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Thread: Blood Flow Integral Problem

  1. #1
    Apr 2010

    Blood Flow Integral Problem

    Blood flows faster the closer it is to the center of a blood vessel because of the reduced friction with the cell walls. According to Poiseuille's laws, the velocity of blood in an idealized cylindrical vessel is given by:


    where R is the radius of the blood vessel, r is the distance of a layer of blood flow from the center of the vessel, and k - p/4Lv involving pressure p, viscosity v, and length L of the vessel (all of which we can assume are constant). Assume here that k is equal to 375.

    Suppose a skier's blood vessel has a constant radius R = 0.08 mm. Find the total blood flow Q, which is given by the formula:

    Integral from 0 to R(2(pi)V(r)rdr)

    What are the units of Q?

    Ultimately, for Q I got 0.024. I'm not sure what units that in though or if I did it correctly. I simply plugged in all of the constants and took the integral from 0 to 0.08.

    Could anyone tell me if I've done this correctly?

    The next problem is a bit more complicated and is based on the first. I havn't made any progress on it because I don't know where to begin.

    Suppose the same skier's blood vessel starts with radius R = 0.08 mm but that cold weather is causing the vessel to contract at a rate of 0.01 mm per minute. How fast is the velocity of the blood changing?

    (This skier will probably be dead after 8 minutes if thats true haha)
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  2. #2
    Senior Member
    Mar 2010
    Q= \pi k R^4/2 \: \: \: units \: are [m^3/s] \: volume \: per \: second.
    If k=375 in [SI] units then R must be in [m] R=8 \cdot  \: 10^{-5} m.
    If so your answer is not correct.

    If the vessel is contracting 0.01 mm/min then
    R(t)=R_0(1- \frac {10^{-5}}{60}t) \: \: \: t[s].

    Q(t)= \frac{dV(t)}{dt}=\pi k R^4(t)/2 = \pi k R_0^4 (1-\frac {10^{-5}}{60}t)^4/2.

    The volume passed from 0 to time t ( seconds ) is

    V= \int_{0}^{t} Q(t) \: dt .
    Last edited by zzzoak; Apr 21st 2010 at 04:55 PM.
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