College Physics Fluid Mechanics PDF

College Physics Chapter 13 Eleventh Edition Fluid Mechanics 1) Density 2) Pressure in a Fluid 3) Archimedes’ s Principle: Buoyancy...

College Physics Chapter 13 Eleventh Edition Fluid Mechanics 1) Density 2) Pressure in a Fluid 3) Archimedes’ s Principle: Buoyancy 4) Surface Tension & Capillarity 5) Fluid Flow 6) Bernoulli’s Equation 7) Applications of Bernoulli’s Equation 8) Viscosity and Turbulence Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Goals for Chapter 13 To study density and pressure in a fluid. To apply Archimedes principle of buoyancy. To describe surface tension and capillary action To study and solve Bernoulli’s equation for fluid flow. To see how real fluids differ from ideal fluids (turbulence and viscosity). Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.1 Density Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Fluids and Density A fluid is a substance that flows. Liquids and gases are fluids. Gases are compressible; the volume of a gas is easily increased or decreased. Liquids are nearly incompressible; the molecules are packed closely, yet they can move around. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Density (1 of 2) The mass density is the ratio of mass to volume: The SI units of mass density are kg / m3. Gasoline has a mass density of 680 kg / m3 , meaning there are 680 kg of gasoline for each 1 cubic meter of the liquid. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Density (2 of 2) Table 13.1 Densities of fluids at 1 atm pressure Rho left parenthesis kilograms Substance ρ(kg per cubic meter m3 ) right/ parenthesis. Hydrogen gas (20°C) 0.083 Helium gas (20°C) 0.166 Air (20°C) 1.20 Air (0°C) 1.28 Gasoline 680 Ethyl alcohol 790 Oil (typical) 900 Water 1000 Seawater 1030 Blood (whole) 1060 Glycerin 1260 Mercury 13,600 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.1 Weighing the Air in a Living Room What is the mass of air in a living room with dimensions 4.0 m  6.0 m  2.5 m? STRATEGIZE: 𝒎 = 𝛒𝑽. Use air density from Table 13.1 (air density at a temperature of 20°C, which is about room temperature. SOLVE The room’s volume is: V = (4.0 m)  (6.0 m)  (2.5 m) = 60 m3 The mass of the air is: m = ρV = (1.20 kg / m3 )(60 m3 ) = 72 kg Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.2 Pressure Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Pressure (1 of 2) Liquids exert forces on the walls of their containers. The pressure is the ratio of the force to the area on which the force is exerted: F p= A The fluid’s pressure pushes on all parts of the fluid itself, forcing the fluid out of a container with holes. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Pressure (2 of 2) We can measure the pressure in a liquid with a simple device. The pressure is everywhere in the fluid; different parts of a fluid are pushing against each other. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Pressure in Liquids (1/4) The force of gravity (the weight of the liquid) is responsible for the pressure in the liquid. The horizontal forces cancel each other out. The vertical forces balance at hydrostatic: pA = p0 A + mg Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Pressure in Liquids (2/4) The liquid is a cylinder of cross- section area A and height d. The mass is m =  Ad. Therefore, the pressure at depth d is: Because we assumed that the fluid is at rest, this pressure is the hydrostatic pressure. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Pressure in Liquids (3 of 4) A connected liquid in hydrostatic equilibrium rises to the same height in all open regions of the container. In hydrostatic equilibrium, the pressure is the same at all points on a horizontal line through a connected liquid of a single kind. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Pressure in Liquids (4 of 4) If we change the pressure at the surface, then the pressure at a point d becomes: p' = p1 + ρgd = ( p0 + p ) + ρgd = ( p0 + ρgd ) + p = p + p That is, the pressure at depth d changes by the same amount as it did at the surface. This is Pascal’s principle: Pascal’s principle If the pressure at one point in an incompressible fluid is changed, the pressure at every other point in the fluid changes by the same amount. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.2 The Pressure on a Submarine (1 of 2) A submarine cruises at a depth of 300 m. What is the pressure at this depth? Give the answer in both pascals and atmospheres. STRATEGIZE use Equation 13.5 to find the pressure at the depth of the submarine. p0 is the pressure at the surface, which is 101 kPa. The density of seawater is given in Table 13.1 as 1030 kg  m3. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.2 The Pressure on a Submarine (2 of 2) SOLVE The pressure at depth d = 300 m is: p = p0 + ρgd = (1.01 103 Pa ) + (1030kg m3 )( 9.80m s2 ) ( 300m ) = 3.13  106 Pa Converting the answer to atmospheres gives: p = ( 3.13  10 Pa )  1 atm 6 = 30.9 atm 1.013  10 Pa 5 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.1 An iceberg floats in a shallow sea. What can you say about the pressures at points 1 and 2? A. p1 > p2 B. p1 = p2 C. p1 < p2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.2 What can you say about the pressures at points 1 and 2? A. p1 > p2 B. p1 < p2 C. p1 = p2 Hydrostatic pressure is the same at all points on a horizontal line through a connected fluid. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.3 What can you say about the pressures at points 1, 2, and 3? A. p1 = p2 = p3 B. p1 = p2 > p3 C. p1 > p2 = p3 D. p1 > p2 > p3 E. p1 < p2 = p3 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.3 Pressure in a Closed Tube (1 of 2) Water fills the tube shown in the figure. What is the pressure at the top of the closed tube? STRATEGIZE This is a liquid in hydrostatic equilibrium. The closed tube is not an open region of the container, so the water cannot rise to an equal height. Nevertheless, the pressure is still the same at all points on a horizontal line. In particular, the pressure at the top of the closed tube equals the pressure in the open tube at the height of the dashed line. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.3 Pressure in a Closed Tube (2 of 2) The top of the container is open to the atmosphere. So, p0 = 1 atm = 101 kPa. SOLVE A point 40 cm above the bottom of the open tube is at a depth of 60 cm. The pressure at this depth is: p = p0 + ρgd = (1.01  105 Pa) + (1000 kg / m3 )(9.80 m / s 2 ) (0.60 m ) = 1.07  105 Pa = 1.06 atm Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Tactics Box 13.1 Measuring and Using Pressure (1 of 2) 1. Draw a picture. Show open surfaces, pistons, boundaries, and other features that affect pressure. Include height and area measurements and fluid densities. Identify the points at which you need to find the pressure. 2. Determine the pressure p0 at surfaces. ▪ Surface open to the air: p0 = patmos, usually 1 atm. ▪ Surface in contact with a gas: p0 = pgas. ▪ Closed surface: p0 = F / A, where F is the force that the surface, such as a piston, exerts on the fluid. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Tactics Box 13.1 Measuring and Using Pressure (2 of 2) 3. Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point along a horizontal line. 4. Allow for gauge pressure. Pressure gauges read pg = p − patmos. 5. Use the hydrostatic pressure equation: p = p0 + ρgd. Text: p 446age Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.4 Pressure in a Tube with Two Liquids (1 of 5) A U-shaped tube is closed at one end; the other end is open to the atmosphere. Water fills the side of the tube that includes the closed end, while oil, floating on the water, fills the side of the tube open to the atmosphere. The two liquids do not mix. The height of the oil above the point where the two liquids touch is 75 cm while the height of the closed end of the tube above this point is 25 cm. What is the gauge pressure at the closed end? Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Example 13.4 Pressure in a Tube with Two Liquids (2 of 5) STRATEGIZE Following the steps in Tactics Box 13.1, we start by drawing the picture shown in the figure. We assume that the pressure at the open surface of the oil is p0 = 1 atm. Pressures p1 and p2 are the same because they are on a horizontal line that connects two points in the same fluid. (The pressure at point A is not equal to p3, even though point A and the closed end are on the same horizontal line, because the two points are in different fluids.) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.4 Pressure in a Tube with Two Liquids (3 of 5) STRATEGIZE Apply the hydrostatic pressure equation twice: once to find the pressure p1 with p0, and again to find the pressure p3 at the closed end once we know p2. Densities of water and oil, which are found in Table 13.1 as ρw = 1000 kg / m3 and ρo = 900 kg / m3. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.4 Pressure in a Tube with Two Liquids (4 of 5) SOLVE The pressure at point 1 is: p1 = p0 + ρo gh = 1 atm + (900 kg / m3 )(9.8 m / s 2 ) ( 0.75 m ) = 1 atm + 6620 Pa (We will keep p0 = 1 atm separate in this result because we’ll eventually need to subtract exactly 1 atm to calculate the gauge pressure.) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.4 Pressure in a Tube with Two Liquids (5 of 5) SOLVE Also use the hydrostatic pressure equation to find P2: p2 = p3 + ρw gd = p3 + (1000 kg / m3 )(9.8 m / s2 ) ( 0.25 m ) = p3 + 2450 Pa Since we know that p2 = p1, so: p3 = p2 − 2450 Pa = p1 − 2450 Pa = 1 atm + 6620 Pa − 2450 Pa ➔ The gauge pressure at point = 1 atm + 4200 Pa 3, the closed end of the tube, is p3 − 1 atm or 4200 Pa. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Pressure in a Tube with Two Liquids ρw = 1000 kg / m3 and ρo = 900 kg / m3. Note p2 = p1: p1 = p0 + ρo gh = 1 atm + (900 kg / m3 )(9.8 m / s 2 ) ( 0.75 m ) = 1 atm + 6620 Pa p2 = p3 + ρw gd = p3 + (1000 kg / m3 )(9.8 m / s2 ) ( 0.25 m ) = p3 + 2450 Pa p3 = p2 − 2450 Pa = p1 − 2450 Pa = 1 atm + 6620 Pa − 2450 Pa = 1 atm + 4200 Pa Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Atmospheric Pressure Gas is compressible, so the air in the atmosphere becomes less dense with increasing altitude. 99% of the air in our atmosphere is below 30 km. Atmospheric pressure varies with altitude and with changes in the weather. Local winds and weather is largely due to air masses of different pressures. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Measuring Atmospheric Pressure A barometer measures the atmospheric pressure patmos. A glass tube is placed in a beaker of the same liquid. Some, but not all liquid leaves the tube. p2 is the pressure due to the weight of the liquid in the tube and p1 = patmos. Equating the two pressures gives patmos = ρgh Atmospheric pressure is measured by measuring the height of the liquid. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Pressure Units In practice, pressure is measured in a number of different units. Table 13.2 Pressure units Unit Abbreviation Conversion to Pa Uses pascal Pa Blank SI unit: 1 Pa = 1 N / m2 used in 1 pascal = 1 newton per square meter. most calculations atmosphere atm 1 atm = 101.3 kPa general millimeters of mm Hg 1 mm Hg = 133 Pa gases and barometric mercury pressure inches of in 1 in = 3.39 kPa barometric pressure in U.S. mercury weather forecasting pounds per psi 1 psi = 6.89 kPa U.S. engineering square inch and industry Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved The Pressure in a Fluid – Figure 13.6 The pressure in any fluid at the same elevation will be the same regardless of the shape or size of the container. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Pascal’s Law – Figure 13.7 In a closed system, pressures transmitted to a fluid are identical to all parts of the container. Variations in the pressure are due only to the depth of the fluid. This principal is vital to mechanical devices like lifts and brakes. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Determining Absolute or Gauge Pressure – Example 13.4 The gauge pressure in exerted by the system. Absolute pressure includes the local atmospheric pressure. Refer to Example 13.4. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.3 Buoyancy Buoyant force = the weight of the fluid displaced by the object Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Archimedes’s Principle – Figure 13.15 An object submersed in a fluid experiences buoyant force equal to the mass of any fluid it displaces. An object can experience buoyant force greater than its mass and float. Even if it sinks, it would weigh measurably less. Refer to Example 13.7. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Buoyancy (1 of 3) The pressure in a liquid increases with depth, so the pressure in a liquid-filled cylinder is greater at the bottom than at the top. The pressure exerts a net upward force on a submerged cylinder. Buoyancy is the upward force of a liquid. Fnet = Fup − Fdown Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Buoyancy (2 of 3). If an isolated parcel of a fluid is in static equilibrium, then the parcel’s weight force pulling it down must be balanced by an upward force: the buoyant force FB. Thus, the buoyant force of the surrounding fluid pushing up matches the fluid weight: FB = w. If we replace the parcel of liquid with an object of the same shape and size, the buoyant force on the new object is exactly the same as before. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Buoyancy (3 of 3) When an object is immersed in a fluid, it displaces the fluid that would fill that region of space. The fluid is called the displaced fluid: Archimedes’ principle: A fluid exerts an upward buoyant force on an object immersed in or floating on the fluid. The magnitude of the buoyant force equals the weight of the fluid displaced by the object. Archimedes’ principle in equation form is: Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.4 (1 of 2) A heavy lead block and a light aluminum block of equal sizes are both submerged in water. Upon which is the buoyant force greater? A. On the lead block B. On the aluminum block C. They both experience the same buoyant force. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.8 (1 of 2) Blocks a, b, and c are all the same size. Which experiences the largest buoyant force? A. Block a B. Block b C. Block c D. All have the same buoyant force. E. Blocks a and c have the same buoyant force, but the buoyant force on block b is different. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.5 (1 of 2) Two blocks are of identical size. One is made of lead and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block B. On the wood block C. They both experience the same buoyant force Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.6 (1 of 2) A barge filled with ore floats in a canal lock. If the ore is tossed overboard into the lock, the water level in the lock will A. Rise. B. Fall. C. Remain constant. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Float or Sink? (1 of 2) Whether an object released underwater will head to the surface or to the bottom depends on whether the upward buoyant force on the object is larger or smaller than the downward weight force. Some objects are not uniform. We therefore define the average density to be  = m /V. avg o o The weight of a compound object can be written as w o = avgVo g. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Float or Sink? (2 of 2) An object will float or sink depending on whether the density of the fluid is larger or smaller than the average density of the object. If the densities are equal, the object is in static equilibrium and hangs motionless. This is called neutral buoyancy. Steel is denser than water, so it sinks. Oil floats because water is denser than oil. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Tactics Box 13.2 Float or Sink? (1 of 2) 1. Object sinks 2. Object floats An object sinks if it weighs An object rises to the surface if more than the fluid it it weighs less than the fluid it displaces—that is, if its displaces—that is, if its average average density is greater density is less than the density than the density of the fluid: of the fluid: avg   f avg   f Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Tactics Box 13.2 Float or Sink? (2 of 2) 3. Object has neutral buoyancy If the densities are equal, the object is in static equilibrium and hangs motionless. This is called neutral buoyancy. avg =  f Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.7 (1 of 2) Which floating block is most dense? A. Block a B. Block b C. Block c D. Blocks a and b are tied. E. Blocks b and c are tied. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.7 Measuring Density of an Unknown Liquid (1 of 4) You need to determine the density of an unknown liquid. You notice that a wooden block floats in this liquid with 4.6 cm of the side of the block submerged. When the block is placed in water, it also floats but with 5.8 cm submerged. What is the density of the unknown liquid? Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.7 Measuring Density of an Unknown Liquid (2 of 4) STRATEGIZE We will write expressions for the volume submerged— the volume of fluid displaced—for the two cases. The block floats in both liquids, so we write Equation 13.9 for the two cases. We will then combine the two equations to solve for the density of the fluid. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.7 Measuring Density of an Unknown Liquid (3 of 4) It’s useful to draw a picture to visualize the situation. 𝐹𝐵 = 𝜌𝑢 𝑉𝑢 𝑔 = 𝑊𝑜 = 𝜌𝑜 𝑉𝑜 g -----(1) 𝐹𝐵 = 𝜌𝑤 𝑉𝑤 𝑔 = 𝑊𝑜 = 𝜌𝑜 𝑉𝑜 g -----(2) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.7 Measuring Density of an Unknown Liquid (4 of 4) SOLVE The block displaces volume Vu = Ahu For unknown liquid: o Vu = Ahu = Vo 𝐹𝐵 = 𝜌𝑢 𝑉𝑢 𝑔 = 𝑊𝑜 = 𝜌𝑜 𝑉𝑜 g -----(1) u Similarly, the block displaces volume Vw = Ahw o Vw = Ahw = Vo 𝐹𝐵 = 𝜌𝑤 𝑉𝑤 𝑔 = 𝑊𝑜 = 𝜌𝑜 𝑉𝑜 g -----(2) w Therefore, we get hw 5.8 cm u = w = 1000 kg / m3 = 1300 kg / m3 hu 4.6 cm Comparison with Table 13.1 shows that the unknown liquid is likely to be glycerin. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Boats and Balloons (1 of 2) The hull of a boat is really a hollow shell, so the volume of water displaced by the shell is much larger than the volume of the hull itself. The boat sinks until the weight of the displaced water exactly matches the boat’s weight. It is then in static equilibrium and floats. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Boats and Balloons (2 of 2) The density of air is low so the buoyant force is generally negligible. Balloons cannot be filled with regular air because it would weigh the same amount as the displaced air and therefore have no net upward force. For a balloon to float, it must be filled with a gas that has a lower density than that of air. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.9 (1 of 2) Blocks a, b, and c are all the same size. Which is the correct order of the scale readings? A. a = b = c B. c > a = b C. c > a > b D. b > c > a E. a = c > b Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.4 Surface Tension & Capillarity Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Surface Tension – Figures 13.21 and 13.23 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Capillary Action Figure 13.26 The tendency of a liquid in a cappary tube to rise or fall as a result of surface tension. Interactions between the fluid and the container walls are significant. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.5 Fluids in Motion Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Fluids in Motion – Figure 13.28 Two kinds of flow exist. – Laminar flow – regular streamlines may be drawn. The smoke in the figure near the incense stick. – Turbulent flow – Irregular and difficult to model. The smoke well away from the incense stick. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Fluids in Motion (1 of 2) For fluid dynamics we use a simplified model of an ideal fluid. We assume 1. The fluid is incompressible. This is a very good assumption for liquids, but it also holds reasonably well for a moving gas, such as air. For instance, even when a 100 mph wind slams into a wall, its density changes by only about 1%. 2. The flow is steady. That is, the fluid velocity at each point in the fluid is constant; it does not fluctuate or change with time. Flow under these conditions is called laminar flow, and it is distinguished from turbulent flow. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Fluids in Motion (2 of 2) The rising smoke begins as laminar flow, recognizable by the smooth contours. At some point, the smoke undergoes a transition to turbulent flow. A laminar-to-turbulent transition is not uncommon in fluid flow. Our model of fluids can only be applied to laminar flow. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  The Equation of Continuity (1 of 3) When an incompressible fluid enters a tube, an equal volume of the fluid must leave the tube. The velocity of the molecules will change with different cross-section areas of the tube: V1 = A1x1 = A1v1t = V2 = A2 x2 = A2v 2 t Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved The Equation of Continuity (2 of 3) Dividing both sides of the previous equation by t gives the equation of continuity: The volume of an incompressible fluid entering one part of a tube or pipe must be matched by an equal volume leaving downstream. A consequence of the equation of continuity is that flow is faster in narrower parts of a tube, slower in wider parts. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  The Equation of Continuity (3 of 3) The rate at which fluid flows through a tube (volume per second) is called the volume flow rate Q. The SI units of Q are m3 / s We can also say that the volume flow rate is constant at all points in the tube. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved QuickCheck 13.10 (1 of 2) Water flows from left to right through this pipe. What can you say about the speed of the water at points 1 and 2? A. v1 > v2 B. v1 = v2 C. v1< v2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.10 Speed of Water through a Hose (1 of 2) A garden hose has an inside diameter of 16 mm. The hose can fill a 10 L bucket in 20 s. a. What is the speed of the water out of the end of the hose? b. What diameter nozzle would cause the water to exit with a speed 4 times greater than the speed inside the hose? STRATEGIZE We are given the volume flow rate; we will use this with Equation 13.12 to determine the speed of the flow. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.10 Speed of Water through a Hose (2 of 2) PREPARE The volume flow rate is Q = V / t = (10 L ) / ( 20 s ) = 0.50 L / s. To convert this to SI units, recall that 1 L = 1000 mL = 103 cm3 = 10 −3 m3. Thus Q = 5.0  10 −4 m3 / s. SOLVE a. The speed of the water is: Q Q 5.0  10 −4 m3 / s v= = 2 = = 2.5 m / s A r  (0.0080 m) 2 b. The quantity Q = vA remains constant as the water flows through the hose. To increase v by a factor of 4, A must be reduced by a factor of 4. The cross-section area depends on the square of the diameter, so the area is reduced by a factor of 4 if the diameter is reduced by a factor of 2. Thus the necessary nozzle diameter is 8 mm. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Representing Fluid Flow: Streamlines and Fluid Elements (1 of 2) A streamline is the path or trajectory followed by a “particle of fluid”. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Representing Fluid Flow: Streamlines and Fluid Elements (2 of 2) A fluid element is a small volume of a fluid, a volume containing many particles of fluid. A fluid element has a shape and volume. The shape can change, but the volume is constant. A1V1 = A2V2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Section 13.6 & 13.7 Fluid Dynamics Bernoulli’s Equation & its Application Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Fluid Dynamics (1 of 3) A fluid element changes velocity as it moves from the wider part of a tube to the narrower part. This acceleration of the fluid element must be caused by a force. The fluid element is pushed from both ends by the surrounding fluid, that is, by pressure forces. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Fluid Dynamics (2 of 3) To accelerate the fluid element, the pressure must be higher in the wider part of the tube. A pressure gradient is a region where there is a change in pressure from one point in the fluid to another. An ideal fluid accelerates whenever there is a pressure gradient. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Fluid Dynamics (3 of 3) The pressure is higher at a point along a stream line where the fluid is moving slower, lower where the fluid is moving faster. This property of fluids was discovered by Daniel Bernoulli and is called the Bernoulli effect. The speed of a fluid can be measured by a Venturi tube. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  QuickCheck 13.11 (1 of 2) Gas flows from left to right through this pipe, whose interior is hidden. At which point does the pipe have the smallest inner diameter? A. Point a B. Point b C. Point c D. The diameter doesn’t change. E. Not enough information to tell. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Applications of the Bernoulli Effect (1 of 2) As air moves over a hill, the streamlines bunch together, so that the air speeds up. This means there must exist a zone of low pressure at the crest of the hill, where the air is moving fastest. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Applications of the Bernoulli Effect (2 of 2) Lift is the upward force on the wing of an airplane that makes flight possible. The wing is designed such that above the wing the air speed increases and the pressure is low. Below the wing, the air is slower and the pressure is high. The high pressure below the wing pushes more strongly than the low pressure from above, causing lift. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Bernoulli’s Equation (1 of 4) We can find a numerical relationship for pressure, height and speed of a fluid by applying conservation of energy: K + U = W As a fluid moves through a tube of varying widths, parts of a segment of fluid will lose energy in its original volume V1 that the other parts of the fluid will gain in the volume it later occupies V2. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Bernoulli’s Equation (2 of 4) The system moves out of cylindrical volume V1 and into V2. The kinetic energies are: 1 1 K1 =  V v12 and K 2 =  V v 22 2 m 2 m The net change in kinetic energy is: 1 1 K = K 2 − K1 =  Vv 22 −  Vv12 2 2 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Bernoulli’s Equation (3 of 4) The net change in gravitational potential energy is: U = U2 − U1 = Vgy 2 − Vgy1 The positive and negative work done are: W1 = F1x1 = ( p1 A1 )x1 = p1( A1x1 ) = p1V W2 = −F2 x2 = −( p2 A2 )x2 = − p2 ( A2 x2 ) = − p2 V The net work is: W = W1 + W2 = p1V − p2 V = ( p1 − p2 ) V Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Bernoulli’s Equation (4 of 4) We combine the equations for kinetic energy, potential energy, and work done: 1 1 Vv 22 − Vv12 + Vgy 2 − Vgy1 = ( p1 − p2 )V 2 2 U W K Rearranged, this equation is Bernoulli’s equation, which relates ideal-fluid quantities at two points along a streamline: Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.11 Pressure in an Irrigation System (1 of 4) Water flows through the pipes shown in the figure. The water’s speed through the lower pipe is 5.0 m/s, and a pressure gauge reads 75 kPa. What is the reading of the pressure gauge on the upper pipe? Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.11 Pressure in an Irrigation System (2 of 4) STRATEGIZE Bernoulli’s equation will allow us to find differences in pressure along the flow. We can also use the equation of continuity to PREPARE The density of water is 1000 kg / m3. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.11 Pressure in an Irrigation System (3 of 4) SOLVE 1) Bernoulli’s equation relates the pressures, fluid speeds, and heights at points 1 and 2. It can be used to solve the pressure p2 at point 2: 1 1 p2 = p1 + v1 − v 2 2 +  gy1 −  gy 2 2 2 2 1 = p1 +  (v12 − v 2 2 ) +  g ( y1 − y 2 ) 2 SOLVE 2) All quantities on the right are known except v2, and that is where the equation of continuity will be used at points 1 and 2 as: v1A1 = v 2 A2 A1 r12 (0.030 m)2 v2 = v1 = 2 v1 = 2 (5.0 m / s) = 11.25 m / s A2 r2 (0.020 m) Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Example 13.11 Pressure in an Irrigation System (4 of 4) SOLVE 3) The pressure at point 1 is We can now use the above expression for p2 to calculate p2 = 105,900 Pa. This is the absolute pressure; the pressure gauge on the upper pipe is: p2 = 105,900 Pa − 1 atm = 4.6 kPa ASSESS We find a lower pressure at point 2; this makes sense. Reducing the pipe size decreases the pressure because it makes v2 > v1. Gaining elevation also reduces the pressure. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Bernoulli’s Equation – Figure 13.29 Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Bernoulli’s Equation Applied – Figure 13.33 I Blood flow characteristics are changed dramatically by plaque. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved  Bernoulli’s Equation Applied II Two Figure 13.34 The venturi tube allows pressure measurement “in line.” Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved Copyright This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Copyright © 2020, 2016, 2012 Pearson Education, Inc. All Rights Reserved