Increase in air speed as it travels down a cone

In summary, the conversation revolves around the speed of air at point A in a diagram involving a venturi. Various approaches are suggested, including using the equation A1v1=A2v2 to find the speed, using professional fluid modeling software, and using calculus to find the speed based on pressure and velocity equations. The conversation also touches on the assumptions of incompressibility and constant volume, as well as the effects of pressure and temperature on the air's movement. There is also a discussion on the mathematical representation of pressure difference.
  • #1
thomas49th
655
0

Homework Statement


Hi, this isn't a homework question, it's for a physics project! Consider the following diagram:

AirCone.png



I was wondering what the speed of the air would be at the point A (orange dotted line).

The red rectangle is a packet of air of height 1m and width d. Presuming that the air is flowing from B to A, it's volume stays the doesn't is speed up? It's a venturi, right? Let's say the angle x = 30°, so the taper is 30°. I have also put some nice numbers in for the lengths.

I presume this will involve a differential equation or two (change in speed with respect to distance)? Do I want to model the air as a packet or am I going about it the wrong way?

Also I presume this falls under fluid dynamics/mechanics and we shall model air as a fluid?

Thanks
Thomas
 
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  • #2
Hi Thomas,

I don't know if you are familiar with mass flow rate, but I believe it should be conserved as the air travels through the pipe, so v1A1 = v2A2, where v1,2 and A1,2 are the velocities and areas at the start and the end of the pipe. This then turns into just a matter of finding the pipe's area at the other end - I don't believe anything complicated is required.
You can find this area by taking the sine of the angle at the edge of the hole at the far end, and multiplying it by the length of the "hypotenuse" part of the pipe, in this case
1m. Then you can subtract 2 times this from the length of the near end, again 1m, to get

v2(1-2sin(x)) = v1A1 = 100,

so v2 = 100 / (1-2sin(x))

In this case, an angle of 30 degrees would result in an invalid answer of infinity, in words A2 would be zero.

This is a very simplistic approach not taking into account the erratic nature of fluids, but for this problem it seems to be what is required. Hope this helps!
 
  • #3
This question can be as simple or as complicated as you want to make it. At the most basic level, you can assume air is incompressible and use the equation A1v1=A2v2, as NT mentioned. At the most advanced level, you can use professional fluid modeling software to model exactly how the fluid is going to behave, with all the associated turbulence, air compression, surface-air interactions, etc.

If you know calculus, you can take this problem one step beyond the flow rates approach. Density is proportional to pressure, so p=C*rho, and the pressure difference across a thin slap of air (of thickness dx, say), multiplied by area, is equal to mass*acceleration of the slap. With that you can get an equation relating pressure to velocity. With Bernoulli's equation, you can get another equation relating pressure to velocity. You can then solve for the speed of air at A.
 
  • #4
ideasrule said:
This question can be as simple or as complicated as you want to make it. At the most basic level, you can assume air is incompressible and use the equation A1v1=A2v2, as NT mentioned. At the most advanced level, you can use professional fluid modeling software to model exactly how the fluid is going to behave, with all the associated turbulence, air compression, surface-air interactions, etc.

If you know calculus, you can take this problem one step beyond the flow rates approach. Density is proportional to pressure, so p=C*rho, and the pressure difference across a thin slap of air (of thickness dx, say), multiplied by area, is equal to mass*acceleration of the slap. With that you can get an equation relating pressure to velocity. With Bernoulli's equation, you can get another equation relating pressure to velocity. You can then solve for the speed of air at A.


Thanks for the reply! I've been thinking about this oneso, are we saying

as the packet of air travels through the cone (ish) object does:
- it's area decreases, therefore it's length must increase to compensate
- volume stay the same but pressure increases?

What exactly happens. If the volume stays the same does that mean the temperature and pressure stay the same as

pv = RTAlso I've highlighted part of a sentence. When you talk of a pressure difference how can you right this mathematically? If p is pressure, is it dp/dx, because it's the pressure difference across a small gap of x.

As Pressure = Cr (where r is density), this means we can write dp/dx as

Cdr/dx,

For a single point, where pressure would remain constant
Force = pressure x area and force = mass x acceleration, so we equate
pressure x Area = mass dv/dx

r = density = mass/volume

so
C mass/volume * area = mass dv/dx
C/volume = Area dv/dx

area is a disc = piR²

volume is this disc * x (x is length)

[tex]\frac{C}{A^{2}x} = \frac{dv}{dx}[/tex]

integrate

v = C/A²x

[tex]v = \frac{C}{A^{2}} ln|x|[/tex]

Is this what you had in mind or have I gone off at a total tangent? I presume I should define x as positive in the left direction of the diagram?
 
  • #5
this now means I need to find a function of A? Right?
 
  • #6
thomas49th said:
as the packet of air travels through the cone (ish) object does:
- it's area decreases, therefore it's length must increase to compensate
- volume stay the same but pressure increases?

What exactly happens. If the volume stays the same does that mean the temperature and pressure stay the same as

pv = RT

If you assume that volume stays the same, you can use A1v1=A2v2 and forget about calculus. If you want more precision, you can assume that mass & temperature both stay the same, which is what you did later in your post. In actuality, the gas's movement is probably more accurately described as abdiatic, in which case PV^gamma = constant (instead of PV=constant in the case of isothermal expansion/compression).

C mass/volume * area = mass dv/dx
C/volume = Area dv/dx

Two problems here. First, C*mass/volume is the air pressure of the air slab, not the pressure difference across the air slab. Since P=C*rho, dP=C*d(rho), which is the pressure difference. However, you're better off leaving the left as area*dP and rewriting mass (rho*dx*A) in terms of pressure.

Second, acceleration is not dv/dx; it's vdv/dx. Do you know why?
[/QUOTE]

Is this what you had in mind or have I gone off at a total tangent? I presume I should define x as positive in the left direction of the diagram?

Yeah, this is what I had in mind.
 
  • #7
How about

[tex]P = maA[/tex]
[tex]P = FA[/tex]
[tex]P = maA[/tex]
[tex]P = m\cdot v\frac{dv}{dx} \cdot A[/tex]

[tex]C\rho= m\cdot v\frac{dv}{dx} \cdot A[/tex]

or have I missed what you tried to tell me in the last post :confused:.I cannot see having a chage in density, without it being with respect to something - such as volume. Could we use the chain rule to make this into change in density with respect to distance, or would this not help?As for acceleration I cannot remember the derivation. I can recall you use the chain rule though! Why don't you show me :-p

Thanks
 

FAQ: Increase in air speed as it travels down a cone

How does air speed change as it travels down a cone?

The air speed increases as it travels down a cone due to the converging shape of the cone. This shape causes the air to become compressed, resulting in an increase in speed.

Why does air speed increase in a cone?

Air speed increases in a cone because the converging shape causes the air to become compressed, which in turn increases its velocity.

Can the increase in air speed in a cone be measured?

Yes, the increase in air speed in a cone can be measured using various instruments such as an anemometer or a Pitot tube. These instruments can measure the air velocity at different points along the cone and calculate the overall increase in speed.

How does the size of the cone affect the increase in air speed?

The size of the cone can affect the increase in air speed, as a larger cone will have a greater surface area for air to travel over and therefore may result in a higher increase in speed compared to a smaller cone.

Is the increase in air speed in a cone consistent?

No, the increase in air speed in a cone is not consistent. It depends on various factors such as the shape and size of the cone, the speed of the air entering the cone, and the fluid dynamics of the air. These factors can cause variations in the increase in air speed along the cone.

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