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Parellel Feeds #167128
08/06/07 06:26 PM
08/06/07 06:26 PM
L
luckyshadow  Offline OP
Member
Joined: Jan 2005
Posts: 315
Maryland USA
I am looking for the calculation that show what the amp draw will be on Each leg of a parellel feeder. Specifically if 1 leg is shorter then the other leg.
I'm sure someone here has that calculation.

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Re: Parellel Feeds [Re: luckyshadow] #167133
08/06/07 07:41 PM
08/06/07 07:41 PM
renosteinke  Offline
Cat Servant
Member
Joined: Jan 2005
Posts: 5,316
Blue Collar Country
I know there's a reason the NEC requires the feeders to be the same size, length, and type of wire ....

Re: Parellel Feeds [Re: renosteinke] #167145
08/06/07 11:51 PM
08/06/07 11:51 PM
Samurai  Offline
Member
Joined: May 2007
Posts: 46
Fl.
I=E/R (R including the conductor resistance)?

Re: Parellel Feeds [Re: Samurai] #167157
08/07/07 09:52 AM
08/07/07 09:52 AM
R
resqcapt19  Offline
Member
Joined: Nov 2000
Posts: 2,148
IL
In general with everything else being equal the current will divide in inverse proportion to the percentage of length. Add the lengths of both (or all) runs, and divide the length of each run by the total. The most current will be on the shortest run. For example, with runs of 48' and 52', 52% of the total current will be on the 48' run and 48% on the 52' run. If you have 3 sets of 31, 33, and 36', you will have 36% of the current on the 31' run, 33% on the 33' one and 31% on the 36' run. However as reno said, the code requires that the runs of parallel conductors be of equal length.
Don


Don(resqcapt19)
Re: Parellel Feeds [Re: resqcapt19] #167170
08/07/07 02:43 PM
08/07/07 02:43 PM
E
earlydean  Offline
Member
Joined: Dec 2003
Posts: 751
Griswold, CT, USA
Use ohm's law, I = E / R, first determine the resistance using Table 8 (or 9) in Chapter 9 of the NEC. In the longer runs, we determine that the conductors do not need to be exactly the same length, in the short runs you had better be very close.


Earl
Re: Parellel Feeds [Re: earlydean] #167174
08/07/07 03:39 PM
08/07/07 03:39 PM
R
resqcapt19  Offline
Member
Joined: Nov 2000
Posts: 2,148
IL
Earl,
There is no need to use ohms law and look up the resistances...the footage method provides the same answer. Much quicker and easier.
Don


Don(resqcapt19)
Re: Parellel Feeds [Re: resqcapt19] #167176
08/07/07 03:52 PM
08/07/07 03:52 PM
R
RODALCO  Offline
Member
Joined: Dec 2005
Posts: 856
Titirangi, Akld, New Zealand
If the cable has a reasonable length, more than 20 metres, 30 cm is not that much of an issue (1.5% of length).
The shorter cable will take a bit more current, heats up a little more, resistance increases I²R and will equalise currents with the longer cable.

Ideally they should be the same length (as per code).


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Re: Parellel Feeds [Re: RODALCO] #167188
08/07/07 08:15 PM
08/07/07 08:15 PM
L
luckyshadow  Offline OP
Member
Joined: Jan 2005
Posts: 315
Maryland USA
Thanks resqcapt19
Thats the formula I was looking for!
I know ohms law but knew there was another one.

Re: Parellel Feeds [Re: luckyshadow] #167235
08/08/07 12:32 PM
08/08/07 12:32 PM
E
earlydean  Offline
Member
Joined: Dec 2003
Posts: 751
Griswold, CT, USA
Don,

Your formula is easy only if (like in your examples) the total length of the parallel paths add to 100 feet.

The question asked was how do you calculate the current flow, not how is the current proportioned.

But, it turns out your answer was what luckyshadow meant to ask.

You two must be on the same wavelength.


Earl
Re: Parellel Feeds [Re: earlydean] #167245
08/08/07 03:03 PM
08/08/07 03:03 PM
R
resqcapt19  Offline
Member
Joined: Nov 2000
Posts: 2,148
IL
Earl,
It works with any combination of lengths, but yes I chose easy numbers so I wouldn't need a calculator. It is always easier than ohms law because you never need to look up the resistance of the wire. Even to use ohms law, you need to know the total current to come up with the current on each of the parallel runs.
Don


Don(resqcapt19)

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