Bernoulli's Equation and Fluid Flow Problems

Every time water speeds up through a narrow pipe, a carburettor draws in fuel or a tank empties through a tap, Bernoulli's equation is at work. It is one of the most useful tools in fluid mechanics, and with the continuity equation beside it you can solve a surprising range of engineering problems in a few lines. This guide explains where the equation comes from, when you may use it, and then works through realistic problems with units at every step.

The Two Equations You Need

The continuity equation

For an incompressible fluid such as water, mass cannot pile up inside a pipe. Whatever volume enters one section per second must leave the next section per second. If \( A \) is the cross-sectional area and \( v \) the mean velocity, then

$$ Q = A_1 v_1 = A_2 v_2 $$

where \( Q \) is the volumetric flow rate in m³/s. Halve the area and the velocity doubles.

Bernoulli's equation

Along a streamline in steady, incompressible, frictionless flow, the total mechanical energy per unit volume stays constant:

$$ p_1 + \tfrac{1}{2}\rho v_1^2 + \rho g z_1 = p_2 + \tfrac{1}{2}\rho v_2^2 + \rho g z_2 $$

Each term has units of pressure (Pa = N/m², which is also J/m³):

• \( p \) is the static pressure,
• \( \tfrac{1}{2}\rho v^2 \) is the dynamic pressure (kinetic energy per unit volume),
• \( \rho g z \) is the hydrostatic or elevation term (potential energy per unit volume).

Engineers often divide by \( \rho g \) to write the equation in heads, each measured in metres:

$$ \frac{p}{\rho g} + \frac{v^2}{2g} + z = \text{constant} $$

When is it valid?

Bernoulli's equation is an idealisation. Use it confidently when:

1. The flow is steady (nothing changes with time at a given point).
2. The fluid is incompressible (liquids, or gases below roughly 30% of the speed of sound).
3. Friction is negligible over the short distance considered.
4. No pump or turbine adds or removes energy between the two points.

When friction matters, engineers add a head-loss term \( h_L \), and when a pump is present they add a pump head \( h_p \). The basic structure stays the same.

Worked Problem 1: Water in a Reducing Pipe

A horizontal pipe carrying water (\( \rho = 1000 \) kg/m³) narrows from a diameter of 100 mm to 50 mm. At the wide section the velocity is 2.0 m/s and the pressure is 200 kPa. Find the velocity and pressure in the narrow section, and the flow rate.

Step 1: Areas.

$$ A_1 = \frac{\pi (0.10\ \text{m})^2}{4} = 7.854 \times 10^{-3}\ \text{m}^2, \qquad A_2 = \frac{\pi (0.05\ \text{m})^2}{4} = 1.963 \times 10^{-3}\ \text{m}^2 $$

Step 2: Velocity from continuity. Since \( A_1/A_2 = (100/50)^2 = 4 \),

$$ v_2 = v_1 \frac{A_1}{A_2} = (2.0\ \text{m/s})(4) = 8.0\ \text{m/s} $$

Step 3: Pressure from Bernoulli. The pipe is horizontal, so \( z_1 = z_2 \) and the elevation terms cancel:

$$ p_2 = p_1 + \tfrac{1}{2}\rho\left(v_1^2 - v_2^2\right) $$

$$ p_2 = 200\,000\ \text{Pa} + \tfrac{1}{2}(1000\ \text{kg/m}^3)\left(4 - 64\right)\ \text{m}^2/\text{s}^2 = 200\,000 - 30\,000 = 170\,000\ \text{Pa} $$

So \( p_2 = 170 \) kPa.

Step 4: Flow rate.

$$ Q = A_1 v_1 = (7.854 \times 10^{-3}\ \text{m}^2)(2.0\ \text{m/s}) = 0.0157\ \text{m}^3/\text{s} \approx 15.7\ \text{L/s} $$

Sanity check. The fluid speeds up in the throat, so its kinetic energy rises and its pressure must fall, which it does (200 kPa to 170 kPa). Checking continuity at the narrow end: \( (1.963 \times 10^{-3})(8.0) = 0.0157 \) m³/s, the same flow rate. This pressure drop is exactly what a venturi meter measures to determine flow rate.

Worked Problem 2: Draining a Water Tank

An open water tank has a 25 mm diameter outlet at its base. The free surface stands 3.2 m above the outlet. Estimate the outlet velocity and the discharge, first ideally and then with a typical discharge coefficient of 0.62 for a sharp-edged orifice.

Step 1: Choose points. Point 1 is on the free surface; point 2 is in the jet just outside the outlet. Both are at atmospheric pressure, so \( p_1 = p_2 \). The tank is large, so the surface falls slowly and \( v_1 \approx 0 \). Take \( z_2 = 0 \) and \( z_1 = h = 3.2 \) m.

Step 2: Apply Bernoulli in head form.

$$ 0 + 0 + h = 0 + \frac{v_2^2}{2g} + 0 \quad \Rightarrow \quad v_2 = \sqrt{2gh} $$

This is Torricelli's theorem.

Step 3: Substitute.

$$ v_2 = \sqrt{2 (9.81\ \text{m/s}^2)(3.2\ \text{m})} = \sqrt{62.78\ \text{m}^2/\text{s}^2} = 7.92\ \text{m/s} $$

Step 4: Ideal discharge.

$$ A = \frac{\pi (0.025\ \text{m})^2}{4} = 4.909 \times 10^{-4}\ \text{m}^2 $$

$$ Q_{\text{ideal}} = A v_2 = (4.909 \times 10^{-4}\ \text{m}^2)(7.92\ \text{m/s}) = 3.89 \times 10^{-3}\ \text{m}^3/\text{s} = 3.89\ \text{L/s} $$

Step 5: Real discharge. The jet contracts (the vena contracta) and loses a little energy to friction, so

$$ Q_{\text{actual}} = C_d Q_{\text{ideal}} = 0.62 \times 3.89\ \text{L/s} \approx 2.41\ \text{L/s} $$

Sanity check. The ideal jet speed equals the speed a stone would reach falling 3.2 m, which makes physical sense because each kilogram of water loses the same potential energy. A domestic tap delivers about 0.1 to 0.2 L/s, and this is a much larger, full-bore outlet under 3.2 m of head, so a couple of litres per second is reasonable.

Common Mistakes Students Make

• Mixing gauge and absolute pressure. Either is fine, but use the same type on both sides.
• Forgetting to square the diameter ratio. Area scales with \( D^2 \), so halving diameter makes velocity four times larger, not two.
• Using kPa with kg/m³ and m/s. The term \( \tfrac{1}{2}\rho v^2 \) comes out in Pa. Convert kPa to Pa before adding.
• Applying Bernoulli across a pump or a long rough pipe without an energy term. Use the extended energy equation instead.
• Neglecting the elevation term in vertical or sloping pipes. A 10 m rise costs about 98 kPa for water.

Where Engineers Use Bernoulli

• Flow measurement: venturi meters, orifice plates and pitot tubes on aircraft all convert a measured pressure difference into velocity.
• Water supply: sizing overhead tanks so that upper floors still receive adequate pressure.
• Ventilation and HVAC: relating duct velocities to static pressure losses.
• Hydropower: estimating jet velocity at the nozzle of a Pelton wheel from the available head.

Did You Know?

• Daniel Bernoulli published his ideas in Hydrodynamica in 1738, but the equation in its modern form owes a great deal to Leonhard Euler's work in the 1750s.
• A pitot tube on an airliner uses the dynamic pressure \( \tfrac{1}{2}\rho v^2 \) to tell pilots their airspeed.
• The popular claim that wings lift only because air "must meet at the trailing edge" is a myth; the air over the top actually arrives earlier, though Bernoulli's relation between speed and pressure still holds.

Key Takeaways

• Continuity \( A_1 v_1 = A_2 v_2 \) gives velocities; Bernoulli then gives pressures.
• Bernoulli's equation balances pressure, kinetic and potential energy per unit volume along a streamline.
• It assumes steady, incompressible, frictionless flow with no pumps or turbines between the points.
• Torricelli's result \( v = \sqrt{2gh} \) follows directly and real orifices need a discharge coefficient.
• Always check units in pascals and confirm that faster flow corresponds to lower pressure.

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