Chapter 7: Applications of Fluid Dynamics
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
This final chapter integrates the principles developed throughout the course into practical, real-world systems involving fluid flow.
Applications include high-speed flow through nozzles, hydrostatic and dynamic pressure loading on large structures like dams, lift generation
on airfoils, and simplified analysis of blood flow in biological vessels. These examples show how fluid dynamics provides critical insight
across engineering and biological domains.
- Use continuity and Bernoulli’s principles to model flow through converging/diverging pipes and nozzles.
- Compute aerodynamic lift based on pressure differences over a wing surface.
- Explain the role of flow velocity and vessel diameter in determining blood pressure and flow rate.
- Apply fluid dynamics to analyze forces acting on dam faces and spillways.
🎯Learning Outcomes
By the end of this chapter, students will be able to:
- Model flow through nozzles and diffusers using continuity and Bernoulli’s equation.
- Understand how velocity and pressure distributions contribute to aerodynamic lift.
- Compute lift forces using the pressure differential across airfoil surfaces.
- Estimate hydrostatic and dynamic fluid loading on dam walls and gates.
- Analyze blood flow in arteries using fluid conservation laws and pressure-flow relationships.
- Evaluate energy conversion and pressure drops in real-world pipe systems with elevation changes.
📖Contents
- Flow Through Nozzles and Diffusers
- Lift on Airfoils and the Bernoulli Principle
- Blood Flow in Arteries
- Loading on Dams
- Hydraulic Lift
Table of Contents
7.1 Flow Through Nozzles and Diffusers
Fluid velocity changes when the cross-sectional area of a pipe changes. In a nozzle (converging section), velocity increases and pressure decreases.
In a diffuser (diverging section), the opposite occurs. These changes are governed by the continuity equation (Q = A v) and
Bernoulli’s principle.
- Continuity: A1v1 = A2v2 (Assumption: Incompressible flow)
- Bernoulli: p + ½ ρ v² + ρ g h = constant (Assumptions: Incompressible, Inviscid, and steady flow)
Fluid Flow Through a Pipe with Different Diameters
7.2 Lift on Airfoils and the Bernoulli Principle
Air traveling faster over the curved upper surface of a wing causes a lower pressure than beneath the wing, generating lift. The magnitude of the lift can be approximated using:
Lift:
Δp = plower − pupper = ½ ρ (vupper² − vlower²)
Approximated total lift = Δp × wing area.
7.3 Blood Flow in Arteries
Blood flow is a real-world example of incompressible fluid motion. When arteries narrow, velocity increases, and pressure drops, as predicted by Bernoulli’s equation.
- Flow rate: Q = A v
- Velocity increases in narrower arteries (smaller A) to keep volume flowrate the same
- Pressure decreases in general where velocity increases
7.4 Loading on Dams
Fluid pressure at a depth h is given by p = ρ g h. The total force on a vertical dam wall is obtained by integrating this pressure from top to bottom:
Total force: F = ½ ρ g h² × wall width
This accounts only for static pressure. In dynamic situations (e.g. flood spillways), Bernoulli’s equation may be used to include kinetic energy.
🤔Puzzle 1: Why Does Dam Wall Thickness Increase More Than Linearly with Depth?
It’s true that fluid pressure increases linearly with depth, according to the hydrostatic relation p = ρgh. One might then expect the dam wall to thicken linearly to match. However, real dams are designed with wall thickness increasing more than linearly. Why?
Answer: The explanation goes beyond fluid mechanics — solid mechanics and structural safety considerations are critical. Here’s why:
- Bending moment increases quadratically with depth:
As pressure increases with depth, the bending moment at the dam base accumulates from the total pressure above.
This bending moment — and the resulting bending stress — grows with the square of depth, not linearly. - Safety factors for extreme conditions:
Extra thickness accounts for possible waves, earthquakes, water seepage, erosion, and uncertainties in material strength or construction. - Greater compressive stress at greater depth:
The base must resist much larger vertical forces, so extra bulk helps prevent shear failure or crushing of the structure.
- Bending moment increases quadratically with depth:
🤔Puzzle 2: Why Are Dams Designed with a Convex Shape Facing Upstream?
Many large dams — especially arch dams — are curved so that the convex side faces upstream (toward the water). Why?
Answer: Again, the reason lies in structural mechanics, particularly how forces distribute through materials like concrete:
- Arch dams curve with the convex side facing upstream and the concave side facing downstream.
- Concrete and masonry have high compressive strength but low tensile strength. Hydrostatic pressure from the water compresses the convex upstream surface.
- This compression reduces tension within the dam wall. As the arch attempts to “straighten,” internal forces are mostly compressive, which concrete can withstand well.
- The curvature also helps redirect forces laterally toward the canyon walls, sharing the load across the structure and reducing the material required.
7.5 Hydraulic Lift
A hydraulic lift operates on Pascal’s principle, which states that a change in pressure applied to a confined incompressible fluid is transmitted undiminished throughout the fluid.
If a small piston with area A1 experiences force F1, the pressure is p = F1/A1. This same pressure acts on a larger piston of area A2, producing a greater force:
F2 = F1 × (A2/A1)
This provides a mechanical advantage, allowing a small input force to lift a heavier object. The trade-off is that the smaller piston must travel a longer distance to lift the load by a small height.
7.6 Key Formula Recap
| Concept | Formula |
|---|---|
| Continuity Equation | Q = A v |
| Bernoulli’s Equation | p + ½ ρ v² + ρ g h = constant |
| Lift Force | L = (pbelow − pabove) × A |
| Hydrostatic Pressure | p = ρ g h |
| Hydrostatic Force on Dam | F = ½ ρ g h² × width |
| Pascal’s Principle (Hydraulic Lift) | F2 = F1 × (A2/A1) |
Return to: Prep4Uni Fluid Mechanics 1
📝EXERCISES
20 Questions & Answers
- What happens to fluid velocity in a converging pipe?
It increases, due to conservation of mass (continuity equation: A1v1 = A2v2). - What effect does increasing velocity have on pressure in an ideal fluid?
Pressure decreases, as stated by Bernoulli’s principle. - What causes lift on an airplane wing?
Faster airflow over the curved top surface reduces pressure compared to the lower surface, generating lift. - What equation is used to estimate lift force?
L = (pbelow – pabove) × A, where A is the wing area. - Why is the flow faster at the throat of a Venturi tube?
The cross-sectional area is smallest there, and by continuity, velocity must increase. - In which part of a nozzle is the pressure lowest?
At the narrowest section, where velocity is highest. - What is dynamic pressure?
It is the kinetic energy per unit volume: (1/2)ρv². - What does Bernoulli’s equation ignore?
Viscosity, turbulence, and unsteady flow effects. - How does blood flow faster in narrower arteries?
Because the total flow rate is constant; smaller area implies higher velocity. - Why do aneurysms form in weakened vessel walls?
Because increased radius and pressure may lead to overstretching due to reduced structural support. - How do engineers account for fluid loading on dam walls?
By integrating the pressure distribution (hydrostatic + dynamic) over the submerged area. - What does Reynolds number help determine?
Whether flow is laminar or turbulent. - What is the primary assumption in modeling blood flow using Bernoulli’s equation?
That blood behaves as an incompressible, inviscid fluid in steady flow. - How does flow rate relate to velocity and area?
Q = A × v - What physical quantity causes lift in a rotating cylinder (Magnus effect)?
Circulation-induced pressure difference. - Why is dynamic pressure useful in Pitot tubes?
Because it allows measurement of airspeed by comparing stagnation and static pressures. - What causes flow separation behind a bluff body?
Adverse pressure gradients that reverse flow direction. - How does Bernoulli’s equation explain reduced pressure under a fast-flowing stream?
Because higher speed corresponds to lower pressure. - What happens to pressure at the center of a whirlpool?
It decreases due to high rotational speed. - What factors limit the accuracy of Bernoulli-based predictions in arteries?
Pulsatility, viscosity, and wall elasticity.
20 Problems & Solutions
- Water flows through a horizontal pipe that narrows from 10 cm to 5 cm diameter. If the velocity in the larger section is 2 m/s, find the velocity in the narrow section.
A1v1 = A2v2 → v2 = v1(A1/A2) = 2 × (10²/5²) = 2 × 4 = 8 m/s. - Air moves at 40 m/s over the top of a wing and 30 m/s under it. If air density is 1.2 kg/m³, find the lift per m².
Δp = ½ρ(vlower² − vupper²) = 0.5×1.2×(900−1600) = −420 Pa → Lift = 420 N/m². - A nozzle reduces pressure from 300 kPa to 240 kPa as water speeds up. What is the change in velocity?
Bernoulli: Δp = ½ρ(v₂² − v₁²) → (v₂² − v₁²) = 2Δp/ρ = 2×60000/1000 = 120 → Δv = √(v₁² + 120) − v₁. - In a horizontal artery, if the velocity doubles due to narrowing, find the change in pressure (blood ρ = 1060 kg/m³).
Δp = −½ρ(v₂² − v₁²) = −0.5×1060×(4v₁² − v₁²) = −1590 v₁² (in Pascals). - Find force on a dam wall 10 m tall and 20 m wide due to water (ρ = 1000 kg/m³).
F = (½)ρgh² × width = 0.5×1000×9.81×100 × 20 = 9.81 × 10⁵ N. - A Venturi meter shows Δp = 3000 Pa between pipe sections. If ρ = 1000 kg/m³, find velocity difference.
Δv² = 2Δp/ρ = 6000/1000 = 6 → Δv = √6 ≈ 2.45 m/s. - Calculate flow rate of water through a 0.1 m² nozzle at 5 m/s.
Q = A × v = 0.1 × 5 = 0.5 m³/s. - Blood flows through an artery with v = 0.2 m/s and diameter = 3 mm. Find volumetric flow rate.
A = πr² = π(0.0015)² ≈ 7.07×10⁻⁶ m² → Q ≈ 0.2 × 7.07×10⁻⁶ ≈ 1.41×10⁻⁶ m³/s. - Water flows out of a tank through a pipe at height 5 m below surface. Find exit speed.
Torricelli’s law: v = √(2gh) = √(2×9.81×5) ≈ 9.9 m/s. - An airplane wing experiences pressure difference of 500 Pa. If area = 10 m², compute total lift.
F = Δp × A = 500 × 10 = 5000 N. - Flow in a tube speeds up from 1 m/s to 3 m/s. If fluid is water, find pressure drop.
Δp = ½ρ(v₂² − v₁²) = 0.5×1000×(9−1) = 4000 Pa. - A dam has a gate 3 m high and 5 m wide submerged fully. Compute hydrostatic force.
F = ½ρg h² × width = 0.5×1000×9.81×9 × 5 = 220725 N. - Air flows over a soccer ball, creating a pressure difference of 100 Pa. If surface area = 0.04 m², find lift.
F = Δp × A = 100 × 0.04 = 4 N. - A vertical pipe reduces in diameter from 10 cm to 5 cm. If flow rate is 0.01 m³/s, find velocities.
A1 = π(0.05)² = 7.85×10⁻³, A2 = π(0.025)² = 1.96×10⁻³ → v1 = Q/A1 ≈ 1.27 m/s, v2 ≈ 5.1 m/s. - Pressure in an artery drops by 100 Pa as blood accelerates. Estimate velocity change (ρ = 1060 kg/m³).
Δv² = 2Δp/ρ ≈ 200/1060 ≈ 0.188 → Δv ≈ 0.43 m/s. - Find flow rate if a 2 cm pipe has water flowing at 3 m/s.
A = π(0.01)² ≈ 3.14×10⁻⁴ m² → Q = A × v ≈ 9.42×10⁻⁴ m³/s. - Explain how Venturi meters help determine flow rate.
They relate pressure difference to velocity via Bernoulli, then use continuity to find Q. - If a nozzle reduces pressure by 20 kPa, what velocity change occurs in water?
Δv² = 2Δp/ρ = 2×20000/1000 = 40 → Δv ≈ 6.32 m/s. - Airplane wings deflect air downward. What principle explains upward lift?
Newton’s third law: downward deflection leads to upward reaction force. - Describe one reason why real fluid behavior deviates from Bernoulli predictions.
Viscosity causes energy loss due to internal friction, not accounted for in ideal flow.