Shear stress for contracting pipe

  • #1
fashara
4
0
Hi, I would like some guidance about a piping problem I have.

I would like to know how to solve this shear stress equation for a narrowing pipe. This equation gives the shear stress on the edges of a cylindrical piece of fluid having radius r and length L within the pipe. I would like to know the total shear stress for the entire pipe. I know that L will be equal to the length of the pipe but the problem is the radius is changing as one moves from left to right down the pipe. The narrowing of the pipe is smooth and hyperbolic in shape. Here is a picture...

24607639535_4a483f248e_q.jpg
hyperboloid-surface by lkjlkj lkjlkjl, on Flickr
The pipe ends are two circles, circle_1 being on the left and circle_2 being on the right, each having a radius of r_1 and r_2, respectively. (I know the picture isn't exactly right, just pretend there are perfect circles on the left and right hand sides) The fluid flows from left to right and r is decreasing from r_1 to r_2 in the direction of the fluid flow.

* Note r_1 > r_2

Intuitively, I would like to integrate the shear stress equation from r_1 to r_2 but I am not sure if this approach is valid.

The equation for the shear stress is given below:

τ=(r/2)[(-dp)/(dz)]

with r = radius of the pipe
and [(-dp)/(dz)] = change in pressure over change in distance

If I integrate in the way that I mention earlier I get:

τ = -(r^2)/4 (dp/dz)

Is this a correct approach to solving for the shear stress along the pipe?

Thanks!
 
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  • #2
Welcome to Physics Forums.

If the pipe radius tapers very gradually from entrance to exit, then you can assume that, locally, the pipe has constant cross radius. Note that, in your case, because the radius actually is varying, the axial pressure gradient will be a function of z. Since the volumetric throughput rate Q is constant, what is the relationship between the pressure gradient, the throughput rate, the local radius R(z), and the viscosity (Poiseulle equation)?

Is this a homework problem?
 
  • #3
Chestermiller said:
Welcome to Physics Forums.

If the pipe radius tapers very gradually from entrance to exit, then you can assume that, locally, the pipe has constant cross radius. Note that, in your case, because the radius actually is varying, the axial pressure gradient will be a function of z. Since the volumetric throughput rate Q is constant, what is the relationship between the pressure gradient, the throughput rate, the local radius R(z), and the viscosity (Poiseulle equation)?

Is this a homework problem?
This is not a homework problem but a thought experiment I am having due to a setup I saw in real life. In the setup there were pressure gauges a little bit to the left and little bit to the right of the pipe in question so that one could measure the pressure drop. In practice the pipe in question only varies a small amount from r_1 to r_2. Under .5 in. Maybe even closer to .25 in or smaller.

I was hoping that I would be able to approximate dp/dz constant because dr is small and the term for shear stress has a much stronger dependence on r than dp/dz. By approximating dp/dz as constant would this mean that dr should also be approximated as constant for the same reason? I do not know the relationship between the pressure gradient and the variables you mentioned above. I also did not get a chance to see what the reading on the pressure gauges were so I do not actually know dp/dz or whether it is valid to say that it is constant.

The most nagging question I have about all of this is mathematical in nature. I am wondering, let's say you do hold dp/dz constant, so that it is not a function of r. Would you integrate the expression for shear stress between r_1 and r_2 to solve for the shear stress between the two ends of the pipe?

I guess what I am basically asking is how do I find the surface area of the hyperbolic shaped pipe between the two circles?

The answer I posted above is slightly incomplete, it should say:

τ = -(r^2)/4 (dp/dz) | Evaluated from r_1 to r_2
Thank you again for any help!
 
  • #4
The key thing about this is that dp/dz is not constant. Let's take the simple example of laminar flow. For this situation, if R is constant, then
$$\frac{dp}{dz}=-\frac{16Q\eta}{\pi R^4}$$and the shear stress at the wall is:
$$\tau_w=\eta\left(\frac{4Q}{\pi R^3}\right)$$
If R is a gradually varying function of z, we simply write:
$$dp=-\frac{16Q\eta}{\pi}\frac{dz}{R^4(z)}$$
For turbulent flow, the math is more complicated, but the basic approach is the same.
 
  • #5
Chestermiller said:
The key thing about this is that dp/dz is not constant. Let's take the simple example of laminar flow. For this situation, if R is constant, then
$$\frac{dp}{dz}=-\frac{16Q\eta}{\pi R^4}$$and the shear stress at the wall is:
$$\tau_w=\eta\left(\frac{4Q}{\pi R^3}\right)$$
If R is a gradually varying function of z, we simply write:
$$dp=-\frac{16Q\eta}{\pi}\frac{dz}{R^4(z)}$$
For turbulent flow, the math is more complicated, but the basic approach is the same.
Okay, I think I know where I went wrong in my analysis. When I was thinking about this I was using the following term for dp/dz:

$$\frac{\partial p}{\partial z}=\frac{P_2-P_1}{z}$$

Where P_2 is the downstream pressure and P_1 is the upstreamHowever, this is only valid for laminar flow in a pipe of constant radius and as you said this cannot be used to approximate the dp/dz in a pipe of changing radius.

The expression:

$$\tau_w=\frac{r}{2}(\frac{-\partial p}{\partial z})$$

was also given under the assumption of a constant pipe radius with fluid in laminar flow.

I was hoping there was some operation(s) that I could use to evaluate this expression for a pipe of changing radius. Or must I use a different equation for shear stress altogether like the one you provided?

Do you know where the expression you used for wall shear stress comes from? Does it come from expressing the navier-stokes momentum equations in terms of stresses and making simplifying assumptions such as assuming only flow in the z direction? I am currently using Fluid mechanics for chemical engineers by WIlkes as my primary guide and I do not think that version of the shear stress equation is in here or if it is I am not recognizing it.

Also, do you have any resource that you can point me towards so that I may further read about this? I hate to keep bothering by asking questions.
 
Last edited:
  • #6
fashara said:
Okay, I think I know where I went wrong in my analysis. When I was thinking about this I was using the following term for dp/dz:

$$\frac{\partial p}{\partial z}=\frac{P_2-P_1}{z}$$

Where P_2 is the downstream pressure and P_1 is the upstreamHowever, this is only valid for laminar flow in a pipe of constant radius and as you said this cannot be used to approximate the dp/dz in a pipe of changing radius.

The expression:

$$\tau_w=\frac{r}{2}(\frac{-\partial p}{\partial z})$$

was also given under the assumption of a constant pipe radius with fluid in laminar flow.
This equation does not only apply to laminar flow. It also applies to turbulent flow.
I was hoping there was some operation(s) that I could use to evaluate this expression for a pipe of changing radius. Or must I use a different equation for shear stress altogether like the one you provided?

Do you know where the expression you used for wall shear stress comes from? Does it come from expressing the navier-stokes momentum equations in terms of stresses and making simplifying assumptions such as assuming only flow in the z direction? I am currently using Fluid mechanics for chemical engineers by WIlkes as my primary guide and I do not think that version of the shear stress equation is in here or if it is I am not recognizing it.

Also, do you have any resource that you can point me towards so that I may further read about this? I hate to keep bothering by asking questions.
This is very interesting. Jim Wilkes, the author of your book, was on my doctoral committee 50 years ago when I was at the University of Michigan.

If you are studying fluid mechanics, then you have been learning about the wall shear stress expressed in terms of the friction factor, and the friction factor expressed in terms of the Reynolds number. So, for turbulent flow, you should be able to express the shear stress at the wall as a function of the key parameters using f = C/Re0.8. I don't remember what the constant C is, but, if I remember correctly it is 0.79. After you make these substitutions, you will be able to integrate dz/R(z)n for turbulent flow.

If you are looking for the best book on all of this, it is still Transport Phenomena by Bird, Stewart, and Lightfoot. This is the book that I referred to more than all the others put together during my long industrial career as a ChE. A new edition was released in about 2001 that featured many of the important developments that had occurred during the previous 4 decades.

Chet
 
  • #7
Chestermiller said:
This equation does not only apply to laminar flow. It also applies to turbulent flow.

This is very interesting. Jim Wilkes, the author of your book, was on my doctoral committee 50 years ago when I was at the University of Michigan.

If you are studying fluid mechanics, then you have been learning about the wall shear stress expressed in terms of the friction factor, and the friction factor expressed in terms of the Reynolds number. So, for turbulent flow, you should be able to express the shear stress at the wall as a function of the key parameters using f = C/Re0.8. I don't remember what the constant C is, but, if I remember correctly it is 0.79. After you make these substitutions, you will be able to integrate dz/R(z)n for turbulent flow.

If you are looking for the best book on all of this, it is still Transport Phenomena by Bird, Stewart, and Lightfoot. This is the book that I referred to more than all the others put together during my long industrial career as a ChE. A new edition was released in about 2001 that featured many of the important developments that had occurred during the previous 4 decades.

Chet
Ah, thank you very much for all of your replies today it has been very helpful! I will definitely check that book out.
 

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