Bernoulli's principle and Bernoulli's equation

Bernoulli’s equation links pressure, speed, and height in steady fluid flow. You’ll see exactly why faster flow means lower pressure (and vice versa) and learn how to apply it confidently in problems.

What You’ll Learn

1. State and use Bernoulli’s equation along a streamline.
2. Identify when Bernoulli applies (steady, incompressible, inviscid flow).
3. Use the continuity equation (A1 v1 = A2 v2) to relate speeds in changing pipes.
4. Compute pressure drops or speed changes in horizontal/vertical flows.
5. Explain static, dynamic, and stagnation pressure (and how Pitot tubes work).
6. Connect Bernoulli to work–energy ideas for intuition and derivation.

Key Concepts Covered

• Bernoulli’s principle, pressure–velocity relationship
• Streamline (laminar) flow; incompressible, inviscid assumptions
• Continuity equation: A1 v1 = A2 v2
• Static pressure, dynamic pressure (½ ρ v²), stagnation pressure
• Elevation/pressure/velocity head; mechanical energy per unit volume
• Work–energy theorem in fluids; hydrostatic limit
• Venturi effect; Pitot tube airspeed
• Validity and common mistakes (turbulence, viscosity, pumps, head loss)

Why This Lesson Matters

From nozzles and Venturi meters to airplane wings and medical devices, Bernoulli’s equation shows how pressure, speed, and elevation trade energy. It’s a staple of AP Physics 2, IB, and JEE/NEET—and the work–energy derivation builds the intuition you need to tackle tricky exam questions without memorizing.

🔗 Prerequisite or Follow-Up Lessons

• Continuity Equation and Mass Flow Rate
• Hydrostatic Pressure and Pascal’s Principle

Full Lesson: Bernoulli’s Equation and Energy in Fluids

1) Plain-English Statement

Along a streamline in steady, incompressible, inviscid flow:
P + ½ ρ v² + ρ g h = constant
Interpretation: pressure energy + kinetic energy per unit volume + gravitational potential energy per unit volume stays the same as the fluid moves.

2) When You Can Use Bernoulli

Use it when all of these hold:

• Steady: flow properties don’t change with time at a point.
• Incompressible: density ρ is (approximately) constant.
• Inviscid: frictional (viscous) losses are negligible.
• Along a streamline: compare points on the same flow line.

3) The Continuity Equation (mass conservation)

For incompressible flow:
A1 v1 = A2 v2
Same volume per second enters and leaves any pipe section. Narrower area ⇒ higher speed.

4) Derivation Sketch (work–energy, no calculus needed)

Setup: Track a fluid “chunk” moving from section 1 to section 2 in time Δt.

• Volumes moved: A1 Δℓ1 (in) and A2 Δℓ2 (out) are equal → A1 Δℓ1 = A2 Δℓ2.
• Divide by Δt → A1 v1 = A2 v2 (continuity).

Work terms:

• Inlet pressure does positive work: W1 = P1 A1 Δℓ1.
• Outlet pressure does negative work: W2 = − P2 A2 Δℓ2.
• Gravity work: W3 = − m g (h2 − h1) (negative if the chunk rises).

Net work: W = W1 + W2 + W3.
Kinetic energy change: ΔKE = ½ m v2² − ½ m v1².
Set W = ΔKE and use m = ρ A1 Δℓ1 = ρ A2 Δℓ2 plus A1 Δℓ1 = A2 Δℓ2 to cancel volumes.
You get:
P2 + ½ ρ v2² + ρ g h2 = P1 + ½ ρ v1² + ρ g h1
or equivalently, P + ½ ρ v² + ρ g h = constant along the streamline.

5) Why “Pressure Is Energy per Volume”

Pressure × volume = work. So pressure = work/volume (units J/m³). That’s why P sits next to ½ ρ v² and ρ g h: all three are energy densities.

6) Special Cases You’ll Use a Lot

• Horizontal flow (h1 = h2):
  P + ½ ρ v² = constant.
  If speed goes up, pressure must drop (Venturi effect).
• No flow (v1 = v2 = 0):
  P2 − P1 = − ρ g (h2 − h1) → the hydrostatic rule.

7) Static, Dynamic, and Stagnation Pressure

• Static pressure (P): what a pressure gauge reads when moving with the flow.
• Dynamic pressure: ½ ρ v² (kinetic energy per volume).
• Stagnation pressure (P0): pressure when the flow is brought to rest isentropically:
  P0 = P + ½ ρ v² (for the Bernoulli conditions).

Pitot tube airspeed:
Measure P0 (total) and P (static), then
v = sqrt( 2 (P0 − P) / ρ ).

Quick check: If ΔP = (P0 − P) = 50 Pa in air (ρ ≈ 1.2 kg/m³),
v ≈ sqrt( 2*50 / 1.2 ) ≈ sqrt(83.33) ≈ 9.1 m/s.

8) Worked Mini-Example (narrowing pipe, same height)

Water (ρ = 1000 kg/m³) flows from area A1 = 4.0 cm² to A2 = 1.0 cm². Upstream speed v1 = 1.0 m/s. Find v2 and pressure drop P1 − P2.

• Continuity: A1 v1 = A2 v2 → v2 = (A1/A2) v1 = 4 × 1.0 = 4.0 m/s.
• Bernoulli (horizontal): P1 + ½ ρ v1² = P2 + ½ ρ v2² →
  P1 − P2 = ½ ρ (v2² − v1²)
  = 0.5 × 1000 × (16 − 1) = 7500 Pa (about 7.5 kPa drop).

9) Problem-Solving Recipe (exams love this)

1. Mark data on a clean diagram. Pick points 1 and 2 on the same streamline.
2. Continuity to relate speeds: A1 v1 = A2 v2.
3. Bernoulli between points: P + ½ ρ v² + ρ g h = constant.
4. Decide if heights cancel (horizontal) or not.
5. Solve for the target (P, v, or h). Check units (Pa, m/s, m).
6. Validate assumptions (steady, incompressible, inviscid; no pumps/turbines unless included).

10) When Bernoulli Fails (and what to use instead)

• High viscosity / long pipes: energy loss matters → include head loss or use Poiseuille (laminar pipe flow).
• Turbulence: use Reynolds number to check; Bernoulli is not reliable across large turbulent regions.
• Pumps/turbines: add work terms. A common engineering form is
  P1 + ½ ρ v1² + ρ g h1 + ρ g H_pump − ρ g h_loss
  = P2 + ½ ρ v2² + ρ g h2.
• Compressible flows (high-speed gases): need compressible-flow relations.

11) Quick “Text Diagrams” You Can Picture

• Venturi: wide → throat → wide again.
  As area ↓ to throat, speed ↑ and static pressure ↓ (throat is the lowest P).
• Pitot probe: forward-facing tube (measures P0) next to static ports on the side (measure P). v from ΔP.

Summary Takeaways

• Bernoulli is just energy conservation per volume along a streamline.
• Pair it with continuity to connect geometry (area) to physics (speed, pressure).
• Always check the assumptions before plugging numbers.

Bernoulli's Principle & Equation: Frequently Asked Questions (FAQ)

Q1. What is Bernoulli’s principle in simple terms, and what is the fundamental pressure-velocity relationship in fluid dynamics?

Bernoulli's principle describes the core pressure-velocity relationship in fluid dynamics. It states that for an ideal fluid in streamline (laminar) flow, an increase in speed occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. The most common takeaway is: where the fluid moves faster, the pressure is lower, assuming height is constant.

Q2. What is Bernoulli's equation, and how does it represent the work-energy principle for fluids?

Bernoulli's equation is the mathematical expression of the principle, representing the conservation of mechanical energy per unit volume for a moving fluid. It is derived from the work–energy in fluids concept: the net work done by pressure forces on a fluid element equals the change in its kinetic and potential energy. The equation P + ½ρv² + ρgy = constant means that the sum of static pressure, dynamic pressure, and gravitational potential energy density is constant along a streamline.

Q3. What do the terms in Bernoulli's equation, P + ½ρv² + ρgy = constant, represent?

Each term represents a type of energy density and has units of pressure:

• P is the static pressure: The pressure exerted by the fluid if it were not moving.
• ½ρv² is the dynamic pressure: The pressure arising from the fluid's motion, representing its kinetic energy density.
• ρgy is the hydrostatic pressure: The pressure due to the fluid's position in a gravitational field, representing the gravitational potential in fluids.
• In engineering, these are sometimes referred to as pressure head, velocity head, and elevation head, respectively, by dividing the entire equation by ρg.

Q4. What are the key assumptions and validity conditions for applying Bernoulli's equation in AP Physics problems?

The equation is only valid under these conditions for the fluid flow:

• Inviscid flow (negligible viscosity): There are no frictional losses. One of the most common mistakes with Bernoulli is ignoring significant viscosity.
• Incompressible flow: The fluid's density (ρ) is constant. This holds for liquids and is a good approximation for gases at speeds well below the speed of sound.
• Steady flow: The velocity at any point in the fluid does not change over time.
• Streamline (laminar) flow: The fluid moves in smooth layers. The equation is not valid for turbulent flow. You must apply the equation along a single streamline.

Q5. What is the difference between Bernoulli’s equation and the continuity equation (A₁v₁ = A₂v₂), and how do they work together?

The Bernoulli vs continuity equation difference is about the conservation laws they represent:

• Continuity Equation (A₁v₁ = A₂v₂): Represents fluid continuity, which is the conservation of mass. It states that the volume flow rate is constant. If the pipe area (A) decreases, the fluid velocity (v) must increase.
• Bernoulli's Equation: Represents the conservation of energy. It explains how pressure and height change when the velocity changes.
• You will often use them together in combined problem setups: use continuity to find the velocities, then use Bernoulli's to find the pressure change.

Q6. Conceptually, why does pressure decrease when velocity increases? What is the intuitive explanation?

This is the central question for understanding Bernoulli's principle. For a fluid to accelerate and speed up, a net force must act on it. This force comes from a pressure gradient, meaning the fluid must flow from a region of higher pressure to a region of lower pressure. This pressure difference provides the unbalanced force needed for the acceleration in a fluid.

Q7. Is the "equal transit time" theory a correct explanation for airfoil lift?

No, this is the widely cited equal transit time myth. This incorrect theory claims air parcels that split at the front of the wing must meet at the back at the same time. In reality, the air over the top moves much faster and arrives at the back much sooner. The real reason for the speed difference is more complex, involving the turning of the air (the Coandă effect) and Newton's third law (the wing pushes air down, so the air pushes the wing up).

Q8. How do you simplify Bernoulli's equation for a horizontal pipe, and how can it be used to calculate a pressure drop?

For a Bernoulli equation for a horizontal pipe, the height is constant (y is constant). This means the ρgy term can be removed from both sides, simplifying the equation to: P₁ + ½ρv₁² = P₂ + ½ρv₂². For a narrowing pipe example, the continuity equation tells you the speed increases in the narrow section. This simplified Bernoulli equation then shows that the pressure (P₂) must decrease as the velocity (v₂) increases.

Q9. How does Bernoulli's equation lead to Torricelli's Law for fluid speed from a tank opening?

This is a classic problem of using Bernoulli to find fluid speed. Let point 1 be the large surface of the water and point 2 be the small opening. At the surface, pressure is atmospheric and velocity is nearly zero. At the opening, pressure is also atmospheric. Bernoulli's equation simplifies to ρgh₁ = ½ρv₂² + ρgh₂. Rearranging this gives Torricelli's Law: v₂ = √(2gΔh), where Δh is the height of the fluid above the opening.

Q10. What is the Venturi effect, and how does a Venturi meter work?

The Venturi effect is the reduction in fluid pressure that occurs when a fluid flows through a constricted section (a choke) of a pipe. It's a direct application of Bernoulli's principle. A Venturi meter uses this effect to measure flow speed. By measuring the pressure difference between the wide and narrow sections of the tube, one can calculate the fluid's velocity.

Q11. Can you provide a simple explanation of dynamic pressure vs. static pressure?

• Static pressure (P) is the pressure of a fluid at rest. It is what you would measure if you were moving along with the fluid. It's an omnidirectional pressure caused by the random motion of molecules.
• Dynamic pressure (½ρv²) is the pressure associated with the bulk movement (kinetic energy) of the fluid. It is not a true pressure in the same sense as static pressure but represents the decrease in static pressure that occurs when the fluid is in motion.

Q12. How do you check the units in P + ½ρv² + ρgh?

Every term in Bernoulli's equation must have units of pressure. The standard unit is the Pascal (Pa). A Pascal is also a Newton per square meter (N/m²) or a Joule per cubic meter (J/m³), reflecting the equation's connection to energy density.

Q13. What is the best strategy for choosing points 1 and 2 when solving combined continuity and Bernoulli problems?

• Choose point 1 at a location where you have the most known information (e.g., the surface of a large, open tank where pressure is atmospheric and velocity is approximately zero).
• Choose point 2 at the location where you need to find an unknown value (e.g., the exit of a pipe where you want to find the velocity).
• When dealing with different elevations, choose a reference height (y=0) that simplifies the problem, such as the height of the lowest point.

Q14. How is Bernoulli's equation a statement of the conservation of mechanical energy?

The equation balances the mechanical energy per unit volume of a fluid. It shows that energy can be converted between three forms: potential energy due to height (ρgy), kinetic energy due to motion (½ρv²), and energy stored in the fluid as pressure (P). In an ideal system, the total amount of this energy remains constant. In real systems, a term for energy loss due to friction (head loss) or energy added by a pump is included.

Q15. What is stagnation pressure, and how do Pitot tubes use it to measure fluid speed?

Stagnation pressure is the pressure at a point in a fluid flow where the fluid is brought to a complete stop (v = 0). A Pitot tube, used on airplanes to measure airspeed, has an opening that faces directly into the airflow, creating a stagnation point. It compares this stagnation pressure (P_stagnation) to the static pressure (P_static) of the surrounding air. Using Bernoulli's equation, the speed can be calculated with the pitot tube formula: v = √(2(P_stagnation − P_static) / ρ).

Q16. Does Bernoulli's equation use gauge pressure or absolute pressure?

Because Bernoulli's equation almost always involves the difference in pressure between two points, you can use either gauge pressure or absolute pressure, as long as you are consistent for both points. The atmospheric pressure term will cancel out. The only time you must be careful is if one pressure is given as absolute (e.g., in a vacuum) and the other is given as gauge; in that case, you must convert them to the same scale before using the equation.

Q17. What are some common mistakes to avoid, and how can you spot an invalid setup for Bernoulli's equation?

To spot invalid Bernoulli setups in exam problems, look for these red flags:

• Significant viscosity or friction: The problem mentions a "long, narrow pipe" or "syrup," indicating friction is important.
• Turbulent flow: The fluid is mixing chaotically, not flowing in smooth streamlines.
• Compressible flow: The problem involves gas moving at very high speeds (near the speed of sound).
• Energy being added or removed: The fluid passes through a pump, fan, or turbine between points 1 and 2.
• Applying across different streamlines: You cannot apply the equation from the top of a stream to the bottom of the stream if they are not part of the same continuous line of flow.

Q18. Why does a shower curtain get sucked in?

This is a classic, everyday example. The spray from the shower head makes the air inside the shower move quickly. This fast-moving, turbulent air has a lower average pressure than the still, higher-pressure air in the bathroom. This pressure difference creates a net inward force, pushing the lightweight curtain into the shower.