2. Physics 207: Lecture 21, Pg 1 Lecture 21 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, Apr. 8 th  Thursday: Read all of Chapter 16

3. Physics 207: Lecture 21, Pg 2 l In many fluids the bulk modulus is such that we can treat the density as a constant independent of pressure: An 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, pressure 1 atm, i.e. car tires) 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

4. Physics 207: Lecture 21, Pg 3 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

5. Physics 207: Lecture 21, Pg 4 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 1 atm = 0.760 m (of Hg)

6. Physics 207: Lecture 21, Pg 5 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

7. Physics 207: Lecture 21, Pg 6 Archimedes’ Principle: A Eureka Moment 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

8. Physics 207: Lecture 21, Pg 7 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

9. Physics 207: Lecture 21, Pg 8 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.

10. Physics 207: Lecture 21, Pg 9 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

11. Physics 207: Lecture 21, Pg 10 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 Expt. 1Expt. 2

12. Physics 207: Lecture 21, Pg 11 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?

13. Physics 207: Lecture 21, Pg 12 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

14. Physics 207: Lecture 21, Pg 13 Home 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

15. Physics 207: Lecture 21, Pg 14 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 Active Figure

16. Physics 207: Lecture 21, Pg 15 Home 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

17. Physics 207: Lecture 21, Pg 16 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. l  Which cup weighs more? Cup I Cup II ( A) Cup I (B) Cup II (C) the same (D) can’t tell

18. Physics 207: Lecture 21, Pg 17 Home 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

19. Physics 207: Lecture 21, Pg 18 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.

20. Physics 207: Lecture 21, Pg 19 Pascal’s Principle in action: Hydraulics, a force amplifier 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

21. Physics 207: Lecture 21, Pg 20 A1A1 A 10 A2A2 M M dBdB dAdA Exercise 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, how do d A and d B compare? 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

22. Physics 207: Lecture 21, Pg 21 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

23. Physics 207: Lecture 21, Pg 22 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

24. Physics 207: Lecture 21, Pg 23 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

25. Physics 207: Lecture 21, Pg 24 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

26. Physics 207: Lecture 21, Pg 25 l Flow obeys continuity equation Volume flow rate (m 3 /s) Q = A·v is constant along flow tube. Follows from mass conservation if flow is incompressible. Mass flow rate is just  Q (kg/s) 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 A 1 A 2 v 1 v 2

27. Physics 207: Lecture 21, Pg 26  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

28. Physics 207: Lecture 21, Pg 27 l For equal volumes in equal times then ½ the diameter implies ¼ the area so the water has to flow four times as fast. l But if the water is moving 4 times as fast then it has 16 times as much kinetic energy. l 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

29. Physics 207: Lecture 21, Pg 28 Recall the standard work-energy relation W net =  K = K f – K i  Apply this principle to a finite fluid volume of  V and mass  m =  V (here W net is work done on the fluid)  Net work by pressure difference over  x with  x 1 = v 1  t) W net = 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 net = (P 1 – P 2 )  V Cautionary note: This work is not internal to the fluid (it is incompressible). It actually reflects a piston or some other moving barrier off to the left displacing the fluid and a second barrier on the right maintaining the lower pressure. For the left barrier on the left the force is in the direction of displacement and on the right the force will be opposite the displacement. In the chapter 17 we will be using an ideal gas as a working fluid and, if the volume of the piston is fixed, then no work is done Conservation of Energy for Ideal Fluid v1v1 v 1/2

30. Physics 207: Lecture 21, Pg 29 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. and with height variations: Bernoulli Equation  P 1 + ½  v 1 2 +  g y 1 = constant Conservation of Energy for Ideal Fluid P1P1 P2P2

31. Physics 207: Lecture 21, Pg 30 Lecture 21 Question to ponder: Does heavy water (D 2 O) ice sink or float? Question to ponder: Does heavy water (D 2 O) ice sink or float? Assignment Assignment  HW9, due Wednesday, Apr. 8 th  Thursday: Read all of Chapter 16