Chapter 10, Problem 21P

To determine

Convert the systolic/diastolic of 115/85 mm of Hg into Pa, psi and $\text{in}\cdot {\text{H}}_{\text{2}}\text{O}$.

Answer to Problem 21P

The systolic pressure from 115 $\text{mm}\cdot \text{Hg}$ to Pa is 15,322 kPa, to psi is 2.2 psi, and to $\text{in}\cdot {\text{H}}_{\text{2}}\text{O}$ is 61.42 $\text{in}\cdot {\text{H}}_{\text{2}}\text{O}$.

The diastolic pressure from 85 $\text{mm}\cdot \text{Hg}$ to Pa is 11.322 kPa, to psi is 1.64 psi, and to $\text{in}\cdot {\text{H}}_{\text{2}}\text{O}$ is 45.27 $\text{in}\cdot {\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

Given data:

A systolic pressure is 115 $\text{mm}\cdot \text{Hg}$.

A diastolic pressure is 85$\text{mm}\cdot \text{Hg}$.

Formula used:

Formula to determine the gauge pressure is,

$P=\rho gh$ (1)

Here,

$\rho$ is the density.

g is the gravitational constant.

h is the height of the fluid column.

Calculation:

Case 1:

Conversion for 115 mm Hg to Pa:

$\begin{array}{c}P=115\times 133.322\text{Pa}\left[\therefore 1\text{mm}\cdot \text{Hg = 133}\text{.322}\text{Pa}\right]\\ =15,332\text{Pa}\\ \text{=15}\text{.322}\times {\text{10}}^3\text{Pa}\\ \text{=15}\text{.322}\text{kPa}\end{array}$

Conversion for 115 mm Hg to psi:

1 mm of Hg is equal to 0.0193368 psi. Therefore,

$\begin{array}{c}P=\left(115\right)\left(0.0193368\text{ psi}\right)\\ \text{=}2.22\text{psi}\end{array}$

Conversion for 115 mm Hg to inch of water:

Substitute $1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ for $\rho$, 15.332 kPa for P, and $9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}$ for g in equation (1) to find the height of the fluid column h.

$\begin{array}{c}15.322\text{kPa}=\left(1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right)\left(9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(h\right)\\ h=\frac{\left(15.322\text{kPa}\right)}{\left(9810\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)}\\ =\frac{\left(15.322\text{kPa}\right)}{\left(9810\frac{1}{{\text{m}}^{\text{3}}}\times \frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right)}\left[\therefore 1\text{N=1}\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =\frac{\left(15.322\text{kPa}\right)}{\left(9810\frac{1}{{\text{m}}^{\text{3}}}\times \text{N}\right)}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}h=\frac{\left(15.322\text{kPa}\right)}{\left(9810\frac{1}{\text{m}}\times \frac{\text{N}}{{\text{m}}^2}\right)}\left[\therefore 1\text{Pa=1}\frac{\text{N}}{{\text{m}}^2}\right]\\ =\frac{\left(15.322\text{kPa}\right)}{\left(9810\frac{1}{\text{m}}\times \text{Pa}\right)}\\ =1.56\text{m}\end{array}$

Converting the height of fluid column $\left(h=1.56\text{m}\right)$ of systolic pressure to inch of water as,

$\begin{array}{c}P=1.56\times 39.37\text{in}\left[\therefore \text{1}\text{m}\text{=}39.37\text{in}\right]\\ =61.42\text{in}\text{.}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Case 2:

Conversion for 85 mm Hg to Pa:

$\begin{array}{c}P=85\times 133.322\text{Pa}\left[\therefore 1\text{mm}\cdot \text{Hg = 133}\text{.322}\text{Pa}\right]\\ =11332\text{Pa}\\ \text{=11}\text{.332}\times {\text{10}}^3\text{Pa}\\ \text{=11}\text{.332}\text{kPa}\end{array}$

Conversion for 85 mm Hg to psi:

1 mm of Hg is equal to 0.0193368 psi. Therefore,

$\begin{array}{c}P=\left(85\right)\left(0.0193368\text{ psi}\right)\\ \text{=}1.64\text{psi}\end{array}$

Conversion for 85 mm Hg to inch of water:

Substitute $1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}$ for $\rho$, 11.332 kPa for P, and $9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}$ for g in equation (1) to find the height of the fluid column h.

$\begin{array}{c}11.332k\text{Pa}=\left(1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}\right)\left(9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(h\right)\\ h=\frac{\left(11,322\text{kPa}\right)}{\left(9810\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)}\\ =\frac{\left(11,322\text{kPa}\right)}{\left(9810\frac{1}{{\text{m}}^{\text{3}}}\times \frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right)}\left[\therefore 1\text{N=1}\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =\frac{\left(11.322\text{kPa}\right)}{\left(9810\frac{1}{{\text{m}}^{\text{3}}}\times \text{N}\right)}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}h=\frac{\left(11.322\text{kPa}\right)}{\left(9810\frac{1}{\text{m}}\times \frac{\text{N}}{{\text{m}}^2}\right)}\left[\therefore 1\text{Pa=1}\frac{\text{N}}{{\text{m}}^2}\right]\\ =\frac{\left(11.322\text{kPa}\right)}{\left(9810\frac{1}{\text{m}}\times \text{Pa}\right)}\\ =1.15\text{m}\end{array}$

Converting the height of fluid column $\left(h=1.15\text{m}\right)$ of diastolic pressure to inch of water as,

$\begin{array}{c}P=1.15\times 39.37\text{in}\left[\therefore \text{1}\text{m}\text{=}39.37\text{in}\right]\\ =45.27\text{in}\text{.}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Conclusion:

Hence, the conversion for systolic/diastolic of 115/85 mm Hg is explained.