Bernoulli's equation expresses energy conservation for a flowing fluid. It relates pressure, velocity, and gravitational potential energy at different points along a streamline, revealing why higher flow speed corresponds to lower pressure and vice versa.
Deriving Bernoulli's equation
Consider a fluid element moving along a streamline (a path that follows the local velocity at each point). Between two points on the streamline, gravity and pressure do work on the element. By the work-energy theorem, the kinetic energy change equals the work done:
\[ \Delta KE = W_{\text{gravity}} + W_{\text{pressure}} \]
\[ \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = -mg(h_2 - h_1) + P_1 A \, s_1 - P_2 A \, s_2 \]
where \( s_1 \) and \( s_2 \) are the distances traveled. Dividing by mass and rearranging (using \( \rho = m/V = m/(A \cdot s) \)):
\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
This is Bernoulli's equation:
\[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]
Each term has dimensions of pressure (or energy per unit volume):
- \( P \) — pressure energy per unit volume
- \( \frac{1}{2} \rho v^2 \) — kinetic energy per unit volume
- \( \rho g h \) — gravitational potential energy per unit volume
The sum of these three terms is constant along a streamline (for an incompressible, non-viscous fluid in steady flow).
Interpreting Bernoulli's equation
High velocity → Low pressure: Where a fluid speeds up (e.g., narrowing pipe, by continuity), the kinetic energy term increases. To keep the sum constant, pressure must decrease. This is the Bernoulli effect, the physical basis for lift on airplane wings and the suction cup principle.
High altitude → Low pressure: As a fluid rises against gravity, \( \rho g h \) increases. Again, pressure decreases to compensate.
Pressure does work: When pressure decreases, the pressure term does positive work on fluid elements, accelerating them.
Worked example: Water flowing from a tank
A large water tank has a small hole 5 meters below the surface. Water flows out through the hole. Neglecting air resistance and assuming the tank is open to atmosphere, what is the speed of water exiting the hole?
Given:
- Depth of hole: \( h_1 - h_2 = 5 \) m
- Atmospheric pressure at surface: \( P_{\text{atm}} = 101,325 \) Pa
- Water density: \( \rho = 1000 \) kg/m³
- \( g = 9.8 \) m/s²
- Tank opening and hole both at atmospheric pressure (open to air)
Setup: Point 1 is at the surface; point 2 is at the hole exit.
- Height 1: \( h_1 = 5 \) m, velocity \( v_1 \approx 0 \) (tank is large, surface falls slowly)
- Height 2: \( h_2 = 0 \) m, velocity \( v_2 = ? \)
- Both at atmospheric pressure: \( P_1 = P_2 = P_{\text{atm}} \)
Apply Bernoulli: \[ P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
Since \( P_1 = P_2 \) and \( v_1 \approx 0 \): \[ \rho g h_1 = \frac{1}{2} \rho v_2^2 \]
\[ v_2 = \sqrt{2 g h_1} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9 \text{ m/s} \]
This result is independent of the tank's cross-section or the hole's size! It depends only on the depth difference, the same as for a ball dropped from height \( h \). This is Torricelli's law: the exit speed of fluid from an orifice equals the speed a particle would have if dropped from the same height.
Practical applications
Airplane wings: The wing shape forces air to move faster over the top surface than the bottom. By Bernoulli, pressure is lower above, creating lift.
Venturi meter: A narrowing in a pipe reduces pressure (by continuity and Bernoulli). Measuring the pressure drop indicates flow rate.
Atomizer/spray bottle: Blowing air across the top of a tube (creating high velocity) lowers pressure there, sucking up liquid from the bottom.
Curve ball in sports: A spinning baseball experiences asymmetric flow. Faster flow on one side (lower pressure) relative to the other (higher pressure) produces sideways force—the Magnus effect.
Special case: Static fluids
When \( v = 0 \) everywhere (no flow), Bernoulli reduces to the hydrostatic pressure relation:
\[ P_1 + \rho g h_1 = P_2 + \rho g h_2 \]
or
\[ P = P_0 + \rho g h \]
This is exactly the hydrostatic formula from §14.2. Bernoulli is thus a generalization that includes both static and dynamic fluids.
Energy dissipation: real fluids
The equation \( P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \) assumes an ideal, non-viscous fluid. In real fluids with viscosity, energy is dissipated as heat due to internal friction. Bernoulli still applies locally between two points close in time, but over longer distances or times, the "constant" decreases—some energy leaks away as heat. §14.6 explores viscosity quantitatively.
What you should be able to do
- Derive Bernoulli's equation from energy conservation
- Identify the three energy terms: pressure, kinetic, and potential
- Apply Bernoulli to find velocities, pressures, or heights in flowing systems
- Explain the Bernoulli effect and relate it to everyday phenomena (lift, spray)
- Compute exit velocity for fluid from a tank (Torricelli's law)
- Recognize that hydrostatic pressure is a special case of Bernoulli
Preview of §14.6
Viscosity: how internal fluid friction opposes flow and leads to the drag force (Stokes' law).
📚 See also: Halliday Vol 2, Ch 14, §14.5.
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