CRUD! I'm not on track

[Titantornado]

 . . . . . . I don't.  Turns out that I forgot to figure for the cut-in speed as I mentioned in my earlier diary entry. So I modified the program to include cut-in speed.  Now, I am WAY off and chasing numbers all over the place, and can't seem to find a match.    Hmmm, something isn't adding up.

What am I missing?  Please look at this snapshot: (sorry for picture size.  It's the best I could do and still be legible)

Note the 26.8 mph speeds match on both calculators, as well as prop dimensions.  But look at "Generator power" on the one calculator, compared to "Total input" on the other.  Not even close.  5549 vs. 2341

I bumped down to a 8 ft prop, with the same windspeed, and the results were 1812 available from the wind, and total alterantor input is now 728.

I bumped down to a 4 ft prop, with the same windspeed, and the results were 453 available from the wind, and total alternator input is now 191.

You can see what I mean by chasing the numbers. I can't get a match no matter how small I make the prop.  The alternater always shows it's too small for the prop. I have a feeling the windspeed calculations must be innaccurate on the right-hand calculator. The way it finds the speed is:

(total input/(0.00508*rotor area*blade efficiency))^(1/3)+cut in windspeed

I have no idea if that's the right formula, but that's what it goes by.

Thoughts?

[SamoaPower]

I think you are trying to compare apples and oranges.

The Warlock calculator is giving you the power available from the rotor and saying nothing about the alternator. The other one is giving you the alternator input required to give you a specified output based on certain stator parameters.

This example says that you have an alternator that is not well matched to the rotor and the big variable is the stator resistance. If you want to use more of the available power, you must reduce the stator resistance.

[scottsAI]

Hello Titantornado,

Have not seen the xls sheet, but it looks right.

Permanent magnet alternator?

If yes, the output voltage of the generator is based on RPM. With 6.9 mph cut in at 26 volts out.

At 13.8mph the output voltage is 26*2 = 52v.

at 26.8mph is 100 volts output. Stator at 0.47 ohm, the current is around 50 amps.

Stator resistance is high but not the only problem.

Looks like your dealing with 3 phase so the numbers work out differently and the fact we are talking AC to DC, the peak voltage and things mess the simple calculations up.

If your striving for the 5kw output then I would reconsider and use an induction motor.

Specially if grid connected is an option, then is overall lower cost option.

Speed up gears are needed and if done right can be very reliable.

Pitch control results in better efficiency, but can be designed without. Breezy 5.5 as example.

Have fun,

Scott.

[willib]

scott i'm afraid you are missing something..

on my blades ( 43" dia)ones

in a six mph wind the rpm was 210-220

in a 9 mph wind they were going 360-370 rpm

close to double the RPM  for a three MPH wind speed increase

[stephent]

Something seems a bit wrong...

having 48.15 amps going into "something" through the coil windings is actually

for a 3 phase-- 48.15/1.73=27.83 amps per phase

per phase loss is the old I^2 x R thing.

so 27.83^2 x .47= 364.08 watts loss per phase.

Total overall loss is simply the per phase loss times 1.73 again.

so 364.08 watts times 1.73 = 629 watts overall loss for the stator.

Doing this 364.08 loss X 3 = 1092.24 isn't correct for a 3 phase loss calculation.

Does that look better now?

Does to me.

But then again I've been wrong a few times today already....My wife told me I was.

[SparWeb]

I found that excel spreadsheet somehow a while ago and played around with it, too.  The formula you quote, about calculating the speed, is probably wrong.  The "apples and oranges" principle also applies - so you have to mind the definitions of the terms.  The problem has been approached from opposite ends of the turbine, and the results aren't consistent because they don't mean the same thing.

Warlock's calculator gives the wind speed where the rotor reaches a certain power level.  It starts with blade properties and goes on to make some conclusions about the properties the electrical generator should have.

Someone has fiddled with the alternator spreadsheet, trying to make it predict the wind speed at which the generator will produce a given current.  This is probably nonsense.  There is not enough data input into the spreadsheet to make that determination.  The only meaning that can be derived from the spreadsheet, with the information available, is the wind speed at which the rotor could capture that energy, but you must delete that silly "+cutin windspeed" first.

[ghurd]

Damn.  I have the same problems!

It's a combination of prediction from one end, and reverse enginering from the other.

Now I start with 3 or 4 working examples, a program or 2, figure a ball park split, flip a coin, and then guess.

Intuition seems to work out fairly close, close enough for my stuff (but my stuff only costs $10).

The point being, I do know what you mean, and you are not alone.  

G-

[Titantornado]

Nah Scott, I never expected 5+ Kw.  1 to 2 would be nice.  I can't imagine how I could get the resistance down lower.  Even if I wired for delta, to get the same cut-in speed, I'd have to add more turns of wire, of a smaller gauge to fit in the space alotted, and would probably end up close to the same resistance when all is said and done.

But you do bring up some curious thoughts.  I know voltage is fairly proportional to windspeed, so going by my 26v at roughly 7 mph, let me figure at 21 mph, which is 78v. (this is rectified voltage)

Using the formula, voltage at windspeed - battery voltage / effective winding resistance x battery voltage = watts out   (effective winding resistance in star, is 1.3x the phase resistance, as recommended by Flux. I call it the Flux Factor) Nando recommends to multiply star phase resistance by 1.73, so we'll try that too.

78. - 26 / (0.47*1.3) * 26 = 2212.8 watts by Flux's recommendation
    
78. - 26 / (0.47*1.73) * 26 = 1669.1 watts by Nando's recommendation
    
    

In either case, the number from the Excel program doesn't even come close.  It reports only 496.6 watts at the same 21 mph.

On to losses.  Using the formula, Amps^2 x effective resistance

85. 1*85.1*(0.47*1.3) = 4424.9 watts loss + 2212.8 watts out = 6637.7 total alternator power
    
64. 2*64.2*(0.47*1.73) = 3338.5 watts loss + 1669.1 watts out = 5007.6 total alternator power
    
    

These numbers indicate alarming losses, and an alternator that's too big for the 14 ft prop. (Warlocks calculator claims there is only 2456.8 watts available at the prop)

The Excel program says 222.9 watts loss and a total alternator power at 719.5, far too small for a 14 ft prop.

So . . . . , about the only thing I can be sure of is my cut-in RPM voltage.  The rest is still a shot in the dark.  Conflicting numbers and plenty of uncertainty.  What an adventure!

[scottsAI]

Hello willib,

Interesting.

http://www.windstuffnow.com/main/blade_design_help.htm

Reading the RPM should go up proportionately with wind speed.

6mph; what is the error as with the other numbers.

For example if it's 5.8 and 9.4 then the numbers are correct.

Or is the RPMs off a bit? Just checking!

TSR = rpm * Pi * D / 60 / V

http://www.fieldlines.com/story/2006/8/26/02131/9547

The TSR of the blade should be linear for a while.

If 6mph is on the low end of the range faster winds can improve the efficiency with higher RPM.

TSR can/will change with wind speed.

What do you think?

Have fun,

Scott.

[scottsAI]

Hello Titantornado,

Using standard wind power equation:

Dia 14' at 21mph 0.38 efficiency = 2755 watts.

The gen is 0.47 ohms, how about the line resistance?

2 volts bridge diode drop. Lets do a loop power equation:

Line/gen power + diode power + battery power = 2755.

0.47I^2 + I(2 + 26) = 2755 watts.

Solving for I = 52.4amps, 1290 gen watts, 1324 w battery.

Add in the 1.3 I = 48.0amps, or 1.73x, I = 43.7amps

Looks like your in the 1k range that you want?

Have fun,

Scott.

[willib]

"TSR can/will change with wind speed."

I agree , because the load from the alt will change as RPMs increase.

"6mph; what is the error as with the other numbers.

For example if it's 5.8 and 9.4 then the numbers are correct.

Or is the RPMs off a bit? Just checking!"

i'm sorry but i dont understand what you are asking.

maybe i'm wrong about doubling the wind speed = double the rpms?

i thought the blade rpm would more than double in twice the wind speed.?

[Flux]

Most of this confusion comes about from not understanding the load matching problem.

The Warlock calculator gives power at the peak of the power curve ( design tsr).

I can't comment on the Excel spreadsheet as I have nothing to check it against.

The formula I gave gives you power into a battery from alternator RPM. Your assumption that alternator rpm increases directly with wind speed is working on the assumption that you have mppt.

Without it you will be lucky to get a doubling in rotational speed from a 3 fold increase in wind speed, in fact if you do manage that then you are about right for good results. The consequences are that the lower full load speed will reduce available alternator output and also you will be stalling the prop so that the Warlock power figure is invalid.

You may see a Cp of .35 at cut in but at full load it will be down to something like .15.

You don't give a rotational speed at cut in so I haven't a basis to compare much.

Power loss figures that you have calculated are probably high, you will get nearer if you use actual winding resistance rather than the effective resistance for power.

The best way to calculate power seems to be to calculate the loss in one phase and multiply by 3. When you say phase resistance I find it confusing, I prefer to regard phase resistance as being the resistance of one phase (.47/2). Let's call it R.

Power per phase in a star winding is I^2R where I is the rms line current.

Total power is 3 x I^2R.  Roughly the rms current will be .77 of the dc battery current for this type of alternator.

I have a feeling that you are fairly well matched.

Flux

[thefinis]

LOL I went back and was checking the thread link and found this quote from Titantorando  and just had to repeat it now cause it seems so apt.

Waaaahhhhhhh!    If everything in life were easy, no one would appreciate their accomplishments.   ;o)

However I feel your pain. The math in trying to make blades, windspeed, and power curves match is almost impossible. One problem I ran into was that the power curves verus the rpms do not run at the same ratio and I wasn't even trying to figure in the losses the alt suffered at increased power generation. Plotting curves on a graph and then comparing them seem to work out better than any thing else I tried.

(Cutin/machine) Losses versus cutin speed should be 2 different things. Cutin speed is how much wind it takes to overcome all losses friction, clogging, inertia, etc. These losses change/vary once running but should be calculated into the efficiency percentage. One problem here is that the efficiency rating will vary with wind speed and rpm changes(losses are based on rpms and power available is based on wind speed cubed). I do not see how one can take out the cutin wind speed and still come out with the right answer unless you are trying to come up with total power made over a period of time like a week.

I spent many a cold wet day playing with figures trying to find a perfect match and then when I went and started building had to throw all of it out except the basics. Too many variables like different friction losses with different bearings, the difference between a perfect blade and the one I built and real wind conditions versus perefect laminar flow.

Finis

[Titantornado]

"I have a feeling that you are fairly well matched.

Flux"

Yea, I do too.  I think I just got too much time to play with numbers, and just get myself all discombobulated.

[scottsAI]

Hello willib,

You have it right, Normally, 2x wind speed should equal 2x RPM.

Over a range the above should be correct. The TSR of a blade has a profile, low speeds it may be lower and high speeds it may be lower. Loading on the blades will change the RPM to add further confusion.

A properly loaded blades RPM vs no load blade the RPM will change by 2x. This is why people are worried about diversion controllers to put a load on the wind gen when the battery is charged.

What I was suggesting; slight errors in measurement on the 6, 9, RPM numbers you listed could account for the RPM and wind speed numbers not being the correct ratio.

9 / 6 = 1.5, 360 / 210 = 1.71 ratio should be same.

Have fun,

Scott.

[Titantornado]

"LOL I went back and was checking the thread link and found this quote from Titantorando  and just had to repeat it now cause it seems so apt.

Waaaahhhhhhh!    If everything in life were easy, no one would appreciate their accomplishments.   ;o)"

HA! You've been savin' that one for a while, eh?  Just goes to show me what goes around, comes around!  LOL, it's OK though.  All in good fun.

Yea, I'm done crunching numbers, I'm just going forward with the original plan and we'll see how it does with practical results from real life tests.

Thanks!