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Bernoulli's Equation and blood

  1. Jun 28, 2005 #1
    not sure where to begin...

    The blood speed in a normal segment of a horizontal artery is 0.13 m/s. An abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fifth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? (See Table 11.1 for appropriate constants.)
     
  2. jcsd
  3. Jun 28, 2005 #2

    dextercioby

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    Are you sure you don't need the Poiseuille-Hagen (1839) equation...?

    Blood is viscous and it can be modelled by a newtonian viscous fluid.

    Daniel.
     
  4. Jun 28, 2005 #3
    I have no idea what that is. It just says Bournelli's equation
     
  5. Jun 28, 2005 #4

    dextercioby

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    Ok.I'm sure it's Daniel Bernoulli.

    Use the continuity equation to find the velocity in the other portion of the artery and then Bernoulli's law to find the pressure difference.

    Daniel.
     
  6. Jun 28, 2005 #5
    Bernoullis eqn says that for incompressible flow we have

    P1 + Q1 = P2 + Q2

    P corresponds to the static pressure
    Q is the dynamic pressure

    Q = 1/2pv^2

    From the continuity equation we have

    mdot1 = mdot2

    mdot = density*area*velocity

    Assuming INCOMPRESSIBLE
    density1 = density2 thus

    A1*v1 = A2*v2

    A2 = A1/5 Thus v2 = 5*v1

    Q1 = 1/2*density of blood*v1^2
    Q2 = 1/2*density of blood*(5*v1)^2

    P1 + Q1 = P2 + Q2

    The change in pressure is P2 - P1

    P2 - P1 = 1/2*density of blood*(v1^2 - 25v1^2)

    delta P = -12*density of blood*v1^2

    notice this value is negative thus the static pressure decreases

    The pressure drop is simply 12*density of blood*v1^2
     
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