Being underground isn't the issue as much as where the responsibility ends for the power company and where you start. As you've started to figure out you will need a larger wire for a long run. Are they, in fact, going to put a transformer on the second pole if they do install that?

Who is paying and who is installing?

The responsibility ends at the meter for the power company. They would put the transformer at the first or second pole depending on which route we go. After the pole they will install 40' of wire to a pedestal mount meter. They check the amps at the meter, that is all they care about. Someone can go as far as they want after the meter as far as they are concerned, if you don't have enough power at the end of the run that is your own fault. After the meter is 100% MY responsibility. I would prefer to only instal one pole and go about 580' if I can. This will save me some money. But I will pay the $3500 to instal the second pole if I have to. What do you guys think?

I understand all that. What I am looking for is can the wire (250 or bigger if needed) handle a 200 amp service after a 580' run. It is a 2400 sf house with central AC. gas stove and electric dryer. No welders or large compressors. One electrician I use says "No problem" on say's he wouldn't really recommend it. just looking for another opinion.

The NEC generally recommends voltage drops of not more than 3-5% for maximum efficiency.

Voltage drop can be calculated using the formula:

E(drop)= I x R

Where:

E(drop)= the actual drop in voltage
I=amperage of the load
R=resistance in ohms per 1000 linear feet

From Table 8 of the NEC ("Conductor Properties") we find that a 250kcmil uncoated copper conductor has a resistance of .0515 ohms per 1000 linear feet.

Plugging the numbers into the formula we get:

E(drop) = I x R
E(drop) = 200amps x (.0515ohms/1000 x 580feet)
E(drop) = 200 x (.0000515 x 580)
E(drop) = 200 x (.02987)
E(drop) = 5.974 volts

The percentage of voltage drop therefore is:

5.974 volts/240 volts=2.489%

So the voltage drop using a 250kcmil conductor for a 580 foot run will yield you less than a 3% voltage drop and will be fine for this application.

[Presuming I did my math properly :) ]

In fact, you could use 4/0 conductors in this case and still end up with less than 3% voltage drop.

3/0 conductors could also be used and still yield less than 4% voltage loss.

The question then becomes what is the price difference in each of these conductor types? (250kcmil vs 4/0 vs 3/0)

That may ultimately decide how you wire the service...

By the way if you used 250kcmil aluminum conductors you would still have about a 4% voltage drop (slightly more than 3/0 copper)...and again...price of aluminum 250kcmil vs the copper 250kcmil vs copper 4/0 vs copper 3/0 could be the deciding factor.

Thanks ALOT for you response. That is exactly the information I was looking for. I will probably go with the 250kcmil aluminum conductor. It will be lighter, cheaper and easier to work with. Thanks again.

The next question you need to resolve, however, is how much 580 feet of 250 kcmil aluminum will cost you.

If you need to buy a 1000 foot coil, paying for the 2nd utility pole may seem like a bargain.... :(

Below is a voltage drop calculator you may find helpful.

It is recommending 400 kcmil aluminum to get below 5%.

Reason is, it is calculating conductor length in both directions (something I did not do in my calculations above) when you plug in single phase, 240 volts, 580 linear feet with a 200 amp load.

You can check to see what the cost of 250 kcmil conductors cost for a 1000 foot coil on the same link: $9,999 for SEU.

And why I mentioned the second pole may seem like a bargain once you see the prices....

Here is a link that might be useful: Voltage Drop Calculator

Circuit resistance is always twice the cable length.

Another thing to consider ids who is responsible for the conductors after installation.

7.2 kV should stay the POCOs responsibility.

A 120/240 V line may devolve to the homeowner, and WILL fall to you if the meter is way out there.

brickeye said:

"Circuit resistance is always twice the cable length."

Yes, that is correct for AC (Alternating Current). AC cycles back and forth first in one direction then the other 60 times per second. (To complete 1 cycle, the current needs to go 2 x 580 feet or 1160 feet in this example.)

Resistance of DC (Direct Current) circuit is measured only in one direction.

The formula should then be for our purposes:

E(drop) = I x R x 2

Where:

E(drop)= actual drop in voltage
I = amperage of the load
R x 2 = Resistance of conductors in ohms per 1000 linear feet in both directions.

To get near the 5% voltage drop then, we would then need to use at the smallest 400kcmil aluminum conductors because:

E(drop) = I x R x 2
E(drop) = 200amps x (.0529ohms/1000 x 580feet) x 2
E(drop) = 200 x (.0529/1000 x 580) x 2
E(drop) = 200 x (.0000529 x 580) x 2
E(drop) = 200 x (.030682) x 2
E(drop) = 12.2728 volts

12.2728 volts/240 volts = 5.1%

500kcmil or larger conductors would be better however.

It comes down to cost of the conductors versus cost of the 2nd pole...

And it is looking like adding the second pole or locating the house closer to the road may be the best economical alternative....

"Yes, that is correct for AC (Alternating Current). AC cycles back and forth first in one direction then the other 60 times per second. (To complete 1 cycle, the current needs to go 2 x 580 feet or 1160 feet in this example.)

Resistance of DC (Direct Current) circuit is measured only in one direction."

It has nothing to do with AC or DC.
For any current to flow you must have a complete circuit: from source to load, and then back to the source.

The voltage drop occurs on the entire circuit based on the current flowing.

It is ALWAYS twice the physical distance sine the current MUST flow out to the load and then back.

A DC circuit would encounter the exact same 2x580 feet of wire resistance.

I met with the power company today. They are going to move the first pole closer to the house and allow more distance to the meter. With all this we are around 500' to the house from the meter. The power company said I should have no problem with 250 kcmil aluminum wire at that distance.

I can buy the wire from the electrical supply house for $2.50/ft so it makes the most sense to go this way.

Someone else also mentioned to me that even though I have a 200 amp service doesn't mean I will use 200 amps. So figuring 200 amps in the formula is overkill. Does this makes sense?

Thanks guys for all the help.

""Yes, that is correct for AC (Alternating Current). AC cycles back and forth first in one direction then the other 60 times per second. (To complete 1 cycle, the current needs to go 2 x 580 feet or 1160 feet in this example.)
Resistance of DC (Direct Current) circuit is measured only in one direction."

It has nothing to do with AC or DC.
For any current to flow you must have a complete circuit: from source to load, and then back to the source.

The voltage drop occurs on the entire circuit based on the current flowing.

It is ALWAYS twice the physical distance sine the current MUST flow out to the load and then back.

A DC circuit would encounter the exact same 2x580 feet of wire resistance."

Voltage drop for AC is measured using the formula E(drop)=I X R x 2.

Voltage drop for DC is measured using the formula E(drop) = I x R.

The reason is that alternating current travels in BOTH directions to create 1 cycle while DC current travels in only 1.

Brickeye is entirely wrong about this one and should do a bit more research on AC vs DC before giving such wrong advice.

But be that as it may...

The voltage drop for 500 feet 250 kcmil aluminum is going to be:

E(drop) = I x R x 2
E(drop) = 200amps x (.0847/1000 X 500) X 2
E(drop) = 200 x (.0000847 x 500) X 2
E(drop) = 200 x (.04235) x 2
E(drop) = 16.94 volts

16.94volts divided by 240volts = 7.05% voltage drop

A 7.05% voltage drop is too much for the run as recommended by the NEC.

To be in the 3-5% recommended range one would still need to use a minimum of 400 kcmil aluminum conductors.

"Brickeye is entirely wrong about this one and should do a bit more research on AC vs DC before giving such wrong advice. ":

manhattean 42 is so far out of anything he knows or understands it is funny.

ALL circuits operate as closed loops. PERIOD>

AC, DC, RF, ALL of them.

In an open circuit there can be NO current flowing.

All the length of conductors MUST be accounted for in a circuit.

The current (AC or DC) MUST return to its source.

This means that the losses in both directions (source to load, then load to source) MUST be accounted for.

The faqct that the voltage reverses 50, 60, or thousands of times aq second does not matter.

You might look up Kirchoff's current law before spouting absoluter nonsense about imagined differences between AC and DC voltage drop.

Current in ALL portions of a circuit MUST be equal.

You cannot accumulate charge carriers.
Every carrier that moves creates a return carrier to the source of the voltage.

The formula for voltage drop is
VD=(2*L*R*I)/1000
where:
L = one way length of circuit,
R = conductor resistance per 1000ft,
I = load current.

This formula applies for any 2 wire DC, AC, or 3 wire AC 1phase circuit.

Brickeye is correct.

I am converting an out building into a home office, building size is 24 x 14 and will have small kitchenette, bathroom(30 gal water heater), window air/heat, I need to run the electrical wire 250 ft to the home meter base...how much and what grade wire will I need?

You really should start another thread rather than glomming on to a three year old one.

An electrician can do a proper load assessment (your list is a bit nebulous). Guessing at 50A load at 250' , #4 Copper. You'll need either 4 conductors rated for wet locations in a conduit (THWN for example) or a cable with three conductors and a ground rated for direct bury if you want to go that way.

You'll also need to install a grounding electrode system at the outbuilding.

"Someone else also mentioned to me that even though I have a 200 amp service doesn't mean I will use 200 amps. So figuring 200 amps in the formula is overkill. Does this makes sense?"

That's correct. An electrician would do a residential load calculation. Yours might come to 150A. Even then that's a thanksgiving or xmas party type load. Most of the time you're only pulling 30-40A.

If there's no breaker at the meter you need three conductors. Cable is usually cheaper than pipe, but needs a deeper trench. One cable like that is called Triplex USE.

If you use pipe, the third wire, neutral, can be smaller than the two hots. The resi. load calc. would tell you how much exactly, but it's usually two wire sizes smaller.

"If there's no breaker at the meter you need three conductors. Cable is usually cheaper than pipe, but needs a deeper trench. One cable like that is called Triplex USE."
What does this mean?? Are you suggesting he run directly from the meter with a second load???

"If you use pipe, the third wire, neutral, can be smaller than the two hots. The resi. load calc. would tell you how much exactly, but it's usually two wire sizes smaller. "
What does pipe have to do with it? The URD I buy has a reduced neutral.

This is also a duplicate thread:
http://ths.gardenweb.com/forums/load/wiring/msg0415472317799.html?2

Here is a link that might be useful: http://ths.gardenweb.com/forums/load/wiring/msg0415472317799.html?2

This post was edited by petey_racer on Thu, Apr 18, 13 at 7:11

"What does this mean??"

It means, from the meter to the first breaker is a three wire feed. After the first breaker are four wire feeds.

"Are you suggesting he run directly from the meter with a second load???"

The meter can feed any number of buildings or structures. One three wire feed is allowed for each. Each building may have it's own "first breaker". As long as there is no other breaker in the line between it and the meter.

"What does pipe have to do with it? The URD I buy has a reduced neutral."

Thank you . You're right. I wrote poorly. In a cable you have no choice. The neutral is what it is.

In pipe the neutral may be as small as the load calculation determines. But no smaller than the EGC (ground) as determined by table 250.122. (i think that's the number.)

"The meter can feed any number of buildings or structures. One three wire feed is allowed for each. Each building may have it's own "first breaker". As long as there is no other breaker in the line between it and the meter. "

Unless it has and/or can accept dual output lugs I have never seen a typical residential meter that can feed more than one load.
One would have to either install a trough or very large box, or install a panel under the meter, or replace the meter with a meter/main/panel.

To simply say "If there's no breaker at the meter you need three conductors." is very misleading.

I took the meter base comment to with a grain of salt. Most often the meter base either contains or in immediately adjacent to, where you'd connect the feeder for the outbuilding.

For instance while I'd have to pull mine off the main panel, I'd have to run the feeder back out of the basement right at the base of the meter anyhow.

Better to put a disconnect (typically with cartridge fuses) after the meter.

At least then you can kill all the laterals without puling the meter) and then run 4-wire.

YOU own the laterals and will be responsible for their maintenance (at some point a likely occurrence).

This is a routine style on larger installations.
Using main lug panels is often only done in a service room or area with the feed panel right there.
That way you can create a completely dead panel for servicing.

It is NOT strictly required by the NEC, but then the code is the MINIMUM you can get away with.

No one is stopping you from making an installation safer for workers by being able to make an entire panel dead.

Arguably it is probably more important outside a commercial/industrial setting since only fully trained workers are usually permitted access.

I cannot think of a pro that blinks at working in a main panel with just the main turned off (we understand what is still hot), and have been required to work in panels with everything on - critical circuits in some places are a really BIG DEAL to interrupt).

Not all that bad if you know what you are doing an use at least some protective gear.

This post was edited by brickeyee on Sat, Apr 20, 13 at 10:58

"Unless it has and/or can accept dual output lugs I have never seen a typical residential meter that can feed more than one load."

In Alberta rural services have a breaker at the pole. If you install multiple service laterals from the pole, you have multiple breakers.

Here our line charge is based on the size of the pole breaker. We have a 100A main breaker at the house. Two years ago we downsized the pole breaker from 50A to 35A, and saved 35 bucks a month. We've yet to trip it.