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poggott 4ft questions


By ralph5, Section Newbies
Posted on Wed Apr 16, 2008 at 02:14:25 AM MST
power output? Pendant cable size? Cut in speed?

I'm building Hugh's 4 foot turbine.

#1. I used his rpm formula from his book for cut in speed of 7mph (3 m/s).
     Rpm = wind speed x tip speed ratio x 60 / circumference
     3 x 7 x 60 / (1.2m x 3.14) = 334 Rpm
     Does this sound like the correct cut in speed?

#2. What information do I use to determine max output power using 24 volts?
     I used the information in the introduction for average wind speed of 11 mph.
     Average power = 20 W
     Energy per day = .5 kWh
     Amp-hours at 24 V = 20 Ah

     I figured 480 watts average using the formula watts = voltage x Amps
     24V x 20A= 480 watts This doesn't seem right.

     Or do i need to find out what amperage the coils will handle and make sure the
     turbine will furl at or just before that amperage? This seems better.

#3. What size of pendant cable? Using 480 watts above, i came to the conclusion
     I need 10 Ga wire for 25 foot run by using:
     Voltage drop = current in amps x resistance in ohms
     I have doubts that this is wrong. Shouldn't there be a figure for 3 phase
     somewhere? (cant find it but i remember seeing a number like .07??? on forum
     someplace.)

Please steer me in the right direction.

poggott 4ft questions | 10 comments (10 topical)

Re: poggott 4ft questions (3.00 / 0) (#1)
by Flux on Wed Apr 16, 2008 at 12:49:39 AM MST

  1.  That is near enough. The prop tsr is nearer 5 but at cut in you need it running fast.
  2.  I don't follow the second question. Maximum output is determined by the wind speed and how quick you can take measurements before it burns out or flies to pieces.
The rest of the argument seems to make no sense. Probably the 20W average at 11mph may be reasonable but I think it will be high. The 20Ah figure is ok but the power stays at 20W ( assuming constant speed).  OH! I see where you are going, that is 480Wh, that's energy not power, don't confuse them.

Yes you do need to furl at a temperature the coils can stand. If you follow the plans it should work out close. I don't know what it will survive at but you will need a lot of wind to get over 200W from a 4ft machine. It may be ok to let it peak at about 250W but I don't think you will see that very often unless you are on a very windy site. In which case I would be tempted to wind with a few less turns of thicker wire for your stronger magnets. I wouldn't expect you to be able to push the output much higher but you would have a cooler stator.

3  Your 480 W is in dreamland, as I say the power out will be about 20W at 11mph and you may be able to push it to 250 in a gale ( I believe Hugh rates it nominally at 100W).

At 24v you can use lighter cables than that, you may be a bit close to stall with cut in at 300 rpm and those strong magnets and going to heavy cables may well be a waste of money, it may reduce your output.

If you are expecting 480W at 11 mph then you are going to be very disappointed and not even me holding your hand is going to fix it.

Flux



Re: poggott 4ft questions (3.00 / 0) (#2)
by Flux on Wed Apr 16, 2008 at 01:04:03 AM MST

Just a few extra thoughts.

The 20W average is more likely to be average yearly power from an area with 11mph average wind speed. That takes in the Rayleigh distribution and gives a higher figure than the expected power at a constant 11 mph wind.

No there is no need to use any factors for the 3 phases when deciding on cable size. The current to a 3 phase bridge running without overlap is carried by a pair of wires at any instant so it is much the same as for 2 wires as far as volt drop goes. If you were looking at cable heating rather than volt drop you would need to consider the rms current and that is about .77 of the dc current, but your only concern is volt drop and in your case a bit of drop is not going to hurt you as I said earlier.

Flux

[ Parent ]



Re: piggott 4ft questions (3.00 / 0) (#3)
by ralph5 on Wed Apr 16, 2008 at 06:07:39 PM MST

OK I'm good with the cut in speed.

I don't know how i can be confused frustrated and have fun at the same time?
I will go about question 2 a different way, this will probably solve some other questions that may arise.
Keep in mind I'm new to this and I would like to build an 8 footer and not have to ask these questions again. here it goes.

Lets say I finished my 4 foot 24volt turbine just like the book said.  I bolt the thing to the top of my truck to test it out. I get going about 7 mph and hit cut in speed approx 300 rpm. Things are working fine. I decide to start going faster. Now by monitoring my voltage and amperage as i go faster; (imagine that the turbine will never fall apart)

How high can my voltage get before it needs to furl? Why?
How high can my amperage get before it needs to furl? Why?
How do i know my generator isn't approaching a critical temperature?

Or I  could just keep going faster until i have a meltdown and go from there.

Hopefully with the answers you give, i will be able to determine the minimum size of fuse i will need, the size pendant cable i will need, and make sure that the turbine wont reach a critical time before it furls. I would like to learn something here rather than exact answers only.
I hope there is enough information.

[ Parent ]



Re: piggott 4ft questions (3.00 / 0) (#4)
by ralph5 on Wed Apr 16, 2008 at 07:36:36 PM MST

update question.

The formula:

Blade power (watts) = 1/2 x air-density x swept-area x wind-speed cubed

Is this blade power in watts (energy) the same possible watts of energy that the generator can produce?

therefor watts = volts x amps, gives me the amperage possible. :)

If this is correct then my next question would be: How do i find out a reliable rpm for the rotor blades?  

[ Parent ]



Re: piggott 4ft questions (3.00 / 0) (#5)
by Flux on Thu Apr 17, 2008 at 01:21:37 AM MST

You amaze me. Most people get bogged down with theory and never produce something or else get on and build something that works and couldn't care less about how it works.

"Blade power (watts) = 1/2 x air-density x swept-area x wind-speed cubed"

Yes that formula gives you the power in the wind. You need to add an extra factor to give you the power into the alternator and that is Cp, the coefficient of performance ( effectively prop efficiency).

According to Betz the maximum power you can extract from the wind is .593 of the power in the wind. Others have worked out figures based on other than simple momentum theory but as yet the Betz figure has not been equalled so let's assume it means 100% prop efficiency. You will not reach that with any real prop and if you manage 50% of the ideal you are doing ok, so a realistic value for Cp for a small prop is about .35.

That brings your available power from the wind down to nearer .3 of the equation you gave.

Now for electrical output we need to include another factor and that is the efficiency of the alternator, rectifier and transmission. At near cut in you may see nearly 80% for a 24v system but with a conventionally loaded battery charging scheme if you attempt to hold electrical efficiency high you will pull the prop off the peak of its power curve and reduce Cp to a very low figure. By compromising electrical efficiency you can hold the prop speed up and gain more from a higher Cp than you loose on the electrical side.

This effectively means that you drop to 50% electrical efficiency but 3 times cut in wind speed. At this point Cp will likely be below .3 so you reach a state where your initial power in the wind now looks rather less impressive.

You will inly see about 15% of the lovely figure that you first calculated.

Now to your practical points. Voltage is no issue, forget it, the battery will hold the volts down ( that is a cause of the low overall efficiency), it doesn't come into the furling issue.

Current is limited by the wire in the stator and the way you can dissipate the heat. You will need to furl probably to limit current long term to about 10A for a 24v machine, short peaks over this will not hurt.

This in real life for a 4ft machine comes down to the fact that you will probably not be furling until near 30 mph and so you will only be limiting power output on a few rare exceptionally windy days and it will have negligible effect on your total power capture.

Unless you have some form of embedded temperature detector you will have no direct way to know your winding temperature. If you make the furling to Hugh's plans and you don't let it exceed 250W for long periods you should be ok ( I suspect the original version with stator on the front is well cooled).

Fuse is only to protect your cables from meltdown in the event of a fault. It had better not be chosen for any other reason. It will not give long term protection to the windings and a blown fuse will result in a runaway machine that is better avoided. a 20A fuse should be fine here.

Prop speed will be determined by how far into stall you drag the prop. It is a relatively slow machine and when reasonably loaded I doubt the speed will ever go over 600 rpm ( unless you blow that fuse and then it may reach 3 times that).

You will probably err on loading it too hard and keeping the speed too low for decent prop output so with your very short line you may need to add a bit of resistance in the line or move the magnets away from the stator to raise the cut in and lower the alternator slope ( probably a better option in a high wind area, resistance may be better in a poor wind area ).

Now you have all the benefits of Hugh's book and all his design work that went into the machine and you have just got a free book from me. Feel free to ask any more questions.

Flux


[ Parent ]



Re: Piggott 4ft questions (3.00 / 0) (#8)
by ralph5 on Thu Apr 17, 2008 at 06:43:28 PM MST

Ah yes! I'm starting to do the happy dance now.

So
.5 x air-density x swept-area x wind-speed cubed
multiply that by .35 And that gives you a realistic value for Blade power in watts.

Now the same formula
.5 x air-density x swept-area x wind-speed cubed
Multiply that by .15 and that gives you a realistic value for electrical power.
(measured on the dc side after the rectifier).

How am i doing?

For the next part.
You seem to make the point that its better to make sure that the prop is turning and doing a good job converting the wind to mechanical power, than to worry about getting all the electricity i can out of the alternator, Or i will stall the turbine.

Am i still on track?

With the information i have now,
I can approximate how many watts in electrical energy i could produce at different wind speeds.
I can approximate how many amps i could produce at different wind speeds.
I can determine which fuses would be too small.
And determine the size of wire needed.

You created another question!
You say "You will probably err on loading it too hard", no doubt i probably will.
I understand how moving the magnets away from the stator will increase cut in. I don't understand how adding resistance to the line will do the same? How is this resistance added? Smaller wire?


[ Parent ]



Re: Piggott 4ft questions (3.00 / 0) (#9)
by Flux on Fri Apr 18, 2008 at 02:04:52 AM MST

"I understand how moving the magnets away from the stator will increase cut in. I don't understand how adding resistance to the line will do the same? How is this resistance added? Smaller wire?"

No adding resistance will not reduce the cut in speed. You need to choose a suitable cut in speed for the blades. Those blades are intended to run at tsr about 5. At cut in you can let them run nearer tsr7 as that is still not far beyond the peak. The tsr will fall very rapidly beyond cut in, so starting high will give you a head start and not let you get below optimum too quickly.

Having got the ideal cut in speed you will still have the tsr falling with load because you are tied to a fixed battery voltage. The ideal load characteristic would let the voltage rise directly with wind speed.

If the alternator is powerful enough ( highly efficient) you will be forced to almost constant speed operation and the prop will stall when tsr gets down around 3.5.

Moving the magnets away from the stator will lower the emf, raise cut in speed and move you up towards the steeper part of the prop curve and you will retain a decent alternator efficiency and match better in the higher wind region. The down side is that your tsr for the lower winds will be higher and you will have a lower prop output in low winds. As the small 4 ft machine produces so little power in very light winds you will be advised not to go too low with cut in.

Now if you raise the cut in to a reasonable speed that gives you useful output in lower winds but you are still stalling in high winds then is the time to add resistance ( yes it can be a bit of thin wire or a physical resistor). This will have negligible effect on the very low currents near cut in but will lower the alternator efficiency sufficiently in high winds to let the prop get back up towards the peak of its curve.

You are right that you trade alternator efficiency to gain blade power.

If you only get very low winds I would go for the slowest cut in that gives you something worthwhile ( don't go for a few mA if it is going to hit you badly higher up the range). Then add resistance to keep away from stall in high winds.

On better wind sites I would just raise the cut in speed by moving the stator until you get out of stall. You may loose all output in the very low winds but you will hit the better winds at much better efficiency and you will more than make up for it in terms of total energy capture.

I think you are just about there now.

Flux

[ Parent ]



Re: Piggott 4ft questions (3.00 / 0) (#10)
by ralph5 on Fri Apr 18, 2008 at 03:08:49 PM MST

Yes very good, This gives me some direction if I need to fine tune it or if i just want to play with it a little bit.

I'm going to keep a hard copy of this post for future reference.  I will probably be back when it comes to charge controllers and dump loads but that will come much later.
Thank you for patience and time.

[ Parent ]



Re: piggott 4ft questions (3.00 / 0) (#7)
by finnsawyer on Thu Apr 17, 2008 at 09:01:50 AM MST

"I bolt the thing to the top of my truck to test it out."  Bad idea.  The speed of the air flow along the top of the truck will be greater than the speed of the truck.  How come no one ever seems to understand that simple fact considering everyone understands how an air foil works?  Ideally, you should mount the turbine above and to the front of the truck to avoid the deflection and speed up of the air flow as much as possible.

Since the current out of the alternator is related to the current in each coil, you should in principle be able to drive the current in an identical test coil to destruction to determine safe operating limits for the alternator current.    
GeoM
[ Parent ]



Re: poggott 4ft questions (3.00 / 0) (#6)
by Gary D on Thu Apr 17, 2008 at 08:31:10 AM MST

 Hi Ralph, your question on #2 is answered by yourself... "Average wattage- 20 watts".
 Your confusion may be the total watt hours for a 24 hour period at the theoretical 11 mph figure. Twenty watts continuous input (joules) is key I think... Someone else may (probably) can explain it better...



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