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Hydro turbine versus wind turbine; air versus water 2

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postTensionBeginner

Mechanical
Feb 20, 2010
6
In a water turbine most (90%-plus) of the kinetic energy contained in a water jet is converted into rotational energy of the turbine. Example, a pelton wheel water turbine that is a cup that turns around a water stream 180 degrees yields very high efficiency because all the momentum change can result in power, when the cup is moving at half the speed of water jet. see
Wind turbine on the other hand uses aerofoil and aerodynamic lift, with efficiency less than 59%, Betz Limit. See
Question: What is the property about air that it cannot be used in a pelton wheel? Is it that the randomness of air particle velocity would increase due to impact with pelton wheel and kinetic energy would be wasted in increasing randomness (heating the air), rather than imparting kinetic energy to the wheel? Is there a more scientific name for this?
 
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Well, they are both fluids. 8<)

But the density of water is 1000 times that of air. (Other that that ... and flow resistance across the surface, end tip turbulence, cavitation of entrapped air or other gasses in the fluid of pump or water turbines, speed of the water and air, and mass flows of both, a water-type turbine could be used in windmills.

But its not the right answer.

A water turbine (or pump) takes a very dense fluid trapped in a very tightly confined space with precisely configured outside ends and entrances, and then adds energy (pump) or extracts energy (turbine) from the fluid's kinetic energy.

A windmill has to "fly" trough very gusty turbulent very low pressure atmospheric flow. Same reason, consider that merchant ships' propellers at 15 knots (also open tipped, unconfined flow) are very short, broad and flat compared to airplane propellers. Gravitating boat props for 25, 45 or more are very different and don't look like pump blades at all. But they are very inefficient at low speeds. (Then again, who cares how efficient a skiboat when next to the dock or under a bridge piers? The aerodynamic flow for propellers is similar across both ships and planes only in the physics book. Not real life.

Because of end-tip and exit turbulence from the preceeding blade into the next blade, it is far, far better to have thin 3-bladed windmill propellers than closely spaced 4, 5, 6 or 28 or 280 blade windmill propellers - even though it appear that "most of the wind energy goes by between the blades".
 
A short answer could be: momentum.
 
Another way to ask the question is: If we were to design a mechanism to funnel the air from the entire rotor diameter into a focused pipe and use pelton wheel like device to impart energy from air, would the efficiency be higher than using aerodynamic blades?

About the previous post, I think high viscosity of water versus air leads to water particles "sticking together"; that is in water flow the viscosity is high so the water particles all have the same speed through out the flow from nozzle to wheel to exit. If air is used, air particles do not "stick together", velocities between layers of air changes easily. So as the air come out of the nozzle, significant energy will be lost in turbulent flows.

Is this the right mechanism?
 
anemometer.gif


A cup anemomenter looks a lot like one.

Itsmoked (Electrical),
I suggest you discharge your body before you touch things.

**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
Cup anemometer is very inefficient--very small volume of air is used to perform work. That is why it is not used for capturing energy; it is used for measurement.

 
Now we're getting somewhere.

Isn't it just the fact that water hits the wheel and is directed out of the way, where air rapidly expands, hangs around on the other side of the wheel and provides drag.



**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
To extend my previous short answer. Pelton wheel is a device that relies on impulse, that by definition is equal to the variation in momentum.
Now the inlet jet velocity is proportional to height being Vi = SQRT(2*g*H) (To be more precise one should also consider a flow coefficient which accounts for some losses).

The force F imposed by the stream on the runner equates the rate of momentum change of the fluid.

F = rho*Q*?V
Being:
rho = fluid density
Q = volume flow rate

As pointed out above density plays a big role. Moreover (from a practical point of view) the generation of fluid kinetic energy is made at expenses of the potential energy of the fluid because it is stored (natural or artificial basins) at a set height. Not easy to get the same amount of potential energy to be converted into kinetic energy, for a mass of air.
 
Thanks everyone. The discussion is leading to better definition of the problem or issue, I think.

ione the question i am asking is why is the efficiency so different between water and air. To continue with your equations:
F exerted on water turbine = 2*rho*Q*?V
Power delivered to turbine = (rho*Q*V^2)/2 = kinetic energy (per second) in the stream of water, when wheel moves at a speed of V/2; where V is the speed of water.

So the theoretical efficiency is 100% (does not matter what rho is).

Now the issue is, if I do the same analysis with air, does conservation of momentum give me large forces like shear and others(?) that do not do any useful work? That is, we know change in momentum is happening but it is not happening all because of the force from impeller; there are other significant forces that are causing momentum change. I believe with air there are significantly large forces that do not perform useful work. What are these forces?

If I were to write the as conservation of energy, would the kinetic energy in air be converted to heat energy?

I need to define the control volume that I am referring to conservation of momentum: It starts from when air leaves the nozzle to when it leaves the edge of the impeller.
 
Compressible or incompressible? This is the question. Air flowing through a nozzle to the atmosphere expands. You cannot exploit the whole energy of the gas impacting the bucket. There other turbine (reactive) which can better exploit compressible fluid.
 
Efficiency is not measured by the applied force. It is the ratio of the power extracted from the device to the power applied. Efficiency can only be 100% when there is no friction, no drag.

While the absolute viscosity of water is greater than air, the kinetic viscosity of air, the ratio of shear to inertia (shear to resistance to change in motion) is actually greater than that of water. Its a common illusion that water is "thicker than air" due to viscosity, but its entirely due to water's higher density.

So I think it is because a greater percentage of total force applied is dissipated in air movement and turbulence in relation to the force applied per unit mass.



**********************
"The problem isn't working out the equation,
its finding the answer to the real question." BigInch
 
I've never seen a Pelton wheel installed in a hydroelectric dam. The ones I have seen look more like a ship's propellor mounted in a tube, with the blades/vanes having the ability to be given variable pitch. Older hydro dams also use inflow turbines (Francis runners), and upstream wicket gates to allow throttling. Both types of turbines can be throttled with little or no loss in efficiency. Pelton turbines have poor throttling ability (low efficiency at off-design flow rates), versus the variable pitch screw turbine, with upstream wicket gates. Both types of machines can achieve peak efficiencies over 90%.

By the same token, a properly designed compressed air turbine (e.g. gas turbine engines) can have quite high efficiency, approaching 85-90% I believe. Of course, these turbines operate with very high inlet pressures relative to a wind turbine.

A wind turbine is an unfair comparison to any of the above machines. The available pressure ratio (and thus energy) from the wind is miniscule in comparison to a typical hydro dam. Higher pressure ratios mean that losses due to fluid dynamic drag become smaller overall. A more "fair" comparison might be from a simple screw turbine immersed in a flowing river.
 
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