Externet Posted April 15, 2012 Posted April 15, 2012 The reverse of a river boat, being the river flow turning the paddle wheel of an anchored boat. How many horsepower would yield a 0.5m/s say-laminar river flow, for 1 m^2 submerged paddles ? Just a pictorial---> http://njscuba.net/zzz_artifacts/paddlewheel.jpg (not homework)
Bignose Posted April 17, 2012 Posted April 17, 2012 Not a straightforward problem. You'd need to calculate the drag on the wheel, and how much resistance the wheel would feel, and hence at what speed the drag from the fluid accelerates the wheel is balanced by the friction deceleration... then you can calculate the possible power. There may be 'rules of thumb' out there is some old design manuals or similar, but again it isn't a straightforward problem. Lots of details to hash out. In reality, I would consider just building the thing and taking measurements rather than making and justifying all the assumptions that would be needed for a reasonable calculation.
Acme Posted July 20, 2014 Posted July 20, 2014 Hydroelectric Power @ Electropedia "Run of River" Power (Kinetic Energy) ... Available Power The maximum power output from a turbine used in a run of river application is equal to the kinetic energy of the water impinging on the blades. Taking the efficiency η of the turbine and its installation into account, the maximum output power Pmax is given by Pmax =½ηρQv2 where v is the velocity of the water flow and Q is the volume of water flowing through the turbine per second. Q is given by Q = A v where A is the swept area of the turbine blades. Thus Pmax =½ηρAv3 This relationship also applies to shrouded turbines used to capture the energy of tidal flows (see below) and is directly analogous to the equation for the theoretical power generated by wind turbines. Note that the power output is proportional to the cube of the velocity of the water. Thus the power generated by one cubic metre of water flowing at one metre per second through a turbine with 100% efficiency will be 0.5 kW or slightly less taking into account the inefficacies in the system. This is only one twentieth of the power generated by the same volume flow from the dam above. To generate the same power with the same volume of water from a run of river installation the speed of the water flow should be √20 metres per second (4.5 m/sec). ...
studiot Posted July 20, 2014 Posted July 20, 2014 (edited) For any rotodynamic machinery there is an engineering trade off between speed, efficiency, power output and torque. Considerations of this lead to the the large radius of the paddle wheel, generating a large moment. Having said that suppose the vanes (this is the correct fluid dynamics word for the paddles) have wetted area a in the sream flow of velocity V and move with tangential velocity v in the direction of the stream. The energy for this velocity v is extracted from the kinetic energy of the stream by momentum exchange. The force on the plate equals the rate of change of momentum of the stream. The mass of fluid striking a stationary plate per second is m = paV where p is the density of the water. Thus the force = change of momentum per second = mV = (paV)V = paV2 Since our wheel is moving we must use the relative velocity between the vane and the stream ie (V-v) So the force on the vane is F = maV(V-v) So the work done per second in moving the vane is maV(V-v)v = (V-v)v per kg of fluid. The energy in the stream = 0.5V2 per kg of fluid. So the efficiency, e, is the work done on the vane per kg divided by the energy of the fluid per kg [math]e = \frac{{\left( {V - v} \right)v}}{{0.5{V^2}}} = \frac{{2\left( {V - v} \right)v}}{{{V^2}}}[/math] Differentiate this with respect to vanve velocity, v and set to zero to find a max [math]\frac{{de}}{{dv}} = V - 2v = 0[/math] Leading to [math]v = \frac{V}{2}[/math] Back substituting [math]\max efficiency = \frac{{2\left( {V - \frac{V}{2}} \right)\frac{V}{2}}}{{{V^2}}} = \frac{1}{2}[/math]That is an undershot paddle wheel can never be more than 50% efficient. This can be improved by shaped vanes as in the Poncelot Wheel https://www.google.co.uk/search?hl=en-GB&source=hp&q=poncelot+wheel&gbv=2&oq=poncelot+wheel&gs_l=heirloom-hp.3..0i10.953.3969.0.4437.14.14.0.0.0.0.125.1486.5j9.14.0....0...1ac.1.34.heirloom-hp..0.14.1486.MUPDmWEQ2nA Best efficiencies are achieved by running the wheel fully submerged, either in a casing as with the Pelton Wheel https://www.google.co.uk/search?q=pelton+wheel&hl=en-GB&gbv=2&oq=&gs_l= or by using a vertical axis wheel such as a Kaplan turbine. https://www.google.co.uk/search?q=kaplan+turbine&hl=en-GB&gbv=2&oq=&gs_l= Edited July 20, 2014 by studiot 3
Externet Posted July 23, 2014 Author Posted July 23, 2014 Thanks, studiot, for your knowledge and dedication to explain and share it.
studiot Posted July 24, 2014 Posted July 24, 2014 My apologies, I seem to have swopped density for mass half way through my analysis. I will put that right when I can.
Externet Posted August 3, 2014 Author Posted August 3, 2014 (edited) Found a calculator, but as 'Murphy lives with me', cannot use it with Linux. Will try another compfuser at some point. Meanwhile, it is by the last paragraph here : ----> http://www.energyalternatives.ca/content/Categories/MicroHydroInfo.asp From ----> http://www.builditsolar.com/Projects/Hydro/hydro.htm More: ---> http://www.builditsolar.com/Projects/Hydro/FlowOfRiver/FlowOfRiver.htm And floating on a tiny river: ----> Edited August 3, 2014 by Externet
Moontanman Posted August 3, 2014 Posted August 3, 2014 I am following this trying to see if i can use some of this in something I am building, a rotating algae scrubber powered by water outflow from an aquarium tank to the sump, it looks like i will have to add some flow from a separate pump to get this to work. My paddle wheel blades will have to be a small percentage of the total area of the paddle wheel due to needing area for algae to grow from what i see so far.. The paddle wheel with be approximately 12" in diameter, and 36 inches across and water flow is about 200 gallons an hour... The paddle wheel will be only slightly submerged to cut down on friction with the sump water, carry on, i am still following...
studiot Posted August 3, 2014 Posted August 3, 2014 It's still an undershot wheel and can never be more than 50% efficient in energy extraction, and then there is the generator efficiency. This one, from the list at the end of the video would be more efficient. http://www.youtube.com/watch?feature=player_embedded&v=dbxLba1EXqs
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