Calculating output force of wind funnel

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SUMMARY

The discussion focuses on calculating the output force of a wind funnel with a circular input radius of 600 inches and an input wind speed of 24 ft/sec. Key calculations include determining the theoretical wind speed at the output and the horsepower (HP) at the output, with 1 HP equating to 560 ft lb/sec. The conversation also addresses the impact of backpressure on airspeed and the necessary output diameter to maintain unrestricted airflow through the funnel. The optimal design for the funnel is crucial for fitting a wind turbine to effectively produce work.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with horsepower calculations
  • Knowledge of air density and weight (2 lbs/yd³)
  • Basic geometry of circular cylinders
NEXT STEPS
  • Calculate the theoretical wind speed at the output of a funnel
  • Explore the effects of backpressure on airflow in funnels
  • Investigate optimal funnel designs for wind turbines
  • Learn about the relationship between air volume and velocity in cylindrical shapes
USEFUL FOR

Engineers, physicists, and designers involved in wind energy applications, particularly those focused on optimizing wind funnel designs for turbines.

ed shaw
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Part one:
Supposed: A wind funnel. Circular Input radius = 600" Windspeed = 24 ft/sec
Air weight = 2 lbs/yd3 Circular Output radius = 60" Disregard = backpressure.
1 HP = 560 ft lb/ sec
Solve for = (1) Theoretical Wind speed at output (2) HP at output.
Part two:
Backpressure will limit air speed at output and restrict input. Using same input size as above, determine the smallest output diameter that would still allow available wind unrestricted flow through funnel.
Solve for = Output diameter.
Solve for = How did you determine Pressure and Back Pressure?

Thank you.
Ed Shaw
 
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what do you think about it yourself? any clues?
 
My drawing of this funnel shows a 12" deep flat collar. When R=600" the volume of the first 12" of the funnel is 13,564,000 in3 or 290 cu yds. Weight of air in that volume is 600 lbs of air. At about 15 mph (2 feet per second) that mass of air generates 23 HP as it moves through the funnel, at the head.
Now, as the funnel narrows, the intake air speed remaining the same, the air inside will begin to gradually speed up. Ignoring back pressure, the same volume and weight of
air must exit the small end of the funnel as is entering the large end, over the same time period. Now, there is a little mistake in the way I stated the problem here. I was trying to keep things simple, but, overlooked that air volume is a square of the radius (V=pi x r2 x Length) of a cylinder. So, my small end of 1/10 radius of the large end does not translate into 1/10 of the volume over that one foot length. So that was a mistake on my part. But, I am guessing that the air speed at the small end is proportional to the respective volumes. If the airspeed going in is 15mph, the airspeed going out of 1/10 sized (in volume) outlet should be 150 mph. I could sit down and figure the volumes.

Now, at a certain point, the pressure of the air moving in is going to be equalized by the backpressure of the air being squeezed. Just like water being poured into a funnel, the air will overflow. Unless I have made an oversight in the first part of the problem, this is where things begin to get tricky for me. The solution to the optimal funnel design is going to depend, I think, on the nature of the medium, in this case, air.
It should be clear to someone working this problem that what I am looking for is the diameter to fit a wind turbine, to produce work.
 
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