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H2bidblog Global Solar, Hydro, PV & Wind Energy Consortium Hydroelectric Power Water Technologies
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Solaris project
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What kind of pipe could be used?

Dear Andrew,
all answers you could found in Russian discussions by proper links, but... :)

Obviously that we have and can get this potential water energy, the question is how?

There are some options, for example:
1. standard pressure pipe - 30 mm, 200 ATM in bottom (your variant) -> Kevlar tube/tether, expensive + 1.5 t of water
2. gravity pipe (waterfall) - variable profile by continuity equation, no pressure (my variant) -> cheap, possible big energy losing
3. aerial lift (practically also replaces the turbine) -> middle price, but complicated

Any variant principally can be used, but demands:
1. bigger balloon - 1500-2000 m3 here - to keep additional water weight
2. R&D for stabilization of flow
3. constructive investigations

Please, ask for details.

Dear Andrew,

I have been thinking about your AirHES concept, having seen your comments on:

The fog harvesting idea looks really interesting - I just can't see how the energy generation works.

The key issue seems to be the pipe. I think I understand your numbers regarding the water generation and power ratings, but I am concerned about the penstock design. For the turbine to work, the potential energy from the water needs to be converted to kinetic energy at the nozzle outlet. As you say, for a 23 kW design (1.16 l/s, 2000 m head), the flow rate through a 3 mm nozzle is 200 m/s. Your 50-100 kg mass of the water in the pipe seems consistent with extending the 3 mm diameter pipe right up to the balloon. But the friction from a 3 mm pipe would surely not allow water to travel at 200 m/s along it?

An alternative would be to use a larger pipe diameter and maintain it in closed channel flow. If you accepted 10% head loss in the pipe, then you would need a pipe of diameter 30 mm (see  data entered with 2000 m length, HDPE, 1.16 litre/s, diameter 30 mm). A full pipe of this dimension would carry water at a more reasonable 1.6 m/s, but hold a tube full of 1413 kg of water, breaking the limit on the balloon's carrying capacity.

Have I missed something here?

Best regards,
Andrew Urquhart.
Student at Loughborough University

How could a gravity pipe work?

Dear Andrew,

Imagine that you have just a waterfall. Obviously, if you allow it to break up into drops of rain, almost all of the potential energy will be spent on the frontal resistance, and there is the usual rain. But if you are able to stabilize the jet, no frontal resistance will be. For sufficiently large jet can also be neglected and the resistance of the air (or the wall) on the side of the border. Then almost all of the potential energy of the fall will be converted into the kinetic energy of the jet. Jet will be narrow and accelerate.

There are several ways to stabilize the jet. For example, if a stream flowing around and along the wire. You can also spin the flow in the pipe, to prevent him disintegrate into droplets. Unfortunately, science does not give the computational model for such free flows. Therefore, more research is needed.

This jet has no pressure. Moreover, it has no weight. Therefore, no matter how much water it was not the weight will be only on the resistance on the lateral boundary...

Dear Andrew,

I will think about your option 2 some more, but my intuitive feeling is that, if you have low pressure in the pipe near the ground, this implies that all of the potential energy from the head has been lost due to pipe ftiction.

By Bernoulli's equation, the sum of the potential, pressure and kinetic energy is a constant along the pipe. At the top of the pipe, all of the energy is potential. At the nozzle, discharging at atmospheric pressure onto the Pelton wheel, we want all the energy to be kinetic. Just before the nozzle, the velocity is low to keep the friction losses down, and at ground level there is no potential energy, so the energy is all in the form of pressure.

If the pressure and potential are both low at the ground end of the pipe, does this not mean that the kinetic energy leaving the nozzle will also be low?

Best regards,
Andrew Urquhart