1. FLUIDS & THERMODYNAMICS AP Physics

2. Fluids Fluids are substances that can flow, such as liquids and gases, and even some solids  We’ll just talk about the liquids & gases Review of Density (remember this from chem?)  ρ = m/V  ρ = density  m = mass (kg)  V = volume (m 3 )  density units: kg/m 3

3. Pressure P = F/A  P = pressure (Pa)  F = force (N)  A = area (m 2 ) Units for pressure: Pascals  1 Pa = 1 N/m 2 Pressure is always applied as a normal force on a surface. Fluid pressure is exerted in all directions and is perpendicular to every surface at every location.

4. Atmospheric pressure Atmospheric pressure is normally about 100,000 Pascals. Differences in atmospheric pressure cause winds to blow

5. Pressure Practice 1 Calculate the net force on an airplane window if cabin pressure is 90% of the pressure at sea level and the external pressure is only 50 % of the pressure at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

6. Pressure Practice 1 solution Atmospheric pressure = 100000 Pa P cabin = 90000 PaP outside = 50000 Pa P = F/A net P = 90000Pa – 50000Pa = 40000Pa A = lw A = 0.3m · 0.43m = 0.129m 2 F = AP F = 0.129m 2 · 40000Pa = 5160N

7. Pressure of a Liquid The pressure of a liquid is sometimes called gauge pressure If the liquid is water, it is called hydrostatic pressure P = ρgh  P = pressure (Pa)  ρ = density (kg/m 3 )  g = 9.81 m/s 2  h = height of liquid column (m)

8. Absolute Pressure Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure P atm + ρgh = P abs 2. The depth of Lake Mead at the Hoover Dam is 180 m. What is the hydrostatic pressure at the base of the dam? What is the absolute pressure at the base of the dam?

9. Absolute Pressure (practice 2 – solution) ρ fresh water = 1000 kg / m 3 P = ρgh P = 1000 kg / m 3 · 9.8m/s 2 · 180 m = 1764000 Pa P abs = P atm + ρgh P abs = 100000Pa + 1764000Pa = 1864000 Pa

10. Buoyancy Force Floating is a type of equilibrium: An upward force counteracts the force of gravity for floating objects The upward force is called the buoyant force Archimedes’ Principle: a body immersed in a fluid is buoyed up by a force that is equal to the weight of the fluid it displaces

11. Calculating Buoyant Force F buoy = ρVg  F buoy : buoyant force exerted on a submerged or partially submerged object  V: volume of displaced fluid  ρ: density of displaced fluid When an object floats, the upward buoyant force equals the downward pull of gravity The buoyant force can float very heavy objects, and acts upon objects in the fluid whether they are floating, submerged, or even resting on the bottom

12. Buoyant force on submerged objects A shark’s body is not neutrally buoyant, so a shark must swim continuously or it will sink deeper Scuba divers use a buoyancy control system to maintain neutral buoyancy (equilibrium) If the diver wants to rise, he inflates his vest, which increases his volume, or the water he displaces, and he accelerates upward

13. Buoyant Force on Floating Objects If the object floats on the surface, we know that Fbuoy = Fg! The volume of displaced water equals the volume of the submerged portion of the object

14. Practice 3 Assume a wooden raft has 80.0 % of the density of water. The dimensions of the raft are 6.0 m long by 3.0 m wide by 0.10 m tall. How much of the raft rises above the level of the water when it floats?

15. Practice 3 solution

16. 1. Determine the density of water by using the buoyant force. Equipment: Beakers String Pulleys Weights/Masses Graduated cylinder (NO BALANCES!) Buoyant Force Labs

17. 2. Balloon Race: Determine the buoyant force on your balloon with the balloon, masses & a balance Without using the balloon, design an apparatus so that when released, your balloon will hit the ceiling LAST.

18. Moving Fluids When a fluid flows, mass is conserved Provided there are no inlets or outlets in a stream of flowing fluid, the same mass per unit time must flow everywhere in the stream The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe We can say this because liquids are generally not compressible, so mass conservation is also volume conservation for a liquid

19. Fluid Flow Continuity V = Avt  V: volume of fluid (m 3 )  A: cross sectional areas at a point in the pipe (m 2 )  v: the speed of fluid flow at a point in the pipe (m/s)  t: time (s) Comparing two points in a pipe: A 1 v 1 = A 2 v 2 A 1, A 2 : cross sectional areas at points 1 and 2 v 1, v 2 : speeds of fluid flow at points 1 and 2

20. Practice 4 & 5 4. A pipe of diameter 6.0 cm has fluid flowing through it at 1.6 m/s. How fast is the fluid flowing in an area of the pipe in which the diameter is 3.0 cm? How much water per second flows through the pipe? 5. The water in a canal flows 0.10 m/s where the canal is 12 meters deep and 10 meters across. If the depth of the canal is reduced to 6.5 m at an area where the canal narrows to 5.0 m, how fast will the water be moving through the narrower region?

21. Practice 4 solution

22. Practice 5 solution

23. Bernoulli’s Theorem The sum of the pressure, the potential energy per unit volume, and kinetic energy per unit volume at any one location in the fluid is equal to the sum of the pressure, the potential energy per unit volume, and the kinetic energy per unit volume at any other location in the fluid for a non-viscous incompressible fluid in streamline flow All other considerations being equal, when fluid moves faster, pressure drops

24. Bernoulli’s Theorem P + ρgh + ½ ρv 2 = constant  P = pressure (Pa)  ρ = density of fluid (kg/m 3 )  g = grav. accel. constant (9.81 m/s 2 )  h = height above lowest point  v = speed of fluid flow at a point in the pipe (m/s) 6. Knowing what you know about Bernoulli’s principle, design an airplane wing that you think will keep an airplane aloft. Draw a cross section of the wing.

25. practice 6 solution

26. Thermodynamics Thermodynamics is the study of heat and thermal energy Thermal properties (heat & temperature) are based on the motion of individual molecules, so thermodynamics overlaps with chemistry Total Energy: E = U + K + E int  U = potential energy  K = kinetic energy  E int = internal or thermal energy

27. Total Energy Potential and kinetic energies are specifically for “big” objects, and represent mechanical energy Thermal energy is related to the kinetic energy of the molecules of a substance

28. Temperature & Heat Temperature is a measure of the average kinetic energy of the molecules of a substance. (like how fast the molecules are moving) The unit is °C or K. Temperature is NOT heat! Heat is the internal energy that is transferred between bodies in contact. The unit is Joules (J) or sometimes calories (cal) A difference in temperature will cause heat energy to be exchanged between bodies in contact. When two bodies are the same temp, they are in thermal equilibrium and no heat is transferred.

29. Laws of Thermodynamics O: When two systems are sitting in equilibrium with a third system, they are also in thermal equilibrium with each other. 1: The first law states that when heat is added to a system, some of that energy stays in the system and some leaves the system. The energy that leaves does work on the area around it. Energy that stays in the system creates an increase in the internal energy of the system. 2: It is impossible to have a cyclic (repeating) process that converts heat completely into work. Also, heat never flows spontaneously from a colder body to a hotter spontaneously from a colder body to a hotter body. 3: It is not possible to reach a temperature of absolute zero (about -273 C). Since temperature is a measure of molecular movement, there can be no temperature lower than absolute zero.

30. Ideal Gas Law P: initial & final pressure (any unit) V: initial & final volume (any unit) T: initial & final temperature (K)  T in Kelvins = T in °C + 273

31. Practice 7 7. Suppose an ideal gas occupies 4.0 L at 23°C and 2.3 atm. What will be the volume of the gas if the temperature is lowered to 0°C and the pressure is increased to 3.1 atm?

32. Practice 7 solution

33. Ideal Gas Equation If you don’t remember this from chem, you shouldn’t have passed! P: pressure (Pa) V: volume (m 3 ) n: number of moles R: gas law constant 8.31 J/(mol K) T: temp (K)

34. Practice 8 8. Determine the number of moles of an ideal gas that occupy 10.0 m 3 at atmospheric pressure and 25°C.

35. Practice 8 solution

36. Ideal Gas Equation P: pressure (Pa) V: volume (m 3 ) N: number of molecules k B : Boltzmann’s constant  1.38 x 10 -23 J/K T: temperature (K) 9. Suppose a near vacuum contains 25,000 molecules of helium in one cubic meter at 0°C. What is the pressure?

37. Practice 9 solution

38. Kinetic Theory of Gases 1. Gases consist of a large number of molecules that make elastic collisions with each other and the walls of their container 2. Molecules are separated, on average, by large distances and exert no forces on each other except when they collide 3. There is no preferred position for a molecule in the container, and no preferred direction for the velocity

39. Average Kinetic Energy of a Gas K ave = 3/2 k B T  Kave = average kinetic energy (J)  kB = Boltzmann’s constant (1.38 x 10-23J/K)  T = Temperature (K) The molecules have a range of kinetic energies, so we take the K ave

40. 10 & 11 10. What is the average kinetic energy and average speed of oxygen molecules in a gas sample at 0C°? 11. Suppose nitrogen and oxygen are in a sample of gas at 100°C: a) What is the ratio of the average kinetic energies for the two molecules? b) What is the ratio of their average speeds?

41. Practice 10 solution

42. Practice 11 solution Suppose nitrogen and oxygen are in a sample of gas at 100°C: a) What is the ratio of the average kinetic energies for the two molecules? b) What is the ratio of their average speeds?

43. Thermodynamics The system boundary controls how the environment affects the system (for our purposes, the system will almost always be an ideal gas) If the boundary is “closed to mass,” that means mass can’t get in or out If the boundary is “closed to energy,” that means energy can’t get in or out What type of boundary does the earth have?

44. First Law of Thermodynamics The work done on a system + the heat transferred to the system = the change in internal energy of the system. ΔU = W + Q  ΔU = E int = thermal energy (NOT potential energy – crazy???)  W = work done on the system (related to change in volume)  Q = heat added to the system (J) – driven by temperature difference – Q flows from hot to cold

45. First Law of Thermodynamics

46. More about “U” U is the sum of the kinetic energies of all the molecules in a system (or gas) U = NK ave U = N(3/2 k B T) U = n(3/2 RT)  since k B = R/N A

47. 12 & 13 12. A system absorbs 200 J of heat energy from the environment and does 100 J of work on the environment. What is its change in internal energy? 13. How much work does the environment do on a system if its internal energy changes from 40,000 J to 45,000 J without the addition of heat?

48. Practice 12 solution

49. Practice 13 solution

50. Gas Process The thermodynamic state of a gas is defined by pressure, volume, and temperature. A “gas process” describes how gas gets from one state to another state Processes depend on the behavior of the boundary and the environment more than they depend on the behavior of the gas

52. Isobaric Process (Constant Pressure)

53. Isometric Process (Constant Volume)

54. Isothermal Process (Constant Temperature)

55. Adiabatic Process (Insulated)

56. Work Calculation of work done on a system (or by a system) is an important part of thermodynamic calculations Work depends upon volume change Work also depends upon the pressure at which the volume change occurs

57. Done BY a gas Done ON a gas Work

58. 14 & 15 14. Calculate the work done by a gas that expands from 0.020 m 3 to 0.80 m 3 at constant atmospheric pressure. How much work is done by the environment when the gas expands this much? 15. What is the change in volume of a cylinder operating at atmospheric pressure if its thermal energy decreases by 230 J when 120 J of heat are removed from it?

59. Practice 14 solution

60. Practice 15 solution

61. Work (Isobaric)

62. Work is Path Dependent

63. 16 & 17 16. One mole of a gas goes from state A (200 kPa and 0.5 m 3 ) to state B (150 kPa and 1.5 m 3 ). What is the change in temperature of the gas during this process? 17. One mole of a gas goes from state A (200 kPa and 0.5 m 3 ) to state B (150 kPa and 1.5 m 3 ). a. Draw this process assuming the smoothest possible transition (straight line) b. Estimate the work done by the gas c. Estimate the work done by the environment

64. Practice 16 solution

65. Practice 17 solution

66. Work Done by a Cycle When a gas undergoes a complete cycle, it starts and ends in the same state. the gas is identical before and after the cycle, so there is no identifiable change in the gas. ΔU = 0 for a complete cycle The environment, however, has been changed

67. Work Done By Cycle Work done by the gas is equal to the area circumscribed by the cycle Work done by the gas is positive for clockwise cycles, and negative for counterclockwise cycles. Work done by the environment is opposite that of the gas

68. 18 Consider the cycle ABCDA, where  State A: 200 kPa, 1.0 m 3  State B: 200 kPa, 1.5 m 3  State C: 100 kPa, 1.5 m 3  State D: 100 kPa, 1.0 m 3 a. Sketch the cycle b. Graphically estimate the work done by the gas in one cycle c. Estimate the work done by the environment in one cycle

69. Practice 18 solution A: 200 kPa, 1.0 m 3 B: 200 kPa, 1.5 m 3 C: 100 kPa, 1.5 m 3 D: 100 kPa, 1.0 m 3

70. 19 Calculate the heat necessary to change the temperature of one mole of an ideal gas from 600 K to 500 K a. At constant volume b. At constant pressure (assume 1 atm)

71. Practice 19 solution Calculate the heat necessary to change the temperature of one mole of an ideal gas from 300 K to 500 K a. At constant volume b. At constant pressure (assume 1 atm) W = P  V so when volume is constant, W = 0 Since V = nRT/P then W = nR(T 2 -T 1 ) when P const W = (1 mol)(8.31 J/mol/K)(500-300) W = 1662 J

72. Second Law of Thermodynamics No process is possible whose sole result is the complete conversion of heat from a hot reservoir into mechanical work (Kelvin-Planck statement) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body (Clausius statement) Basically, heat can’t be completely converted into useful energy

73. Heat Engines Heat engines can convert heat into useful work According to the 2 nd Law of Thermodynamics, Heat engines always produce some waste heat Efficiency can be used to tell how much heat is needed to produce a given amount of work

74. Work and Heat Engines Q H = W + Q C  Q H : Heat that is put into the system and comes from the hot reservoir in the environment  W: Work that is done by the system on the environment  Q C : Waste heat that is dumped into the cold reservoir in the environment

75. Heat Transfer

76. Heat Engines

77. Adiabatic vs. Isothermal Expansion

78. Carnot Cycle an ideal heat- engine cycle in which the working substance goes through the four successive operations of isothermal expansion to a desired point, adiabatic expansion to a desired point, isothermal compression, and adiabatic compression back to its initial state.

79. Carnot Cycle

80. 20 20. A piston absorbs 3600 J of heat and dumps 1500 J of heat during a complete cycle. How much work does it do during the cycle?

81. Practice 20 solution

82. Efficiency of Heat Engine In general, efficiency is related to what fraction of the energy put into a system is converted to useful work In the case of a heat engine, the energy that is put in is the heat that flows into the system from the hot reservoir Only some of the heat that flows in is converted to work. The rest is waste heat that is dumped into the cold reservoir

83. Efficiency of Heat Engine Efficiency = W/Q H = (Q H – Q C ) / Q H  W: Work done by the engine on the environment  Q H : Heat absorbed from hot reservoir  Q C : Waste heat dumped into cold reservoir Efficiency is often given as percent efficiency HOW COULD YOU: find the efficiency of your hair dryer?

84. 21 A coal-fired steam plant is operating with 33% thermodynamic efficiency. If this is a 120 MW plant, at what rate is heat energy used?

85. Practice 21 solution A coal-fired steam plant is operating with 33% thermodynamic efficiency. If this is a 120 MW plant, at what rate is heat energy used? Efficiency = W/Q H 0.33 = 120MW/Q H Q H = 363.6 MW

86. Carnot Engine Cycle

87. Efficiency of Carnot Cycle For a Carnot engine, the efficiency can be calculated from the temperatures of the hot and cold reservoirs. Carnot Efficiency = (T H – T C ) / T H  T H : temperature of hot reservoir (K)  T C : temperature of cold reservoir (K)

88. 22 & 23 22. Calculate the Carnot efficiency of a heat engine operating between the temperature of 60 and 1500°C. 23. For #22, how much work is produced when 15 kJ of waste heat is generated?

89. Practice 22 solution

90. Practice 23 solution 23. For #22, how much work is produced when 15 kJ of waste heat is generated? 15 kJ (Q c = the amount of heat wasted) is 19% of Q H (the amount of heat to begin with) since it has 81% efficiency (see last slide) Therefore…15kJ/.19 = Q H = 78.95 kJ Q H = Q C + W 78.95 = 15 + W W = 63.95 kJ Or use Efficiency = (Q H – Q C ) / Q H & solve for Q H then W

91. Entropy Entropy is disorder, or randomness The entropy of the universe is increasing. Ultimately, this will lead to what is affectionately known as “Heat Death of the Universe.”

92. Entropy ΔS = Q/T  ΔS: change in entropy (J/K)  Q: heat going into the system (J)  T: temperature (K) If change in entropy is positive, randomness or disorder has increased Spontaneous changes involve an increase in entropy Generally, entropy can go down only when energy is put into the system