In this problem we derive the algebraic expression for the velocity profile in a

Question

In this problem we derive the algebraic expression for the velocity profile in a cylindrical tube. Start with the expression for the pressure gradient. - dP / dz r / 2 = tau r, z Assume that the pressure gradient is uniform and that the input (Po) and output (Pl) pressures arc across a vessel of length L. Write the algebraic expression above by substituting in an expression for the pressure gradient (dP/dz) in terms of Pa. Pl and L. Assume that blood is Newtonian with a viscocity of mu, re-write the algebraic expression found in Pan A for the shear stress ( )in terms of the shear rate (dvz / dr). By requiring the velocity at the wall to be zero (Vz=0), integrate the Equation in Part B to find, If the viscosity of blood is mu = 3.5 times 10-3 Pa s, and the pressure gradient is 0.01 mmHg/cm, and the diameter of the vessel is 0.05mm. Find the value of the maximum shear stress in this vessel. Where is the maximum shear stress found? What in the minimum shear stress (in dynes/cm2) and at what radius (in mm) is the minimum found?

Explanation / Answer

A.dP/dz P/z = (PL - Po )/(L-0)

rz = (Po - PL )r / 2L

B. From Newton's law of viscosity rz = -dvz /dr

dvz /dr = -(Po - PL )r / 2L

C.Integration gives

vz = [-(Po - PL )r2 / 4L]+C

We have vz =0 at r=d/2.....Since wall is stationary

C= (Po - PL )(d/2)2 / 4L

vz = (Po - PL )[(d/2)2 -r2 ] / 4L

Take d/2 out of the bracket we get

vz = (Po - PL )d2[1-(2r/d)2 ] / 16L

D. rz = (Po - PL )r / 2L   suggest shear stress is a linear function of r

The maximum is found at maximum value of r which is r=d/2 while minimum is found at r=0

max = (Po - PL )d / 4L =0.01*1.3328*10-3 *0.5/4=1.66*10-6 dynes/cm2

min =0

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