Fluids Day #1 Introducing Fluids Hydrostatic Pressure

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

Review: Density  = m/V  = m/V   : density (kg/m 3 )  m: mass (kg)  V: volume (m 3 ) Units: Units:  kg/m 3 You should remember how to do density calculations from chemistry!

Pressure P = F/A P = F/A  P : pressure (Pa)  F: force (N)  A: area (m 2 ) Pressure unit: Pascal ( 1 Pa = 1 N/m 2 ) Pressure unit: Pascal ( 1 Pa = 1 N/m 2 ) 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. 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. balloon

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. 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.

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

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

The Pressure of a Liquid P =  gh P =  gh  P: pressure (Pa)   : density (kg/m 3 )  g: acceleration constant (9.8 m/s 2 )  h: height of liquid column (m) This is often called hydrostatic pressure if the liquid is water. It excludes atmospheric pressure. 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. 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 Absolute pressure is obtained by adding the atmospheric pressure to the hydrostatic pressure  p abs = p atm +  gh

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

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?

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.

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

Sample Problem Calculate the increase in hydrostatic pressure experienced by the levee base for an expected (SPH Design: 11.5 ft) storm surge. How does this compare to the increase that occurred during Hurricane Katrina, where the water rose to the top of the levee (17.5ft)? (Normal Lake height is 1.0 ft)

Fluids Day #2 Buoyancy Force

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. An upward force counteracts the force of gravity for these objects. This upward force is called the buoyant force. F buoy mg

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. Archimedes’ Principle: a body immersed in a fluid is buoyed up by a force that is equal to the weight of the fluid displaced. F buoy =  Vg F buoy =  Vg  F buoy : 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. 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. 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.

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

Buoyant force on submerged object mg  Vg SCUBA divers use a buoyancy control system to maintain neutral buoyancy (equilibrium!).

Buoyant force on submerged object mg  Vg If the diver wants to rise, he inflates his vest, which increases the amount of water he displaces, and he accelerates upward.

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

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?

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. 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.

Parking in St. Bernard Parish after Hurricane Katrina

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

Estimation problem Estimate the mass of this house in kg.

Fluids Day #3 Buoyancy Force Lab: “Sink or Swim” from Conceptual Physics Lab Manual

Fluids Day # 4 Moving Fluids

Fluid Flow Continuity The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe. The volume per unit time of a liquid flowing in a pipe is constant throughout the pipe. V = Avt V = Avt  V: volume of fluid (m 3 )  A: cross sectional areas at a point in the pipe (m 2 )  v: speed of fluid flow at a point in the pipe (m/s)  t: time (s) A 1 v 1 = A 2 v 2 A 1 v 1 = A 2 v 2  A 1, A 2 : cross sectional areas at points 1 and 2  v 1, v 2 : speed of fluid flow at points 1 and 2 oulli.html oulli.html oulli.html oulli.html

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? 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?

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

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?

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

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

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. 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. All other considerations being equal, when fluid moves faster, the pressure drops.

Bernoulli’s Theorem P +  g h + ½  v 2 = Constant P +  g h + ½  v 2 = Constant  P : pressure (Pa)   : density of fluid (kg/m 3 )  g: gravitational acceleration constant (9.8 m/s 2 )  h: height above lowest point (m)  v: speed of fluid flow at a point in the pipe (m/s)

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. 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.

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. In a hurricane or tornado, the high winds traveling across the roof of a building can actually lift the roof off the building. cid= &q=Hurri cane+Roof&hl=en cid= &q=Hurri cane+Roof&hl=en cid= &q=Hurri cane+Roof&hl=en cid= &q=Hurri cane+Roof&hl=en

Applications of Fluids Concepts

Storm Surges in Hurricanes   /mtr/hurr/home.rxml /mtr/hurr/home.rxml /mtr/hurr/home.rxml  ne.htm ne.htm ne.htm 

Bernoulli Effect in Design opics/Aerodynamics.html opics/Aerodynamics.html opics/Aerodynamics.html opics/Aerodynamics.html frankfurt.de/~weltner/Flight/PHYSIC4.ht m frankfurt.de/~weltner/Flight/PHYSIC4.ht m frankfurt.de/~weltner/Flight/PHYSIC4.ht m frankfurt.de/~weltner/Flight/PHYSIC4.ht m andrews.ac.uk/Mathematicians/Bernoulli_D aniel.html andrews.ac.uk/Mathematicians/Bernoulli_D aniel.html andrews.ac.uk/Mathematicians/Bernoulli_D aniel.html andrews.ac.uk/Mathematicians/Bernoulli_D aniel.html

Building for Hurricanes

Building in the Wetlands

Hydrostatic Pressure: Dams /dam/basics.html /dam/basics.html /dam/basics.html /dam/basics.html  (This has information on failed dams at end of article). om/build.html om/build.html om/build.html om/build.html /dams/foun/press.htm /dams/foun/press.htm

Hydrostatic Pressure: Levees gy/0, html gy/0, html gy/0, html gy/0, html

Thermo Day #1 Introduction to Thermodynamics Gas Laws

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

Total energy E = U + K + E int E = U + K + E int  U: potential energy  K: kinetic energy  E int : internal or thermal energy Potential and kinetic energies are specifically for “big” objects, and represent mechanical 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. Thermal energy is related to the kinetic energy of the molecules of a substance.

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 O C or K. 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 O C or K. Temperature is NOT heat! 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). 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. 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.

Ideal Gas Law P 1 V 1 / T 1 = P 2 V 2 / T 2 P 1 V 1 / T 1 = P 2 V 2 / T 2  P 1, P 2 : initial and final pressure (any unit)  V 1, V 2 : initial and final volume (any unit)  T 1, T 2 : initial and final temperature (in Kelvin!) Temperature in K is obtained from temperature in o C by adding 273. Temperature in K is obtained from temperature in o C by adding 273.

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

Ideal Gas Equation P V = n R T P V = n R T  P: pressure (in Pa)  V: volume (in m 3 )  n: number of moles  R: gas law constant  8.31 J/(mol K)  T: temperature (in K)

Sample problem Determine the number of moles of an ideal gas that occupy 10.0 m 3 at atmospheric pressure and 25 o C. Determine the number of moles of an ideal gas that occupy 10.0 m 3 at atmospheric pressure and 25 o C.

Thermo Day 2 Kinetic Theory of Gases

Ideal Gas Equation PV = n R T (using moles) PV = n R T (using moles) P V = N k B T (using molecules) P V = N k B T (using molecules)  P: pressure (Pa)  V: volume (m 3 )  N: number of molecules  k B : Boltzman’s constant  1.38 x J/K  T: temperature (K)

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

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

Kinetic Theory of Gases 1. Gases consist of a large number of molecules that make elastic collisions with each other and the walls of the 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.

Simulations _dyn/KinThI.html _dyn/KinThI.html _dyn/KinThI.html _dyn/KinThI.html 00/Laboratory/GLP.htm 00/Laboratory/GLP.htm 00/Laboratory/GLP.htm 00/Laboratory/GLP.htm

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

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

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

Thermo Day 3 Systems and the First Law of Thermodynamics

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

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

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

First Law of Thermodynamics U (E int ) W Q

Awkward notation WARNING! We all know that U is potential energy in mechanics. However… We all know that U is potential energy in mechanics. However… U is E int (thermal energy) in thermodynamics! U is E int (thermal energy) in thermodynamics! Yuk. 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. 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.

More about U U is the sum of the kinetic energies of all molecules in a system (or gas). U is the sum of the kinetic energies of all molecules in a system (or gas). U = N K ave U = N K ave U = N (3/2 k B T) U = N (3/2 k B T) U = n (3/2 R T) U = n (3/2 R T)  Since k B = R /N A

First Law of Thermodynamics  U = Q + W  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”. This law is sometimes paraphrased as “you can’t win”.

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? 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?

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? 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?

Walker, 18.5 A  B B  C C  A QW UU -53 Ja)b) -280 J-130 Jc) d)150 Je)

Walker, 18.6 One mole of an ideal monatomic gas is initially at 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. One mole of an ideal monatomic gas is initially at 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.

Thermo Day 4 Gas Processes

Gas Process The thermodynamic state of a gas is defined by pressure, volume, and temperature. 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. 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. Processes depend on the behavior of the boundary and the environment more than they depend on the behavior of the gas.

T1T1 T2T2 T3T3 Gas “isotherms” Isothermal Process (constant temperature) P V PV = nRT  T = 0 (constant T) Initial State of Gas Final State of Gas Isothermal Process

Isobaric Process (constant pressure) P V T1T1 T2T2 T3T3 PV = nRT  P = 0 (constant P) Isobaric Expansion Isobaric Contraction

Isometric Process (constant volume) P V T1T1 T2T2 T3T3  V = 0 (constant V) PV = nRT

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

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

p Work done BY gas VV p W gas = p  V W env = -p  V Positive work Negative work

p Work done ON gas VV p W gas = p  V W ext = -p  V Negative since  V is negative Positive since  V is negative

Sample problem Calculate the work done by a gas that expands from m 3 to 0.80 m 3 at constant atmospheric pressure. Calculate the work done by a gas that expands from 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? How much work is done by the environment when the gas expands this much?

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? 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?

W CD = p 1  V Work (isobaric) P V P2P2 V1V1 A V2V2 B P1P1 C D W AB > W CD Where we are considering work done BY the gas W AB = p 2  V

W ACD Work is path dependent P V P2P2 V1V1 A V2V2 B P1P1 C D W ABD > W ACD Where we are considering work done BY the gas W ABD

Problem 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 ). 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) What is the change in temperature of the gas during this process?

Problem 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 ). 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 ).  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.

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

Thermo Day 5 Gas Cycles 2 nd Law of Thermodynamics

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. 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.  U = 0 for a complete cycle.  U = 0 for a complete cycle. The environment, however, has been changed. The environment, however, has been changed.

Work done by cycle P V P2P2 V1V1 A V2V2 B P1P1 C D Work done by the gas is equal to the area circumscribed by the cycle. W ABCD Work done by gas is positive for clockwise cycles, negative for counterclockwise cycles. Work done by environment is negative of work done by gas.

Sample Problem Consider the cycle ABCDA, where 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. A) Sketch the cycle. B) Graphically estimate the work done by the gas in one cycle. B) Graphically estimate the work done by the gas in one cycle. C) Estimate the work done by the environment in one cycle. C) Estimate the work done by the environment in one cycle.

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

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

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 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.) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. (Clausius statement.)

Thermo Day 6 Heat Engines

Heat engines can convert heat into useful work. Heat engines can convert heat into useful work. According to the 2 nd Law of Thermodynamics. Heat engines always produce some waste heat. 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. 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. 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.

Heat Transfer Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC Q H = Q C

Heat Engines Engine Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC W Q H = Q C + W

Carnot Cycle P V Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression Q H = Q C + W Efficiency = W/Q H

Work and Heat Engines Q H = W + Q C 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. Engine Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC W

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? 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?

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

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

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? 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?

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

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

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

Entropy… Entropy is disorder, or randomness. 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”. 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? Does the entropy in your room tend to increase or decrease?

Entropy  S = Q/T  S = Q/T   S: 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. If change in entropy is positive, randomness or disorder has increased. Spontaneous changes involve an increase in entropy. Spontaneous changes involve an increase in entropy. Generally, entropy can go down only when energy is put into the system. Generally, entropy can go down only when energy is put into the system.

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

Thermo Extra Credit Home Lab Hair Dryer Efficiency Lab – Parts I and II

Efficiency Lab – part 1 Efficiency is defined as the fraction of work or energy put into a system is actually useful. Efficiency is defined as the fraction of work or energy put into a system is actually useful. Use a T-shirt, water, plastic basket/bucket, mass balances (may use at school by request), and a hair drier. (Hair dryers typically run on 1800 Watts, but you will need to look for the power used by yours) Develop a procedure to determine the efficiency of the hair drier. Use a T-shirt, water, plastic basket/bucket, mass balances (may use at school by request), and a hair drier. (Hair dryers typically run on 1800 Watts, but you will need to look for the power used by yours) 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. This is the “latent heat of vaporization” for water Hint: You must use the fact that it takes 2260 J of heat energy to vaporize one gram of water. This is the “latent heat of vaporization” for water Write a procedure to figure out the efficiency of your hair drier. Write a procedure to figure out the efficiency of your hair drier. If working in a group, each person in the group must perform at least one step in the procedure. Specify who did what with the person’s name written by their step. If working in a group, each person in the group must perform at least one step in the procedure. Specify who did what with the person’s name written by their step.

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

Thermo Review Review for Exam: Free Response Questions from website and Ranking Tasks