Chapter 6: Bernoulli’s Equation

Course: Prep4Uni Fluid Mechanics 1

Chapter 1: Pressure
Chapter 2: Variation of Pressure with Depth
Chapter 3: Pressure Measurement
Chapter 4: Buoyant Forces & Archimedes’ Principle
Chapter 5: Fluid Dynamics
Chapter 6: Bernoulli’s Equation
Chapter 7: Applications of Fluid Dynamics

🚁Overview

Starting from energy conservation for an inviscid, incompressible, steady flow, we derive Bernoulli’s equation
p + ½ρv² + ρgh = constant
along a streamline. We then apply it to practical devices such as Pitot tubes and Venturi meters, and discuss the equation’s assumptions and limitations.

Derive p + ½ρv² + ρgh = constant along a streamline.
Solve Bernoulli problems involving changes in height and velocity.
Recognize the underlying assumptions (inviscid, steady, incompressible) and limitations.

📖Contents

Derivation of Bernoulli’s Equation
Applications of Bernoulli’s Equation
Assumptions & Limitations of Bernoulli’s Equation
Worked Examples on Bernoulli’s Equation
Practice Problems on Bernoulli’s Equation
Key Formula Recap

🎯Learning Outcomes

Derive Bernoulli’s equation p + ½ρv² + ρgh = constant from energy conservation along a streamline.
Identify and distinguish between static pressure, dynamic pressure (½ρv²), and hydrostatic head (ρgh).
Apply Bernoulli’s equation to Pitot‐tube measurements to determine flow velocity from stagnation and static pressures.
Use Bernoulli’s equation together with continuity to solve Venturi‐meter flow‐rate problems.
Account for changes in elevation (h) and velocity (v) when applying Bernoulli’s equation between two points.
Recognize the key assumptions (steady, inviscid, incompressible flow; no energy losses) and limitations of Bernoulli’s equation.

6.1 Derivation of Bernoulli’s Equation

(Using Energy Conservation)

Consider a fluid element of volume V moving steadily along a streamline from point 1 to point 2 in an incompressible, inviscid flow. We apply the principle of conservation of mechanical energy:

i. Work Done by Pressure Forces

The fluid at the upstream end (point 1) pushes the element with pressure p₁, doing work p₁V.
The fluid at the downstream end (point 2) resists the motion with pressure p₂, doing negative work −p₂V.

Net work done on the fluid:
W = p₁V − p₂V = V(p₁ − p₂)

ii. Change in Kinetic Energy

ΔKE = ½ ρ V (v₂² − v₁²)

iii. Change in Potential Energy

ΔPE = ρ V g (h₂ − h₁)

iv. Energy Conservation Equation

Net work input = Change in kinetic energy + Change in potential energy:

V(p₁ − p₂) = ½ ρ V (v₂² − v₁²) + ρ V g (h₂ − h₁)

Divide through by V:

p₁ − p₂ = ½ ρ (v₂² − v₁²) + ρ g (h₂ − h₁)

v. Final Rearrangement

Rearranged into Bernoulli’s form:

p + ½ ρ v² + ρ g h = constant

Assumptions:

Steady flow
Incompressible fluid (constant volume)
Inviscid (no friction losses)
Along a single streamline

6.2 Applications of Bernoulli’s Equation

Pitot Tube

A Pitot tube measures stagnation pressure p₀ where the flow is brought to rest. Applying Bernoulli between free-stream (point 1) and stagnation (point 0):

p₀ = p₁ + ½ρv₁² → v₁ = √[2(p₀ − p₁)/ρ].

Venturi Meter

A Venturi constricts flow from area A₁ to A₂, causing a pressure drop from p₁ to p₂. From Bernoulli and continuity (A₁v₁ = A₂v₂):

p₁ + ½ρv₁² = p₂ + ½ρv₂²,
v₂ = v₁·(A₁/A₂) ⇒
Q = A₂·v₂ = A₂·√{[2(p₁−p₂)/ρ]·[A₁²/(A₁²−A₂²)]}.

Venturi Tube for Measuring Fluid Flowrate

6.3 Assumptions & Limitations

Inviscid: Neglects viscosity; no frictional losses.
Steady: Flow properties do not change with time.
Incompressible: Constant density (valid for liquids).
No energy losses: Ignores turbulence, heat transfer, pumps.
Within these bounds, Bernoulli’s equation accurately predicts pressure–velocity–height relationships.

6.4 Worked Examples

Example 1: A Pitot tube in air (ρ = 1.2 kg/m³) reads a stagnation pressure 1200 Pa above static. Find the flow speed.

Solution:
v = √[2·1200/1.2] ≈ 44.7 m/s.

Example 2: Water flows through a horizontal Venturi with A₁ = 0.05 m², A₂ = 0.02 m², and p₁−p₂ = 5000 Pa. Compute Q.

Solution:
Q = A₂·√{[2·5000/1000]·[0.05²/(0.05²−0.02²)]} ≈ 0.058 m³/s.

6.5 Practice Problems

Problem 1. A Pitot tube measures Δp = 2500 Pa in water (ρ = 1000 kg/m³). Find the flow speed.

Solution: v = √[2·2500/1000] ≈ 2.24 m/s.

Problem 2. Air (ρ = 1.2 kg/m³) flows from a duct of diameter 0.3 m into a 0.15 m nozzle. If static pressures are equal, find v₂ given v₁ = 5 m/s.

Solution: v₂ = v₁·(D₁/D₂)² = 5·(0.3/0.15)² = 20 m/s.

Problem 3. Water flows down a vertical pipe; what is the pressure change per meter of drop, neglecting velocity change?

Solution: From p + ρgh = constant, Δp/Δh = −ρg ≈ −9810 Pa/m.

6.6 Key Formula Recap

Formula	Use
`p + ½ρv² + ρgh = constant`	Energy conservation along a streamline
`v = √[2(p₀ − p)/ρ]`	Pitot tube speed from dynamic pressure
`Q = A₂·√{[2(p₁−p₂)/ρ]·[A₁²/(A₁²−A₂²)]}`	Venturi meter flow rate

Proceed to: Chapter 7: Applications of Fluid Dynamics

Return to: Prep4Uni Fluid Mechanics 1

📝EXERCISES

25 Questions & Answers on Bernoulli’s Equation

1. What is Bernoulli’s equation for steady incompressible flow?

p + ½ρv² + ρgh = constant along a streamline for inviscid, steady, incompressible flow.

2. What does the term ½ρv² represent?

It is the dynamic pressure, the kinetic energy per unit volume of the fluid.

3. What is the physical meaning of the term ρgh?

It is the hydrostatic pressure head, the potential energy per unit volume due to elevation h.

4. Under what assumptions is Bernoulli’s equation valid?

Inviscid (no viscosity), incompressible (constant ρ), steady flow, along a single streamline, and no energy losses.

5. What is stagnation pressure?

The pressure a fluid attains when brought to rest isentropically: p₀ = p + ½ρv².

6. How do you compute fluid speed from a Pitot tube reading?

Use v = √[2(p₀ − p)/ρ], where p₀ is stagnation and p static pressure.

7. What is the Venturi effect?

A constriction in a pipe accelerates fluid, lowering static pressure: p₁ + ½ρv₁² = p₂ + ½ρv₂², with v₂ > v₁.

8. State the continuity equation for incompressible flow.

A₁v₁ = A₂v₂, ensuring mass conservation when ρ is constant.

9. How does elevation difference enter Bernoulli’s equation?

Via the term ρgh: a rise in elevation decreases pressure and/or velocity head accordingly.

10. Why can we neglect viscous losses in Bernoulli’s derivation?

Because the fluid is assumed inviscid; in real flows, head loss terms must be added to the equation.

11. What modification is needed to include pump work?

Add a work term + W_pump/V to the right side of Bernoulli’s equation.

12. Why is Bernoulli’s equation applied along a streamline?

Because energy conservation holds for a single path; different streamlines need their own constants.

13. How do compressibility effects alter Bernoulli’s equation?

For compressible flows, the pressure–density relation must be integrated: ∫dp/ρ + ½v² + gh = constant.

14. What is dynamic head?

Another name for ½ρv², representing the kinetic energy term.

15. What is static head?

The sum p + ρgh, representing potential energy per unit volume.

16. How does a Pitot‐static tube measure flow velocity?

By comparing stagnation pressure (Pitot) to static pressure (static ports) and using Bernoulli’s relation.

17. What is total head?

H = (p/ρg) + (v²/2g) + h, constant along a streamline for ideal flow.

18. Why does pressure drop in a nozzle?

Because velocity increases (dynamic head increases), so static pressure must decrease to keep the sum constant.

19. What happens if Bernoulli’s assumptions are violated?

Predictions become inaccurate; viscous losses, unsteady effects, or compressibility must be accounted for.

20. How can Bernoulli’s equation explain lift on an airfoil?

Faster flow over the top surface yields lower pressure, creating a net upward force (lift).

21. What is the role of potential energy in Bernoulli’s equation?

The term ρgh represents gravitational potential energy per unit volume.

22. How do you derive Bernoulli’s equation from work–energy?

Equate work done by pressure and gravity to changes in kinetic energy of a fluid element.

23. Why is Bernoulli’s equation not valid across a shock wave?

Because the flow is no longer inviscid or adiabatic and discontinuities violate steady assumptions.

24. What is the stagnation enthalpy?

For compressible flow: h₀ = h + v²/2, analogous to stagnation pressure energy.

25. How is Bernoulli’s equation used in Venturi flow measurement?

Measure the pressure difference between wide and narrow sections, then compute Q via Bernoulli + continuity.

25 Problems & Solutions on Bernoulli’s Equations

Problem 1: A Pitot tube in air (ρ = 1.2 kg/m³) records Δp = 800 Pa. Find the speed.

Solution:
v = √[2 Δp/ρ] = √[2×800/1.2] ≈ 36.5 m/s.

Problem 2: Water flows horizontally past a venturi: p₁−p₂ = 5000 Pa, A₁/A₂ = 4. Compute v₁ if v₂ = 8 m/s.

Solution:
Continuity: v₁ = (A₂/A₁)v₂ = 8/4 = 2 m/s.
Bernoulli check: p₁−p₂ = ½ρ(v₂²−v₁²) = 0.5×1000×(64−4) = 30000 Pa, inconsistent with given, so revisit data.

Problem 3: In a vertical pipe, velocity is constant. Find Δp over Δh = 5 m.

Solution:
From p+ρgh=const, Δp = −ρgΔh = −1000×9.81×5 = −4.91×10⁴ Pa.

Problem 4: A horizontal Venturi meter has A₁ = 0.1 m², A₂ = 0.05 m², and p₁−p₂ = 2000 Pa. Find Q.

Solution:
Q = A₂√{[2(p₁−p₂)/ρ]·[A₁²/(A₁²−A₂²)]} = 0.05√{[2×2000/1000]·[0.01/(0.01−0.0025)]} ≈ 0.05×2×√[1.333] ≈ 0.115 m³/s.

Problem 5: Air (ρ = 1.18 kg/m³) flows at 15 m/s in a 0.2 m diameter pipe. What is the stagnation pressure?

Solution:
Δp = ½ρv² = 0.5×1.18×15² ≈ 132.4 Pa; stagnation pressure = static + 132.4 Pa.

Problem 6: Water flows from 0.2 m pipe into 0.1 m pipe; v₁ = 3 m/s. Find v₂.

Solution:
v₂ = v₁(A₁/A₂) = 3×(0.2/0.1)² = 3×4 = 12 m/s.

Problem 7: A Pitot‐static tube in water measures p₀−p = 5000 Pa. Find v.

Solution:
v = √[2×5000/1000] ≈ 3.16 m/s.

Problem 8: Oil (ρ = 850 kg/m³) flows through a Venturi; p₁−p₂ = 1500 Pa, A₁/A₂ = 3. Find Q.

Solution:
Q = A₂√{[2×1500/850]·[9/(9−1)]} = A₂√{[3.529]·[1.125]} ≈ A₂×2≈2A₂.

Problem 9: In a horizontal pipe pressure drops 800 Pa when velocity goes from 4 to 6 m/s. Check Bernoulli consistency.

Solution:
Δ(½ρv²)=0.5×1000×(36−16)=10000 Pa, so Δp should be 10000 Pa, not 800 Pa; energy loss present.

Problem 10: A water jet issues from a hole 2 m below free surface. Find exit velocity.

Solution:
Torricelli: v = √[2gh] = √[2×9.81×2] ≈ 6.26 m/s.

Problem 11: A Venturi meter reading gives p₁−p₂ = 2500 Pa in water. With A₁/A₂ = 4, compute v₁.

Solution:
From Bernoulli + continuity: p₁−p₂ = ½ρv₁²[(A₁²/A₂²)−1] → v₁ = √[2·2500/(1000·(16−1))] ≈ 0.577 m/s.

Problem 12: Find the head loss if measurement shows p₁ + ½ρv₁² + ρgh₁ > p₂ + ½ρv₂² + ρgh₂ by 1000 Pa.

Solution:
Head loss = Δ(total pressure)/ρg = 1000/(1000×9.81) ≈ 0.102 m.

Problem 13: A fluid pipe rises 10 m; velocity unchanged. Calculate pressure decrease.

Solution:
Δp = −ρgΔh = −1000×9.81×10 = −9.81×10⁴ Pa.

Problem 14: Air (ρ=1.2) flows at 20 m/s; what is dynamic pressure?

Solution:
½ρv² = 0.5×1.2×400 = 240 Pa.

Problem 15: A nozzle accelerates flow from 5 to 15 m/s; find Δp in water.

Solution:
Δ(½ρv²) = 0.5×1000×(225−25)=100000 Pa; so static pressure drop = 100000 Pa.

Problem 16: Why is Bernoulli’s equation not valid inside a boundary layer?

Solution:
Viscous effects dominate, violating inviscid assumption.

Problem 17: In open channel flow, how is Bernoulli modified?

Solution:
Include velocity head v²/2g and depth y: y + v²/2g + p/ρg = constant.

Problem 18: A tank discharges at 4 m/s; pressure at exit is atmospheric. Find pressure at depth 5 m.

Solution:
Use p₁+ρgh₁ = p₂+ρgh₂: p₁ = ρg(h₂−h₁) =1000×9.81×5 =4.905×10⁴ Pa.

Problem 19: A flow meter measures static and total heads as 10 m and 12 m. Find v.

Solution:
Δh = 2 m → v = √(2gΔh) ≈ √(2×9.81×2)=6.26 m/s.

Problem 20: Explain why Bernoulli’s equation must be adjusted for hydraulic losses.

Solution:
Real flows have friction; add head‐loss term h_L on RHS: p+½ρv²+ρgh = constant + ρgh_L.

Problem 21: A fluid issues vertically downward at 5 m/s from height 10 m. Find impact pressure at exit.

Solution:
Convert kinetic + potential: p = ρgΔh + ½ρv² =1000×9.81×10 +0.5×1000×25 ≈ 1.006×10⁵ Pa.

Problem 22: For a Venturi with throat area half inlet, derive p₁−p₂ in terms of ρ and v₁.

Solution:
v₂=2v₁, so p₁−p₂ = ½ρ(v₂²−v₁²)=½ρ(4v₁²−v₁²)=1.5ρv₁².

Problem 23: Why is Bernoulli’s equation applied between points on the same streamline?

Solution:
Energy constant differs between streamlines in rotational or unsteady flows.

Problem 24: Air flows 10 m/s at 20°C; find dynamic pressure.

Solution:
ρ≈1.204 kg/m³ → ½ρv²=0.5×1.204×100=60.2 Pa.

Problem 25: A nozzle expands from 0.05 m to 0.1 m diameter; if p₁−p₂=1000 Pa, find v₁.

Solution:
Continuity: v₂=v₁/4. Bernoulli: 1000=0.5ρ(v₁²−(v₁/4)²)=0.5ρ(15/16)v₁² → v₁≈5.16 m/s.

Prepare for University Studies & Career Advancement

Bernoulli’s Equation