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Relation of cylinder wall pressure and flow inside

  1. Apr 3, 2013 #1
    What is the relation between the pressure that is exerted on the internal walls of a cylinder and the characteristics of a flow of water running inside it?

    Let's assume that the flow velocity is constant and that we have an inviscid flow of pipe flow type. Furthermore lets assume that the flow is laminar.

    From what I know, the pressure in a flow is described both by its static pressure (ρ*g*h) and its dynamic pressure (0.5*ρ*v^2).

    But how are these two pressure variables related to the pressure exerted on the internal wall of the cylinder containing the flow?

    Dynamic pressure is supposed to act only in the direction parallel to the flow direction, does this mean that the dynamic pressure does not affect the pressure exerted on the cylinder wall?

    I have been scratching my head about this problem for ages now and thought it was time to bring it to the experts.

    I hope my question is clear enough and that some bright person out there can point me in the right direction! =)

    /M
     
  2. jcsd
  3. Apr 3, 2013 #2
    You really want the fluid to be inviscid? How about you change the fluid to a newtonian fluid.
     
  4. Apr 4, 2013 #3
    Hi Starcus. Welcome to Physics Forums.

    As a guy with over 40 years of fluid mechanics experience, I cant tell you that there aren't two different pressures. There is only one pressure, "the pressure." This is also the force per unit area acting on the wall of the pipe (although it may vary with position along the pipe).

    You can get a better understanding of what is going on by focusing on specific problems, with increasing level of complexity. If you are willing to do some analysis, I can guide you through this.

    Consider a vertical pipe of constant cross sectional area, running from x = 0 at the bottom to x = L at the top. It is filled with an incompressible fluid of density ρ. The the top of the pipe is open to the atmosphere, where the gage pressure is zero. The upward velocity of the fluid in the pipe is zero. Derive an equation for how the gage pressure p varies with x, and for the pressure at the bottom of the pipe.

    If you get a solution to this that we agree on, I will give you the next more complicated version of the problem to work on. Are you game to try?

    Chet
     
  5. Apr 5, 2013 #4
    Chet, this was exactly the answer I was hoping for! That there is "only one pressure" and I would be ever thankful if you could guide me through this.

    From my limited knowledge of physics, the answer to the problem you described is the expression for static pressure p=ρgh. Where h is the height of the fluid column, g is the earth acceleration and ρ the density of the fluid. If L denotes the highest point of the pipe then (L-x) would denote the height of the water column h.

    Thus:

    The pressure p is expressed by the equation p=ρg(L-x) and the pressure at the bottom of the pipe would then simply be p=ρgL. Is this correct?

    Looking forward to your continued teaching on the subject!

    EDIT: actually, when looking at the expression I just wrote, I came up with a theory for how to determine the pressure in a moving fluid. Is it so, that the pressure of a fluid column in motion, is in fact not related to the flow itself (which depends on the fluids velocity) but rather its acceleration? Since the expression for static pressure contains the earth-acceleration, is it possible to just add the acceleration of a fluid in motion to calculate what the pressure in this moving fluid would be? Im just speculating here and I have very limited experience in physics but I thought I would have a go at it anyway. Thanks again for your help!

    /M
     
    Last edited: Apr 5, 2013
  6. Apr 5, 2013 #5
    Yes. You are definitely on the right track, and have begun to anticipate where this is headed. I'm going to take your result, and express it in a slightly different way, if that's OK.
    [tex](p+\rho g x)_{x=L}=(p+\rho g x)_{x=0}=(p+\rho g x)_{x= x}[/tex]
    This is exactly the same as your relationships if you also note that p = 0 at x = L.

    Now let's do the problem a little differently, by using a differential mass balance. Consider the parcel of liquid situated between locations x and x + Δx in the pipe. The mass of this material is [itex]\rho A\Delta x[/itex], and the force of gravity acting on this mass is [itex]\rho g A\Delta x[/itex], where A is the cross sectional area of the pipe. Also let p(x) be the pressure at location x, and p(x + Δx) be the pressure at location x + Δx. If we do a force balance on the mass, we obtain:
    [tex]p(x)A-p(x+\Delta x)A=\rho g A\Delta x[/tex]
    If we divide this equation by AΔx, and take the limit as Δx approaches zero, we obtain:
    [tex]-\frac{dp}{dx}=\rho g [/tex]
    or, equivalently,
    [tex]\frac{d(p+\rho g x)}{dx}=0 [/tex]
    The solution to this differential equation is:
    [tex]p+\rho g x= Constant [/tex]
    You can see how this equation is related to the equation [itex](p+\rho g x)_{x=L}=(p+\rho g x)_{x=0}=(p+\rho g x)_{x= x}[/itex] that we wrote earlier. Note also that it is beginning to look a little like the Bernoulli equation.

    Now for the next level of complexity in the analysis. Same setup, but this time our ideal incompressible fluid is flowing upward at a uniform velocity v. It exits into the air at the top of the pipe. Redo the differential force balance, and obtain the relationship for the pressure variation in this situation. (You should have no problem with this, since you already articulated what would happen in this situation).

    Chet
     
  7. Apr 6, 2013 #6
    Ah, this is starting to look interesting! From what I can tell, we now want to know how the pressure varies depending on its velocity, or [itex]\frac{dp}{dv}[/itex], but the change in pressure with change in velocity is already inherent in the force balance? Since it contains the acceleration a?

    F=ρgAΔx

    [itex]\frac{F}{A}[/itex]=ρgΔx

    p(g)=ρgΔx

    p(Δv)=ρgΔx

    In our case, where the velocity of the fluid is uniform, the difference in velocity would be zero. Thus:

    p(0)=0

    I think I might not have understood the complexity fully here and made it a little bit to simple for me. In that case please let me know and give me a hint of how to rethink the problem!

    /M
     
  8. Apr 6, 2013 #7
    Hi Starcus,

    Problem #2 was a trick problem. Since the fluid is not accelerating, and it is an ideal inviscid fluid (so there is no viscous drag), the pressure distribution in the pipe is exactly the same as it was in Problem #1, when the fluid velocity was zero.

    p(x) + ρgx = const.
    p(L) = 0
    p(0) = ρgL
    p(x) = ρg(L-x)

    Now for Problem #3:
    Reset the fluid velocity back to zero, but, in this case, let the cross sectional area of the pipe (i.e., the pipe radius R) vary as a function of x: A=A(x). Redo the differential force balance on a wafer of fluid contained between locations x and x + Δx, and determine, for this case, how the pressure is varying with vertical location x. Hint: When the cross sectional area is varying like this, the wall of the pipe exerts an axial force in the fluid wafer.
     
  9. Apr 11, 2013 #8
    Starcus hasn't responded to my previous post, so I guess he might have lost interest. Anyway, I thought it could be of value to continue the analysis further, for anyone else who had been following the thread.

    For Problem #3,

    Upward axial force on wafer from fluid below = [itex]p(x)A(x)[/itex]
    Downward axial force on wafer from fluid above = [itex]p(x+\Delta x)A(x+\Delta x)[/itex]

    Upward axial force from pipe sidewall on wafer = [itex]\frac{p(x)+p(x+\Delta x)}{2}(A(x+\Delta x)-A(x))[/itex]

    where [itex]\frac{p(x)+p(x+\Delta x)}{2}[/itex] is the average pressure along the edge of the wafer, and [itex](A(x+\Delta x)-A(x))[/itex] is the difference in projected area of the side wall.

    Gravitational force on wafer = [itex]\rho g\frac{A(x)+A(x+\Delta x)}{2}\Delta x[/itex]
    Therefore, the force balance on the wafer is:

    [itex]p(x)A(x)-p(x+\Delta x)A(x+\Delta x)+\frac{p(x)+p(x+\Delta x)}{2}(A(x+\Delta x)-A(x))-\rho g\frac{A(x)+A(x+\Delta x)}{2}\Delta x =0[/itex]

    Taking the limit as Δx approaches zero, we get:
    [tex]-\frac{d(pA)}{dx}+p\frac{dA}{dx}-\rho gA=0[/tex]
    But, if we differentiate the first term by parts and combine terms, we get:
    [tex]-A\frac{dp}{dx}-\rho gA=0[/tex]
    Finally, dividing this equation by A, we get:
    [tex]-\frac{dp}{dx}=\rho g[/tex]
    This is exactly the same equation we got when the cross section area of the pipe was constant. It shows that, in a vessel in which the cross sectional area varies with depth, the hydrostatic equilibrium equation is the same as for the case in which the cross sectional area does not vary.

    Now, finally, for Problem #4: SAME AS PROBLEM 3, except that the fluid is now flowing upward in the variable diameter pipe, with a specified volumetric flow rate.

    In this case, the fluid velocity v(x) is not a constant along the length of the pipe, but varies inversely with the cross sectional area (we assume here that the diameter variations with x are very gradual, so that the upward velocity can, at all locations, be approximated by a flat velocity profile). Because the axial velocity is varying with x, the fluid is accelerating, and this needs to be taken into account in the force balance on the wafer:

    mass of wafer = [itex]\rho g\frac{A(x)+A(x+\Delta x)}{2}\Delta x[/itex]
    acceleration of wafer =[itex]v\frac{dv}{dx}[/itex]

    Therefore, for Problem #4, the force balance becomes:
    [itex]p(x)A(x)-p(x+\Delta x)A(x+\Delta x)+\frac{p(x)+p(x+\Delta x)}{2}(A(x+\Delta x)-A(x))-\rho g\frac{A(x)+A(x+\Delta x)}{2}\Delta x =\rho g\frac{A(x)+A(x+\Delta x)}{2}v\frac{dv}{dx}\Delta x[/itex]
    If we divide this equation by Δx, take the limit as Δx approaches zero, and divide the resulting equation by A(x), we obtain:
    [tex]\frac{dp}{dx}+\rho g+ρv\frac{dv}{dx}=0[/tex]
    or, equivalently,
    [tex]\frac{d(p+\rho gx+\rho\frac{ v^2}{2})}{dx}=0[/tex]
    Integrating with respect to x gives:
    [tex]p+\rho gx+\rho\frac{ v^2}{2}=Const[/tex]
    This can be recognized as what we would obtain if we applied the Bernoulli equation to upward flow of an incompressible fluid in a vertical pipe of variable cross section.
     
  10. Apr 11, 2013 #9
    Hi again Chet,

    thank you very much for your detailed solutions. The reason for me not responding in a while is that the calculus is simply a little bit to advanced for me and that made me loose track. However, I understand the steps in your analysis briefly and I will try to think it through to see if I can answer my original question. Is it so that the Bernoulli equation can be used to find out what pressure is exerted upon the walls of a cylinder containing a flow?

    I also have another question regarding problem no. 4 if you have the time to answer it:

    When we say that the fluid is accelerating, we say that this is beacuse the wafers axial velocity is varying with x. But what the velocity is really doing is varying with time t right?
    But since x also varies with time t as the fluid-wafer is in motion, we have [itex]\frac{dx}{dt}[/itex] which is the same as the fluids velocity v and this is why we can express the wafers acceleration as v[itex]\frac{dv^{2}}{dx}[/itex] by applying the chain rule?

    Thanks again.
     
    Last edited: Apr 11, 2013
  11. Apr 11, 2013 #10
    Yes. That is exactly right.

    Chet
     
  12. Apr 25, 2013 #11
    Chestermiller,

    could you maybe guide me through putting up a model for the following situation:

    We have a horizontally oriented pipe of radius r and length l containing a column of fluid with density ρ. This fluid is accelerated by a piston acting on the fluid with the force F.

    How can I determine the pressure exerted on the pipe walls?

    I have been trying to solve the problem myself based on your earlier lessons but I cant make it fall into place. I think my calculus skills are simply to bad.

    Many thanks in advance for your help!

    /Starcus
     
  13. Apr 25, 2013 #12
    Solving this problem won't require any calculus. First of all, I'd like to point out that this is an unsteady state problem, so we are not going to get what we want using the Bernoulli equation. Also, since the amount of fluid in the pipe is going to be decreasing (as fluid exits downstream), I want to focus exclusively on the situation at time zero.

    What is your guess for the fluid pressure at the piston x = 0 (in terms of your parameters)? What is your guess for the fluid pressure at the exit cross section, x = l ?

    Without any further information, what is your guess for the pressure distribution between x = 0 and x = l ?

    What is the mass of fluid in the pipe at time zero?

    What is its acceleration?

    Is the acceleration the same for all parcels of fluid mass in the pipe?

    Get back with me when you have answers to these questions.
     
  14. May 5, 2013 #13
    Hi there, sorry for replying slowly again I got a burst of school assignments.

    I think the fluid pressure at points x=0 and x=l should be the same at time t=0. If the piston is applying the force F to the column of fluid, my guess for the pressure in the fluid would be [itex]p=\frac{F}{πr^{2}}+ρgh [/itex] I feel very uncertain on this problem though.

    The mass of the fluid would be [itex]m_{fluid}=πr^{2}lρ [/itex]

    The fluids acceleration would be [itex] a=\frac{F}{πr^{2}lρ} [/itex]

    Intuitively I think that the acceleration should be the same for all parcels of fluid mass.
     
  15. May 5, 2013 #14
    You said that the tube is horizontal, so there is no ρgh involved. The gage pressure at x = 0 is is just [itex]p=\frac{F}{πr^{2}}[/itex]. The pressure at x = 0 is atmospheric, so the gage pressure is zero. So you were correct to determine that the net force acting on the column of fluid was F. This gave you the acceleration. You were also correct to think that the acceleration should be the same for all parcels of the fluid mass (since the fluid is incompressible).

    The pressure varies linearly with distance, so that [itex]p=\frac{F}{πr^{2}}(1-\frac{x}{l})[/itex]
     
  16. May 6, 2013 #15
    Thank you Chet! Your help is invaluable to me.

    I have a couple follow up questions if you have the time:

    actually I have noted in similar problems that the static component does not affect the overall pressure in the fluid if the pipe is horizontal, but why is this? Gravity is still acting on the fluid? If the pipe is elevated by an angle? Does the static pressure affect the system again?

    Why does the pressure vary linearly with distance if the force and acceleration is distributed uniformely in the fluid? Is it beacuse you take friction into account?

    As you might see, I am not very well versed with fluid mechanics but I have an urge to understand these things.

    Many thanks and have a nice day!
     
  17. May 6, 2013 #16
    Gravity only works in the vertical direction. There is a pressure variation across the diameter of a horizontal pipe from the bottom to the top as a result of static pressure. This pressure difference is ρgD, but it is typically negligible, and doesn't affect the pressure variations in the horizontal direction.

    No. For an ideal (inviscid) fluid, friction is not present. To accelerate each little parcel of fluid along the axis of the pipe, the force on the left side of the parcel has to be a little higher than the force on the right side. These forces are equal to the pressure times the area. So the pressure on the left side of each little parcel has to be a little higher than on the right side. In order of this to happen, the pressure has to be decreasing along the axis of the pipe. Since the mass of the fluid is uniformly distributed along the length of the pipe, the pressure gradient has to be linear.

    There is another way of looking at all this, but I have been hesitant to present it for fear of confusing you. But, anyway, here goes. It actually is possible to analyze this problem using the Bernoulli equation, but you need to be very careful. Right now, you are looking at the problem from a stationary (inertial) frame of reference, where the fluid and piston are accelerating. It is also possible to analyze this problem using a non-inertial frame of reference, in which your frame of reference is accelerating together with the piston and the fluid along the pipe. In this case, the piston and the fluid appear to be stationary, but, because the piston and fluid are accelerating, they are experiencing the equivalent of artificial gravity in horizontal direction (negative x-direction). The acceleration associated with this artificial gravity is equal to the acceleration of the piston and cylinder. Now you can apply the Bernoulli equation to this "stationary" system (from your frame of reference), if you accept the idea that gravity is -a. So,

    p-ρax=C=F/A

    p = F/A - ρax

    So you can see that the pressure is decreasing linearly with x. At x = l, the pressure is equal to zero, so
    F=ρaAl
     
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