Question

Blood plasma (at 37.0°C) is to be supplied to a patient at the rate of 2.80 × 10−6 m3/s. If the tube connecting the plasma to the patient’s vein has a radius of 2.00 mm and a length of 52.5 cm, what is the pressure difference between the plasma and the patient’s vein? Viscosity of blood plasma is 1.30 × 10−3 Pa·s.

Answer 1

ANSWER:

304.3 Pa

STEP-BY-STEP EXPLANATION:

We have the poiseuille law, which would be the following equation:

$v=\frac{\pi\cdot\Delta P\cdot r^4\cdot t}{8\cdot\eta\cdot L}$

Where,

v = volume of the liquid

r = radius

t: time

n = coefficiente of viscosity

Δp = change of pressure

L : lenght

We solve for Δp, and we would have:

$\begin{gathered} \Delta P=\frac{8\cdot\eta\cdot L\cdot v}{\pi\cdot\cdot r^4\cdot t} \\ \frac{v}{t}=Q \\ \text{ therefore:} \\ \Delta P=\frac{8\cdot\eta\cdot L\cdot Q}{\pi\cdot r^4} \\ \text{ replacing:} \\ L=52.5\text{ cm = 0.525 m} \\ r=2\text{ mm = 0.002 m} \\ \Delta P=\frac{8\cdot1.3\cdot10^3\cdot0.525\cdot2.8\cdot10^{-6}}{3.14\cdot(0.02)^4} \\ \Delta P=304.3\text{ Pa} \end{gathered}$

The pressure difference is 304.3 Pa