Increase in air speed as it travels down a cone

[thomas49th]

Homework Statement

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

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

 

[NT123]

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!

 

[ideasrule]

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.

 

[thomas49th]

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)

\frac{C}{A^{2}x} = \frac{dv}{dx}

integrate

v = C/A²x

v = \frac{C}{A^{2}} ln|x|

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?

 

[thomas49th]

this now means I need to find a function of A? Right?

 

[ideasrule]

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.

 

[thomas49th]

How about

P = maA
P = FA
P = maA
P = m\cdot v\frac{dv}{dx} \cdot A

C\rho= m\cdot v\frac{dv}{dx} \cdot A

or have I missed what you tried to tell me in the last post .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

Thanks