Energy available to a wind turbine

  • Thread starter pgunderson
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HI folks, new to this forum.

I'm in a Fluid Dynamics class and need to make a presentation on Bernoulli's conservation of energy equation as it applies to wind turbines.

I think I understand the different energies involved here: kinetic(v^2/2), pressure(P/rho), potential(zg)... I want to make sure I am applying the equation right and comprehending the concepts.

It seems to me that in an airstream we've basically only got KE operating for us. The velocity entering the blade assembly is fast, energy is extracted by the turbine, and the air leaving the assembly is then slower.

But someone pointed out to me that if you think about the area really close to the plane of the swept area, there is also a pressure differential.

Is this small enough to be ignored? It seems the problem I'm to use as an example doesn't consider this possibility. The only givens are: steady windspeed of 12 m/s, blade assembly of 50m diameter, and use air density of 1.25kg/m^3.

I'm to find the mech. energy of air/unit mass (ok) and the power generation potential (ok) and the actual power assuming 30% efficiency (ok).

Is my assumption correct that the only available energy is the KE of the wind's velocity? What happens with turbulence? If there's a pressure differential at some point, can that be causing the airstream to take on a shape that is not similar to a tube with cross-sectional area equivalent to the swept area of the blades? What rules can I apply to ensure that this really is a conservation problem?

Any conceptual help will be very much appreciated!

patti
 

Answers and Replies

  • #2
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Yes, your benoulis approach is correct.
there is a pressure rise near the blade (wheel).
the Z diff=0
upsteram air V > down stream air V
Work done/mass flow rate = KE1-KE2
Power = massflow rate x (V1^2 -V2^2)/2g
= 1/2g * rho * Area * Vtur *(V1^2-V2^2)
Pmax (possible) = 8/27g *rho * Area * V1^3 (optimised one after differentiation)
 

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