Fluids Physics Notes PDF

Solids and Fluids NMY A241 1  Atoms § The scale of the size of an atom is 10-10 m = 0.1 nm, sometimes called an Angstrom, Å. § The number of atoms in a given amount of matter has been measured to be 6.022·1023 atoms in 12 grams of 12...

Solids and Fluids NMY A241 1  Atoms § The scale of the size of an atom is 10-10 m = 0.1 nm, sometimes called an Angstrom, Å. § The number of atoms in a given amount of matter has been measured to be 6.022·1023 atoms in 12 grams of 12C. § The number is called Avogadro’s number, NA = 6.022·1023. § Let’s take diamonds as an example: § Diamonds are composed of carbon atoms. § ρ = 3.51 g/cm3 (3510 kg/m3). § The mass of a diamond is often classified in terms of its mass. § 1 carat = 200 mg = 0.200 g = 2.00 ·10-4 kg. § A typical diamond might be 1 carat. 6.02⋅1023 atoms N C = 0.200 g = 1.00⋅1022 atoms 12 g NMY A241 2  Structures of Carbon Atoms § New structures composed entirely of carbon atoms have been produced. § Fullerenes are composed of 60 carbon atoms arranged in a truncated icosahedron. § Buckyballs § Carbon nanotubes consist of interlocking hexagons of carbon atoms. § Graphene is a 2-dimensional sheet of carbon atoms arranged in a hexagonal lattice. § Fullerenes, carbon nanotubes, and graphene are early products of nanotechnology. § These new materials may revolutionize material science. NMY A241 3  Moles § The SI unit for the amount of mass is the gram-mole, or usually just the mole (mol). § The mass of one mole of a substance is given by the atomic mass number of that element times one gram. § For 12C, 12 grams of carbon have 6.02·1023 atoms. § For 197Au, 197 grams of gold have 6.02·1023 atoms. § For molecules, we add the atomic mass number for all the atoms in the molecule. § Oxygen normally exists as O2. § Oxygen has mass number 16, so oxygen molecules have a relative atomic mass number of 32. § 32 grams of gaseous oxygen contain 6.02·1023 O2 molecules. NMY A241 4  Concept Check § The volume of a mole of gas is 22.4 liters at standard temperature and pressure. Assume that air is composed of N2 (relative mass 28). What is the density of air at standard temperature and pressure? (Hints: 1 liter has a volume of 1000 cm3, density is mass per unit volume.) A. 0.678 mg/cm3 B. 1.25 mg/cm3 m = 28 g C. 5.67 mg/cm3 V = 22.4 L D. 10.1 mg/cm3 m 28 g L 3 E. 28.0 mg/cm3 ρ= = = 0.00125 g/cm V 22.4 L 1000 cm 3 ρ = 1.25 mg/cm 3 NMY A241 5  States of Matter § A gas is a system in which each atom or molecule moves through space as a free particle. § A gas is compressible and fills its container. § A gas can be treated as a fluid because it can flow and exert pressure in the walls of its container. § A liquid is nearly incompressible. § When a liquid is placed in a container, it fills only the volume corresponding to its initial volume. § A solid does not require a container but defines its own shape. § A solid can be compressed or deformed slightly. NMY A241 6  Other States of Matter § Not everything is a solid, liquid, or gas. § The state of a substance depends on its temperature. § Matter in stars is a plasma. § Beaches are made of sand, which is a granular medium. § Glasses can be thought of as a liquid with a very high viscosity. § Foams and gels are another state of matter. NMY A241 7  Other States of Matter § Less that 20 years ago, the existence of a new form of matter called Bose-Einstein condensates was verified. § This form of matter can assume an ordered state in which all the atoms tend to have the same energy and momentum. § The matter in our bodies does not fit into any of these classifications. § Biological material consists mostly of water, yet it can keep or change its shape. § We will not treat these other states of matter, but instead we will concentrate on the properties of solids and fluids. NMY A241 8  Elasticity of Solids § Metallic object are composed of atoms arranges in a three- dimensional lattice. § Each atom has a well-defined distance from its neighbors. § We can model the forces that hold this lattice together as springs. § The lattice is very stiff so we imagine that the springs are very stiff. § Rubber balls are solid but do not seem to be rigid. § We can squish them easily with our hands. § These objects are composed of atoms arranged in long chains rather than a lattice. NMY A241 9  Elasticity of Solids § All rigid objects are somewhat elastic. § We can deform (stretch, bend, etc.) a solid slightly, and it will return to its original shape. § However, if we deform that solid past its elastic limit, it will remain permanently deformed. § If we deform the solid object further, it will break. § Our discussion of the deformation of solids will always refer to deformation below the elastic limit. § We will classify the deformation of solids into three classes: § Stretching or tension § Compression § Shear NMY A241 10  Elasticity of Solids § Example of these deformations are: § Stretching of power lines § Compression of the Hoover Dam § Shearing by scissors NMY A241 11  Stress and Strain § Although we define three types of solid deformation, we can treat these different phenomena with two concepts: § Stress § Deforming force per unit area § Strain § Unit deformation § A stress produces a strain. § We relate stress to strain through a constant called the modulus of elasticity: § stress = modulus · strain § We will define a different modulus for each type of deformation. § Note again that these concepts only apply if we do not exceed the elastic limit. NMY A241 12  Tension § In the case of tension, or stretching, we define a force F that is applied to both ends of an object with length L and the object stretches to a new length of L+ΔL. NMY A241 13  Tension § For tension, the stress is defined as the force F per unit area A applied to the end of an object, perpendicular to the area on the end and parallel to the stretching direction. F ΔL =Y A L § Y is called Young’s modulus. § The table shows some representative average values for Y. NMY A241 14  Compression § In the case of volume compression, we define a force F that is applied to the entire surface of an object with volume V and the object compresses to a new volume of V – ΔV. NMY A241 15  Compression § For compression, the stress is defined as the pressure p (force per unit area) applied to the entire surface. § The resulting strain is the fractional change in the volume of the object, V/ΔV: F ΔV p= = B A V § B is called the bulk modulus. § The table shows some representative average values for B. NMY A241 16  Shear § In the case of shearing, we define a force F that is parallel to an area A, rather that perpendicular as was the case for tension. NMY A241 17  Shear § For shearing, the stress is defined as force per unit area F/A applied parallel to the cross-sectional area. § For this case we push on the end of an object with length L. § The resulting strain is the fractional deflection of the object Δx/L: F Δx =G A L § G is called the shear modulus. § The table shows some representative average values for G. NMY A241 18  Wall Mount for Flat-Panel TV § You’ve just bought a new flat-panel TV and you want to mount it on the wall with four bolts, each with diameter 0.50 cm. § The TV must be mounted 10.0 cm from the wall to allow for cooling. PROBLEM 1: § If the mass of your new TV is 42.8 kg, what is the shear stress on the bolts? SOLUTION 1: § The combined cross sectional area of the bolts is: ⎛ πd 2 ⎞⎟ ⎜ A = 4⎜⎜ ⎟⎟ = π (0.0050 m) = 7.85⋅10−5 m2 2 ⎜⎝ 4 ⎟⎠ NMY A241 19  Wall Mount for Flat-Panel TV § One force acting on the bolts is the force of gravity on the TV, exerted at one end of each bolt. § The force is balanced by a force due to the wall, acting on the other end of the bolts. § This wall force holds the TV in place. § It points in the opposite direction from gravity. § The force entering into the stress is: F = mg = (42.8 kg)(9.81 m/s2 ) = 420 N § The shear stress on the bolts is: F 420 N 6 2 = = 5.35⋅10 N/m A 7.85⋅10−5 m2 PROBLEM 2: § The shear modulus of the steel used is 9.0∙1010 N/m2. § What is the resulting vertical deflection of the bolts? NMY A241 20  Wall Mount for Flat-Panel TV § The deflection is: F Δx F L =G ⇒ Δx = A L AG F L 0.1 m Δx = = (5.35⋅10 N/m ) 6 2 AG 9.0⋅1010 N/m2 Δx = 5.94⋅10−6 m § Even though the shear stress is more than 5 million N/m2, the resulting sag of the your new flat-panel TV is only about 0.006 mm. § This is undetectable to the naked eye. § Monday night football!! NMY A241 21  Stretched Wire PROBLEM: § A 1.50 m long steel wire of diameter 0.400 mm is hanging vertically. § A spotlight with mass m = 1.50 kg is attached to the wire. § How much does the wire stretch? SOLUTION: Think § The diameter of the wire gives us the cross-sectional area. § The weight of the spotlight gives us the force. § The stress is the weight divided by the area. § The strain is the change in the length of the wire divided by its original length. NMY A241 22  Stretched Wire Sketch Research § The cross-sectional area of the wire is: 2 ⎛ d ⎞⎟ πd 2 A = π ⎜⎜ ⎟⎟ = ⎜⎝ 2 ⎟⎠ 4 NMY A241 23  Stretched Wire § The force on the wire is: F = mg § Relating the stress and the strain: F ΔL =Y A L Simplify Combining out equations gives us: F mg 4mg ΔL = 2 = 2 =Y ⇒ A πd πd L 4 4mgL ΔL = Yπd 2 NMY A241 24  Stretched Wire Calculate § Putting in our numerical values gives us: 4mgL 4(1.50 kg)(9.81 m/s 2 )(1.50 m) ΔL = = = 0.00087824 m (200⋅109 N/m 2 )π (0.400⋅10−3 m) 2 2 Yπd Round ΔL = 0.000878 m = 0.878 m Double-check § The units are correct. § The wire stretched less than 1 mm compared with it original length of 1.50 m, which is less than 0.1%. NMY A241 25  Elastic Limit § The stress is proportional to the strain as as long as the elastic limit is not exceeded. § Here is a typical stress-strain diagram for a ductile metal drawn into a wire: NMY A241 26  Pressure § Pressure is defined as force per unit area: F p= A § The SI unit of pressure is N/m2, which has been given the name pascal, abbreviated Pa: 1N 1 Pa = 1 m2 § The average pressure of the Earth’s atmosphere, 1 atm, is: 1 atm = 1.013⋅105 Pa § Atmospheric pressure is often quoted in non-SI units: 1 atm = 760 torr = 760 mm Hg = 14.7 lb/in 2 = 29.92 in Hg NMY A241 27  Weighing Earth’s Atmosphere § The Earth’s atmosphere is composed of 78.08% nitrogen, 20.95% oxygen, 0.93% argon, 0.25% water vapor, and traces of other gases, including carbon dioxide (CO2). § The CO2 content of the atmosphere currently is 0.039% = 390 ppm, but it varies 6-7 ppm with the seasons. § The CO2 content of the atmosphere has been rising since the Industrial Revolution, mainly as a result of burning fossil fuels. PROBLEM: § What is the mass of Earth’s atmosphere and what is the mass of the CO2 in Earth’s atmosphere? SOLUTION: Think § Earth’s atmospheric pressure is 1.01·105 Pa. NMY A241 28  Weighing Earth’s Atmosphere Sketch § In the sketch we show a column of air with weight mg above an area A on the Earth’s surface. § This air exerts a pressure p on the surface. Research § We can relate pressure, force, and area: F p= A § The surface area of the Earth is: A = 4πR 2 with R = 6370 km NMY A241 29  Weighing Earth’s Atmosphere Simplify § Combining our equations gives us: F mg p= = 2 ⇒ A 4πR 4πR 2 p m= g § Putting in our numerical values: 4π (6.37 ⋅10 m) (1.01⋅105 Pa ) 2 6 m= = 5.24978⋅1018 kg 9.81 m/s 2 Round m = 5.25⋅1018 kg NMY A241 30  Weighing Earth’s Atmosphere Double-check § To get the mass of 1 ppm of CO2 in the atmosphere, remember that the molar mass of CO2 is 12 + (2·16) = 44 g. § The average mass of a mole of the atmosphere is: 0.78(2·14) + 0.21(2·16) + 0.01(40) = 28.96 g. § The mass of 1 ppm of CO2 is: −6 44 m1 ppm of CO = 10 ⋅ m = 7.97 ⋅1012 kg = 8.0 billion tons 2 28.96 § We add about 2 ppm of CO2 to the atmosphere each year by burning fossil fuels. § About the same amount dissolves in the Earth’s oceans. NMY A241 31  Pressure-Depth Relationship § Consider a container of water open to the Earth’s atmosphere. § Imagine a cube of water. § The top of the cube is horizontal and is at a depth of y1. § The bottom of the cube is also horizontal and is at a depth of y2. § The other sides are vertical. § The vertical forces on the cube must sum to zero: F2 − F1 − mg = 0 NMY A241 32  Pressure-Depth Relationship § The pressure at depth y1 is p1. § The pressure at depth y2 is p2. § Assuming the area of each face of the cube is A, we can write: F1 = p1 A F2 = p2 A § Assuming the volume of the cube is V and ρ is the density of water, then the mass of the cube of water is: m = ρV NMY A241 33  Pressure-Depth Relationship § Remembering: F2 − F1 − mg = p2 A− p1 A−ρVg = 0 § The volume of the cube of water is: V = A( y1 − y2 ) § So we can rearrange and substitute for the volume: p2 A = p1 A+ ρ A( y1 − y2 ) g § So we get for the pressure as function of depth: p2 = p1 + ρ ( y1 − y2 ) g NMY A241 34  Pressure-Depth Relationship § A common problem involves the pressure as a function of depth below the surface of a liquid. § We define the pressure at the surface of a liquid (y1 = 0) to be p0. § We define the pressure at depth h (y2 = 0) to be p. § We get the expression for the pressure at a given depth of a liquid: p = p0 + ρgh § This equation holds for any shape of the vessel containing the liquid. NMY A241 35  Submarine § A U.S. Navy submarine of the Los Angeles class is 110 m long and has a hull diameter of 10. m. § Assume that submarine has a flat top with an area of A = 1100 m2. § The density of seawater is 1024 kg/m3. PROBLEM: § What is the total force pushing down in the top of this submarine at a depth of 250. m? NMY A241 36  Submarine SOLUTION: § The pressure inside the submarine is p0. § The pressure difference between the inside and outside of the submarine is: Δp = ρgh = (1024 kg/m 3 )(9.81 m/s 2 )(250. m) = 2.8⋅109 N § Multiply the pressure times the area to get the force: F = ΔpA = 2.8⋅109 N § This force is 40 times the weight of the submarine! NMY A241 37  Concept Check § Three containers of water have different shapes. § The distance between the top surface of the water and the bottom of each container is the same. § Which of the following statements correctly characterizes the pressure at the bottom of the containers? A. The pressure at the bottom of container 1 is the highest. B. The pressure at the bottom of container 2 is the highest. C. The pressure at the bottom of container 3 is the highest. D. The pressure at the bottom of all three containers is the same. NMY A241 38  Gauge Pressure and Barometers § The pressure p in the pressure depth relationship p = p0 + ρgh is the absolute pressure. § The difference between an absolute pressure and the atmospheric air pressure is called a gauge pressure. § A tire gauge measures gauge pressure, ρgh. § A barometer is a device that is used to measure atmospheric pressure. § We will discuss two types of barometers: § Mercury barometer § Water barometer NMY A241 39  Mercury Barometer § A mercury barometer can be constructed by taking a tube full of mercury and inverting it in a container of mercury. § The distance between the top of the mercury in the tube and the level of the mercury in the container can be related to the atmospheric pressure: p0 = ρgh § The mercury barometer leads to the often quoted mm of mercury measurement of the atmospheric pressure: 1 atm 1 atm = = 760 29.92 mmofHg inches mercury NMY A241 40  Water Barometer § Suppose we wanted to construct a barometer, but we did not want to use mercury. § We decided to use water instead, how tall would the column of water be? p0 = ρgh p0 1.01⋅105 Pa h= = ρwater g 1000 kg/m 3 ⋅9.81 m/s 2 h = 10.3 m = 33.8 feet for a water barometer for mercury, h = 0.760 m = 29.92 inches = 2.49 feet NMY A241 41  Manometer § An open tube manometer measures the gauge pressure of a gas. § It consists of a U-shaped tube partially filled with mercury. § The closed end is connected to a container of gas whose gauge pressure is to be measured. § The other end is open and experiences atmospheric pressure. § The gauge pressure of the gas in the container is: pg = p − p0 = ρgh § The gauge pressure can be positive or negative. NMY A241 42  Barometric Altitude Relation for Gases § In the derivation of the depth pressure relationship, we have made use of the incompressibility of liquids. § However, if our fluid is a gas, we cannot make this assumption. § We start again with a thin layer of fluid in a fluid column. § The pressure difference between bottom and top surface is still given by the weight of the thin layer of fluid divided by the area: F mg ρVg ρ (ΔhA) g Δp = − = − =− =− = −ρgΔh A A A A NMY A241 43  Barometric Altitude Relation for Gases § The negative sign comes from the fact that the pressure decreases with increasing altitude, because the weight of the fluid column above is reduced. § So far nothing is different from the derivation of the incompressible case. § However, for compressible fluids we find that the density is proportional to the pressure: ρ p = ρ0 p0 § Strictly this relationship is only true for ideal gases. NMY A241 44  Barometric Altitude Relation for Gases § Combining our two equations gives us: Δp gρ0 =− p Δh p0 § Taking the limit of Δh → 0, we find the equation: dp gρ0 =− p dh p0 § This equation is a differential equation. § The solution of this differential equation is: p(h) = p0 e−hρ0 g/ p0 § This equation is known as the barometric pressure formula. § It relates altitude to pressure in gases. § It applies as long as the temperature does not change, with altitude and gravitation constant. NMY A241 45  Air Pressure on Mount Everest PROBLEM: § What is the air pressure on the top of Mount Everest? SOLUTION: § We use the barometric pressure formula: p(h) = p0 e−hρ0 g/ p0 § We can rewrite the barometric pressure formula as: ⎛ p ⎞⎟ −h/⎜⎜⎜ o ⎟⎟⎟ ⎜⎝ ρo g ⎟⎠ p(h) = p0 e−hρ0 g/ p0 = p0 e § The density of and pressure of air at sea level are: ρ0 = 1.229 kg/m 3 p0 = 1.01⋅105 Pa § We can find the constant term in the barometric pressure formula: p0 1.01⋅105 Pa = = 8377 m ρ0 g (1.229 kg/m )(9.81 m/s ) 3 2 NMY A241 46  Air Pressure on Mount Everest § For the Earth’s atmosphere we can write an expression for the pressure at a given height as: p(h) = p0 e−h/(8377 m) § The height of Mount Everest is 8850 m (29,035 ft) so the pressure at the summit is: p(8850 m) = p0 e−8850/8377 = 0.347 p0 = 35 kPa § This is 35% of the pressure at sea level. § This low air pressure is dangerous if you are not acclimated. § Everest climbers take months to acclimate and many still have trouble. NMY A241 47  How Close? § Because: ρ p = ρ0 p0 § The barometric pressure formula also means: ρ(h) = ρ0 e−hρ0 g/ p0 NMY A241 48  Concept Check § As you descend into a mine shaft below sea level, the air pressure A. decreases linearly. B. decreases exponentially. C. increases linearly. D. increases exponentially. NMY A241 49  Concept Check § You are skiing at Winter Park at 10,000 ft (3.05 km). § You buy a bottle of water, drink it all, and then put the lid back on the bottle and toss it in the back seat of your car. § You drive down to Denver which is at an altitude of 5000 ft (1524 km). § When you look at the water bottle in the back seat, what do you see? A. The water bottle has exploded. B. The water bottle has been crushed. C. The water bottle is not changed from the way it looked at Winter Park. NMY A241 50  Pascal’s Principle § If we exert pressure on any part of an incompressible fluid, this pressure will be transferred to the rest of the fluid with no losses. § This concept is called Pascal’s Principle and can be stated: “When a change in pressure occurs at any point in a confined fluid, an equal change in pressure occurs at every point in the fluid.” § Pascal’s Principle is the basis for many modern hydraulic devices in use today. NMY A241 51  Pascal’s Principle § Consider a cylinder partially filled with water. § Place a massless piston on the surface of the water. § Place a weight on the piston that exerts a pressure pt on the top of the water. § The pressure p at a depth h is then: p = pt + ρgh § If we add a second weight, the change in pressure at depth h can only be due to the Δpt caused by the addition of the second weight: Δp = Δpt NMY A241 52  Pascal’s Principle § Now let’s consider two connected pistons partially filled with an incompressible fluid. § One piston has area Ain. § The second piston has area Aout so that Ain < Aout. § We exert a force Fin on the first piston producing a change in the pressure in the fluid. NMY A241 53  Pascal’s Principle § This change will be transmitted throughout the fluid, including areas near the second piston, so we can write: Fin Fout Aout Δp = = ⇒ Fout = Fin Ain Aout Ain § We can see that we have magnified the force by a factor equal to the ratio of the areas of the pistons. NMY A241 54  Archimedes’ Principle § Let’s imagine a cube of water in a volume of water. § The weight of this cube of water is supported by the buoyant force FB resulting from the pressure differential between the top and bottom of the cube: F2 − F1 = mg = FB § For our imaginary cube of water, the buoyant force equals the weight: FB = mf g NMY A241 55  Archimedes’ Principle § Now let’s replace the cube of water with a cube of steel. § The steel cube weighs more than the cube of water, so now there is a net force in the y-direction given by: Fnet,y = FB − msteel g < 0 § This force is downward so the metal cube will sink. NMY A241 56  Archimedes’ Principle § Let’s replace our steel cube with a wooden cube. § Now the weight of the wood is less than the weight of the water that the wood replaced, so the net force is upward. § The wooden block would rise toward the surface. § If we place an object less dense than water in water, the object will float. § The object will sink into the water until the weight of the object equals the weight of the water displaced: FB = mf g = mobject g NMY A241 57  Archimedes’ Principle § A floating object displaces its own weight of fluid. § This is true independent the amount of fluid present. § Let’s look at a ship in a lock. § In both positions, the ship floats with the same fraction below the water level. § What matters is the amount of water displaced, not the total amount of water in the lock. § If an object with a higher density than water is submerged, it will experience an upward buoyant force that is less than its weight: Actual weight - buoyant force = apparent weight NMY A241 58  Floating Iceberg PROBLEM: § What fraction of an iceberg floating in seawater is visible above the surface? SOLUTION: § Let Vt be the total volume of the iceberg and Vs be the volume of the iceberg that is submerged. NMY A241 59  Floating Iceberg § The fraction above the water surface is: Vt −Vs Vs f= = 1− Vt Vt § Because the iceberg is floating, the submerged part of the iceberg must displace a volume of water with the same weight as the entire iceberg. § We can calculate the weight of the iceberg knowing the density of freshwater ice: ρice = 917 kg/m 3 § The weight of the iceberg is: miceberg g = ρiceVt g NMY A241 60  Floating Iceberg § The density of seawater is: ρseawater = 1024 kg/m 3 § The weight of the displaced seawater is: mseawater g = ρseawaterVs g § The weight of the iceberg is equal to the weight of the displaced seawater: ρiceVt g = ρseawaterVs g ⇒ ρiceVt = ρseawaterVs § We can rearrange this formula to get: Vs ρice = Vt ρseawater § So the fraction above water is: Vs ρice 917 kg/m 3 f = 1− = 1− = 1− = 0.104 = 10.4% Vt ρseawater 1024 kg/m 3 NMY A241 61