Question

Consider arterial blood viscosity µ = .0 025 poise. If the length of the artery is 5.1 cm, radius
3
8 10−
× cm. and 3
1 2 P = P − P = 4×10 dynes/ 2
cm then find (i) maximum peak velocity of
blood and (ii) the shear stress at the wall.

Answer 1

Here, viscosity of blood is given as $\mu=0.0025 poise$ , length of capillary = 5.1 cm, radius=$3\times10^{-8}$ cm and $P_1-P_2= 4\times10^{12} \frac{dyne}{cm^2}$

we know that maximum velocity in case of fluid flow

Vmax=$\frac{\Delta P r^2}{4 \mu l}= \frac{4\times10^{12}\times9\times10^{-16}}{4\times0.0025\times5.1}$ = 235 $cm/s$

and for shear stress we know that,

$\tau=\frac{\Delta P r}{2l}= \frac{4\times10^{12}\times3\times10^{-8}}{2\times5.1}=1.17 \times 10^4 dyne/cm^2$