Fluid Dynamics Notes

Chapter 3 Fluid Dynamics

Objectives

• Ideal fluid
• Steady flow
• The continuity equation: $A<em>1v</em>1 = A<em>2v</em>2$
• 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:

  • $m<em>1 = \rho V</em>1 = \rho L<em>1 A</em>1 = \rho (v<em>1 \Delta t) A</em>1 = \rho A<em>1 v</em>1 \Delta t$
  • $m<em>2 = \rho V</em>2 = \rho L<em>2 A</em>2 = \rho (v<em>2 \Delta t) A</em>2 = \rho A<em>2 v</em>2 \Delta t$
  • $m<em>1 = m</em>2$
  • $\rho A<em>1 v</em>1 \Delta t = \rho A<em>2 v</em>2 \Delta t$
  • $A<em>1 v</em>1 = A<em>2 v</em>2$

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:
  • $Av_1 = 9.00 \times 10^{-5} m^3/s$
  • $A_2 = \pi r^2 = 3.14 (7.0 \times 10^{-3})^2$
  • $V<em>2 = \frac{Av</em>1}{A_2} = 0.58 m/s$

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:
  • $A<em>1 = \pi r</em>1^2 = \pi (0.008)^2 = 2.01 \times 10^{-4} m^2$
  • $A<em>2 = \pi r</em>2^2 = \pi (0.0025)^2 = 1.96 \times 10^{-5} m^2$
  • $A<em>1 V</em>1 = A<em>2 V</em>2$
  • $V<em>2 = \frac{A</em>1}{A<em>2} V</em>1 = \frac{2.01 \times 10^{-4}}{1.96 \times 10^{-5}} \times 3 = 30.8 m/s$

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

• $P<em>1 + \frac{1}{2} \rho v</em>1^2 + \rho g h<em>1 = P</em>2 + \frac{1}{2} \rho v<em>2^2 + \rho g h</em>2 = constant$

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, $h<em>1 = h</em>2$, 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: $P<em>1 + \rho g h</em>1 = P<em>2 + \rho g h</em>2$
• 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

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:
  • $V<em>1 = \frac{A</em>2}{A<em>1} V</em>2 = \frac{3}{1} \times 2 = 6 m/s$
  • $P<em>1 + \frac{1}{2} \rho v</em>1^2 = P<em>2 + \frac{1}{2} \rho v</em>2^2$
  • $P<em>1 = P</em>0 + \frac{1}{2} \rho (v<em>2^2 - v</em>1^2) = 8.4 \times 10^4 Pa$
  • Because $P<em>1 < P</em>0$, 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