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University Physics I · Optional: Fluids · 15.9

Fluids problem studio

Eight problems, from pressure with depth to a real pipe's viscous losses. Each names the model it is allowed to use — hydrostatic or flowing, ideal or dissipative — and holds you to three significant figures.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Hydrostatics, hydraulics, buoyancy, continuity, Bernoulli flow, and assumption checks. Treat this course-map statement as a claim to test rather than an invitation to import a familiar equation. In Optional: Fluids, begin from pressure modelling, then state the system, observable, assumptions, and evidence before calculating.

Simple definition
Hydrostatics, hydraulics, buoyancy, continuity, Bernoulli flow, and assumption checks.
Example
A strong response uses streamline and control-volume sketches and states where the model stops being reliable.
01

The subsection's claim

Hydrostatics, hydraulics, buoyancy, continuity, Bernoulli flow, and assumption checks.

02

How to work with it

Start from pressure modelling. Then declaring the fluid model before selecting hydrostatic or flow equations. Select an equation only after its variables and assumptions match the stated system.

03

What evidence would decide

Does the buoyant force equal the weight of displaced fluid within uncertainty? Useful evidence includes force and volume measurements, fluid density, prediction comparison, and uncertainty.

04

Keep the boundary visible

This GioPhysics course map is an adaptable teaching sequence, not a claim of accreditation or a universal university syllabus. Departments can adjust the order, mathematical depth, laboratory hours, and optional fluids endpoint to match local requirements. A result should be checked against units, signs, limiting cases, and the conditions under which its model was derived.

02

Change one variable at a time

Make the relationship visible.

Interactive model
1.45 m s⁻¹
26 mm
0.26

Narrow the throat: v₂ climbs as (d₁/d₂)² and the dashed Bernoulli line falls as that square again. Then lift K off zero — the solid line drops below the dashed one, and that gap is the dissipation Bernoulli has no term for.

Interactive physics modelA horizontal water pipe steps from a 62 mm bore to a 26 mm bore. Continuity lifts the mean speed from 1.45 to 8.25 m s⁻¹, Bernoulli takes 32.9 kPa off the gauge pressure, and dissipation takes a further 8.8 kPa that no ideal-fluid model predicts.p₁ = 245 kPa gaugeideal p₂ = 212.1 kPareal p₂ = 203.2 kPad₁ = 62 mm · v₁ = 1.45 m s⁻¹d₂ = 26 mm · v₂ = 8.25 m s⁻¹

THROAT SPEED v₂8.25 m s⁻¹

BERNOULLI DROP32.9 kPa

DISSIPATED8.8 kPa

THROAT Re214 ×10³

Live interpretationTHROAT SPEED v₂: 8.25 m s⁻¹. BERNOULLI DROP: 32.9 kPa. DISSIPATED: 8.8 kPa. THROAT Re: 214 ×10³

03

Catch the common trap

Explain before calculating.

What should a Fluids problem studio solution make visible?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

01 · EasyA diver works at a depth of 12.5 m in fresh water (ρ = 1000 kg m⁻³). Atmospheric pressure at the surface is 101.3 kPa and g = 9.81 m s⁻². Find the gauge pressure and the absolute pressure at that depth. A circular viewport of diameter 0.180 m is set flush in the diver's housing, whose interior is held at atmospheric pressure; find the net inward force on that viewport.
  1. Gauge pressure is the hydrostatic increase below the free surface: pgauge = ρgh = (1000)(9.81)(12.5) = 122 625 Pa = 123 kPa.
  2. Absolute pressure adds the atmosphere pressing on that surface: pabs = 101.3 + 122.625 = 223.925 kPa = 224 kPa.
  3. Viewport area: A = πr² = π(0.0900)² = 2.5447 × 10⁻² m².
  4. Water pushes in with pabs A while the housing air pushes out with patm A, so the net load is the gauge pressure alone: F = pgauge A = (122 625)(2.5447 × 10⁻²) = 3120.4 N.
  5. The viewport is flat and pressure varies linearly with depth, so the resultant is exactly the centre-depth pressure times the area — the uniform treatment costs nothing here. What it hides is the line of action: for a vertical port the pressure changes by ρg(0.180) = 1.77 kPa across it, putting the centre of pressure r²/(4h) = 0.16 mm below its centre.

Answerpgauge = 123 kPa, pabs = 224 kPa, F = 3.12 kN inward.

02 · EasyA hydraulic jack has an input piston of diameter 30.0 mm and an output piston of diameter 240 mm, joined by oil in a sealed circuit. Both pistons sit at the same height. A worker pushes the input piston down with a force of 180 N through a stroke of 0.240 m. Find the pressure increment in the oil, the lifting force at the output piston, and how far the output piston rises.
  1. Areas: A₁ = π(0.0150)² = 7.0686 × 10⁻⁴ m², A₂ = π(0.120)² = 4.5239 × 10⁻² m². The area ratio is (0.120/0.0150)² = 64.0.
  2. Pascal's principle: a pressure increment applied to an enclosed fluid is transmitted undiminished. Δp = F₁/A₁ = 180/(7.0686 × 10⁻⁴) = 2.5465 × 10⁵ Pa = 255 kPa.
  3. The same Δp acts on the output piston: F₂ = ΔpA₂ = F₁(A₂/A₁) = 180 × 64.0 = 11 520 N.
  4. Treat the oil as incompressible, so the swept volumes match: A₁d₁ = A₂d₂, giving d₂ = 0.240/64.0 = 3.75 × 10⁻³ m.
  5. Model check: work in = (180)(0.240) = 43.2 J; work out = (11 520)(0.00375) = 43.2 J. Force is multiplied 64 times, energy is not — and any oil compressibility or seal friction would make the output work smaller, never larger.

AnswerΔp = 255 kPa, F₂ = 11.5 kN, output rise d₂ = 3.75 mm.

03 · MediumA rectangular timber raft measures 2.40 m × 1.80 m × 0.220 m and has density 620 kg m⁻³. It floats flat in fresh water (1000 kg m⁻³). Find its draft — the depth of the submerged part — then find the extra load, in kilograms and in newtons, that would bring the top face exactly level with the water surface.
  1. Raft volume: V = (2.40)(1.80)(0.220) = 0.95040 m³. Mass: m = ρV = (620)(0.95040) = 589.25 kg.
  2. Floating equilibrium is FB = W, so ρwater(A d)g = ρwood(A t)g. The plan area A and g cancel: d = t(ρwood/ρwater) = (0.220)(0.620) = 0.1364 m.
  3. Fully submerged, the raft would displace ρwater V = (1000)(0.95040) = 950.40 kg of water.
  4. The load rides on the deck and displaces nothing itself, so it is the difference in displaced mass: Δm = 950.40 − 589.25 = 361.15 kg.
  5. As a weight: W = Δm g = (361.15)(9.81) = 3542.9 N. This is the load at which the raft is on the point of swamping — stability, not just flotation, would fail first in any real sea state.

AnswerDraft = 0.136 m; extra load = 361 kg, weight 3.54 kN.

04 · MediumA machine part hangs from a spring balance. In air the balance reads 7.85 N; with the part fully submerged in water (1000 kg m⁻³) it reads 6.85 N. Find the buoyant force, the volume of the part, and its density. Then state the assumption that lets you treat the air reading as the true weight.
  1. The balance reads the tension, which is the apparent weight. Vertical equilibrium gives T = W − FB, so FB = 7.85 − 6.85 = 1.00 N.
  2. Archimedes: FB = ρwater g Vdisplaced, so V = 1.00/[(1000)(9.81)] = 1.0194 × 10⁻⁴ m³. Fully submerged means Vdisplaced = Vobject.
  3. Mass from the air reading: m = W/g = 7.85/9.81 = 0.80020 kg.
  4. Density: ρ = m/V = 0.80020/(1.0194 × 10⁻⁴) = 7850 kg m⁻³. The shortcut ρ = ρwater Wair/(Wair − Wapp) = 1000 × 7.85/1.00 gives the same number without finding V.
  5. Assumption: air buoyancy on the part is neglected. Air displaced is about 1.2 kg m⁻³ against 7850 kg m⁻³, so the air reading is low by roughly 0.015% — far below the balance resolution.

AnswerFB = 1.00 N, V = 1.02 × 10⁻⁴ m³, ρ = 7850 kg m⁻³ — consistent with steel.

05 · MediumWater flows steadily through a horizontal pipe whose internal diameter narrows from 62.0 mm to 26.0 mm. In the wide section the mean speed is 1.45 m s⁻¹. Find the cross-sectional area of each section, the volume flow rate, and the mean speed in the narrow section.
  1. Areas from the internal radii: A₁ = π(0.0310)² = 3.0191 × 10⁻³ m² and A₂ = π(0.0130)² = 5.3093 × 10⁻⁴ m².
  2. Volume flow rate at the wide section: Q = A₁v₁ = (3.0191 × 10⁻³)(1.45) = 4.3777 × 10⁻³ m³ s⁻¹, about 4.38 litres per second.
  3. Continuity for an incompressible fluid in steady flow: the same Q passes every cross-section, so A₁v₁ = A₂v₂.
  4. v₂ = Q/A₂ = (4.3777 × 10⁻³)/(5.3093 × 10⁻⁴) = 8.245 m s⁻¹. Equivalently v₂ = v₁(d₁/d₂)² = 1.45 × (62.0/26.0)² = 1.45 × 5.6864.
  5. Note the leverage: halving the diameter quadruples the speed, so a modest 2.38× diameter ratio gives a 5.69× speed ratio.

AnswerA₁ = 3.02 × 10⁻³ m², A₂ = 5.31 × 10⁻⁴ m², Q = 4.38 × 10⁻³ m³ s⁻¹, v₂ = 8.25 m s⁻¹.

06 · MediumContinue with the pipe from the previous problem: v₁ = 1.45 m s⁻¹ in the 62.0 mm section and v₂ = 8.245 m s⁻¹ in the 26.0 mm section, both at the same height. A gauge on the wide section reads 245 kPa. Treating the water as an ideal fluid in steady flow along a streamline, find the gauge pressure in the narrow section.
  1. Bernoulli along a streamline: p₁ + ½ρv₁² + ρgy₁ = p₂ + ½ρv₂² + ρgy₂. The pipe is horizontal, so y₁ = y₂ and the ρgy terms cancel.
  2. Dynamic pressure in the wide section: ½ρv₁² = (500)(1.45²) = (500)(2.1025) = 1051.3 Pa.
  3. Dynamic pressure in the narrow section: ½ρv₂² = (500)(8.245²) = (500)(67.980) = 33 990 Pa.
  4. Pressure drop: p₁ − p₂ = ½ρ(v₂² − v₁²) = 33 990 − 1051 = 32 939 Pa = 32.9 kPa.
  5. p₂ = 245 − 32.9 = 212 kPa gauge. The water is accelerated by that pressure difference, not by the walls — which is why the fast section is the low-pressure one, the reverse of most students' first guess.

Answerp₂ = 212 kPa gauge, a drop of 32.9 kPa from the wide section.

07 · HardA siphon of uniform bore, primed and running, carries water out of a wide open tank. The outlet is 1.85 m below the tank's free surface, and the tube's highest point (the crest) is 0.950 m above that surface. Take patm = 101.3 kPa, ρ = 1000 kg m⁻³, g = 9.81 m s⁻². Find the outlet speed and the absolute pressure at the crest, then say whether the siphon can run and what sets the limit.
  1. Bernoulli from the free surface (p = patm, v ≈ 0, take y = 0) to the outlet (p = patm, y = −1.85 m). The atmospheric terms cancel and the surface speed is negligible because the tank is wide, leaving ½ρv² = ρg(1.85).
  2. v = √(2gh) = √(2 × 9.81 × 1.85) = √36.297 = 6.0247 m s⁻¹. This is Torricelli's result: tube length and shape drop out, only the surface-to-outlet drop matters.
  3. The bore is uniform, so continuity forces the same speed at the crest: vcrest = 6.0247 m s⁻¹.
  4. Surface to crest: pcrest = patm − ρg(0.950) − ½ρvcrest² = 101 300 − 9319.5 − 18 148.5 = 73 832 Pa.
  5. Model check: the siphon runs while pcrest stays above the vapour pressure of water, about 2.3 kPa at 20 °C. At 73.8 kPa there is ample margin, so the flow holds.
  6. Raising the crest does not change v, which the surface-to-outlet drop alone fixes, so the ceiling is the crest height that drives pcrest down to the vapour pressure: hmax = (101 300 − 2340 − 18 148.5)/9810 = 8.24 m at this speed, and (101 300 − 2340)/9810 = 10.1 m in the zero-flow limit — the barometric ceiling no water siphon beats.

Answerv = 6.02 m s⁻¹; pcrest = 73.8 kPa absolute, far above water's ~2.3 kPa vapour pressure, so it runs.

08 · HardReturn to the 62.0 mm / 26.0 mm pipe, with v₁ = 1.45 m s⁻¹ and v₂ = 8.245 m s⁻¹. Water has density 1000 kg m⁻³ and dynamic viscosity η = 1.00 × 10⁻³ Pa s. Find the Reynolds number in each section. A pair of gauges then measures a real pressure drop of 41.8 kPa between the two sections, against the 32.9 kPa Bernoulli predicted. Find the missing pressure and express it as a percentage of the measured drop.
  1. The Reynolds number compares inertial to viscous effects: Re = ρvd/η, with d the internal diameter.
  2. Wide section: Re₁ = (1000)(1.45)(0.0620)/(1.00 × 10⁻³) = 8.99 × 10⁴.
  3. Narrow section: Re₂ = (1000)(8.245)(0.0260)/(1.00 × 10⁻³) = 2.14 × 10⁵. Both sit far above the ~4000 pipe transition, so the flow is turbulent throughout.
  4. Missing pressure: Δploss = 41.8 − 32.939 = 8.861 kPa, or 8.86 kPa to three significant figures.
  5. As a fraction of the measurement: 8.861/41.8 = 0.212, so 21.2% of the real drop went into viscous and turbulent dissipation rather than into speeding the water up.
  6. Model check: Bernoulli conserves mechanical energy per unit volume along a streamline and carries no dissipation term, so it always underestimates the pressure drop in a real pipe. The gap widens with pipe length, wall roughness, and flow speed.

AnswerRe₁ = 8.99 × 10⁴, Re₂ = 2.14 × 10⁵, both turbulent; Δploss = 8.86 kPa, about 21.2% of the measured drop.