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) (Pa.s)/(m^(3))`

Step-by-step explanation:

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

`R=(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=(8 (3(10)^(-3) Pa.s)(1000(10)^(-6)m))/(\pi (25(10)^(-6)m)^(4))`

`R=1.955(10)^(13) (Pa.s)/(m^(3))`