Subsection 2.7.3 Pressure Unit Conversions
In interpreting pressure measurements, a great deal of confusion arises because the zero point on most pressure gages represents atmospheric pressure rather than zero absolute pressure, and several different pressure and head scales are used. Thus it is important to specify the kind of pressure being measured under given conditions as well as the desired units.
To clarify the numerous meanings of the word pressure, the relationship among gage pressure, atmospheric pressure, vacuum, and absolute pressure is illustrated in
Figure 2.7.20 .
Figure 2.7.20. Pressure Relationship
If you can remember these values, which are all equivalent expressions for atmospheric pressure, you can use them to create unit-factors as needed.
\begin{align}
P_\text{atm} \amp = 1 \text{ atm} = 14.7 \text{ psi} = 101.3 \textrm{ kPa} =1.01 \textrm{ bar}\\ \amp= 29.92 \textrm{ in Hg} = 760 \textrm{ mm Hg} = 33.9 \textrm{ ft H}_2\textrm{O}\tag{2.7.3}
\end{align}
Absolute pressure measures up from zero.
Gage pressure measures up or down from atmospheric pressure.
Vacuum measures down from atmospheric pressure.
\begin{align*}
P_\text{abs} \amp = P_\text{atm} + P_\text{gage}\\
P_\text{gage} \amp = P_\text{abs} - P_\text{atm}\\
P_\text{vacuum} \amp = P_\text{atm}- P_\text{abs}
\end{align*}
Example 2.7.21 . Absolute pressures.
Express a barometric pressure of 27.5 in. Hg, in each of the following units of pressure: a) atm b) mm Hg c) kPa d) psi
Answer .
\begin{equation*}
27.5 \textrm{ in Hg} = 0.919 \textrm{ atm} = 698.5 \textrm{ mm Hg}= 93.3 \textrm{ kPa} = 13.5 \textrm{ psia}
\end{equation*}
Solution .
Barometers measure the absolute pressure of the atmosphere, so all results are also absolute pressures.
For each conversion, we create an appropriate unit-factor from
(2.7.3) .
\begin{align*}
27.5 \textrm{ in Hg} \amp = 27.5 \textrm{ in Hg} \times \left[\frac{1 \textrm{ atm}}{29.92 \textrm{ in Hg}}\right] \amp = 0.919 \textrm{ atm}\\
\amp = 27.5 \textrm{ in Hg} \times \left[\frac{760 \textrm{ mm Hg}}{29.92 \textrm{ in Hg}}\right] \amp = 698.5 \textrm{ mm Hg}\\
\amp = 27.5 \textrm{ in Hg} \times \left[\frac{\kPa{101.3}}{29.92 \textrm{ in Hg}}\right] \amp = 93.3 \textrm{ kPa}\\
\amp = 27.5 \textrm{ in Hg} \times \left[\frac{\psinch{14.7}}{29.92 \textrm{ in Hg}}\right] \amp = 13.5 \textrm{ psia}
\end{align*}
Example 2.7.22 . psia to mm Hg conversion.
Convert from 10.4 psia to mm Hg Vacuum.
Answer . 10.4 psia = 222.3 mm Hg (vacuum)
Solution .
Set up a ratio to convert psia to mm Hg Absolute
\begin{align*}
\frac{10.4 \textrm{ psia}}{14.7 \textrm{ psia}} \amp= \frac{p}{760 \textrm{ mm Hg} }\\
\amp = 537.7 \textrm{ mm Hg (absolute)}
\end{align*}
Subtract from atmospheric pressure to get vacuum
\begin{equation*}
V = 760 - p = 222.3 \textrm{ mm Hg (vacuum)}
\end{equation*}
Example 2.7.23 . in Hg to psi conversion.
Convert 10 in Hg vacuum to pounds per square inch absolute (psia).
Answer .
\begin{equation*}
10 \textrm{ in Hg (vac)} = \psi{9.79}
\end{equation*}
Solution .
Vacuum measurements are measured down from atmospheric pressure.
First convert the given value to an absolute pressure.
\begin{align*}
P_\text{abs} \amp = p_\text{atm} - p_\text{vac}\\
\amp = 29.92 \textrm{ in Hg} - 10 \textrm{ in Hg} \amp = 19.92 \textrm{ in Hg abs} \text{ (absolute)}\\
\amp = 19.92 \textrm{ in Hg}\times \left[\frac{\psi{14.7}}{29.92 \textrm{ in Hg}}\right]\\
\amp = \psi{9.79}
\end{align*}
Example 2.7.24 . Guage pressure.
To convert 5 bar to psi and to kPa. All pressures are gauge values.
Answer .
\begin{equation*}
5 \mathrm{\ bar} = \psi{72.7} = \kPa{500}
\end{equation*}
Solution .
\begin{align*}
5 \textrm{ bar} \amp = 5 \textrm{ bar} \times \left[\frac{\psi{14.7}}{1.01 \textrm{ bar}}\right] \amp= \psi{72.7}\\
\amp = 5 \textrm{ bar} \times \left[\frac{\kPa{100}}{1 \textrm{ bar}}\right] \amp = \kPa{500}
\end{align*}
Example 2.7.25 . psig to kPa absolute.
Convert 35 psig to kPa (absolute)
Answer .
\begin{equation*}
\psi{35} \text{ (gage)} = \kPa{343} \text{ (absolute)}
\end{equation*}
Solution .
\begin{align*}
\psi{35} \text{ (gage)} \amp = (35 + 14.7) \textrm{ psi (absolute)} = 49.7 \textrm{psia}\\
\amp = \psi{49.7} \times \left[\frac{\kPa{101.3}}{\psi{14.7}}\right] = \kPa{343} \textrm{ (absolute)}
\end{align*}
