small LENZ 2 with bicylce hub dynamo

[marquee moon]

Hello,

I'm interested in building a small LENZ2 VAWT.

I have a 6v/3w bicycle hub dynamo (not a rim dynamo). Ben at Gotwind.org suggested that due to my urban, turbulent wind, a small HAWT (as described on his website) may not work efficiently, if at all.   I started looking at other possibilities to use my  dynamo hub. I've read that VAWT's are better at accepting wind from rapidly changing dirrections, and whilst they're most efficient in clean air, can handle some turbulence.  So I'm going to give the LENZ2 a go. I'd like to post my calculations, and ask for comment on them. Please be critical (I'll live to regret saying that...)

Firstly, I don't have ( nor can I find) an output vs RPM rating for this dynamo. I've assumed that this hub should be optimised for around 13mph (my normal speed) on a touring bicycle (700mm diameter wheel). So, first things first > converting this speed into RPM at the hub:

13 mph =  20.8 km/h = 20,800 m/h

a 700mm wheel = 2.2m circumference, so for each revolution, 2.2m is travelled.

(20800/ 2.2)/ 60 minutes = 160 RPM

However, Ben from Gotwind.org suggests on this page "The main feature of the dynohub was that it reached it's rated output at only 12 mph (with a 26" wheel) this equates to a mere 60 rpm". Ben was talking about the Sturmey Archer dynohub in this statement, but I'd think the design spec would be similar.  

I'm going to split the difference and say 110 rpm is an optimum rotational speed.

Basing my wind speed on the UK's wind speed database (at 10m above ground level), I get 6 mph.

I'll assume a reduction to 5mph nearer ground level (say, 3 - 4 m agl).

To calculate the diameter:

By re-aranging Windstuffnow Ed's formulas,  I find that 5 feet (1.5 m) is an adequate circumference:

Windspeed x 88/ circumference (ft) x 0.8= RPM

(5 x 88)/ (5 x 0. = 110RPM

This equates to (150/pi) 47.7 cm diameter, or 23.8 cm radius

Now to calculate the size, based on the wattage. The dynamo is rated at 6v, 3w.

Another of Ed's useful formula:

Watts= (0.0508 x windspeed^3 x (efficiency of blades x efficiency of motor) x area (ft^2)

Readingh a few cycle forums, this dynamo has an efficiency of 0.59. No idea about the blades, but I've taken a value off the windstuffnow website of 0.4

3. w= (0.0508 x 5^3 x (0.4 x 0.59) x A
   
3. = (0.0508 x 125 x 0.236) x A
   
3. = 1.5A
   
3. /1.5 =A
   
   

A = 2 ft^2

Area = 1800 cm2

Height = 1800/ 47.7 = 37.7 cm

rounding up the measurements....

Diameter = 50 cm

Height = 40 cm

Does this sound OK?  Am I on the right track?

Am I correct in thinking that if I slightly increase the hight (say to equal the diameter),  this will increase the torque, resulting in a greater likelihood of the turbine starting at lower windspeeds (ie overcoming the inertia caused by the cogging) and also resulting in a net increase in power output? Advisable or not?

I intend to make a centrally mounted turbine, with the dynamo sitting in the center. One of the spoke mounting rings will have a plywood disk mounted to it, onto which will be screwed the blade arms. The bolt which normally attaches to the bikes' forks will be attached to a steel plate on a pole.

Sorry for long post, but I'd like to get all the calculations and design clear in my head prior to starting....

[electrondady1]

perhaps ed will respond

i think it would be prudent to physically test the alternator at specific speeds.

at the relatively low rpm of a vawt it's possible

 i use a musical metronome

you can adjust the beats per minute(rpm)

then turn the alt in sync and observe the output .

when you put an electrical load on the alternator the speed of the mill will be reduced.

if it is reduced to the point were the alternator is no longer producing a usable  voltage you will need increased height.  

     

[marquee moon]

thanks, I'll give the metronome idea a go. when you use a metronome, do you hand crank the motor? (I can imagine that at higher revs this may get fun...)  

[windstuffnow]

Sounds like a fun project !  Your rpm calculations on a 5ft diameter turbine in a 5mph wind is a little off though... It would run at or around 22 rpm.  You measured diameter and it should have been circumference

5mph x 88 / ( 5 x 3.14 ) x .8 = 22.4 rpm

Finding the size of turbine for a given wind speed would be...

Area = Watts / ( .00508 x windspeed^3 x efficiency)

So...

3 watts / ( .00508 x 5mph^3 x (.4 x .59) ) =  20.01 sq ft

It might be better to go with a smaller diameter to increase the rpm's and increase the height to make up the area.  This would depend on the attatchment of the genny and the rpm's needed to achieve the 3 watts.  ( direct drive or geared ?)

.    

[marquee moon]

thank you very much for your advice, ed. I'll re-do calculations accordingly.

Its direct drive, no gears.

[gotwind2]

Hi Marquee,

The statement on my site is correct,

The Sturmey archer dynohub (all hub dynamos are similar) reach the rated voltage of 6 VOLTS at 60 rpm, double this for 12 volts, 120 rpm.

I would go for 6v battery charging personaly, so your vawt needs to spin at around 60- 80 rpm - 5 watts would be your peak power in my experience.

Good luck with the cogging - they are very bad in that respect.

Ben.

[marquee moon]

OK, I've put the calculations into a spreadsheet to play around with the different parameters. Very interesting.

By reducing the diameter, I can increase the RPM, but the trade off is that I need to increase the height to keep the total area the same. Likewise, by increasing the diameter, I can reduce the RPM, so a higher windspeed is required to attain the 60rpm mentioned by Ben above.

So it becomes a trade off- an optimization.  

To get anything worthwhile out of a 5mph wind with this dynamo hub, I'll need a turbine that's 90cm(3ft) diameter x 135cm(4.5ft) tall . At this size, at 5mph, I should get 2 watts (4v) at 58rpm

For such a large turbine, it hardly seems worth it, because without a voltage doubler, I wouldn't even be able to charge a 6v battery in average wind speeds (I'd only be charging on the windy days)

If I'm learning nothing else, its that the speed of the wind has a huge impact on the amount of energy that can be extracted. At 5mp with the above dimensions, I'd get 4v, but at 6mph I'd get 7v, and at 7mph I'd get over 11v

I see that there's no easy way around low wind speed.