2. Physics 207: Lecture 21, Pg 1 Physics 207, Lecture 21, Nov. 12 Goals: Chapter 15 Chapter 15  Use an ideal-fluid model to study fluid flow.  Investigate the elastic deformation of solids and liquids Chapter 16 Chapter 16  Recognize and use the state variables that characterize macroscopic phenomena.  Understand the idea of phase change and interpret a phase diagram.  Use the ideal-gas law.  Use pV diagrams for ideal-gas processes. Assignment Assignment  HW9, Due Wednesday, Nov. 19 th  Monday: Read all of Chapter 17

3. Physics 207: Lecture 21, Pg 2 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

4. Physics 207: Lecture 21, Pg 3  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

5. Physics 207: Lecture 21, Pg 4  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

6. Physics 207: Lecture 21, Pg 5 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 next chapter 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

7. Physics 207: Lecture 21, Pg 6 Human circulation: Vorp et al. in Computational Modeling of Arterial Biomechanics l This (plaque) is a serious situation, because stress concentration within the plaque region increases the probability of plaque rupture, which can lead to a sudden, catastrophic blockage of blood flow. As atherosclerosis progresses, the buildup of plaque can lead to a stenosis, or partial blockage, of the arterial lumen. Blood flowing through a stenosis experiences a pressure decrease due to the Bernoulli effect, which can cause local collapse of the artery and further stress concentration within the artery wall.

8. Physics 207: Lecture 21, Pg 7 Cavitation In the vicinity of high velocity fluids, the pressure can gets so low that the fluid vaporizes.

9. Physics 207: Lecture 21, Pg 8 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 VV Conservation of Energy for Ideal Fluid

10. Physics 207: Lecture 21, Pg 9 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

11. Physics 207: Lecture 21, Pg 10 Torcelli’s Law l The flow velocity v = (2gh) ½ where h is the depth from the top surface P +  g h + ½  v 2 = const A B P 0 +  g h + 0 = P 0 +  + ½  v 2 2g h = v 2 P 0 = 1 atm A B

12. Physics 207: Lecture 21, Pg 11 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 l In truth Bernoulli’s relationship has a minor impact. Momentum transfer of the air with the wing is primary higher velocity lower pressure lower velocity higher pressure Note: density of flow lines reflects velocity, not density. We are assuming an incompressible fluid.

13. Physics 207: Lecture 21, Pg 12 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

14. Physics 207: Lecture 21, Pg 13

15. Physics 207: Lecture 21, Pg 14 Unusual properties of water If 4 ° C water starts to freeze in a closed, rigid container what is the net pressure that develops just before it freezes? B = 0.2 x 10 10 N/m 2 and  V / V 0 = -0.0001 l 0.2 x 10 10 N/m 2 = P / 0.0001 l 2 x 10 5 N/m 2 = P = 2 atm l Note: Ice B = 9 x 10 9 N/m 2 and the density is 920 Kg/m 3 P = 0.08 x 9 x 10 9 N/m 2 or 7 x 10 8 N/m 2 = 7000 atm (which quite sufficient to burst a brittle metal container)

16. Physics 207: Lecture 21, Pg 15 Fluids: A tricky problem l A beaker contains a layer of oil (green) with density ρ2 floating on H2O (blue), which has density ρ3. A cube wood of density ρ1 and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure. l What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water? l Hint: The magnitude of the buoyant force (directed upward) must exactly equal the magnitude of the gravitational force (directed downward). The buoyant force depends on d. The total buoyant force has two contributions, one from each of the two different fluids. Split this force into its two pieces and add the two buoyant forces to find the total force

17. Physics 207: Lecture 21, Pg 16 Fluids: A tricky problem l A beaker contains a layer of oil (green) with density ρ 2 floating on H2O (blue), which has density ρ 3. A cube wood of density ρ 1 and side length L is lowered, so as not to disturb the layers of liquid, until it floats peacefully between the layers, as shown in the figure. l What is the distance d between the top of the wood cube (after it has come to rest) and the interface between oil and water? l Soln: (ρ 1 - ρ 3 ) L = ( ρ 2 - ρ 3 ) d For Monday

18. Physics 207: Lecture 21, Pg 17 Thermodynamics: A macroscopic description of matter l Recall “3” Phases of matter: Solid, liquid & gas l All 3 phases exist at different p,T conditions l Triple point of water: p = 0.06 atm T = 0.01°C l Triple point of CO 2 : p = 5 atm T = -56°C

19. Physics 207: Lecture 21, Pg 18 Modern Definition of Kelvin Scale l The temperature of the triple point on the Kelvin scale is 273.16 K l Therefore, the current definition of the Kelvin is defined as 1/273.16 of the temperature of the triple point of water The triple point of water occurs at 0.01 o C and 4.58 mm (0.06 atm) of Hg

20. Physics 207: Lecture 21, Pg 19 Special system: Water l Most liquids increase in volume with increasing T  Water is special  Density increases from 0 to 4 o C !  Ice is less dense than liquid water at 4 o C: hence it floats  Water at the bottom of a pond is the denser, i.e. at 4 o C Water has its maximum density at 4 degrees.  (kg/m 3 ) T ( o C) l Reason: Alignment of water molecules

21. Physics 207: Lecture 21, Pg 20 Exercise l Not being a great athlete, and having lots of money to spend, Bill Gates decides to keep the pool in his back yard at the exact temperature which will maximize the buoyant force on him when he swims. Which of the following would be the best choice? (A)0 o C (B) 4 o C (D) 32 o C(D) 100 o C (E) 212 o C

22. Physics 207: Lecture 21, Pg 21 Temperature scales l Three main scales 212 Farenheit 100 Celcius 320273.15 373.15 Kelvin Water boils Water freezes 0-273.15-459.67 Absolute Zero

23. Physics 207: Lecture 21, Pg 22 Some interesting facts l In 1724, Gabriel Fahrenheit made thermometers using mercury. The zero point of his scale is attained by mixing equal parts of water, ice, and salt. A second point was obtained when pure water froze (originally set at 30 o F), and a third (set at 96 ° F) “when placing the thermometer in the mouth of a healthy man”.  On that scale, water boiled at 212.  Later, Fahrenheit moved the freezing point of water to 32 (so that the scale had 180 increments). l In 1745, Carolus Linnaeus of Upsula, Sweden, described a scale in which the freezing point of water was zero, and the boiling point 100, making it a centigrade (one hundred steps) scale. Anders Celsius (1701-1744) used the reverse scale in which 100 represented the freezing point and zero the boiling point of water, still, of course, with 100 degrees between the two defining points. T (K) 10 8 10 7 10 6 10 5 10 4 10 3 100 10 1 0.1 Hydrogen bomb Sun’s interior Solar corona Sun’s surface Copper melts Water freezes Liquid nitrogen Liquid hydrogen Liquid helium Lowest T ~ 10 -9 K

24. Physics 207: Lecture 21, Pg 23 Ideal gas: Macroscopic description l Consider a gas in a container of volume V, at pressure P, and at temperature T l Equation of state  Links these quantities  Generally very complicated: but not for ideal gas PV = nRT R is called the universal gas constant In SI units, R =8.315 J / mol·K n = m/M : number of moles l Equation of state for an ideal gas  Collection of atoms/molecules moving randomly  No long-range forces  Their size (volume) is negligible  Density is low  Temperature is well above the condensation point

25. Physics 207: Lecture 21, Pg 24 Boltzmann’s constant l In terms of the total number of particles N l P, V, and T are the thermodynamics variables PV = nRT = (N/N A ) RT k B is called the Boltzmann’s constant k B = R/N A = 1.38 X 10 -23 J/K PV = N k B T l Number of moles: n = m/M  One mole contains N A =6.022 X 10 23 particles : Avogadro’s number = number of carbon atoms in 12 g of carbon m=mass M=mass of one mole

26. Physics 207: Lecture 21, Pg 25 What is the volume of 1 mol of gas at STP ? T = 0 °C = 273 K p = 1 atm = 1.01 x 10 5 Pa The Ideal Gas Law

27. Physics 207: Lecture 21, Pg 26 Example l A spray can containing a propellant gas at twice atmospheric pressure (202 kPa) and having a volume of 125.00 cm 3 is at 22 o C. It is then tossed into an open fire. When the temperature of the gas in the can reaches 195 o C, what is the pressure inside the can? Assume any change in the volume of the can is negligible. Steps 1. Convert to Kelvin 2. Use P/T = nR/V = constant 3. Solve for final pressure http://www.thehumorarchives.com/joke/WD40_Stupidity

28. Physics 207: Lecture 21, Pg 27 Example problem: Air bubble rising l A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm. The bubble rises to the surface, where the pressure is p 2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T 1 = 4°C, T 2 = 23°C l What is the ratio of the volume of the bubble as it reaches the surface,V 2, to its volume at the bottom, V 1 ? (Ans.V 2 /V 1 = 3.74) l Is it safe for the diver to ascend while holding his breath? No! Air in the lungs would expand, and the lung could rupture. This is addition to “the bends”, or decompression sickness, which is due to the pressure dependent solubility of gas. At depth and at higher pressure N 2 is more soluble in blood. As divers ascend, N 2 dissolved in their blood stream becomes gaseous again and forms N2 bubbles in blood vessels, which in turn can obstruct blood flow, and therefore provoke pain and in some cases even strokes or deaths. Fortunately, this only happens when diving deeper than 30 m (100 feet). The diver in this question only went down 25 meters. How do we know that?

29. Physics 207: Lecture 21, Pg 28 Example problem: Air bubble rising l A diver produces an air bubble underwater, where the absolute pressure is p 1 = 3.5 atm. The bubble rises to the surface, where the pressure is p 2 = 1 atm. The water temperatures at the bottom and the surface are, respectively, T 1 = 4°C, T 2 = 23°C l What is the ratio of the volume of the bubble as it reaches the surface,V 2, to its volume at the bottom, V 1 ? (Ans.V 2 /V 1 = 3.74) l Use the ideal gas law….For Monday… If thermal transfer is efficient. [More than likely the expansion will be “adiabatic” and, for a diatomic gas, PV  = const. where  = 5/3 ….See chapters 17 & 18]

30. Physics 207: Lecture 21, Pg 29 Physics 207, Lecture 20, Nov. 10 Assignment Assignment  HW9, Due Wednesday, Nov. 19 th  Monday: Read all of Chapter 17