💧 Fluid Mechanics Basics — Pressure, Density & Flow
Fluid mechanics underpins hydraulics, pneumatics, pipe systems, pumps, dam design and ship stability. The key equations — hydrostatic pressure, Archimedes' principle and the continuity equation — let engineers calculate forces on submerged surfaces, predict whether objects float, and size pipework.
2Topics
8Examples
4Calculators
12Practice Qs
💧 Level 3 context: ρ (rho) is the standard symbol for density. Water density = 1000 kg/m³ and g = 9.81 m/s² appear in nearly every fluid mechanics calculation — memorise both. Pressure increases linearly with depth: every 10 m of water adds approximately 98.1 kPa.
💧Pressure & DensityP=ρgh, ρ=m/V, P=F/A
⚓️Buoyancy & FlowArchimedes, continuity
⚙️Applications4 fluid calculators
🎮Game12 randomised questions
📝Practice12 graded questions
💧
Pressure & Density
Pressure is force per unit area. In a static fluid, pressure increases with depth according to P = ρgh. Density links mass to volume and determines how fluids layer and whether objects sink or float.
🎮Live explorer — hydrostatic pressure P = ρghDrag sliders — pressure updates instantly
Gauge pressure
49.1 kPa
ρgh
Absolute pressure
150.4 kPa
P𝑎𝑡𝑚 + ρgh
Force on 1 m²
49 050 N
F = P × A
💡 Every 10 m of water depth adds approximately 98.1 kPa (about 1 atmosphere) of pressure.
📋Try it — Pressure & Density CalculatorUpdates as you type
Select what to find and enter values above.
📚 Key vocabulary
- Pressure (P)
- force per unit area. P = F/A. Unit: pascal (Pa = N/m²). 1 kPa = 1000 Pa.
- Hydrostatic pressure
- pressure in a static fluid due to the weight of fluid above. P = ρgh. Increases linearly with depth.
- Density (ρ)
- mass per unit volume. ρ = m/V. Unit: kg/m³. Water = 1000 kg/m³.
- Gauge pressure
- pressure above atmospheric. Measured by most gauges. Absolute pressure = gauge + atmospheric (101.3 kPa).
- Pascal's Law
- pressure applied to a confined fluid is transmitted equally in all directions — the basis of hydraulic systems.
- Specific gravity
- ratio of a substance's density to water. Dimensionless. Steel SG ≈ 7.8; wood SG ≈ 0.6 (floats).
👁️ Visual aid — pressure increases with depth in a static fluid. All points at the same depth have equal pressure regardless of container shape.
✍️Worked example 1 — pressure at depth step through it
QuestionCalculate the gauge pressure at a depth of 8 m in seawater. (ρ = 1025 kg/m³, g = 9.81 m/s²)
Formula
P = ρghSubstitute
P = 1025 × 9.81 × 8Calculate
= 1025 × 9.81 = 10 055.25 × 8 = 80 442 PaAnswerP = 80 442 Pa = 80.4 kPa ✓
Step 1 of 5
✍️Worked example 2 — density of a material
QuestionA steel block has mass 7.8 kg and volume 0.001 m³. Calculate its density.
Formula
ρ = m ÷ VSubstitute
ρ = 7.8 ÷ 0.001 = 7800Answerρ = 7800 kg/m³
✍️Worked example 3 — force on a submerged surface
QuestionA rectangular hatch (0.5 m × 0.4 m) is 3 m below the surface of fresh water. Find the gauge pressure and total force on it.
Pressure
P = ρgh = 1000 × 9.81 × 3 = 29 430 PaArea
A = 0.5 × 0.4 = 0.2 m²Force
F = P × A = 29 430 × 0.2 = 5886 NAnswerP = 29.4 kPa F = 5886 N
✍️Worked example 4 — depth from known pressure
QuestionA pressure gauge reads 196.2 kPa gauge in fresh water. Find the depth.
Rearrange
h = P ÷ (ρg) = 196 200 ÷ (1000 × 9.81)Calculate
h = 196 200 ÷ 9810 = 20Answerh = 20 m
⚓️
Buoyancy & Fluid Flow
Archimedes' principle gives the upthrust on any submerged or floating object. The continuity equation ensures mass is conserved in a flow: where a pipe narrows, velocity must increase.
🎮Live explorer — Archimedes' principleDrag sliders — upthrust and net force update
Weight W
490.5 N
ρ𝑜𝑏𝑗gV
Upthrust F𝐟
981.0 N
ρ𝑓gV
Net force
+490.5 N
F𝐟 − W
Result
FLOATS ▲
💡 An object floats when its average density is less than the fluid density — even if made of steel, if it contains enough air (like a ship).
📋Try it — Buoyancy & Flow CalculatorUpdates as you type
Select what to find and enter values above.
📚 Key vocabulary
- Upthrust (F𝐟)
- the upward buoyant force on an object in a fluid, equal to the weight of fluid displaced. F𝐟 = ρgV.
- Archimedes' Principle
- any object fully or partially submerged in a fluid experiences an upthrust equal to the weight of the displaced fluid.
- Continuity equation
- A₁v₁ = A₂v₂ — for incompressible flow, the volume flow rate Q is constant. A smaller pipe means faster flow.
- Volume flow rate (Q)
- Q = A × v. Unit: m³/s. Also expressed in litres/s (1 m³/s = 1000 L/s).
- Streamline flow
- smooth, orderly flow in parallel layers. Applies at low velocities. At high velocities, flow becomes turbulent.
- Reynolds number
- dimensionless ratio of inertial to viscous forces. Re < 2000 → laminar flow; Re > 4000 → turbulent flow.
👁️ Visual aid — continuity equation: pipe narrows from A₁ = 0.05 m² to A₂ = 0.02 m². v₂ = A₁v₁/A₂ = 5 m/s.
✍️Worked example 5 — Archimedes' principle step through it
QuestionA metal block (volume = 0.004 m³, density = 8000 kg/m³) is fully submerged in fresh water. Find the upthrust, weight and net force.
Upthrust
F𝐟 = ρ𝑓 × g × V = 1000 × 9.81 × 0.004 = 39.24 NWeight
W = ρ𝑜𝑏𝑗 × g × V = 8000 × 9.81 × 0.004 = 313.9 NNet force
F𝑛𝑒𝑡 = F𝐟 − W = 39.24 − 313.9 = −274.7 NAnswerNet = −274.7 N downward — block sinks ✓
Step 1 of 5
✍️Worked example 6 — continuity equation
QuestionWater flows at 3 m/s through a pipe of diameter 0.2 m. It enters a narrower section of diameter 0.1 m. Find the velocity in the narrow section.
Areas
A₁ = π(0.1)² = 0.0314 m² A₂ = π(0.05)² = 0.007 854 m²Continuity
v₂ = A₁v₁ ÷ A₂ = 0.0314 × 3 ÷ 0.007 854Calculate
v₂ = 0.09 424 ÷ 0.007 854 = 12 m/sAnswerv₂ = 12 m/s
✍️Worked example 7 — volume flow rate
QuestionWater flows at 2.5 m/s through a pipe of cross-sectional area 0.04 m². Find the volume flow rate in m³/s and in litres/s.
Flow rate
Q = A × v = 0.04 × 2.5 = 0.1 m³/sConvert
Q = 0.1 m³/s × 1000 = 100 litres/sAnswerQ = 0.1 m³/s = 100 L/s
✍️Worked example 8 — floating object, fraction submerged
QuestionA wooden log (ρ = 600 kg/m³) floats in fresh water. What fraction of its volume is submerged?
Principle
Weight = Upthrust → ρ𝑜𝑏𝑗gV = ρ𝑓gV𝑠𝑢𝑏Rearrange
V𝑠𝑢𝑏/V = ρ𝑜𝑏𝑗 ÷ ρ𝑓 = 600 ÷ 1000Answer0.6 (60%) submerged
⚙️
Engineering Applications
Four fluid mechanics calculators for hydraulic, pipe flow and buoyancy scenarios.
💧 Hydraulic pressure — dam or tank wall
Find the gauge pressure and total force on a rectangular submerged surface at a given depth.
P = ρgh F = P × A
🧮 Calculator
Enter depth, width and surface height (fresh water assumed).
⚓️ Buoyancy — will it float?
Calculate upthrust, weight and net force for a fully submerged object in any fluid.
F𝐟 = ρ𝑓gV W = ρ𝑜𝑏𝑗gV F𝑛𝑒𝑡 = F𝐟 − W
🧮 Calculator
Enter object density, fluid density and volume.
🔌 Pipe flow — continuity
Find velocity in a pipe section from inlet flow rate and the ratio of diameters.
v₂ = A₁v₁ ÷ A₂ A = πr²
🧮 Calculator
Enter both diameters and inlet velocity.
📈 Volume flow rate
Calculate the volume flow rate and mass flow rate for a pipe given its diameter and fluid velocity.
Q = Av mass flow = ρQ
🧮 Calculator
Enter diameter, velocity and fluid density.
🎮
Fluid Mechanics Challenge
12 randomised questions — pressure, density, buoyancy and flow.
💧 Quick-fire round
Question 1 / 12
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📝
Practice Questions
12 graded questions — type your answer to check, then reveal full working.
💧 Constants: ρ𝑣𝑎𝑡𝑒𝑟 = 1000 kg/m³, g = 9.81 m/s². Show formula, substitution and every step.
Q1 · P = ρgh1 mark
Calculate the gauge pressure at 4 m depth in fresh water.
P = ρgh = 1000×9.81×4 = 39 240 Pa
Q2 · Density1 mark
An object has mass 2.7 kg and volume 0.001 m³. Calculate its density.
ρ = m/V = 2.7/0.001 = 2700 kg/m³
Q3 · Force on surface2 marks
A circular hatch of diameter 0.6 m is 5 m below the surface of fresh water. Find the force on it.
P = 1000×9.81×5 = 49 050 Pa
A = π×0.3² = 0.2827 m²
F = 49 050×0.2827 = 13 868 N ≈ 13 883 N
A = π×0.3² = 0.2827 m²
F = 49 050×0.2827 = 13 868 N ≈ 13 883 N
Q4 · Find depth2 marks
A pressure gauge reads 147.15 kPa gauge in fresh water. Find the depth.
h = P/(ρg) = 147 150/(1000×9.81) = 15 m
Q5 · Upthrust2 marks
A block with volume 0.05 m³ is fully submerged in fresh water. Find the upthrust.
F𝐟 = ρgV = 1000×9.81×0.05 = 490.5 N
Q6 · Floats or sinks?2 marks
An object (ρ = 850 kg/m³, V = 0.2 m³) is submerged in seawater (ρ = 1025 kg/m³). Find the net force and state whether it rises or sinks.
W = 850×9.81×0.2 = 1667.7 N
F𝐟 = 1025×9.81×0.2 = 2011.05 N
Net = 2011.05−1667.7 = 343.35 N upward (rises)
F𝐟 = 1025×9.81×0.2 = 2011.05 N
Net = 2011.05−1667.7 = 343.35 N upward (rises)
Q7 · Flow rate1 mark
Water flows at 3 m/s through a pipe of cross-section 0.08 m². Find the volume flow rate.
Q = Av = 0.08×3 = 0.24 m³/s
Q8 · Continuity2 marks
Water enters a pipe (A₁ = 0.1 m²) at 2 m/s. It exits through a smaller pipe (A₂ = 0.04 m²). Find the exit velocity.
v₂ = A₁v₁/A₂ = 0.1×2/0.04 = 5 m/s
Q9 · Fraction submerged2 marks
A block of wood (ρ = 750 kg/m³) floats in fresh water. What fraction of its volume is submerged?
Fraction = ρ𝑜𝑏𝑗/ρ𝑓 = 750/1000 = 0.75 (75% submerged)
Q10 · P = F/A2 marks
A hydraulic piston of area 0.02 m² applies a force of 8000 N. Calculate the pressure generated in the hydraulic fluid.
P = F/A = 8000/0.02 = 400 000 Pa = 400 kPa
Q11 · Combined3 marks
A solid steel ball (ρ = 7800 kg/m³, V = 0.002 m³) is suspended in seawater (ρ = 1025 kg/m³). Find (a) upthrust, (b) weight in air, (c) apparent weight in seawater.
F𝐟 = 1025×9.81×0.002 = 20.12 N
W = 7800×9.81×0.002 = 153.04 N
Apparent W = 153.04−20.12 = 132.92 N
W = 7800×9.81×0.002 = 153.04 N
Apparent W = 153.04−20.12 = 132.92 N
Q12 · Combined3 marks
Water flows at 4 m/s through a 200 mm diameter pipe. It narrows to 100 mm. Find (a) flow rate Q, (b) velocity in the narrow section, (c) mass flow rate (ρ = 1000 kg/m³).
A₁ = π×0.1² = 0.03142 m²
Q = 0.03142×4 = 0.12566 m³/s
A₂ = π×0.05² = 0.007 854 m²
v₂ = 0.12566/0.007854 = 16 m/s
Mass flow = 1000×0.12566 = 125.66 kg/s
Q = 0.03142×4 = 0.12566 m³/s
A₂ = π×0.05² = 0.007 854 m²
v₂ = 0.12566/0.007854 = 16 m/s
Mass flow = 1000×0.12566 = 125.66 kg/s