Question

(a) For a nonsmoker, with blood viscosity of 2.50 10-3 Pa·s, normal blood flow requires a pressure difference of 8 mm Hg between the two ends of a certain artery. If this person were to smoke regularly, his blood viscosity could increase to 2.80 10-3 Pa·s (a 12% increase), and the diameter of his artery would constrict by 11%. What pressure difference between the ends of the artery would then be needed to maintain the same blood flow rate? mm of Hg

Answer 1

Answer:

13.6565mm of Hg

Explanation:

We only need to use the from Hagen poiseuille equation to be able to find the pressure change.  That is

$\Delta P = \frac{128\eta LQ}{\pi D^4}$

Once the equation is defined, we clear based on the pressure and replace

$\frac{\Delta P_2}{\Delta P_1} = (\frac{\eta_2}{\eta_1})(\frac{D_1}{D_2})^4$

$\frac{\Delta P_2}{\Delta P_1} = \frac{2.8*10^{-3}}{2.5*10^{-3}}(\frac{D_1}{(1-0.1)D_1})^4$

$\frac{\Delta P_2}{8}=1.707$

Finally we have,

$\Delta P_2 = 13.6565mm of Hg$