Calculating Wind Speed for Suspended Cylinder

[Saitama]

Homework Statement

I don't have the exact problem statement but I remember the data given.

I have a cylinder suspended in mid-air.The question asks to calculate the wind speed required to keep the cylinder in this position. The cylinder is of height h and diameter d. The pressure of air is P and temperature is T. Molar mass of air M and density of cylinder is $\rho$. Assume that the wind gust that keeps the cylinder suspended is blowing straight upwards, and that the air molecules bounce off the cylinder elastically.

Homework Equations

The Attempt at a Solution

I think I have to equate the lift force due to air with the weight of cylinder. The lift force is given by $\frac{1}{2}\rho' v^2A$ where $\rho'$ is density of air and A is the projected (or effective?) area.

Here, $A=hd$ and $\rho'=\frac{PM}{RT}$. The weight of cylinder is $\frac{(\pi d^2 h)\rho g}{4}$. Equating them
\frac{1}{2}\frac{PM}{RT} v^2 hd=\frac{(\pi d^2 h)\rho g}{4}

Solving for v gives the wrong answer.

Any help is appreciated. Thanks!

 

[CWatters]

The Attempt at a Solution

I think I have to equate the lift force due to air with the weight of cylinder. The lift force is given by $\frac{1}{2}\rho' v^2A$ where $\rho'$ is density of air and A is the projected (or effective?) area.

Not my field but I think there should be a drag coefficient in there.

 

[Saitama]

Not my field but I think there should be a drag coefficient in there.

I don't think the lift force formula would work here. I guess I will have to apply the Bernoulli equation here.

Let $P_1$ be the pressure below the cylinder and $P_2$ above the cylinder. From bernoulli equation,
P_1+\frac{1}{2}\rho' v_1^2=P_2+\rho' gd+\frac{1}{2}\rho' v_2^2
where $v_1$ and $v_2$ are the speed of wind below and above the cylinder.

I can calculate the pressure difference from the above equation but I feel that there's something wrong in the above equation. I have to calculate speed of wind and I have two variables $v_1$ and $v_2$. I need a few hints.

 

[voko]

I think you need to find out how much momentum the molecules that strike the cylinder due to the wind must carry so as to cancel out the weight of the cylinder.

This should be very similar to how the pressure of gas is found given its temperature (or the root mean square velocity of its molecules).

 

[Saitama]

I think you need to find out how much momentum the molecules that strike the cylinder due to the wind must carry so as to cancel out the weight of the cylinder.

This should be very similar to how the pressure of gas is found given its temperature (or the root mean square velocity of its molecules).

I am a bit confused. When an air molecule collide with the curved surface of cylinder, won't it bounce at an angle?

 

[voko]

I am a bit confused. When an air molecule collide with the curved surface of cylinder, won't it bounce at an angle?

I thought that the cylinder had its flat sides parallel to the ground.

But it is not really much different if it is the other way around, just find out how a ball would collide with a circular wall.

 

[Saitama]

I thought that the cylinder had its flat sides parallel to the ground.

But it is not really much different if it is the other way around, just find out how a ball would collide with a circular wall.

Please see the attachment. Is it correct?

 

[voko]

Correct. You just need to integrate over the lower half.

 

[Chestermiller]

Not my field but I think there should be a drag coefficient in there.

CWatters is correct. You need to google Drag Coefficient Cylinder. This will give you the drag coefficient for air flow past a cylinder, and will enable you to calculate the upward force exerted by the upward flowing air on the cylinder. To do this, you will first need to calculate the Reynolds number, which involves the air velocity, density, and viscosity, and the diameter of the cylinder. From this you can determine the drag coefficient from the characteristic graph. Calculations like this are really simple.

chet

 

[haruspex]

CWatters is correct. You need to google Drag Coefficient Cylinder.

Based on the assumption stated in the OP, I would say voko's approach is right.
Pranav-Arora, if you change the 1/2 on the left of your equation to 2/3 do you get the right answer?

 

[Chestermiller]

Based on the assumption stated in the OP, I would say voko's approach is right.
Pranav-Arora, if you change the 1/2 on the left of your equation to 2/3 do you get the right answer?

It would be interesting to see how the answers compare using the two different approaches.

Chet

 

[Saitama]

I am really sorry for the late reply. I had a test so I couldn't pay attention to this problem. Sorry.

Based on the assumption stated in the OP, I would say voko's approach is right.
Pranav-Arora, if you change the 1/2 on the left of your equation to 2/3 do you get the right answer?

Nope, that doesn't give the right answer. I think that's because in the original problem statement, it was not the cylinder. It was a shark which was to be considered as a cylinder of given dimensions. I posted what was necessary.

Let the mass striking the cylinder be $dm$. Initial momentum $p_i=dm v$ in vertical direction. Final momentum in vertical direction $p_{fv}=dmv\cos(2\theta)$ and in horizontal direction, $p_{fh}=dm v\sin(2\theta)$.

Change in momentum in vertical direction, $dp_v=-2dmv\sin^2\theta$ and for horizontal direction $dp_h=dmv\sin(2\theta)$. I think the force due to change in horizontal momentum will be zero due to symmetry.

The problem is how to express dm in terms of other variables?

Thank you for your time.

 

[voko]

How much air collides with the cylinder over $dt$?

 

[Saitama]

How much air collides with the cylinder over $dt$?

$\rho' v(dt) h(d/2)\cos \alpha d\alpha$?

 

[voko]

What is $\alpha$?

 

[Saitama]

What is $\alpha$?

I meant $\theta$. Sorry.

 

[voko]

If it is the angle with the horizontal, then at the very bottom of the cylinder you have dm = 0, which can't be right.

 

[Saitama]

If it is the angle with the horizontal, then at the very bottom of the cylinder you have dm = 0, which can't be right.

Then what should be the expression for dm? Can you give me a few hints?

 

[voko]

Then what should be the expression for dm? Can you give me a few hints?

I would rather suggest that you explain how you derived the formula for $dm$.

 

[Saitama]

I would rather suggest that you explain how you derived the formula for $dm$.

Deriving that was more of a hit and trial method.

I took a slice of cylinder of area $h(d/2) d\theta$. Since the air molecules strike vertically, the effective area where they collide is $h(d/2) cos\theta d\theta$. Mass is density times volume. Calculating volume here was a bit of trouble for me. I had the area, so multiplying it by $v(dt)$ looked appropriate. So I get the volume to be $h(d/2)v(dt) cos\theta d\theta$. Now dm can be calculated. I hope I explained my point.

 

[voko]

All of that is correct - almost, because if $\theta$ is the angle with the horizontal, then the effective area at the very bottom of the cylinder is zero. And that is where one should expect maximum action from the air. So you need one small modification.

 

[Saitama]

All of that is correct - almost, because if $\theta$ is the angle with the horizontal, then the effective area at the very bottom of the cylinder is zero. And that is where one should expect maximum action from the air. So you need one small modification.

Should it be $\sin\theta$ instead of $\cos\theta$?

 

[voko]

Should it be $\sin\theta$ instead of $\cos\theta$?

Yep.

Or measure the angle from the vertical.

 

[Saitama]

Yep.

The force due to $dm$ mass is
$$dF=\frac{dp_v}{dt}=\frac{-2dmv\sin^2\theta}{dt}=-hdv^2 \sin^2\theta cos\theta d\theta$$
(See post#12 for $dp_v$).

Looks good?

 

[voko]

Hmm. You diagram in #12 has the angle measured from the horizontal, and the incoming air is also horizontal. The problem says the incoming air is vertical, and you clearly regard that direction as vertical, because that is how you compute the diff in the momentum. If that is so, then I should withdraw my previous remarks - you should have $\cos \theta$ in the effective area just like you did originally, even though the discrepancy between the sketch and the description is unfortunate.

In #12, I am not sure I understand how you got the minus sign. Other than that. I think your formula is correct.

 

[haruspex]

Nope, that doesn't give the right answer.

I went through it again, and now I'm getting 4/3, not 2/3.

Let the mass striking the cylinder be $dm$. Initial momentum $p_i=dm v$ in vertical direction. Final momentum in vertical direction $p_{fv}=dmv\cos(2\theta)$ and in horizontal direction, $p_{fh}=dm v\sin(2\theta)$.

Your diagram seems to be on its side. So we agree you want $\int _{-\frac \pi 2}^{\frac \pi 2} dmv(1+\cos(2\theta))$, where dm/dt = ρ hr cos(θ)vdθ?

 

[Saitama]

Hmm. You diagram in #12 has the angle measured from the horizontal, and the incoming air is also horizontal. The problem says the incoming air is vertical, and you clearly regard that direction as vertical, because that is how you compute the diff in the momentum. If that is so, then I should withdraw my previous remarks - you should have $\cos \theta$ in the effective area just like you did originally, even though the discrepancy between the sketch and the description is unfortunate.

In #12, I am not sure I understand how you got the minus sign. Other than that. I think your formula is correct.

I am extremely sorry for misleading words. I tried my best to write post #12 in a proper way but I was already expecting this type of problem to occur. Sorry.

And my formula is wrong. I forgot to take into account the vector nature of momentum. Hence $dp_v=dmv(1+\cos2\theta)$.

Thanks for the help and your time voko, I have reached the correct answer.

The correct answer is:
v=\sqrt{\frac{3g\rho \pi dRT}{16PM}}

I went through it again, and now I'm getting 4/3, not 2/3.

Yes, it does give the right answer. Did you search for the drag coefficient?

Your diagram seems to be on its side. So we agree you want $\int _{-\frac \pi 2}^{\frac \pi 2} dmv(1+\cos(2\theta))$, where dm/dt = ρ hr cos(θ)vdθ?

Yes. :)

 

[haruspex]

Yes, it does give the right answer. Did you search for the drag coefficient?

No, I just treated it as molecules bouncing off the curved surface, as you did, and got the cos(θ)(1+cos(2θ)) integral.