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Roman W.

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PHYSICS 220 Lecture 18 Fluid Dynamics Dynamics Textbook Sections 10.4 – 10.7 Lecture 17 Purdue University, Physics 220 1 Overview • Last Lecture Pif tdbll“bi”fti– Pressure is orce exer e y mo ecu es “bounc ng off con a ner •P = F/A – Pascal’s Principle • A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid Gra it / eight effects press re– vity w pressu •P = P 0 + ρgd Archimedes’ Principle: Buoyant force is “weight” of displaced fluid– is fluid •F = ρ g V • Today – Moving Fluids – Continuity Lecture 17 Purdue University, Physics 220 2 – Bernoulli's Principle Archimedes’ Principle • Buoyant Force F B Wiht ffliddi l d– Weight o fluid disp ace –F B = ρ fluid V disp g F V– g = mg = ρ object object g – Object sinks if ρ object > ρ fluid – Object floats if ρ object < ρ fluid • If object floats… –F B = F g – Therefore ρ fluid V disp g = ρ object V object g – Therefore V disp /V object = ρ object /ρ fluid Lecture 17 Purdue University, Physics 220 3 Exercise 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 yp the water is at the very brim. When the ice melts, the level of the water in the glass will: A) Go up, causing the water to spill out of the glass B) Go down C) Stay the same CORRECT W ice = F B = ρ W gV displaced W =ρ g V water W water W ice = W water V displaced = V water Lecture 17 Purdue University, Physics 220 4 Exercise Which weighs more: A) A large bathtub filled to the brim with water)g B) A large bathtub filled to the brim with water with a battle-ship floating in it C) Th i h th Tub of water + ship CORRECT ey we g e same Weight of ship = Buoyant force = Tub of water Weight of displaced water Lecture 17 Purdue University, Physics 220 5 Overflowed water Fluids in Motion • The general theory of fluids in motion is very complex since the velocity of a fluid usually varies the fluid from place to place within the fluid Aidilif• Assumptions ma e to s mplify: – The fluid density is constant • The fluid is incompressible – The flow is steady Th l it i i d d t f ti• e ve ocity s n epen en o time • It can still vary from point to point in space There is no friction– is no friction • No viscosity – There are no complex flow patterns are flow • Such as turbulence Lecture 17 6Purdue University, Physics 220 Ideal Fluids • A fluid that satisfies the four assumptions is called an ideal fluid • An ideal fluid can be used to illustrate some key ffl idd iaspects of fluid dynam cs Lecture 17 7Purdue University, Physics 220 Fluid Flow Concepts A P P ρ Vl fl t ΔV/Δt A ( 3 /) 1 1 A 2 2 v 1 v 2 • Volume flow ra e: = v m /s) • Mass flow rate: Δm/Δt = ρAv (kg/s) • Continuity: ρA 1 v 1 = ρA 2 v 2 mass flow rate the same everywhere same everywhere • The flow rate is the product of the area and speed • Q = vA v A • The equation of continuity says the incoming flow rate equals the outgoing flow rate Lecture 17 Purdue University, Physics 220 8 • SI units for flow rate are m 3 / s Continuity A four-lane highway merges down to a two-lane highway. The officer in the police car four to two The in the car observes 8 cars passing every second, at 30 mph. How many cars does the officer on the motorcycle observe passing every second? A) 4 B) 8 C) 16 How fast must the cars in the two-lane section be going? All cars stay on the road… Must pass motorcycle too Must go faster Lecture 17 Purdue University, Physics 220 9 fast must the two be A) 15 mph B) 30 mph C) 60 mph , else pile up! Continuity of Fluid Flow • Watch “plug” of fluid moving through the narrow part of the tube (A 1 ) • Time for “plug” to pass point Δt = x 1 / v 1 • Mass of fluid in “plug” m 1 = ρ V 1 =ρ A 1 x 1 or m 1 = ρA 1 v 1 Δt • Watch “plug” of fluid moving through the wide part of the tube (A 2 ) • Time for “plug” to pass point Δt = x 2 / v 2 • Mass of fluid in “plug” m 2 = ρ V 2 =ρ A 2 x 2 or m 2 = ρA 2 v 2 Δt • Continuity Equation says m 1 = m 2 fluid isn’t building up or disappearing Lecture 17 Purdue University, Physics 220 10 •A 1 v 1 = A 2 v 2 Faucet A stream of water gets narrower as it falls from a faucet (try it & see). (y Explain this phenomenon using the equation of continuity. A 1 V 1 A 2 V 2 From the continuity equation, we know that A 1 V 1 =A 2 V 2 . Therefore as the velocity increases the cross-, , the cross sectional area has to decrease. Since the water is in free fall gravity causes the speed to increase as it goes down. Because the speed is increasing the cross sectional area has to decrease in order to keep the product of the two equal. Lecture 17 Purdue University, Physics 220 11 Flow Speed and Pressure • Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening • Acceleration due to change in pressure: P 1 > P 2 – Smaller tube has FASTER flow and LOWER pressure Where the velocity is high, the pressure is low and where the velocity is low the pressure is high Lecture 17 Purdue University, Physics 220 12 velocity is low pressure Bernoulli’s Equation • Consider tube where both area and height change W=ΔK+ΔUW = K + (P 1 -P 2 ) V = ½ m (v 2 2 –v 1 2 ) + mg(y 2 -y 1 ) (P P )V=½ V(v 2 v 2 )+ Vg(y y ) 1 - 2 ) V = ½ ρV (v 2 – 1 ) + ρ 2 - 1 P 1 +ρgy 1 + ½ ρv 1 2 = P 2 +ρgy 2 + ½ρv 2 2 Lecture 17 Purdue University, Physics 220 13 Bernoulli’s Equation 22 11 2 2 11 P gy v P gy vρ ρρρ++ =++ Lecture 17 Purdue University, Physics 220 14 Bernoulli’s Principle • How can a sail boat move against the wind? • Bernoulli’s principle states: – Where the velocity is high, the pressure is low and where the velocity is low the pressure is high Lecture 17 Purdue University, Physics 220 15 Airplane Wing • There is a bulge at the top of the wing • This causes the speed of the air to be slightly higher over the top of gp the wing compared to the bottom of the wing • An increase in speed causes a decrease in in pressure • F =(P P )A total = (P bot – top ) A Lecture 17 Purdue University, Physics 220 16 Dynamics of Ideal Fluids • Conservation of mass ⇒ Continuity equation : constantVolumeFlow rate Av = The same volume of fluid passes per unit time • Conservation of energy ⇒ Bernoulli’s equation 2 1 constant 2 Pvgyρ ρ++= which relates height, velocity and pressure of the fluid Lecture 17 Purdue University, Physics 220 17 Question What will happen when I “blow” air between the tlt?two pla es A) Move Apart B) Come Together C) Nothing There is air pushing on both sides of plates. If we get rid of the air in the middle, then just have air on the outside pushing them together. Lecture 17 Purdue University, Physics 220 18 Bernoulli ACT Through which hole will the water come out fastest? P + +½ 2 P + +½ 2 A 1 ρgy 1 + ½ ρv 1 = 2 ρgy 2 + ½ρv 2 Nt Alth hl h B Note: All three o es ave same pressure P=1 Atmosphere C ρgy 1 + ½ ρv 1 2 = ρgy 2 + ½ρv 2 2 gy 1 + ½ v 1 2 = gy 2 + ½v 2 2 Lecture 17 Purdue University, Physics 220 19 Smaller y gives larger v. Hole C is fastest Example A garden hose with inner diameter 2 cm, carries water at 2.0 m/s. To spray your friend, you place your thumb over the nozzle giving an effective opening diameter of 0.5 cm. What is the speed of the water exiting the hose? What is the pressure difference between inside the hose and outside? Continuity Equation A v =A v 1 1 = A 2 2 v 2 = v 1 ( A 1 /A 2 ) =v(r 2 /r 2 ) Bernoulli Equation = v 1 ( r 1 / r 2 = 2 m/s x 16 = 32 m/s P 1 +ρgy 1 + ½ ρv 1 2 = P 2 +ρgy 2 + ½ρv 2 2 P 1 – P 2 = ½ ρ (v 2 2 – v 1 2 ) Lecture 17 Purdue University, Physics 220 20 = ½ x (1000 kg/m 3 ) (1020 m 2 /s 2 ) = 5.1x10 5 Pa Lift a House Calculate the lift force on a 15 m x 15 m house when a 30 m/s wind (1 29 kg/m 3 ) blows over the top 30 m/s . blows top. P 1 +ρgy 1 + ½ ρv 1 2 = P 2 +ρgy 2 + ½ρv 2 2 P 1 –P 2 = ½ ρ (v 2 2 –v 1 2 ) ½ ( 2 2 )= ρ v 2 –v 1 = ½ (1.29) (30 2 ) N / m 2 = 581 N/ m 2 F=PAF = P = 581 N/ m 2 (15 m)(15 m) = 131,000 N Lecture 17 Purdue University, Physics 220 21 = 29,000 pounds! ACT A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bk h hihb id d hh hb fbucket near t e top, whic ent s ownwar suc t at t e ottom o this pipe even with the other hole, like in the picture below: Through which drain is the water spraying out with the highest gp speed? A) The hole B) Th i e p pe C) Same The speed of the water depends on the pressure at the opening through which the water flows, which depends on the depth of the opening from the surface of the water. Since both the hole and the opening of the downspout are at [the same] depth below the surface, water from the Lecture 17 Purdue University, Physics 220 22 two will be discharged with the same speeds. Quiz 1) Two drinking glasses, A and B, are filled with water to the same depth Glass A has twice the diameter of glass same . Glass the B. Compare the weight of water in the two glasses. A) The weight of water in glass A is greater B) The weight is the same in both the glasses C) The weight of water in glass B is greater Lecture 17 Purdue University, Physics 220 23 Quiz 2) Two drinking glasses, A and B, are filled with water to the same depth Glass A has twice the diameter of glass B . Glass the . Compare the pressure at the bottom of the two glasses. A) The pressure in glass B is greater pressure B) The pressure is greater in glass A C) The pressure is the same in both cases pressure same in cases Lecture 17 Purdue University, Physics 220 24 Real Fluids • Real fluids have frictional forces internal to the fluid and between the fluid and the walls of the and fluid walls container Th ll d i f• ese are ca e v scous orces • These forces depend on the speed of the fluid – They are often negligible for small speeds • In many cases, viscosity is important Lecture 17 25Purdue University, Physics 220 Viscosity • Consider the fluid flowing through a tube • The fluid molecules near the wall of the tub itffthllithexperience strong orces rom the mo ecu es n the wall • The molecules close to the wall move relatively slowly • The motion of each layer in the fluid is strongly influenced by the adjacent layers • Since the adjacent layers are moving at different speeds, each layer exerts a force on the layers next to it • “Thick” fluids experience large forces and have a large viscosity viscosity – Honey is an example • “Thin” fluids have much smaller viscosities – Water is an example Lecture 17 Purdue University, Physics 220 26 Poiseuille’s Law • For a viscous fluid to move through a tube the pressure tube, pressure difference between the ends of the tube must overcome the tube viscous force • The average fluid speed is given gp by Poiseuille’s Law −PP 2 12 • The viscosity of the fluid is given the symbol η η = avg vr L8 • This average velocity can be used to find the average flow rate of the fluid: Q = v avg A Lecture 17 Purdue University, Physics 220 27 Stoke’s Law • The drag force on a spherical object of radius r that moves slowly through a fluid with a viscosity η is fluid viscosity This is the same viscosity as in Poiseuille’s Law πη=− G G drag r6Fv • same viscosity • The negative sign indicates that the drag force is opposite to the velocity velocity • Two things may happen at the free surface of a fluid fluid – Which happens depends on the intermolecular forces • In most liquids, the molecules form the shape giving the smallest possible surface area Lecture 17 28Purdue University, Physics 220 – That shape is a sphere Capillary Pressure • The forces between the liquid’s ll d l l fmolecules an a wall can a so affect the shape of a free liquid surface The s rface of the liq id in a capillar• u of the u capillary tube is not flat • The water molecules are attracted to water are to the walls of the tube and this pulls the water up a small amount at the walls • This is a concave meniscus • The mercury molecules are attracted mercury are more strongly to each other than to the walls • This leads to a convex meniscus Lecture 17 Purdue University, Physics 220 29 Turbulence • When fluid velocities are large, the flow patterns can be very complex and fluctuate with time • This behavior is called turbulence • The picture shows vortices produced in in the baseball’s wake Lecture 17 Purdue University, Physics 220 30 Summary of Concepts • Mass flow rate: A(k/)– ρAv (kg s • Volume flow rate: –Av (m 3 /s) • Continuity: – ρA 1 v 1 = ρA 2 v 2 • Bernoulli: –P 1 + 1/2 ρv 1 2 + ρgh 1 = P 2 + 1/2 ρv 2 2 + ρgh 2 A 1 P 1 A 2 P 2 v 1 v 2 ρ Lecture 17 Purdue University, Physics 220 31 Administrator Microsoft PowerPoint - Lecture17bw [Compatibility Mode]

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