Question

A blood vessel is 0.10m in length and has a radius of `1.5 * 10^-3` m. Blood of viscosity, `4.0 * 10^-3` `Pa*s` flows at a rate of `1.0*10^-7``m^3``/``s`. What is the pressure difference between the two ends of the vessel?

Answer 1

`20.12Pa`

Explanation:

Pouiseulle's Law governs the pressure difference between the ends of a pipe to regulate the fluid flow in the pipe, and may be given by

`DeltaP=(8QetaL)/(pir^4)`.

Since the density of the blood was not given in the question, I am gong to take an average value of `1050kg//m^3` for it.
Notice that this value is very close to the density of water which is realistic since blood is about 90% water.

However, this law is only valid for laminar flow so we first have to find the Reynold's number for the blood flow in this case to ensure that we are allowed to apply Pouiselle's Law.

`R_e=(vrhod)/eta`

`=((Qt)/(pir^2)*rhod)/eta` , since `Q=V/t and V=pir^2h` for cylindrical vessel.

`=((1xx10^(-7))(1050)(3xx10^(-3)))/(pi(1,5xx10^(-3))^2(4xx10^(-3))`

`=11,14` `<1000`

Since the Reynold's number is less than 1000, it implies the fluid flow is laminar and hence Pouiseulle's Law is valid to find the pressure difference required across the ends of the blood vessel in order to maintain this flow rate :

`DeltaP=(8QetaL)/(pir^4)`

`=(8(1xx10^(-7))(4xx10^(-3))(0.10))/(pi(1,5xx10^(-3))^4)`

`=20.12Pa`.

(For illustration purposes we may convert this value into mmHg by using the conversion factor that `1atm=101,3kPa=760mmHg` to obtain that `20.12Pa=0.151mmHg`.
Now since the normal systolic blood pressure is around 120mmHg and the normal diastolic pressure is around 80mmHg in most blood vessels of a healthy person, then such a low blood pressure must either mean it is the pressure in a very insignificant blood vessel in the body, far from the heart probably, or else the person is just about dead!)