普通物理 General Physics 14 - Fluid Statics and Dynamics 郭艷光Yen-Kuang Kuo 國立彰化師大物理系暨光電科技研究所 電子郵件: ykuo@cc.ncue.edu.tw 網頁: http://ykuo.ncue.edu.tw

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Outline 14-1 What Is Physics? 14-2 What Is a Fluid? 14-3 Density and Pressure 14-4 Fluids at Rest 14-5 Measuring Pressure 14-6 Pascal’s Principle 14-7 Archimedes’ Principle 14-8 Ideal Fluids in Motion 14-9 The Equation of Continuity 14-10 Bernoulli’s Equation 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-1 What Is Physics? A nuclear engineer might study the fluid flow in the hydraulic system of an aging nuclear reactor, while a medical engineer might study the blood flow in the arteries of an aging patient. A naval engineer might be concerned with the dangers faced by a deep-sea diver or with the possibility of a crew escaping from a downed submarine. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-1 What Is Physics? Hydraulic engineering is also applied in many Broadway and Las Vegas shows, where huge sets are quickly put up and brought down by hydraulic systems. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-2 What Is a Fluid? Fluids conform to the boundaries of any container in which they are placed. A fluid cannot exert a force tangential to its surface. It can only exert a force perpendicular to its surface. Liquids and gases are classified together as fluids to contrast them from solids. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure Density: It has a mass Δm and volume ΔV. The density (symbol ρ) is defined as the ratio of the mass over the volume. Δm ΔV SI unit: kg/m3 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure In theory, the density at any point in a fluid is the limit of this ratio as the volume element V at that point is made smaller and smaller. In practice, we assume that a fluid sample is large relative to atomic dimensions and thus is “smooth” (with uniform density), rather than “lumpy” with atoms. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure If the fluid is homogeneous the above equation has the form: (uniform density) 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 8

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure Pressure : The SI unit for pressure is N/m2 is known as the Pascal (symbol: Pa). 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure We define the pressure on the piston from the fluid as: In theory, the pressure at any point in the fluid is the limit of this ratio as the surface area ∆A of the piston, centered on that point, is made smaller and smaller. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-3 Density and Pressure 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-1 A living room has floor dimensions of 3.5 m and 4.2 m and a height of 2.4 m. (a) What does the air in the room weigh when the air pressure is 1.0 atm? (b) What is the magnitude of the atmosphere’s force on the floor of the room? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-1 Key Idea: (a) The air’s weight is equal to mg, where m is its mass. (b) Mass m is related to the air density and the air volume V by ρ= m/V. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-1 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-4 Fluids at Rest 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-4 Fluids at Rest The pressure at a point in a fluid in static equilibrium depends on the depth of that point but not on any horizontal dimension of the fluid or its container. The figure shows four containers of olive oil. Rank them according to the pressure at depth h, greatest first. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-4 Fluids at Rest If we take and and The equation above takes the form: Note: The difference is known as “gauge pressure”. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-2 A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. He ignores instructions and fails to exhale during his ascent. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-2 When he reaches the surface, the difference between the external pressure on him and the air pressure in his lungs is 9.3kPa. From what depth does he start? What potentially lethal danger does he face? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-2 Key Idea: The pressure at depth h in a liquid of density ρis given by p = p0 + ρgh, where the gauge pressure ρgh is added to the atmospheric pressure p0. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-2 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-3 The U-tube in Fig. 14-4 contains two liquids in static equilibrium: Water of density ( 998 kg/m3) is in the right arm, ρw and oil of unknown density ρx is in the left. Measurement gives l = 135 mm and d = 12.3 mm. What is the density of the oil? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-3 Key Idea: The pressure pint at the level of the oil–water interface in the left arm depends on the density x and height of the oil above the interface. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-3 The water in the right arm at the same level must be at the same pressure pint . The reason is that, because the water is in static equilibrium, pressures at points in the water at the same level must be the same even if the points are separated horizontally. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-3 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-5 Measuring Pressure Toricelli observed that the mercury column drops so that its length is equal to h. If we take y1 = 0 and y2= h then p1 = p0 (p2－p1) =ρg(y1 － y2) → p0 =ρgh The height h does not depend on the cross sectional area A of the tube. The average height of the mercury column at sea level is equal to 760 mm. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-5 Measuring Pressure The open tube manometer At level 1: and At level 2: and 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-6 Pascal’s Principle Pascal’s principle can be formulated as follows: A change in the pressure applied to enclosed incompressible liquids transmitted undiminished to every portion of the fluid and to the walls of the container. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-6 Pascal’s Principle The hydraulic lever, energy considerations Note: Since . The output work: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 14-6 Pascal’s Principle With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force applied over a smaller distance. The product of force and distance remains unchanged so that the same work is done. However, there is often tremendous advantage in being able to exert the larger force. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-7 Archimedes’ Principle The gravitational force . known as “buoyant force”. is the mass of the water in the bag. If V is the bag volume, we have: The magnitude of the buoyant force: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-7 Archimedes’ Principle This force is directed upwards and its magnitude is equal to the weight mƒg of the fluid that has been displaced by the body. The submerged body in Fig. (a) is at equilibrium with Fg = Fb. In Fig. (b), Fg>Fb and the stone accelerates downwards. In Fig. (c), Fb>Fg and the wood accelerates upwards. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 A surfer rides on the front side of a wave, at a point where a tangent to the wave has a slope of . The combined mass of surfer and surfboard is m = 83.0 kg, and the board has a submerged volume of V = 2.50×103. The surfer maintains his position on the wave as the wave moves at constant speed toward shore. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 What are the magnitude and direction (relative to the positive direction of the x axis. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 Key Idea: The buoyancy force on the surfer has a magnitude Fb equal to the weight of the seawater displaced by the submerged volume of the surfboard. The direction of he force is perpendicular to the surface at the surfer’s location. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 By Newton’s second law, because the surfer moves at constant speed toward the shore, the (vector) sum of the buoyancy force , the gravitational force , and the drag force must be 0. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-4 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-5 In Fig. 14-12, a block of density ρ= 800 kg/m3 floats face down in a fluid of density ρf = 1200 kg/m3. The block has height H = 6.0 cm. (a) By what depth h is the block submerged? (b) If the block is held fully submerged and then released, what’s the magnitude of its acceleration? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-5 Key Idea: (a) Floating requires that the upward buoyant force on the block match the downward gravitational force on the block. (b) The buoyant force is equal to the weight mfg of the fluid displaced by the submerged portion of the block. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-5 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-8 Ideal Fluids in Motion Steady flow: The velocity of the moving fluid at any fixed point does not change with time. This type of flow is known as “laminar”. Incompressible flow: The assumption is made that the moving fluid is incompressible i.e. its density is uniform and constant. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-8 Ideal Fluids in Motion No viscous flow: Viscosity in fluids is a measure of how resistive the fluid is to flow. Viscosity in fluids is the analog of friction between solids. Both mechanisms convert kinetic energy into thermal energy (heat). An object moving in a non-viscous fluid experiences no drag force. Irrigational flow: A small particle that moves with the fluid will not rotate about an axis through its center of mass. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-9 The Equation of Continuity 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-9 The Equation of Continuity Volume flow rate: Mass flow rate: The continuity equation can be written in the from: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-9 The Equation of Continuity The figure shows a pipe and gives the volume flow rate (in cm3/s) and the direction of flow for all but one section. What are the volume flow rate and the direction of flow for that section? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-6 Figure 14-17 shows how the stream of water emerging from a faucet “necks down” as it falls. The indicated cross-sectional areas are Ao = 1.2 cm2 and A = 0.35 cm2. The two levels are separated by a vertical distance h = 45 mm. What is the volume flow rate from the tap? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-6 Key Idea: the volume flow rate through the higher cross section must be the same as that through the lower cross section. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-6 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-10 Bernoulli’s Equation The work-kinetic energy theorem: (eqs.1) The change in kinetic energy: (eqs.2) 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-10 Bernoulli’s Equation eqs.3 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-10 Bernoulli’s Equation eqs.1 eqs.2 eqs.3 For the special case in which 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

14-10 Bernoulli’s Equation Water flows smoothly through the pipe shown in the figure, descending in the process. Rank the four numbered sections of pipe according to (a) the volume flow rate RV through them, (b) the flow speed v through them, and (c) the water pressure p within them, greatest first. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-7 Ethanol of density ρ= 791 kg/m3 flows smoothly through a horizontal pipe that tapers in cross- sectional area from A1 = 1.20 × 103 m2 to A2 = A1/2. The pressure difference between the wide and narrow sections of pipe is 4120 Pa. What is the volume flow rate RV of the ethanol? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-7 Key Idea: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-7 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-8 In the old West, a desperado fires a bullet into an open water tank, creating a hole a distance h below the water surface. What is the speed v of the water exiting the tank? 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-8 Key Idea: This situation is essentially that of water moving (downward) with speed v0 through a wide pipe (the tank) of cross-sectional area A and then moving (horizontally) with speed v through a narrow pipe (the hole) of cross- sectional area a. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-8 Because the water flowing through the wide pipe must entirely pass through the narrow pipe, the volume flow rate RV must be the same in the two “pipes.” 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-8 Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-9 As a race car moves forward at 27.25 m/s, air is forced to flow over and under the car The air forced to flow under the car enters through a vertical cross-sectional area A0 = 0.0330 m2 at the front of the car and then 0 flows beneath the car where the vertical cross-sectional area is A1 = 0.0310 m2. Treat this flow as steady flow through a stationary horizontal pipe that decreases in cross-sectional area from A0 to A1. 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-9(a) At the moment it passes through A0, the air is at atmospheric pressure p0. At what pressure p1 is the air as it moves through A1? Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 Example 14-9(b) If the horizontal cross-sectional area of the car is Ah = 4.86 m2, what is the magnitude of the net vertical force F net,y on the car due to the air pressures above and below the car? Solutions: 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授

普通物理講義-14/國立彰化師範大學物理系/郭艷光教授 End of chapter 14! 2018/11/19 普通物理講義-14/國立彰化師範大學物理系/郭艷光教授