Current in Paralleled Conductors

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bphgravity

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In another post, ( Here )two different sized conductors are paralelled and serving a load. I began to perform a calculaton to better describe the result of this installation but got stuck on what I thought was going to be a simple equation. I know its a simple ratio, however I can't seem to figure it out.

Here's the setup. We have a #1 in parallel with a #1/0. For example lets suppose each is exactly 100'. These two conductors are connected to a 200-ampere load at 480 volts. How much current is carried in each conductor?
 
Re: Current in Paralleled Conductors

I come up with about 80 amps on the 1 AWG and 120 amps on the 1/0.

I sure will be interested if I am even close, I just took a stab at it based on Table 8 resistance.
 
Re: Current in Paralleled Conductors

Ignoring that this installation violates 310-4, the current will divide in inverse proportion to the conductor impedance. Let's assume copper conductors, and for the sake of simplicity assume that the ratio of the impedances for the conductors is the same as the ratio of the DC resistances. From Table 8 (using the 1999 NEC) #1 is 0.154 ohms per 1000' and #1/0 is 0.122 ohms per 1000'. The resistance of the parallel set is = (1 / ((1 / 0.0154) + (1 / 0.0122))) = 0.00681 ohms. The system voltage doesn't matter. 200A through 0.00681 ohms = 1.361 volts drop across the conductors. For the #1, 1.361V / 0.0154 ohms = 88.4A. For the #1/0, 1.361V / 0.0122 ohms = 111.6A.

Martin
 
Re: Current in Paralleled Conductors

The currents (I1) and (I2) which pass through the respective resistances (R1) and (R2) are:


I1 = VP / R1 = (IP)(RP) / R1 = (IP)(R2) / (R1 + R2)

I2 = VP / R2 = (IP)(RP) / R2 = (IP)(R1) / (R1 + R2)


I didn't do anything but take this out of the text book. Without checking it's accuracy or quality this should work just fine.

It isn't as simple as a series voltage divider. And being as how I don't do this calculation often I don't have a standard approach that I use for it.

Edit: P = Paralell.

[ May 15, 2005, 08:58 PM: Message edited by: physis ]
 
Re: Current in Paralleled Conductors

bphgravity
Is this one of your trick questions? If you do the math most gave you an good anwser and that's what I came up with but again if you think about it I believe that the 1/0 will carry most of the load due to less resistance and the #1 will carry some but not 80 amps. Couldn't really tell the prefect answer but this is a theory.
Jim
 
Re: Current in Paralleled Conductors

James
Do not doubt it. Look at the resistance.
1/0 = 0.1222 ohms/1000 and #1 = 0.154 ohms/1000.
There is only .031 ohms difference.

.1222/(.1222 + .154)= 0.44 x 200 = 88 amps
.154/(.1222 + 154) = .558 x 200 = 111 amps
Since the current flow is inverse to the resistance, the 111 flows thru the 1/0 and 88 thru the #1.

[ May 15, 2005, 11:50 PM: Message edited by: bob ]
 
Re: Current in Paralleled Conductors

Using the same resistances that hmspe used (because I get the same thing from the 2002 and multiplying by 0.1 for 100').

R1 = 0.0154 Ohms (#1)
R2 = 0.0122 Ohms (1/0)
RP = 0.00681 Ohms (so far so good hmspe)
IP = 200 amps

Using the two third equations I posted:

I1 = (IP)(R2) / (R1 + R2)
I2 = (IP)(R1) / (R1 + R2)

For I1: (#1)

I1 = (200)(0.0122) / ((0.0154) + (0.0122))
88.4 = 2.44/0.0276

For I2: (1/0)

I2 = (200)(0.0154) / ((0.0154) + (0.0122))
111.6 = 3.08/0.0276

Provided I haven't made any mistakes, at 200 amps:

The #1 is carrying 88.4 amps.
The 1/0 is carrying 111.6 amps.
 
Re: Current in Paralleled Conductors

Another way to look at this problem is to use conductance instead of resistance.

G1 = 1/0.122 Ohms = 8.20 Mhos
G2 = 1/0..154 Ohms = 6.49 Mhos

The ratio of the currents is simply:

I1/I2 = 8.20/6.49 = 1.26
I1 = 1.26*I2

Now write and solve two simultaneous equations:

-I1 + 1.26*I2 = 0
I1 + I2 = 200A
----------------------
2.26*I2 = 200A
I2 = 88.5A
I1 = 111.5A

BTW, the same result would be obtained for any length
 
Re: Current in Paralleled Conductors

No tricks here. I was working the problem as a ratio similar to how rattus worked it out to establish a percent that would be carried on each conductor. What I found interesting is that in this particular situation, neither conductor exceeds their 75? ampacities. So, I was trying to figure out why the failure took place. I assumed the figures I was comming up with must have been wrong.
 
Re: Current in Paralleled Conductors

Originally posted by bphgravity: What I found interesting is that in this particular situation, neither conductor exceeds their 75? ampacities. So, I was trying to figure out why the failure took place.
Click to expand...
That should not be a surprise. The combined 75C ampacity of the two conductors (disregarding the related code violation) is 130 + 150, or 280 amps. This is far above the load that you described as 200 amps. If the conductors are rated at 90C, then their combined ability to handle currents will be higher still. So if a failure were to take place, it would not be a failure of the conductors? insulation systems.

In reading the other thread, it appears that ?insulation breakdown? was not a factor in the equipment failure. The failure was a melted and broken lug. I would say that the most likely cause is a loose connection. The lug might not have been adequately crimped onto the conductor, or the screw holding the lug in place might not have been tightened (or might have vibrated loose). It is clear that this installation was not performed by a skilled electrician, so any number of poor installation practices might have been employed.
 
Re: Current in Paralleled Conductors

I don't know why I didn't use this one.

I1 = VP / R1
I2 = VP / R2

R1 = 0.0154 Ohms
R2 = 0.0122 Ohms
RP = 0.00681 Ohms
IP = 200 amps
VP = 1.362 volts

1.362/0.0154=88.4
1.362/.0122=111.6

Way easier.
 
Re: Current in Paralleled Conductors

Even simpler; the current divider is directly proportional to the cross-sectional areas of the conductors.

I1 = 200A x 83690cm / (83690cm + 105600cm) = 88.4A

I2 = 200A x 105600cm / (83690cm + 105600cm) = 111.6A

Of course this assumes everything else (like length) is constant. Also skin effect would affect much larger conductors; but if the relative size of the conductors were fairly close (not more than a size or two different) it would still be a very good approximation.
 
Re: Current in Paralleled Conductors

Bob, did you do that using, basically, only circular mils?

This looks interesting, I'm gonna have to check this out.

Edit: Using K without using K. :D

Did you come up with this or did you find it?

[ May 16, 2005, 02:23 PM: Message edited by: physis ]
 
Re: Current in Paralleled Conductors

Originally posted by physis:
Bob, did you do that using, basically, only circular mils?

...

Did you come up with this or did you find it?
Click to expand...
Yep; to me it was obvious.
 
Re: Current in Paralleled Conductors

Well, I haven't checked it, but I don't doubt it works.

It's a pretty cool approach really.

Except that I can't think of a single good reason to use K.

But still cool.
 
Re: Current in Paralleled Conductors

Originally posted by rbalex: Even simpler; the current divider is directly proportional to the cross-sectional areas of the conductors. . . .
Click to expand...
Originally posted by rbalex: Yep; to me it was obvious.
Click to expand...
It?s obvious to me as well, but it takes a three-step thought process that had not occurred to me before. Step 1 is to recall, as others have already said, that current divides in inverse proportion to resistance. Step 2 is to recognize that the resistance of a wire is inversely proportional to area. Step 3 is to combine those two to arrive at the conclusion, as Bob did first, that current divides in direct proportion to the cross-sectional area.

That?s one Attaboy to rbalex! :D
 
Re: Current in Paralleled Conductors

Sam, what rbalex has done is reduce the formulas for current division,

I1 = G1/(G1 + G2)*Itotal

I2 = G2/(G1 + G2)*Itotal

to their simplest form.

You will note that these formulas take the form of the familiar voltage divider formula and can be applied to any current division problem.

In this case the lengths and resistivities of the two wires are equal and have been divided out leaving only the areas of the wires.

Most likely though, he just looked at the problem and realized that current divides proportionally according to the areas.
 
Re: Current in Paralleled Conductors

Physis
Since Resistance is directly related to CM, you can ratio the CM or the resistance and get the same answer.
 
Re: Current in Paralleled Conductors

Originally posted by rattus:
...
Most likely though, he just looked at the problem and realized that current divides proportionally according to the areas. [/QB]
Click to expand...
Yeah, that's pretty much what I did :D

It is important to specify all other conditions are constant and cross-sectional area is the only variable.

Also, if there were a significant difference in cross-sectional area skin effect would begin to exert a much greater influence; i.e., the ampacity / cm is lower for larger diameter conductors in AC service.

For DC service, theoretically it wouldn't make any difference; conductors would share current proportionally if they were the same length and other factors the same.
 
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