Wind, Narrowing Cones and Temperature

[roger kryless]

Imagine that i have a source of _constant_ wind. It travels, with negligible variance, at 50 mph. The problem is that at the accessible point, the source of the wind has an opening of only 1 meter across. (Basically, at this point, imagine a small wind tunnel).

The question is, how much energy can i generate from this source of wind (lets assume that the efficiency of any wind turbine i use is 30%). The standard equations can be used for this, i believe. But beyond this is where i get confused:


(1) imagine that i put a cone on the wind source, to reduce it to 0.5 meters across. Thereby increasing wind speed. How do i calculate the wind velocity change? Is this is a good thing to do / in theory/practice?
- will this increase the density of the air (do i need to factor that in)
- will this increase the temperature?


(2) imagine that the source of the wind was 10 degrees (F) hotter than the surrounding air. Does that help or hurt the performance? How can i use that advantageously? It may wind up that hotter air will hurt performance?

thanks for any advice.
rk

 

[boneh3ad]

I am not 100% sure what you are talking about earlier on with generating energy from your source of wind. However, I can give you some insight into the other problems.

In general, compressibility of air is not important until you get to about Mach 0.3. Since you are absolutely nowhere near that, you will not have to worry about density changes and you really won't have to worry about temperature changes either as a result of the contraction. It is really a pretty simple algebra problem. I don't know how familiar you are with fluid mechanics, but it is simply the conservation of mass. The mass going into your "cone" has to equal the mass coming out. It basically boils down to conservation of mass:

$\rho_{in} A_{in} V_{in} = \rho_{out} A_{out} V_{out}$

Where$\rho$ is the density, A is the duct area and V is the fluid velocity. The subscripts just mean the inlet or outlet of the cone. Since there is no compressibility in this case, you know that $\rho_{in}=\rho_{out}$ so you are left with the simple equation:

$ V_{out}=V_{in} \frac{A_{in}}{A_{out}}$

That should pretty much cover your question labeled (1).

For question (2), that depends on how you define performance. What are you trying to do with this air? In general, a 10 degree difference isn't going to do anything to your fan performance, so the only concern is your end goal, which you haven't told us about.