Voltage Drop vs. Circular Mils Calculation

[luckyboy886]

Hi guys,

I came across the followng problem associated with the voltage drop and the circular mils calculation.

Can anyone give me some help? Thank you.

Question: What is the Circular Mils (CM) rating for the branch-circuit in illustration below?

(picture link: http://www.flickr.co...N06/6720463707/)

I used the following formula to calculate the CM:

CM = 1.732 x K x I x D / VD

(From EC&M Article: http://ecmweb.com/ne...nt_let_voltage/)

In this case, I determined K = 12.9, I = 82.5A, D = 125', VD = 3% x 480V = 14.4V. Then,

CM = 1.732 x 12.9 x 82.5 x 125 / 14.4 = 16,000 circular mils

Per NEC Chapter 9 Table 8, #8 copper should be used.

But the problem is, the allowable ampacity of #8 copper is only 50A per Table 310.16, which is smaller than the continuous load requirement in this example (82.5A).

Something is wrong in my above calculation.

But I couldn't figure out whether I applied the wrong formula or used the wrong parameters.

Can anyone give me some help?

Thank you.

 

[sc57]

Start with load current first. Minimum wire size should be 1.25 X load current, which is 1.25 x 82.5 = 103.16 amps. Which requires # 2 wire as per NEC for THHW, 75 degree centigrade wires.

Then in some cases if run (distance) is very long you have to accommodate for voltage drop.

 

[luckyboy886]

Start with load current first. Minimum wire size should be 1.25 X load current, which is 1.25 x 82.5 = 103.16 amps. Which requires # 2 wire as per NEC for THHW, 75 degree centigrade wires.

Then in some cases if run (distance) is very long you have to accommodate for voltage drop.

SC57,

I agree with you #2 wire should be used based on 125% x load current. I also noticed that the #2 wire will give me less than 3% voltage drop.

However, this is not the purpose of my post.

I couldn't understand why the formula CM = 1.732 x K x I x D / VD doesn't lead me to the right size of wire. Even worse, this formula tells me to use #8 wire, which is less than the 82.5A load requirement.

Am I missing something by applying this formula?

 

[sc57]

Given formula is not appropriate to calculate wire size. It is good for calculating voltage drop.

Look at the last example in the article. Calculate wire size without exceeding voltage drop.

As per the calculation I (amps) comes to 172 amps for #1 wire, but as per NEC table #1 wire is good for 130 amps.

In example we are calculating wire size for given voltage drop instead of voltage drop for minimum wire size needed as per NEC.

 

[luckyboy886]

Given formula is not appropriate to calculate wire size. It is good for calculating voltage drop.

Look at the last example in the article. Calculate wire size without exceeding voltage drop.

As per the calculation I (amps) comes to 172 amps for #1 wire, but as per NEC table #1 wire is good for 130 amps.

In example we are calculating wire size for given voltage drop instead of voltage drop for minimum wire size needed as per NEC.

SC57,

You made a point here.

It seems that after sizing the wire (to satisfy the VD requirement) according to the formula CM = 1.732 x K x I x D / VD, we still need to check the allowable current for the selected wire.

In this example, we could use #8 wire (according to the given formula) without exceeding the 3% voltage drop IF there is no current limitation on the wire. But obviously, it will lead to a violation of the NEC code. (82.5A curent on #8 wire vs. 50A allowable current per NEC code)

A similar example will be the third example in the article (with Figure 3).

The calculation CM is 10,892 circular mils, and therefore 8 AWG wire is chosen.

Then we should verify that the 8 AWG is capable to handle the load (18A), which is obviously true. 18A x 125% = 22.5A, which is less than the 50A allowable current for 8 AWG wire.

Is my statement correct?

 

[sc57]

luckyboy886,

In actual practice, what happens is first you calculate wire size from load current x 1.25. Then check for voltage drop, make sure it is below 3%. You don't use CM. For conduit fill calculation you need CM, but instead you can use NEC tables to find out how many wires (of the same size) can be run into the conduit without exceeding 40% of cross section area. If it is different size you need to add CM's of different wires and do the conduit fill calc. (NEC table 1 chapter 9)

Most of the time you leave some room for expansion. You don't want to get close to 3%. When runs are smaller <200' usually you don't have voltage drop issue. When they get longer >300', you will encounter these problems.

For your exam use NEC tables and don't leave room for spare or expansion, they want you to follow codes.