part C correctly. |
| tony thrower | Answers in the order of each question: (a) Violates 310.4 (1) (b) The current will be equal on the two 100 ft. runs and will be 1.0526 times greater on the 95 ft. run. Current takes the path of least resistance. (c) NEC 240.4c and 240.6a: 1000 amperes (d) A. 100 ft. run 327.386 amps B. 100 ft. run 327.386 amps C. 95 ft. run 344.827 amps (e) If the load is non-continuous there is no violation. If the load is continuous the code is violated. (f) Although the 1000 ampere OCPD will not allow these currents the feeder could carry a maximum load of 1101.55 amperes for non-continuous loads: A. 100 ft. run 360.779 amps B. 100 ft. run 360.779 amps C. 95 ft. run 380 amps If this load is continuous the maximum load will be 881.247 amps: A. 100 ft. run 288.623 amps B. 100 ft. run 288.623 amps C. 95 ft. run 304 amps (g) 1000 amperes (h) I don't know. (i) I don't know. (j) No! The NEC sets minimum standards which are considered to be necessary for safety. |
| cs409 | how much of this question is theory and how much of it is related to an actual install? |
| LIVEWIRE | cs409, The installation is real The overcurrent protection questions are theoretical. The "soft-hearted" AHJ is a figment of my imagination. |
| LIVEWIRE | Tony, I like your answers as follows: A. Sounds good to me. It's by the book. B. Sounds good to me. Amperage should split according to equations above. C. Ryan's answer of 1140A is the "theoretical maximum level of overcurrent protection" answer I was looking for. This shows consideration for NEC 240.4(C) and the understanding of how 75 deg C lugs factor into the design. Tony, I agree that a 1000A fuse is the chosen fuse, when looking at NEC 240.6(A). In addition to that fuse size, a 1200A breaker with the Longtime set at .9 would give protection of 1,080A. D. If we use the "theoretical maximum" of 1140A the amperage level would be 393A in the 95' leg and 373.4A in the 100' runs. E. To answer part of this question, the 1140A load places 393A in the 95' leg. This violates the 380A rating of the conductor run. The 1000A fuse puts 345A in the 95' leg and the 1080A circuit breaker setting puts 372A in the 95' leg. We will assume that this load is non-continuous. Tony has added this new element of consideration. Tony, please elaborate on continuous/non-continuous use ratings. F. Your answer sounds good. Would like it even more if the the load amps created 381A amps in the 95' run. G. 1000A fuse sounds good to me. H. I like 1200A breaker with Longtime set at .9 or 1080A I. I think we'll have a separate discussion on 80% vs.100% ratings J. It doesn't sound like we've found a way to violate the 380A rating of the 95' conductor unless you installed an overcurrent device greater than approximately 1100A. If you could get a adjustable trip breaker to dail to .95, you would have a problem. I'm not sure I've used one that does. Should the AHJ approve this install? Why or why not. |
| Dave Nix | quote: How come the run with less resistance has more amp draw?? |
| Dave Nix | quote: How does this line-up with the code below? quote: |
| tony thrower | Dave, If current is given (3) paths to choose to take it will take the path of least resistance; which literally means most of the current will take the path that has the least resistance. The amount of current will be inversely proportional with the resistance.If the paths all have the same resistance the current will divide evenly and will be equal through each path.This is ideal when using parallel conductors.If one conductor has twice the resistance of another conductor that is in parallel with it it will carry half as much current. Using NEC table 8 chapter 9under uncoated copper conductors: 500 kc mil resistance is .0000258 ohms per foot: 100 feet x .0000258 ohms = .00258 ohms 95 feet x .0000258 ohms = .002451 ohms Using the Reciprocal Method for parallel resistors My total conductor resistance is .000845172 ohms. Conductor resistance .000845172 x 1000 amps = .845172414 volts dropped across each conductor. volts/ohms=amps .845172414 volts / .0000258 ohms =327.586 amps per 100 ft. run .845172414 volts / .002451 ohms = 344.827 amps in 95 ft. run |
| Dave Nix | quote: Good method for total resistance. However, we want to know the individual conductor resistance and how the one with less resistance relates to the other two with higher resistance. What I mean is, the voltage drop will be more with a longer conductor because the amp draw is constant. The amp draw does not change in the equation. We need to calculate each one on it's properties and compare them as individual conductors to see the amount of system performance. Right? |
| tony thrower | The uncoated resistance for 500 kc mil is .0258 ohms per 1000 ft. The voltage drop across each conductor is the same because they are in parallel,therefore, the current will be 1.05263 times higher through the 95 ft. run than the 100 ft. runs because the resistance of the 100 ft. runs is 1.05263 times greater than the resistance of the 95 ft. run. I'm sorry I don't know how to draw the circuit using a computer terminal. |
| LIVEWIRE | Each conduit run has a Phase A conductor, Phase B conductor, Phase C conductor, neutral and grounding conductor. For simplification, we will comment only on the Phase A conductors. The two 100' conductors are in parallel with the 95' conductor. They are in separate conduits but they are lugged together as "one" both at the source of power and at the load. This puts the three conductors in parallel. One of the principles of conductors in parallel is that the voltage across all three of them is equal (regardless of length). We know voltage is a "constant" and we know that voltage=amps x resistance. If resistance is lower in the shorter conductor, then the amperage must go up. Try paralleling a 10' #12 AWG with a 5' #12 AWG to the hot side of a 150W porcelin. use 2 equal length #12 for the neutral. With a clamp-on, measure the difference in current between the 10' and 5' conductors. The 5' conductor should be twice the amperage level. Measure the amperage in the two neutrals. They should be equal. Alright... neutrals vs grounded conductors... save the responses.. |
| Dave Nix | OK, for my feeble brain: Say we have two 12AWG conductors connected in paralell. They are powering a 300W incandescent light bulb. One conductor is 300' long and the other is 30' long. The load is a fixed value. It will draw 300W no matter how you slice it. What is the amp reading on each conductor? What is the voltage in each? |
| JimmyDee | quote: Can't do. NEC says only 1/0 and larger. quote: Jim |
| tony thrower | In rural Alabama you might see 2 #12 in parallel. If they are used to connect a 300 watt bulb to a 120 volt power source and one # 12 is 30 ft. long and the other is 300 ft. long the 30' # 12 would carry 2.272727 amps and the 300' # 12 would carry .227273 amps. Gone Fishing, Tony |
| renosteinke | I think that we get a little sidetracked when we focus on length and amp draw. What really matters is the voltage and resistance. Why? because it is the voltage difference that will make the point where the wires connect a load in and of itself. This difference is caused by the differing voltage drop. This "load" can only express itself as heat. In the confines of a connection, it doesn't take a lot of watts to start cooking the insulation. I suppose, if we wanted to make things "perfect," we would have some way of simulating the full load expected, and adjust the length of our wires until all showed the same voltage at the end of the run. In the real world, this isn't really feasable. So, we do the next best thing; we try to eliminate variables (length, size of wires, etc.), make our connections, and see what happens. If there is a major difference, the heating will be obvious even under no-load conditions. More likely, one of the connections will heat up. An IR scanner, or temperature (non-contact) probe is useful for spotting such problems. |
| LIVEWIRE | Reno, Voltage is equal across parallel branches of a circuit. Per your comments, Yes, voltage matters, but it not a variable in the example of the 100' conductors in parallel with the 95' conductor. Yes, resistance matters, but resistance is purchased by the foot in copper wiring. i.e. Resistannce is proportional to length. Therefore, amperage divides proportionally to length as it does to resistance. We have chosen to use length, rather than resistance for this discussion. Not sure about your other comments |
| Ed MacLaren | Using Tony's resistance values for 500 kcmil copper, and assuming two conductors in parallel, each 100 amps of load would divide approximately as shown on the sketch. Ed 
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