VD & Distance

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Alwayslearningelec

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So I'm not quite getting what the "max distance" column is suppose to imply and why they listed it.

E.G. have a 400A 3P 480v breaker feeding a 3P 480v 400A panel. For this example they show the feeder tag 4004 which is 1 set of 4#600 and 1 #3.
The distance is 610'. Doing VD calc I come up with 1.51%. Am I missing something. Is this correct?
 

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For feeder #4004 I came up with 409' using the Southwire App on my phone.

A maximum distance of 409.20 feet will limit the voltage drop to 2% or less with 1 parallel (set) of 600 kcmil Copper conductor delivering a total of 320 amps on a 480 volt three phase system.
Click to expand...
 
The online calculator I use shows a 2% drop at 320 amps for 335' 600 kcmil in steel conduit with a conductor temperature of 40°. If I change that to 75°C, the distance for a 2% VD goes to 325'.
 
infinity said:
For feeder #4004 I came up with 409' using the Southwire App on my phone.
Click to expand...
These are the voltage drops I get from an engineers spreadsheet calculator. This is for Copper #600 going 600' and 400A(actually 320A- 80%) - VD is 1.49% A 3P 480V
 

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horsegoer said:
So I'm not quite getting what the "max distance" column is suppose to imply and why they listed it.

E.G. have a 400A 3P 480v breaker feeding a 3P 480v 400A panel. For this example they show the feeder tag 4004 which is 1 set of 4#600 and 1 #3.
The distance is 610'. Doing VD calc I come up with 1.51%. Am I missing something. Is this correct?
Click to expand...
It appears they have already completed the calculation for such that you should get no greater than 2% voltage drop at 80% of the current rating of the conductor used. My calculation gets a voltage drop of 2.61% at 455 feet at 320 amps assuming non-steel conduit. Do they spaek to any power factor levels? Of course, a decrease in PF only makes it worse.
 
NewtonLaw said:
It appears they have already completed the calculation for such that you should get no greater than 2% voltage drop at 80% of the current rating of the conductor used. My calculation gets a voltage drop of 2.61% at 455 feet at 320 amps assuming non-steel conduit. Do they spaek to any power factor levels? Of course, a decrease in PF only makes it worse.
Click to expand...
My engineer is adamant on his calculator attached. 1.54% VD for #600 3P 480v going 620' to 400A load( 80% 320A). It's from Uglys book.
 

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horsegoer said:
My engineer is adamant on his calculator attached. 1.54% VD for #600 3P 480v going 620' to 400A load( 80% 320A). It's from Uglys book.
Click to expand...
My guess is that his calculator is wrong. Not sure why but when I input these numbers into the Southwire calculator it gets an error because one set of any size conductor will not work. When I use Don's calculator with your numbers I get 3.9% VD.
 
infinity said:
My guess is that his calculator is wrong. Not sure why but when I input these numbers into the Southwire calculator it gets an error because one set of any size conductor will not work. When I use Don's calculator with your numbers I get 3.9% VD.
Click to expand...
Same here. But he also said he did it by hand as well and got same result.
 
horsegoer said:
Same here. But he also said he did it by hand as well and got same result.
Click to expand...
Maybe he's correct, who knows, the results of the various calculations in this thread are all over the place. Here is Mike Holt's explanation, note the bold part:

VD = 1.732 x K x Q x I x D/CM - Three Phase


“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.


“K” = Direct Current Constant: This is a constant that represents the direct current resistance for a one thousand circular mils conductor that is one thousand feet long, at an operating temperature of 75º C. The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors. The “K” constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0.


“Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.


“I” = Amperes: The load in amperes at 100 percent, not 125 percent for motors or continuous loads.

“CM” = Circular-Mils: The circular mils of the circuit conductor as listed in Chapter 9, Table 8
Click to expand...
“D” = Distance: The distance the load is located from the power supply, not the total length of the circuit conductors.
Click to expand...
 
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infinity said:
Maybe he's correct, who knows, the results of the various calculations in this thread are all over the place. Here is Mike Holt's explanation, note the bold part:
Click to expand...
And since the conductors are 600kcmil, the Q factor would apply.

If I calculated it correctly, that would be about a 6.5% increase in resistance? (Using PVC conduit)
 
horsegoer said:
can you elaborate ?
Click to expand...
VD = 1.732 x K x Q x I x D/CM - Three Phase

“Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

So I looked up the AC resistance (0.075 for PVC) and the DC resistance (0.0704). Then do the math as stated above 0.075/0.0704 = 1.065.

So if I use 1.065 for Q in the formula above I will be increasing the rest of the formula by 6.5%.

If I did that wrong, please let me know.
 
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