Measuring Pressure

Pressure in fluids can be reported relative to a absolute vacuum or relative to ambient atmospheric pressure. Devices like manometers and barometers utilize hydrostatic equilibrium to measure these pressure values.

Gauge Pressure vs. Absolute Pressure

Absolute Pressure ( $p_{\text{abs}}$ ): The total physical pressure relative to an absolute vacuum ( $p = 0$ ).
Gauge Pressure ( $p_g$ ): The pressure relative to ambient atmospheric pressure ( $p_{\text{atm}}$ ).

p_{\text{abs}} = p_g + p_{\text{atm}}

(1)

Gauges measure $p_g$ , reading zero at atmospheric pressure. Fluids push rather than pull, so the minimum possible absolute pressure is zero, corresponding to a lower limit on gauge pressure of $p_g = -p_{\text{atm}}$ .

Manometers

A manometer is a U-shaped tube containing fluid. An open-tube manometer has one side exposed to $p_{\text{atm}}$ and the other connected to absolute pressure $p_{\text{abs}}$ :

p_g = p_{\text{abs}} - p_{\text{atm}} = h \rho g

(2)

Barometers

A mercury barometer consists of a tube inverted in a mercury pool with a near-vacuum above the column, measuring atmospheric pressure directly:

p_{\text{atm}} = h \rho g

(3)

Standard atmospheric pressure supports a column height of $h = 760 \text{ mm Hg}$ .

Units of Pressure

Common conversions for atmospheric pressure:

SI Unit: Pascal ( $1 \text{ Pa} = 1 \text{ N/m}^2$ )
Bar: $1 \text{ bar} = 10^5 \text{ Pa}$ , $1 \text{ mbar} = 100 \text{ Pa}$
Torr / mm Hg: $1 \text{ torr} = 1 \text{ mm Hg} = 133.3 \text{ Pa}$
Standard Atmosphere: $1 \text{ atm} = 1.013 \cdot 10^5 \text{ Pa} = 760 \text{ mm Hg} = 14.7 \text{ psi} = 1013 \text{ mbar}$

Example 1

A U-tube with both ends open is filled with liquid of density $\rho_1$ to equal heights. An immiscible liquid of lower density $\rho_2 < \rho_1$ is poured into one side, settling on top to a height $h_2$ above the interface. Derive an expression for the height $h_1$ of Liquid 1 on the opposite side measured from the interface level.

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Equate the hydrostatic pressures at the interface level on both sides of the U-tube: $p_{\text{side 1}} = p_{\text{side 2}}$ .

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The pressure at the interface level in arm 2 must match the pressure at the same horizontal level in arm 1:

p_{\text{atm}} + \rho_1 g h_1 = p_{\text{atm}} + \rho_2 g h_2

Subtract $p_{\text{atm}}$ and divide by $g$ :

\rho_1 h_1 = \rho_2 h_2 \implies h_1 = \left(\frac{\rho_2}{\rho_1}\right) h_2

The height difference between the two top surfaces is $h_2 - h_1 = \left(1 - \frac{\rho_2}{\rho_1}\right) h_2$ .