USAF_Dentman

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Here's 2 questions i'm having trouble getting the right answers to..

1)Angioplasty is a technique in which arteries partially blocked with plaque are dilated to increase blood flow. By what factor must the radius of an artery be increased in order to increase blood flow by a factor of 10?


2)When physicians diagnose arterial blockages, they quote the reduction in flow rate. If the flow rate in an artery has been reduced to 50% of its normal value due to plaque formation, and the average pressure difference has increased by 30%, by what factor has the plaque reduced the radius of the artery?


What I know is that you use Poiseuille's Law. And then you write equations for the normal and for the reduced flow rates..But then what?


Thanks
 

deuist

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First, some definitions:
A=Area=(pi)*r^2
t=time

Flow=A/t

If I want to increase the flow by a factor of 10, I need to increase the radius by a factor of sqrt(10) (recall that the formula for area requires you to square the radius). Therefore, the answer is 3.16.
 
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USAF_Dentman

USAF_Dentman

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deuist said:
First, some definitions:
A=Area=(pi)*r^2
t=time

Flow=A/t

If I want to increase the flow by a factor of 10, I need to increase the radius by a factor of sqrt(10) (recall that the formula for area requires you to square the radius). Therefore, the answer is 3.16.
Not correct..

This is the help it gave..

HELP: Use Poiseuille's Law. Write the flow for two different radii, the original and the increased radius. Divide the two equations and solve for the ratio of the two radii in terms of the ratio of the two flows.

but i dont get what they are trying to say..
 

Mbkcd

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Provided that the equation is given to you, what else is there to get?

Assume identical tube length, viscosity, and pressure drop. These factors, along with the 8 and pi terms, will all cancel out and leave you with:

1 = (Qi / Ri^4) / (Qf / Rf^4)

Where i and f denote initial and final, R is tube radius, Q is flowrate, and Qf = 10*Qi.

This can be simplified to:

10 = (Rf^4) / (Ri^4)

4th-root(10) * Ri = Rf
 

Mbkcd

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2nd question:

1 = ( (1.3/0.5) / (Ri^4) ) / (1 / Rf^4)

0.5/1.3 = Rf^4 / Ri^4

0.385*Ri^4 = Rf^4

4th-root(0.385)*Ri = Rf

Rf = 0.788 * Ri


I believe that's right, though I could be way off.. notepad is a bad way to solve algebra problems.
 
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USAF_Dentman

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Mbkcd said:
2nd question:

1 = ( (1.3/0.5) / (Ri^4) ) / (1 / Rf^4)

0.5/1.3 = Rf^4 / Ri^4

0.385*Ri^4 = Rf^4

4th-root(0.385)*Ri = Rf

Rf = 0.788 * Ri


I believe that's right, though I could be way off.. notepad is a bad way to solve algebra problems.

Yup, i got 78.7% and it was correct..