I was posed this question the other day and it stumped me.An existing 2" conduit about 350' long.Customer wants to feed 3 40 HP.motors @ 480 volts.We increasd the wire to #3 for more than 3 current carrying conductors in a single raceway.An apprentice asked me if the increase in wire size would also take care of voltage drop if it was a problem at this length.I think it would and have searched the nec but have not found anything that would apply,but sometimes when you think you are right you won't look very hard to prove yourself wrong.Any input on this would be appriciated.Because as we all know this is a continual learning process.As a wise man once said "show me a man who didn't learn something today,and I'll show you a man who didn't pay attention"

According to my calculations: You have about 1.6% voltage drop. 2xohm x L x I x .866 .000245ohm x 2 x 350 ft. x 52Amps x .866 = 7.7 Volts. Maybe someone else could do check this over.

#16675 - 11/15/0211:24 AMRe: Derating and voltage drop

When the wire size is increased by the derating (more than 3), one is reducing the resistance of the wire, per foot, to reduce the heat per foot, to protect the insulation from being overheated.

The derating is kind of a red herring, that is, difficult not to think about once it comes to mind.

The derating protects the insulation by reducing the resistance per unit length of current carrying conductor.

When the resistance is lower, the conductor voltage drop is lower. Colateral effect. A good one, especially when the run is long. Derating is not concerned with length of run, or the type of load. In your case, the motor start current and heavy load current won't drag the voltage down as much. In this case, the derating of the conductor introduces a little more headroom for the motor's nonlinear use of voltage and current.

Al

Al Hildenbrand

#16676 - 11/15/0212:51 PMRe: Derating and voltage drop

Fedup: There is an NEC requirement for VD; total of 5% (3% feeder; 2% branch) Not the exact wording, but the #'s are correct. BTW, there is a nifty calculator available (over on the left border of this page) If I get back to the desk later, with the calc, I'll run your numbers thru. We just did 2-25 HP pumps, 1675' run, 480 volt, and installed 4" w/500 mcm Cu. John

John

#16677 - 11/15/0212:58 PMRe: Derating and voltage drop

First off, the size you have used for <assuming> each Motor will pretty much cover the voltage drop. The 5% max quotation in the NEC is a FPN [Fine Print Note], which is more of a suggestion - not a mandatory "Design" limit [NEC is not a design manual... yadda yadda yadda].

Figuring the volts lossed would be done using the normal apparent items [such as distance, amperage and etc.], but could also be done using more elaborate methods - which include total Reactance of Conductors, Ambient Temperature adjustments and the like. Which ever method you choose to do depends on how "Strict" you are and / or how "Strict" your Client is. In the long run, what you measure at the Load's end is what matters.

Conflict... derating and applying Voltage Drop calcs together... Solutions... Apply simple calcs, size one AWG size larger, complile advanced and complex calcs, "experiment"; all will be NEC compliant but will this satisfy the Client?

So what I am getting at (in a nutshell) is you have many choices on how to address this situation, but the simplest way to do everything is to figure a wire size that works with the Derated Value, then check it to see what a simple voltage drop calculation turns up with that wire size. If that size "Borderlines" any suggested value (3% branch circuit, 5% from service to load per NEC FPN, or something a Client / Vendor suggests), then upsize until the numbers are better.

Using your numbers, here's what I see: <OL TYPE=A>

[*]3 - 40 HP 3 phase Induction Motors,

[*]40 HP 3 phase Induction Motor @ 480 VAC = 52 Amps,

[*]3 - #6 THHN cu minimum per Motor,

[*]At least 15 volts lossed per Motor -52 amps, 350' run,

[*]9 current carrying conductors = 70% derating

[*]75 amps x 0.7 = 52.5 amps - derated value of #6 THHN cu </OL>

Using the #3 cu conductors, I came up with the same voltage drop figure that Redsy did - using the "Simple Formula" (Resistance only), which is 7.8 volts lossed. This equals out to 1.625% of 480 VAC.

The "Best" approach here is to make sure the Complete Voltage Drop is not excessive. This would be from point of service to the load, covering what the Actual Voltage is at the Motor[s] by measuring the voltage at the service and / or the MCC, applying voltage drop calcs to THAT voltage, then find out if it will "Work" with the nameplate voltage of the motor.

Wish I could give a clearer message and direct answers, but you might be able to use some of this for a good answer!

Scott s.e.t.

Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!

#16680 - 11/16/0208:38 AMRe: Derating and voltage drop

the voltage drop did not apply to this situation. It was a what if question.If the run was 1000' would the increased wire size for more than 3 conductors also cover voltage drop.It is obvious that it does but it was one of those simple questions that lock up the old brain matter.Thank you all for the replies

#16681 - 11/16/0203:06 PMRe: Derating and voltage drop

i screwed this up, so you probably should skip to the next message, but i'll leave it here in case someone wants to see what's wrong with it

i think you just have to do both calculations because voltage drop will have different reductions than derating reductions.

430.6 >>> table 430.150 >>> 52a since fedup started with #3 i assume this is single motor each b.c. so.. 430.22(A) says to use 125% of flc for continuous duty from T430.150 430.22(E) uses other than 125% if not continuous duty so it could be different here

430.150 >>> 52a FLC 430.22 >>> 52x1.25= 65a 310.13 assuming 350 ft is underground or piped underground so it's wet >>> xhhw/90*C 310.16 >>> 6awg Cu Chapter 9 Table 9 6awg Cu in pvc is .49ohms/1000ft but upsized once to 4awg is .31ohms/1000ft

for 6awg Cu: L-N drop [.49] [350ft/1000] [65] = 11.15v L-L drop [11.1475] [1.732] = 19.3v 19.3v/480v = 4% is more than the fpn 3%/b.c.

for 4awg Cu: L-N drop [.31] [350ft/1000] [65] = 7.05v L-L drop [7.0525] [1.732] = 12.21v 12.21v/480v = 2.5% is less than the fpn 3%.

210.19 fpn4 says 3% is ok for VD on bc's

so 480v x .03 = 14.4v would be an allowable VD for this b.c. using 4awg Cu..... but if you use 3+ bundled wires then adjustment factors are more restrictive than VD.....

using 70% from Table 310.15(B)(2)(a) Adjustment Factors for More Than Three Current-Carrying Conductors in a Raceway or Cable.......

75*C column wire for 6awg is 65amps, so.. [.7] [65] = 45.5amps is the adjusted maximum ampacity for 6awg. too low, we need 65amps [to meet the 52amps x 1.25, per 430.22]

75*C column wire for 4awg is 75amps, so.... [.7] [75] = 52.2amps, still too low.

75*C column wire for 3awg is 100amps, so.. [.7] [100] = 70amps is ok.

90*C column wire for 4awg is 95amps, so.. [.7] [95] = 66.5amps, also works.

so my choice is 4awg Cu XHHW [ok to use 90*C column at 95amps if underground], or in a dry location then XHHW-2 at 90*C

[This message has been edited by Cindy (edited 11-16-2002).]

#16682 - 11/16/0207:10 PMRe: Derating and voltage drop

Although we ultimately ended up selecting the same size conductor in the end, I'd like to point something out to you.

Derating for adjacent conductor adjustment and ambient temperature correction can be calculated at 100% of the actual load as per 210.19. You can apply these derating factors before calculating the 125% of continuous plus noncontinuous loads. Also, when figuring voltage drop there is no need to use 125% of full load.

Here is the method that I learned and use to calculate wire size:

STEP 1: Compute the load at 125% of continuous load (unless the assembly is rated for 100% operation) plus 100% of noncontinuous load.

STEP 2: Go to table 310.16 and select the conductor from the correct temperature column. The equipment rating dictates the correct temperature column, and is covered in 110.14c.

Having completed steps 1 and 2, write the conductor size down; if their aren't any adjacent conductor or temperature factors involved, you are done. If adjacent conductor or temperature correction factors are involved, go to steps 3 and/or 4.

Steps 3 and 4 are a completely seperate calculation and applied to the actual load, not 125% of continuous load plus the noncontinuous load.

Step 3: Apply when there are more than 3 current carrying conductors in a raceway or cable. Use table 310.15b2a, the adjustment factor table.

Step 4: Apply when the conductors are going to be installed in an elevated temperature such as a boiler room or rooftop. The correction factor table at the bottom of table 310.16 is used to determine the corrected ampacity of the conductor. If using THHN, which is rated at 90 degree C operation, apply steps 3 and 4 at the actual load and select the conductor from the 90 degree column of table 310.16.

Step 5: Compare the conductor selected in steps 1 and 2 with the conductor selected in steps 3 and 4; the larger of the two is the correct size.

Now do your voltage drop calculation, if neccesary, increase the size of the conductor to compensate.

So, step 1: a 40 HP 3 phase motor @ 460 volts = 52 amps X 1.25 = 65 amps.

Step 2: If we go by the book, we cannot assume that all equipment is rated at 75 degee C, and must use the 60 degree column of 310.16 which dictates a #4 copper conductor (110.14c). In reality though, we realize that this equipment is probably rated at 75C, which dictates a #6 copper conductor.

Step 3: Adjacent conductor factor for 9 conductors is 70%. If we divide 52 amps by 70%, we see that our conductor must carry 75 amps. Article 430.22a dictates 125% of 52 amps for continuous duty. 125% X 52 = 65 amps. As you can see, the adjustment for adjacent conductors dictates 75 amps, already more than 125% of full load that is required for continuous duty. A trip back to the 90 degree column of 310.16 we select a #6 copper conductor.

Step 4 does not apply.

Step 5: Comparing the conductor selected from steps 1 and 2 to the one selected in step 3, we can determine that if the equipment temperature rating is known to be 75 degree C or higher, a #6 THWN or better can be used. If the equipment temperature rating cannot be determined, a #4 THWN or better must be used.

Now voltage drop. I like to use the reciprical formula for 3 phase: Cmils needed = 1.73 X K X I X L/desired volts dropped.

1.73 X 12 X 52 amps (no need to calculate at 125%) X 350'/3% of 480V is roughly 14 volts = 26,988 Cmils. #6 = 26,240 Cmils, #4 = 41740 Cmils. The #4 would be required to maintain 3% or less voltage drop.

Matt

[This message has been edited by Matt M (edited 11-16-2002).]