Voltage Drop

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tacomafc

tacomafc

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Im sorry the guy in the field gave me the wrong info. The load is 300 watts and the distance is 1600ft @ 120 volts.

2 x 12.9 x 2.5 x 1600 = 103200/cma?

103200/10380(cma for 8awg)=9.94 What do I do with this number
 
Re: Voltage Drop

10380 is the circle mil area of #10 AWG wire (see table 8 in the 2005 NEC - Chapter 9). The Circle Mil Area of #8 is 16,510.
You divide 103200 by 16510 and get a VOLTAGE DROP of 6.25 volts. Amperage has nothing to do with it at this point. Since your supply voltage is 120-volts, 6.25 volts is a drop of a little over 5%. If this is not acceptable you will need to shorten the distance or install larger wiring.
 
Re: Voltage Drop

Here again, my spreadsheet tells me that a #6 will give you a 4 volt, or 3%, voltage drop.
 
Re: Voltage Drop

Originally posted by tacomafc: a number 8 awg wire @ 75 deg. is 50 amps. 3 percent of the is 1.5 so my voltage drop has to be less than 1.5 if I use a number 8 wire
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You are confusing your terms. You are taking 3% of a current value (50 amps) and calling that your limit for voltage drop.

You are also using the ampacity of a conductor in a discussion of voltage drop. The two are not related in any way. What matters to voltage drop is the current drawn by the load, not the current that the conductor is capable of handling. You said the load was 300 watts at 120 volts. That is equivalent to 2.5 amps. I see that you have used the 2.5 amp value correctly in your calculation.
 
Re: Voltage Drop

Originally posted by haskindm:You divide 103200 by 16510 and get a VOLTAGE DROP of 6.25 volts.
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If you used this same process with a #6, and divide 103200 by its area of 26240, you will get a voltage drop of 3.93. That is the same value that my spreadsheet gave me.
 
Re: Voltage Drop

charlie, could you send me a copy of your spread sheet? Thank you.

(Moderator?s Note: Edited to remove email address. If you wish to send an email message to this person, then first send a Private Message via this Forum, and ask for the email address.)

[ January 20, 2006, 03:34 PM: Message edited by: charlie b ]
 
Re: Voltage Drop

An accpetable voltage drop is whatever is designed or preferred by the designer. Typically 3% for branch circuits and 2% for feeders.

This is the maximum permitted voltage drop on feeder and branch circuits in any non-dwelling building in Florida.
 
Re: Voltage Drop

Originally posted by tacomafc: charlie, could you send me a copy of your spread sheet? Thank you.
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Sorry, but I can't. I didn't create it, I can't certify that it is accurate, and I don't have the rights to distribute it. I treat it as a "first estimate" for calculation purposes. But if I were to perform a formal calculation, I would not use it.

However, there are similar spreadsheets available for a free download. Someone might be able to provide a link.
 
Re: Voltage Drop

charlie, when you say that you get a number 3.93 is that 3.93% and if it is should I then use a numer 4 awg wire so that I will be under 3%? Thank you, sorry still a little confused
 
Re: Voltage Drop

charlie, when you say that you get a number 3.93 is that 3.93% and if it is should I then use a numer 4 awg wire so that I will be under 3%? Thank you, sorry still a little confused
 
Re: Voltage Drop

Originally posted by tacomafc:
charlie, when you say that you get a number 3.93 is that 3.93% and if it is should I then use a numer 4 awg wire so that I will be under 3%? Thank you, sorry still a little confused
Click to expand...
That is 3.93 volts. By dividing that by 120 volts you'll get the percentage of VD.

3.93/120=0.03275 or 3.275%
 
Re: Voltage Drop

You are correct, Infinity. That is what I intended, and is in fact what I said in my first post of this thread.

It should be noted that 3.275% can be rounded up to 3.3%, or rounded down to 3%, depending on your desired precision. Since the NEC's FPNs just say 3% and 5%, I presume that the nearest whole percent is adequate precision. That is why I said a #6 would give you 3% VD.
 
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