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) | |
|---|---|---|---|---|---|---|---|---|
| Arterial | 12.67 | 12.67 | 0.67 | 12.67 | 12.67 | 24.40 | 0.67 | 12.40 |
| Venous | 6.80 | -5.20 | -5.20 | -5.20 | 24.40 | 12.40 | 12.40 | 12.40 |
| Head arterial pressure | 0.67 | -5.20 | ||||||
| Head venous pressure | -5.20 | -5.20 | ||||||
| Foot arterial pressure | 12.67 | 24.40 | ||||||
| Foot venous pressure | 6.80 | 12.40 |
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