Fluid properties, pressure, hydrostatics and buoyancy

Properties describe different responses

A fluid continues to deform under a sustained shear stress; ordinary liquids and gases are fluids. Density ρ is mass per volume, in kg/m³. Dynamic viscosity μ, in Pa·s, relates shear stress to deformation rate for a Newtonian fluid. Density describes how much mass occupies space; viscosity describes resistance to shearing motion. A dense fluid need not be especially viscous. Both properties depend on conditions, including temperature.

Compressibility describes volume or density change with pressure. A constant-density liquid model can work over a limited range, while a gas often needs an equation of state. Surface tension acts at interfaces and can dominate small droplets or narrow passages. A model that ignores it should be used where its forces are small compared with the effects being studied.

Pressure needs a reference

Pressure p is normal force per area, measured in pascals: 1 Pa = 1 N/m². In an ordinary fluid at rest, pressure at a point acts equally in all directions; the pressure force on a small surface acts perpendicular to that surface. A curved or large surface needs the contributions over its area combined.

Absolute pressure is referenced to vacuum. Gauge pressure is referenced to a stated surrounding pressure, often local atmospheric pressure: pg = pabs − patm. A zero gauge reading therefore need not mean zero absolute pressure. For fluid at rest with constant density and gravity, p = p₀ + ρgh, where h is depth below a surface of pressure p₀. Greater depth supports more fluid weight per area. The equation does not apply unchanged to an accelerating tank or a strongly varying density.

Buoyancy follows the pressure differences

Pressure on the lower parts of an immersed body usually exceeds pressure on the upper parts. For a body surrounded by a static fluid, the resulting buoyant force equals the weight of displaced fluid: B = ρgVdisplaced when density is uniform. A floating body displaces enough fluid for buoyancy to balance its weight. This force balance does not by itself establish whether the floating orientation is stable.

Constructed tank and floating-body case

Assume a liquid with constant ρ = 1,000 kg/m³, g = 9.81 m/s² and surface absolute pressure 100 kPa. At depth 2 m, the hydrostatic increase is 1,000 × 9.81 × 2 = 19,620 Pa = 19.62 kPa. Thus the local absolute pressure is 119.62 kPa, and its gauge pressure relative to that surface pressure is 19.62 kPa.

Now consider a rigid sealed body of total volume 0.003 m³ and mass 2 kg, freely floating with surface tension and air buoyancy neglected. Weight is 19.62 N. Required displaced volume is B/(ρg) = 19.62/9,810 = 0.002 m³, or two-thirds of the body volume. Its fully submerged buoyancy would be 29.43 N. Shape affects the waterline and stability; the balance alone gives displaced volume.

Try the reasoning

1. Would doubling the tank width double pressure at the same depth?

   Show the answer and reasoning

   No. Under the same static, uniform-density conditions and surface pressure, ρgh is unchanged. The total force on a larger bottom area increases, because force also depends on area.

2. Can a negative gauge pressure describe an ordinary positive absolute pressure?

   Show the answer and reasoning

   Yes. If local absolute pressure is 80 kPa and the reference atmosphere is 100 kPa, the gauge pressure is −20 kPa. The reference must be stated.

3. Why does sinking not imply zero buoyancy?

   Show the answer and reasoning

   A fully immersed body still displaces fluid. It sinks when the downward weight exceeds buoyancy and any other upward forces, not because buoyancy disappears.

What to carry forward

State the fluid properties, static conditions and pressure reference before calculating pressure or displaced volume.