parallel feeders
- Thread starter radelli
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websparky
Senior Member
- Location
- Cleveland, Ohio
Re: parallel feeders
Sounds unusual.
Chances are if you think that you are pretty close in length, you should have no noticable differences in a resistance test. It should be in your favor.
Now if your conduit runs have a lot of concentric bends in the same direction where the combined radius would make a difference....??
Sounds unusual.
Chances are if you think that you are pretty close in length, you should have no noticable differences in a resistance test. It should be in your favor.
Now if your conduit runs have a lot of concentric bends in the same direction where the combined radius would make a difference....??
Re: parallel feeders
thanks websparky, i have 3-4inch rigid conduits coming out of the side of a 1200 amp breaker enclosure and 90ing down and changing to flex into the side of a 500kva x-former, of course each 90 is a little longer from the enclosure, with 600mcm, how would it be possible to make the difference up in such tight quarters? tia
thanks websparky, i have 3-4inch rigid conduits coming out of the side of a 1200 amp breaker enclosure and 90ing down and changing to flex into the side of a 500kva x-former, of course each 90 is a little longer from the enclosure, with 600mcm, how would it be possible to make the difference up in such tight quarters? tia
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: parallel feeders
With all due respect to the Inspectors who participate in this forum, this particular request is patently and laughably absurd, and shows this particular Inspector?s total lack of understanding of basic scientific principles. The DC resistance of 600 MCM copper is 0.0353 ohms per 1000 feet. Even if we use the Effective AC impedance in rigid conduit, the number is still small: 0.047 ohms per 1000 feet. So one inch of this stuff will have an impedance no larger than 0.000004 ohms. There does not exist a measuring device in the toolkit of any electrician or Inspector with that degree of precision. It cannot be done, so therefore it cannot be requested without making the requestor look foolish in the eyes of the entire profession.
With all due respect to the Inspectors who participate in this forum, this particular request is patently and laughably absurd, and shows this particular Inspector?s total lack of understanding of basic scientific principles. The DC resistance of 600 MCM copper is 0.0353 ohms per 1000 feet. Even if we use the Effective AC impedance in rigid conduit, the number is still small: 0.047 ohms per 1000 feet. So one inch of this stuff will have an impedance no larger than 0.000004 ohms. There does not exist a measuring device in the toolkit of any electrician or Inspector with that degree of precision. It cannot be done, so therefore it cannot be requested without making the requestor look foolish in the eyes of the entire profession.
Re: parallel feeders
I did a quick estimate, and came up with a difference 0.02 Amps for a 10' difference in length two feeders. (I ignored inductance and capactitance.) I don't think the inspector realizes that the resistance of the feeders are very small compared to the resistance of the load, and the difference between two feeders is even smaller.
If "Inspector Milliamp" still insists, you might suggest using an ammeter to measure the current in each feeder as proof the load is equally shared. That would also check for bad or loose connections.
To actually measure the resistance, you could disconnect the supply and load, jumper two lines at one end, and measure between two lugs at the other end (you will actually be measuring the resistance of your jumper wire and the test leads of the meter, so make sure you get good connections).
Steve
I did a quick estimate, and came up with a difference 0.02 Amps for a 10' difference in length two feeders. (I ignored inductance and capactitance.) I don't think the inspector realizes that the resistance of the feeders are very small compared to the resistance of the load, and the difference between two feeders is even smaller.
If "Inspector Milliamp" still insists, you might suggest using an ammeter to measure the current in each feeder as proof the load is equally shared. That would also check for bad or loose connections.
To actually measure the resistance, you could disconnect the supply and load, jumper two lines at one end, and measure between two lugs at the other end (you will actually be measuring the resistance of your jumper wire and the test leads of the meter, so make sure you get good connections).
Steve
- Location
- Illinois
- Occupation
- retired electrician
Re: parallel feeders
Steve,
How can you make that calculation without knowing the total length of the run? To make the calculation you need to know the length of both runs, solve for the parallel resistance, then solve for the voltage drop based on the load current and the parallel resistance. Using the voltage drop number and the resistance you then solve for the current in each of the paths based on the resistance of the individual paths. An easier way would be to take the total conductor length of the three parallel runs and divide the individual lengths by the total length to get a percentage. The percentage length of the longest of the three conductors multiplied by the total current will equal the current in the shorter conductor and the percentage length of the shortest conductor multiplied by the total current will be the current in the longer conductor.
Example: total current 1200 amps
run 1 = 19'
run 2 = 20'
run 3 = 21'
total = 60'
run 1 = 31.667% of total length
run 2 = 33.333% of total length
run 3 = 35% of total length
run 1 current = 35%(1200) = 420 amps
run 2 current = 33.333%(1200) = 399.99 amps
run 3 current = 31.667%(1200) = 380.01 amps
600 kcmil has a 75?C ampacity of 420 amps. If there was a little more difference in the length the permitted ampacity of the shortest run would be exceeded.
Don
Steve,
How can you make that calculation without knowing the total length of the run? To make the calculation you need to know the length of both runs, solve for the parallel resistance, then solve for the voltage drop based on the load current and the parallel resistance. Using the voltage drop number and the resistance you then solve for the current in each of the paths based on the resistance of the individual paths. An easier way would be to take the total conductor length of the three parallel runs and divide the individual lengths by the total length to get a percentage. The percentage length of the longest of the three conductors multiplied by the total current will equal the current in the shorter conductor and the percentage length of the shortest conductor multiplied by the total current will be the current in the longer conductor.
Example: total current 1200 amps
run 1 = 19'
run 2 = 20'
run 3 = 21'
total = 60'
run 1 = 31.667% of total length
run 2 = 33.333% of total length
run 3 = 35% of total length
run 1 current = 35%(1200) = 420 amps
run 2 current = 33.333%(1200) = 399.99 amps
run 3 current = 31.667%(1200) = 380.01 amps
600 kcmil has a 75?C ampacity of 420 amps. If there was a little more difference in the length the permitted ampacity of the shortest run would be exceeded.
Don
Re: parallel feeders
Don
I know the figures you gave is based just on the cable length but in a real world situation the load resistance would be much greater than the resistance of the cables thus negating most of the difference in amperage due to the difference in cable length. Not all but most.
Don McLain
Don
I know the figures you gave is based just on the cable length but in a real world situation the load resistance would be much greater than the resistance of the cables thus negating most of the difference in amperage due to the difference in cable length. Not all but most.
Don McLain
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: parallel feeders
As to the example in the original post, let us presume, just for the sake of discussion, that the overall length is 10 feet. The Inspector is looking for a difference of no more than 1 inch. That represents a difference of 0.83%. Viewed from this perspective, I reiterate that the Inspector?s requirement is patently and laughably absurd.
Don: Your example has a difference of 1 foot, in an overall length of 20 feet. That represents a 5% difference in length. I would not argue that 5% is pushing the reasonable limit of ?the same length.? I would assert strongly that a 10% difference in length would be a violation of 310.4.
As to the example in the original post, let us presume, just for the sake of discussion, that the overall length is 10 feet. The Inspector is looking for a difference of no more than 1 inch. That represents a difference of 0.83%. Viewed from this perspective, I reiterate that the Inspector?s requirement is patently and laughably absurd.
- Location
- Illinois
- Occupation
- retired electrician
Re: parallel feeders
Don,
Charlie,
I agree that 1" is absurd and impossible to obtain. The code should set a maximum permitted difference based on a percentage of the total length. Maybe 5% would be a good limit. Many of the types of runs like this are short enough that a 1' difference would overload the cables. In my previous example if the conductor lengths were 9', 10', 11', the short conductor would carry 440 of the 1200 amps. Of course if the 1200 amps is a result of an Article 220 load calculation, it will be very unlikely that the conductors will ever see a 1200 amp load.
Don
Don,
I don't see how the load at the load end of the parallel conductors changes anything except the total amps that the parallel conductors see. The total current will still be split among the parallel conductors based on the resistance of each individual conductor.
Charlie,
I agree that 1" is absurd and impossible to obtain. The code should set a maximum permitted difference based on a percentage of the total length. Maybe 5% would be a good limit. Many of the types of runs like this are short enough that a 1' difference would overload the cables. In my previous example if the conductor lengths were 9', 10', 11', the short conductor would carry 440 of the 1200 amps. Of course if the 1200 amps is a result of an Article 220 load calculation, it will be very unlikely that the conductors will ever see a 1200 amp load.
Don
Re: parallel feeders
Don, you're right. My calculation above is wrong. I was apparently thinking the same thing Don McLain was. Then I drew it out and was supprised to find the load resistance doesn't matter.
I see you have taken the current divider formula and simplified it to use raw lengths instead of resistances.
Steve
Don, you're right. My calculation above is wrong. I was apparently thinking the same thing Don McLain was. Then I drew it out and was supprised to find the load resistance doesn't matter.
I see you have taken the current divider formula and simplified it to use raw lengths instead of resistances.
Steve
ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: parallel feeders
I red-tagged a 400 amp installation about a week ago for this. They had a CT can chase nippled to a service disconnect. They used parrellel service entrance conductors for a length of about six feet. On each phase, one conductor was about 10-12" shorter than the other conductor. I got a call from his boss about it, but I held him to it. He couldn;t get past the fact there was less than a foot of difference. I couldn't get past the fact there was more than a 10% difference.
I red-tagged a 400 amp installation about a week ago for this. They had a CT can chase nippled to a service disconnect. They used parrellel service entrance conductors for a length of about six feet. On each phase, one conductor was about 10-12" shorter than the other conductor. I got a call from his boss about it, but I held him to it. He couldn;t get past the fact there was less than a foot of difference. I couldn't get past the fact there was more than a 10% difference.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: parallel feeders
Just in case anyone has an interest in the exact formulas, I will give them to you. Let?s call the three cable lengths L1, L2 and L3. Lets call the three currents I1, I2, and I3, and call the total current I-tot. In Don?s example, L1 was 19 feet, L2 was 20 feet, and L3 was 21 feet. Don used the following simplification:
</font>
</font>
Actually, he simplified it even further than that. I would call Don?s calculation process a good ?rule of thumb,? but it has its limits. And it is not, strictly speaking, the ?current divider formula.? You can use this simplified process to estimate the currents in three parallel legs, and get an answer within 1% of the answer you would get by using the long-hand formulas for three parallel resistors and the current divider formula. But I did not analyze the rule of thumb, to see how accurate it could be for four or more parallel legs.
Just in case anyone has an interest in the exact formulas, I will give them to you. Let?s call the three cable lengths L1, L2 and L3. Lets call the three currents I1, I2, and I3, and call the total current I-tot. In Don?s example, L1 was 19 feet, L2 was 20 feet, and L3 was 21 feet. Don used the following simplification:
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I1 = I-tot times (L3) / (L1 + L2 + L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I1 = 35.000% of I-tot</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I2 = I-tot times (L2) / (L1 + L2 + L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I1 = 33.333% of I-tot</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I3 = I-tot times (L1) / (L1 + L2 + L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I3 = 31.667% of I-tot</font>
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I1 = (L2 times L3) / (L1 L2 plus L1 L3 plus L2 L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I1 = 35.029% of I-tot</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I2 = (L1 times L3) / (L1 L2 plus L1 L3 plus L2 L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I2 = 33.278% of I-tot</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">I3 = (L1 times L2) / (L1 L2 plus L1 L3 plus L2 L3)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Result: I3 = 31.693% of I-tot</font>
Re: parallel feeders
thanks for all the input, the total lenght of the 12 conductors was 27 feet each before i landed them. if i hadn't trimmed them, this inspector would probally have tagged me for not being in a neat workman like manner, just the kind of guy he is. i've never been asked to perform this test before, i can't believe that this was the intent of the code. b.t.w., the neutral bar is at least 6 inches lower than the top of the breaker in this enclosure, with 24 600mcm in this can and limited space, i'm not sure it is even possible. thanks for all your help.
thanks for all the input, the total lenght of the 12 conductors was 27 feet each before i landed them. if i hadn't trimmed them, this inspector would probally have tagged me for not being in a neat workman like manner, just the kind of guy he is. i've never been asked to perform this test before, i can't believe that this was the intent of the code. b.t.w., the neutral bar is at least 6 inches lower than the top of the breaker in this enclosure, with 24 600mcm in this can and limited space, i'm not sure it is even possible. thanks for all your help.
Re: parallel feeders
Don
You are right. I didn?t read the post correctly. I seen three conductors and my mind jumped to three phase. With three paralleled cables connected to the same load the only variance is the resistance of the cables and it works out exactly as you stated. I will try and put my brain in gear next time before I post.
Don McLain
Don
You are right. I didn?t read the post correctly. I seen three conductors and my mind jumped to three phase. With three paralleled cables connected to the same load the only variance is the resistance of the cables and it works out exactly as you stated. I will try and put my brain in gear next time before I post.
Don McLain
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: parallel feeders
Anyone read this differently?
This is from 310.4:
As I read this, the phase conductors do not have to be the same length as the grounded conductor. In fact, the phase A conductors do not have to be the same length as the phase B conductors. What has to be the same length is all three conductors that you use for phase A. In their turn, the three conductors that you use for phase B have to be the same length as each other, but they can be a different length as those of phase A.
Anyone read this differently?
Re: parallel feeders
i agree with charlie b 100percent on 310.4...the intent of paralleled conductors is to use a number of smaller conductors to replace the use of one larger conductor,,,in doing so, 2 or more conductors in parallel need to be the same lenght to become the one larger conductor.........
i agree with charlie b 100percent on 310.4...the intent of paralleled conductors is to use a number of smaller conductors to replace the use of one larger conductor,,,in doing so, 2 or more conductors in parallel need to be the same lenght to become the one larger conductor.........
ryan_618
Senior Member
- Location
- Salt Lake City, Utah
Re: parallel feeders
Heres a question I was asked the other day.
A person has an existing 4-inch conduit going from the CT to an MDP (service equipment). They need to add another conduit parrallel to obtain 800 amps. Could a person install a 3" conduit along side the 4 inch? Assuming that each raceway will contain one of each phase and the grounded condutor.
What would the effects of reactance be on these conductors?
[ April 10, 2004, 10:20 AM: Message edited by: ryan_618 ]
Heres a question I was asked the other day.
A person has an existing 4-inch conduit going from the CT to an MDP (service equipment). They need to add another conduit parrallel to obtain 800 amps. Could a person install a 3" conduit along side the 4 inch? Assuming that each raceway will contain one of each phase and the grounded condutor.
What would the effects of reactance be on these conductors?
[ April 10, 2004, 10:20 AM: Message edited by: ryan_618 ]
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