Calculating Energy Output of Moving Water Wheels

[TertiusS]

I have a engineering question that hopefully you guys find interesting enough to help me with. It works as follows.

I have a steadily moving column of water moving at 1.4m per second. I have two water wheels with diameter of 2000mm and 3000mm. Each wheel has 5 flat boards which measure 2000mm x 500mm. These boards are fixed to the water wheel and makes the wheel turn by being pushed by the water. I need to know how much energy preferably in kilowatt is delivered on the output shafts which is fixed to the center of each water wheels.

I hope I have explained it correctly, please ask me for more info if you need it. Thanks for everybody willing to help me with this.

 

[minger]

You don't have enough information to get a 'real' estimate. You do however have enough for a rough guess. You have a fluid, of known density, moving at a known rate. Thus, you know the kinetic energy per unit time, or the available power.

Now as far as what kind of efficiencies you'll be seeing, well that's a completely different question. You could perhaps look up some journal articles on Pelton wheels to determine some sort of empirical relation for an efficiency.

 

[russ_watters]

Not enough information: we need flow rate and pressure (height?) drop.

 

[minger]

Wouldn't the pressure drop be a function of energy extracted from the flow? In theory the maximum energy available from the flow should be the kinetic energy of the flow?

Now efficiency is a totally different monster.

 

[russ_watters]

Wouldn't the pressure drop be a function of energy extracted from the flow? In theory the maximum energy available from the flow should be the kinetic energy of the flow?

Yes: The pressure drop plus flow rate tells you the maximum amount of energy that can be extracted from the flow. So you need the existing pressure drop. Power is just flow rate times pressure drop. If there is no static pressure (such as with a wind turbine), then the only pressure available is the velocity pressure, which is tiny - but in a hydro plant you typically have oodles of static pressure.

 

[TertiusS]

Thanks for the replies. It seems more complex than I thought. Even the engineering for dummies book would be to difficult for me to grasp so hoping to just get a answer seems out of the question. However I will do what you suggested and try solve this problem for myself.

I include some drawings of the problem. Perhaps this can help you guys to give a estimate on the kilowatts

Any new readers to this thread please continue to give some input.

 

[russ_watters]

Ok, when you said "column of water", I envisioned something vertical with static pressure like a regular hydroelectric dam. Minger is right: this is like a wind turbine and there is essentially no static pressure, so you can just calculate the kinetic energy of the water to get the energy available. Then probably start with 50% efficiency for an estimate. Kinetic energy is just .5 mv^2. Can you take a shot at doing that calculation yourself?

 

[minger]

OK OK, I was seriously wondering, what pressure drop you were talking about, haha. "Column" of water was a bit of a misnomer though.

Tertius, as we mentioned before, the diagram helps in no way. You're going to have to do some research, and hopefully find some sort of empirical relation. Ideally, you'll then base your design on the relation so you have a better idea of what you're getting.

 

[TertiusS]

Thanks to you both.

I believe i am not very suited to solving this problem one reason my lack of understanding and secondly English not being my home language, I would not know what to do with the sentence "and hopefully find some sort of empirical relation. Ideally, you'll then base your design on the relation so you have a better idea of what you're getting". Thanks to Microsoft Office a can at least sort out the grammar and spelling when typing.

Essentially I want to generate electricity using a canal 3m x 1.5m deep. This canal runs for 50km through and around my town and I believe its allot of potential energy going to waist. I need to know the force I have in the water wheel so that i can match a gearbox and generator.

I can manage the Kinetic energy formula .5 mv^2 formula, although I am not sure what value to use for mass. Any idea on the mass. ? Would the kinetic energy formula still be of use.

Perhaps I should find a professor of sorts here in South Africa and ask for help in a common language. But I appreciate your help allot. Thanks Again.

 

[russ_watters]

The mass is the mass flow rate of water through the part of the canal with the water wheel in it.

 

[minger]

Empirical relation means that perhaps someone has already done the work for you. It means that for your design, there may be a chart or table that says, "For a n-bladed straight paddle water wheel, in a-meters deep, b-meters deep of water, the efficiency (or power output) is
<br /> \eta = \dot{m} a^x b^y n^z \gamma<br />
Where x,y,z are these "empirical" relations you might be able to find from charts/tables.

Now, that's just an example, a guess, but you'll have to do some research to find out.

 

[russ_watters]

Let me give a broader treatment of the issue than I provided before:

Most of the time, when I hear "hydro power", I think of a hydroelectric dam. A hydroelectric dam's average capacity is set by the flow rate of the river feeding it and the height of the dam. The calculation is so simple that it is a first-week fluid mechanics example, with several different variations. The most basic is just pressure times volumetric flow rate (or potential energy flow rate, which is essentially the same equatin). That problem can be varied to cover things like a tank with a hole in it. The size of the hole will tell you the flow rate, but you need to apply Bernoulli's equation to get it.

For a flowing river, the problem is much more difficult. The kinetic energy of the water is very small compared to the potential energy due to the height change, which is why you see dams instead of paddle-wheels in rivers. The kinetic energy is kept down by friction with the sides and bottom of the river. Nevertheless, the kinetic energy calculation isn't very difficult and hopefully the OP has tried it...

But what makes a river difficult is what happens when you put a paddle wheel in it. In a typical hydro dam, essentially all of the potential energy in the water is available to do work and the efficiency of the turbine might be 75%. But that's not the case in a river. Even if your turbine is 75% efficient, putting it into the river disturbs the flow in a way that makes a lot of the energy un-harnessable without vastly altering the calculation and the reality of what is happening. Most obviously, a turbine will create drag which will slow the flow. But the flow rate has to be constant, so what will end up happening is water will "pile up" behind the turbine until it gains enough potential energy to overcome the drag of the turbine and reach a new equilibrium - basically making it more similar to the dam. So if you want to harness more than just a tiny fraction of the water's energy, you'll need to consider altering the shape of the canal, raising the sides and creating a little bit of a dam.

Ultimately, if the dimensions of the canal are the same before and after the turbine, it isn't kinetic energy you are harnessing but potential energy. Since the flow rate is fixed and the dimensions are fixed the velocity and therefore the kinetic energy are also constant.