# Biophysics Textbook Questions - Chapter 12

#### Problem 12-6

Height $h$ will be determined by pressure $P_2 = \rho gh\quad [1]$

Find $P_2$ using Bernoulli's Equation $P_1 + \rho gy_1+ \frac{1}{2}\rho v_1{^2}= P_2 + \rho gy_2 + \frac{1}{2}\rho v_2{^2}$

But $y_1 = y_2$ (the centre of the tube has constant elevation)

$\therefore P_1 + \frac{1}{2}\rho v_1{^2} = P_2 + \frac{1}{2} \rho v_2{^2} \quad [2]$

$P_1, \rho$ and $v_1$ are known. $\therefore$ to find $P_2$ you need $v_2$ first.

Use the equation of continuity: $A_1v_1 = A_2 v_2 \quad [3]$

$\therefore$ you must determine $A_2$ (to find $v_2$)

From right to left, tube area increases from $A/2$ to $A$ in distance of $0.15 \; \text {ml}$ (from point 3 to point 1).

$\therefore$ the rate of change of area with distance is:

$\frac{A-A/2}{0.15\; \text{ml }} = \frac{A/2}{0.15\; \text{ml}} = \frac{A}{0.30} \text m^{-1}$

$\therefore$ the increase in area in distance of $0.05\; \text m$ (from point 3 to point 2) is:

$\bigg( \frac{A}{0.30} \text m^{-1} \bigg) (0.05 \; \text m) = \frac{A}{6} \\ \therefore A_2 = \frac{A}{2} + \frac{A}{6} = \frac{2A}{3}\quad (\text{where}\; A_2\; \text{at point 2}\; \text {and}\; \frac{A}{2} \text{at point 3}) \\ \therefore \text{in}\; [3], \; A_v = \frac{2A}{3}v_2\; \therefore\; v_2 = \frac{3}{2} v_1 = 0.15\; \text{m/s} \\ \therefore \; \text{in}\;[2], \; 50+ \frac{1}{2} (1000)(0.10)^2 = P_2 + \frac{1}{2}(1000)(0.15)^2 \\ \Rightarrow P_2 = 43._8 \; Pa \\ \therefore \; \text{in}\; [1], \;43._8= (1000)(9.8)h\\ \therefore \; h = 0.0045\; \text m$

#### Problem 12-9

Subscripts: Healthy, Diseased

"same volume of blood ... per unit time" $\Rightarrow Q_D = Q_H$

\begin {align} \therefore \frac{\pi R_D{^4}}{8 \eta}\bigg( \frac{\Delta P}{L} \bigg)_D &= \frac{\pi R_H{^4}}{8 \eta} \bigg( \frac{\Delta P}{L} \bigg)_H \\ \therefore \bigg( \frac{\Delta P}{L} \bigg)_D & = \frac{R_H{^4}}{R_D{^4}}\bigg( \frac{\Delta P}{L} \bigg)_H \quad [1] \end {align}

Areas:

\begin {align} A_D & = 0.600 \; A_H \\ \therefore\; \pi R_D{^2} & = 0.600\; \pi R_H {^2}\\ \therefore \frac{R_D{^2}}{R_H{^2}} & = 0.600 \quad [2] \end {align}

In [1], need $\frac{R_H{^4}}{R_D{^2}}$

\begin {align} \therefore \; \text{square}\; [2] \rightarrow \frac{R_D{^4}}{R_H{^4}} & = (0.600)^2 = 0.360 \\ \therefore \; \frac{R_H{^4}}{R_D{^4}} & = \frac{1}{0.360} = 2.78 \end {align}

Subst. in $[1]$.

$\therefore \bigg( \frac{\Delta P}{L} \bigg)_D = 2.78 \; \bigg( \frac{\Delta P}{L}\bigg)_H$

#### Problem 12-13

(a)

$360^\circ = 0.36\; \text s \\ 120^\circ - 15^\circ = 105 ^\circ \\ \text{Then} \; \frac{105^\circ}{360^\circ}= \frac{x}{0.36\; \text s} \Rightarrow x \approx 0.105 \; \text s \; \text{or}\; x \approx 0.11 \; \text s$

(b)

\begin {align} \text{"area" of triangle} & = \frac{1}{2}\;b\; h \\ & \approx \; \frac{1}{2} (0.11 \text s)(6.5\; \text{cm}^3/\text s) \\ & \approx \; 0.36 \text{cm} ^3 \end {align}

(c) $0.36\; \text{cm}^3$ flows in $1$ beat, and there are $2.75$ beats per second.

\begin {align} \therefore \text {in}\; 1.0\; \text{min., the volume is}\; & \approx 0.36 \text{cm}^3 \times 2.75 \times 60 \\ & \approx 59\; \text{cm} ^3 \end {align}