Greetings, Jahangir
Kinetic energy conversion is practical to do only less efficient. The
structures required to produce the energy are larger and the forces larger.
For instance the power out in watts is K A V^3 where K is a factor based on
the type of turbine, A is the swept area in m^2 and V is the velocity in
m/s. This power formula is well documented and a K of .42 for Darrieus
turbines and .15 or so for undershot water wheels. Propeller turbines are
somewhere in between depending on whether they were specifically designed
for the flow speed.
These types of turbines are more expensive per watt than conventional hydro,
however, only they can be used in some locations.
regards,
steve gregory, M.A.Sc.,P.Eng.
Alternative Hydro Solutions Ltd.

> -----Original Message-----
> From: Mohammad Jahangir Khan [mailto:[EMAIL PROTECTED]
> Sent: Thursday, November 04, 2004 10:23 AM
> To: [EMAIL PROTECTED]
> Subject: [microhydro] Re: Newbie question: energy estimation from flow
> rat e
>
>
>
>
>
> Hi Martin
> Thans for your reply and sorry for my late  response.
>
> In fact I am looking into 'Kinetic energy conversion' and
> not 'Potential/head energy conversion'. I know that's a bit
> unconventional, but do you think that is totally
> impossible/impractical to do ?
>
> Thanks
> Jahangir.
>
>
>
>
>
>
> --- In [EMAIL PROTECTED], "Martin.Leahy"
> <[EMAIL PROTECTED]> wrote:
> >
> > Most waterpower systems require that you concentrate most of the
> practical
> > head (fall) between the entry (high) point and the exit back to
> the stream
> > (low point). i.e. one is attempting to harness the potential
> energy of the
> > water falling from high point to low point. [Harnessing the
> kinetic (speed)
> > energy is a different and less used approach]
> >
> > Power available is then:
> >
> > Power (kW) = flow (m3/s) x head (m) x g (m/s^2) x overall
> efficiency
> >
> > where g is the acceleration due to gravity.
> >
> > or say power (kW) = flow (m3/s) x head (m) x 7
> >
> > The units one buys and sells electricity are kWh (i.e. the number
> of kW
> > times the hours of operation at that power level). Cost effective
> systems
> > will usually have a 30-40% capacity factor (i.e. the power output
> in an
> > average year would be nameplate power x hours in the year x 0.4).
> >
> > You need to measure the head, and the more (financial/time) risk
> you take
> > the more confident you need to be about the accuracy of the head
> and flow
> > you measure!
> >
> > Regards,
> >
> > Martin
> >
> > -----Original Message-----
> > From: Mohammad Jahangir Khan [mailto:[EMAIL PROTECTED]
> > Sent: 01 November 2004 14:08
> > To: [EMAIL PROTECTED]
> > Subject: [microhydro] Newbie question: energy estimation from flow
> rate
> >
> >
> >
> >
> >
> > Hi
> >
> > Is it possible to measure the energy/power available from a free-
> > flowing water stream ? The flow-rate is given in m3/s.
> >
> > Thanks.
> > J. Khan
> >
> >
> >
> >
> >
> >
> >
> >
> >
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>
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