2. Physics 207: Lecture 20, Pg 1 Chapter 15, Fluids l This is an actual photo of an iceberg, taken by a rig manager for Global Marine Drilling in St. Johns, Newfoundland. The water was calm and the sun was almost directly overhead so that the diver

3. Physics 207: Lecture 20, Pg 2 Physics 207, Lecture 20, Nov. 10 Goals: Chapter 15 Chapter 15  Understand pressure in liquids and gases  Use Archimedes’ principle to understand buoyancy  Understand the equation of continuity  Use an ideal-fluid model to study fluid flow.  Investigate the elastic deformation of solids and liquids Assignment Assignment  HW9, Due Wednesday, Nov. 19 th  Wednesday: Read all of Chapter 16

4. Physics 207: Lecture 20, Pg 3 Fluids (Ch. 15) l At ordinary temperature, matter exists in one of three states  Solid - has a shape and forms a surface  Liquid - has no shape but forms a surface  Gas - has no shape and forms no surface l What do we mean by “fluids”?  Fluids are “substances that flow”…. “substances that take the shape of the container”  Atoms and molecules are free to move.  No long range correlation between positions.

5. Physics 207: Lecture 20, Pg 4 Fluids l An intrinsic parameter of a fluid  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

6. Physics 207: Lecture 20, Pg 5 Fluids 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 pressure (a scalar) as: l Another parameter: Pressure

7. Physics 207: Lecture 20, Pg 6 What is the SI unit of pressure? A. Pascal B. Atmosphere C. Bernoulli D. Young E. p.s.i. 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)

8. Physics 207: Lecture 20, Pg 7 l When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid l For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! p(y) = p 0 - y g  p 0 + d g  l Gauge pressure (subtract p 0, usually 1 atm) Pressure vs. Depth Incompressible Fluids (liquids) F 2 = F 1 + m g = F 1 +  Vg F 2 /A = F 1 /A +  Vg/A p 2 = p 1 -  g y

9. Physics 207: Lecture 20, Pg 8 Pressure vs. Depth l For a uniform fluid in an open container pressure same at a given depth independent of the container l Fluid level is the same everywhere in a connected container, assuming no surface forces

10. Physics 207: Lecture 20, Pg 9 Pressure Measurements: Barometer l Invented by Torricelli l A long closed tube is filled with mercury and inverted in a dish of mercury  The closed end is nearly a vacuum l Measures atmospheric pressure as One 1 atm = 0.760 m (of Hg)

11. Physics 207: Lecture 20, Pg 10 Exercise Pressure l What happens with two fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: Compare the densities of the liquids: (A)  1 <  2 (B)  1 =  2 (C)  1 >  2 11 22 dIdI

12. Physics 207: Lecture 20, Pg 11 Exercise Pressure l What happens with two fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: At the red arrow the pressure must be the same on either side.  1 x =  2 (d 1 + y)  Compare the densities of the liquids: (A)  1 <  2 (B)  1 =  2 (C)  1 >  2 11 22 dIdI y

13. Physics 207: Lecture 20, Pg 12 Archimedes’ Principle l Suppose we weigh an object in air (1) and in water (2). How do these weights compare? W2?W2? W1W1 W 1 < W 2 W 1 = W 2 W 1 > W 2 l Buoyant force is equal to the weight of the fluid displaced

14. Physics 207: Lecture 20, Pg 13 The Golden Crown l In the first century BC the Roman architect Vitruvius related a story of how Archimedes uncovered a fraud in the manufacture of a golden crown commissioned by Hiero II, the king of Syracuse. The crown (corona in Vitruvius’s Latin) would have been in the form of a wreath, such as one of the three pictured from grave sites in Macedonia and the Dardanelles. Hiero would have placed such a wreath on the statue of a god or goddess. Suspecting that the goldsmith might have replaced some of the gold given to him by an equal weight of silver, Hiero asked Archimedes to determine whether the wreath was pure gold. And because the wreath was a holy object dedicated to the gods, he could not disturb the wreath in any way. (In modern terms, he was to perform nondestructive testing). Archimedes’ solution to the problem, as described by Vitruvius, is neatly summarized in the following excerpt from an advertisement:VitruviusHiero IIadvertisement l The solution which occurred when he stepped into his bath and caused it to overflow was to put a weight of gold equal to the crown, and known to be pure, into a bowl which was filled with water to the brim. Then the gold would be removed and the king’s crown put in, in its place. An alloy of lighter silver would increase the bulk of the crown and cause the bowl to overflow. l From http://www.math.nyu.edu/~crorres/Archimedes/Crown/CrownIntro.html

15. Physics 207: Lecture 20, Pg 14 Archimedes’ Principle l Suppose we weigh an object in air (1) and in water (2).  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.

16. Physics 207: Lecture 20, Pg 15 Sink or Float? l The buoyant force is equal to the weight of the liquid that is displaced. l If the buoyant force 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:  Object is in equilibrium

17. Physics 207: Lecture 20, Pg 16 Bar Trick piece of rock on top of ice What happens to the water level when the ice melts? A. It rises B. It stays the same C. It drops

18. Physics 207: Lecture 20, Pg 17 Exercise V 1 = V 2 = V 3 = V 4 = V 5 m 1 < m 2 < m 3 < m 4 < m 5 What is the final position of each block?

19. Physics 207: Lecture 20, Pg 18 Exercise V 1 = V 2 = V 3 = V 4 = V 5 m 1 < m 2 < m 3 < m 4 < m 5 What is the final position of each block? Not this But this

20. Physics 207: Lecture 20, Pg 19 Exercise Buoyancy l A small 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? styrofoam Pb (A) It sinks (C)(B)(D) styrofoam Pb Active Figure

21. Physics 207: Lecture 20, Pg 20 Exercise Buoyancy l A small 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 (assuming density of Pb > water)? styrofoam Pb (A) It sinks (C)(B)(D) styrofoam Pb

22. Physics 207: Lecture 20, Pg 21 Exercise More Buoyancy l Two identical cups are filled to the same level with water. One of the two cups has plastic balls floating in it.  Which cup weighs more? Cup I Cup II ( A) Cup I (B) Cup II (C) the same (D) can’t tell

23. Physics 207: Lecture 20, Pg 22 Exercise More Buoyancy l Two identical cups are filled to the same level with water. One of the two cups has plastic balls floating in it.  Which cup weighs more? Cup I Cup II ( A) Cup I (B) Cup II (C) the same (D) can’t tell

24. Physics 207: Lecture 20, Pg 23 Exercise Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (  oil <  ball <  water ) is slowly added to the container until it just covers the ball.  Relative to the water level, the ball will: Hint 1: What is the buoyant force of the part in the oil as compared to the air? water oil (A) move up (B) move down (C) stay in same place

25. Physics 207: Lecture 20, Pg 24 Exercise Even More Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (  oil <  ball <  water ) is slowly added to the container until it just covers the ball.  Relative to the water level, the ball will: Hint 1: What is the buoyant force of the part in the oil as compared to the air? water oil (A) move up (B) move down (C) stay in same place

26. Physics 207: Lecture 20, Pg 25 Pascal’s Principle l So far we have discovered (using Newton’s Laws):  Pressure depends on depth:  p =  g  y l 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.

27. Physics 207: Lecture 20, Pg 26 Pascal’s Principle in action l Consider the system shown:  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.  Pascal’s Principle says that increased pressure from F 1 (F 1 /A 1 ) is transmitted throughout the liquid. l F 2 > F 1 with conservation of energy

28. Physics 207: Lecture 20, Pg 27 A1A1 A 10 A2A2 M M dBdB dAdA Exercise Hydraulics: A force amplifier Aka…the lever l Consider the systems shown on right.  In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d i in the liquid levels.  If A 2 = 2  A 1, compare d A and d B V 10 = V 1 = V 2 d A A 1 = d B A 2 d A A 1 = d B 2A 1 d A = d B 2

29. Physics 207: Lecture 20, Pg 28 A1A1 A2A2 M d1d1 Home Example Hydraulics: A force amplifier Aka…the lever l Consider the system shown on right.  Blocks of mass M are placed on the piston of both cylinders.  The small & large cylinders displace distances d 1 and d 2  Compare the forces on disks A 1 & A 2 ignoring the mass and weight of the fluid in this process W 2 = - M g d 2 = - F 2 d 2 M W 1 = M g d 1 = F 1 d 1 W 1 + W 2 = 0 = F 1 d 1 - F 2 d 2 F 2 d 2 = F 1 d 1 F 2 = F 1 d 1 / d 2 = F 1 A 2 / A 1 V 1 = V 2 = A 1 d 1 = A 2 d 2 d 1 / d 2 = A 2 / A 1 implying P 1 = P 2 = F 1 /A 1 = F 2 /A 2

30. Physics 207: Lecture 20, Pg 29 Fluids in Motion l To describe fluid motion, we need something that describes flow:  Velocity v l There are different kinds of fluid flow of varying complexity  non-steady / steady ç compressible / incompressible ç rotational / irrotational ç viscous / ideal

31. Physics 207: Lecture 20, Pg 30 Types of Fluid Flow l Laminar 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 l Turbulent flow  An irregular flow characterized by small whirlpool like regions  Turbulent flow occurs when the particles go above some critical speed

32. Physics 207: Lecture 20, Pg 31 Types of Fluid Flow l Laminar 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 l Turbulent flow  An irregular flow characterized by small whirlpool like regions  Turbulent flow occurs when the particles go above some critical speed

33. Physics 207: Lecture 20, Pg 32 Onset of Turbulent Flow The SeaWifS satellite image of a von Karman vortex around Guadalupe Island, August 20, 1999

34. Physics 207: Lecture 20, Pg 33 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 l Fluid dynamics is very complicated in general (turbulence, vortices, etc.) l Consider the simplest case first: the Ideal Fluid  No “viscosity” - no flow resistance (no internal friction)  Incompressible - density constant in space and time

35. Physics 207: Lecture 20, Pg 34 l Flow obeys continuity equation Volume flow rate Q = A·v is constant along flow tube. Follows from mass conservation if flow is incompressible. streamline A 1 v 1 = A 2 v 2 Ideal Fluids l Streamlines do not meet or cross l Velocity vector is tangent to streamline l Volume of fluid follows a tube of flow bounded by streamlines l Streamline density is proportional to velocity

36. Physics 207: Lecture 20, Pg 35  Assuming the water moving in the pipe is an ideal fluid, relative to its speed in the 1” diameter pipe, how fast is the water going in the 1/2” pipe? Exercise Continuity l A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” 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

37. Physics 207: Lecture 20, Pg 36  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 then 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 P  V = (F/A) (A  x) = F  x ) Exercise Continuity v1v1 v 1/2 (A) 2 v 1 (B) 4 v 1 (C) 1/2 v 1 (D) 1/4 v 1

38. Physics 207: Lecture 20, Pg 37 Lecture 20, Nov. 10 Question to ponder: Does heavy water (D 2 0) ice sink or float? Question to ponder: Does heavy water (D 2 0) ice sink or float? Assignment Assignment  HW9, Due Wednesday, Nov. 19 th  Wednesday: Read all of Chapter 16 Next slides are possible for Wednesday

39. Physics 207: Lecture 20, Pg 38 Recall the standard work-energy relation W =  K = K f – K i  Apply the principle to a section of flowing fluid with volume  V and mass  m =  V (here W is work done on fluid)  Net work by pressure difference over  x (  x 1 = v 1  t) W = F 1  x 1 – F 2  x 2 = (F 1 /A 1 ) (A 1  x 1 ) – (F 2 /A 2 ) (A 2  x 2 ) = P 1  V 1 – P 2  V 2 and  V 1 =  V 2 =  V (incompressible) W = (P 1 – P 2 )  V VV Conservation of Energy for Ideal Fluid

40. Physics 207: Lecture 20, Pg 39 W = (P 1 – P 2 )  V and W = ½  m v 2 2 – ½  m v 1 2 = ½ (  V) v 2 2 – ½ (  V)  v 1 2 (P 1 – P 2 ) = ½  v 2 2 – ½  v 1 2 P 1 + ½  v 1 2 = P 2 + ½  v 2 2 = const. Bernoulli Equation  P 1 + ½  v 1 2 +  g y 1 = constant VV Conservation of Energy for Ideal Fluid

41. Physics 207: Lecture 20, Pg 40 This leads to… P 1 + ½  v 1 2 = P 2 + ½  v 2 2 = const. and with height variations Bernoulli’s Equation  P 1 + ½  v 1 2 +  g y 1 = constant VV Conservation of Energy for Ideal Fluid

42. Physics 207: Lecture 20, Pg 41 Bernoulli’s Principle l A housing contractor saves some money by reducing the size of a pipe from 1” diameter to 1/2” diameter at some point in your house. 2) What is the pressure in the 1/2” pipe relative to the 1” pipe? (A) smaller(B) same(C) larger v1v1 v 1/2

43. Physics 207: Lecture 20, Pg 42 Cavitation In the vicinity of high velocity fluids, the pressure can gets so low that the fluid vaporizes. Venturi result

44. Physics 207: Lecture 20, Pg 43 Applications of Fluid Dynamics l Streamline flow around a moving airplane wing l Lift is the upward force on the wing from the air l Drag is the resistance l 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.

45. Physics 207: Lecture 20, Pg 44 l Young’s modulus: measures the resistance of a solid to a change in its length. l Bulk modulus: measures the resistance of solids or liquids to changes in their volume. Some definitions L0L0 LL F V0V0 V 0 -  V F l Elastic properties of solids : elasticity in length volume elasticity

46. Physics 207: Lecture 20, Pg 45

47. Physics 207: Lecture 20, Pg 46 EXAMPLE 15.11 An irrigation system QUESTION:

48. Physics 207: Lecture 20, Pg 47 EXAMPLE 15.11 An irrigation system

49. Physics 207: Lecture 20, Pg 48 EXAMPLE 15.11 An irrigation system

50. Physics 207: Lecture 20, Pg 49 EXAMPLE 15.11 An irrigation system

51. Physics 207: Lecture 20, Pg 50 EXAMPLE 15.11 An irrigation system

52. Physics 207: Lecture 20, Pg 51 Elasticity

53. Physics 207: Lecture 20, Pg 52 Elasticity F/A is proportional to ΔL/L. We can write the proportionality as The proportionality constant Y is called Young’s modulus. The quantity F/A is called the tensile stress. The quantity ΔL/L, the fractional increase in length, is called strain. With these definitions, we can write

54. Physics 207: Lecture 20, Pg 53 EXAMPLE 15.13 Stretching a wire QUESTIONS:

55. Physics 207: Lecture 20, Pg 54 EXAMPLE 15.13 Stretching a wire

56. Physics 207: Lecture 20, Pg 55 EXAMPLE 15.13 Stretching a wire

57. Physics 207: Lecture 20, Pg 56 Volume Stress and the Bulk Modulus

58. Physics 207: Lecture 20, Pg 57 Volume Stress and the Bulk Modulus A volume stress applied to an object compresses its volume slightly. The volume strain is defined as ΔV/V, and is negative when the volume decreases. Volume stress is the same as the pressure. where B is called the bulk modulus. The negative sign in the equation ensures that the pressure is a positive number.

59. Physics 207: Lecture 20, Pg 58 The figure shows volume flow rates (in cm 3 /s) for all but one tube. What is the volume flow rate through the unmarked tube? Is the flow direction in or out? A. 1 cm 3 /s, in B. 1 cm 3 /s, out C. 10 cm 3 /s, in D. 10 cm 3 /s, out E. It depends on the relative size of the tubes.

60. Physics 207: Lecture 20, Pg 59 Rank in order, from highest to lowest, the liquid heights h 1 to h 4 in tubes 1 to 4. The air flow is from left to right. A. h 1 > h 2 = h 3 = h 4 B. h 2 > h 4 > h 3 > h 1 C. h 2 = h 3 = h 4 > h 1 D. h 3 > h 4 > h 2 > h 1 E. h 1 > h 3 > h 4 > h 2