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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, Congress:Advisor:Kenneth M. 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 = mgh (Potential Energy in Joules)

? = 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 = mgh = 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

See also

Directory:Hydro

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