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Fluids At ordinary temperature, matter exists in one of three states (5 if you include plasma and Bose-Einstein condensate). At ordinary temperature,

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Presentation on theme: "Fluids At ordinary temperature, matter exists in one of three states (5 if you include plasma and Bose-Einstein condensate). At ordinary temperature,"— Presentation transcript:

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2 Fluids At ordinary temperature, matter exists in one of three states (5 if you include plasma and Bose-Einstein condensate). At ordinary temperature, matter exists in one of three states (5 if you include plasma and Bose-Einstein condensate). Solid - has a shape and forms a surface Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surface Gas - has no shape and forms no surface What do we mean by “fluids”? What do we mean by “fluids”? Fluids are “substances that flow”…. “substances that take the shape of the container” Fluids are “substances that flow”…. “substances that take the shape of the container” Atoms and molecules are free to move. Atoms and molecules are free to move. No long range correlation between positions. No long range correlation between positions.

3 Fluids What parameters do we use to describe fluids? What parameters do we use to describe fluids? Density Density units : kg/m 3 = 10 -3 g/cm 3  (water) = 1.000 x 10 3 kg/m 3 = 1.000 g/cm 3  (ice) = 0.917 x 10 3 kg/m 3 = 0.917 g/cm 3  (air) = 1.29 kg/m 3 = 1.29 x 10 -3 g/cm 3  (Hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3 If any substance is more dense than water, it will sink

4 Fluids What parameters do we use to describe fluids? What parameters do we use to describe fluids? Pressure Pressure A n l Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface.  Force (a vector) in a fluid can be expressed in terms of hydrostatic pressure (a scalar) as: units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 10 5 Pa 1 mbar = 10 2 Pa 1 torr = 133.3 Pa 1 atm = 1.013 x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in 2 (=PSI)

5 Practice An 80 kg metal cylinder, 2.0 m long and with each end having an area of 25 cm 2. It stands vertically on one of the ends. What pressure does the cylinder exert on the floor?

6 Solution

7 Practice A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. Waterbed

8 Solution A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. Let’s rearrange for Mass, M Now, we write weight in terms of density, volume and g Water density is 1000 kg/m 3

9 Solution A waterbed is 2m on a side and 30cm deep. Find a) Its weight. b) The pressure the waterbed exerts on the floor. The force is the weight of the bed. Surface area of contact between the floor and the bed. Weight is 1.18 x 10 4 N

10 Specific Gravity An old-fashioned but still very common way of expressing the density of a substance is to relate it to the density of water. The ratio of the density of any substance to the density of water is known as the specific gravity of the substance. If the specific gravity is less than 1, it will float in water, if it is greater than 1, it will sink.

11 Act 3 A cork has a volume of 5 cm 3 and weighs 0.03N. What is the specific gravity of cork?

12 Act 4 Solution A cork has a volume of 5 cm 3 and weighs 0.03N. What is the specific gravity of cork? First let’s find the mass Now for the density Finally the specific gravity

13 l When the pressure is much less than the bulk modulus of the fluid, we treat the density as a constant independent of pressure: 1. LIQUID: incompressible (density almost constant) 2. GAS: compressible (density depends a lot on pressure) l For an incompressible fluid, the density IS the same everywhere, but the pressure is NOT Pressure vs. Depth Incompressible Fluids (liquids) Bulk Modulus The Bulk Modulus of a substance measures the substance’s resistance to uniform compression. It is defined as the pressure increase needed to cause a relative decrease in volume.

14 If the pressures were different, fluid would flow in the tube! However, if fluid did flow, then the system was NOT in equilibrium, since no equilibrium system will spontaneously leave equilibrium. The horizontal forces acting at the left and right side of a cylinder at specific depth are PA L and PA R. Since the cylinder is in equilibrium PA L = PA R Therefore The pressure, P at either side is the same. Pressure vs. Depth For a fluid in an open container: pressure same at a given depth independent of the container fluid level is the same everywhere in a connected container (assuming no surface forces) Why is this so? Why, in equilibrium, does the pressure below the surface depend only on depth? Imagine a tube that would connect two regions at the same depth.

15 ACT 5 1. What happens with two different fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: è Compare the densities of the liquids: A)  1 <  2 B)  1 =  2 C)  1 >  2 I 11 22 dIdI

16 ACT 5 Solution C)  1 >  2 At the depth of the interface, the pressures in each side must be equal. Since there’s more liquid above this depth on the left side, that liquid must be less dense! I 11 22 dIdI p

17 Pressure Measurements: Barometer Invented by Torricelli Invented by Torricelli A long closed tube is filled with mercury and inverted in a dish of mercury A long closed tube is filled with mercury and inverted in a dish of mercury The closed end is nearly a vacuum The closed end is nearly a vacuum Measures atmospheric pressure as Measures atmospheric pressure as One 1 atm = 0.760 m (of Hg)

18 l Due to gravity, the pressure depends on depth in a fluid l Consider an imaginary fluid volume (a cube, each face having area A) è The sum of all the forces on this volume must be ZERO as it is in equilibrium. »There are three vertical forces: n The weight (mg) n The upward force from the pressure on the bottom surface (F 2 ) n The downward force from the pressure on the top surface (F 1 ) n Since the sum of these forces is ZERO, we have: F 2 -F 1 =mg Same fluid) Pressure vs. Depth Incompressible Fluids (liquids)

19 Same fluid Now since Force=pressure × Area Pressure vs. Depth Incompressible Fluids (liquids) Now since Mass=Density × Volume The pressure at the bottom of a vertical column is equal to the pressure at the top of the column (P o, Usually atmospheric), plus the pressure due to all the water in the column of height, h. (density x gravity x height). Volume is area x height

20 GAUGE Pressure One must be careful when we are talking about pressure. Do we want to include atmospheric pressure in our calculations? One must be careful when we are talking about pressure. Do we want to include atmospheric pressure in our calculations? If we want the pressure over and above atmospheric pressure, we have P 0 =0. This is called the Gauge Pressure. If we want the pressure over and above atmospheric pressure, we have P 0 =0. This is called the Gauge Pressure. If we need to include atmospheric pressure we use: P 0 =1x10 5 N/m 2 = 100 kPa If we need to include atmospheric pressure we use: P 0 =1x10 5 N/m 2 = 100 kPa Most pressure gauges register the pressure over and above atmospheric pressure. Most pressure gauges register the pressure over and above atmospheric pressure. For example a tire gauge registers 220 kPa, the actual pressure within the tire is 220 kPa +100 kPa = 320 kPa For example a tire gauge registers 220 kPa, the actual pressure within the tire is 220 kPa +100 kPa = 320 kPa

21 What is the pressure at point A? At point B? At point A: At point B: Gauge Pressure in a Tank Filled with Gasoline and Water 10m 3 m

22 Pascal’s Principle So far we have discovered (using Newton’s Laws): So far we have discovered (using Newton’s Laws): Pressure depends on depth:  P =  g  y Pressure depends on depth:  P =  g  y Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. l Pascal’s Principle explains the working of hydraulic lifts  i.e., the application of a small force at one place can result in the creation of a large force in another.  Will this “hydraulic lever” violate conservation of energy?  No

23 Pascal’s Principle Consider the system shown: Consider the system shown: A downward force F 1 is applied to the piston of area A 1. A downward force F 1 is applied to the piston of area A 1. This force is transmitted through the liquid to create an upward force F 2. This force is transmitted through the liquid to create an upward force F 2. Pascal’s Principle says that increased pressure from F 1, [P=(F 1 /A 1 )], is transmitted throughout the liquid. Pascal’s Principle says that increased pressure from F 1, [P=(F 1 /A 1 )], is transmitted throughout the liquid. Since A 2 is larger than A 1, we have a force multiplier. Yes, but what gives? Free Energy? Where did this extra force come from? Yes, but what gives? Free Energy? Where did this extra force come from?

24 Pascal’s Principle Displaced volumes are the same, so… Therefore energy is conserved Therefore energy is conserved With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force applied over a smaller distance. Let’s check the if the work done by F 1 equal the work done by F 2

25 ACT In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm 2. The cylinder on the right has a cross sectional area of 10.0 cm 2 a)Determine the amount of weight (F) you must apply on the right side of the system to hold the system in equilibrium (Stop the left side from moving down). b)If the left cylinder is pushed down 5.0cm, determine the distance F will move. c)Determine the mechanical advantage of the system

26 ACT Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm 2. The cylinder on the right has a cross sectional area of 10.0 cm 2 a)Determine the weight (F) required to hold the system in equilibrium b)If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will move. c)Determine the mechanical advantage of the system Apply Pascal’s Principle for F 1

27 ACT 7 Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 a)Determine the weight (F) required to hold the system in equilibrium b)If the left cylinder is pushed down 5.0cm, determine the distance the right cylinder will move.. c)Determine the mechanical advantage of the system The volume of fluid displaced by the 100 cm 2 cylinder equals the change in the volume of fluid in the right hand cylinder or

28 ACT 7 Solutions In the hydraulic system shown below in the diagram, the 200kg cylinder has a cross sectional area of 100 cm2. The cylinder on the right has a cross sectional area of 10.0 cm2 a)Determine the weight (F) required to hold the system in equilibrium b)If the left cylinder is pushed down 5.0cm, determine the distance F will move. c)Determine the mechanical advantage of the system Determine the Ideal Mechanical Advantage (IMA) of the system. IMA=(output Force/ Input Force)

29 Archimedes’ Principle Suppose we weigh an object in air (1) and in water (2). Suppose we weigh an object in air (1) and in water (2). How do these weights compare? How do these weights compare? W2?W2? W1W1 W 1 < W 2 W 1 = W 2 W 1 > W 2  Why? Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.

30 Archimedes’ Principle W2?W2? W1W1 l The buoyant force is equal to the difference in the pressures times the area. Therefore, the buoyant force is equal to the weight of the fluid displaced.

31 A 60 kg rock lies at the bottom of a pool. Its volume is 3.0 x 10 4 cm 3. What is its apparent weight? The buoyant force on the rock due to the water is equal to the weight of 3.0 x 10 4 cm 3 = 3.0 x 10 -2 m 3 of water: The weight of the rock is: This is as if the rock had a mass of 31kg Archimedes’ Principle

32 Sink or Float? The buoyant force is equal to the weight of the liquid that is displaced. The buoyant force is equal to the weight of the liquid that is displaced. If the buoyant force of a fully submerged object is larger than the weight of the object, it will float; otherwise it will sink. If the buoyant force of a fully submerged object is larger than the weight of the object, it will float; otherwise it will sink. Fmg B y l We can calculate how much of a floating object will be submerged in the liquid: è Object is in equilibrium

33 Sink of Float? The Tip of The Iceberg: What fraction of an iceberg is submerged? If the Density of ice is 917 kg/m 3 and the density of water is 1024 kg/m 3 Fmg B y Object is in equilibrium

34 Archimedes Archimedes is said to have discovered his principle in his bath while thinking how he might determine whether the king’s new crown was pure gold or fake. Gold has a specific gravity of 19.3, somewhat higher than most metals, but a determination of specific gravity or density is not easy with a irregularly shaped object. Archimedes realised that if the crown is weighed in air and weighed in water, the density can de determined.

35 Archimedes A 14.7 kg crown has an apparent weight of 13.4 kg when submerged in water. Is it gold? The apparent weight of the submerged crown, w`, equals its actual weight, w, minus the buoyant force, F B. We can find the specific gravity : This corresponds to a density of 11,300 kg/m 3 (that of lead) Alternate solution

36 Let’s find the volume of the displaced water (this is also the volume of the crown) We can now determine the density of the crown

37 Practice A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the Styrofoam block as shown. A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the Styrofoam block as shown. If you turn the Styrofoam + Pb upside down, what happens? If you turn the Styrofoam + Pb upside down, what happens? styrofoam Pb A) It sinks C) B) D) styrofoam Pb

38 Solution styrofoam Pb l If the object floats right-side up, then it also must float upside-down. l It displaces the same amount of water in both cases l However, when it is upside-down, the Pb displaces some water. l Therefore the styrofoam must displace less water than it did when it was right-side up (when the Pb displaced no water). A) It sinks C) B) styrofoam Pb D)

39 Question Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle- ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water Weight of ship = Buoyant force =Weight of displaced water 15

40 Act 8 Suppose you float a large ice-cube in a glass of water, and 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 out of the glass. 2. Go down. 3. Stay the same. F B =  Water g V displaced F B = Weight of ice =  ice g V ice   W g V melted_ice Must be same!

41 Example Problems At what depth is the water pressure two atmospheres? (It is one atmosphere at the surface. 1.01  10 5 Pa ) What is the pressure at the bottom of the deepest oceanic trench (about 10 4 meters)? Solution: For d = 10 4 m : This assumes that water is incompressible. If water were compressible, would the pressure at the bottom of the ocean be greater or smaller than the result of this calculation?

42 Have you ever tried to submerge a beach ball (r = 50 cm) in a swimming pool? It’s difficult. How big a downward force must you exert to get it completely underwater? Solution: I’m ignoring the weight of the beach ball. The force is the weight of a 523 kg object. Example Problems (2) We are displacing this much water

43 More Fun With Buoyancy Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. Which cup weighs more? è Archimedes principle tells us that the cups weigh the same. è Each plastic ball displaces an amount of water that is exactly equal to its own weight. Cup I Cup II

44 Still More Fun! A plastic ball floats in a cup of water with half of its volume submerged. Oil (  oil <  ball <  water ) is slowly added to the container until it just covers the ball. Relative to the water level, the ball moves up. Why? è For oil to cover the ball, the ball must have “displaced” some water. è Therefore, the buoyant force on the ball increases. è Therefore, the ball moves up (relative to the water). è Note that we assume the buoyant force of the air on the ball is negligible (it is!); the buoyant force of the oil is not. water oil

45 Archimedes Summary Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. If the weight of the water displaced is less than the weight of the object (buoyant force is less than the weight of object), the object will sink. Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.

46 Fluids in Motion Up to now we have described fluids in terms of their static properties: Up to now we have described fluids in terms of their static properties: Density  Density  Pressure P Pressure P To describe fluid motion, we need something that can describe flow: To describe fluid motion, we need something that can describe flow: Velocity v Velocity v l There are different kinds of fluid flow of varying complexity non-steady / steady compressible / incompressible rotational / irrotational viscous / ideal

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

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

49 Onset of Turbulent Flow The SeaWifS satellite image of a von Karman vortex around Guadalupe Island, August 20, 1999

50 l Simplest situation: consider ideal fluid moving with steady flow - velocity at each point in the flow is constant in time l In this case, fluid moves on streamlines streamline Ideal Fluids Fluid dynamics is very complicated in general (turbulence, vortices, etc.) Fluid dynamics is very complicated in general (turbulence, vortices, etc.) Consider the simplest case first: the Ideal Fluid Consider the simplest case first: the Ideal Fluid No “viscosity” - no flow resistance (no internal friction) No “viscosity” - no flow resistance (no internal friction) Incompressible - density constant in space and time Incompressible - density constant in space and time

51 l Flow obeys continuity equation Volume flow rate Q = A·v is constant along flow tube. Follows from mass conservation if flow is incompressible. A 1 v 1 = A 2 v 2 Ideal Fluids Streamlines do not meet or cross Streamlines do not meet or cross Velocity vector is tangent to streamline Velocity vector is tangent to streamline Volume of fluid follows a tube of flow bounded by streamlines Volume of fluid follows a tube of flow bounded by streamlines Streamline density is proportional to velocity Streamline density is proportional to velocity

52 The Equation of Continuity Q: Have you ever used your thumb to control the water flowing from the end of a hose? A: When the end of a hose is partially closed off, thus reducing its cross-sectional area, the fluid velocity increases. This kind of fluid behavior is described by the equation of continuity.

53 Continuity A four-lane highway merges down to a two-lane highway. The officer in the police 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? A) 15 mph B) 30 mph C) 60 mph All cars stay on the road… Must pass motorcycle too Must go faster, else pile up!

54  Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1cm diameter pipe, how fast is the water going in the ½ cm pipe? Exercise Exercise A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. v1v1 v 1/2 (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1

55 Continuity of Fluid Flow Watch “plug” of fluid moving through the narrow part of the tube (A 1 ) Time for “plug” to pass point  t = x 1 / v 1 Mass of fluid in “plug” m 1 =  Vol 1 =  A 1 x 1 or m 1 =  A 1 v 1  t Watch “plug” of fluid moving through the wide part of the tube (A 2 ) Time for “plug” to pass point  t = x 2 / v 2 Mass of fluid in “plug” m 2 =  Vol 2 =  A 2 x 2 or m 2 =  A 2 v 2  t Continuity Equation says m 1 = m 2 fluid isn’t appearing or disappearing A 1 v 1 = A 2 v 2 Fluid In= Fluid Out

56 Faucet Preflight A stream of water gets narrower as it falls from a faucet (try it & see). Explain this phenomenon using the equation of continuity A1A1 A2A2 V1V1 V2V2 “Velocity and area are inversely proportional. Velocity increases, so area decreases.” “Since the area inside the faucet is large, the velocity of the water will be low. Due to the equation of continuity, the area of the water has to be lower due to the velocity increasing outside of the faucet. ”

57 Fluid Flow Concepts Continuity:  A 1 v 1 =  A 2 v 2 Mass flow rate:  Av (kg/s) Volume flow rate: Av (m 3 /s) i.e., mass flow rate the same everywhere e.g., flow of river 24 Since we have an ideal fluid (incompressible), then we can simply use:

58 Pressure, Flow and Work Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening Continuity Equation says fluid speeds up going to smaller opening, slows down going to larger opening But, if the fluid is moving faster, then it has more kinetic energy. But, if the fluid is moving faster, then it has more kinetic energy. Therefore something did work on the fluid to accelerate it (apply a force) Therefore something did work on the fluid to accelerate it (apply a force) Acceleration due to change in pressure. P 1 > P 2 Acceleration due to change in pressure. P 1 > P 2 Smaller tube has faster water and LOWER pressure Smaller tube has faster water and LOWER pressure Change in pressure does work! Change in pressure does work! W = F 1  x 1 –F 2  x 2 = P 1 A 1  x 1 - P 2 A 2  x 2 = (P 1 – P 2 )V W = F 1  x 1 –F 2  x 2 = P 1 A 1  x 1 - P 2 A 2  x 2 = (P 1 – P 2 )V In essence, Bernoulli’s Principle states that where the velocity of a fluid is high, the pressure is low and where the velocity is low, the pressure is high.

59 Pressure ACT What will happen when I “blow” air between the two plates? What will happen when I “blow” air between the two plates? A) Move Apart B) Come Together C) Nothing A) Move Apart B) Come Together C) Nothing

60 Bernoulli’s Equation For steady flow, the speed, pressure, and elevation of an incompressible and nonviscous fluid are related by an equation discovered by Daniel Bernoulli (1700–1782). It is the application of the conservation of energy. P is the pressure of the fluid at the specified spot. y is the vertical height at the specified spot, v is the velocity at the specified spot.

61 Pressure, Flow and Work Work done = Gain in KE + Gain in PE or Work is done by the difference in pressure This is the work done to move the coloured fluid from left to right Fluid to left gives pressure.

62 Bernoulli’s Equation External Pressure Kinetic Energy per Volume Potential energy per unit Volume + + This sum has the same value at all points along a streamline. The Fluid is pushed from Point 1 to Point 2

63 Question The container below is filled to a height h with liquid of density ρ. The atmospheric pressure is P atm, determine the pressure of the fluid at each labelled point. Compare the pressure at point 3 with that at point 5 with A 3 =1/2A 5 h 12 3 45 (Cross sectional area are the same, v 1 =v 2 =v 3, height are the same, h 1 =h 2 =h 3 ) Let’s see how

64 Question The container below is filled to a height h with liquid of density ρ. The atmospheric pressure is P atm, determine the pressure of the fluid at each labelled point. Compare the pressure at point 3 with that at point 5 with A 3 =1/2A 5 h 12 3 45 (Cross sectional area are the same, v 1 =v 2 =v 3, height are the same, h 1 =h 2 =h 3 )

65 Bernoulli ACT Through which hole will the water come out fastest? Through which hole will the water come out fastest? A B C P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v 2 2 Note: Outside all three holes the external pressure is the same: P=1 Atm (the ρgy term accounts for the pressure of water in column) y 2 > y 1  v 1 > v 2 Smaller y gives larger v. Hole C is fastest y2y2 y1y1

66  For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast.  But if the water is moving four times as fast the it has 16 times as much kinetic energy. Something must be doing work on the water (the pressure drops at the neck and we recast the work as W=P  V = (F/A) (A  x) = F  x ) Exercise A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. v1v1 v 1/2 (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1

67 Practice 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 bucket near the top, which bent is downward such that the bottom of this pipe is even with the other hole, like in the picture below: Though which drain is the water spraying out with the highest speed? 1. The hole 2. The pipe 3. Same Speed is determined by pressure difference where water meets the atmosphere. Relevant height is where the hole is. Both are exiting at the same height!

68 Example A garden hose w/ 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? Bernoulli Equation P 1 +  gy 1 + ½  v 1 2 = P 2 +  gy 2 + ½  v 2 2 P 1 – P 2 = ½  (v 2 2 – v 1 2 ) = ½  (1000 kg/m 3 ) (1024 m 2 /s 2 – 4 m 2 /s 2 ) = 5.1  10 5 Pa Continuity Equation A 1 v 1 = A 2 v 2 v 2 = v 1 ( A 1 /A 2 ) = v 1 ( π r 1 2 / π r 2 2 ) = 2 m/s  16 = 32 m/s

69 Bernoulli’s Principle A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. A housing contractor saves some money by reducing the size of a pipe from 1cm diameter to 1/2cm diameter at some point in your house. 1) What is the pressure in the 1/2cm pipe relative to the 1cm pipe? (A) smaller(B) same(C) larger v1v1 v 1/2

70 Applications of Bernoulli's Equation The tarpaulin that covers the cargo is flat when the truck is stationary but bulges outward when the truck is moving.

71 Household Plumbing In a household plumbing system, a vent is necessary to equalize the pressures at points A and B, thus preventing the trap from being emptied. An empty trap allows sewer gas to enter the house.

72 Curveball Pitch

73 Applications of Fluid Dynamics Streamline flow around a moving airplane wing Streamline flow around a moving airplane wing Lift is the upward force on the wing from the air Lift is the upward force on the wing from the air Drag is the resistance 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 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 higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

74 The constriction in the Subclavian artery causes the blood in the region to speed up and thus produces low pressure. The blood moving UP the LVA is then pushed DOWN instead of down causing a lack of blood flow to the brain. This condition is called TIA (transient ischemic attack) or “Subclavian Steal Syndrome. Applications of Fluid Dynamics

75 Flash Files Pressure Buoyancy Fluid Flow Bernoulli’s Equation

76 Summary These are the formulas that you need to know:

77 Bernoulli’s Equation: Continuity Equation: Av is flow rate Pascal’s Principle : Archimedes’ Principle:


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