Introduction to Fluid Mechanics

Why it matters

Fluid mechanics is essential in designing and analyzing systems where fluids play a critical role, such as in pipelines, pumps, and HVAC systems. Understanding fluid behavior helps engineers optimize these systems for efficiency and safety.

Key ideas

• Fluid Properties: Fluids are substances that can flow and do not have a fixed shape. They are characterized by properties such as density, viscosity, and surface tension.
• Density (ρ): Mass per unit volume of a fluid, typically measured in kg/m³.
• Viscosity (μ): A measure of a fluid's resistance to flow, with units of Pa·s (Pascal-second).
• Pressure: The force exerted by a fluid per unit area, measured in Pascals (Pa).
• Continuity Equation: Based on the conservation of mass, it states that for steady flow through a pipe with no branches or leaks, ρ₁A₁v₁ = ρ₂A₂v₂, using cross-sectional mean velocities.
• Bernoulli's Equation: Relates the pressure, velocity, and elevation in a flowing fluid, along a streamline for steady, incompressible, inviscid flow under gravity, without a pump or turbine between the points.

Additional property relations

A fluid continuously deforms under sustained shear stress. For a Newtonian fluid in simple shear, τ = μ du/dy. Dynamic viscosity μ has units Pa·s; kinematic viscosity ν = μ/ρ has units m²/s. Do not confuse them.

For a static constant-density fluid under uniform gravity, pressure increases with downward depth by Δp = ρgΔh. Distinguish gauge pressure from absolute pressure: p_abs = p_gauge + p_atmospheric.

Formulas

• ρ = m / V
  • ρ: Density (kg/m³)
  • m: Mass (kg)
  • V: Volume (m³)
• Q = A·v
  • Q: Volumetric flow rate (m³/s)
  • A: Cross-sectional area (m²)
  • v: Cross-sectional mean normal velocity for Q = Av (m/s)
• P + 0.5·ρ·v² + ρ·g·h = constant
  • P: Pressure (Pa)
  • ρ: Density (kg/m³)
  • v: Velocity (m/s)
  • g: Acceleration due to gravity (9.81 m/s²)
  • h: Height above a reference point (m)

Worked example

Given: A constant-diameter pipe of diameter 0.1 m carries steady incompressible water flow (ρ = 1000 kg/m³) at mean speed 2 m/s. Point 2 is 5 m above point 1. Neglect losses, take equal kinetic-energy correction factors at both sections, and assume no pump or turbine. Calculate P₁ − P₂.

1. Calculate the cross-sectional area of the pipe.
   • A = π·(d/2)² = π·(0.1/2)² = 0.00785 m²
2. Apply Bernoulli's equation between the two points.
   • P1 + 0.5·ρ·v1² + ρ·g·h1 = P2 + 0.5·ρ·v2² + ρ·g·h2
3. Assume velocity is constant (v1 = v2 = 2 m/s) and solve for pressure difference.
   • P1 - P2 = ρ·g·(h2 - h1)
   • P1 - P2 = 1000 kg/m³ · 9.81 m/s² · 5 m
   • P1 - P2 = 49050 Pa

Final Answer: 49050 Pa

Common mistakes

• Confusing mass flow rate with volumetric flow rate.
• Ignoring units, leading to incorrect calculations.
• Misapplying Bernoulli's equation by not considering all terms or assuming conditions that do not apply.

For GATE ME

Questions often involve applying Bernoulli's equation, calculating flow rates, and understanding fluid properties. Practice problems on pressure differences, flow through pipes, and energy conservation in fluid systems.

Quick check

1. What is the unit of viscosity?
2. State Bernoulli's equation.
3. How does density affect fluid pressure?

Answers: 1. Pa·s, 2. P + 0.5·ρ·v² + ρ·g·h = constant, 3. For a static fluid at the same depth below the same surface pressure, a larger density produces a larger hydrostatic pressure increase: ΔP = ρgΔh. Density alone does not determine absolute pressure.