Pressure variation in a static fluid

(a) Pascal's law and pressure terms

Pascal's law: the pressure intensity at a point in a fluid at rest is the same in all directions, i.e. $p_x = p_y = p_z$. Any pressure applied to a confined fluid is transmitted equally in all directions.

• Atmospheric pressure: pressure due to the weight of the air above; standard value $101.325\ \text{kPa}$ at sea level.
• Absolute pressure: measured from absolute zero (perfect vacuum).
• Gauge pressure: measured above local atmospheric pressure; $p_{gauge} = p_{abs} - p_{atm}$.
• Vacuum pressure: pressure below atmospheric; $p_{vac} = p_{atm} - p_{abs}$.

 p_abs ^
       |   ______ point A (above atm)
       |  |  gauge pressure
 p_atm |--|-------------------------
       |  |  vacuum pressure
       |  |______ point B (below atm)
       |
 0     +--- absolute zero ----------

$$p_{abs} = p_{atm} + p_{gauge}$$

(b) Derivation of hydrostatic equation

Consider a vertical cylinder of fluid of cross-section $dA$ and height $dz$ in equilibrium. Take $z$ upward. Pressure at the bottom is $p$ and at the top is $p + dp$.

Forces in the vertical direction:

• Upward force on bottom: $p\,dA$
• Downward force on top: $(p + dp)\,dA$
• Weight: $\rho g\, dA\, dz$ (downward)

For equilibrium, $\sum F_z = 0$:

$$\begin{aligned} p\,dA - (p + dp)\,dA - \rho g\,dA\,dz &= 0 \\ \frac{dp}{dz} &= -\rho g = -\gamma \end{aligned}$$

The negative sign shows that pressure decreases as $z$ increases. For horizontal directions there is no net force, so $\partial p/\partial x = \partial p/\partial y = 0$: pressure is constant on a horizontal plane.

For an incompressible fluid ($\rho$ constant), integrating between two points 1 and 2:

$$p_2 - p_1 = -\rho g (z_2 - z_1) \quad \text{or} \quad \frac{p}{\rho g} + z = \text{constant}$$

At depth $h$ below a free surface where the pressure is $p_0$:

$$p = p_0 + \rho g h$$

Here $p/\rho g$ is the pressure head. The sum of pressure head and elevation head is constant throughout a static fluid.