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Bernoulli's Equation and blood

  • Thread starter texasgrl05
  • Start date
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.)
 

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.
 
I have no idea what that is. It just says Bournelli's equation
 

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.
 
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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