Physics for ME · Module 13 of 16 · Fluids preview
Fluids: Pressure, Buoyancy, and Flow Intuition
Pressure, Archimedes, continuity, and a first feel for Bernoulli. Deliberately introductory: the full Fluid Mechanics course comes later.
Readiness check
From Modules 1, 4, and 6. Tick only what you can do closed-notes.
- Work with pressure units: Pa, kPa, bar.
- Compute forces from F = PA with unit care.
- Use density ρ = m/V fluently.
- Run an energy audit (for Bernoulli intuition).
- Draw an FBD with a pressure force on a surface.
The core idea
Pressure is force spread over area, it grows with depth, and enclosed fluids transmit it everywhere.
P = F/AP = P₀ + ρghFb = ρfluidgVPascal's principle (pressure transmitted undiminished) powers every hydraulic machine. Archimedes' buoyancy is hydrostatics applied to a submerged body. For moving fluid, continuity (Av constant) and Bernoulli (pressure-velocity trade) supply the first intuition.
The skills, taught in order
Fluids split into two stories: fluid at rest, where depth sets pressure, and fluid in motion, where speed trades against pressure. Six skills cover the introductory half of both, enough for hydraulics and a first feel for flow.
13.1 Pressure and its units
Pressure is force spread over area, P = F/A, measured in pascals (1 Pa = 1 N/m²). Useful sizes: 1 bar = 10⁵ Pa and 1 atmosphere ≈ 101 kPa. Gauge pressure is measured above atmospheric; absolute pressure includes it.
13.2 Hydrostatics
In a still fluid, pressure rises with depth as P = P₀ + ρgh. It depends only on depth, density, and gravity, never on the container's width or shape, which is why a thin standpipe can load a dam like a lake.
Micro-example. Ten metres of water adds ρgh = 1000 × 9.81 × 10 ≈ 98 kPa, about one atmosphere, so every 10 m of depth roughly doubles the absolute pressure.
13.3 Pascal's principle
Pressure applied to an enclosed fluid is transmitted undiminished to every part of it. A small force on a small piston becomes a large force on a large one, F₂ = F₁·A₂/A₁: the hydraulic lever of Worked Example 1.
13.4 Buoyancy
A submerged body feels an upward force equal to the weight of the fluid it displaces: Fb = ρfluidgV. It depends on the fluid's density and the displaced volume, not on the object's weight; comparing the two decides float or sink.
Micro-example. Ice floats because its density (about 920 kg/m³) is below water's, so it displaces its own weight while still poking above the surface.
13.5 Continuity
An incompressible fluid that speeds up must have somewhere narrower to flow: A₁v₁ = A₂v₂. Halving the diameter quarters the area and so quadruples the speed.
13.6 Bernoulli intuition
Along an ideal (frictionless, steady) streamline, P + ½ρv² + ρgy stays constant. Where a flow speeds up its pressure drops, the trade behind venturi meters and aerofoil lift. Real pipes add losses, the work of the full Fluid Mechanics course.
| Law | Relation | Regime |
|---|---|---|
| Pressure | P = F/A | any |
| Hydrostatic pressure | P = P₀ + ρgh | fluid at rest |
| Pascal (hydraulics) | F₁/A₁ = F₂/A₂ | fluid at rest |
| Buoyancy | Fb = ρfluidgV | fluid at rest |
| Continuity | A₁v₁ = A₂v₂ | fluid in motion |
| Bernoulli | P + ½ρv² + ρgy = const | ideal flow |
Engineering connection: a preview, not the full course; it prepares hydraulics, pumps, and flow meters, and feeds the Fluid Mechanics course.
Worked example 1: the hydraulic lift
A workshop lift supports a 12 000 N car on a 200 cm² piston. The effort piston has an area of 10 cm². Find the required effort force, the working pressure, and the distance trade-off.
- ProblemFind the effort force and system pressure in Figure 1, and the stroke ratio.
- Given / findLoad 12 000 N on A₂ = 200 cm² = 0.02 m²; effort area A₁ = 10 cm² = 0.001 m².
- AssumptionsIncompressible fluid, negligible piston weights and friction, equal piston heights.
- ModelPascal: the pressure under both pistons is the same, so F₁/A₁ = F₂/A₂.
- EquationsP = F₂/A₂ F₁ = P·A₁
- SolveP = 12 000/0.02 = 600 kPa (6 bar). F₁ = 600 000 × 0.001 = 600 N: a 20:1 multiplication. Volume conservation makes the effort piston travel 20 times farther than the car rises.
- CheckEnergy audit (Module 6): F₁d₁ = 600 × 20h = 12 000 × h = F₂d₂. Force is multiplied, work is not: no free lunch, just a force-distance trade like a lever.
- ConclusionSix bar, a hand-scale force, and geometry hold a car: that is all of fluid power. Brakes, presses, and excavators scale this same triangle of P, A, and stroke.
Worked example 2: flow through a constriction
Water (ρ = 1000 kg/m³) flows through a horizontal pipe that narrows from 50 mm to 25 mm diameter. In the wide section the speed is 2.0 m/s and the pressure is 300 kPa. Find the speed and pressure in the narrow section.
- ProblemFind the speed and pressure in the narrow section of the pipe in Figure 2.
- Given / findρ = 1000 kg/m³; wide d₁ = 50 mm at v₁ = 2.0 m/s and P₁ = 300 kPa; narrow d₂ = 25 mm; horizontal. Find v₂ and P₂.
- AssumptionsIncompressible, steady, effectively frictionless flow along a horizontal streamline (the ideal Bernoulli case).
- ModelContinuity fixes the speed from the area change; Bernoulli then trades that speed for pressure.
- EquationsA₁v₁ = A₂v₂ P₁ + ½ρv₁² = P₂ + ½ρv₂²
- SolveArea scales as diameter squared, so the 2:1 diameter gives a 4:1 area and v₂ = 4 × 2.0 = 8.0 m/s. Then P₂ = P₁ + ½ρ(v₁² − v₂²) = 300 000 + 500(4 − 64) = 270 kPa, a 30 kPa drop.
- CheckFaster flow sits at lower pressure, as Bernoulli demands. Units: ½ρv² = (kg/m³)(m²/s²) = Pa, matching the pressures. The drop is real and measurable, which is how a venturi meter reads flow rate.
- ConclusionSqueeze a flow and it speeds up (continuity) while its pressure falls (Bernoulli). That paired trade runs venturi meters, carburettors, and aerofoil lift, and it is the first idea the full Fluid Mechanics course makes rigorous once losses are added.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Pressure depends on tank width | Wide reservoirs assumed to press harder | "What does P = ρgh contain?" | Depth, density, gravity: width never appears. A thin standpipe loads a dam's base like a lake. |
| Buoyancy linked to the object's weight | "Heavy things get less lift" | "Whose density sits in ρgV?" | The fluid's. Buoyancy depends on displaced volume only; weight decides sink or float. |
| Hydraulics as free energy | 20× force celebrated without the stroke cost | "What happened to the distances?" | Work in = work out (minus losses): the force gain is paid in travel. |
| Bernoulli everywhere | Ideal pressure-velocity trades in long, narrow, real pipes | "Are losses negligible on this path?" | Bernoulli is the frictionless ideal. Real pipe networks need the loss terms of Fluid Mechanics. |
Practice ladder
Find the gauge pressure 3.5 m below the surface of a water tank, and the force on a 20 × 20 cm inspection hatch at that depth.
Show answer
P = 1000 × 9.81 × 3.5 = 34.3 kPa. F = PA = 34 300 × 0.04 = 1373 N: a hatch the size of a sheet of paper carries a person's weight and a half.
A solid aluminium part (ρ = 2700 kg/m³) of volume 0.002 m³ hangs from a crane scale, fully submerged in water. What does the scale read?
Show answer
Weight = 2700 × 0.002 × 9.81 = 53.0 N. Buoyancy = 1000 × 0.002 × 9.81 = 19.6 N. Scale: 33.4 N. Submerged weighing measures volume: Archimedes' original trick.
Water flows at 2 m/s in a 50 mm pipe that necks down to 25 mm. Find the speed in the neck and state (Bernoulli intuition) what happens to the pressure there.
Show answer
Area ratio 4:1, so v = 8 m/s in the neck. Faster means lower pressure in the ideal trade: the venturi principle behind carburetors and flow meters.
Examine one real hydraulic device (car jack, brake system, log splitter spec sheet). Identify both areas or bore sizes, compute the force ratio and working pressure at rated load, and check the pressure against the hose or seal rating.
What good work looks like
Bore data sourced, the F/A arithmetic shown, the system pressure compared with a component rating, and the stroke trade-off mentioned.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Verify Archimedes at home: weigh an object, then weigh it submerged (hang it from a kitchen scale into water). Compute its volume and density from the difference and identify the material.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the hydrostatic pressure law and its three ingredients.
P = P₀ + ρgh: surface pressure, fluid density, depth. Geometry of the container is absent.
2. State Pascal's principle and its machine consequence.
Pressure applied to an enclosed fluid transmits undiminished: F₂ = F₁·A₂/A₁, the hydraulic lever.
3. State Archimedes' principle precisely.
The buoyant force equals the weight of the displaced fluid: Fb = ρfluidgVdisplaced, upward.
4. What does continuity conserve, and what follows in a neck?
Volume flow Av for incompressible fluid: smaller area, faster flow.
5. What is Bernoulli's trade, and its standing assumption?
Along an ideal streamline, pressure + ½ρv² + ρgh stays constant: valid only when losses are negligible.