Calculating Power in a Fluid Stream: Kinetic Energy Equation for Turbine Design

[XIX]

Hi all, I am working on designing a simple turbine for my senior design project. What I am trying to figure out is what the available power in a fluid stream is. I am thinking of the kinetic energy equation but instead of a given mass, I have mass flow rate, which gives units of Watts.

Power = 1/2 * (mass flowrate)* (velocity)^2

I don't know if this is the right equation to use. If not, what would it be?

 

[russ_watters]

Welcome to PF.

In general, you would want to use a potential energy equation because the kinetic energy of the fluid must typically be constant in order for conservation of mass to hold. In other words, the flow rate of the fluid can't decrease through the turbine unless the fluid is ceasing to exist. The only way kinetic energy could be part of it is if you have a small pipe in and a lage pipe out - then it would be partly kinetic and partly potential.

For most applications, the vast majority of the energy comes from the pressure change across the turbine. An exception would be a steam turbine where you have a change in state.

 

[minger]

That's not entirely true russ. What you're referring to is a reaction turbine (uses a difference in pressure). Pelton wheel impulse turbines have and continue to be in use across the world.

If you do the math, you'll see that if you follow Pelton's original design such that the wheel's tangential speed is 1/2 of the fluid jet velocity, the fluid will leave with very little energy left.

In a situation where you have a high fluid velocity, and perhaps little elevation to use, why not use a Pelton wheel?

 

[XIX]

I should have specified that this is in fact an impulse turbine. Where energy is extracted by way of changing the kinetic energy of the fluid. It is not a Pelton wheel but, a Boundary Layer Turbine (aka Tesla Turbine). They are relatively simple in design and construction. The principle is still the same.

The idea is that high speed fluid passes between closely spaced discs. The fluid drags the discs along with it because of the boundary layer effect.

I ask this thread so I could get an idea about how much power can be extracted if I had a 100% efficient turbine. The boundary layer types aren't known for their efficiency maybe ~12%. So I am thinking for a rough estimate of the power I could use that equation I listed before.

I see what you mean about conservation of mass. The inlet jet is a much smaller area than the discharge.

 

[minger]

I mean, don't worry about conservation of mass; the fluid can be stopped. Imagine you have a jug of water and a tube coming off the bottom such that the fluid has enough energy to make it all the way down the hall. If you put a Pelton wheel upstream, if you convert all the energy the fluid will "stop". Well, where does it go? Probably all over your floor.

I mean, the water will go down. If you're damming a river this may be something you need to consider. However, if you have some sort of experimental setup, you can always use gravity to route the "fully spent" water away from the turbine.

p.s. Yes, I am vaguely familiar with Tesla Turbines (the only parts I know is the rough workings and the terrible efficiencies.

 

[russ_watters]

That's not entirely true russ. What you're referring to is a reaction turbine (uses a difference in pressure). Pelton wheel impulse turbines have and continue to be in use across the world.

If you do the math, you'll see that if you follow Pelton's original design such that the wheel's tangential speed is 1/2 of the fluid jet velocity, the fluid will leave with very little energy left.

What creates the jet? Accoring to the wiki, there is a power plant in Germany that uses a Pelton wheel turbine and it is just a different turbine on a regular hydroelectric dam. The energy stored in the dam is potential energy and it is converted to kinetic energy by the jet. So the analysis of the energy available is still a potential energy analysis.

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

In a situation where you have a high fluid velocity, and perhaps little elevation to use, why not use a Pelton wheel?

What would create such a situation? What would it look like?

 

[russ_watters]

I should have specified that this is in fact an impulse turbine. Where energy is extracted by way of changing the kinetic energy of the fluid.

As with my post above: where is this high speed jet coming from? What does the system look like?

I can't envision a scenario where kinetic energy in a fluid someone is tryint to extract power from exists without a conversion from potential energy.

 

[minger]

Let's say he's doing this in a laboratory; he has access to a faucet. If he hooks a hose up, then he has a fast moving fluid at low pressure. Does it really make sense to have to pump up some sort of column of water to try and get a decent head on it? How high would he need to pump it? Why not just use the energy as it is?

In any hydroelectic dam, you start with kinetic energy. The "dam" part is there to convert that energy into potential energy. This is done because I will assume that a reaction type turbine has been discovered that it can be more efficient. When money is no option, you'll spend it for a few extra points of efficiency.

In any case, you can't argue that it would be more accurate to derive a power equation using the current state of the fluid, whatever it is. So, regardless of whether or not the fluid "was" at a high head, low flow state, it isn't anymore. We've had friction and the assuming it's measurable, it would be more accurate to use the current state.

Now, I won't argue that it would be helpful to get a better idea of the OPs setup...

 

[XIX]

My setup as requested goes a little something like this:

Pressurized Air tank ===> Nozzle ===> Turbine ===>Atmosphere

I am doing this in a "lab" type setting so, all conditions are controllable.

You are correct in that the kinetic energy is coming from my potential "dam" of compressed air. I am not concerned so much about efficiency because I want to see the results I get after testing this turbine. What I would like to know is if it is viable to calculate the kinetic power (not energy) in the fluid stream via the equation I listed before.

You (russ) seem to be suggesting that there is a way to measure energy inside the pressure vessel. I am guessing through bernoulli. I know the velocity of the fluid entering the turbine but I have not built it yet so I don't know what the outlet velocity will be.

 

[minger]

Well, power is really just energy per unit time, they're nearly one in the same.

The problem with using the upstream properties in this situation is that you'd then have to calculate the losses through the nozzle and through the pipe and so forth. You'd make a good choice to put pressure and flow meters just upstream of the turbine. Measure velocity and pressure, from those properties you can calculate the available internal and kinetic energy. You must account for both.

Actually, now that I'm talking, rather than a flow meter and pressure tap, you could put a hot wire in for flow measurement, and a total pressure tap. The total pressure measurement will give you your "energy", and the mass flow will allow you do get an available power.

 

[XIX]

Ah OK, that makes sense. Thank you!