Poiseuille's Law

• Apr 2nd 2011, 08:27 AM
smray7
Poiseuille's Law
R sub 0 - normal radius

P sub 0 - normal pressure

R and P are the constricted values.

Use Poiseuille's Law:
F(flux- rate of blood past a given point) = (1/8) (pi PR^4/nl)
to show that P and R are related by the equation:

(P/Po) = (Ro/R)^4

note: Deduce that if the radius of an artery is reduced to 3/4 of it's former value, then the pressure is more than tripled.

I don't even know where to begin with this problem. Like, I can't even get started. What should I do? Help please!
• Apr 2nd 2011, 08:43 AM
TheEmptySet
Quote:

Originally Posted by smray7
R sub 0 - normal radius

P sub 0 - normal pressure

R and P are the constricted values.

Use Poiseuille's Law:
F(flux- rate of blood past a given point) = (1/8) (pi PR^4/nl)
to show that P and R are related by the equation:

(P/Po) = (Ro/R)^4

note: Deduce that if the radius of an artery is reduced to 3/4 of it's former value, then the pressure is more than tripled.

I don't even know where to begin with this problem. Like, I can't even get started. What should I do? Help please!

The question is a little hard to read but I think this is what you are looking for

$\displaystyle F(R,P)=\left( \frac{1}{8}\pi P R^4 \right)$

Now $\displaystyle F(R_0,P_0)=\left( \frac{1}{8}\pi P_0 R_0^4 \right)$

and

$\displaystyle F\left( \frac{3}{4}R_0,P\right)=\left( \frac{1}{8}\pi P_0\left( \frac{3}{4}R_0 \right)^4\right)=\left( \frac{1}{8}\pi P_0 \frac{81}{256}R_0^4 \right)$

Now if we take their ratio we get

$\displaystyle \frac{F(R_0,P_0)}{F\left( \frac{3}{4}R_0,P\right)}=\frac{256}{81}> 3$
• Apr 4th 2011, 06:29 AM
smray7
I don't understand how all the work you showed me proves P/Po = (Ro/R)^4.
• Apr 4th 2011, 07:22 AM
TheEmptySet
Quote:

Originally Posted by smray7
I don't understand how all the work you showed me proves P/Po = (Ro/R)^4.

$\displaystyle F(R,P)=V=F(R_0,P_0)$