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Homework Help: Fluid problem -- Piston moving in an oil-filled cylinder

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A piston 10 is coaxially in diameter moves within a cylinder whose inner diameter is 10.006 in;
    The annular space between piston and cylinder is filled with oil having a kinematic viscosity is 0.004 feet ^ 2 / s and specific gravity of 0.85

    If the piston makes its way with a speed of 30 ft / min and the length of the piston within the cylinder is 10 feet, calculate the shear force and the friction opposing the motion of the piston

    2. Relevant equations

    (tangencial force) / area = (dynamic viscosity) (velocity/space betwteen the two surfaces)

    3. The attempt at a solution
    I can know the dynamic viscosity of the equation kinematic viscosity = dynamic viscosity/density and the density it can be known from the specific gravit.

    So if I draw a Diagram of piston, I have two tangent forces and the weight. But I know that (tangencial force) / area = (dynamic viscosity) (velocity/space betwteen the two surfaces) and the area of contact it can be calculated from 2(pi)(radio) (Lenght)???
    Last edited by a moderator: Mar 1, 2015
  2. jcsd
  3. Mar 1, 2015 #2
    It looks like you're on the right track, but where are there two tangent forces acting on the piston? I see only one.

    What do you get for the dynamic viscosity?

    What do you get for the distance between the two surfaces?

    What do you get for the velocity/space between the two surfaces (in reciprocal seconds)?

    What do you get for the tangential shear stress (i.e., (tangential force) / area)?

    What do you get for the tangential force?

    I don't understand what they mean by "friction opposing the motion of the piston"

  4. Mar 1, 2015 #3
    With you answer I realized I don't have understand where act the tangent force, It acts in the area of the piston like a circle (bases of the piston) ????
  5. Mar 1, 2015 #4
    No. It acts on the cylindrical surface of the piston. That's why you use 2πrL.

  6. Mar 1, 2015 #5
    Why it's only one?? Because there are many tangent forces acting in the contour of the cylinder?? or do you refer to the resultant??
    Last edited: Mar 1, 2015
  7. Mar 1, 2015 #6
    I calculated the dynamic viscosity it gives me 3.16 g / (cm -s). So Should I apply directly the equation
    (tangencial force) / area = (dynamic viscosity) (velocity/space betwteen the two surfaces)

    Because the area=2pi(R)(L)= 2pi (25.4 cm / 2) (304.8 cm)
    velocity=15.24 cm/s
    space between= 25.4152 cm - 25.4 =0.01520 cm
  8. Mar 1, 2015 #7
    Yes. Just the resultant.
  9. Mar 1, 2015 #8
    Ok, thank you!
  10. Mar 1, 2015 #9
    The space between is half of that. You took the difference between the diameters, and it should be the clearance, which is the difference between the radii. The velocity gradient in the gap should be 2000 reciprocal seconds.

  11. Mar 3, 2015 #10
    You are right! Thank you!!
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