Pelton Turbine Power Calculation

h_jets might have been "folded in" with h_{minor losses} or with overall η_tr.

Or a 6% discrepancy might just have been ignored.

Practically the power house is to operate under; maximum loading, part load and under variable loading conditions during various parts of a day.

Load factor of the plant is the determining factor and varies from plant to plant.

Stream flow to the reservoir varies during various seasons of the year, Head cannot be practically maintained constant.

Areas like unit speed, specific speed and methods to reduce cavitation need to be taken care of.

Hello everybody:

Thanks Tornado and mountk2 for your reply.

In brief, I submit the data for one small hydropower plant:

As you can see, the losses in the jets are of some important magnitude compared with the rest of losses.

Interesting. My earlier reply was only a guess, and not a particularly good one, seeing how large the h_jets value is in this example.

Maybe the calculation methods that omit the h_jets value are meant for other types of turbines where this feature is absent or minute?

Your original post refers to "small hydro" the data sheet you have provided says 9328 kW installed capacity based on two units of Pelton horizontal!

Overall efficiency of 85% to 90% may usually be obtained in large machines of the order you have specified.

Technically the overall efficiency reaches its peak when the ratio U/V1 is slightly less than the theoretical value of 0.5, say about 0.46.

As for the generators optimum output is stringently binding so the BHP of the prime mover is a prime concern.

As for the jets, with me it goes insignificant since the schemes are dual purpose (irrigation) accordingly from my viewpoint, standard procedures are to be followed as a general practice only.

Hi.

This is also a guess, but I would have included the head loss from the nozzles in with the headloss for the manifold, which would, of course, be different at different loads.

As for efficiency, I've seen that for <10MW peltons, generally considered small hydro in Norway, the turbine efficiency for a new turbine is usually guaranteed at around 92%. This is at least the case for the 4 small hydro plants we're working on now.

This efficiency is generally followed up during commissioning with turbine efficiency measurements for warrantying the turbine, which my company specializes in.

I'm new to hydro, but I'm pretty sure the basis for calculating efficiency in turbines is experience, measuring and modelling. They go backwards, starting with the measurements and then finding the headloss coefficient for specific nozzels and manifolds and so on.

Are you comparing your calculations to measured loss, to others' calculations of the same turbine, or recommended calculations? The 6% loss- I'm not sure where you believe you are not finding headloss for the jets.

Regards,

You might try looking for jets of a different and more efficient design if that is your only truly controllable variable.

Hello everybody:

Once again, thanks for your replies.

hooplah; yes, I am comparing my own calculations to other´s calculations of the same turbine.

My calculations of the losses can be accounted as follows:

Minor head losses in penstock = 1,81 m ------> 0,40%

Friction losses in penstock = 9,53 m ----------> 2,13%

Manifold losses = 0,20 m ----------------------> 0,04%

Nozzles losses = 26,48 m ----------------------> 5,91%

Total losses = 8,48%. This gives to me a Net Head of ~ 410 m, instead of 435,54 m calculated by others; where clearly can be seen that they have omitted the nozzles losses.