4003-03. Fluid Mechanics

💧 Fluid Mechanics

Fluid mechanics describes how liquids and gases behave under forces. You will apply hydrostatic pressure and Archimedes' principle to floating and submerged bodies, then use continuity and Bernoulli's equation to analyse flowing fluids in pipes, nozzles and Venturi meters.

P = ρghArchimedes F=ρgVManometryContinuity A₁v₁=A₂v₂BernoulliTorricelli

2Topics

8Examples

4Calculators

12Practice Qs

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What you'll cover

Two focused topics building from static fluids to flowing fluids.

⛯️Hydrostatics & BuoyancyPressure, Archimedes, manometry, flotation

💨Fluid DynamicsContinuity, Bernoulli, Torricelli, flow rate

⚙️ApplicationsFour engineering calculators

🎮Quick-fire game12 randomised questions

📝Practice questions12 graded questions with full working

⛯️

Hydrostatics & Buoyancy

Pressure in a static fluid increases linearly with depth. This governs manometers, dams, submerged structures and floating bodies via Archimedes' principle.

P = ρghGauge & absolute PF = ρgVManometryFlotation

📜 Key rules — hydrostatics & buoyancy

Hydrostatic pressure: P = ρgh (Pa), independent of the shape of the container — depends only on fluid density and vertical depth.

Gauge vs absolute pressure: P_abs = P_gauge + P_atm. Gauge pressure is measured relative to atmospheric; absolute pressure includes it.

Archimedes' principle: F_B = ρ_fluidgV_submerged. A body floats when F_B = Weight; it sinks when the maximum possible F_B (fully submerged) is less than Weight.

Flotation: fraction submerged = ρ_object ÷ ρ_fluid (for a uniform floating body). Apparent weight when submerged = Weight − F_B.

Manometer: a column of manometer fluid balances the pressure being measured — P = ρ_{manometer fluid}gh_{manometer fluid}. Mercury (ρ=13600 kg/m³) is used for high pressures because it needs a shorter column.

💡 What every symbol means, rule by rule:

P = ρgh — P is pressure (Pa), ρ ("rho") is the fluid's density (kg/m³), g is 9.81 m/s², h is vertical depth below the surface (m). No other shape or width term appears — that's what "independent of container shape" means.
P_abs = P_gauge + P_atm — P_atm is atmospheric pressure pressing down on the fluid's open surface (normally 101325 Pa). Most pressure gauges are zeroed against the atmosphere, so they read P_gauge — the pressure on top of atmospheric. Add the atmosphere back in to get the true, total pressure, P_abs.
F_B = ρ_fluidgV_submerged — F_B is the upward buoyancy force (N). ρ_fluid is the density of the fluid the object sits in (not the object's own density). V_submerged is only the volume of the object that is actually under the fluid surface — for a fully submerged object that's its whole volume; for a floating object it's just the submerged portion.
fraction submerged = ρ_object ÷ ρ_fluid — ρ_object is the density of the floating body itself. Dividing it by the fluid's density tells you what fraction of the object's volume must go underwater to displace enough fluid to support its weight. A less dense object (small ρ_object/ρ_fluid) floats higher.
P = ρ_{manometer fluid}gh_{manometer fluid} — this is just the same P = ρgh rule, applied specifically to the liquid inside the manometer tube (often mercury or water). h_{manometer fluid} is the height difference between the two liquid columns in the tube, which is what you actually read off the scale.

📚 Key vocabulary

Fluid: a substance that deforms continuously under shear stress — liquids and gases. Cannot resist shear when at rest.
Hydrostatic pressure: pressure due to the weight of fluid above a point: P = ρgh. Acts equally in all directions at a given depth.
Gauge & absolute pressure: gauge pressure is relative to atmospheric (0 at atmosphere); absolute pressure = gauge + atmospheric. Most engineering gauges read gauge pressure.
Archimedes' principle: a body submerged (fully or partially) in a fluid experiences an upward buoyancy force equal to the weight of fluid displaced.
Flotation: a floating body sinks until the weight of displaced fluid equals its own weight. Fraction submerged = ratio of densities.
Manometer: a U-tube instrument that measures pressure by balancing it against a column of liquid of known density.
Freeboard: the height of a floating body that sits above the waterline (total height − submerged depth) — the vessel's safety margin against waves or extra load pushing it under.

👁️ Visual aid (3D view) — a floating body: buoyancy force F_B (up) balances the force due to gravity F_g (down) at equilibrium. The submerged depth sets how much fluid is displaced.

✍️Worked example 1 — Archimedes: will it float? step through it

QuestionGiven: a steel block (mass 800 kg, volume 0.12 m³) is lowered into seawater (ρ = 1025 kg/m³). Find: (a) the buoyancy force, (b) the apparent weight.

WeightW = mg = 800 × 9.81 = 7848 N

BuoyancyF_B = ρgV = 1025 × 9.81 × 0.12 = 1206.63 N

Apparent wtW_app = 7848 − 1206.63 = 6641.4 N

Float?F_B (1207 N) < W (7848 N) → sinks. Fraction submerged if floating = 800/1025 = 0.780 (78.0%)

AnswerF_B = 1207 N, apparent weight = 6641 N, object sinks ✓

Step 1 of 6

✍️Worked example 2 — hydrostatic pressure in a reservoir

👁️ Visual aid (3D view) — open reservoir 8 m deep: gauge pressure builds with depth; add atmospheric pressure for absolute

QuestionGiven: a reservoir 8 m deep (ρ = 1000 kg/m³), atmospheric pressure = 101.325 kPa. Find: (a) the gauge pressure at the bottom, (b) the absolute pressure at the bottom.

Gauge PP = ρgh = 1000×9.81×8 = 78480 Pa = 78.5 kPa

Absolute PP_abs = 78480 + 101325 = 179805 Pa = 179.8 kPa

AnswerGauge: 78.5 kPa Absolute: 179.8 kPa

✍️Worked example 3 — manometer: measuring gauge pressure

👁️ Visual aid (3D view) — mercury manometer: the column height difference h balances the pipe's gauge pressure

QuestionGiven: a mercury manometer connected to a pipe shows a height difference of 0.45 m (ρ_Hg (mercury) = 13600 kg/m³). Find: the gauge pressure in the pipe.

ManometerP = ρ_Hg (mercury) × g × h = 13600×9.81×0.45 = 60037 Pa = 60.0 kPa

AnswerP = 60.0 kPa (gauge)

✍️Worked example 4 — floating pontoon: depth submerged & freeboard

👁️ Visual aid (3D view) — pontoon (1 m tall) floats with 0.585 m submerged and 0.415 m freeboard above the waterline. F_B and F_g act through the same central line, pulling in opposite directions, and pulse to show they're balanced.

QuestionGiven: a rectangular pontoon (ρ = 600 kg/m³, 4 m × 2 m base, 1 m tall) floats in seawater (ρ = 1025 kg/m³). Find: (a) the submerged depth, (b) the freeboard (height above water).

VolumeV = 4×2×1 = 8 m³

Why ρV?Density is mass per unit volume (ρ = m/V). Rearranged, that gives m = ρV — multiply the pontoon's own density by its total volume to get its mass.

Massm = ρV = 600×8 = 4800 kg

WeightF_g = mg = 4800×9.81 = 47088 N

EquilibriumF_B = F_g — floating means the buoyancy force exactly balances the force due to gravity (weight)

Substituteρ_swgV_sub = mg

Why it cancelsg appears on both sides of the equation, so it cancels — you can solve using mass directly without ever calculating weight in Newtons.

Solve for volumeV_sub = m ÷ ρ_sw = 4800 ÷ 1025 = 4.683 m³

Depthd = V_sub ÷ base area = 4.683 ÷ 8 = 0.585 m

AnswerSubmerged depth = 0.585 m Freeboard (height above water) = 1 − 0.585 = 0.415 m

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⛯️Hydrostatics & buoyancy calculatorSelect calculation type

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Fluid Dynamics — Continuity & Bernoulli

Once a fluid moves, mass conservation (continuity) and energy conservation (Bernoulli) let you relate velocity, pressure and height at different points in a flow.

Continuity A₁v₁=A₂v₂BernoulliTorricelliFlow rate Q=AvMass flow ṁ=ρAv

📜 Key rules — fluid dynamics

Continuity: A₁v₁ = A₂v₂ (incompressible flow). Volumetric flow rate Q = Av (m³/s). Mass flow rate ṁ = ρAv (kg/s).

Bernoulli: P + ½ρv² + ρgh = constant along a streamline for ideal (inviscid, incompressible) flow. Trades pressure, kinetic and potential energy per unit volume.

Venturi effect: where a pipe narrows, continuity forces velocity up, so Bernoulli forces pressure down — the basis of venturi meters and carburettors.

Torricelli's theorem: v = √(2gh) for a free jet from an orifice a depth h below a reservoir surface — Bernoulli applied between the surface and the jet.

Streamline flow assumes no energy loss to friction; real pipe flow includes head losses, but Bernoulli gives the ideal baseline used throughout this unit.

📚 Key vocabulary

Streamline: a line tangent to the velocity of every fluid particle — no flow crosses a streamline. Bernoulli applies along a streamline.
Continuity: conservation of mass for incompressible flow: the volume flow rate is the same at every cross-section of a pipe, A₁v₁=A₂v₂.
Volumetric & mass flow rate: Q = Av (m³/s) is volume passing per second; ṁ = ρAv (kg/s) is mass passing per second.
Bernoulli's equation: conservation of energy per unit volume along a streamline: P + ½ρv² + ρgh = constant. Applies to ideal (inviscid, incompressible) flow.
Viscosity: internal resistance of a fluid to shear. Dynamic viscosity μ (Pa·s). Newtonian fluids have constant viscosity.
Torricelli's theorem: the velocity of a jet leaving an orifice a depth h below a free surface equals that of a body falling freely through height h: v = √(2gh).

👁️ Visual aid (3D view) — Bernoulli: high-pressure wide section slows down; narrow section speeds up, pressure drops (Venturi effect)

✍️Worked example 5 — Bernoulli in a pipe step through it

💡 Area of a circle: pipes are round, so their cross-sectional area comes from A = πr², where r is the radius — half the diameter (r = d ÷ 2), not the diameter itself. Here, diameter 0.2 m gives radius 0.1 m, so A₁ = π×(0.1)² = 0.03142 m². Sanity check: because r is squared, doubling a pipe's diameter roughly quadruples its area — small diameter changes have a big effect on flow.

QuestionGiven: water flows in a horizontal pipe (ρ = 1000 kg/m³). Section 1: diameter 0.2 m, velocity 1.5 m/s, gauge pressure 80 kPa. Section 2: diameter 0.1 m. Find: (a) the velocity v₂ at section 2, (b) the gauge pressure P₂ at section 2.

AreasA₁ = π(0.1)² = 0.03142 m² A₂ = π(0.05)² = 0.007854 m²

ContinuityA₁v₁ = A₂v₂ → v₂ = A₁v₁/A₂ = 0.03142×1.5/0.007854 = 6 m/s

Bernoulli (h₁=h₂)P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₂P₂ = 80000 + ½×1000×(1.5²−6²) = 80000 − 16875 = 63125 Pa

Answerv₂ = 6 m/s, P₂ = 63.1 kPa ✓

Step 1 of 6

✍️Worked example 6 — Torricelli: free-jet velocity

👁️ Visual aid (3D view) — orifice 3.5 m below the surface: the jet leaves as if it had fallen freely through h

QuestionGiven: a large tank has a small orifice 3.5 m below the water surface. Find: the velocity of the emerging jet (Torricelli's theorem).

Bernoulliv = √(2gh) = √(2×9.81×3.5) = √68.67 = 8.287 m/s

Answerv = 8.29 m/s

✍️Worked example 7 — continuity: pipe diameter reduction

👁️ Visual aid (3D view) — pipe reduces from 250 mm to 100 mm: the same mass must pass, so velocity rises where area falls

QuestionGiven: water flows in a pipe that reduces from 250 mm to 100 mm diameter, with inlet velocity 1.2 m/s. Find: (a) the outlet velocity, (b) the volumetric flow rate.

Why πr²?Radius = diameter ÷ 2, not the diameter itself: 250 mm → r = 0.125 m; 100 mm → r = 0.05 m.

AreasA₁ = π(0.125)² = 0.04909 m² A₂ = π(0.05)² = 0.007854 m²

Continuityv₂ = A₁v₁/A₂ = 0.04909×1.2/0.007854 = 7.50 m/s

Flow rateQ = A₁v₁ = 0.04909×1.2 = 0.0589 m³/s

Answerv₂ = 7.50 m/s Q = 0.0589 m³/s

✍️Worked example 8 — venturi meter: combined continuity & Bernoulli

👁️ Visual aid (3D view) — venturi meter: continuity and Bernoulli combine to link the 35 kPa pressure drop to throat velocity

QuestionGiven: a venturi meter has inlet diameter 200 mm and throat diameter 100 mm (ρ = 1000 kg/m³), with a pressure difference of 35 kPa between inlet and throat. Find: (a) the throat velocity, (b) the mass flow rate.

Area ratioA₂/A₁ = (50/100)² = 0.25 → v₁ = 0.25v₂ (continuity)

BernoulliP₁−P₂ = ½ρv₂²(1−0.25²) = 468.75v₂²

Solve v₂35000 = 468.75v₂² → v₂² = 74.67 → v₂ = 8.64 m/s

Why πr²?Throat diameter 100 mm gives radius r = 0.05 m, so A₂ = πr², not π×diameter².

Mass flowA₂ = π(0.05)² = 0.007854 m² ṁ = ρA₂v₂ = 1000×0.007854×8.64 = 67.9 kg/s

Answerv₂ = 8.64 m/s ṁ = 67.9 kg/s

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⚙️

Engineering Applications

Four calculators covering hydrostatics, manometry, continuity and Bernoulli flow analysis.

⛯️ Hydrostatic pressure & buoyancy

Find the pressure at depth and the buoyancy force on a submerged body.

P = ρgh F_B = ρgV

🧮 Calculator

Fluid ρ (kg/m³)

Depth h (m)

Volume V (m³)

Enter fluid density, depth and volume.

📊 Manometer → gauge pressure

Convert a manometer column reading into a gauge pressure.

P = ρ_mgh_m

🧮 Calculator

Manometer fluid ρ (kg/m³)

Column height h (m)

Atmospheric P (Pa) — optional

Enter manometer fluid density and column height.

💨 Continuity & flow rate

Find the outlet velocity and the volumetric / mass flow rate for a pipe of changing area.

A₁v₁ = A₂v₂ Q = Av ṁ = ρAv

🧮 Calculator

A₁ (m²)

v₁ (m/s)

A₂ (m²)

ρ (kg/m³) — optional

Enter A₁, v₁ and A₂.

📈 Bernoulli: pipe flow analysis

Given conditions at point 1, find velocity at point 2 using Bernoulli and continuity (horizontal pipe).

P₁+½ρv₁² = P₂+½ρv₂² A₁v₁=A₂v₂

🧮 Calculator

ρ (kg/m³)

P₁ (Pa)

v₁ (m/s)

P₂ (Pa)

Enter P1, v1, P2 and fluid density.

📜

History & Discovery

Every formula in this unit was worked out by someone, usually centuries before pipes, pumps or pressure gauges existed. Here's where the ideas came from.

Archimedes · c.250 BCPascal · 1653Bernoulli · 1738Torricelli · 1643

🧤 Archimedes of Syracuse (c. 287–212 BC, Ancient Greece)

Asked to check whether a king's crown was pure gold without melting it down, Archimedes noticed — supposedly while lowering himself into a bath and watching the water level rise — that an object displaces its own volume of fluid, and that the fluid pushes back with a force equal to the weight of what it displaced. He is said to have run through the streets shouting "Eureka!" ("I have found it!").

💡 The principle, plainly: any object submerged, fully or partly, in a fluid feels an upward push equal to the weight of the fluid it has shoved out of the way. This depends only on the volume displaced, not on the object's own weight — if that upward push is bigger than the object's weight, it floats; if it's smaller, it sinks.

F_B = ρ_fluidgV_submerged — used in Worked examples 1 & 4 and the Hydrostatics calculator

Archimedes' principle is the reason ships, submarines and hot-air-balloon-style pontoons can be designed at all — every flotation and submerged-weight calculation in this unit traces straight back to a 2,200-year-old bath.

⚖️ Blaise Pascal (1623–1662, France)

Pascal, better known today for his work in mathematics and probability, showed in 1653 that pressure applied anywhere to a fluid confined in a closed container is transmitted equally and undiminished in every direction. This is why a small force on a narrow piston can be turned into a huge force on a wide one — the working principle behind every hydraulic jack, press and brake system.

💡 The principle, plainly: pressure applied anywhere to a fluid sealed inside a container spreads out equally and undiminished in every direction. Push on the fluid at one point, and every other point in the container feels that same extra push — not just the spot you pressed.

P_abs = P_gauge + P_atm — underlies gauge/absolute pressure work in Worked example 2 and the pressure unit, Pascal (Pa), named after him

Every SI unit of pressure (the pascal, Pa = N/m²) carries his name. Hydraulic vehicle brakes, car jacks and excavator arms all rely directly on Pascal's law.

💫 Daniel Bernoulli (1700–1782, Switzerland)

In his 1738 book Hydrodynamica, Bernoulli showed that for a flowing fluid, pressure, speed and height trade off against each other so that their combined energy per unit volume stays constant along a streamline. Faster-moving fluid exerts lower pressure — the same effect that generates lift under an aircraft wing and draws fuel into a carburettor.

💡 The principle, plainly: along a smooth streamline, a moving fluid's pressure energy, kinetic (motion) energy and height energy always add up to the same total. Speed the fluid up, and its pressure has to drop to make room in that total; slow it down, and pressure rises again.

P + ½ρv² + ρgh = constant — used in Worked examples 5 & 8 and the Bernoulli calculator

Venturi meters, carburettors, aircraft wings and even how a shower curtain gets sucked inward all come down to Bernoulli's trade-off between pressure and velocity.

🌌 Evangelista Torricelli (1608–1647, Italy)

A student of Galileo, Torricelli built the first mercury barometer in 1643, proving that air itself has weight and exerts pressure — disproving the old belief that "nature abhors a vacuum." He also worked out that a jet of liquid escaping from a hole below a free surface leaves with the same speed as an object that had fallen freely through that same height.

💡 The principle, plainly: a jet of liquid leaving a hole a depth h below the surface shoots out at exactly the speed a dropped object would reach after falling that same height h — both are just gravity converting the same stored energy into speed.

v = √(2gh) — used in Worked example 6 and the Torricelli calculator

The unit "torr" for pressure (still used in vacuum engineering) is named after him. Torricelli's theorem is the starting point for analysing tank drainage, sprinkler jets and emergency spillway design.

💡 Spot the pattern: every rule in this unit is 60–2,300 years old, yet still describes fluids exactly — from a leaking tank to a jet engine intake. Good physics doesn't go out of date.

🎮

Fluid Mechanics Challenge

12 randomised questions — hydrostatics, buoyancy, continuity, Bernoulli and Torricelli.

⛯️ Quick-fire round

Question 1 / 12Score: 0Streak: 0 🔥

Press Start to begin!

📝

Practice Questions

12 graded questions — type your answer, check, then reveal full working.

⛯️ g = 9.81 m/s². State equations used. Give answers to 3 s.f. unless stated.

Q1 · Hydrostatic P1 mark

Find gauge pressure at the base of a 12 m water column. (ρ = 1000 kg/m³)

P = ρgh = 1000×9.81×12 = 117720 Pa = 117.7 kPa

Q2 · Archimedes2 marks

A wooden block (mass 50 kg, volume 0.1 m³) floats in water (ρ = 1000 kg/m³). Find the buoyancy force and fraction submerged.

W = 50×9.81 = 490.5 N. Floating: F_B = W = 490.5 N
Fraction submerged = 50/1000/0.1 = 0.5 (50%)

Q3 · Continuity2 marks

Water flows in a pipe. Diameter reduces from 0.3 m to 0.15 m. Inlet velocity = 2 m/s. Find exit velocity.

A₁ = π(0.15)² = 0.07069 m² A₂ = π(0.075)² = 0.01767 m²
v₂ = A₁v₁/A₂ = 0.07069×2/0.01767 = 8 m/s

Q4 · Bernoulli2 marks

Horizontal pipe: P₁ = 120 kPa, v₁ = 2 m/s, v₂ = 8 m/s. (ρ = 1000) Find P₂.

P₂ = 120000 + ½×1000×(4−64) = 120000 − 30000 = 90000 Pa = 90 kPa

Q5 · Torricelli1 mark

An orifice is 6 m below the water surface. Find the jet velocity.

v = √(2×9.81×6) = √117.72 = 10.85 m/s

Q6 · Archimedes apparent weight2 marks

An aluminium component (density 2700 kg/m³, volume 0.005 m³) is fully submerged in oil (ρ = 900 kg/m³). Find apparent weight.

W = 2700×9.81×0.005 = 132.435 N
F_B = 900×9.81×0.005 = 44.145 N
W_app = 132.435−44.145 = 88.29 N

Q7 · Manometer2 marks

A mercury manometer shows a column difference of 0.38 m (ρ_Hg = 13600 kg/m³). Find the gauge pressure.

P = 13600×9.81×0.38 = 50698 Pa ≈ 50.7 kPa

Q8 · Absolute pressure2 marks

Find the absolute pressure at a depth of 15 m in seawater (ρ = 1025 kg/m³), given atmospheric pressure = 101.325 kPa.

P_gauge = 1025×9.81×15 = 150806 Pa
P_abs = P_gauge + P_atm = 150806 + 101325 = 252131 Pa ≈ 252.1 kPa

Q9 · Flotation fraction1 mark

A block of density 720 kg/m³ floats in water (ρ = 1000 kg/m³). Find the fraction submerged.

Fraction submerged = ρ_object/ρ_fluid = 720/1000 = 0.72 (72%)

Q10 · Mass flow rate2 marks

Water flows through a pipe of diameter 0.12 m at 3 m/s. Find the mass flow rate. (ρ = 1000 kg/m³)

A = π(0.06)² = 0.011310 m² Q = Av = 0.033929 m³/s
ṁ = ρQ = 1000×0.033929 = 33.93 kg/s

Q11 · Continuity1 mark

A pipe area reduces from 0.05 m² to 0.02 m². Inlet velocity = 4 m/s. Find the outlet velocity.

v₂ = A₁v₁/A₂ = 0.05×4/0.02 = 10 m/s

Q12 · Combined Venturi4 marks

A horizontal venturi has A₁ = 0.08 m², A₂ = 0.02 m², P₁ = 150 kPa, v₁ = 2 m/s (ρ = 1000 kg/m³). Find v₂ and P₂.

v₂ = A₁v₁/A₂ = 0.08×2/0.02 = 8 m/s
P₂ = P₁+½ρ(v₁²−v₂²) = 150000+500×(4−64) = 150000−30000 = 120000 Pa = 120 kPa