elliottr1 wrote:Heron - Thanks for the formula - looks good. No doubt when looking at electrical power supply, this formula could be extended to include loss of power through transfer distance and transformation etc.
For an example I am thinking of a 10 square Km lagoon with an average tidal range of 5 metres, and a head of 2.5 metres running for a 12 hour period per day.
How would you see the flow rate being calculated for this example. Hope this gives a pointer for you?
I am looking for the units of measurement of Max Power to be in MW in order to perform a comparison with other systems.
Have a good dinner.
Heron wrote:head*flow*eff? doesnt give J/s?
This should do it with out looking in my ref.
Power = npghQ
Where n = turbine efficiency
Q= flow rate per sec m^3/s
pgh = potential energy per unit volume.
Note h = head in meters.
p = 10^3 kgm^-3 for water.
g= 9.81 m/s^2
And pay close attention to the units.
This will only predict max power. You'll want a design that allows for maximum rate of flow. Since power will be dependent on the flow rate.
elliottr1 wrote:OK got that and it works well. Now to extend the problem slightly.
What would the impact on max power potential if I was to divide the lagoon into 2 sections and allow the water to flow through the first into the second and then in reverse on the ebb tide? Would it be possible to generate more than the max power that could be generated if the lagoon was not divided. If not then what are the limiting factors?
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