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Lift Drag Fluid Flow

Reference data and engineering information about lift drag fluid flow for fluid mechanics applications.

liftdragfluidflow

Overview

Engineering reference data for Lift Drag Fluid Flow in fluid mechanics.

Key Formulas

Reynolds Number

Re=ρvDμRe = \frac{\rho v D}{\mu}

Ratio of inertial to viscous forces — determines flow regime.

Bernoulli's Equation

P+12ρv2+ρgh=constP + \frac{1}{2}\rho v^2 + \rho g h = \text{const}

Conservation of energy for steady, inviscid, incompressible flow.

Continuity Equation

A1v1=A2v2A_1 v_1 = A_2 v_2

Conservation of mass for incompressible flow.

Darcy-Weisbach

ΔP=fLDρv22\Delta P = f \frac{L}{D} \frac{\rho v^2}{2}

Pressure drop due to friction in a pipe.

Variables

SymbolDescriptionUnit
ReReReynolds number
ρ\rhoFluid densitykg/m³
vvFlow velocitym/s
DDCharacteristic dimensionm
μ\muDynamic viscosityPa·s
PPPressurePa
ffDarcy friction factor

Worked Example: Aircraft Aerodynamics

The following example illustrates the application of the lift, drag, and power equations for an aircraft wing.

Given Parameters:

  • Velocity, v=100 m/sv = 100 \text{ m/s}
  • Wing Area, A=20 m2A = 20 \text{ m}^2
  • Drag Coefficient, cD=0.06c_D = 0.06
  • Lift Coefficient, cL=0.7c_L = 0.7
  • Air Density, ρ=1.2 kg/m3\rho = 1.2 \text{ kg/m}^3

Calculations:

1. Lifting Force (FLF_L): Using the lift equation:

FL=cL12ρv2AF_L = c_L \cdot \frac{1}{2} \rho v^2 A

Substituting the values:

FL=0.712(1.2 kg/m3)(100 m/s)2(20 m2)=84,000 N=84 kNF_L = 0.7 \cdot \frac{1}{2} \cdot (1.2 \text{ kg/m}^3) \cdot (100 \text{ m/s})^2 \cdot (20 \text{ m}^2) = 84,000 \text{ N} = 84 \text{ kN}

2. Drag Force (FDF_D): Using the drag equation:

FD=cD12ρv2AF_D = c_D \cdot \frac{1}{2} \rho v^2 A

Substituting the values:

FD=0.0612(1.2 kg/m3)(100 m/s)2(20 m2)=7,200 N=7.2 kNF_D = 0.06 \cdot \frac{1}{2} \cdot (1.2 \text{ kg/m}^3) \cdot (100 \text{ m/s})^2 \cdot (20 \text{ m}^2) = 7,200 \text{ N} = 7.2 \text{ kN}

3. Required Thrust Power (PP): The power required to overcome drag is the product of drag force and velocity.

P=FDvP = F_D \cdot v

Substituting the values:

P=(7,200 N)(100 m/s)=720,000 W=720 kWP = (7,200 \text{ N}) \cdot (100 \text{ m/s}) = 720,000 \text{ W} = 720 \text{ kW}

References