Chapter 2, Problem 54P

To determine

The systolic and diastolic pressures of the healthy person in kPa, psi, and in meter water column.

Explanation of Solution

Given:

Height of the systolic pressure of the healthy person $\left(h_{\text{high}}\right)$ is $120\text{mmHg}$.

Height of the diastolic pressure of the healthy person $\left(h_{\text{low}}\right)$ is $80\text{mmHg}$.

Density of the mercury $\left({\rho }_{\text{mercury}}\right)$ is $13600\text{kg}/{\text{m}}^{\text{3}}$.

Density of the water $\left({\rho }_{\text{water}}\right)$ is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Acceleration due to gravity is $\left(g\right)$$9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the systolic pressure of the healthy person in $\text{kPa}$$\left(P_{\text{high}}\right)$.

  $\begin{array}{c}{P}_{\text{high}}={\rho }_{\text{mercury}}g{h}_{\text{high}}\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(120\text{mm}\text{of}\text{Hg}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(120\text{mm}\text{of}\text{Hg}\times \frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.120\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =16.0\text{kPa}\end{array}$

Thus, the systolic pressure of the healthy person in $\text{kPa}$ is $\underset{\_}{16.0\text{kPa}}$.

Calculate the diastolic pressure of the healthy person in $\text{kPa}$$\left(P_{\text{low}}\right)$.

  $\begin{array}{c}{P}_{\text{low}}={\rho }_{\text{mercury}}g{h}_{\text{low}}\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(80\text{mm}\text{of}\text{Hg}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(80\text{mm}\text{of}\text{Hg}\times \frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.080\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =10.7\text{kPa}\end{array}$

Thus, the diastolic pressure of the healthy person in $\text{kPa}$ is $\underset{\_}{10.7\text{kPa}}$.

Calculate the systolic pressure of the healthy person in $\text{psi}$$\left(P_{\text{high}}\right)$.

  $\begin{array}{c}{P}_{\text{high}}=16.0\text{kPa}\\ =16.0\text{kPa}\times \frac{1\text{psi}}{6.895\text{kPa}}\\ =2.32\text{psi}\end{array}$

Thus, the systolic pressure of the healthy person in $\text{psi}$ is $\underset{\_}{2.32\text{psi}}$.

Calculate the diastolic pressure of the healthy person in $\text{psi}$$\left(P_{\text{low}}\right)$.

  $\begin{array}{c}{P}_{\text{low}}=10.7\text{kPa}\\ =10.7\text{kPa}\times \frac{1\text{psi}}{6.895\text{kPa}}\\ =1.55\text{psi}\end{array}$

Thus, the diastolic pressure of the healthy person in $\text{psi}$ is $\underset{\_}{1.55\text{psi}}$.

Calculate the height of the systolic pressure of the healthy person in $\text{m}$ water column $\left(h_{\text{high}}\right)$.

  $\begin{array}{c}{h}_{\text{water, high}}=\frac{{\rho }_{\text{mercury}}}{{\rho }_{\text{water}}}{h}_{\text{high}}\\ =\left(\frac{13,600\text{kg}/{\text{m}}^{\text{3}}}{1000\text{kg}/{\text{m}}^{\text{3}}}\right)\left(0.120\text{m}\right)\\ =1.63\text{m}\end{array}$

Thus, the height of the systolic pressure of the healthy person in $\text{m}$ water column is $\underset{\_}{1.63\text{m}}$.

Calculate the height of the diastolic pressure of the healthy person in $\text{m}$ water column $\left(h_{\text{low}}\right)$.

  $\begin{array}{c}{h}_{\text{water, low}}=\frac{{\rho }_{\text{mercury}}}{\rho }{h}_{\text{low}}\\ =\left(\frac{13,600\text{kg}/{\text{m}}^{\text{3}}}{1000\text{kg}/{\text{m}}^{\text{3}}}\right)\left(0.080\text{m}\right)\\ =1.09\text{m}\end{array}$

Thus, the height of the diastolic pressure of the healthy person in $\text{m}$ water column is $\underset{\_}{1.09\text{m}}$.