3.5 Bernoulli's equation

Bernoulli's Equation Derivation

Bernoulli's equation connects pressure, velocity, and elevation along a streamline in a flowing fluid. It comes directly from the conservation of energy applied to a fluid element, and it's one of the most used relationships in fluid dynamics. The trade-off: it only holds under specific assumptions (steady, incompressible, inviscid flow along a streamline), so knowing when it breaks down matters just as much as knowing the equation itself.

The standard form is:

$P + \frac{1}{2}\rho v^2 + \rho g z = \text{constant along a streamline}$

where $P$ is static pressure, $\rho$ is fluid density, $v$ is flow velocity, $g$ is gravitational acceleration, and $z$ is elevation.

Steady Flow Energy Balance

Steady flow means fluid properties at any fixed point don't change with time. Under this condition, you can track a fluid element moving along a streamline and apply the work-energy theorem between two points. The work done by pressure forces on the element equals the sum of changes in its kinetic and potential energy. That balance is exactly what Bernoulli's equation expresses.

Incompressible Flow Assumption

Incompressible flow means $\rho$ stays constant throughout the flow field. This is a good approximation for all liquids and for gases at Mach numbers below about 0.3. Holding density constant is what lets you write the pressure term simply as $P$ rather than needing to integrate a varying $\rho$ through the energy equation.

Negligible Viscous Effects

The derivation assumes the fluid is inviscid, meaning frictional losses between fluid layers are zero. This works well for high Reynolds number flows where inertial forces dominate. Without viscous dissipation, mechanical energy is conserved (none is converted to heat), which is what makes the "constant along a streamline" statement possible.

Along a Streamline

Bernoulli's equation applies along a single streamline only. A streamline is the path a fluid particle traces through the flow field. The constant on the right side of the equation can differ from one streamline to another, so you cannot apply the equation between two arbitrary points in the flow unless the flow is also irrotational (more on that below).

Bernoulli's Equation Components

Each term in Bernoulli's equation represents a form of energy per unit volume. Their sum stays constant along a streamline, so an increase in one term forces a decrease in another.

Pressure Term

$P$ is the static pressure, the actual thermodynamic pressure at a point in the fluid. When you divide through by $\rho g$, this becomes $P/(\rho g)$, called the pressure head (units of length). Don't confuse static pressure with absolute or gauge pressure in a problem; always check which reference is being used.

Velocity Term

$\frac{1}{2}\rho v^2$ is the dynamic pressure, representing kinetic energy per unit volume. Divided by $\rho g$, it becomes $v^2/(2g)$, the velocity head. This term is what a Pitot tube measures indirectly.

Elevation Term

$\rho g z$ is the hydrostatic pressure contribution from elevation, representing gravitational potential energy per unit volume. Divided by $\rho g$, it becomes simply $z$, the elevation head. The reference level for $z$ is arbitrary, but you must use the same reference for both points in a calculation.

Constant Along a Streamline

The sum $P + \frac{1}{2}\rho v^2 + \rho g z$ stays fixed as you move along a given streamline. If velocity increases (say, through a constriction), pressure must drop to compensate. This is the core physical insight behind most Bernoulli applications. Remember: the constant can be different on neighboring streamlines unless the flow is irrotational.

Applications of Bernoulli's Equation

Pitot Tubes for Velocity Measurement

A Pitot tube measures flow velocity by comparing two pressures:

• Stagnation pressure $P_0$: measured at a point where the flow is brought to rest (the tube opening faces directly into the flow)
• Static pressure $P$: measured at a tap flush with the flow direction

Applying Bernoulli between the free stream and the stagnation point:

$P + \frac{1}{2}\rho v^2 = P_0$

Solving for velocity:

$v = \sqrt{\frac{2(P_0 - P)}{\rho}}$

This is how airspeed indicators in aircraft work.

Venturi Meters for Flow Rate Measurement

A Venturi meter narrows a pipe to a smaller cross-section (the throat) and measures the pressure drop between the wider upstream section and the throat.

1. Write Bernoulli's equation between the upstream section (subscript 1) and the throat (subscript 2), assuming the pipe is horizontal so elevation terms cancel: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

2. Apply the continuity equation: $A_1 v_1 = A_2 v_2$, so $v_1 = (A_2/A_1)v_2$.

3. Substitute and solve for $v_2$: $v_2 = \sqrt{\frac{2(P_1 - P_2)}{\rho\left(1 - (A_2/A_1)^2\right)}}$

4. The volume flow rate is then $Q = A_2 v_2$.

In practice, a discharge coefficient (typically 0.95–0.99 for a well-designed Venturi) is applied to account for small viscous losses.

Lift Force on Airfoils

The curved upper surface of an airfoil accelerates air to a higher velocity than the flatter lower surface. By Bernoulli's equation, higher velocity means lower pressure on top. The net pressure difference between the lower and upper surfaces produces an upward lift force.

This is a useful first-order explanation, but the full picture also involves circulation and the Kutta condition. Bernoulli alone doesn't explain why the velocity differs; it just relates the velocity difference to a pressure difference.

Pressure Drops in Pipes

For a pipe with changing diameter or elevation, Bernoulli's equation between two points gives:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g z_2$

You can solve for the pressure at point 2 if you know the velocities (from continuity) and the elevations. Keep in mind that real pipes have friction losses, so this result is only accurate for short sections or flows where viscous losses are small.

Siphons and Aspirators

A siphon moves liquid over a barrier and down to a lower outlet using only gravity. Bernoulli's equation applied between the liquid surface and the outlet shows that the pressure difference from the elevation drop drives the flow. At the highest point of the siphon, pressure drops below atmospheric; if it drops to the vapor pressure, the siphon breaks (cavitation).

An aspirator (or ejector) forces a high-speed fluid through a nozzle, creating a low-pressure region that draws in a second fluid. Bernoulli's principle explains why the fast-moving stream produces the suction effect.

Limitations of Bernoulli's Equation

Steady Flow Requirement

Any flow where conditions change with time violates this assumption. Pulsating flows (like those from a reciprocating pump) or transient startup conditions require the unsteady Bernoulli equation, which adds a time-dependent term $\rho \int \frac{\partial v}{\partial t} \, ds$ along the streamline.

Incompressible Flow Assumption

For gases above Mach 0.3, density changes become significant and Bernoulli's equation gives increasingly inaccurate results. High-speed aerodynamics and flows with large temperature gradients require compressible flow relations (isentropic flow equations or the full energy equation).

Inviscid Flow Assumption

All real fluids have viscosity. Bernoulli's equation fails in regions where viscous effects dominate: inside boundary layers, in fully developed pipe flow over long distances, and in any low Reynolds number flow. In these cases, energy is dissipated as heat, and the "constant" in Bernoulli's equation decreases along the streamline.

Irrotational Flow Assumption

The standard form of Bernoulli's equation (applied between any two points, not just along a single streamline) requires irrotational flow, meaning zero vorticity everywhere. Flows with shear layers, vortices, or separation regions are rotational. In rotational flow, you can still use Bernoulli along a streamline, but you cannot use it across streamlines.

Along a Streamline Restriction

Even when all other assumptions hold, the equation only connects points on the same streamline. For complex flow patterns with multiple streamlines at different energy levels, you need either the full energy equation or computational methods.

Bernoulli's Equation vs. the General Energy Equation

Similarities

Both are rooted in conservation of energy. Both track the exchange between pressure energy, kinetic energy, and potential energy in a moving fluid. For an ideal flow that meets all of Bernoulli's assumptions, the two equations give identical results.

Differences in Assumptions

Differences in Applicability

Bernoulli's equation is a special case of the energy equation. Use it when the assumptions are reasonable: short flow paths in clean, fast-moving, low-Mach flows with no energy addition or removal. When you have pumps, turbines, significant friction, heat transfer, or compressibility, switch to the general energy equation, which adds head loss ($h_L$) and shaft work ($h_s$) terms:

$\frac{P_1}{\rho g} + \frac{v_1^2}{2g} + z_1 + h_s = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} + z_2 + h_L$

Examples and Problem-Solving

Bernoulli's Equation in Different Scenarios

Pipe with a constriction: Water flows through a horizontal pipe that narrows from $A_1 = 0.01 \, \text{m}^2$ to $A_2 = 0.005 \, \text{m}^2$. If the upstream velocity is $v_1 = 2 \, \text{m/s}$ and upstream pressure is $P_1 = 200 \, \text{kPa}$, find $P_2$.

1. Continuity: $v_2 = v_1 (A_1/A_2) = 2 \times (0.01/0.005) = 4 \, \text{m/s}$
2. Bernoulli (horizontal, so $z_1 = z_2$): $P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) = 200{,}000 + \frac{1}{2}(1000)(4 - 16) = 200{,}000 - 6{,}000 = 194 \, \text{kPa}$

The pressure drops by 6 kPa at the constriction, which makes sense: velocity went up, so pressure went down.

Step-by-Step Problem-Solving Approach

1. Draw the system and label the two points where you'll apply Bernoulli's equation. Confirm they're on the same streamline.
2. Check assumptions: Is the flow steady, incompressible, and approximately inviscid between these points?
3. Choose a reference elevation ($z = 0$) and stick with it.
4. Write Bernoulli's equation between the two points. Cancel any terms that are equal or zero (e.g., elevation terms for a horizontal pipe).
5. Apply continuity ($A_1 v_1 = A_2 v_2$) if you have two unknowns for velocity.
6. Substitute known values and solve for the unknown.
7. Check your answer: Does the pressure drop where velocity increases? Are the magnitudes physically reasonable?

Common Mistakes and Misconceptions

• Applying Bernoulli across streamlines in rotational flow. Always confirm the two points share a streamline, or that the flow is irrotational.
• Ignoring elevation changes. In vertical or inclined flows, the $\rho g z$ terms matter and can dominate.
• Forgetting to use consistent units. Mixing kPa with Pa, or gauge with absolute pressure, is a common source of errors.
• Assuming higher velocity always means lower pressure. This is only true when elevation is constant. If a fluid speeds up while also dropping in elevation, pressure could stay the same or even increase.
• Using Bernoulli where viscous losses are significant. Long pipes, flows through porous media, or flows in boundary layers need the energy equation with a head loss term.

Practice Problems

Work through these before checking solutions:

1. Water exits a large tank through a small hole 3 m below the surface. Find the exit velocity. (Hint: this is Torricelli's theorem, a direct application of Bernoulli.)
2. A Venturi meter in a horizontal pipe has an upstream diameter of 10 cm and a throat diameter of 5 cm. If the pressure difference is 20 kPa and the fluid is water, find the volume flow rate.
3. A Pitot tube in an air stream ($\rho = 1.2 \, \text{kg/m}^3$) reads a stagnation pressure of 101,800 Pa and a static pressure of 101,325 Pa. What is the air velocity?