Recall that the transverse velocity for Poiseuille flow is Др u(r) + cj In r + c2 , 0 < r < R, 4µL where Ap/L is the (constant) pressure gradient, µ is the dynamic viscosity, and Lis the length of the tube. Since the solution can't blow up at r = 0, c = 0, and c2 is determined by the no-slip condition u(R) = 0. Integrating the resulting solution over the cross-section of the tube gives the flux ARª Ap Qo 8µ L Now imagine that the tube has a central catheter so that -R

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Recall that the transverse velocity for Poiseuille flow is
Др
u(r)
+ cj
In r + c2 ,
0 < r < R,
4µL
where Ap/L is the (constant) pressure gradient, µ is the dynamic viscosity, and Lis the length of the tube. Since the solution can't blow
up at r = 0, c = 0, and c2 is determined by the no-slip condition u(R) = 0. Integrating the resulting solution over the cross-section of
the tube gives the flux
ARª Ap
Qo
8µ L
Now imagine that the tube has a central catheter so that -R <r < R. Use the solution for u above and apply no-slip conditions at
r = -R and r = R to compute the flux Q for the catheterized tube.
3
Qo
Transcribed Image Text:Recall that the transverse velocity for Poiseuille flow is Др u(r) + cj In r + c2 , 0 < r < R, 4µL where Ap/L is the (constant) pressure gradient, µ is the dynamic viscosity, and Lis the length of the tube. Since the solution can't blow up at r = 0, c = 0, and c2 is determined by the no-slip condition u(R) = 0. Integrating the resulting solution over the cross-section of the tube gives the flux ARª Ap Qo 8µ L Now imagine that the tube has a central catheter so that -R <r < R. Use the solution for u above and apply no-slip conditions at r = -R and r = R to compute the flux Q for the catheterized tube. 3 Qo
Expert Solution
Step 1

The cross-section of the tube gives the expression for flow

Qo=πR48μ·pL  ..........(1)

Now for r=R3 we have Qo=Q

So

Q=πR48μ34·pL=πR48μ×81·pLfrom equation (1)Q=181QoQQo=0.012

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