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Homework Help: Fluid Dynamics-Drag Force Problem.

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider the flow of a fluid with viscosity μ through a circular pipe. The velocity profile in the pipe is given as U(r)=Umax(1-rn/Rn), where Umax is the maximum flow velocity which occurs at the center line; r is the radial distance from the centerline; and U(r) is the flow velocity at any position r. Develop a relation for the drag force exerted in the pipe wall by the fluid in the flow direction per unit length of pipe.
    Problem Image:

    2. Relevant equations
    Shear stress: τ=μ(du/dy) where μ=Viscosity coefficient and (du/dy)=Velocity gradient. Units are (N/m2)
    Shear Force: F=τA=μ(du/dy)A; where A=Area of contact. Units are in N.

    3. The attempt at a solution

    I can't seem to grasp the idea of drag force for this problem, this problem is from Cengel and Cimbala's Fluid Mechanics 1st Edition. Problem number is 2-44. The problem I'm having is that no where in the theoretical part of the chapter do they mention the relationship between drag force and shear force, are they the same? The most the chapter mentions is the following: "The force a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity". The solution manual also states that du/dr is negative, but I can't understand why.
    Solution Manual image: SolutionUmax.PNG
  2. jcsd
  3. Dec 27, 2015 #2


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    You are given that the velocity profile U(r) = Umax(1-rn/Rn), where r = 0 at the center of the pipe.

    What does dU/dr represent in terms of this velocity profile?
    What is its physical significance?
    What happens to the velocity of the flow as r increases so that r → R?
  4. Dec 28, 2015 #3
    The drag force and the shear force are synonymous for this problem. Regarding the sign of dU/dr, did you try differentiating the equation they gave you for U with respect to r? If so, please show us your work.

  5. Dec 28, 2015 #4

    The velocity is reduced to 0, because the plates are fixed and the no slip condition allows us to assume the velocity the fluid has at R is equal to 0.

    I read the solution manual again in hopes of understanding the issue. So it says that du/dr=-du/dy because y=R-r, where y would be the distance from the top plate to a given distance. And when we differentiate y=R-r we obtain -dy=dr and can substitute that in the original. But it still seems to elude me why it says that in pipe flow du/dr is negative, because in the shear stress equation they use du/dr instead of -du/dy. Are they the same?
  6. Dec 28, 2015 #5
    Yes, I differentiated in respect to r. The problem is that the sear stress equation has an added - sign at the beginning when the original equation doesn't, the differential result is the same for both cases, except for the sign. I just can't picture why du/dr is negative in pipe flow? Isn't du/dr going in the same direction as the flow? Or does it mean that it's negative because from the centerline to R the parabola decreases?
  7. Dec 28, 2015 #6


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    For the time being, forget the shear stress. That's not helping you to understand the significance of what du/dr being negative means.

    What is the definition of du/dr? How would you describe this quantity in words to someone who has no knowledge of differential calculus?

    You still should answer these questions:
  8. Dec 28, 2015 #7
    The axial velocity at the center of the pipe is a maximum, and it decreases to zero at the wall of the pipe. So the derivative of the axial velocity with respect to radial position is negative.

  9. Dec 28, 2015 #8
    I did respond them, but the stayed inside the quotation marks.
    What does dU/dr represent in terms of this velocity profile?
    It is the rate of change that the velocity experiments as the distance from the centerline changes too.
    What is its physical significance?
    It allows us to see the shape of the fluid, which is a parabola, its maximum velocity is achieved at the center line because of the no-slip condition at both plates where the fluid assumes the velocity of the plates 0.
    @Chestermiller Has answered this question, stating that because the maximum is at the centerline and it decreases as it reaches R, it is a negative rate of change, meaning the velocity decreases as r reaches R. And it is also assumed for this problem that the drag force is equal to the shear force, because it is the only force which acts against the walls of the tube. I guess my doubts are solved, at least to the point where the steps seem logical now. I'll do some more problems and see if I grasped the idea right. Is there any resource you would recommend I read on the topic of shear force and viscocity?
  10. Dec 28, 2015 #9
    Transport Phenomena by Bird, Stewart, and Lightfoot is a gem.
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