Fluid Statics and Pressure

01 / UNDERSTAND & APPLY

1. Hydrostatic pressure increases with depth because it supports fluid above

Pressure is normal force per unit area in the local limit, p = dF_normal/dA. In a fluid at rest without shear stress, pressure at a point is the same in every direction, while the force it exerts on a small surface acts perpendicular to that surface. Pressure itself is a scalar; the direction of the resulting force comes from the surface normal.

Take upward coordinate z and a thin horizontal fluid slice of area A and thickness dz. Force balance gives p(z)A − p(z + dz)A − ρgA dz = 0, so dp/dz = −ρg. For constant density and uniform gravity, pressure at depth h below a surface with pressure p₀ is p = p₀ + ρgh. If density varies significantly with height, integrate the differential equation rather than use one constant ρ.

At the same height in one connected static fluid of uniform density, pressure is equal. Container shape does not appear in ρgh; a narrow tube can generate the same bottom pressure as a broad vessel with the same liquid depth. The total force on a base also depends on its area, and sidewall forces explain why base force need not equal the weight of the liquid alone.

Absolute pressure is measured from vacuum; gauge pressure is measured relative to a specified ambient pressure, usually atmospheric. Thus p_absolute = p_atmosphere + p_gauge. A negative gauge reading means below ambient, not automatically a negative absolute pressure. Gas-law and thermodynamic calculations normally need absolute pressure.

02 / UNDERSTAND & APPLY

2. Walk through pressure differences and integrate surface loads

In a manometer, start at a point of known pressure and move through the connected static columns. Moving down by height h in fluid density ρ adds ρgh; moving up subtracts it. Across a level interface, pressure is continuous when surface-tension curvature effects are negligible. Equal-height comparison applies within the same connected static fluid, not indiscriminately between isolated fluids.

For a simple U-tube with one side open to atmosphere and the other connected to a gas, a height difference Δh in a dense manometric liquid gives |p_gas − p_atm| = ρgΔh when gas-density corrections are negligible. The gas side with the lower liquid level has the higher pressure. The measured Δh is the difference between both levels, not just the displacement of one arm from its original level.

Pressure on a submerged vertical plate is not uniform: p_gauge(y) = ρgy below the free surface. The net horizontal force when the other side is at atmosphere is ∫ρgy dA. For a rectangular plate of width b from the surface to depth H, F = ρgbH²/2. The force acts below the geometric centre, at depth 2H/3, because the deeper parts experience greater pressure.

To locate the centre of pressure, equate moments: y_CP F = ∫y p(y) dA. This is different from simply applying the pressure at the deepest point over the whole area. If both sides of a wall contain fluids or unequal gas pressures, compute the pressure difference at each height before integrating. Atmospheric contributions cancel only when they act equally on both sides.

03 / UNDERSTAND & APPLY

3. A hydraulic machine trades displacement for force

Pascal’s principle states that an applied pressure change in a confined fluid is transmitted throughout the static fluid under the ideal incompressible model. For two pistons at the same height with negligible friction and matched external ambient pressures, F₁/A₁ = F₂/A₂. The force can grow with piston area even though the transmitted pressure change is the same.

Volume conservation requires A₁s₁ = A₂s₂ for piston displacement magnitudes, with one piston moving inward and the other outward. Therefore the larger-force piston moves a shorter distance. Ideally F₁s₁ = F₂s₂, so mechanical advantage does not create energy. Different piston heights add a hydrostatic pressure difference; unequal external pressures must also be accounted for.

Compressibility, trapped gas, seals, friction and flow losses affect real machines. The static area-ratio law is a model of pressure transmission, not a guarantee of unlimited force or instantaneous motion. A hydraulic system’s useful output power cannot exceed its input power without another energy source.

04 / UNDERSTAND & APPLY

4. Buoyancy is the net pressure force, equal to displaced fluid weight

A submerged body experiences larger pressure on lower surfaces than on upper surfaces. Integrating pressure over its closed wetted surface gives an upward resultant equal to the weight of the displaced static fluid: B = ρ_fluid g V_displaced for uniform density. This is Archimedes’ principle. It applies to gases as well as liquids under the corresponding hydrostatic assumptions.

For a floating body in vertical equilibrium with negligible air buoyancy and no other vertical forces, B = mg. If a uniform solid of volume V and density ρ_body floats in uniform fluid, the submerged fraction is V_sub/V = ρ_body/ρ_fluid. This result requires the body to be able to float without taking on fluid or changing its effective volume; a hollow vessel uses its total mass and displaced exterior volume rather than the material density alone.

A fully submerged rigid object in an incompressible fluid has the same buoyant force at different depths when fluid density and g are unchanged. Absolute pressure increases, but the pressure difference across the object stays tied to its displaced volume. A compressible object or a fluid with variable density can behave differently. A body sealed against the bottom with no fluid underneath needs a fresh force analysis rather than a blindly applied full-surrounding-fluid formula.

If a dense body hangs from a scale while fully submerged and stationary, tension is T = mg − B under the usual vertical-force model. If the body is less dense and held down by a tether, the downward tether force instead balances excess buoyancy. The centre of buoyancy is the centroid of displaced volume for uniform fluid; rotational stability of floating shapes depends on how this point shifts with tilt, not just on vertical force balance.

05 / UNDERSTAND & APPLY

5. Continuity accounts for mass passing through a flow tube

For steady flow through a streamtube without leaks or sources, mass flow rate is constant: ρAv = constant when v denotes the section-averaged normal speed. If density is constant, volume flow rate Q = Av is constant. A narrower section therefore has greater mean speed for the same incompressible throughput.

Area is proportional to diameter squared. Halving a circular pipe’s diameter makes its area one quarter, so the same volume flow needs four times the mean speed. Halving the area instead only doubles speed. Confusing diameter with area is one of the most common numerical mistakes in continuity problems.

A streamline is tangent to the local velocity at each point at an instant. In steady flow, the velocity at a fixed location does not change with time, although a moving fluid parcel can accelerate as it enters a narrowing or bends. “Steady” does not mean “the same speed everywhere.” Streamlines cannot cross where the velocity field is single-valued and nonzero, because that would assign two directions to one point.

For compressible flow, use ρAv rather than Av. In a branched system, inflowing mass rates equal outgoing rates in steady state. Continuity alone supplies no pressure relation; it is mass conservation. To relate pressure and speed, a dynamical or energy equation with its own assumptions is required.

06 / UNDERSTAND & APPLY

6. Bernoulli is a mechanical-energy balance with conditions

For steady, incompressible, inviscid flow along a streamline, with no pump or turbine energy added or removed between the points, p + ρv²/2 + ρgz is constant. The terms represent pressure work, kinetic energy and gravitational potential energy per unit volume. Dividing by ρg gives pressure head, velocity head and elevation head, each measured in length.

The equation follows by considering an equal volume of fluid moving between two cross sections. Pressure at the inlet does work p₁ΔV, pressure at the outlet does opposing work p₂ΔV, and their difference changes the parcel’s kinetic and gravitational energy. This explains why pressure is an energy term rather than an extra force that must be added again after the balance.

At equal height on the same ideal streamline, higher speed corresponds to lower static pressure. But “faster fluid always has lower pressure” is not a universal statement between arbitrary flows, heights or separate streamlines. A pump can raise both pressure and speed, while viscous losses can reduce total mechanical head. State the comparison that permits the conclusion.

In a horizontal Venturi constriction, continuity fixes the speed increase and Bernoulli gives p₁ − p₂ = ρ(v₂² − v₁²)/2. A wall pressure tap measures static pressure, not automatically the stagnation pressure obtained by bringing a streamline to rest. A Pitot-type stagnation measurement ideally gives p_stagnation − p_static = ρv²/2 under the same incompressible lossless assumptions.

In a real pipe, viscosity transfers mechanical energy to heat, so a head-loss term is needed. If pressure predicted by a liquid-flow model approaches its vapour pressure, cavitation may invalidate the assumed continuous liquid stream. An impossible or implausible pressure is a signal to revisit the model rather than accept any algebraic output.

07 / UNDERSTAND & APPLY

7. Free surfaces and jets provide useful boundary pressures

For a large open tank discharging through a small opening at depth h below the free surface, both the free surface and emerging jet are at atmospheric pressure. If surface speed is negligible and losses are ignored, Bernoulli gives exit speed v = √(2gh), Torricelli’s result. It resembles free fall because the same gravitational energy per unit mass is converted to kinetic energy.

The small-opening assumption allows the free-surface speed to be neglected. For finite tank area A and opening area a in a simple quasisteady model, continuity gives surface speed (a/A)v and Bernoulli gives v² = 2gh/[1 − (a/A)²] before adding real discharge effects. A contracting jet and viscosity generally require a discharge coefficient for actual flow rate.

If a horizontal jet leaves an opening at height y above level ground, and the free surface is at height H, ideal exit speed is √[2g(H − y)]. Fall time is √(2y/g), giving horizontal range 2√[y(H − y)]. The range is greatest at y = H/2 for this fixed-surface, negligible-air-drag setup. A deeper opening has greater speed but less fall time.

For a tank whose free-surface height decreases, speed and flow rate change during discharge. Treat the height as a function of time and use volume conservation to derive the draining equation rather than use the initial flow rate for the whole process. The underlying quasisteady approximation assumes the local adjustment is fast compared with the overall fall of the surface.

08 / UNDERSTAND & APPLY

8. Original practice with pressure and flow checks

Use SI units, especially when converting square centimetres, litres and diameters. Keep the absolute/gauge choice consistent and identify where the fluid is static before using hydrostatic equality.

09 / UNDERSTAND & APPLY

9. Match the equation to the physical regime

First classify the region as static or flowing. In a static column, use pressure balance and add density-dependent vertical increments. For a supported or floating body, draw a force balance that includes buoyancy and any tether or contact. For a steady flow, write continuity before applying an energy relation.

Before using Bernoulli, list the streamline, density, viscosity, height and pump assumptions. A smaller pipe area determines speed through continuity, but pressure follows only after the correct energy or force model is supplied. Check that predicted flow directions, support forces and absolute pressures are physically compatible with the scenario.

• Pressure is scalar; its force acts normal to a surface.
• Hydrostatic equal-height pressure requires the appropriate connected static fluid.
• Gauge pressure and absolute pressure are different reference choices.
• Buoyancy equals displaced fluid weight under the surrounding hydrostatic-fluid model.
• Continuity conserves mass, while Bernoulli accounts for mechanical energy under stated conditions.
• Area ratios involve diameter squared.
• A boundary pressure at a free jet often simplifies an efflux calculation.