parallel feeder lengths

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hardworkingstiff said:
I would think the amps would divide so that the voltage drop would be equal on the sets of wire, no matter how many sets you have.
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If I have five 1/0 AWG copper conductors for one particular phase;

98', 99', 100', 101' and 102'

and the load is 750 amps how is the current distributed between these 5 conductors.

I have no idea where to start on this problem. :oops: :?
 
1st, remember that I quit college after 1.5 years in 1972. I did OK in math until I got to Calculus, then road block.

My answer would be:
98' = 153.03061 amps
99' = 151.48484 amps
100' = 149.97 amps
101' = 148.48514 amps
102' = 147.02941 amps

The way I did it (don't laugh everyone) was:

Calculate the resistance in each wire.

Then add the resistances together.

Calculate the % of each wire of that total (yes I know that is not the total resistance in the parallel circuit).

Take the reciprocal of the % calculated in the previous step.

Multiply the reciprocal by the total amps.

I believe the answer to be correct, the method used is questionable.
 
Bob,
If I have five 1/0 AWG copper conductors for one particular phase;

98', 99', 100', 101' and 102'

and the load is 750 amps how is the current distributed between these 5 conductors.
Click to expand...
You can do it the hard way by working out the impedance of each of the conductors, solve the total parallel impedance, and use that total impedance to solve for the parallel circuit voltage drop. Now because the conductors are in parallel the voltage drop across each is the same. You take the voltage drop and use it along with the impedance to solve for current in each of the conductors. I don't really want to take the time to do that, but if someone does, they will find that they get almost the same answer that I will get using the percentage of length method.
Total length...500'
98'=19.6% of length=20.4% of current = 153A
99'=19.8% of length=20.2% of current = 151.5A
100'=20% of length=20% of current=150A
101= 20.2% of length=19.8% of current =148.5A
102=20.4% of length = 19.6% of current = 147A

Don
 
Hardworking,
I see you beat me to it, but there is no need to even look at the actual resitance of the conductors...just the percentage of length.
Don
 
hardworkingstiff said:
1st, remember that I quit college after 1.5 years in 1972.
Click to expand...

No problem, I was tossed right out of high school in my sophomore year and have never gone back.

It looks like you nailed it as you closely match Don's numbers.
 
hardworkingstiff said:
Don,

I like your method. What I don't understand is the method of making the transition from the % of the length to the % of the current. Can you help?

Thanks.
Click to expand...

After further review, I found a way (don't know if it was the same as you).
 
hardworking,
I like your method. What I don't understand is the method of making the transition from the % of the length to the % of the current. Can you help?
Click to expand...
The percentage of length is directly related to the resistance of the conductor, while the current is inversely related. That is the longest wire will have the largest percentage of length, the largest resistance and the smallest percentage of current. The shortest will have the smallest percentage of length, the least resistance and the largest percentage of current. It really does the same thing as when you took the reciprocal of the percentage.

Don
 
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