Author Topic: Typical Stats from a rotor-style turbine  (Read 1606 times)

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juiced

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Typical Stats from a rotor-style turbine
« on: August 18, 2004, 02:55:25 PM »
Hi. I wanted to know a few basic questions.


  My first one would be; how much does it cost to build a simple volvo-wheel rotor turbine? Can i use other cars? Why vovlo?


  Secondly, how much does one generate on an average basis? In terms of usage, what could i power? Is this just a battery charger? How long to charge a battery from dead to full?


 The reason im asking like this, is because honestly my math sux, and i still have a hard time grasping certain concepts with electricity. i.e. if you say 18 mega-watt hours; i truly have no idea what that is equivalent to in electric items powered.


  Could i power a PC with one volvo-rotor mill? I would have to say the PC is on all day.

« Last Edit: August 18, 2004, 02:55:25 PM by (unknown) »

tecker

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Re: Typical Stats from a rotor-style turbine
« Reply #1 on: August 19, 2004, 03:31:07 AM »


   Instead of remaining in a quandry over this just purchase some solar panels and

batteries . Then add an alternator , mabe an auto alternator take the pulley off at the parts store devise a regulator circuit then work on the mechanical prop assembly ,

 then the power functions will begin to make sense at needed power and available power.

 Not tring to evade the question . I think you can be a faster study this way . You can run a Pc on a battery bank and an inverter is pretty much plug and play with a coulple

of marine batteries.  

« Last Edit: August 19, 2004, 03:31:07 AM by tecker »

DanB

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Re: Typical Stats from a rotor-style turbine
« Reply #2 on: August 19, 2004, 07:16:16 AM »
Hi Juiced:


"  My first one would be; how much does it cost to build a simple volvo-wheel rotor turbine? Can i use other cars? Why vovlo?"


I like them because I used to work on them... and I have access to gobs of free, or nearly free parts.  They also have nice large wheel bearings, larger than some cars.  They are also fairly common, and fairly large compared to many cars.  Their weight means they run with larger brake rotors.  Lots of newer cars are getting larger and have big brake rotors - so I think about anything you choose would work OK.  The old volvos were some of the last to have nice disk brakes and rear wheel drive, so the front strut is handy.  But you could use any wheel hub off the rear, of a front wheel drive car.  Mainly - I use them because they are available to me, our local mechanic (who used to employ me) has volvo front struts piled up like cordwood.


"Secondly, how much does one generate on an average basis?"


Depends on how well it's built - how windy it is... how turbulant the site is...how tall the tower is etc...

In a 10mph wind I hope for about 100 watts output.  On winter days up here, we can often see sustained winds around 10mph all day.. so we might expect about 2KWH/day on breezy winter days.  Windy days... perhaps a good bit more.  Calm days.. perhaps near 0.  It really mostly depends on your location and tower height.


 Tower height is very important, because electricity runs downhill... the higher your tower the faster the power will run down into your batteries.  If its high enough...  the electricity running down hill could actually spin the wind turbine on a totally calm day!

(just kidding)

tower height is very important though...


"In terms of usage, what could i power? Is this just a battery charger? How long to charge a battery from dead to full?"


For most practical purposes.. it is just a battery charger.  Folks who live off the power grid almost always live of batteries, and charging them up is the fun part!  How fast will it charge a battery... depends on how big the battery is!


Juiced... you should scrounge up a copy of windpower workshop!  Super book.. it'd answer almost every question you have.

« Last Edit: August 19, 2004, 07:16:16 AM by DanB »
If I ever figure out what's in the box then maybe I can think outside of it.

juiced

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Re: Typical Stats from a rotor-style turbine
« Reply #3 on: August 19, 2004, 11:30:58 AM »
O.k. so, using 2 12/V car batteries (just example, i know they arent the best) and having a 2KWH per day would mean:


   the batteries charge in X hours


   5 x 100W lightbulbs would last X hours


     if im starting to understand this... (not the battery question) i should be able to run the "light array" for 4 hours off 2 KWH?


      or am i still in space? lol (BTW, thanks ALOT (TONS) for the patience)


 

« Last Edit: August 19, 2004, 11:30:58 AM by juiced »

TomW

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Re: Typical Stats from a rotor-style turbine
« Reply #4 on: August 19, 2004, 12:20:26 PM »
juuiced;


Well, assuming no losses in the battery or cables and the battery capacity is such that it can absorb and discharge these levels just for the sake of simplicity. If you pump in 2KW of power you should be able to draw 2KW of power from them.


Obviously in our lossy world of reality there will be losses in each stage which will draw their share of tax on that power. 500 watts for 4 hours would be 2KW so with no losses the answer is "yes". This also should apply more or less to other loads say 50 watts for 40 hours. Lots of things affect this however such as safe discharge rates for the battery over a given time. My experience is that you can use an 80% to 85% factor to get a handle on the real amount you can get back after you put it in.


In general to make it easier to understand you can just forget about losses to get an understanding of the way it works as long as you keep in mind it may need to be derated by some amount to get exact numbers.


Bottom line, in my opinion, while trying to learn this stuff is to ignore losses, etc but keep in mind they are there and once you understand the basics you can get deeper into the intricacies of losses.


Just my opinion, subject to misinterpretation and may not be held by others.


Cheers.


TomW

« Last Edit: August 19, 2004, 12:20:26 PM by TomW »

Ungrounded Lightning Rod

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Re: Typical Stats from a rotor-style turbine
« Reply #5 on: August 19, 2004, 04:02:17 PM »
You'll probably want deep-cycle batteries.  Those come in a number of sizes (and for a serious home system you'd want to get big ones, like golf-cart / forklift batteries).  But let's assume you've gotten a couple of the car-sized ones used for travel trailers and the like.


Those come in two sizes, which I think are about 85 and 105 amp-hours respectively.  You only want to discharge them 80% max, so assuming you got two of the larger ones (and neglecting charging efficiencies) you're looking at 168 amp-hours at 13.5 V to go from minimum to full charge on the pair.  That's about 2 1/4 KW HR.


Another way to look at this is charging them 24 hours at 7 amps (about 95 watts) brings them from don't-go-lower to full.


And running a single hundred watt bulb will run them back down over that same period.  Which is why you want a bigger mill and bigger batteries if you're running a house off 'em. A "couple car batteries" - or their deep-cycle equivalent - won't power even a seriously energy-starred house for a single calm day.

« Last Edit: August 19, 2004, 04:02:17 PM by Ungrounded Lightning Rod »

juiced

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Thanks alot guys.
« Reply #6 on: August 19, 2004, 10:58:27 PM »
I think i am going to invest in some courses to really set me straight. I understand alot more now, with your help.


    I dont quite get amps yet. i alwas thought amps was "speed" of electricity and volts were "volume" (think like water).


   I wonder how many people are cracking up on my threads. oh well.

« Last Edit: August 19, 2004, 10:58:27 PM by juiced »

tecker

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Re: Thanks alot guys.
« Reply #7 on: August 20, 2004, 02:13:20 AM »


Ok the voltage is a measurement of electrical pressure and current is a amount of electron flow that follows the path or completed circuit. Ohms law  V = IxR. The potential energy (voltage)  and kinetic energy (current)  . Break it down It won't take long  to get going Your intrest level is moving you now .
« Last Edit: August 20, 2004, 02:13:20 AM by tecker »

tecker

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Re: Thanks alot guys.
« Reply #8 on: August 20, 2004, 02:34:34 AM »


  One Company that caters to people with a need to know is Lindsay publications

  most of the books are project analyzed with accessible material lists .
« Last Edit: August 20, 2004, 02:34:34 AM by tecker »

Ungrounded Lightning Rod

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Re: Thanks alot guys.
« Reply #9 on: August 20, 2004, 11:23:25 AM »
The potential energy (voltage) and kinetic energy (current).


Actually, voltage (like pressure) is "potential energy per unit of current flow across the voltage difference" and current is a measure of flow of charge from one place to another (like a flow of water from high pressure to low), not of kenetic energy or inertia.


When he's just getting started we have to be careful not to give him an answer that's ALMOST right.  That could keep him confused for longer.

« Last Edit: August 20, 2004, 11:23:25 AM by Ungrounded Lightning Rod »