Fluid Dynamics Notes

Chapter 3 Fluid Dynamics

Objectives

  • Ideal fluid
  • Steady flow
  • The continuity equation: A1v1 = A2v2
  • Bernoulli's equation: P + \frac{1}{2}\rho v^2 + \rho gh = constant
  • Reynold number: Re = \frac{\rho vr}{\eta}

Fluid

  • Liquids and gases are fluids.

Specific Properties of Fluids

  • Flow
  • Viscosity
  • Compressibility

Ideal Fluid

  • An ideal fluid is a simplified model used to replace real fluids to make problems simpler.
  • Properties:
    • Non-viscous: no internal friction.
    • Incompressible: density remains constant.
  • Non-viscous means viscosity can be neglected, implying no shearing forces within the fluid.
  • Incompressible means the fluid's density remains constant.

Streamline

  • At any given moment, lines can be drawn such that the tangent at every point on the line has the same direction as the velocity.

Steady Flow

  • In steady flow, the speed at every point on a streamline does not change over time.

Continuity Equation

  • If a pipe has different cross-sections, the velocity varies at different places.

  • Derivation of the Continuity Equation:

    • m1 = \rho V1 = \rho L1 A1 = \rho (v1 \Delta t) A1 = \rho A1 v1 \Delta t
    • m2 = \rho V2 = \rho L2 A2 = \rho (v2 \Delta t) A2 = \rho A2 v2 \Delta t
    • m1 = m2
    • \rho A1 v1 \Delta t = \rho A2 v2 \Delta t
    • A1 v1 = A2 v2

Volume Flow Rate

  • Volume of fluid moving through a pipe.
  • Volume flow rate = area of pipe × velocity of fluid.
  • Volume flow rate must be constant throughout the pipe.

The Continuity Equation

  • States that the cross-sectional area of the pipe and the velocity of the fluid are inversely proportional.

Application of Continuity Equation

  • The question of why the aorta has a large cross-sectional area but great velocity, while blood capillaries have a small cross-sectional area but small velocity is addressed.

Sample Problem: Aorta

  • In a normal resting result, the heart pumps blood into the aorta at an average volume flow rate (Av) of about 9.0 \times 10^{-5} m^3/s. Calculate the average speed of the blood in an aorta of inside diameter 1.40 cm.
  • Solution:
    • Av_1 = 9.00 \times 10^{-5} m^3/s
    • A_2 = \pi r^2 = 3.14 (7.0 \times 10^{-3})^2
    • V2 = \frac{Av1}{A_2} = 0.58 m/s

Sample Problem: Garden Hose

  • Water enters a typical garden hose of diameter 1.6 cm with a velocity of 3m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.5 cm is attached to the end of the hose.
  • Solution:
    • A1 = \pi r1^2 = \pi (0.008)^2 = 2.01 \times 10^{-4} m^2
    • A2 = \pi r2^2 = \pi (0.0025)^2 = 1.96 \times 10^{-5} m^2
    • A1 V1 = A2 V2
    • V2 = \frac{A1}{A2} V1 = \frac{2.01 \times 10^{-4}}{1.96 \times 10^{-5}} \times 3 = 30.8 m/s

Pressure and Height

  • As altitude increases, pressure decreases.

Pressure and Velocity

  • When blowing air between two pieces of paper, the papers move closer together.
  • As the velocity of a fluid increases, the pressure exerted by that fluid decreases.

Bernoulli's Equation

  • P1 + \frac{1}{2} \rho v1^2 + \rho g h1 = P2 + \frac{1}{2} \rho v2^2 + \rho g h2 = constant

Bernoulli’s Equation

  • Describes the behavior of a fluid under varying conditions of flow and height.
    • Where P is the static pressure (in Pa), \rho is the fluid density (in kg/m^3), v is the velocity of fluid flow (in m/s), and h is the height above a reference surface.

Simplified Bernoulli's Equation for Horizontal Pipe

  • If the pipe is horizontal, h1 = h2, then the Bernoulli equation simplifies to: P + \frac{1}{2} \rho v^2 = constant
  • When the velocity of fluid in a horizontal flow tube decreases, the pressure increases.

Applications of Bernoulli's Equation

  • Wing Airfoil
  • Automotive spoilers

Body Position and Blood Pressure

  • When v = 0 or v = constant, Bernoulli's Equation simplifies to: P1 + \rho g h1 = P2 + \rho g h2
  • Arterial pressure is affected by body position

Measuring Blood Pressure

  • When measuring blood pressure, the arm and the heart must be at the same level.
    • Arm higher than the heart results in a lower blood pressure measurement.
    • Arm below the heart results in a higher blood pressure measurement.

Blood Pressure Values

Lie (kPa)Stand (kPa)Lie (kPa)Stand (kPa)Lie (kPa)Stand (kPa)Lie (kPa)Stand (kPa)
Arterial12.6712.670.6712.6712.6724.400.6712.40
Venous6.80-5.20-5.20-5.2024.4012.4012.4012.40
Head arterial pressure0.67-5.20
Head venous pressure-5.20-5.20
Foot arterial pressure12.6724.40
Foot venous pressure6.8012.40

Blood Pressure

  • Gauge pressure
  • Absolute pressure = P_0 + Gauge pressure

Sample Problem: Water Flow in a Horizontal Pipe

  • Water flows through a horizontal pipe with different cross-sectional areas; the outlet cross-sectional area is three times the smallest area. If the outlet velocity equals 2m/s, calculate the pressure at the smallest area of the pipe. If there is a small hole in the smallest area, determine whether water will escape from this hole.
  • Solution:
    • V1 = \frac{A2}{A1} V2 = \frac{3}{1} \times 2 = 6 m/s
    • P1 + \frac{1}{2} \rho v1^2 = P2 + \frac{1}{2} \rho v2^2
    • P1 = P0 + \frac{1}{2} \rho (v2^2 - v1^2) = 8.4 \times 10^4 Pa
    • Because P1 < P0, water will not escape from this hole.

Laminar Flow

  • When a fluid flows in a tube, the flow velocity is different at different points of a cross-section.

Turbulent Flow

  • Mechanical energy dissipated is typically much larger in turbulent flow than in laminar flow.

Reynolds Number

  • Used to determine whether the flow is laminar or turbulent.
  • Consider a fluid of viscosity \,eta and density \,ho. If it is flowing in a tube of radius r and has an average velocity \,nu, then the Reynolds number is defined by:
    • Re = \frac{\rho v r}{\eta}

Reynolds Number and Flow Type

  • In tubes, it is found experimentally that:
    • If Re < 1000, flow is laminar.
    • If Re > 1500, flow is turbulent.
    • If 1000 < Re < 1500, flow is unstable (may change from laminar to turbulent or vice versa).

Sample Problem: Artery Flow

  • An artery has an inner diameter of 2cm, the average velocity of the blood \,nu = 0.25 m/s, \,eta = 3.0 \times 10^{-3} Pas, \rho = 1.05 \times 10^3 kg \cdot m^{-3}, find Reynolds number and determine the state of the fluid flow.
  • Solution:
    • Re = \frac{\rho v r}{\eta} = \frac{1.05 \times 10^3 \times 0.25 \times \frac{0.02}{2}}{3.0 \times 10^{-3}} = 875
    • Re < 1000, flow is laminar.

Summary

  • Definition of fluid
  • Properties of real fluid
  • Streamlines
  • Ideal fluid
  • Steady flow
  • The continuity equation
  • Bernoulli’s equation
  • Reynold number
  • Master and be familiar with these concepts