Bernoulli’s Equation: MDCAT Physics notes
Bernoulli’s Equation notes for MDCAT: P + ½ρv² + ρgh = constant, energy per unit volume terms, fast flow means low pressure, venturi and Torricelli.
The equation
For steady, non-viscous, incompressible flow along a streamline, Bernoulli's equation states:
$$P + \tfrac12\rho v^2 + \rho g h = \text{constant}$$
It is the law of conservation of energy applied to a moving fluid. Each term has units of pressure ($\text{N m}^{-2}$), which is the same as energy per unit volume ($\text{J m}^{-3}$).
| Term | Meaning |
|---|---|
| $P$ | Pressure (work done by pressure per unit volume) |
| $\tfrac12\rho v^2$ | Kinetic energy per unit volume |
| $\rho g h$ | Potential energy per unit volume |
These follow from dividing $\tfrac12 mv^2$ and $mgh$ by volume $V$, since $m/V = \rho$.
Speed and pressure
For a horizontal pipe, $h$ is the same everywhere, so
$$P + \tfrac12\rho v^2 = \text{constant}$$
Where the speed is high, the pressure is low. Combining with continuity ($A_1v_1 = A_2v_2$): as fluid enters a narrower part of the pipe, its velocity increases and its pressure decreases. The increase in kinetic energy associated with the drop in pressure is a consequence of Bernoulli's principle.
Example: water moving at $1\ \text{m s}^{-1}$ in the wide end reaches $3\ \text{m s}^{-1}$ in the narrow end. Pressure is lower at the narrow end:
$$P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2) = \tfrac12(1000)(9 - 1) = 4000\ \text{Pa}$$
Applications
- Venturi meter: measures flow speed from the pressure difference between the wide and narrow sections.
- Aeroplane wing (aerofoil): air moves faster over the curved upper surface, lowering pressure there and producing lift.
- Carburettor, spray gun, filter pump: fast-moving air or water creates low pressure that draws in fuel, liquid or air.
- Swing of a spinning ball: unequal air speeds on the two sides produce a pressure difference.
- Blood flow: at a narrowed (constricted) artery, blood speeds up and pressure drops.
Torricelli's theorem
For liquid escaping from a small hole at depth $h$ below the open surface of a large tank:
$$v = \sqrt{2gh}$$
the same speed a body gains falling freely through $h$.
Key formulas
- $P + \tfrac12\rho v^2 + \rho gh = \text{constant}$
- KE per unit volume $= \tfrac12\rho v^2$; PE per unit volume $= \rho gh$
- Horizontal: $P_1 - P_2 = \tfrac12\rho(v_2^2 - v_1^2)$
- Torricelli: $v = \sqrt{2gh}$
Common MDCAT traps
- Narrow region: high velocity with low pressure, not high pressure.
- $\tfrac12\rho v^2$ is KE per unit volume, not KE or KE per unit area.
- PE per unit volume is $\rho gh$, not $mgh$ or $gh$.
- Speed-pressure trade-off is Bernoulli; the speed-area relation alone is continuity.
Quick revision
- Bernoulli = energy conservation for fluids.
- Faster flow, lower pressure.
- Each term has units of pressure.
- Lift on wings comes from lower pressure above.