Voltage Drop

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wattup

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I need to maintain no more than a 3% voltage drop with the following information. Can you tell me what wire size to run using THHN?
Single phase with neutral feeder to 100 amp panel ( assume they want to load it to the max.) Voltage is 208 and the footage is 550 feet.
 
to find the voltage drop use this formula:

CM = 2KLI/VD

CM = circular mils
K = 12.9
L = length
I = current
VD = voltage drop for 208 use 6.24

For 3 phase change formula to CM = 1.732KLI/VD
 
buckofdurham said:
I'll need an other cup of coffee, first.:wink:
Click to expand...

208 3phase 3wire 2.8 percent drop
1 awg cu 202 volts when you get to the end

120 - 208 3 phase 4- wire 2.8 percent drop
1 awg cu , 116.6 volts when you get there:wink:
 
080909-0803 EST

wattup:

I do not think you provided sufficient information.

So I will guess you have a 208 three phase source Y connected. This means 120 V line to neutral. 3% of 120 = 3.6 V drop. Your total wire loop length is 2*550 = 1100 ft. The resistance needs to be less than 3.6/100 = 0.036 ohms. Assume copper wire. Look in a wire table for resistance. This will require something larger than 0000. This may not be a practical approach vs a higher voltage and a stepdown transformer at the destination.

If for some strange reason you have a 208 single phase source the allowed voltage drop would be 208*0.03 = 6.24 V and maximum loop resistance is 0.062 ohms.

brian john's reference web site will do the calculation for you. But you need to ask the correct question and I am not sure you have done that.

Probe some of these questions ---
Is the voltage of interest at the destination 208 or 120?
Do you have two 120 circuits sharing a common neutral?
If neutral is shared what is maximum current in the neutral?
Might it be possible that under full load conditions a higher percent drop might be allowed because of low probability.
Are there only a few critical devices at the destination such that other means might exist to solve the voltage variation for these units? For example some Sola constant voltage transformers.
Is it more economic to transfer power at a higher voltage and reduce wire cost?

.
 
kenaslan said:
550' is not that long. I just did a 36000' 14 mile conveyor for a gold mine. .15A strobes every 600' each half of conveyor 4/0 wire for .15A @120V NOW THATS VOLTAGE DROP
Click to expand...

That's crazy. Last strobe @ 85vac with 29% VD. #3cu feeder @ 480v, w/120v current-xfmr donuts for each outlet, gets similar efficiency.
 
ramsy said:
That's crazy. Last strobe @ 85vac with 29% VD. #3cu feeder @ 480v, w/120v current-xfmr donuts for each outlet, gets similar efficiency.
Click to expand...
I think he said 0.15a each, not 15a.
 
LarryFine said:
I think he said 0.15a each, not 15a.
Click to expand...

Yes, each strobe 0.15 x 3600ft/600ft = 9a.
480/120 CT's may also work for 15a receptacles. Figuring 60 of them.
 
iwire said:
Roger was your figure arrived at using the point to point method?
Click to expand...

The 2005 NFPA Handbook has a VD example following Chap.9 Tbl.9, based on wire impedance, modified by ambient, power factor and inductive reactance. My result should match this example, but using a Power factor = 1, then plugging in 9 amps over 36000 feet. This method doesn't spread loads across 14 miles, so my 85vac figure is only valid for the last strobe at the End Of Line. Available Short Circuit Current from xfmr (point-to-point) is not part of this VD calc.

Using the DC method from Brian John's link, the results for that 4/0 14 mile long 120v cable is 82vac at EOL. So, the results are similar.
 
Last edited:
080912-0945 EST

My quick estimate is 16.1 V drop at 36000 ft.

This is from
Resistance of 1200 ft loop = 0.0490*1.2 = 0.0588 loop resistance ohms per 600 ft segment.
The current thru the xth segment is (N+1-x)*0.15 A.
Where N is the total number of segments.
The voltage drop of the first segment is 60*0.15*0.0588, the second is 59*0.15*0.0588. and so on.
This can be written as 0.15*0.0588* the sum of 60 .... 1, or
0.0088 ( 60+59+58 ... 2+1).
The sum of this series is 1830 if I did not make a mistake.
Thus, the drop to the end of the 36000 ft is 0.0088*1830 = 16.1 volts.

A very rough way is to assume 1/2 the total current or 4.5 A at 36000 ft or 2*36*0.049*4.5 = 15.9 volts. This saves calculating the value of the series and is more than adequate.

.
 
My results agree with Gar's.

IMHO this would have been a perfect application for a MWBC. 4 #4 conductors (assuming 208/120V three phase) would have given better performance than 2 #4/0 conductors in terms of voltage drop, with less than 1/2 the copper.

-Jon
 
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