# Study Guide 14 - Fluid Dynamics

1. An artery of radius $\mathrm {R_1}$ carries blood with an average linear speed $\mathrm {v_1.}$ It divides into 3 smaller arteries, each of radius
$\mathrm {R_2 = ½R_1.}$ What will be the speed $\mathrm {v_2}$ of the blood in the smaller arteries, relative to $\mathrm {v_1?}$

The equation of continuity says that for an incompressible fluid the volume per unit time must be constant, i.e.,
$\mathrm {V_1/t = V_2/t}$

If we represent the volume as an area times a distance:
$\mathrm {A_1 \times 1/t = A_2 \times 2/t}$

But x/t is a velocity, so
$\mathrm {A_1v_1 = A_2v_2}$

In this problem $\mathrm {A_2 = 3A´_2,}$ where $\mathrm {A´_2}$ is the cross-section area of one of the 3 sub-arteries.

$\mathrm {A_1v_1 = 3A´_2v_2}$

For circular cross-sections, $\mathrm {A \sim R^2}$
$\mathrm{R_1{^2}v_1 = 3R_2{^2}v_2}$

But $\mathrm {R_2 = ½R_1}$
$\mathrm {R_1{^2}v_1 = 3(¼R_1{^2})v_2 \\ v_2 = (4/3)v_1}$

2. The capillary tube delivers water by viscous flow from a reservoir and is 0.50 mm in diameter. The time taken to collect $\mathrm {50\; cm^3}$ of water was 7.0 minutes. How much longer would it take to collect $\mathrm {50\; cm^3}$ of water if the only change is to decrease the tube diameter to 0.30 mm?

Since this is a problem of viscous flow which law is involved?

This problem involves Poiseuille"s equation, i.e.,

$\mathrm{Q = \frac{\pi r^4 \Delta P}{8 \eta \ell}\\ \frac{V}{t_1} = \frac{\pi r_1{^4} \Delta P}{8 \eta \ell} \\ \frac{V}{t_2} = \frac{\pi r_2{^4} \Delta P}{8 \eta \ell} \\ \frac{t_2}{t_1} = \bigg ( \frac{r_1}{r_2} \bigg)^4 \\ t_2 = \bigg( \frac{r_1}{r_2}\bigg)^4 t_1 = \bigg (\frac{5}{3} \bigg) ^4 \times 7 = 54 \; min}$

The increase in time = 54 - 7 = 47 min.

3. The aorta in the diagram has a constriction which reduces its cross-sectional area from A to A/3. Assuming the blood has zero viscosity, calculate the pressure difference $\mathrm {(P_L - P_S)\; in \;Nm^{-2},}$ between the large and small regions of the aorta.

This is a problem of non-viscous flow. Which law is involved?

Bernoulli's equation which states:
$\mathrm {P + ½\rho v^2 +\rho gh = constant}$
or relating the conditions at two places:
$\mathrm {P_1 + ½\rho v_1{^2} + \rho gh_1 = P_2 + ½\rho v_2{^2} +\rho gh_2}$
For a level pipe $\mathrm {h_1 = h_2}$
$\mathrm {P_1 + ½\rho v_1{^2} = P_2 + ½\rho v_2{^2}}$

In this case $\mathrm {A_2 = (1/3)A_1}$ and so the equation of continuity gives:
$\mathrm {A_1v_1 = A_2v_2 = (1/3)A_1v_2 \\ v_2 = 3v_1}$

$\mathrm {P_1 + ½\rho v_1{^2} = P_2 + ½\rho (9v_1)^2 \\ P_1 - P_2 = ½\rho(9v_1{^2} - v_1{^2}) = ½\rho 8v_1{^2} = 4\rho v_1{^2} \\ = 4(1000)(0.3)^2 = 360\; Pa}$