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November 1, 2005, 22:57 |
hydro power
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#1 |
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id like to produce power out of a 6.5m wide stream trouble is there isnt much fall and can only dam it to 2m heigh could anyone advise me, if thats enough to turn a 100 kw turbine
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November 2, 2005, 20:34 |
Re: hydro power
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#2 |
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Hi Owen,
interesting question, although you don't need CFD to do the calculation. First law of thermodynamics says: W_out = mdot_in(h + 1/2 V^2 + gz)_in - mdot_out(h + 1/2V^2 + gz)out With: mdot_in = mdot_out h_in = h_out V_in = V_out Efficiency = 100% W_out = mdot*gz in metric units: W_out[W] = mdot[Kg/s]*9.81[m/s^2]*2[m] W_out[W] = mdot[Kg/s]*19.6[J/Kg] So for 100[kW] you would need to pass 5000[Kg/s] through your turbine. That is about 5 m^3 of water/s. That seems like a lot for a 6.5[m] wide stream. Can you measure the current flow? Check out the web there are some resources on backyard power-set ups and such that may have some useful info. Good Luck, Bak_Flow |
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November 3, 2005, 01:14 |
Re: hydro power
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#3 |
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thanks BaK_Flow. understood 5 m3 could you explain calculation abreivations please.would daming river come into your calculations, more pressure less nozzle size to accomadate 5m3s. actualy river is 30m wide was only going to harness 6.5m, could channel more water if needed. at the moment average flow would be approx 10m3s.apprieciate your input. thanks owen.
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