1. Fluids
Fluids are substances that can flow, such as liquids and gases, and even a few solids.
In Physics B, we will limit our discussion of fluids to substances that can easily flow, such as liquids and gases.

2. Review: Density  = m/V Units:
: density (kg/m3)
m: mass (kg)
V: volume (m3)
Units:
kg/m3
You should remember how to do density calculations from chemistry!

3. Pressure P = F/A Pressure unit: Pascal ( 1 Pa = 1 N/m2)
balloon
P = F/A
P : pressure (Pa)
F: force (N)
A: area (m2)
Pressure unit: Pascal ( 1 Pa = 1 N/m2)
The force on a surface caused by pressure is always normal (or perpendicular) to the surface. This means that the pressure of a fluid is exerted in all directions, and is perpendicular to the surface at every location.

4. Sample Problem
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 that at sea level. Assume the window is 0.43 m tall and 0.30 m wide.

5. Atmospheric Pressure
Atmospheric pressure is normally about 100,000 Pascals.
Differences in atmospheric pressure cause winds to blow.
Low atmospheric pressure inside a hurricane’s eye contributes to the severe winds and the development of the storm surge.

6. Hurricane Katrina’s Storm Surge
Mississippi River Gulf Outlet
New Orleans East/St. Bernard Parish

7. The Pressure of a Liquid
P = gh
P: pressure (Pa)
: density (kg/m3)
g: acceleration constant (9.8 m/s2)
h: height of liquid column (m)
This is often called hydrostatic pressure if the liquid is water. It excludes atmospheric pressure.
It is also sometimes called gauge pressure, since a diver’s pressure gauge will read hydrostatic pressure. Gauge pressure readings never include atmospheric pressure, but only the pressure of the fluid.
Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure
pabs = patm + gh

8. Sample Problem
Calculate for the bottom of a 3 meter (approx 10 feet) deep swimming pool full of water:
(a) hydrostatic pressure
(b) absolute pressure
Which one of these represents the gauge pressure?

9. Hydrostatic Pressure in Dam Design
The depth of Lake Mead at the Hoover Dam is 600ft. What is the hydrostatic pressure at the base of the dam?

10. Hydrostatic Pressure in Levee Design
Hurricane Katrina, August 2005
A hurricane’s storm surge can overtop levees, but a bigger problem can be increasing the hydrostatic pressure at the base of the levee.

11. New Orleans Elevation Map
New Orleans is largely below sea level, and relies upon a system of levees to keep the lake and the river at bay

12. Sample Problem
Calculate the increase in hydrostatic pressure experienced by the levee base for an expected (SPH Design) storm surge. How does this compare to the increase that occurred during Hurricane Katrina, where the water rose to the top of the levee?

13. Floating is a type of equilibrium
An upward force counteracts the force of gravity for these objects. This upward force is called the buoyant force.
Fbuoy mg

14. 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 displaced.
Fbuoy = Vg
Fbuoy: the buoyant force exerted on a submerged or partially submerged object.
V: the volume of displaced liquid.
: the density of the displaced liquid.
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 water whether they are floating, submerged, or even sitting on the bottom.

15. Buoyant force on submerged object
A sharks body is not neutrally buoyant, and so a shark must swim continuously or he will sink deeper. mg
Fbuoy = rVg

16. Buoyant force on submerged objectmg
rVg
SCUBA divers use a buoyancy control system to maintain neutral buoyancy (equilibrium!).

17. Buoyant force on submerged objectmg
rVg
If the diver wants to rise, he inflates his vest, which increases the amount of water he displaces, and he accelerates upward.

18. Buoyant force on floating object
rVg
If the object floats on the surface, we know for a fact Fbuoy = mg! The volume of displaced water equals the volume of the submerged portion of the ship. mg

19. Sample problem
Assume a wooden raft has 80.0% of the density of water. The dimensions of the raft are 6.0 meters long by 3.0 meters wide by 0.10 meter tall. How much of the raft rises above the level of the water when it floats?

20. Sample problem
You want to transport a man and a horse across a still lake on a wooden raft. The mass of the horse is 700 kg, and the mass of the man is 75.0 kg. What must be the minimum volume of the raft, assuming that the density of the wood is 80% of the density of the water.

21. Parking in St. Bernard Parish after Hurricane Katrina

22. Parking in St. Bernard Parish after Hurricane Katrina

23. Parking in St. Bernard Parish after Hurricane Katrina

24. “Mobile” Homes in St. Bernard Parish after Hurricane Katrina

25. “Mobile” Homes in St. Bernard Parish after Hurricane Katrina

26. Estimation problem
Estimate the mass of this house in kg.

27. Buoyancy Lab
Using the equipment provided, verify that the density of water is 1,000 kg/m3.
Report must include:
Free body diagrams.
All data.
Calculations.
air
water
Note: established value for the density of pure water is 1,000 kg/m3.

28. Reading a Venier Caliper

29. Reading a Venier Caliper

30. Reading a Venier Caliper

31. Reading a Venier Caliper

32. Fluid Flow Continuity
The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe.
V = Avt
V: volume of fluid (m3)
A: cross sectional areas at a point in the pipe (m2)
v: speed of fluid flow at a point in the pipe (m/s)
t: time (s)
A1v1 = A2v2
A1, A2: cross sectional areas at points 1 and 2
v1, v2: speed of fluid flow at points 1 and 2

33. Sample problem
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?

34. Flash flooding can be explained by fluid flow continuity.
Natural Waterways
Flash flooding can be explained by fluid flow continuity.

35. Sample problem
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 meters at an area where the canal narrows to 5.0 meters, how fast will the water be moving through this narrower region? What will happen to the water if something prevents it from flowing faster in the narrower region?

36. Artificial Waterways
Flooding from the Mississippi River Gulf Outlet was responsible for catastrophic flooding in eastern New Orleans and St. Bernard during Hurricane Katrina.

37. Fluid Flow Continuity in Waterways
Mississippi River Gulf Outlet levees are overtopped by Katrina’s storm surge.
A hurricane’s storm surge can be “amplified” by waterways that become narrower or shallower as they move inland.

38. Bernoulli’s Theorem
The sum of the pressure, the potential energy per unit volume, and the 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, the pressure drops.

39. Bernoulli’s Theorem P +  g h + ½ v2 = Constant P : pressure (Pa)
 : density of fluid (kg/m3)
g: gravitational acceleration constant (9.8 m/s2)
h: height above lowest point (m)
v: speed of fluid flow at a point in the pipe (m/s)

40. Sample Problem
Knowing what you know about Bernouilli’s principle, design an airplane wing that you think will keep an airplane aloft. Draw a cross section of the wing.

41. Bernoulli’s Principle and Hurricanes
In a hurricane or tornado, the high winds traveling across the roof of a building can actually lift the roof off the building.

42. Big City Evacuation
These URLs, some of which are disputed, are provided for the Big City Evacuation project

43. Storm Surges in Hurricanes

44. Bernoulli Effect in Design

45. Building for Hurricanes

46. Building in the Wetlands

47. Hydrostatic Pressure: Dams
(This has information on failed dams at end of article).

48. Hydrostatic Pressure: Levees

49. Thermodynamics Thermodynamics is the study of heat and thermal energy.
Thermal properties (heat and temperature) are based on the motion of individual molecules, so thermodynamics is a lot like chemistry.

50. Total energy E = U + K + Eint
U: potential energy
K: kinetic energy
Eint: internal or thermal 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.

51. Temperature and Heat
Temperature is a measure of the average kinetic energy of the molecules of a substance. Think o it as a measure of how fast the molecules are moving. The unit is OC 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 at the same temperature, no heat is transferred. This is called Thermal Equilibrium.

52. Ideal Gas Law P1 V1 / T1 = P2 V2 / T2
P1, P2: initial and final pressure (any unit)
V1, V2: initial and final volume (any unit)
T1, T2: initial and final temperature (in Kelvin!)
Temperature in K is obtained from temperature in oC by adding 273.

53. Sample problem
Suppose an ideal gas occupies 4.0 liters at 23oC and 2.3 atm. What will be the volume of the gas if the temperature is lowered to 0oC and the pressure is increased to 3.1 atm.

54. Ideal Gas Equation P V = n R T P: pressure (in Pa) V: volume (in m3)
n: number of moles
R: gas law constant
8.31 J/(mol K)
T: temperature (in K)

55. Sample problem
Determine the number of moles of an ideal gas that occupy 10.0 m3 at atmospheric pressure and 25oC.

56. Ideal Gas Equation PV = n R T (using moles)
P V = N kB T (using molecules)
P: pressure (Pa)
V: volume (m3)
N: number of molecules
kB: Boltzman’s constant
1.38 x J/K
T: temperature (K)

57. Sample problem
Suppose a near vacuum contains 25,000 molecules of helium in one cubic meter at 0oC. What is the pressure?

58. Sample problem
Can you find a relationship between the Universal Gas Law constant R (8.31 J/mol K) and Boltzman’s constant kB (1.38 x J/K) ?

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

60. Simulations http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html

61. Average Kinetic Energy of a Gas
Kave = 3/2 kBT
Kave: average kinetic energy (J)
kB: Boltzmann’s Constant (1.38 x J/K)
T: Temperature (K)
The molecules have a range of kinetic energies; Kave is just the average of that range.

62. Sample Problem
What is the average kinetic energy and the average speed of oxygen molecules in a gas sample at 0oC?

63. Sample Problem
Suppose nitrogen and oxygen are in a sample of gas at 100oC.
A) What is the ratio of the average kinetic energies for the two molecules?
B) What is the ratio of their average speeds?

64. System Boundary Environment System
For our purposes, the system will almost always be an ideal gas.
System

65. System Boundary
The system boundary controls how the environment affects the system.
If the boundary is “closed to mass”, that means that mass can’t get in or out.
If the boundary is “closed to energy”, that means energy can’t get in or out.

66. Discussion
Consider the earth as a system. What type of boundary does it have?

67. First Law of Thermodynamics
System U
(Eint)
Work W
Heat Q

68. Awkward notation WARNING!
We all know that U is potential energy in mechanics. However…
U is Eint (thermal energy) in thermodynamics!
Yuk.
This means when we are in thermo, U is thermal energy, which is related to temperature. When we are in mechanics, it is potential energy, which is related to configuration or position.

69. More about U
U is the sum of the kinetic energies of all molecules in a system (or gas).
U = N Kave
U = N (3/2 kBT)
U = n (3/2 R T)
Since kB = R /NA

70. First Law of Thermodynamics
U = Q + W
U: change in internal energy of system (J)
Q: heat added to the system (J). This heat exchange is driven by temperature difference. It occurs only if the boundary allows energy transfer.
W: work done on the system (J). Work will be related to the change in the system’s volume. It occurs only if the boundary can change shape in some way.
This law is sometimes paraphrased as “you can’t win”.

71. Problem
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?

72. Problem
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?

73. Walker, 18.5 Q W DU A  B -53 J a) b) B  C -280 J -130 J c) C  A e)

74. Walker, 18.6
One mole of an ideal monatomic gas is initially a a temperature of 323 K. Find the temperature of the gas if 2250 J of heat are added and it does 834 J of work.

75. 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.

76. Isothermal Process (constant temperature)
Gas “isotherms” P V
PV = nRT
Initial State of Gas
Isothermal Process
Final State of Gas
DT = 0 (constant T)

77. Isobaric Process (constant pressure)V T1 T2 T3
PV = nRT
Isobaric Expansion
Isobaric Contraction
DP = 0 (constant P)

78. Isometric Process (constant volume)
PV = nRT
DV = 0 (constant V)

79. Adiabatic process (insulated)
isotherm P V
PV = nRT
adiabat
Temperature, pressure, and volume all change in an adiabatic process.
Q = 0 (no heat enters or leaves)

80. 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.

81. Wgas = pV Wenv = -pV Work done BY gas p Positive work p
Negative work

82. Wgas = pV Wext = -pV Work done ON gas p
Negative since V is negative p
Wext = -pV
Positive since V is negative

83. Sample problem
Calculate the work done by a gas that expands from m3 to 0.80 m3 at constant atmospheric pressure.
How much work is done by the environment when the gas expands this much?

84. Sample problem
What is the change in volume of a cylinder operating at atmospheric pressure if its internal energy decreases by 230 J when 120 J of heat are removed from it?

85. P Work (isobaric) WAB > WCD A B C D V V1 V2 P2
Where we are considering work done BY the gas
WAB = p2V C D P1
WCD = p1V V

86. Work is path dependent WABD > WACD P V1 A V2 B P2
Where we are considering work done BY the gas
WABD C D P1
WACD V

87. Problem
One mole of a gas goes from state A (200 kPa and 0.5 m3) to state B (150 kPa and 1.5 m3).
A) What is the change in temperature of the gas during this process?

88. Problem
One mole of a gas goes from state A (200 kPa and 0.5 m3) to state B (150 kPa and 1.5 m3).
B) Draw this process, assuming the smoothest possible transition (straight line) for the process.
C) Estimate the work done by the gas during the process.
D) Estimate the work done by the environment during the process.

89. Sample Problem
Draw the gas process from state A (200 kPa and 0.5 m3) to state B (150 kPa and 1.5 m3).

90. 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.
DU = 0 for a complete cycle.
The environment, however, has been changed.

91. Work done by cycle P V1 A V2 B P2 WABCD C D P1 V
Work done by the gas is equal to the area circumscribed by the cycle.
Work done by gas is positive for clockwise cycles, negative for counterclockwise cycles.
Work done by environment is negative of work done by gas. P V1 A V2 B P2
WABCD C D P1 V

92. Sample Problem Consider the cycle ABCDA, where A) Sketch the cycle.
State A: 200 kPa, 1.0 m3
State B: 200 kPa, 1.5 m3
State C: 100 kPa, 1.5 m3
State D: 100 kPa, 1.0 m3
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.

93. A: 200 kPa, 1.0 m3 B: 200 kPa, 1.5 m3 C: 100 kPa, 1.5 m3 D: 100 kPa, 1.0 m3

94. Problem
Calculate the heat necessary to change the temperature of one mole of an ideal gas from 300K to 500K
A) at constant volume.
B) at constant pressure (assume 1 atmosphere).

95. 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.)

96. Efficiency Lab – part 1
Efficiency is defined as the fraction of work or energy put into a system is actually useful.
At your disposal are T-shirts, water, plastic baskets, mass balances, and an 1800 Watt hair drier. Develop a procedure to determine the efficiency of the hair drier.
Hint: You must use the fact that it takes 2260 J of heat energy to vaporize one gram of water.
Write a procedure to figure out the efficiency of a hair drier in your “mega group”. Each person in the group must perform one and only one step in the procedure.

97. Heat Engines Heat engines can convert heat into useful work.
According to the 2nd 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.
NOTE: A heat engine is not something that produces heat. A heat engine transfers heat from hot to cold, and does mechanical work in the process.

98. QH = QC Heat Transfer QH QC Heat Source (High Temperature)
Heat Sink (Low Temperature) QH
QH = QC QC

99. Heat Engines QH W QH = QC + W QC Heat Source (High Temperature) Engine
Heat Sink (Low Temperature) QH
Engine W
QH = QC + W QC

100. Carnot Cycle Heat in Work Heat out P V QH = QC + W Efficiency = W/QH
Adiabatic compression P V
Isothermal expansion
Adiabatic expansion
Work
Isothermal compression
Heat out

101. Work and Heat Engines QH W QC QH = W + QC
Heat Source (High Temperature)
Heat Sink (Low Temperature)
QH = W + QC
QH: 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.
QC: Waste heat that is dumped into the cold reservoir in the environment. QH
Engine W QC

102. Sample Problem
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?

103. Efficiency of Heat Engine
In general, efficiency is related 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.

104. Efficiency of Heat Engine
Efficiency = W/QH = (QH - QC)/QH
W: Work done by engine on environment
QH: Heat absorbed from hot reservoir
QC: Waste heat dumped to cold reservoir
Efficiency is often given as percent efficiency.

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

106. Efficiency of Carnot Cycle
For a Carnot engine, the efficiency can be calculated from the temperatures of the hot and cold reservoirs.
Carnot Efficiency = (TH - TC)/TH
TH: Temperature of hot reservoir (K)
TC: Temperature of cold reservoir (K)

107. Sample Problem
Calculate the Carnot efficiency of a heat engine operating between the temperatures of 60 and 1500 oC.

108. Sample Problem
For the problem described in the previous example, how much work is produced when 15 kJ of waste heat is generated.

109. 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”.
Does the entropy in your room tend to increase or decrease?

110. Entropy
DS = Q/T
DS: Change in entropy (J/K)
Q: Heat going into system (J)
T: Kelvin 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.

111. Sample problem
You vaporize 8 kg of liquid water at its boiling point. What is the entropy change?

112. Efficiency Lab – part 2
Determine the % efficiency of your hair drier using the procedure from the previous day.
Keep a record of all data collected.
Calculate the % efficiency.
Turn in your mega-group report, with all calculations clearly illustrated.
Heat of vaporization of water is 2260 J/g.