Continuity & Bernoulli: practice problems with step-by-step solutions

Free Continuity & Bernoulli practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: continuity, horizontal reducer, rising pipe.

Free practice problemEasy
12d1​ = 75 mm, v1​ = 0.75 m/sd2​ = 25 mm

Water flows at 0.75 m/s through the wide section of the pipe, which has a diameter of 75 mm. Find the volume flow rate Q.

d1 = 75 mmd2 = 25 mmv1 = 0.75 m/s
L/s
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What you practise

  1. Easycontinuity
  2. Mediumhorizontal reducer
  3. Hardrising pipe

Key formulas

Continuity
A1v1 = A2v2
Bernoulli
p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
Flow rate
Q = A·v
Circle area
A = πd²/4

Learn Continuity & Bernoulli from scratch

Before this: Hydrostatic force

1What continuity and Bernoulli are about

Put your thumb over the end of a garden hose and the water shoots out faster. The same amount of water must get through a smaller opening, so it has to move faster.

That idea is continuity: in a full pipe, water cannot pile up or disappear. What flows in each second must flow out each second.

The second idea, Bernoulli’s equation, links speed, pressure and height. Where the water speeds up, its pressure drops. Where it climbs, its pressure drops too.

Together these two ideas let you predict the speed and the pressure anywhere along a pipe. In this topic the fluid is incompressible (like water) and losses by friction are ignored.

Self-check Water flows through a pipe that gets narrower. What happens to its speed in the narrow part?
  1. It decreases
  2. It increases
  3. It stays the same
  4. The water stops

Answer: It increases. The same volume per second must pass through a smaller area, so the water must move faster.

2Flow rate and continuity

The volume flow rate Q is the volume of fluid that passes per second, in m³/s. One litre per second (L/s) is 0.001 m³/s.

For a fluid moving at a mean speed v through a cross-section of area A:

Q = A · v A = π · d² / 4

The mass flow rate is ṁ = ρ · Q, in kg/s, where ρ is the density (1000 kg/m³ for water).

Continuity says Q is the same at every section of the pipe:

12v1​v2​d1​d2​p1​p2​
A1 · v1 = A2 · v2

For round pipes the π/4 cancels, so v2 = v1 · (d1 / d2)². Halving the diameter makes the water four times faster.

Self-check A pipe narrows from 80 mm to 40 mm. The water moves at 1 m/s in the wide part. How fast does it move in the narrow part?
  1. 0.25 m/s
  2. 2 m/s
  3. 4 m/s
  4. 8 m/s

Answer: 4 m/s. v2 = 1 · (80/40)² = 4 m/s. The area depends on the diameter squared. Using the diameter ratio without squaring gives 2 m/s.

3Bernoulli’s equation

Along a streamline without losses, the energy per cubic metre of fluid stays the same. It has three parts: pressure, motion and height.

p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
  • p: the pressure, in Pa.
  • ½ρv²: the kinetic energy per m³, also in Pa.
  • ρgz: the height energy per m³, where z is the height above any reference level you choose.

So if one part grows, another must shrink. Faster water has lower pressure. Higher water has lower pressure.

A useful special case is a tank that drains through a small hole a depth h below the open surface. The surface hardly moves and both points are at air pressure, which gives Torricelli’s law:

v = √(2gh)
Self-check An open water tank drains through a small hole 5 m below the water surface. Ignoring losses, how fast does the jet leave the hole? (g = 9.81 m/s²)
  1. 49.1 m/s
  2. 7.00 m/s
  3. 9.90 m/s
  4. 98.1 m/s

Answer: 9.90 m/s. v = √(2 · 9.81 · 5) = √98.1 = 9.90 m/s. Forgetting the 2 gives 7.00 m/s, and forgetting the square root gives 98.1.

4The method step by step

  1. Sketch the pipe and number two sections: 1 where you know most, 2 where the unknown is.
  2. Convert units: mm to m, kPa to Pa, L/s to m³/s.
  3. Find the areas A = π·d²/4 and use continuity to get the missing velocity.
  4. Write Bernoulli between sections 1 and 2. For a horizontal pipe z1 = z2, so the height terms cancel.
  5. Solve for the unknown, then convert back to kPa if needed.

Keep the units consistent. With ρ in kg/m³ and v in m/s, the term ½ρv² comes out in Pa, not kPa.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Water (ρ = 1000 kg/m³) flows through a horizontal pipe that narrows from 100 mm to 50 mm. In the wide section the pressure is 200 kPa and the velocity is 1.5 m/s. Ignoring losses, find the pressure in the narrow section.

1

Draw the pipe

121.5 m/sv2​100 mm50 mm200 kPap2​ = ?

Section 1 is the wide part, where everything is known. Section 2 is the narrow part.

2

Velocity in the narrow part

The diameter halves, so the speed grows by (100/50)² = 4:

v2 = 1.5 · (100 / 50)² = 6 m/s
3

Write Bernoulli

The pipe is horizontal, so the height terms cancel:

p2 = p1 + ½ρ(v1² − v2²)
4

Fill in, in pascals

½ · 1000 · (1.5² − 6²) = 500 · (2.25 − 36) = −16 875 Pa
p2 = 200 000 − 16 875 = 183 125 Pa ≈ 183.1 kPa
5

Check the answer

The water is faster in the narrow part, so its pressure must be lower. 183.1 kPa is lower than 200 kPa, as expected.

The flow rate is Q = 0.007854 · 1.5 = 11.8 L/s in both sections.

Answer

p2 ≈ 183.1 kPa in the narrow section (v2 = 6 m/s).

6Common mistakes

  • Using the diameter ratio instead of the area ratio: the speeds change with the square of the diameter.
  • Using the radius as if it were the diameter in A = π·d²/4, or forgetting the /4.
  • Mixing kPa and Pa. Work in Pa in Bernoulli, then convert at the end.
  • Reporting m³/s as L/s. Multiply m³/s by 1000 to get L/s.
  • Forgetting the height term ρ·g·Δz when the pipe rises or falls.
Self-check Water flows at 2 m/s through a pipe with an inside diameter of 100 mm. What is the volume flow rate?
  1. 0.0157 L/s
  2. 15.7 L/s
  3. 20 L/s
  4. 62.8 L/s

Answer: 15.7 L/s. A = π · 0.1²/4 = 0.007854 m², so Q = 0.007854 · 2 = 0.0157 m³/s = 15.7 L/s. Forgetting the /4 gives 62.8, and forgetting π/4 gives 20.

7How exam questions look

  • Easy: continuity. Find a flow rate, a mass flow rate, the velocity in a narrower section, or the pipe diameter needed for a given flow. Some questions use Torricelli’s law for a draining tank.
  • Medium: a horizontal reducer. Find the pressure in one section from the other. Variations use a venturi meter, a pitot tube (v = √(2Δp/ρ)) or an orifice with a discharge coefficient.
  • Hard: a rising pipe. The height term matters: each metre of rise lowers the pressure by ρ·g·1 = 9.81 kPa for water. Variations include a siphon, a pressurised tank and a fountain.

In the example above, a rise of 5 m between the sections would lower p2 by another 1000 · 9.81 · 5 = 49 050 Pa, to 134.1 kPa.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
12d1​ = 100 mm, v1​ = 2.75 m/sd2​ = 50 mm, v2​ = ?

Water flows through a pipe that narrows from 100 mm to 50 mm in diameter. The velocity in the wide section is 2.75 m/s. Find the velocity v2 in the narrow section.

d1 = 100 mmd2 = 50 mmv1 = 2.75 m/s

Find v2 (m/s)

Show hints
  1. The same volume of water passes every section each second: Q = A·v.
  2. A₁v₁ = A₂v₂, and the area scales with d², so v₂ = v₁·(d₁/d₂)².
  3. A1 = 0.007854 m².
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

The flow rate is the same in both sections.

A1v1 = A2v2
v2 = v1·(d1/d2)² = 2.75·(100/50)² = 11 m/s

Answer

v2 = 11 m/s
Example 2Medium
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
Example 3Hard
pair​ = 190 kPaQ = ?d = 25 mmh = 4 m

A closed water tank has compressed air above the water at a gauge pressure of 190 kPa. Water leaves through an opening with a diameter of 25 mm, 4 m below the water surface, into the open air. Ignoring losses, find the flow rate Q. Use g = 9.81 m/s².

pair = 190 kPa (gauge)h = 4 md = 25 mmρ = 1000 kg/m³g = 9.81 m/s²

Find Q (L/s)

Show hints
  1. Bernoulli from the water surface (gauge pressure p, at rest) to the jet (atmospheric).
  2. p/ρ + gh = ½v², so v = √(2(p/ρ + gh)) with p in Pa.
  3. p/ρ + gh = 229.2 m²/s².
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.

Bernoulli from the surface to the jet

v = √(2·(190000/1000 + 9.81·4)) = 21.41 m/s

Flow rate

Q = A·v = 0.0004909·21.41 = 0.01051 m³/s = 10.51 L/s

Answer

Q = 10.51 L/s

Questions and answers

What is Continuity & Bernoulli?

Put your thumb over the end of a garden hose and the water shoots out faster. The same amount of water must get through a smaller opening, so it has to move faster.

That idea is continuity: in a full pipe, water cannot pile up or disappear. What flows in each second must flow out each second.

How do I solve Continuity & Bernoulli problems step by step?

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.

What are common mistakes in Continuity & Bernoulli?
  • The air pressure pushes the water out too: include p/ρ in the Bernoulli equation.
  • The water height also adds to the velocity: v = √(2(p/ρ + gh)).
What do the Continuity & Bernoulli exercises cover?
  • Easy: continuity
  • Medium: horizontal reducer
  • Hard: rising pipe

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.

How practising works

  • 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.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Hydrostatic force

Continue with

Pipe friction & head loss

Calculators

Bernoulli equation calculator

Courses with this topic

Fluid MechanicsPhysics for EngineersHydraulic EngineeringFE Mechanical exam practice

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Printable worksheet

Continuity & Bernoulli worksheet with answers
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