Physics 201 Chapter 13 Lecture 1

Physics 201 Chapter 13 Lecture 1 Fluid Statics Pascal s Principle Archimedes Principle (Buoyancy) Fluid Dynamics Continuity Equation Bernoulli Equation 11/30/2009 Physics 201, UW-Madison 1

Fluids Density = Mass/Volume ρ = M / V units = kg/m 3 Pressure (P) P = Force/Area [N/m 2 ] 1 N/m 2 = 1 Pascal (Pa) Pressure variation with depth P = ρ g h Atmospheric Pressure Even when there is no breeze air molecules are continuously bombarding everything around - results in pressure normal atmospheric pressure = 1.01 x 10 5 Pa (14.7 lb/in 2 ) 11/30/2009 Physics 201, UW-Madison 2

Densities of substances 11/30/2009 Physics 201, UW-Madison 3

Compressiblity Density & Pressure are related by the Bulk Modulus LIQUID: incompressible (density almost constant) GAS: compressible (density depends a lot on pressure) B = Δp ( ΔV / V ) 11/30/2009 Physics 201, UW-Madison 4

Variation of pressure with depth m = ρv; V = Ah m = ρah P = F A = mg A ( ρah)g ; i.e., P = A P = hρg True for all shapes of containers 11/30/2009 Physics 201, UW-Madison 5

Pressure difference of 3 m of water compares to the change of descending 3000m in air 11/30/2009 Physics 201, UW-Madison 6

Pascal s Principle A change in pressure in an enclosed fluid is transmitted undiminished to all the fluid and to its container. This principle is used in hydraulic system P 1 = P 2 (F 1 / A 1 ) = (F 2 / A 2 ) F 2 11/30/2009 Physics 201, UW-Madison 7

Pascal s Principle This principle is used in hydraulic system F 1 P 1 = P 2 (F 1 / A 1 ) = (F 2 / A 2 ) Can be used to achieve a mechanical advantage F 2 = F 1 (A 2 / A 1 )» Work done is the same: height by which the surface A 2 rises is smaller than the change in the height of surface with area A 1. A 1 F 2 A 2 11/30/2009 Physics 201, UW-Madison 8

Using Fluids to Measure Pressure Use Barometer to measure Absolute Pressure Top of tube evacuated (p=0) Bottom of tube submerged into pool of mercury open to atmosphere (p=p 0 ) Pressure dependence on depth: h = p 0 ρg Barometer Use Manometer to measure Gauge Pressure Measure pressure of volume (p 1 ) relative to the atmospheric pressure ( gauge pressure ) p 1 Manometer p 0 The height difference (Δh) measures the gauge pressure: Δh = (p p ) 1 0 1 atm = 760 mm (29.9 in) Hg ρg Δh = 10.3 m (33.8 ft) H 2 0 11/30/2009 Physics 201, UW-Madison 9

Measurement of Pressure Manometer If both sides of an U-tube are open to atmosphere the levels of the fluid are the same on both sides If one side is connected to a pressurized side the level difference between the two sides can be used to measure pressure. 11/30/2009 Physics 201, UW-Madison 10

Measuring the tire pressure: Is this a manometer or a barometer? 11/30/2009 Physics 201, UW-Madison 11

Measuring Blood Pressure Blood pressure is quite high, 120/80 mm of Hg Use higher density fluid in a manometer: Mercury 11/30/2009 Physics 201, UW-Madison 12

Atmosphere - pressure vs height P = P 0 e ρ 0 P 0 gh --> whiteboard 11/30/2009 Physics 201, UW-Madison 13

Pressure in a fluid Impulse to wall: F x!t =!p x =!(Mv x ) F x =!(Mv x )/!t Force is perpendicular to surface Force proportional to area of surface pressure (p) p = Force/area [N/m 2 ] 1 N/m 2 = 1 Pascal (Pa) molecule v v F x wall 11/30/2009 Physics 201, UW-Madison 14

Pressure y average vertical force = f y =!p y!t Atmospheric Pressure Even when there is no breeze air molecules are continuously bombarding everything around - results in pressure normal atmospheric pressure = 1.01 x 10 5 Pa (14.7 lb/in 2 ) 11/30/2009 Physics 201, UW-Madison 15 = "! mv $ y #!t % & '

Archimedes Object immersed in a fluid is subject to a buoyant force. Force on sides cancel Force on top F t = ρgh T A Force on bottom F b = ρgh B A ΔF = ρg A Δh F B = (mg) disp 11/30/2009 Physics 201, UW-Madison 16

The pressure is a function of the depth only (for a given density of the fluid and of g) 11/30/2009 Physics 201, UW-Madison 18

Float Weight of object = ρ 0 gv Buoyant force is the weight of the displaced fluid Weight of fluid = ρ f gv Displace just enough fluid such that forces = 0! 11/30/2009 Physics 201, UW-Madison 19

Buoyant Force (B) Archimedes Principle weight of fluid displaced (P=F/A, P=ρgh)» B = ρ fluid g V displaced» W = ρ object g V object» object sinks if ρ object > ρ fluid» object floats if ρ object < ρ fluid» Eureka! If object floats.» B=W» Therefore ρ fluid g V displaced = ρ object g V object» Therefore V displaced /V object = ρ object / ρ fluid 11/30/2009 Physics 201, UW-Madison 20

Float Buoyant force is the weight of the displaced fluid Weight of object = ρ Ice V total g Weight of fluid = ρ SeaWater gv submersed General solution: V displaced /V object = ρ object / ρ fluid Displace just enough fluid such that forces = 0! 11/30/2009 Physics 201, UW-Madison 21

The weight of a glass filled to the brim with water is W b. A cube of ice is placed in it, causing some water to spill. After the spilled water is cleaned up, the weight of the glass with ice cube is W a. How do the weights compare: 1. W b > W a. 2. W b < W a. 3. W b = W a. Archimedes Principle Archimedes Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice 11/30/2009 Physics 201, UW-Madison 22

Question Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up causing the water to spill. 2. Go down. 3. Stay the same. Archimedes Principle: The buoyant force on an object equals the weight of the fluid it displaces. Weight of water displaced = Buoyant force = Weight of ice When ice melts it will turn into water of same volume 11/30/2009 Physics 201, UW-Madison 23

Buoyancy Two cups hold water at the same level. One of the two cups has plastic balls (projecting above the water surface) floating in it. Which cup weighs more? 1) Cup I 2) Cup II 3) Both the same Cup I Cup II Archimedes principle tells us that the cups weigh the same. Each plastic ball displaces an amount of water that is exactly equal to its own weight. 11/30/2009 Physics 201, UW-Madison 24

Sunken Balls Two identical glasses are filled to the same level with water. Solid steel balls are at the bottom in one of the glasses. Which of the two glasses weighs more? 1. The glass without steel balls 2. The glass with steel balls 3. Both glasses weigh the same The steel balls sink. The buoyant force equal to the weight of the displaced water is not sufficient to counter the weight of the steel balls. Therefore, the glass with steel balls weighs more. 11/30/2009 Physics 201, UW-Madison 25

Buoyant force and depth Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greater depth. The force needed to hold brick B in place is: 1. larger 2. the same as 3. smaller than the force required to hold brick A in place. The buoyant force on each brick is equal to the weight of the water it displaces and does not depend on depth. 11/30/2009 Physics 201, UW-Madison 26

Fluid Flow Fluid flow without friction Volume flow rate: ΔV/Δt = A Δd/Δt = Av (m 3 /s) Continuity: A 1 v 1 = A 2 v 2 i.e., flow rate the same everywhere e.g., flow of river 11/30/2009 Physics 201, UW-Madison 27

Problem Two hoses, one of 20-mm diameter, the other of 15-mm diameter are connected one behind the other to a faucet. At the open end of the hose, the flow of water measures 10 liters per minute. Through which pipe does the water flow faster? 1. The 20-mm hose 2. The 15-mm hose 3. Water flows at the same speed in both cases 4. The answer depends on which of the two hoses comes first in the flow When a tube narrows, the same volume occupies a greater length. For the same volume to pass through points 1 and 2 in a given time, the velocity must be greater at point 2. The process is reversible. 11/30/2009 Physics 201, UW-Madison 28

Faucet A stream of water gets narrower as it falls from a faucet (try it & see). V 1 A 1 A 2 V 2 The velocity of the liquid increases as the water falls due to gravity. If the volume flow rate is conserved, them the cross-sectional area must decrease in order to compensate The density of the water is the same no matter where it is in space and time, so as it falls down and accelerates because of gravity,the water is in a sense stretched, so it thins out at the end. 11/30/2009 Physics 201, UW-Madison 29

Types of Fluid Flow Laminar flow Steady flow Each particle of the fluid follows a smooth path The paths of the different particles never cross each other The path taken by the particles is called a streamline Turbulent flow An irregular flow characterized by small whirlpool like regions Turbulent flow occurs when the particles go above some critical speed 12/01/2009 30

Viscosity Characterizes the degree of internal friction in the fluid This internal friction, viscous force, is associated with the resistance that two adjacent layers of fluid have to moving relative to each other It causes part of the kinetic energy of a fluid to be converted to internal energy 12/01/2009 31

Ideal Fluid Flow There are four simplifying assumptions made to the complex flow of fluids to make the analysis easier 1. The fluid is nonviscous internal friction is neglected 2. The flow is steady the velocity of each point remains constant 3. The fluid is incompressible the density remains constant 4. The flow is irrotational the fluid has no angular momentum about any point 12/01/2009 32

Streamlines The path the particle takes in steady flow is a streamline The velocity of the particle is tangent to the streamline A set of streamlines is called a tube of flow 12/01/2009 33

Streamlines 11/30/2009 Physics 201, UW-Madison 34

Continuity equation Δm 1 = ρ 1 ΔV 1 = ρ 1 Av 1 Δt Volume Flow rate Mass flow rate I V = ΔV Δt I M 1 = Δm 1 Δt = Av = ρ 1 A 1 v 1 In steady state Δm 1 Δt = Δm 2 Δt ρ 2 A 2 v 2 = ρ 1 A 1 v 1 General case: mass may be accumulated or decreased in the volume between A1 and A2 I M 2 I M 1 = dm 2 dt dm 1 dt = dm 12 dt Continuity equation 11/30/2009 Physics 201, UW-Madison 35

Continuity equation Volume Flow rate I V = ΔV Δt = Av Mass flow rate I M 1 = Δm 1 Δt = ρ 1 A 1 v 1 In steady state ρ 2 A 2 v 2 = ρ 1 A 1 v 1 Case of incompressible fluid: density constant A 2 v 2 = A 1 v 1 11/30/2009 Physics 201, UW-Madison 36

Bernoulli s Equation As a fluid moves through a region where its speed and/or elevation above the Earth s surface changes, the pressure in the fluid varies with these changes Consider the two shaded segments The volumes of both segments are equal The net work done on the segment is W =(P 1 P 2 ) V Part of the work goes into changing the kinetic energy and some to changing the gravitational potential energy 12/01/2009 Physics 201, UW-Madison

Bernoulli s Equation The change in kinetic energy: ΔK = 1/2 mv 2 2-1/2 mv 1 2 The masses are the same since the volumes are the same The change in gravitational potential energy: ΔU = mgy 2 mgy 1 The work also equals the change in energy Combining: W = (P 1 P 2 )V =1/2 mv 2 2-1/2 mv 1 2 + mgy 2 mgy 1 Rearranging and expressing in terms of density: P 1 + 1/2 ρv 1 2 + mgy 1 = P 2 + 1/2 ρv 2 2 + mgy 2 12/01/2009 Physics 201, UW-Madison 38

Bernoulli s Equation Pressure drops in a rapidly moving fluid whether or not the fluid is confined to a tube For incompressible, frictionless fluid: P + 1 2 ρv2 + ρgh = constant 1 2 ρv2 = 1 1 2 mv2 V = KE V ρgh = mgh V = PE V Bernoulli equation states conservation of energy For Static Fluids:P 1 + ρgh 1 = P 2 + ρgh 2 Bernoulli's Principle (constant depth):p 1 + 1 2 ρv 2 1 = P 2 + 1 2 ρv 2 2 12/01/2009 39

What is the pressure of an incompressible fluid in the constricted region? Continuity equation gives velocity in the constricted region (increases with A1/A2)): A 2 v 2 = A 1 v 1 Bernoulli equation P 1 + 1 2 ρv 2 1 = const 12/01/2009 says that pressure drops as P 1 v 2 40

Applications of Bernoulli s Principle Wings and sails Higher velocity on one side of sail versus the other results in a pressure difference that can even allow the boat to sail into the wind Entrainment Reduced pressure in high velocity fluid pulls in particles from static or lower velocity fluid» Bunsen burner, Aspirator, Velocity measurement 12/01/2009 41

Problem (a) Calculate the approximate force on a square meter of sail, given the horizontal velocity of the wind is 6 m/s parallel to its front surface and 3.5 m/s along its back surface. Take the density of air to be 1.29 kg/m 3. (b) Discuss whether this force is great enough to be effective for propelling a sail boat. Bernoulli eq.for constant height P 1 + 1 2 ρv 2 1 = P 2 + 1 2 ρv 2 2 Force, F = (P 1 P 2 )A = 1 2 ρ(v 2 2 v 2 1 )A = 15.3 N The force is small. However, when the sails are large, the force can be high enough to propel a sail boat. For larger boats, one can add more than one sail to increase the surface area. One can even sail into the wind, where (P 1 P 2 ) is small. 12/01/2009 42

Applications of Fluid Dynamics Streamline flow around a moving airplane wing Lift is the upward force on the wing from the air Drag is the resistance The lift depends on the speed of the airplane, the area of the wing, its curvature, and the angle between the wing and the horizontal 12/01/2009 43

Lift General In general, an object moving through a fluid experiences lift as a result of any effect that causes the fluid to change its direction as it flows past the object Some factors that influence lift are: The shape of the object The object s orientation with respect to the fluid flow Any spinning of the object The texture of the object s surface 12/01/2009 44

Golf Ball The ball is given a rapid backspin The dimples increase friction Bernoulli says: Higher relative velocity will reduce the pressure. Increases lift It travels farther than if it was not spinning 12/01/2009 45

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Problem (a) What is the pressure drop due to Bernoulli effect as water goes into a 3 cm diameter nozzle from a 9 cm diameter fire hose while carrying a flow of 40 L/s? (b) To what maximum height above the nozzle can this water rise neglecting air resistance. v 1 = F 1 = 40 10 3 m 3 /s A 1 π(0.045) 2 v 2 = F 2 = 40 10 3 m 3 /s A 2 π(0.015) 2 = 6.29 m/s = 56.6 m/s P 1 P 2 = 1 2 ρ(v 2 2 v 2 1 ) = 1.58 10 6 N/m 2 h = v2 2g = (56.6)2 2 9.8 m=163 m 12/01/2009 47

P 1, v 1, h 1 Torricelli s Theorem h1 P 2 =P 1, v 2, h 2 h2 Bernoulli's equation at constant pressure (P 1 = P 2 ) P 1 + 1 2 ρv 2 1 + ρgh 1 = P 2 + 1 2 ρv 2 2 + ρgh 2 1 2 ρv 2 1 + ρgh 1 = 1 2 ρv 2 2 + ρgh 2 v 2 2 = v 2 1 + 2g(h 1 h 2 ) Density is constant h = h 1 h 2 Same as kinematics equation for any object falling with negligible friction. 12/01/2009 48

F Friction in fluids Viscosity L V 0 η = F A v L shearing stress strain V=0 Newton s law Laminar flow -- no turbulence Pressure, τ = η dv dz η is the coefficient of viscosity 'A' is the moving surface 12/01/2009 49

Real fluid flow At constant velocity net force is zero. F = ( P 1 P 2 )πr 2 and the area on which the force is acting is A = 2πrL τ = F A = ΔPr 2L τ η = Δv or dv = ( P P 1 2 )r dr 2Lη Δr 12/01/2009 50

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More Viscosity for a given situation P 1, P 2,η, and L are constant let b = ΔP 2ηL and dv = brdr and integrating v = br 2 2 + C v = 0 at the boundary r = R and, substituting, ( v = P 1 P ) 2 4ηL ( R 2 r 2 ) 12/01/2009 52

Flow and Viscosity ΔV = ( vt)2πrδr and again let B = πtb and, then dv = BR 2 rdr Br 3 dr V = BR 2 Finally, rdr B r 3 dr 0 R V t 0 R = π ( P 1 P ) 2 R 4 8ηL = B R4 2 B R4 4 = B R4 4 Poiseuille s Eqn. no turbulence no sized particles constant η 12/01/2009 53