[bobmurray1]

I keep confusing myself on this question, basically i want to know how you would determine the amount of potential energy you could capture by using a turbine, if water is flowing down a pipe into a river.

The change in elevation from the start to end of pipe is 30 meters.
The pipe is 100 meters long.
Water is flowing at a rate of 10 liters per second.

Not assuming the efficiency of the water turbine itself, is the formula just the potential energy minus the friction in the pipe? Thus:
Power = (Mass Flowrate X Gravitational constant X Change in Elevation) - Friction losses
= ((10 L/s X 0.001 m3/L X 1000 kg/m3) X (9.81 m/s2) X (30 m)) - Friction
= (2943 J/s or 2.94 kJ/s) - Friction

Also second question, I know how to calculate the friction in a pipe, but the problem i cannot get my head around is that because the pipe is nearly a meter in diameter, the pipe is not full of water, thus i dont know how to calculate Reynolds number etc? Any help here would also be appreciated,

Thanks Guys

[curiouscat]

Your basic idea sounds good.

About predicting friction in a partially full pipe: that's a bit messier but not impossible. I'll post details in a bit.

[curiouscat]

http://en.wikipedia.org/wiki/Darcy_friction_factor_formulae

Read Free Surface flow here.

Or try this:

http://books.google.co.in/books?id=hNel9Xc0NRMC&lpg=PA56&ots=aqqaHJRsaU&dq=Colebrook-White%20equation%20free%20surface%20flow&pg=PA57#v=onepage&q=Colebrook-White%20equation%20free%20surface%20flow&f=false

[Enthalpy]

To extract power from the flowing liquid, the pipe must be filled, in order to get pressure at the turbine. You don't need to calculate a free surface flow hence.

It means that the turbine (usualy its injector) must resist the flow to fill the tube. A reservoir upstream is appreciated to keep the tube full despite throughput variations.

Since the free surface flow suffices to evacuate the throughput, the filled tube will lose little pressure at that same throughput.

Check if the tube must evacuate surges as well; this implies a spillway in dams.

Also check if the tube survives the pressure. 3 bar isn't much, but how good are all joints? Beware quick-closing valves, as they create pressure spikes in long tubes.

The water must be clean enough for the turbine.

30 m make almost 3 bar or 300 kPa. Multiply by 10*10^-3 m³/s, get 3 kW water power minus losses. Or 2927W, with 9.806 m/s² and 995 kg/m³ for pure water at RT.

[curiouscat]

To extract power from the flowing liquid, the pipe must be filled, in order to get pressure at the turbine.

This didn't make sense to me at all. Are you saying fluid in a partially filled conduit won't have KE / PE? Why can't it turn a turbine then?

Maybe I'm wrong.

[Enthalpy]

I imagine - I may be wrong - that the liquid flows gently downwards, following the bottom of the pipe section all the way. That would mean that it loses its potential energy as it flows.

Water has no overpressure when arriving at the low point, if indeed it doesn't displace the atmosphere and fill the tube's section there.

If really water acquires nearly 24m/s during its trip down, it has kinetic energy, but in a form that turbines dislike. Most turbines (except the Pelton, inadequate for 30m head) need to be fully submerged to run properly, without cavitation: Francis, Kaplan.
http://en.wikipedia.org/wiki/Water_turbine
http://en.wikipedia.org/wiki/Francis_turbine
http://en.wikipedia.org/wiki/Kaplan_turbine
their blades run faster than the water, in order to make the generator smaller.

A turbine could run from a free jet yet being less slow than a Pelton, but it would be bulky and not very efficient.
http://en.wikipedia.org/wiki/Pelton_wheel

Hence the standard design of dams, where pipes are filled with water, and with inlet vanes that control the throughput at the turbine. A fast turbine (faster than the flow, desired with such a low head) is not only submerged, it also operates meters deeper than the oulet so that hydrostatic pressure avoids cavitation.