Themes > Science > Physics > Mechanics > Fluid Mechanics > Worked Examples > Laminar flow in pipes examples

7.1
The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore tube of 0.025 m bore follows the law, u = 2.5 - kr2. Where k is a constant. The flow is laminar and the velocity at the pipe surface is zero. The fluid has a coefficient of viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m3/s (b) the shearing force between the fluid and the pipe wall per metre length of pipe.
[6.14x10-4 m3/s, 8.49x10-3 N]

The velocity at distance r from the centre is given in the question:

u = 2.5 - kr2

  • Also we know: m = 0.00027 kg/ms 2r = 0.025m
    •

    We can find k from the boundary conditions:

    when r = 0.0125, u = 0.0 (boundary of the pipe)

  • 0.0 = 2.5 - k0.01252
    •
  • k = 16000
    •

    u = 2.5 - 1600 r2

  • a)
  • Following along similar lines to the derivation seen in the lecture notes, we can calculate the flow dQ through a small annulus dr:
    •

    b)

    The shear force is given by F = t (2pr)

    From Newtons law of viscosity

    7.2
    A liquid whose coefficient of viscosity is m flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity u. Show that the pressure loss in a length of pipe is 32um/d2.
    Oil of viscosity 0.05 kg/ms flows through a pipe of diameter 0.1m with a velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.
    [11 520 N/m2]

    See the proof in the lecture notes for

    Consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe

    The fluid is in equilibrium, shearing forces equal the pressure forces.

    Newtons law of viscosity ,

    We are measuring from the pipe centre, so

    Giving:

    In an integral form this gives an expression for velocity,

    The value of velocity at a point distance r from the centre

    At r = 0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0;

    At a point r from the pipe centre when the flow is laminar:

    The flow in an annulus of thickness dr

    So the discharge can be written

    To get pressure loss in terms of the velocity of the flow, use the mean velocity:

    1. From the question m= 0.05 kg/ms d = 0.1m

    u = 0.6 m/s L = 120.0m

    7.3
    A plunger of 0.08m diameter and length 0.13m has four small holes of diameter 5/1600 m drilled through in the direction of its length. The plunger is a close fit inside a cylinder, containing oil, such that no oil is assumed to pass between the plunger and the cylinder. If the plunger is subjected to a vertical downward force of 45N (including its own weight) and it is assumed that the upward flow through the four small holes is laminar, determine the speed of the fall of the plunger. The coefficient of velocity of the oil is 0.2 kg/ms.
    [0.00064 m/s]

    Flow through each tube given by Hagen-Poiseuille equation

    There are 4 of these so total flow is

    Force = pressure area

    Flow up through piston = flow displaced by moving piston

    Q = Avpiston

    3.2410-6 = p0.042vpiston

    vpiston = 0.00064 m/s

    7.4
    A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of 0.076metres internal diameter. Oil fills the space between them to a depth of 0.2m. The rotque required to rotate the cylinder in the drum is 4Nm when the speed of rotation is 7.5 revs/sec. Assuming that the end effects are negligible, calculate the coefficient of viscosity of the oil.
    [0.638 kg/ms]

    From the question r-1 = 0.076/2 r2 = 0.075/2 Torque = 4Nm, L = 0.2m

    The velocity of the edge of the cylinder is:

    ucyl = 7.5 2pr = 7.52p0.0375 = 1.767 m/s

    udrum = 0.0

    Torque needed to rotate cylinder

    Distance between cylinder and drum = r1 - r2 = 0.038 - 0.0375 = 0.005m

    Using Newtons law of viscosity:


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