voltage drop

[calvin1]

I feel stupid asking this question, I have done the math over and over and come up with the same answer, so I am missing something obvious.
here is what I have

574 amp load, 3 phase, 240 volt, distance to subpanel, 350'
copper conductors, thhn, 5% vd. this is not a real situation, I am doing it for practice.

1.732x12.9x574x350 = 4488668.52 / 12 = 374,055.71 cmil so I go to table 8 which gives me 400kcmil
400 is only good for 335 amps.
what am I missing here? thanks for the help calvin

 

[calvin1]

when I use a calcutor it comes up with parallel 350's which makes sense

 

[david luchini]

What you are missing is that voltage drop has nothing to do with conductor ampacity.

Code requires that you have a conductor with sufficient ampacity for the load served. For instance, two sets of 350mcm.

Your voltage drop would then be 1.732*12.9*350*574/700,000=6.4 volt drop => 2.67% drop.

 

[calvin1]

yes but if I use the formua cmil = 1.732 x k x I x D / vd I should get the wire size shouldn't I ?

 

[gar]

120104-0806 EST

Calvin:

You can not just blindly use formulas and expect to get the correct answer for a problem you have. You must understand how the formula applies to your problem. So what is your real question or problem?

Fundamentally your problem for most applications has to be concerned with the ability of the wire to handle the current thru the wire. Therefore, voltage drop is not the primary first concern.

.

 

[calvin1]

My original problem is I have a 574 amp load, 240V, 3 Phase, which is 350' away., and I want to figure out what size wire I need if I have a 5% voltage drop. When I do the cmil calculation it come up with 400 kcmil, If I just go to 310 15-b with out voltage drop, it comes out to 1250 kcmil. Why is the calculation coming out that way? I have used this calculation many times with out any problem, for smaller wires. I keep thinking I am missing a decimal point, or somehow not reading and transfering the cmil to the right table. thanks.

 

[kwired]

My original problem is I have a 574 amp load, 240V, 3 Phase, which is 350' away., and I want to figure out what size wire I need if I have a 5% voltage drop. When I do the cmil calculation it come up with 400 kcmil, If I just go to 310 15-b with out voltage drop, it comes out to 1250 kcmil. Why is the calculation coming out that way? I have used this calculation many times with out any problem, for smaller wires. I keep thinking I am missing a decimal point, or somehow not reading and transfering the cmil to the right table. thanks.

You calculated the voltage drop for that particular size conductor.

You did not calculate the ampacity of the conductor at a specific temperture.

If you operate the load on that size conductor it will work at approximately the voltage drop you calculated - you will likely also find the rise in heat in the conductor is higher than the insulation and/or termination points can handle. Will probably work for quite some time but eventually termination failure will probably happen.

When figuring size of conductor needed you must figure out what is needed for both termination temperature rating as well as conductor insulation rating. Which ever is largest is the minimum size conductor needed.

Then you look at voltage drop. If the amount of drop is acceptable you need to do nothing. If it is not acceptable then you increase conductor size accordingly.

In your case you needed a larger conductor to start with before you even consider voltage drop.

 

[david luchini]

My original problem is I have a 574 amp load, 240V, 3 Phase, which is 350' away., and I want to figure out what size wire I need if I have a 5% voltage drop. When I do the cmil calculation it come up with 400 kcmil, If I just go to 310 15-b with out voltage drop, it comes out to 1250 kcmil. Why is the calculation coming out that way? I have used this calculation many times with out any problem, for smaller wires. I keep thinking I am missing a decimal point, or somehow not reading and transfering the cmil to the right table. thanks.

You have a load of 574A. You need a conductor that will carry 574A. From T310.16, two sets of 350mcm will carry 574A (or one set of 1250mcm.)

Once you have a conductor that will safely carry the load, you can figure the voltage drop for that conductor. It will be 6.4V (2.67%) for two sets of 350mcm, or it will be 3.59V (1.5%) for one set of 1250mcm.

Trying to figure out a required conductor size by voltage drop rather than ampacity for the load is putting the cart before the horse.

 

[charlie b]

. . . if I use the formua cmil = 1.732 x k x I x D / vd I should get the wire size shouldn't I ?

You should get the wire size that will result in the amount of VD you want. But that formula does not give you the wire size that will result in the ampacity that you want.

Let me put it this way: Ampacity (i.e., the maximum allowable current) is based on the wire's cross-sectional area, and is also based on the insulation system?s ability to withstand the heat generated within the wire. Voltage drop is based on the resistance of the wire (which is related to cross-sectional area), the total distance, and the amount of current that is actually flowing (neither of the later two being related to cross-sectional area). So given that ampacity and VD are based on different things, why should you expect the two to be in any way related to each other?

 

[calvin1]

here is what is throwing me off
from mike holts nec exam prep 2008

" a 15 kva, 18 A , 3 phase load rated 460V, is located 100' from the panelboard, what size conductor is required to prevent the voltage drop from exceeding 3%"

k = 12.90 ohms
I = 18A
d = 100'
Evd = 480 x .03 = 14.40 V

Cmil = (1.732 x 12.9 x 18 x 100 / 14.4V

cmil = 2793 cimi 14 AWG

In my first post I think I did the same thing with a different voltage so why am I not coming up with a right answer.

 

[calvin1]

You should get the wire size that will result in the amount of VD you want. But that formula does not give you the wire size that will result in the ampacity that you want.

Let me put it this way: Ampacity (i.e., the maximum allowable current) is based on the wire's cross-sectional area, and is also based on the insulation system?s ability to withstand the heat generated within the wire. Voltage drop is based on the resistance of the wire (which is related to cross-sectional area), the total distance, and the amount of current that is actually flowing (neither of the later two being related to cross-sectional area). So given that ampacity and VD are based on different things, why should you expect the two to be in any way related to each other?

First let me thank all of you for taking the time to explain this to me,

In the formula cmil = 1.732 x k x I x d / vd what is the cmil that I come up with , shouldn't it reflect the amperage that I put into

but in my formula isn't " I " the ampacity that I want to end up with? I guess I am not understanding what the cmil in the formula stands for, I thought it was for the cross sectional area of the ampacity that I put in the formula, as realted to the voltage I wanted to have at that amperage and the end of the line.

sorry for being so thick on this but I would like to understand it.

 

[calvin1]

You calculated the voltage drop for that particular size conductor.

You did not calculate the ampacity of the conductor at a specific temperture.

If you operate the load on that size conductor it will work at approximately the voltage drop you calculated - you will likely also find the rise in heat in the conductor is higher than the insulation and/or termination points can handle. Will probably work for quite some time but eventually termination failure will probably happen.

When figuring size of conductor needed you must figure out what is needed for both termination temperature rating as well as conductor insulation rating. Which ever is largest is the minimum size conductor needed.

Then you look at voltage drop. If the amount of drop is acceptable you need to do nothing. If it is not acceptable then you increase conductor size accordingly.

In your case you needed a larger conductor to start with before you even consider voltage drop.

but when I get my cmil, and use thhn of the same cmil won't that take care of the temperature?

 

[ptonsparky]

No. Kcmil has nothing to do with insulation. It could be bare.

 

[kwired]

but when I get my cmil, and use thhn of the same cmil won't that take care of the temperature?

When you calculate voltage drop you are calculating the resistance of the conductor and how much voltage is going to drop across it.

If you do same calculation but at a different voltage you still come up with same amount of drop. But it will be a different percent of the applied voltage. (notice that applied voltage is not in the calculation)

You may find if you calculate the voltage drop for only a 10 foot long circuit you will come up with an even smaller conductor. But you still need to have a minimum size conductor based on sections in 310.

 

[calvin1]

that does make sense what you are saying, but isn't this formula here the same thing that I am asking, what size conductor is needed for a 3 % voltage drop? I guess this is what is confusing me.

here is what is throwing me off
from mike holts nec exam prep 2008

" a 15 kva, 18 A , 3 phase load rated 460V, is located 100' from the panelboard, what size conductor is required to prevent the voltage drop from exceeding 3%"

k = 12.90 ohms
I = 18A
d = 100'
Evd = 480 x .03 = 14.40 V

Cmil = (1.732 x 12.9 x 18 x 100 / 14.4V

cmil = 2793 cimi 14 AWG

In my first post I think I did the same thing with a different voltage so why am I not coming up with a right answer.

 

[david luchini]

In the formula cmil = 1.732 x k x I x d / vd what is the cmil that I come up with , shouldn't it reflect the amperage that I put into

It does. It is "I".

but in my formula isn't " I " the ampacity that I want to end up with?

No the "I" in the formula has nothing to do with the ampacity that you want to end up with. The "I" is the actual current flowing based on the load.

I guess I am not understanding what the cmil in the formula stands for, I thought it was for the cross sectional area of the ampacity that I put in the formula, as realted to the voltage I wanted to have at that amperage and the end of the line.

cmil is the cross sectional area of the conductor that you use. It is a physical size, and has nothing to do with ampacity. For instance, in your original post, two sets of 350mcm in separate conduits would give you an ampacity of 620, large enough for your 574A load. It would give you a voltage drop of 2.67%.

But if you used two sets of 350mcm in the same conduit (assuming 6 current carrying conductors) then the ampacity would be 496, NOT large enough for your load. The voltage drop would still be 2.67%

sorry for being so thick on this but I would like to understand it.

You first need to size a conductor with an ampacity sufficient for the load. Then you can figure the voltage drop for that conductor, and if desired, increase the conductor size to reduce the voltage drop.

 

[calvin1]

from Mike holt exam prep 2008

?The size of a conductor (actual resistance) affects voltage drop. If we want to decrease the voltage drop of a circuit, we can increase the cross-sectional area of the conductor (reduce its resistance). When sizing conductors to prevent excessive voltage drop, use the following formula:?


3 phase circuit voltage drop
Cmil, = 1.732 x K x I x D ) / e VD

then they give this example

" a 15 kva, 18 A , 3 phase load rated 460V, is located 100' from the panelboard, what size conductor is required to prevent the voltage drop from exceeding 3%"

k = 12.90 ohms
I = 18A
d = 100'
Evd = 480 x .03 = 14.40 V

Cmil = (1.732 x 12.9 x 18 x 100 / 14.4V

cmil = 2793 cimi 14 AWG

In my first post I think I did the same thing with a different voltage so why am I not coming up with a right answer.

My confused understanding of this is : if you want a certain ampacity with a certain voltage drop you can find the Cmil of the conductor with this formula.

 

[calvin1]

It does. It is "I".



No the "I" in the formula has nothing to do with the ampacity that you want to end up with. The "I" is the actual current flowing based on the load.



cmil is the cross sectional area of the conductor that you use. It is a physical size, and has nothing to do with ampacity. For instance, in your original post, two sets of 350mcm in separate conduits would give you an ampacity of 620, large enough for your 574A load. It would give you a voltage drop of 2.67%.

But if you used two sets of 350mcm in the same conduit (assuming 6 current carrying conductors) then the ampacity would be 496, NOT large enough for your load. The voltage drop would still be 2.67%



You first need to size a conductor with an ampacity sufficient for the load. Then you can figure the voltage drop for that conductor, and if desired, increase the conductor size to reduce the voltage drop.

thanks that does make perfect sense, but what about the mike holt formula I quoted above, they did not figure out the ampacity first, they found the cmil, buy using the desired amperage that would be available at the load at a certain Vd, ?

 

[david luchini]

from Mike holt exam prep 2008

?The size of a conductor (actual resistance) affects voltage drop. If we want to decrease the voltage drop of a circuit, we can increase the cross-sectional area of the conductor (reduce its resistance). When sizing conductors to prevent excessive voltage drop, use the following formula:?

I'm not sure why you are not getting this when several of us have explained the same thing in several different ways. So lets try this: Read what the exam prep quotation says in red. The formula is ONLY for sizing conductors to prevent excessive voltage drop. The formula has NOTHING to do with size conductors to have the proper ampacity for the load to be served.

My confused understanding of this is : if you want a certain ampacity with a certain voltage drop you can find the Cmil of the conductor with this formula.

This is what we have been trying to tell you - the formula has nothing to do with a certain ampacity, only with a certain voltage drop.

You can use the formula to find the conductor size to limit the voltage drop with a known load current over a known distance. In your original example, you find that a 400mcm will limit the voltage drop to less than 5% (it will give a 11.22 volt drop, or 4.68%.) The ampacity of 400mcm is only 335 Amps per table 310.16 ---

So if you put your 574 Amp load through 400mcm conductors, you will see an 11.22 Volt drop over 350 feet...and you will overheat the conductors, destroy the conductor insulation and start a fire...

 

[david luchini]

thanks that does make perfect sense, but what about the mike holt formula I quoted above, they did not figure out the ampacity first, they found the cmil, buy using the desired amperage that would be available at the load at a certain Vd, ?

That's because the question in the Mike Holt exam didn't ask you to determine if the ampacity of the conductor was sufficient for the load, they only asked what conductor size would be need to limit the voltage drop...

from Mike holt exam prep 2008

?The size of a conductor (actual resistance) affects voltage drop. If we want to decrease the voltage drop of a circuit, we can increase the cross-sectional area of the conductor (reduce its resistance). When sizing conductors to prevent excessive voltage drop, use the following formula:?

" a 15 kva, 18 A , 3 phase load rated 460V, is located 100' from the panelboard, what size conductor is required to prevent the voltage drop from exceeding 3%"

k = 12.90 ohms
I = 18A
d = 100'
Evd = 480 x .03 = 14.40 V

Cmil = (1.732 x 12.9 x 18 x 100 / 14.4V

cmil = 2793 cimi 14 AWG

But in their example, they give a load current of 18A, and end up with a conductor (#14) that has an ampacity of 20 from T310.16. So their example doesn't have the same problem that yours did. (They don't address the proper OCP for the #14)