Fluid Mechanics

Fluid mechanics studies the behavior of liquids and gases at rest and in motion. It is the basis 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 are nearly incompressible (constant density). Gases are 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 a 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 the 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)

A fluid speeds up when the pipe narrows.

Example: water flows in a pipe (D₁ = 100 mm) at 2 m/s, then enters a 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 is constant (energy conserved).

Example: Venturi Meter

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

Water flows through a 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 the narrower section: pressure trades off for velocity.

Torricelli's Theorem (Tank Draining)

Velocity of fluid exiting a tank at depth h:

V = sqrt(2gh)

Same as the 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 the 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 to 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 a 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 to 0.35
• Flat plate perpendicular: 1.28

Drag goes 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 plus 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 is proportional to V² (quadratic with velocity).
• Reynolds number: determines laminar vs turbulent flow.

Next Steps

Continue to 07-machine-design.md for gears, bearings, and power transmission.