How to calculate if wind through a funnel can open a check valve?

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In summary, the conversation discusses the use of a rotating funnel to capture wind and compress it through a check valve into a soil column. The equation for calculating the pressure at the check valve is provided, but it is noted that this is not affected by any geometric factors. The idea of using a funnel to capture more air is questioned and it is suggested to do research on check valve pressure drop. The concept of the anemometer is also mentioned as a similar idea.
  • #1
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Imagine a wind speed of 'W' (variable), rotating funnel of 'X' diameter that narrows and then conveys flow through a 'Y' diameter, 'Z' length pipe until it hits a 'Y' diameter check valve. What is the minimum cracking pressure needed of the check valve, provided the variables above?
Imagine a wind speed of 'W' (variable), rotating funnel of 'X' diameter that narrows and then conveys flow through a 'Y' diameter, 'Z' length pipe until it hits a 'Y' diameter check valve. What is the minimum cracking pressure needed of the check valve, provided the variables above?

I am hoping to understand the equation that can be used to understand if/how the captured wind could be compressed via a funnel type apparatus against the check valve, and exceed the cracking pressure thus allowing flow to occur.
 
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  • #2
What is a rotating funnel? Cracking Pressure? What is the fluid ? Speed regime? Sizes? This depends very much on scale.
 
  • #3
If the air at the inlet has pressure ##p##, its velocity Mach number is ##M##, and the stagnation pressure - i.e. when the flow is completely stopped by the check valve - is ##p_t##, then the pressure at the check valve is:
$$p_t = p \left(1+\frac{M^2}{5}\right)^{3.5}$$
This is regardless of any geometric factor. (If the airflow hits the bottom of a bucket, it will stop no matter what.)

If there is flow going through the check valve then it gets more complicated - since the flow is not fully stopped - and the static pressure (that keeps the check valve opened) will be lower according to the flow velocity.
 
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  • #4
hutchphd said:
What is a rotating funnel? Cracking Pressure? What is the fluid ? Speed regime? Sizes? This depends very much on scale.
The funnel (lets say inlet diameter 18", 20 degree reducing to 2" outlet) spins freely to capture the prevailing wind (fluid is air, wind speeds of 10 - 20 mph). the outlet connects to a one-way check-valve with low (i.e., 0.5 psi) cracking or opening pressure. The question is can the low pressure wind be funneled towards the check valve with enough velocity/pressure to open it?
 
  • #5
jack action said:
If the air at the inlet has pressure ##p##, its velocity Mach number is ##M##, and the stagnation pressure - i.e. when the flow is completely stopped by the check valve - is ##p_t##, then the pressure at the check valve is:
$$p_t = p \left(1+\frac{M^2}{5}\right)^{3.5}$$
This is regardless of any geometric factor. (If the airflow hits the bottom of a bucket, it will stop no matter what.)

If there is flow going through the check valve then it gets more complicated - since the flow is not fully stopped - and the static pressure (that keeps the check valve opened) will be lower according to the flow velocity.

Hmmm
So having a funnel shape to "capture" more air has no effect? Feels weird. My intuition suggests to me that if the entry of the bucket is bigger than its end (cone-shaped bucket) then the geometry should have some impact but it wouldn't be the first time intuition doesn't take me all the way to the right answer.

Mach number ##M## only changes with the velocity at the entry and the speed of sound in the gas and I guess those are independent of the geometry of the bucket. But it's so weird!

The wiki link shows the derivation for stagnation pressure in the incompressible case (Bernoulli) but for the compressible case, it only tells where the formula comes from. Do you recommend some source to read the derivation? Of course, assuming a closed valve to keep things simple as you said.

I gotta hit the bed but I'll keep my eyes on the thread.
 
  • #6
mw11 said:
The funnel (lets say inlet diameter 18", 20 degree reducing to 2" outlet) spins freely to capture the prevailing wind (fluid is air, wind speeds of 10 - 20 mph).
Wait, you mean like an anemometer?

1691012417716.png

https://www.thoughtco.com/history-of-the-anemometer-1991222
 
  • #7
The concept of the anemometer is similar; however, put a funnel on the one end, add a hollow stem, and check valve. If it helps, the objective is to inject air (wind assisted, very low flow) through a check valve (trying to figure out design parameters) and into the soil column.
 
  • #8
mw11 said:
If it helps, the objective is to inject air (wind assisted, very low flow) through a check valve (trying to figure out design parameters) and into the soil column.
A sketch of your proposed device would help a lot. Thanks. :smile:
 
  • #9
And pending your sketch, it's sounding like you just want to make a wind-powered air pump. There are simpler and much more efficient ways to do that...
 
  • #10
You very well may be right... however, this is my current scenario.. quick sketch attached!
 

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  • #11
Before you get too enamored with your idea, I suggest spending some time using search terms check valve pressure drop.

mw11 said:
to understand if/how the captured wind could be compressed via a funnel type apparatus against the check valve, and exceed the cracking pressure thus allowing flow to occur.
Stagnation pressure is stagnation pressure, and no funnel device will increase it.
 
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  • #12
mw11 said:
If it helps, the objective is to inject air (wind assisted, very low flow) through a check valve (trying to figure out design parameters) and into the soil column.
Sorry if I'm a bit slow on this, but why do you want to inject air into the dirt?
 
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  • #13
Why do you need a check valve?
If the wind does not blow, there will be no flow.

The air pressure on the soil will be similarly increased upwind of a tree or a wall. Wind pressures are normally extremely low, the flap valve will be very delicate.

Code:
  Speed    Ram
   kph     psi
    10    0.0007
    20    0.0027
    30    0.0062
    40    0.0110
    50    0.0171
    60    0.0247
    70    0.0336
    80    0.0439
    90    0.0555
   100    0.0686
 
  • #14
Juanda said:
Do you recommend some source to read the derivation?
The starting point is actually the derivation of stagnation temperature. The stagnation pressure can then be found with the isentropic relations for an ideal gas.

Basically, the energy contained in the moving fluid is the sum of the fluid enthalpy and its kinetic energy. If the fluid speed is reduced to zero, then the enthalpy must increase accordingly.

Juanda said:
So having a funnel shape to "capture" more air has no effect? Feels weird.
What will happen is that the funnel will basically be filled with air (at stagnation pressure), and the incoming will just spill over.

If you imagine your theory right - The bigger the funnel entrance, the bigger the pressure - it would mean that at some point the pressure would be so high that it would want to exit the funnel, against the wind. So what would happen? The funnel would empty itself until total vacuum is achieved? That wouldn't make sense.
 
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  • #15
To be fair to your intuition:

The funnel size does matter once gas starts to flow. for a given gas flow, a larger funnel/system will operate nearer the stagnation pressure.
 
  • #16
Dullard said:
To be fair to your intuition:

The funnel size does matter once gas starts to flow. for a given gas flow, a larger funnel/system will operate nearer the stagnation pressure.
To be fair, it is not how large a funnel/system is that will determine this, but the inlet/outlet areas ratio.
 
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  • #17
True. If you ignore 'for a given flow.'
 
  • #18
berkeman said:
Sorry if I'm a bit slow on this, but why do you want to inject air into the dirt?

Bueller? Bueller?

1691095785447.png

https://www.aveleyman.com/FilmCredit.aspx?FilmID=6184

Okay, does anybody else know of a good reason to inject air into dirt? Inquiring minds want to know...
 
  • #20
mw11 said:
Haha. The purpose would be to stimulate/enhance aerobic microorganisms that break down contaminants

Interesting. How do you guide the ##O_2## throughout the soil? Would it be more efficient to just water the soil with oxygenated ##H_2O##?
 
  • #21
mw11 said:
Haha. The purpose would be to stimulate/enhance aerobic microorganisms that break down contaminants
There are farmers who plough their land, and inject the diesel engine exhaust deep into the soil during the process. The CO2, H2O and diesel particulate soot do not cause air pollution, and may actually do some good. There is also excess oxygen in the diesel exhaust, along with some NOx and sulphur from cheap fuel.
http://www.theconsciousfarmer.com/exhaust-emissions/
 
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Related to How to calculate if wind through a funnel can open a check valve?

How do you determine the wind speed required to open a check valve?

The wind speed required to open a check valve can be determined by calculating the force exerted by the wind on the valve and comparing it to the cracking pressure of the valve. This involves using the Bernoulli equation to find the dynamic pressure of the wind and then converting this pressure into a force that acts on the valve area.

What role does the funnel shape play in opening a check valve?

The shape of the funnel affects the velocity and pressure of the wind as it passes through. A converging funnel can increase the wind speed and pressure at the exit, potentially making it easier to open the check valve. The specific geometry of the funnel will influence how much the wind speed is amplified.

How do you calculate the dynamic pressure of the wind through a funnel?

Dynamic pressure can be calculated using the formula \( P_d = \frac{1}{2} \rho v^2 \), where \( \rho \) is the air density and \( v \) is the wind speed. If the wind is funneled, the speed at the narrowest point can be found using the continuity equation, which states that \( A_1 v_1 = A_2 v_2 \), where \( A \) is the cross-sectional area and \( v \) is the velocity.

What is the cracking pressure of a check valve and how is it relevant?

The cracking pressure is the minimum pressure difference needed to open the valve. It is crucial in determining whether the wind pressure generated through the funnel will be sufficient to open the valve. This value is typically provided by the valve manufacturer.

How do you factor in air density when calculating wind force on a check valve?

Air density (\( \rho \)) is a key component in calculating dynamic pressure and, consequently, the force exerted by the wind. It can be affected by temperature, altitude, and humidity. Standard air density at sea level and 15°C is approximately 1.225 kg/m³. This value is used in the dynamic pressure formula to find the force exerted on the valve.

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