Answer
(a) $25cm^3$
(b) 2.2
Work Step by Step
(a) We know that
$\Delta V=\frac{(P_1-P_2)\pi r^4\Delta t}{8\eta L}$
We plug in the known values to obtain:
$\Delta V=\frac{(450Pa)\pi(2.4.\times 10^{-3}m)^4(1.0s)}{8(0.0027Ns/m^2)(8.5\times 10^{-2}m)}$
$\Delta V=25cm^3$
(b) We can find the required reduced blood flow rate as follows:
$(\frac{\Delta V}{\Delta t})_2/(\frac{\Delta V}{\Delta t})_1=\frac{(P_1-P_2)\pi r_2^4}{8\eta L}/\frac{(P_1-P_2)\pi r_1^4}{8\eta L}=(\frac{r_2}{r_1})^4$
$(\frac{\Delta V}{\Delta t})_2/(\frac{\Delta V}{\Delta t})_1=(\frac{0.82r_1}{r_1})^2=0.45$
Thus, the blood rate is reduced by a factor of 2.2.
