Download presentation

Presentation is loading. Please wait.

Published bySamantha Walton Modified over 2 years ago

1

2
Chapter 15B - Fluids in Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

3
Fluids in Motion All fluids are assumed in this treatment to exhibit streamline flow. Streamline flow is the motion of a fluid in which every particle in the fluid follows the same path past a particular point as that followed by previous particles.

4
Assumptions for Fluid Flow: Streamline flowTurbulent flow All fluids move with streamline flow. The fluids are incompressible. There is no internal friction. All fluids move with streamline flow. The fluids are incompressible. There is no internal friction.

5
Rate of Flow The rate of flow R is defined as the volume V of a fluid that passes a certain cross-section A per unit of time t. The volume V of fluid is given by the product of area A and vt: Rate of flow = velocity x area vt Volume = A(vt) A

6
Constant Rate of Flow For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases: A1A1 A2A2 R = A 1 v 1 = A 2 v 2 v1v1 v2v2 v2v2

7
Example 1: Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What must be the diameter of the nozzle in order that the water emerge at 16 m/s? The area is proportional to the square of diameter, so: d 2 = cm

8
Example 1 (Cont.): Water flows through a rubber hose 2 cm in diameter at a velocity of 4 m/s. What is the rate of flow in m 3 /min? R 1 = m 3 /s R 1 = m 3 /min

9
Problem Strategy for Rate of Flow: Read, draw, and label given information. The rate of flow R is volume per unit time. When cross-section changes, R is constant. Read, draw, and label given information. The rate of flow R is volume per unit time. When cross-section changes, R is constant. Be sure to use consistent units for area and velocity.

10
Problem Strategy (Continued): Since the area A of a pipe is proportional to its diameter d, a more useful equation is: The units of area, velocity, or diameter chosen for one section of pipe must be consistent with those used for any other section of pipe.

11
The Venturi Meter The higher velocity in the constriction B causes a difference of pressure between points A and B. P A - P B = gh h A B C

12
Demonstrations of the Venturi Principle The increase in air velocity produces a difference of pressure that exerts the forces shown. Examples of the Venturi Effect

13
Work in Moving a Volume of Fluid P1P1 A1A1 P1P1 A1A1 P2P2 A2A2 A2A2 P2P2 h Volume V Note differences in pressure P and area A Fluid is raised to a height h. F1F1, F 2

14
Work on a Fluid (Cont.) Net work done on fluid is sum of work done by input force F i less the work done by resisting force F 2, as shown in figure. Net Work = P 1 V - P 2 V = (P 1 - P 2 ) V F 1 = P 1 A 1 F 2 = P 2 A 2 v1v1 v2v2 A1A1 A2A2 h2h2 h1h1 s1s1 s2s2

15
Conservation of Energy Kinetic Energy K: Potential Energy U: Net Work = K + U also Net Work = (P 1 - P 2 )V F 1 = P 1 A 1 F 2 = P 2 A 2 v1v1 v2v2 A1A1 A2A2 h2h2 h1h1 s1s1 s2s2

16
Conservation of Energy Divide by V, recall that density m/V, then simplify: Bernoullis Theorem: v1v1 v2v2 h1h1 h2h2

17
Bernoullis Theorem (Horizontal Pipe): h 1 = h 2 v1v1 v2v2 Horizontal Pipe (h 1 = h 2 ) h Now, since the difference in pressure P = gh, Horizontal Pipe

18
Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = 12 cm, what is the velocity of the water in the constriction? v 1 = 4 m/s v2v2 h h = 12 cm Bernoullis Equation (h 1 = h 2 ) 2gh = v v 1 2 Cancel, then clear fractions: v 2 = 4.28 m/s Note that density is not a factor.

19
Bernoullis Theorem for Fluids at Rest. For many situations, the fluid remains at rest so that v 1 and v 2 are zero. In such cases we have: P 1 - P 2 = gh 2 - gh 1 P = g(h 2 - h 1 ) h = 1000 kg/m 3 This is the same relation seen earlier for finding the pressure P at a given depth h = (h 2 - h 1 ) in a fluid.

20
Torricellis Theorem h1h1 h2h2 h When there is no change of pressure, P 1 = P 2. Consider right figure. If surface v 2 and P 1 = P 2 and v 1 = v we have: Torricellis theorem: v 2

21
Interesting Example of Torricellis Theorem: v v v Torricellis theorem: Discharge velocity increases with depth. Holes equidistant above and below midpoint will have same horizontal range. Maximum range is in the middle.

22
Example 4: A dam springs a leak at a point 20 m below the surface. What is the emergent velocity? h Torricellis theorem: Given: h = 20 m g = 9.8 m/s 2 v = 19.8 m/s 2

23
Strategies for Bernoullis Equation: Read, draw, and label a rough sketch with givens. The height h of a fluid is from a common reference point to the center of mass of the fluid. In Bernoullis equation, the density is mass density and the appropriate units are kg/m 3. Write Bernoullis equation for the problem and simplify by eliminating those factors that do not change. Read, draw, and label a rough sketch with givens. The height h of a fluid is from a common reference point to the center of mass of the fluid. In Bernoullis equation, the density is mass density and the appropriate units are kg/m 3. Write Bernoullis equation for the problem and simplify by eliminating those factors that do not change.

24
Strategies (Continued) For a stationary fluid, v 1 = v 2 and we have: P = g(h 2 - h 1 ) For a horizontal pipe, h 1 = h 2 and we obtain: h = 1000 kg/m 3

25
For no change in pressure, P 1 = P 2 and we have: Strategies (Continued) Torricellis Theorem

26
General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 200 kPa, and the point B is 8 m higher than point A. The lower section of pipe has a diameter of 16 cm and the upper section narrows to a diameter of 10 cm. Find the velocities of the stream at points A and B. 8 m A B R=30 L/s A A = (0.08 m) 2 = m 3 A B = (0.05 m) 2 = m 3 v A = 1.49 m/sv B = 3.82 m/s R = 30 L/s = m 3 /s

27
General Example (Cont.): Next find the absolute pressure at Point B. 8 m A B R=30 L/s Consider the height h A = 0 for reference purposes. Given: v A = 1.49 m/s v B = 3.82 m/s P A = 200 kPa h B - h A = 8 m P A + gh A +½ v A 2 = P B + gh B + ½ v B 2 0 P B = P A + ½ v A 2 - gh B - ½ v B 2 P B = 200,000 Pa + ½ 1000 kg/m 3 )(1.49 m/s) 2 – (1000 kg/m 3 )(9.8 m/s 2 )(8 m) - ½ 1000 kg/m 3 )(3.82 m/s) 2 P B = 200,000 Pa Pa –78,400 Pa – 7296 Pa P B = 115 kPa

28
Summary Bernoullis Theorem: Streamline Fluid Flow in Pipe: P A - P B = gh Horizontal Pipe (h 1 = h 2 ) Fluid at Rest: Torricellis theorem:

29
Summary: Bernoullis Theorem Read, draw, and label a rough sketch with givens. The height h of a fluid is from a common reference point to the center of mass of the fluid. In Bernoullis equation, the density r is mass density and the appropriate units are kg/m 3. Write Bernoullis equation for the problem and simplify by eliminating those factors that do not change. Read, draw, and label a rough sketch with givens. The height h of a fluid is from a common reference point to the center of mass of the fluid. In Bernoullis equation, the density r is mass density and the appropriate units are kg/m 3. Write Bernoullis equation for the problem and simplify by eliminating those factors that do not change.

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google