Hydrostatic Pressure

When Fluids are defined to be ‘at rest’ (that is, all forces are balanced) they are called hydrostatic. In hydrostatic fluids, the pressure increases the ‘deeper’ we go.

Definition §

Formula - 1 §

Hydrostatic pressure

$$\Delta P=-{\rho }_{fluid}g\Delta y={\rho }_{fluid}gd$$

$$P_{y_2}=P_{y_1}+{\rho }_{fluid}\cdot g(y_2-y_1)$$

Formula - 2 §

Hydrostatic Pressure of Water at a Depth

$$P_h=P_0+{\rho }_{water}gh$$

Formula - 3 §

Hydrostatic Pressure of Air at an altitude

$$P_h=P_0-{\rho }_{air}gh$$

Terms §

• $P$ = Pressure (in $\text{Pa}$)
  • $P_h$ = Pressure at depth/altitude $h$. This is also referred to as the absolute pressure
  • $P_0$ =Atmospheric Pressure $\approx 100\text{ kPa}$
• $\rho$ = Density (in ${\text{kg/m}}^3$)
• $g$ = Acceleration due to gravity $\approx 9.8{\text{ m/s}}^2$
• $y$ = Height relative to sea level. (in $\text{m}$)
  • $d$ = Depth (negative height relative to sea level)
  • $\Delta y$ = $y_2-y_1$ = Change in y position.
• $h$ = Change in height from sea level (always positive!)#tosee#todo is it?

Gauge pressure

Pressure gauges (devices that measure pressure) don’t actually measure $P_h$, the absolute pressure. They measure the gauge pressure: ${\rho }_{fluid}\cdot gh$

Derivation §

The formula for hydrostatic pressure is derived by using Newton’s first law of motion and Pascal’s Law.

Recall that for a body at rest, the sum of all forces must be zero:

$$\sum \vec{F}=0$$

Let’s start by looking at an incompressible fluid that starts from sea level, and extends downwards. We know that for the forces to balance out, the force from the water must balance that from the atmosphere.

Furthermore, the atmospheric force is from the weight of all the air molecules:

$$F_{atmosphere}=m_{atmos}\cdot g$$

Then the force from the fluid comes from the molecules inside the fluid.

$$F_{fluid}=F_{atmosphere}$$

Again, we tend to talk about things in terms of pressure, so for the rectangular area that is contact with the air, the pressures are equal:

$$P_{atmosphere}=\frac{F_{atmos}}{A}$$

$$P_{y=0}=\frac{F_{fluid}}{A}$$

$$P_{atmosphere}=P_{fluidaty=0}$$

Now, since we defined the fluid to be incompressible and it is confined, we can apply Pascal’s Law, which says the pressure is equal in a fluid without gravity.

If we take a circular disc of area, $C$ at height $y=-h$ inside the fluid, due to Pascal’s Law, it must have the same pressure, meaning for each side of the disc (top and bottom), the forces must be equal:

Hydrostatic Pressure _1.excalidraw

Makes sense so far?

Now let’s introduce Gravitational Forces into the system. Pascal’s Law, unfortunately, doesn’t quite behave the same anymore.

At $y=0$, we have the top surface experiencing a force due to the weight of air:

$$P_{fluid}=P_{air}=P_{y=0}$$

On the disk, $C$ with area $A$ we have cylindrical column of water above it, which adds extra force on top of the pressure force!

$$\sum F_C=F_{fluid}+F_{weightofcolumn}$$

Hydrostatic Pressure _2.excalidraw

In terms of pressure and density:

$$F_{weightofcolumn}=m_{column}g={\rho }_{fluid}{V}_{column}\cdot g$$

$$P_C\cdot A=P_{fluid}\cdot A+\rho V_{column}\cdot g$$

Since $V_{column}=A\cdot h$, where h is the difference in y, we have:

$$P_C\cdot A=P_{fluid}\cdot A+\rho Ahg$$

dividing by $A$:

$$P_C=P_{fluid}+\rho hg$$

Now express everything in terms of y level:

$$P_{-h}=P_0+\rho hg$$