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Talk:Directory:Beck Mickle Hydro Ltd. Micro Hydro Generator
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Scottish inventor, Ian Gilmartin, has invented a mini water wheel capable of supplying enough electricity to power a house from as little as an eight-inch water fall. The contraption is designed to be used in small rivers or streams.
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B.S.
On Jan. 4, 2006, New Energy Congress member, Ken Rauen wrote:
"I rarely assess a technology this way, but here goes: B___S___!!!!! If the inventor has something here, it is not hydroelectric!"
Waterturbine in Nimbin
On Jan. 4, 2006, New Energy Congress member, Mark Dansie wrote:
My sentiments as well Ken.
In Australia a company based in Nimbin (the hippy town) has been manufacturing and selling a small waterturbine for many years. It's a great unit, and many have been sold to the Pacific Islands and local alternative lifestylers. They work for years trouble free, but require a reasonable head and flow of water. The graeter the head, the greater the power. However not all of us have a water fall nearby.
Performance Calculations
On Jan. 4, 2006, Danny Miller wrote:
The scenario described is full of easily calculable factors. One cannot get more energy out than the difference in kinetic and potential energy provided.
Since the provided scenario is a "stream", I am considering the difference in velocity (kinetic energy) from the inlet and outlet to be negligible and thus neglecting it in the calculation.
2000W is 2000 joules/sec. 1 joule is the energy of raising or lowering 0.7376 lbs over 1 ft. So at 100% efficiency, you'd need to move 2212.8 lbs/sec over 8", 276.6 gal/sec. 16,596 gal/min.
Water turbines have had 80%-90% efficiency for over 100 years when used under ideal circumstances. As such it's notable that while a small, economical, low draw waterwheel may be something new, it cannot possibly produce much more power than turbines have in the past. The 70% specified is a somewhat low performer but operating on such low head may be something new.
So let's take 70% specified in this Wiki. Then we need 23,709 gpm through the turbine to generate 2KW. That's a pretty powerful stream @ 8" of head! This would fill a 50m by 25m by 2m Olympic swimming pool in 27.8 minutes. I have to note that since the speed of water in a natural stream is usually limited to a few feet per sec, the width of the device depicted is perhaps a meter, and the height of the water channel's cross sectional area must be only a small fraction of the 8" head then I don't see how such a volume could flow through a device of the width depicted. I get 89.9 cu meter/min through a 1 meter wide by 2cm high cross section (10% of head) requires 74.8 m/sec flow rate, or 167.3 mph!
Perhaps he meant the device could begin turning at only 8" of head, but achieved 2KW at a higher head which would require a lower flow rate?
Performance Calculations: Another look
On Nov. 30, 2011, Bearsrule86 wrote:
Thank you for the calculations, I had started the calculations when I found your articles.
I had similar findings, but used the 1kW per hour assumption.
Resulting in needing 10,605 gpm or 669 L/s
Task: Find gpm with a drop of 8" to make 32 kWh a day
Given:
32 kWh= 24 kWh/day @ 70% efficiency (stated on main article)
1 kWh = 3,600,000 J
1 W = 1 J/s
8" = 0.2032 m
g = 9.807 m/s^2
u = m*g*h (Potential Energy in Joules)
ρ = 1000 kg/m^3
1 day = 1440 min
1 m^3 = 264.172 gal
1 L/s = 15.85 gpm
Calculations:
32 kWh/day * (3,600,000 J/kWh) = 115,800,000 J/day
115,800,000 J/day = m*g*h = m*(9.807 m/s^2)*(0.2032m)
Solving for m results in m = 57,810,683 kg/day
Dividing by density of water (1000 kg/m^3) results in m = 57,810 m^3/day
Converting from day to min (1 day = 1440 min) results in m = 40.15 m^3/min
Converting to US gal (1 m^3 = 264.172 gal) results in m = 10,605 gpm Converting to Liters per sec (1 L/s = 15.85 gpm) results in m = 699 L/s
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