Bernoulli's Equation and the Venturi Effect for the MCAT

Bernoulli's equation, P + ½ρv² + ρgh = constant along a streamline, is conservation of energy for a moving ideal fluid, with every term written as energy per unit volume. Pair it with the continuity equation, A₁v₁ = A₂v₂, and most flow problems fall out: a narrower section forces faster flow, and at the same height faster flow means lower pressure. That pressure drop at a constriction is the Venturi effect.

Step one: continuity sets the speed

For an incompressible fluid in steady flow, no volume can pile up or vanish between two points in a pipe. The volume flow rate Q must be the same everywhere:

Q = Av = constant, so A₁v₁ = A₂v₂

Q is in m³/s. A is the cross-sectional area, and for a round pipe A = πr². The radius enters squared, which is where most errors start:

When a pipe splits into branches, the rule applies to the total area: whatever volume per second enters the junction must leave through all the branches combined. This explains why blood moves slowly through capillaries even though each capillary is tiny: there are so many of them that their combined cross-sectional area is far larger than the aorta's, so the average speed there is far lower.

Step two: Bernoulli sets the pressure

Bernoulli's equation compares two points on the same streamline:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Read each term as an energy density in J/m³, which is the same unit as pascals:

• P is the static pressure, the push the fluid exerts on the pipe wall.
• ½ρv² is kinetic energy per unit volume, sometimes called dynamic pressure. Compare it with ½mv²: the mass has been replaced by density.
• ρgh is gravitational potential energy per unit volume. Compare it with mgh.

The sum stays constant, so if one term rises another must fall. In a horizontal pipe the ρgh terms cancel, and the trade is purely between pressure and speed. If the fluid is not moving at all, the speed terms vanish and the equation reduces to the hydrostatic result, P_deep = P_shallow + ρgΔh.

Why does pressure fall where flow speeds up? A fluid parcel entering a constriction must accelerate, and something must push it forward. The only push available is a pressure difference, higher behind the parcel than ahead of it. So the pressure has to be lower in the fast, narrow section.

The assumptions behind it

Bernoulli's equation as written assumes an ideal fluid:

1. Incompressible (constant density). Good for liquids and for slow gas flows.
2. Nonviscous, so no energy is lost to internal friction.
3. Steady flow, with velocity at each point not changing in time.
4. Laminar flow along a streamline, without turbulence.

Real fluids in long, narrow tubes violate assumption 2. In that case pressure also falls steadily along the tube even where the area is constant, and Poiseuille's law describes the drop instead: Q = ΔPπr⁴/(8ηL). Halving the radius in that regime raises resistance 16-fold. Keep the two laws separate. Bernoulli tracks energy trades in an ideal fluid; Poiseuille tracks energy lost to viscosity.

Worked example 1: a horizontal Venturi

Water (ρ = 1000 kg/m³) flows at 0.80 m/s through a horizontal pipe of radius 3.0 cm, which narrows to a radius of 1.5 cm. Find the speed in the narrow section and the pressure difference.

1. The radius ratio is 2, so the area ratio is 2² = 4 and v₂ = 4 × 0.80 = 3.2 m/s.
2. The pipe is horizontal, so P₁ - P₂ = ½ρ(v₂² - v₁²).
3. P₁ - P₂ = 0.5 × 1000 × (10.24 - 0.64) = 0.5 × 1000 × 9.6 = 4800 Pa, about 4.8 kPa.

The flow rate is Q = π(0.030)²(0.80), about 2.3 × 10^-3 m³/s, or 2.3 liters per second, in both sections. A Venturi meter uses exactly this logic in reverse: measure the pressure difference with two vertical tubes, then solve for the speed and the flow rate.

Worked example 2: pushing water uphill

Water flows through a pipe of constant diameter from a basement up to a faucet 4.0 m higher. How much lower is the pressure at the faucet while the water is flowing?

1. Same diameter means same area, so continuity gives v₁ = v₂. The speed terms cancel.
2. Bernoulli leaves P₁ - P₂ = ρg(h₂ - h₁) = 1000 × 9.8 × 4.0 = 39,200 Pa, about 39 kPa.

The water is not slowing down. The pressure, not the speed, pays for the climb. This is the case students miss when they memorize "faster means lower pressure" without the words "at the same height."

Worked example 3: Torricelli's theorem

An open water tank has a small hole 0.45 m below the water surface. The tank is wide, so the surface drops so slowly that its speed is close to zero. How fast does water leave the hole?

Apply Bernoulli between the surface (point 1) and the hole (point 2). Both are open to the atmosphere, so P₁ = P₂, and v₁ is about 0. What remains is ρgh = ½ρv₂², so

v = √(2gh) = √(2 × 9.8 × 0.45) = about 3.0 m/s

That is the same speed an object would reach falling 0.45 m, and density cancels, so oil would leave at the same speed. If the hole has an area of 2.0 cm² (2.0 × 10^-4 m²), the outflow is Q = Av, about 5.9 × 10^-4 m³/s, or 0.59 L/s, ignoring the slight narrowing of a real jet.

Blood vessels: the classic applications

Two clinical setups show up again and again in physiology-flavored physics questions.

• Narrowed artery (stenosis). Continuity says blood speeds up through the narrow segment. In the ideal Bernoulli picture, the faster flow comes with lower static pressure on the wall at that spot. Real blood is viscous, so the narrowing also raises resistance sharply, following the r⁴ dependence in Poiseuille's law.
• Widened artery (aneurysm). The opposite trade. The wide section has slower flow, and by Bernoulli its static pressure is higher than in the normal segments nearby, which pushes outward on an already weakened wall.

A careful answer names both principles: continuity for the speed, Bernoulli for the pressure at a given height.

Common traps

• Holding v constant instead of Q. Flow rate is what is conserved. Speed changes with area.
• Using radius ratios as area ratios. Area goes as r². A pipe with one third the diameter has one ninth the area and nine times the speed.
• Dropping the height term. "Faster flow means lower pressure" is only true at the same height. In a vertical or sloped pipe, include ρgh.
• Using ρ = 1 for water. In SI units, water is 1000 kg/m³. Using 1 g/cm³ with meters per second gives an answer a thousand times too small.
• Thinking density matters for Torricelli. Exit speed depends only on the depth of the hole below the surface.
• Applying Bernoulli to long viscous flow. If friction matters, pressure falls along a uniform tube, and Poiseuille's law is the right tool.
• Confusing static pressure with total energy. The pressure on the wall drops in a constriction, but the fluid has not lost energy. It has converted pressure energy into kinetic energy, and it converts it back downstream.

One-minute recap

Continuity: Q = Av is constant, so narrowing a pipe speeds the flow, and speed scales with 1/r². Bernoulli: P + ½ρv² + ρgh is constant along a streamline for an ideal fluid, each term an energy per volume in pascals. At constant height, faster flow means lower static pressure (Venturi effect). At constant speed, higher means lower pressure by ρgΔh. With the fluid at rest it becomes the hydrostatic equation, and for a hole in an open tank it gives Torricelli's v = √(2gh). Viscous losses in long tubes belong to Poiseuille's law.