# Thread: Blood Flow Integral Problem

1. ## 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:

V=k(R^2-r^2)

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)

2. $Q= \pi k R^4/2 \: \: \: units \: are [m^3/s] \: volume \: per \: second.$
If k=375 in [SI] units then R must be in [m] $R=8 \cdot \: 10^{-5}$ m.

If 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

$V= \int_{0}^{t} Q(t) \: dt$.