On 2 Feb I wrote:
> In the second case, the turbine impedes the flow of water and so less energy
> is
> lost as friction or needed for kinetic energy. However the water must flow
> for
> the turbine to work and so some of the potential energy is still consumed as
> friction or converted to kinetic energy. In the best case the turbine is able
> to extract up to two thirds of the potential energy.
As written, this is not correct and I humbly apologize to the group for possibly
adding to the confusion rather than helping to alleviate it.
Typically in micro-hydro applications the objective is to extract as much power
as possible from the water stream. Under these conditions the "two thirds"
property is correct. However if a lesser amount of power is required the flow
rate can be reduced (with consequent reduction in frictional losses and kinetic
energy) so the proportion of energy available rises above the two thirds value.
The values for the proportion available to the turbine for different flow rates
are:
flow rate as %age of open flow proportion of potential energy
available to the turbine
57.7% 67% (this provides
maximum power)
40% 84%
20% 96%
The situation resembles the discharge of a lead acid battery - the slower the
rate of discharge, the greater is the total amount of energy that can be
recovered.
So in theory, if maximum energy is to be extracted from the loss of potential as
a fixed amount of water falls through a fixed height then use as low a flow rate
as possible. However, if there is adequate continuous flow then maximum power
(thus maximum energy over a given interval of time) is obtained at 57.7% of open
flow, even though one third is lost as friction and unrecoverable kinetic
energy.
In practise though it is not quite this simple. As the flow rate drops any
turbine will be operating at such a low part flow value that the reduced
efficiency of the turbine will hardly compensate for the increased proportion of
available potential energy. Also in my experience, the flow rate through a
penstock seldom exceeds 25% of its open flow value. So in considering the total
picture of converting the power of flowing water to delivered electricity,
penstock losses are usually of minor concern.
Dominic Read replied with:
> Following your best case that we could extract 2/3 of this energy for
> electrical power then each litre of water with a head
> of 10m could produce 980 Joules.
>
> So a flow of 1 litre per second with a 10 meter head could theoretically
> produce nearly 1000 Watts.
>
> I wonder how close to that ideal any micro-hydro project has ever come?
I haven't followed your calculations in detail, but at some point you are out by
a factor of 10. The potential energy of 1 litre of water at 10m head is 98
joules. If this falls at a rate of 1 litre/sec the hydraulic power is 98 watts.
The water-to-wire efficiency of micro-hydro projects varies greatly, however for
systems delivering less than 1kW it usually lies between 35% and 60%.
Frank Leslie replied with:
> Is "two thirds limit to the power " related to the Betz Limit of 59.3% of
> available power in the wind power domain?
No, it is quite unrelated. The "two thirds" limit relates to the proportion of
hydraulic power available to a turbine and says nothing about limits to which
the turbine is able to convert the hydraulic power to mechanical (shaft) power.
Over the past two or three years there have been some insteresting postings to
this group regarding turbine (and also waterwheel) efficiency limits. I would
think that the theoretical efficiency limit for the Gorlov helical water current
turbine would come conceptually closest to the Betz limit for wind turbines.
Regards,
Max Enfield
Planetary Power
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