Bernoulli's Equation Explained with Solved Numerical

On this page

The concept
Assumptions
Where Bernoulli’s equation is used
Common mistakes

Bernoulli’s equation is the energy balance for a flowing fluid. It says that along a streamline, pressure, speed and height trade off against each other: where a fluid speeds up, its pressure drops; where it climbs, its pressure drops too. It explains how a venturi measures flow, how an aircraft wing and a racing car’s wings make force, and how water pours out of a tank.

The concept

Think of every kilogram of fluid as carrying three kinds of energy:

Pressure energy, from the pressure pushing it along: $p/\rho$
Kinetic energy, from its speed: $V^2/2$
Potential energy, from its height: $g z$

If no energy is added by a pump, taken out by a turbine or lost to friction, the total stays the same from one point to the next:

p_1 + \tfrac{1}{2}\rho V_1^2 + \rho g z_1 = p_2 + \tfrac{1}{2}\rho V_2^2 + \rho g z_2

Dividing every term by $\rho g$ gives the same equation in heads (metres of fluid), which is how pump curves and hydraulic systems are usually described:

\frac{p}{\rho g} + \frac{V^2}{2g} + z = \text{constant} \qquad \text{(pressure head + velocity head + elevation head)}

Bernoulli’s equation almost always works together with the continuity equation, $A_1V_1 = A_2V_2$ , which gives the velocity change from the area change.

Water flows from a wide, low section 1 to a narrow, high section 2. It speeds up because the area is smaller, and it climbs, so both changes take energy from the pressure: the gauge at 2 reads lower than the gauge at 1.

Assumptions

Bernoulli’s equation is only valid when:

The flow is steady: nothing changes with time at a given point.
The fluid is incompressible: true for liquids, and for gases below about 30 % of the speed of sound.
There is no friction: viscous losses are ignored. For real flow over a distance, add a head loss $h_L$ on the downstream side.
It is applied along a streamline (or anywhere in an irrotational flow).
No shaft work: there is no pump or turbine between the two points. If there is, add the pump head on the upstream side or the turbine head on the downstream side.

Solved numerical 1: pressure in a rising, narrowing section

Water (ρ = 1000 kg/m³) flows at 0.02 m³/s through the section in the diagram. At section 1 the diameter is 150 mm and the pressure is 250 kPa. Section 2 is 3 m higher, with a diameter of 75 mm. Ignoring losses, find the pressure at section 2.

Step 1: areas.

A_1 = \frac{\pi \times 0.150^2}{4} = 0.01767\ \text{m}^2, \qquad A_2 = \frac{\pi \times 0.075^2}{4} = 0.004418\ \text{m}^2

Step 2: velocities from continuity.

V_1 = \frac{Q}{A_1} = \frac{0.02}{0.01767} = 1.13\ \text{m/s}, \qquad V_2 = \frac{0.02}{0.004418} = 4.53\ \text{m/s}

Step 3: rearrange Bernoulli’s equation for p₂.

p_2 = p_1 + \tfrac{1}{2}\rho\,(V_1^2 - V_2^2) - \rho g\,(z_2 - z_1)

Step 4: substitute.

p_2 = 250\,000 + 500 \times (1.28 - 20.49) - 1000 \times 9.81 \times 3

p_2 = 250\,000 - 9\,610 - 29\,430 = 210\,960\ \text{Pa}

Answerp₂ ≈ 211 kPa: about 10 kPa is lost to the speed increase and 29 kPa to the 3 m climb

Solved numerical 2: water flowing out of a tank (Torricelli)

A large open tank has a 20 mm diameter hole in its side, 2 m below the water surface. Find the speed of the jet and the flow rate. Take a discharge coefficient of 0.62 for a sharp-edged hole.

Step 1: choose the points. Point 1 is on the water surface: pressure atmospheric, velocity almost zero (the tank is large), height 2 m above the hole. Point 2 is in the jet just outside the hole: pressure atmospheric, height 0.

Step 2: apply Bernoulli. The pressures cancel and $V_1 \approx 0$ , leaving:

g z_1 = \tfrac{1}{2} V_2^2 \quad\Rightarrow\quad V_2 = \sqrt{2 g h} = \sqrt{2 \times 9.81 \times 2} = 6.26\ \text{m/s}

Step 3: flow rate. The jet contracts and loses a little energy, which the discharge coefficient accounts for:

Q = C_d\,A\,V_2 = 0.62 \times \frac{\pi \times 0.020^2}{4} \times 6.26 = 1.22 \times 10^{-3}\ \text{m}^3/\text{s}

AnswerJet speed 6.26 m/s; flow rate 1.22 L/s

Where Bernoulli’s equation is used

Flow measurement: venturi meters, orifice plates and pitot tubes measure a pressure difference and convert it to velocity or flow rate.
Aerodynamics: air flows faster over the curved upper surface of a wing, so pressure there is lower, creating lift. A racing car’s wings are mounted upside down to create downforce.
Pumps and hydraulic systems: the total head a pump must supply is the change in Bernoulli head plus the losses.
Carburettors and spray nozzles: low pressure in a fast-moving stream draws in fuel or liquid.

Common mistakes

Mixing gauge and absolute pressure between the two points. Use the same at both.
Forgetting continuity. The velocity at each point must come from the actual area there.
Ignoring losses over long distances. Over a long duct with bends and valves, friction losses can be larger than every other term; see Reynolds number and fluid flow.
Using it across a pump or turbine without adding the head they add or remove.

Key points

Along a streamline: p + ½ρV² + ρgz is constant for steady, incompressible, frictionless flow.
Where flow speeds up or climbs, its pressure falls.
Use continuity (A₁V₁ = A₂V₂) to get the velocities first.
Add head losses for real flow and pump or turbine head where there is shaft work.

Bernoulli's equation explained with a solved numerical

The concept

Assumptions

Where Bernoulli’s equation is used

Common mistakes

See it in CAD

Formula 1 Car

Read next

Reynolds number and fluid flow: laminar vs turbulent

Fluid mechanics formulas with units and examples

The first law of thermodynamics explained with solved problems

Need this designed or explained for your team?