Fluid Mechanics

Fluid Mechanics studies the behavior of liquids and gases at rest and in motion. Essential for designing pumps, pipelines, aircraft, and hydraulic systems.

Fluid Properties

Density (ρ)

ρ = m/V

Units: kg/m³

Typical values:
- Water: 1000 kg/m³
- Air (20°C, 1 atm): 1.2 kg/m³
- Gasoline: 750 kg/m³
- Mercury: 13,600 kg/m³

Specific Gravity (SG)

Ratio of fluid density to water density:

SG = ρ_fluid / ρ_water

Viscosity (μ)

Resistance to flow: internal friction in a fluid.

Dynamic viscosity (μ): Units Pa⋅s or N⋅s/m²

Kinematic viscosity (ν):

ν = μ/ρ

Units: m²/s

Examples:

• Water: μ ≈ 0.001 Pa⋅s (low viscosity, flows easily)
• Honey: μ ≈ 10 Pa⋅s (high viscosity, flows slowly)
• Air: μ ≈ 1.8 × 10⁻⁵ Pa⋅s

Temperature effect: Viscosity of liquids decreases with temperature (easier to flow when hot).

Compressibility

Liquids: Nearly incompressible (constant density)
Gases: Highly compressible (density varies with pressure)

For most liquid flows: assume incompressible.

Fluid Statics (Fluids at Rest)

Pressure in Fluids

Pressure increases with depth:

P = P₀ + ρgh

Where:
- P = pressure at depth h (Pa)
- P₀ = pressure at surface (Pa)
- ρ = fluid density (kg/m³)
- g = 9.81 m/s²
- h = depth below surface (m)

Example: Pressure 10 m underwater

P = P_atm + ρgh
P = 101,325 + 1000 × 9.81 × 10
P = 101,325 + 98,100 = 199,425 Pa ≈ 2 atm

Manometers

Measure pressure difference using fluid column:

ΔP = ρgh

Where h = height difference in manometer

Buoyancy (Archimedes' Principle)

A body immersed in fluid experiences an upward force equal to the weight of displaced fluid.

F_buoyancy = ρ_fluid × V_displaced × g

Where:
- ρ_fluid = fluid density
- V_displaced = volume of fluid displaced

Floating condition:

F_buoyancy = Weight of object
ρ_fluid × V_submerged × g = m_object × g

Example: Will it float?

A wooden block (ρ = 600 kg/m³, V = 0.1 m³) in water.

Weight: W = ρ_wood × V × g = 600 × 0.1 × 9.81 = 588.6 N

Maximum buoyancy (fully submerged):
F_b = ρ_water × V × g = 1000 × 0.1 × 9.81 = 981 N

Since F_b > W, it floats!

Volume submerged: V_sub = W / (ρ_water × g) = 588.6 / 9810 = 0.06 m³
Fraction submerged: 0.06/0.1 = 60%

Fluid Dynamics (Fluids in Motion)

Flow Types

Laminar Flow: Smooth, orderly layers (low velocity, high viscosity)
Turbulent Flow: Chaotic, mixing (high velocity, low viscosity)

Reynolds Number determines flow type:

Re = ρVD/μ = VD/ν

Where:
- ρ = density
- V = velocity
- D = characteristic length (pipe diameter)
- μ = dynamic viscosity
- ν = kinematic viscosity

For pipe flow:

• Re < 2300: Laminar
• Re > 4000: Turbulent
• 2300 < Re < 4000: Transition

Continuity Equation (Conservation of Mass)

For incompressible flow:

A₁V₁ = A₂V₂ = Q

Where:
- A = cross-sectional area (m²)
- V = average velocity (m/s)
- Q = volumetric flow rate (m³/s)

Conclusion: Fluid speeds up when pipe narrows!

Example: Water flows in pipe (D₁ = 100 mm) at 2 m/s, enters narrower pipe (D₂ = 50 mm). Find V₂.

A₁ = πD₁²/4 = π(0.1)²/4 = 0.00785 m²
A₂ = πD₂²/4 = π(0.05)²/4 = 0.00196 m²

V₂ = V₁(A₁/A₂) = 2 × (0.00785/0.00196) = 8 m/s

Bernoulli's Equation

For steady, incompressible, frictionless flow along a streamline:

P₁ + ½ρV₁² + ρgh₁ = P₂ + ½ρV₂² + ρgh₂

Or:
P + ½ρV² + ρgh = constant

Where:
- P = static pressure
- ½ρV² = dynamic pressure
- ρgh = hydrostatic pressure

Energy interpretation:

• P/(ρg) = pressure head
• V²/(2g) = velocity head
• h = elevation head

Total head = constant (energy conserved)

Example: Venturi Meter

   P₁, V₁, A₁         P₂, V₂, A₂
      ____           ____
   __|    |_________|    |__
              ↓
           (narrower)

Water flows through venturi: A₁ = 0.01 m², A₂ = 0.005 m², P₁ = 200 kPa, V₁ = 2 m/s. Find P₂.

From continuity:
V₂ = V₁(A₁/A₂) = 2 × (0.01/0.005) = 4 m/s

From Bernoulli (same elevation):
P₁ + ½ρV₁² = P₂ + ½ρV₂²
200,000 + ½(1000)(2)² = P₂ + ½(1000)(4)²
200,000 + 2,000 = P₂ + 8,000
P₂ = 194,000 Pa = 194 kPa

Pressure drops in narrower section (trades pressure for velocity).

Torricelli's Theorem (Tank Draining)

Velocity of fluid exiting tank at depth h:

V = √(2gh)

Same as free-fall velocity!

Pipe Flow and Losses

Real flows have friction losses.

Darcy-Weisbach Equation

Head loss due to friction:

h_L = f × (L/D) × (V²/2g)

Where:
- h_L = head loss (m)
- f = friction factor (dimensionless)
- L = pipe length (m)
- D = pipe diameter (m)
- V = average velocity (m/s)

Friction factor (f):

• Laminar (Re < 2300): f = 64/Re
• Turbulent: Use Moody chart or correlations (depends on Re and roughness)

Minor Losses

Losses from bends, valves, expansions, etc.:

h_L = K × (V²/2g)

Where K = loss coefficient (from tables)

Pump Power

Power required to pump fluid:

Power = ρgQH

Where:
- ρ = density (kg/m³)
- g = 9.81 m/s²
- Q = flow rate (m³/s)
- H = total head rise (m)

Including efficiency:

Power_input = ρgQH / η

Where η = pump efficiency (typically 60-80%)

Example: Pump water at 0.05 m³/s up 20 m elevation (η = 70%).

Power_output = ρgQH = 1000 × 9.81 × 0.05 × 20 = 9,810 W ≈ 9.8 kW
Power_input = 9,810 / 0.7 = 14,014 W ≈ 14 kW

Drag and Lift

Drag Force

Resistance to motion through fluid:

F_D = ½ρV²C_D A

Where:
- F_D = drag force (N)
- ρ = fluid density (kg/m³)
- V = velocity (m/s)
- C_D = drag coefficient (dimensionless)
- A = reference area (m²)

Typical C_D values:

• Sphere: 0.47
• Cylinder: 1.0
• Streamlined car: 0.25-0.35
• Flat plate perpendicular: 1.28

Drag increases with V²: doubling speed quadruples drag.

Lift Force

Force perpendicular to flow (airfoils, wings):

F_L = ½ρV²C_L A

Where C_L = lift coefficient

Airfoils generate lift through pressure difference (Bernoulli effect + circulation).

Example: Car Drag

A car (frontal area 2.5 m², C_D = 0.3) at 100 km/h (27.8 m/s) in air.

F_D = ½ρV²C_D A
F_D = ½ × 1.2 × 27.8² × 0.3 × 2.5
F_D = 348 N

Power to overcome drag:
P = F_D × V = 348 × 27.8 = 9,674 W ≈ 9.7 kW ≈ 13 hp

Dimensional Analysis

Use dimensionless numbers to characterize flow:

Applications:

• Re: Pipe flow, drag prediction
• Fr: Ship design, open channel flow
• Ma: Compressible flow, aerodynamics

Practice Problems

Problem 1: Pressure at Depth

Find pressure 5 m below water surface (P_atm = 101 kPa).

<details> <summary>Solution</summary>

P = P₀ + ρgh = 101,000 + 1000 × 9.81 × 5
P = 101,000 + 49,050 = 150,050 Pa = 150 kPa

</details>

Problem 2: Continuity

Water flows at 3 m/s in 100 mm pipe, enters 50 mm pipe. Find new velocity.

<details> <summary>Solution</summary>

A₁/A₂ = (D₁/D₂)² = (100/50)² = 4
V₂ = V₁ × 4 = 3 × 4 = 12 m/s

</details>

Problem 3: Drag Force

A sphere (diameter 0.2 m, C_D = 0.47) moves at 10 m/s through water. Find drag force.

<details> <summary>Solution</summary>

A = πD²/4 = π(0.2)²/4 = 0.0314 m²
F_D = ½ρV²C_D A = ½ × 1000 × 10² × 0.47 × 0.0314
F_D = 73.8 N

</details>

Key Takeaways

✓ Pressure: P = P₀ + ρgh (increases with depth)
✓ Buoyancy: F = ρ_fluid × V_displaced × g
✓ Continuity: A₁V₁ = A₂V₂ (mass conservation)
✓ Bernoulli: P + ½ρV² + ρgh = constant (energy conservation)
✓ Drag: F ∝ V² (quadratic with velocity)
✓ Reynolds number: Determines laminar vs turbulent flow

Next Steps

Apply fluid mechanics knowledge to machine design:

• Chapter 07: Machine Design: gears, bearings, power transmission
• Study computational fluid dynamics (CFD)
• Learn about compressible flow and aerodynamics

Pro Tip: Always check if flow is compressible (Ma > 0.3) or incompressible before applying equations!