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

Transcription

1 Chapter 15B - Fluids in Motion A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 007

2 Paul E. Tippens Fluid Motion The lower falls at Yellowstone National Park: the water at the top of the falls passes through a narrow slot, causing the velocity to increase at that point. In this chapter, we will study the physics of fluids in motion.

3 Objectives: After completing this module, you should be able to: Define the rate of flow for a fluid and solve problems using velocity and cross- section. Write and apply Bernoulli s s equation for the general case and apply for (a) a fluid at rest, (b) a fluid at constant pressure, and (c) flow through a horizontal pipe.

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

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

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

7 Constant Rate of Flow For an incompressible, frictionless fluid, the velocity increases when the cross-section section decreases: R v A v A 1 1 vd 1 1 vd A 1 R = A 1 v 1 = A v A v 1 v v

8 Example 1: Water flows through a rubber hose 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: vd vd 1 1 d vd (4 m/s)( cm) 1 1 v (0 cm) d = cm

9 Example 1 (Cont.): Water flows through a rubber hose cm in diameter at a velocity of 4 m/s.. What is the rate of flow in m 3 /min? R v A v A R v1a1; A1 d 4 d1 (4 m/s) (0.0 m) R1 v 1 R = m 3 /s R 1 3 m 1 min min 60 s R 1 = m 3 /min

10 Problem Strategy for Rate of Flow: Read, draw, and label given information. The rate of flow R is is volume per unit time. When cross-section section changes, R is is constant. R v A v A 1 1 Be sure to use consistent units for area and velocity.

11 Problem Strategy (Continued): Since the area A of of a pipe is is proportional to to its its diameter d,, a more useful equation is: is: vd 1 1 vd The units of of area, velocity, or or diameter chosen for one section of of pipe must be consistent with those used for any other section of of pipe.

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

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

14 Work in Moving a Volume of Fluid P 1 A 1 F P ; F PA A1 F 1 Volume V P 1 A 1 A P Note differences in pressure P and area A F P ; F PA A A P h Fluid is raised to a height h., F

15 Work on a Fluid (Cont.) F 1 = P 1 A 1 v 1 v F = P A A 1 A h h 1 s 1 s Net work done on fluid is is sum of of work done by input force F i i less the work done by resisting force F,, as as shown in in figure. Net Work = P 1 V - P V = (P 1 -P ) V

16 Conservation of Energy Kinetic Energy K: K ½mv ½mv 1 Potential Energy U: U mgh mgh1 Net Work = K + U F 1 = P 1 A 1 v 1 v F = P A A 1 A h h 1 s 1 s also Net Work = (P 1 -P )V ( P P) V (½mv ½ mv ) ( mgh mgh ) 1 1

17 Conservation of Energy ( P P ) V (½mv ½ mv ) ( mgh mgh ) 1 1 Divide by V, recall that density m/v, then simplify: P gh ½ v P gh ½ v Bernoulli s Theorem: P gh ½ v Const v 1 v h 1 h

18 Bernoulli s s Theorem (Horizontal Pipe): P gh ½ v P gh ½ v Horizontal Pipe (h 1 = h ) P P ½ v ½ v 1 1 v 1 h v h 1 = h Now, since the difference in pressure P = gh, Horizontal Pipe P gh ½ v ½ v 1

19 Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = 1 cm,, what is the velocity of the water in the constriction? Bernoulli s Equation (h 1 = h ) P gh ½ v ½ v 1 h v 1 = 4 m/s v h= 6 cm Cancel, then clear fractions: gh = v -v 1 v gh v (9.8 m/s )(0.1 m) (4 m/s) 1 v = 4.8 m/s Note that density is not a factor.

20 Bernoulli s s Theorem for Fluids at Rest. For many situations, the fluid remains at rest so that v 1 and v are zero. In such cases we have: P gh ½ v P gh ½ v P 1 -P = gh - gh 1 P P = g(h -h 11 )) This is the same relation seen earlier for finding the pressure P at a given depth h = (h -h 1 ) in a fluid. h = 1000 kg/m 3

21 Torricelli s s Theorem When there is no change of pressure, P 1 = P. P gh ½ v P gh ½ v Consider right figure. If surface v and P 1 = P and v 1 = v we have: Torricelli s theorem: v gh h h h 1 v v gh

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

23 Example 4: A dam springs a leak at a point 0 m below the surface. What is the emergent velocity? Torricelli s theorem: v gh h v gh Given: h = 0 m g = 9.8 m/s v (9.8 m/s )(0 m) v = 19.8 m/s

24 Strategies for Bernoulli s s Equation: Read, draw, and label a rough sketch with givens. The height h of of a fluid is is from a common reference point to to the center of of mass of of the fluid. In In Bernoulli s equation, the density is mass density and the appropriate units are kg/m 3.. Write Bernoulli s equation for the problem and simplify by eliminating those factors that do not change. P gh ½ v P gh ½ v 1 1 1

25 Strategies (Continued) P gh ½ v P gh ½ v For a stationary fluid, v 1 = v and we have: P P = g(h -h 11 )) h = 1000 kg/m 3 For a horizontal pipe, h 1 = h and we obtain: P P ½ v ½ v 1 1

26 Strategies (Continued) P gh ½ v P gh ½ v For no change in pressure, P 1 = P and we have: Torricelli s Theorem v gh

27 General Example: Water flows through the pipe at the rate of 30 L/s. The absolute pressure at point A is 00 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. R = 30 L/s = m 3 /s D A R ; R A A = (0.08 m) = m 3 A B = (0.05 m) = m 3 8 m A R=30 L/s B v A R 1.49 m/s; 3.8 m/s A A m /s R m /s v m A m v A = 1.49 m/s v B = 3.8 m/s

28 General Example (Cont.): Next find the absolute pressure at Point B. Given: v A = 1.49 m/s v B = 3.8 m/s P A = 00 kpa h B -h A = 8 m 8 m A R=30 L/s Consider the height h A = 0 for reference purposes. 0 P A + gh A +½ v A = P B + gh B + ½ v B P B = P A + ½ v A - gh B -½ v B P P B = 00,000 B = 00,000 Pa Pa Pa + ½ 1000 kg/m 3 78,400 Pa )(1.49 m/s) 796 Pa (1000 kg/m 3 )(9.8 m/s )(8 m) -½ 1000 kg/m 3 )(3.8 m/s) P B = 115 kpa B

29 R v A v A Summary Streamline Fluid Flow in Pipe: 1 1 vd vd 1 1 Fluid at Rest: P A -P B = gh Horizontal Pipe (h 1 = h ) P P ½ v ½ v 1 1 Bernoulli s Theorem: P gh ½ v Constant Torricelli s theorem: v gh

30 Summary: Bernoulli s s Theorem Read, draw, and label a rough sketch with givens. The height h of of a fluid is is from a common reference point to to the center of of mass of of the fluid. In In Bernoulli s equation, the density r is is mass density and the appropriate units are kg/m 3.. Write Bernoulli s equation for the problem and simplify by eliminating those factors that do not change. P gh ½ v P gh ½ v 1 1 1

31 CONCLUSION: Chapter 15B Fluids in Motion