Chapter 1.11, Problem 71P

Blood pressure is usually measure by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of a person at the level of the heart. Using a mercury manometer and a stethoscope, the systolic pressure (the maximum pressure when the heart is pumping) and the diastolic pressure (the minimum pressure when the heart is resting) are measured in mmHg. The systolic and diastolic pressures of a health person are about 120 mmHg and 80 mmHg, respectively, and are indicated as 120/80. Express both of these gage pressures in kpa, psi, and meter water column.

To determine

The systolic and diastolic pressure of a healthy person is expressed in a kPa water column for high.

The systolic and diastolic pressure of a healthy person is expressed in a kPa water column for low.

The systolic and diastolic pressure of a healthy person is expressed in a psi water column for high.

The systolic and diastolic pressure of a healthy person is expressed in a psi water column for low.

The systolic and diastolic pressure of a healthy person is expressed in a meter water column for high.

The systolic and diastolic pressure of a healthy person is expressed in a meter water column for low.

Answer to Problem 71P

The systolic and diastolic pressure of a healthy person is expressed in a kPa water column for high is $\underset{\_}{16.0\text{kPa}}$.

The systolic and diastolic pressure of a healthy person is expressed in a kPa water column for low is $\underset{\_}{10.7\text{kPa}}$.

The systolic and diastolic pressure of a healthy person is expressed in a psi water column for high is $\underset{\_}{2.32\text{psi}}$.

The systolic and diastolic pressure of a healthy person is expressed in a psi water column for low is $\underset{\_}{1.55\text{psi}}$.

The systolic and diastolic pressure of a healthy person is expressed in a meter water column for high is $\underset{\_}{1.63\text{m}}$.

The systolic and diastolic pressure of a healthy person is expressed in a meter water column for low is $\underset{\_}{1.09\text{m}}$.

Explanation of Solution

Show the free body diagram of the systolic and diastolic pressure of a healthy person.

Write the expression of gauge pressure.

$P=\rho gh$ (I)

Here, the density is $\rho$, the height of the mercury column above the free surface is $h$, and acceleration of gravity is $g$.

Write the expression of gauge pressure for water.

$P={\rho }_{\text{water}}g{h}_{\text{water}}$ (II)

Write the expression of gauge pressure for mercury.

$P={\rho }_{\text{mercury}}g{h}_{\text{mercury}}$ (III)

Substitute Equation (III) into Equation (II)

$\begin{array}{l}{\rho }_{\text{mercury}}g{h}_{\text{mercury}}={\rho }_{\text{water}}g{h}_{\text{water}}\\ h_{\text{water}}=\frac{{\rho }_{\text{mercury}}}{{\rho }_{\text{water}}}{h}_{\text{mercury}}\end{array}$ (IV)

Conclusion:

Write the unit conversion from mm of Hg to m of Hg.

For $P_{\text{systolic}}$,

$\begin{array}{c}{P}_{\text{systolic}}=120\text{mm}\text{of}\text{Hg}\\ =120\text{mm}\text{of}\text{Hg}\times \left(\frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =0.12\text{m}\text{of}\text{Hg}\end{array}$

For $P_{\text{diastolic}}$,

$\begin{array}{c}{P}_{\text{diastolic}}=80\text{mm}\text{of}\text{Hg}\\ =80\text{mm}\text{of}\text{Hg}\times \left(\frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =0.08\text{m}\text{of}\text{Hg}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0.12\text{m}$ for $h$ in Equation (I).

$\begin{array}{c}{P}_{\text{high}}=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.12\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =16.0\text{kPa}\end{array}$.

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0.08\text{m}$ for $h$ in Equation (I).

$\begin{array}{c}{P}_{\text{low}}=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.08\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =10.7\text{kPa}\end{array}$.

Write the unit conversion from kPa to psi.

For $P_{\text{high}}$,

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

For $P_{\text{low}}$,

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

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{water}}$, and $0.12\text{m}$ for $h$ in Equation (IV).

$\begin{array}{c}{h}_{\text{water,high}}=\frac{13600\text{kg}/{\text{m}}^{\text{3}}}{1000\text{kg}/{\text{m}}^{\text{3}}}\times \left(0.12\text{m}\right)\\ =1.63\text{m}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{water}}$, and $0.08\text{m}$ for $h$ in Equation (IV).

$\begin{array}{c}{h}_{\text{water,high}}=\frac{13600\text{kg}/{\text{m}}^{\text{3}}}{1000\text{kg}/{\text{m}}^{\text{3}}}\times \left(0.08\text{m}\right)\\ =1.09\text{m}\end{array}$

Thus, systolic and diastolic pressure of a healthy person is expressed in a kPa water column for high is $\underset{\_}{16.0\text{kPa}}$, in a kPa water column for low is $\underset{\_}{10.7\text{kPa}}$, in a psi water column for high is $\underset{\_}{2.32\text{psi}}$, in a psi water column for low is $\underset{\_}{1.55\text{psi}}$, in a meter water column for high is $\underset{\_}{1.63\text{m}}$, and in a meter water column for low is $\underset{\_}{1.09\text{m}}$.