Wire Sizing

[George Stolz]

From a PM:

I Asked this question previously but i did not understand the answer because i might have asked it wrong so i am trying the question again I have a chiller with a 700amp rating on the name plate with 480volts and the MDP 230' away please tell me the formula for getting the correct size feeders so far i used 2 x 230 x 12.9 x 700 x .866 = 3597190.8/14.4 = 249804.91 which is 250mcm feeders 310.16 says 250mcm is only good for 255 amps, please tell me what i did wrong in my calculations.

I have no idea what this formula is. I would simply use the Table and be done with it.

 

[kwired]

From a PM:



I have no idea what this formula is. I would simply use the Table and be done with it.

Looks like an attempt to determine ampacity based on voltage drop.

He needs to start out with 700 amp conductors if that is minimum size needed and then figure out if the conductor selected can provide desired level of voltage drop.

 

[Dennis Alwon]

He sent me the same pm. He did a VD formula incorrectly -- I think I set him straight. I also asked him to post on the forum instead of PM'ing members. I think the calculation doesn't tell you you need to parallel and when he did the calc it came out lower than the amp of the chiller.

 

[kwired]

2 x 230 x 12.9 x 700 x .866 = 3597190.8/14.4 = 249804.91 which is 250mcm feeders

In his formula the 2 at the beginning should have probably been 1.73 since it is a three phase load, and I assume the .866 multiplier is maybe the power factor of the load.

He did somewhat successfully come up with an acceptable sized conductor for voltage drop purposes.

He did not come up with minimum ampacity of conductor required.

When he does come up with minimum ampacity required he shouldn't need to worry about excessive voltage drop.

 

[drbond24]

Even though it looks a bit weird, the formula is correct. It is the approach to the problem that caused the PM'er to have trouble. Determine the wire size you need first with the handy-dandy table in the NEC (or in my case an Ugly's book) and then use this formula to check if that wire size gives you an acceptable voltage drop.

What the PM'er did was calculate the miminum circular mils they needed to provide a 3 percent voltage drop on the circuit. This is where the 14.4 came from; that is 3 percent of 480.

0.866 is the square root of 3 divided by 2. This actually cancels the 2 at the beginning of the equation, leaving the square root of 3 which is correct.

 

[Dennis Alwon]

Okay I see now that it is 480V but the equation comes out at 250KCM which clearly is not correct for 700amps. Now if you parallel 3 feeders that will work but the formula does not work here.

255 amps *3= 765 amps but how is someone to know that. Is there another method that would work more accurately.

It looks like I didn't help him.

 

[kwired]

Okay I see now that it is 480V but the equation comes out at 250KCM which clearly is not correct for 700amps. Now if you parallel 3 feeders that will work but the formula does not work here.

255 amps *3= 765 amps but how is someone to know that. Is there another method that would work more accurately.

It looks like I didn't help him.

If three conductors are paralleled each one only carries 1/3 of the load (if they are all equal size, length, etc) You would have to do the calculation with total cross sectional area of all three conductors if they are paralleled.

Bottom line is the guy was attempting to calculate conductor size based on voltage drop and not on 310.15 ampacity values, which must be fulfilled before even thinking about voltage drop.

 

[drbond24]

I just said the formula was correct. The use of the formula in this case was not correct.

Used in this way, all the formula is saying is this: In order to achieve 3 percent voltage drop at this voltage, current, and line length, your wire size must be at least 250 MCM.

The formula doesn't care about NEC tables so it doesn't know that 250 MCM is too small for 700 amps. All it knows is voltage drop.

If there is a formula that will accurately calculate wire size, I don't know what it is. This formula is just used after you've already selected wire size some other way, presumably the NEC tables, to verify that size has acceptable voltage drop for your application.

 

[drbond24]

Bottom line is the guy was attempting to calculate conductor size based on voltage drop and not on 310.15 ampacity values, which must be fulfilled before even thinking about voltage drop.

Exactly.

 

[Dennis Alwon]

Even using the online formulas that calculate the wire size you get the same. There it asks for amperage, length, voltage etc and yet the same 250kcm. I think these formulas are dangerous because they aren't accurate.

Say you have a 20 amp load and the distance is 500 feet and it is 120V. The online calc tells us use a #1. However reverse the distance and load and you still get #1- thus a 500 amp circuit running 20 feet is still saying #1. I see why people get confused.

 

[Dennis Alwon]

Bottom line is the guy was attempting to calculate conductor size based on voltage drop and not on 310.15 ampacity values, which must be fulfilled before even thinking about voltage drop.

Agreed so how does he get the answer he needs for a 700 amp load going 230 feet or 500 feet, etc. Is there no way to do it other than trial and error working with cm and 3%

 

[kwired]

Agreed so how does he get the answer he needs for a 700 amp load going 230 feet or 500 feet, etc. Is there no way to do it other than trial and error working with cm and 3%

He only needs to do one voltage drop calculation - in this case he has found out that minimum of 250 kcmil fulfills his voltage drop needs. He then has to go to art 310 to determine minimum ampacity needed.

Minimum ampacity determined in 310 is not what the conductor material itself can carry it is how much current the conductor can carry without overheating the conductor insulation. The conductor itself can carry somewhat unlimited amount of current but will have a resistance and if current is high enough will reach a point where it melts, burns, or whatever.

 

[Dennis Alwon]

He only needs to do one voltage drop calculation - in this case he has found out that minimum of 250 kcmil fulfills his voltage drop needs. He then has to go to art 310 to determine minimum ampacity needed.

Minimum ampacity determined in 310 is not what the conductor material itself can carry it is how much current the conductor can carry without overheating the conductor insulation. The conductor itself can carry somewhat unlimited amount of current but will have a resistance and if current is high enough will reach a point where it melts, burns, or whatever.

Thank you- I understood that but the results confuse people. Basically if the size wire comes out smaller than the table requires then there is no VD to worry about. I did forget that the formula does not look at insulation. I assumed it did.