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 0.8) = 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
- w= (0.0508 x 5^3 x (0.4 x 0.59) x A
- = (0.0508 x 125 x 0.236) x A
- = 1.5A
- /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....
