Fluid Mechanics: practice problems and exam questions

Fluid statics and flow: pressure in liquids, forces on submerged surfaces, continuity and Bernoulli’s equation. Exam problems often chain these steps: continuity gives the speed, and Bernoulli then gives the pressure.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Hydrostatic forcevertical wall · submerged gate · centre of pressurecalculations
  2. Continuity & Bernoullicontinuity · horizontal reducer · rising pipecalculations
  3. Pipe friction & head lossReynolds number · Darcy–Weisbach head loss · minor losses and total headcalculations
  4. Pumps & pump powerhydraulic power · pump efficiency · system head and powercalculations

Exam tips for Fluid Mechanics

  • Pressure in a liquid at rest depends only on depth: p = ρ·g·h.
  • Apply continuity first: A₁·v₁ = A₂·v₂.
  • Use gauge pressure or absolute pressure, but the same one on both sides of Bernoulli’s equation.

Worked examples

Hydrostatic forceEasy
h = 4 m

Water stands 4 m deep against a vertical wall. Find the height ȳ above the base at which the resultant force acts.

h = 4 mb = 2 mρ = 1000 kg/m³g = 9.81 m/s²

Find ȳ (m)

Show hints
  1. Pressure in a still liquid is p = ρ·g·z.
  2. The pressure diagram on the wall is a triangle. F is its area times the width; it acts through its centroid.
  3. The pressure at the base is 39.24 kPa.
Show the answer and the step-by-step solution

Hydrostatic force: on a plane surface, F = ρ·g·yc·A, where yc is the depth of the centroid. It acts below the centroid, at the centre of pressure.

h = 4 mF

Pressure varies linearly with depth

p = ρ·g·z, from zero at the surface to ρ·g·h at the base, so the load on the wall is a triangle.

pbase = ρgh = 1000·9.81·4 = 39.24 kPa

Line of action

A triangular load acts at its centroid, one third of the height from the base
ȳ = h/3 = 4/3 = 1.333 m above the base

Answer

ȳ = 1.333 m
Continuity & BernoulliMedium
12d1​ = 150 mm, p1​ = 250 kPad2​ = 50 mm, p2​ = ?

Water flows through a horizontal pipe that narrows from 150 mm to 50 mm. In the wide section the pressure is 250 kPa and the velocity is 0.5 m/s. Ignoring losses, find the pressure p2 in the narrow section.

d1 = 150 mmd2 = 50 mmp1 = 250 kPav1 = 0.5 m/sρ = 1000 kg/m³

Find p2 (kPa)

Show hints
  1. First use continuity to find the velocity in the narrow section.
  2. Bernoulli for a horizontal pipe: p₁ + ½ρv₁² = p₂ + ½ρv₂².
  3. v2 = 4.5 m/s.
Show the answer and the step-by-step solution

Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.

Continuity and Bernoulli

Work in Pa, then divide by 1000 to get kPa.

v2 = v1·(d1/d2)² = 0.5·(150/50)² = 4.5 m/s
½ρ(v1² − v2²) = ½·1000·(0.5² − 4.5²)/1000 = −10 kPa
p2 = p1 + ½ρ(v1² − v2²) = 250 + (−10) = 240 kPa

Answer

p2 = 240 kPa

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Bernoulli equation calculatorPipe head loss calculator (Darcy–Weisbach)Pump power calculatorReynolds number calculatorDensity calculator (mass, volume and density)

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