Question

An arteriole has a radius of 25 μm and it is 1000 μm long. The viscosity of blood is 3 x 10-3 Pa s and its density is 1.055 g cm-3 . Assume the arteriole is a right circular cylinder. A. Assuming laminar flow, what is the resistance of this arteriole?

Answer 1

Answer: $1.955(10)^{13} \frac{Pa.s}{m^{3}}$

Explanation:

This can be solved by the Poiseuille’s law for a laminar flow:

$R=\frac{8 \eta L}{\pi r^{4}}$

Where:

$R$ is the resistance of the arteriole

$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

$L=1000 \mu m=1000(10)^{-6}m$ is the length of the arteriole

$r=25 \mu m=25(10)^{-6}m$ is the radius of the arteriole

$R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}$

$R=1.955(10)^{13} \frac{Pa.s}{m^{3}}$