Topic 6.513 min read
Bernoulli's Equation
Bernoulli's equation applies conservation of mechanical energy to steady ideal-fluid flow. The sum of pressure, kinetic energy per unit volume and gravitational potential energy per unit volume remains constant along a streamline.
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Beginner Level
A moving fluid can carry energy in three forms: pressure energy, energy of motion and energy due to height. Bernoulli's equation says that their total remains constant for ideal steady flow.
- P
- pressure (Pa)
- ρ
- density (kg m⁻³)
- v
- speed (m s⁻¹)
- h
- height (m)
At the same height, faster ideal-fluid flow is associated with lower static pressure.
Recommended for Class 11
Bernoulli's Equation
Overview
Bernoulli's equation is an approximate relation between pressure, speed and elevation for an ideal fluid. It is obtained by applying conservation of mechanical energy to fluid moving between two points of a pipe.
Learning Outcomes
- State the conditions for Bernoulli's equation.
- Derive the equation from pressure work and energy changes.
- Interpret every term as energy per unit volume.
- Use the two-point and constant forms.
- Explain pressure-speed and pressure-height relations.
- Solve standard numerical problems.
Intermediate Level
Conditions
- Flow is steady.
- Fluid is incompressible, so density is constant.
- Viscosity and frictional energy loss are neglected.
- The two points lie along the same streamline.
- No pump or turbine adds or removes mechanical energy between the selected points.
Derivation: Work Done by Pressure
Consider equal fluid volume V moving from a lower section 1 to an upper section 2 during time Δt. The distances moved are Δx₁ = v₁Δt and Δx₂ = v₂Δt.
At section 1, pressure force F₁ = P₁A₁ acts with the motion, so it does positive work. At section 2, F₂ = P₂A₂ acts opposite to the motion of the selected fluid volume, so its work is negative.
Continuity gives A₁v₁Δt = A₂v₂Δt = V. Therefore the net pressure work is:
- V
- equal fluid volume transferred (m³)
Applying Conservation of Energy
The net pressure work changes the kinetic and gravitational potential energy of the fluid volume.
Divide by m, multiply by ρ and rearrange to obtain the two-point form:
Meaning of the Three Terms
| Term | Meaning | SI unit |
|---|---|---|
| P | pressure energy per unit volume | Pa = J m⁻³ |
| ½ρv² | kinetic energy per unit volume | Pa = J m⁻³ |
| ρgh | gravitational potential energy per unit volume | Pa = J m⁻³ |
Useful Special Cases
For a horizontal pipe, h₁ = h₂, so the height terms cancel.
If speed is also unchanged, pressure decreases as height increases.
At equal height, continuity makes fluid faster in a narrow section. Bernoulli then predicts lower static pressure there under ideal conditions.
Solved Examples
Example 1: Horizontal Pipe
Water speeds up from 2.0 m s⁻¹ to 5.0 m s⁻¹ in a horizontal pipe. If P₁ = 1.50 × 10⁵ Pa and ρ = 1000 kg m⁻³, find P₂.
- v₁ = 2.0 m s⁻¹
- v₂ = 5.0 m s⁻¹
- P₁ = 1.50 × 10⁵ Pa
- ρ = 1000 kg m⁻³
Use P₁ + ½ρv₁² = P₂ + ½ρv₂².
P₂ = P₁ + ½ρ(v₁² − v₂²).
P₂ = 1.50 × 10⁵ + 500(4 − 25).
P₂ = 1.50 × 10⁵ − 1.05 × 10⁴.
Final answer: P₂ = 1.395 × 10⁵ Pa
Example 2: Rise in Height
Water moves with unchanged speed from point 1 to a point 4.0 m higher. If P₁ = 2.00 × 10⁵ Pa, find P₂. Take ρ = 1000 kg m⁻³ and g = 9.8 m s⁻².
- h₂ − h₁ = 4.0 m
- v₁ = v₂
- P₁ = 2.00 × 10⁵ Pa
Equal speeds cancel kinetic terms.
P₂ = P₁ − ρg(h₂ − h₁).
P₂ = 2.00 × 10⁵ − (1000)(9.8)(4.0).
Final answer: P₂ = 1.608 × 10⁵ Pa
Example 3: Narrow Section
Explanation: This conclusion depends on Bernoulli's ideal-flow conditions.
Common Mistakes
- Writing ½ρv instead of ½ρv².
- Omitting density from ρgh.
- Using different height reference levels without consistency.
- Adding height in metres directly to pressure in pascals.
- Claiming faster flow always means lower pressure in every situation.
- Applying the equation across large frictional losses without correction.
Short-Question Preparation
Q. State Bernoulli's equation.
A. P + ½ρv² + ρgh = constant along a streamline for steady ideal-fluid flow.
Q. What principle is used in its derivation?
A. Conservation of mechanical energy.
Q. What does ½ρv² represent?
A. Kinetic energy per unit volume.
Q. Why is pressure work at section 2 negative?
A. The pressure force there acts opposite to the displacement of the selected fluid volume.
Q. What happens to pressure when horizontal ideal flow speeds up?
A. Static pressure decreases.
Long-Question Preparation
Q. Derive Bernoulli's equation for ideal-fluid flow.
A. 1. Select equal volume V moving between two sections. 2. Write pressure work W₁ = P₁A₁Δx₁ and W₂ = −P₂A₂Δx₂. 3. Use continuity to set A₁Δx₁ = A₂Δx₂ = V. 4. Obtain W = (P₁ − P₂)V. 5. Substitute V = mρ. 6. Set pressure work equal to ΔK + ΔU. 7. Substitute kinetic and potential energy changes. 8. Rearrange to obtain P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. 9. State assumptions and interpret all three terms.
Sticky Notes
- Bernoulli = energy conservation.
- P is pressure energy per volume.
- ½ρv² is kinetic energy per volume.
- ρgh is potential energy per volume.
- Every term has unit Pa.
- At equal height: faster flow → lower pressure, ideally.
60-Second Revision
Summary
Bernoulli's equation balances pressure, kinetic and gravitational potential energy per unit volume along an ideal-fluid streamline. Pressure work supplies the change in kinetic and potential energy.
Ask-the-Tutor Modes
- Explain Bernoulli simply.
- Show the derivation step by step.
- Explain each energy term.
- Solve a horizontal-pipe problem.
- Solve a height-change problem.
- Check whether assumptions apply.
- Test me with an MCQ.
Common Mistakes
Common Mistakes
- Mistake
State the ideal-flow assumptions.
- Mistake
Use v² in the kinetic term.
- Mistake
Keep pressure and energy-per-volume units consistent.
- Mistake
Do not treat the pressure-speed statement as universal.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
P + ½ρv² + ρgh is constant.
All terms have unit Pa.
State the ideal-flow conditions.
At equal height, faster flow has lower static pressure.
Bernoulli's equation expresses conservation of mechanical energy per unit volume in ideal-fluid flow.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
- Definition
- Bernoulli's equation expresses conservation of mechanical energy per unit volume in ideal-fluid flow.
- Formula
- P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
- Example
- In a horizontal narrowing, speed rises and static pressure falls.
- Critical warning
- Do not omit the square on speed.
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Exam Practice
Important Short Question
Q1. State Bernoulli's equation and its conditions.
Important Short Question
Q2. Explain the three terms in Bernoulli's equation.
Conceptual Question
Q3. Why does pressure fall when horizontal ideal flow speeds up?
Long Question Preparation
Q4. Derive Bernoulli's equation from pressure work.