Chapter 11 - Fluids in Motion. Sections 7-9

Transcription

1 Chapter - Fluids in Motion Sections 7-9

2 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. Paul E. Tippens

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 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 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.

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

6 Rate of Flow The rate of flow 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 Δl, but Δl = vδt, so V = A Δl, or V Avt A Δl = the distance traveled Volume = A(vt) V m m Avt m t Av Which is the equation for the MASS FLOW RATE.

7 Constant Rate of Flow The MASS that flows into a region = The MASS that flows out of a region. For an incompressible, frictionless fluid, the velocity increases when the cross-section decreases: Av t A v t Av Av,The Equation of Continuity A R = A v = A v A v v v

8 Example - 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 6 m/s? A v v r v A v r v r r r v v 0. 5 ( cm ) ( 4m / s ) ( 6m / s ) cm ; r 0. 5 cm 0. 5cm

9 Example (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 A v R m t A v r v R r v ( 0. 0 m ) ( 4 m / s ) R = m 3 /s R m s 3 x 60 s min R = m 3 /min

10 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. A v A v Be sure to use consistent units for area and velocity.

11 Problem Strategy (Continued): Since the area A of a pipe is proportional to its diameter d, a more useful equation is: v d v d 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.

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

13 .9 Bernoulli's Principle The Swiss Physicist Daniel Bernoulli, was interested in how the velocity changes as the fluid moves through a pipe of different area. He especially wanted to incorporate pressure into his idea as well. Conceptually, his principle is stated as: "If the velocity of a fluid increases, the pressure decreases and vice versa

14 Work in Moving a Volume of Fluid We will break it up into small sections: WORK is equal to FORCE times DISPLACEMENT, (in this case, the length of the section the fluid travels. The grey area.) W Fx WORK is done by the Section One flowing FORWARD W F l As well as Section Two working AGAINST it, W F l The work is negative due to Newton's Third Law and working against gravity. Work must go into or onto moving it.

15 The Pressure formula can be inserted for the force and solving for F. P F F A PA Substituting PA in for F, you will get the following two equations: W P A l and W P Al Since WORK is energy we have to ask ourselves if there is any other energy. Since the water rises it has HEIGHT and thus POTENTIAL ENERGY. W mg( y y 3 )

16 Setting all three WORKS equal to the NET WORK. W W W net net net W W W W P A l W W 3 3,, so, so, P A l mgy mgy WHAT DOES THE NET WORK EQUAL TO? A CHANGE IN KINETIC ENERGY! Wnet KE, KE mv mv KE P A l P A l mgy mgy mv mv P A l P A l mgy. mgy

17 Consider that Density = Mass per unit Volume, AND that VOLUME is equal to AREA time LENGTH, then substituting and solving for m, we get: m A l We will now substitute for the MASS in the equation with energy. mv mv P Al P Al mgy mgy ρaδlv ρaδlv o P A Δl P A Δl ρalgy To get: ρalgy

18 We can now cancel out the AREA and LENGTH in all cases because AREA times LENGTH equals VOLUME, and volume remains constant in the pipe. ρalv P ρaδlv P A Δl P A Δl ρalgy ρal gy Leaving: ρv ρv P P ρgy ρgy Moving everything related, initials and finals together, to one side results in: ρv ρgy P ρv ρgy

19 What this basically shows is that Conservation of Energy holds true within a fluid and that if you add the PRESSURE, the KINETIC ENERGY(in terms of density) and POTENTIAL ENERGY(in terms of density) you get the SAME VALUE anywhere along a streamline. P ρv ρgy Constant Which is known a Bernoulli s Equation!

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

21 Example 3: Water flowing at 4 m/s passes through a Venturi tube as shown. If h = cm, what is the velocity of the water in the constriction? Bernoulli s Equation (h = h ) ½ P gh ½v v h v = 4 m/s v h = 6 cm Cancel ρ, then clear fractions: gh = v - v v gh v (9.8 m/s )(0. m) (4 m/s) v = 4.8 m/s Note that density is not a factor.

22 Bernoulli s Theorem for Fluids at Rest. For many situations, the fluid remains at rest so that v and v are zero. In such cases we have: P gh ½v P gh ½v P - P = gh - gh P = g(h - h ) This is the same relation seen earlier for finding the pressure P at a given depth h = (h - h ) in a fluid. h = 000 kg/m 3

23 Torricelli s Theorem When there is no change of pressure, P = P. P gh ½v P gh ½v Consider right figure. If surface v 0 and P = P and v = v we have: Torricelli s theorem: v gh h h h v 0 v gh

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

25 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 = 9.8 m/s

26 Strategies for Bernoulli s 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 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.

27 Conceptual Applications In general, things tend to move TOWARD areas of low pressure. When the velocity of a fluid increases, this creates a NET FORCE toward the low pressure area. Most of the time it happens when the AREA decreases. Example: Putting your thumb over a water hose. Another example is an airfoil.

28 The shape of the top of the airfoil causes the streamlines to scrunch together and thus increasing their velocity. The airfoil then moves upward to the low pressure area created.

29 Example Problem A large storage tank filled with water develops a small hole in its side at a point 6 m below the water level. If the rate of flow from the leak is.5 x 0-3 m 3 /min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole. Answer: a. 7.7 m/s; b m or.73 mm

30 Example Problem Water flows through a fire hose of diameter 6.35 cm at a rate of 0.0 m 3 /s. The fire hose ends in a nozzle of inner diameter. cm. What is the velocity with which the water exits the nozzle? Answer: 3.6 m/s

31 Assignment Pages , #54, 55, 56, 6, 67, and 7