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# Voltage drop on the service entrance

## Voltage drop on the service entrance

#1
04-19-03, 08:06 AM
Member
Join Date: May 2001
Posts: 251
Voltage drop on the service entrance

I have a 100 amp service on a house built in 72. The meter is stubed up in the back yard and the Panel in the house is fed with an underground line from the meter. The line from the meter to the house is about 110 feet. My question is does it make sence from a payback point of view to move the meter to the house and get rid of the 100 foot gap between the meter and the panel. The power company said if the service is upgraded to 200 amp they would want the meter moved to the house. They said it cost 160 dollars for the wire and they want a trench open for them for the new wire. I think I should be able to get a meter socket for around \$250 dollars. The question is how much electricity is being wasted that I am paying for in that 100 foot run and is it enough it would make cents?

I also read on a site that oversizing the wire by one size larger than the NEC requires is a good idea because of the electricity saved is this true?

Thanks

#2
04-19-03, 10:55 AM
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Join Date: Feb 2002
Location: port chester n y
Posts: 1,983
The minimum size conductor for a 100 amp residentail service is #4 copper. The minimum size conductor for an 80 amp load @ 240 volts in order to comply with NEC voltage-drop requirements is also #4 copper,so I doubt that you are "wasting" power in the Service Conductors.

If you were to "up-grade" from 100 amps to 200 amps you'll need a 200 amp meter-socket and a 200 amp main Main Circuit-Breaker panel with #00 copper conductors between the meter and the panel.

Don't expect the utilty co. to use 200 amp conductors between the utilty-pole and the meter location at the house.They will use a much smaller conductor.

Please know that the "over-all" distance for a Service voltage-drop calculation is the distance between the utilty co. transformer on the utility- pole and the Main-Breaker at the Service panel.

#3
04-19-03, 11:33 AM
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Join Date: Dec 2001
Location: New York
Posts: 1,287
You have a valid thought, regarding wasted power between the meter and you house panel. If you need to upgrade for ampacity reasons, then the story is different and do it anyway.

For example (with lots of assumptions):
If your existing service is 100A, 240V with #2 aluminum for 110' and your overall average load (including time when you are sleeping or not at home) in the house was 30A, you pay .10 cents per kWH, then the instantaneous loss would be 2.11V or approx \$55/year.
The calc was (2.11*30*24*365/1000*.1)

This gives you a relatively short payback, assuming the assumptions are correct. It makes sense to keep the losses on the utility side of the meter.

Hope the calcs are fair (accurate).

The calc has been edited due to a screwup (thanks texdiyguy)

Last edited by HandyRon; 04-19-03 at 05:29 PM.
#4
04-19-03, 11:39 AM
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Join Date: Dec 2001
Location: New York
Posts: 1,287
In addition, as mentioned by PATTBAA, the utility may not use 200A service conductors for your upgrade, but as long as that is on their side of the meter, and you don't have voltage sag problems when you draw current, who cares, it's their losses.

#5
04-19-03, 05:15 PM
texdiyguy
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You have a units problem in your calculation.

volts*volts*hours/1000 doesn't result in kilowatt hours

However:

((2.11*30*24*365) / 1000) * .1 is:

volts * amps * hours / 1000 which does yield kilowatt-hours

Therefore, the losses associated with the voltage drop would be more in the neighborhood of \$55/year.

#6
04-19-03, 05:27 PM
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Join Date: Dec 2001
Location: New York
Posts: 1,287
texdiyguy, Thanks
You are correct. Similar point, less dramatic.

#7
04-19-03, 05:44 PM
texdiyguy
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Yes, and it's even less dramatic when you consider that, with the reduced voltage that gets to the house, many applications will consume less energy, especially resistive loads such as lighting.

#8
04-21-03, 07:32 AM
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Join Date: Feb 2002
Location: port chester n y
Posts: 1,983
P = I X I x R. The resistance of 200 ft. of #4 conductor = .06 Ohms = R

For I = 30, I x I = 900

P = 900 X .06 = 54 watts = power loss in conductor.

Total power = 240 volts X 30 amps = 7200 watts

Power loss in % = 54/7200 =.07%