Fluid Pressure and Depth: Physics with Calculus Study Notes

Fluid Mechanics

Pressure Variation with Depth in a Fluid

Understanding how pressure changes with depth in a fluid is fundamental in fluid mechanics. This section explains the relationship between pressure, depth, and density, and provides methods for calculating pressure at various depths.

• Pressure in Fluids: Pressure in a fluid at rest increases with depth due to the weight of the fluid above.

• Key Variables:

  • \( \rho \) (rho): Density of the fluid (kg/m3).

  • \( g \): Acceleration due to gravity (9.8 m/s2).

  • \( d \): Depth below the fluid surface (m).

• Atmospheric Pressure: For fluids open to the air, atmospheric pressure (\( P_{atm} \)) acts on the surface. Standard atmospheric pressure at sea level is \( P_{atm} = 101.3 \) kPa.

Calculating Pressure at Depth

The pressure at a depth \( d \) in a fluid can be calculated using the following equations:

• For fluids open to the atmosphere:

$P = P_{atm} + \rho g d$

• For fluids with no surface pressure (e.g., sealed and evacuated containers):

$P = \rho g d$

• Key Points:

  • Pressure increases linearly with depth.

  • Pressure at a given depth is independent of the shape or volume of the container.

Worked Example: Diver in the Ocean

• Given:

  • Depth, \( d = 10 \) m

  • Density of water, \( \rho = 1000 \) kg/m3

  • Acceleration due to gravity, \( g = 9.8 \) m/s2

  • Atmospheric pressure, \( P_{atm} = 101.3 \) kPa

• Find: Total pressure experienced by the diver.

• Solution:

  • Calculate the pressure due to the water column:

  $P_{water} = \rho g d = 1000 \times 9.8 \times 10 = 98,000 \ \text{Pa} = 98 \ \text{kPa}$

  • Total pressure at depth:

  $P_{total} = P_{atm} + P_{water} = 101.3 \ \text{kPa} + 98 \ \text{kPa} = 199.3 \ \text{kPa}$

• Conclusion: The total pressure 10 m below the surface is approximately double the atmospheric pressure. There is an approximate rise in total pressure of 100 kPa for every 10 m depth of ocean.

Worked Example: Hydrostatic Paradox (Glasses of Water)

• Problem: Three vessels of different shapes but same base area and same water level (depth), with air pumped out (surface pressure = 0).

• Question: Are the water pressures at the bottom of each vessel the same?

• Solution: Yes, the pressure at the bottom depends only on the depth and density, not the shape of the vessel.

• Equation:

  $P = \rho g d$

• Conclusion: All vessels experience the same pressure at the bottom if the depth and fluid are the same, regardless of vessel shape.

Hydrostatic Equilibrium

In a state of hydrostatic equilibrium (fluid at rest), the pressure at all points at the same depth in a connected fluid is equal. This principle is essential for understanding fluid statics and is the basis for many practical applications, such as hydraulic systems.

Derivation: Pressure at Depth

The equation for pressure at depth can be derived by considering a column of fluid:

• Imagine a cylinder of fluid with cross-sectional area \( A \) and height \( d \).

• Mass of the fluid column: $m = \rho V = \rho A d$

• Weight of the fluid column: $W = m g = \rho A d g$

• Pressure is force per unit area. The force exerted by the fluid column is its weight:

$P = \frac{W}{A} = \frac{\rho A d g}{A} = \rho g d$

• If atmospheric pressure acts on the surface, add \( P_{atm} \):

$P = P_{atm} + \rho g d$

Summary Table: Pressure in Fluids

Situation	Equation for Pressure at Depth	Notes
Open to atmosphere	$P = P_{atm} + \rho g d$	Atmospheric pressure acts on surface
Sealed, evacuated container	$P = \rho g d$	No surface pressure
Same fluid, same depth, different shapes	$P = \rho g d$	Pressure at bottom is the same

Key Takeaways

• Fluid pressure increases with depth due to gravity.

• Pressure at a given depth depends only on the fluid's density, gravitational acceleration, and depth, not on the container's shape or volume.

• Atmospheric pressure must be included when the fluid is open to the air.

• Hydrostatic equilibrium ensures equal pressure at equal depths in a connected fluid.