Wiring Size Calculations?
I have installed a micro hydro electrical system in central America. The turbine is in the river from where I have brought #8 cable about 1000', bringing the DC to an AC inverter (3500 watt capacity). I plan to change the size of this cable for better performance in the future. Now I have to bring cable from the inverter 1200' to the cabin area where I expect to use about 2500 watts (if everything were turned on at once) spread over six one room cabins and one kitchen/dining area. I was thinking of doing this all on one circuit. So my question is what size cable will I need and also can the cable that goes in the cabins be smaller that the cable that brings the power to the cabins. Also, if I bring the cable 700' to the kitchen area and set up a sub circuit there, can I then use smaller wire to go the remaining distance of 500' to the cabins? I am really new to all of this so please bear with me. Thanks.
Swami 
Swami Need a little more info
Yes, youre right in thinking about using smaller size cables as you step down to the various loads. To answer you more thoroughly though, I need some more information. First, when you say #8 cable, are you talking about AWG #8 (American Wire Gauge)? Also, the watts will only be used to determine the wire size if we also know the voltage used. So let me know the AC voltage. By the way, #8 AWG, if thats what you are using, seems pretty small for this application of 1000 feet. The resistance of that length of wire is going to drop your allowable ampacity of the wire at LEAST one wire size and probably TWO wire sizes, maybe even THREE! (dont have a book in front of me) That takes you down to the #12 or #14 ampacity range which is a mere 1520 amps approximately. Barely enough to run a good vacuum cleaner. Make sure you use the proper overcurrent protection in front of any wire you run. Hope this helps.
When in doubt, Hire an Electrician! 
The current is 110V. 12001400 ft of AC current to go from the inverter to the cabins where we will be using as much as 2500 watts.
Regarding the #8, remember that this is DC current from the turbine to the inverter. Someone else suggested using a step up transformer to boots the AC current to 240 and then step it down at the cabins in order to save on wire expense. Could I size the AC wire at 50% loss of voltage and avoid having to buy a second transformer to step the 240V current back down to 110V? Yes, I am talking about AWG wire 
If you do the stepup/stepdown design the transformers have to be well regulated to prevent lights dimming under varying loads.
Without doing the calculations, the Return On Investment for this arrangement might take 50 years. You post the proposed design and parts list, we critique. . . 
Outback, the company I bought our inverter from, sells a transformer.
See here: (http://store.altenergystore.com/Encl...rwFan/p4089/) If I bought 2 of them, one for stepping up and one for stepping the current back down, I would be out about $900. Then I would have to by the wire, but the transformers could cut the wire size in half, no? At 1200 plus feet to provide a maximum of 2500 watts I am getting the impression that the cost of wire will be very high. I would hire an electrician, but this is in the jungle in Central America, and the electrician I hired has given me some real bad advice thus far. For this project he suggested #6 AWG. It seems to me that this size wire would be no where near the right size. 
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You may want to go higher than 240v depending on the initial costs and breakeven point for the electricity you save, but you lose the safety benefit of skin resistance at voltages above 600v [the skin punctures]. The wire's I^2 R loss will tell you how much watts you are wasting. #10 awg is 1 ohm/1000' so if it carries 30A it's losing, round trip, 900x2= 1800 watts, so at 10 cents/kwh you'd spend $1576 in one year to heat the wire. 
Swami,
First of all you need to install the inverter as close as possible to the DC circuit of the turbine. What is the DC voltage of your turbine? If it is 24 volts then your 3000 watts from the turbine would be 125 amps DC. You would need a MEGA gauge wire to transmit 125 amps of DC current 1000 feet! Convert it to 120 VAC and then you stand a chance of having something left at the end of the line! Would be better to use a step up transformer to 240 AC but it would put a constant load on the inverter at all times. (your idle current would be quite large) You would need to step down at each cabin and you might have a chance. Need to get a pro involved on a project when you run cable this great a distance. 
Quote:
You may be misunderstanding me. The inverter is 1000' from the turbine because the turbine is down in the river. I am not having any problem brining up current and from the turbine to the inverter. It's a 48 volt inverter and the turbine generates 500 continuous watts. By the time the power reaches the inverter it has lost 17% of those watts, given the size of the present wire. If I convert the wire run to #4 awg wire I will expect to loose only about 5% of the watts generated. So once I do that I will be fine in terms of bringing DC current to the inverter. So the only issue I am concerned with is bringing AC current 1200', where I will need 2500 watts of power. 
Swami,
now I am really confused. If your turbine is generating 500 watts then unless your are using some type of storage battery bank then you only have 500 watts to deal with. You charging up batteries at the inverter site and then powering the inverter via 48VDC? If so then if you run #8 wire for the 1000 ft distance you will have a 14% voltage drop at 2500 watts@120 volts. WAY too much. You would have to have #0 gauge wire to get you in the ~2% drop that is the recommended max. per NEC. Check out this link to help you calculate your needs. http://www.stealth316.com/2wireresistance.htm 
napper; for a 120 volt circuit at 20 amps (that would be 2.4kW), and 1000 feet, would result in more like a 26% voltage drop with #8.
He also needs to run 1200 feet from his inverter to the builidings. At 20 amp, he would need more like 250 mcm to keep the voltage drop to 2%. actually, he does not need to limit is that tight. 5% would be no problem and depending on what he is running, up to 10% may be acceptable. A #3 would give just under 10% voltage drop for those parameters. 
Sorry. Yes, we are taking the 500 watts from the river and storing it in batteries. It's a 48V system. I am loosing too many watts from the river to the inverter, but it's working for the kind of loads I have at present. Later I will address the loss and increase the wire size from the turbine to the inverter.
On the AC size wI will be running lights (compact florescent), and fans in each cabin. At the central kitchen lights fan and an occasional blender and other such kitchen supplies. Nap, do you think the 10% loss is acceptable ad safe for these needs? 
10% is generally acceptable for all loads. As a matter of fact, our power comapnies are required to maintain voltage within that 10%. NEMA requires that electrical appliances be able to tolerate 10% variance.
The problem I see, well, one of the problems, 2500 watts is only a 20 amps at 120 volts, total. You are really going to have to figure out what each device uses and determine if this is even a realistic installation. If everybody used power equally at the same time, that would allow less than 4 amps per cabin. Can't do very much on 4 amps. as to the circuit from the river. Calculations show you are losing about 25% of your power. From 500 watts the drop puts it at about 370 watts but if you are charging batteries, that may be all you need. What I would do is hunt around on the internet for sites that deal with "green power", photovoltaics, solar power, or any other self produced power for a home. They have already engineered through what you are trying to do right now. On top of everything else, money is a limiting factor for everybody I know. What I would do is emulate what a power company (POCO) does which is produce high voltage for transmission and the transforming it down to usable at the other end. If you could do a 48 volt to 480 and then 480 to 120 transformation, you could reduce the wire size a great deal. Somwhere down to around a #4 wire would then be large enough to carry the load but don;t get too set on any of these numbers yet. You are desiging a system and getting set on any one fact at this point becomes very limiting. Wire that is rated for 600 volts or less is quite typical. Once you breach that rating, it starts getting very expensive and harder to deal with. That is why I would limit the voltages to under 600. 480 is a very common voltage so much equipment is avaiable for this range or voltages. I still don;t see enough power to run much of anything on the other end though regardless how you get it there. 
Here to help!
OK Swami
I feel your pain. Not being able to find qualified people in the middle of nowhere. I guess, only because you need help and I'm a nice guy, fly me down there, and I'll help you put your system in for $30 bucks an hour cash. I'm just here to help! 
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You will have the 2500 watts. What I am saying is 2500 watts at 120 volts will result in approximately 20 amps current. Since you have 5 or 6 buildings to feed, that would split up to around 4 amps per builidng IF everybody was using power at once.
what I said is to calculate what the actual current draw is of what you are going to run. Quote:
2500 watts is not a lot of power when you are running a motor or any sort of heating unit. 
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