Chapter 9, Problem 67P

(a)

To determine

The pressure difference of the blood flows through an artery.

Answer to Problem 67P

The pressure difference of the blood flows through an artery is $50\text{Pa}$.

Explanation of Solution

Write the equation for volume flow rate.

$\frac{\Delta V}{\Delta t}=Av$ (I)

Here, $\Delta V/\Delta t$ is the volume flow rate of the blood, $A$ is the area of cross section of an artery, and $v$ is the speed of the blood flows in the artery.

Write the equation from poiseuille’s law for viscous flow.

$\frac{\Delta V}{\Delta t}=\frac{\pi \Delta {\mathrm{Pr}}^4}{8\eta L}$ (II)

Here, $\Delta P$ is the change in pressure, $r$ is the radius of the artery, $\eta$ is the coefficient of viscosity of blood, and $L$ is the length of the artery.

Compare the equation (I) and (II).

$\frac{\pi \Delta {\mathrm{Pr}}^4}{8\eta L}=Av$

Solve the above equation for $\Delta P$.

$\Delta P=\left(\frac{8\eta L}{\pi r^4}\right)Av$

Replace $\pi r^2$ for $A$ in the above equation.

$\begin{array}{c}\Delta P=\left(\frac{8\eta L}{\pi r^4}\right)\pi r^{2}v\\ =\frac{8\eta Lv}{r^2}\end{array}$ (III)

Conclusion:

Substitute $2.1\times {10}^{-3}\text{Pa}\cdot \text{s}$ for $\eta$, $0.20\text{m}$ for $L$, $6.0\text{cm/s}$ for $v$, and $2.0\text{mm}$ for $r$ in equation (III).

$\begin{array}{c}\Delta P=\frac{8\left(2.1\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(0.20\text{m}\right)\left(6.0\text{cm/s}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)}{{\left[\left(2.0\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)\right]}^2}\\ =\frac{8\left(2.1\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(0.20\text{m}\right)\left(0.060\text{m/s}\right)}{{\left(0.0020\text{m}\right)}^2}\\ =50400\times {10}^{-3}\text{Pa}\\ \simeq 50\text{Pa}\end{array}$

Therefore, the pressure difference of the blood flows through an artery is $50\text{Pa}$.

(b)

To determine

Find the pressure difference of the blood flows through a capillary.

Answer to Problem 67P

The pressure difference of the blood flows through a capillary is $1100\text{Pa}$.

Explanation of Solution

Write the relation between pressure difference from the equation (III).

$\Delta P^{\prime }=\frac{8\eta Lv}{r^2}$

Here, $\Delta P^{\prime }$ is the change in pressure, $r$ is the radius of the capillary, $\eta$ is the coefficient of viscosity of blood, $L$ is the length of the capillary, and $v$ is the speed of the blood flows in the capillary.

Conclusion:

Substitute $2.1\times {10}^{-3}\text{Pa}\cdot \text{s}$ for $\eta$, $1.0\text{mm}$ for $L$, $0.60\text{mm/s}$ for $v$, and $3.0\text{μm}$ for $r$.

$\begin{array}{c}\Delta P^{\prime }=\frac{8\left(2.1\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(1.0\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)\left(0.60\text{mm/s}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right)}{{\left[\left(3.0\text{μm}\right)\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right]}^2}\\ =\frac{8\left(2.1\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(0.0010\text{m}\right)\left(0.00060\text{m/s}\right)}{{\left(3.0\times {10}^{-6}\text{m}\right)}^2}\\ =1120\text{Pa}\\ \simeq 1100\text{Pa}\end{array}$

Therefore, the pressure difference of the blood flows through an capillary is $1100\text{Pa}$.

(c)

To determine

Compare the average blood pressure in 100torr.

Answer to Problem 67P

The average blood pressure in 100torr is $13\text{kPa}$.

Explanation of Solution

The average blood pressure of the human being is about 100torr.

Compare the answers in part (a) and (b) to average blood pressure.

$\left(100\text{torr}\right)\left(\frac{101.3\text{kPa}}{760.0\text{torr}}\right)=13\text{kPa}$

Thus, the 100torr is approximately $13\text{kPa}$ which is much larger than both pressures found in part (a) and (b).

Hence the pressure change in part (b) is more than 20 times greater than the pressure change in (a). The great arteries and wide veins present little resistance to blood flow compared to capillaries.