Chapter 19. Heat Engines - PDF Free Download

Chapter 19 Heat Engines

Thermo Processes Eint = Q+ W Adiabatic No heat exchanged Q = 0 and E int = W Isobaric Constant pressure W = P (V f V i ) and E int = Q + W Isochoric Constant Volume W = 0 and E int = Q Isothermal Constant temperature E int = 0 and Q = -W W V i = nrt ln V f

C P and C V Note that for all ideal gases: where R = 8.31 J/mol K is the universal gas constant. Slide 17-80

Heat Engines

Refrigerators

Important Concepts

2 nd Law: Perfect Heat Engine Can NOT exist! No energy is expelled to the cold reservoir It takes in some amount of energy and does an equal amount of work e = 100% It is an impossible engine No Free Lunch! Limit of efficiency is a Carnot Engine

Reversible and Irreversible Processes The reversible process is an idealization. All real processes on Earth are irreversible. Example of an approximate reversible process: The gas is compressed isothermally The gas is in contact with an energy reservoir Continually transfer just enough energy to keep the temperature constant The change in entropy is equal to zero for a reversible process and increases for irreversible processes. Section 22.3

The Maximum Efficiency Q T T = and ec = 1 Q T T c c c h h h COP C Q Tc W T T c = = h c COP H Q Th W T T h = = h c

Heat Engines In a steam turbine of a modern power plant, expanding steam does work by spinning the turbine. The steam is then condensed to liquid water and pumped back to the boiler to start the process again. First heat is transferred to the water in the boiler to create steam, and later heat is transferred out of the water to an external cold reservoir, in the condenser. This steam generator is an example of a heat engine. 2013 Pearson Education, Inc. Slide 19-33

Heat Engine Eint = 0 for the entire cycle A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work A heat engine carries some working substance through a cyclical process The working substance absorbs energy by heat from a high temperature energy reservoir (Q h ) Work is done by the engine (W eng ) Energy is expelled as heat to a lower temperature reservoir (Q c )

Thermal Efficiency of a Heat Engine Eint = 0 for the entire cycle Weng = Qh Qc Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature e W Q Q Q 1 Q Q Q eng h c c = = = h h h

Analyze this engine to determine (a) the net work done per cycle, (b) the engine s thermal efficiency and (c) the engine s power output if it runs at 600 rpm. Assume the gas is monatomic and follows the idealgas process above.

From Last Week.. A 4.00-L sample of a nitrogen gas confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and at 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Find the number of moles of the gas. (c) Find the volumes and temperatures at the end of each process (d) Find the Work and heat for each process. (e) What was the net work done on the gas for this cycle?

The Brayton Cycle Many ideal-gas heat engines, such as jet engines in aircraft, use the Brayton Cycle, as shown. The cycle involves adiabatic compression (1-2), isobaric heating during combustion (2-3), adiabatic expansion which does work (3-4), and isobaric cooling (4-1). The efficiency is: 2013 Pearson Education, Inc. Slide 19-58

Otto Cycle The Otto cycle approximates the processes occurring in an internal combustion engine If the air-fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle is 1 e = 1 ( V ) 1 1 V γ 2 γ is the ratio of the molar specific heats V 1 / V 2 is called the compression ratio Typical values: Compression ratio of 8 γ = 1.4 e = 56% Efficiencies of real engines are 15% to 20% Mainly due to friction, energy transfer by conduction, incomplete combustion of the air-fuel mixture

No Perfect Heat Engines A perfect heat engine connected to a refrigerator would violate the second law of thermodynamics. 2013 Pearson Education, Inc. Slide 19-46

Rank in order, from largest to smallest, the work W out performed by these four heat engines. e W Q Q Q 1 Q Q Q eng h c c = = = h h h A. W b > W a > W c > W d B. W b > W a > W b > W c C. W b > W a > W b = W c D. W d > W a = W b > W c E. W d > W a > W b > W c

QuickCheck 19.8 How much heat is exhausted to the cold reservoir? A. 7000 J. B. 5000 J. C. 3000 J. D. 2000 J. E. 0 J. 2013 Pearson Education, Inc. Slide 19-54

QuickCheck 19.8 How much heat is exhausted to the cold reservoir? A. 7000 J. B. 5000 J. C. 3000 J. D. 2000 J. E. 0 J. 2013 Pearson Education, Inc. Slide 19-55

QuickCheck 19.9 Which heat engine has the larger efficiency? A. Engine 1. B. Engine 2. C. They have the same efficiency. D. Can t tell without knowing the number of moles of gas. 2013 Pearson Education, Inc. Slide 19-56

Heat Pumps and Refrigerators Heat engines can run in reverse This is not a natural direction of energy transfer Must put some energy into a device to do this Devices that do this are called heat pumps or refrigerators COP = heating energy transferred at high temp work done by heat pump = Qh W COP = cooling energy transferred at low temp = work done by heat pump Q W C

Refrigerators In a sense, a refrigerator or air conditioner is the opposite of a heat engine. In a heat engine, heat energy flows from a hot reservoir to a cool reservoir, and work W out is produced. In a refrigerator, heat energy is somehow forced to flow from a cool reservoir to a hot reservoir, but it requires work W in to make this happen. 2013 Pearson Education, Inc. Slide 19-41

Refrigerators Shown is the energytransfer diagram of a refrigerator. All state variables (pressure, temperature, thermal energy, etc.) return to their initial values once every cycle. The heat exhausted per cycle by a refrigerator is: Q H = Q C +W in 2013 Pearson Education, Inc. Slide 19-42

Refrigerators The purpose of a refrigerator is to remove heat from a cold reservoir, and it requires work input to do this. We define the coefficient of performance K of a refrigerator to be: If a perfect refrigerator could be built in which W in = 0, then heat would move spontaneously from cold to hot. This is expressly forbidden by the second law of thermodynamics: 2013 Pearson Education, Inc. Slide 19-43

QuickCheck 19.6 The coefficient of performance of this refrigerator is A. 0.40. B. 0.60. C. 1.50. D. 1.67. E. 2.00. 2013 Pearson Education, Inc. Slide 19-44

QuickCheck 19.6 The coefficient of performance of this refrigerator is A. 0.40. B. 0.60. C. 1.50. D. 1.67. E. 2.00. 2013 Pearson Education, Inc. Slide 19-45

Refrigerators Understanding a refrigerator is a little harder than understanding a heat engine. Heat is always transferred from a hotter object to a colder object. The gas in a refrigerator can extract heat Q C from the cold reservoir only if the gas temperature is lower than the cold-reservoir temperature T C. Heat energy is then transferred from the cold reservoir into the colder gas. The gas in a refrigerator can exhaust heat Q H to the hot reservoir only if the gas temperature is higher than the hot-reservoir temperature T H. Heat energy is then transferred from the warmer gas into the hot reservoir.

Refrigerators

Heat Pumps

Coefficient of Performance The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) In heating mode, the COP is the ratio of the heat transferred in to the work required COP = energy transferred at high temp work done by heat pump = Q h W

A heat pump, is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room. Suppose that the ratio of the actual energy entering the room to the work done by the device s motor is 10.0% of the theoretical maximum ratio. Determine the energy entering the room per joule of work done by the motor, given that the inside temperature is 20.0 C and the outside temperature is 5.00 C. energy transferred at high temp Qh COP heating = = work done by heat pump W Q W h Qh = 0.100 W Carnot cycle Qh Th 293 K = 0.100 = 0.100 = 1.17 W T T 293 K 268 K h c 1.17 joules of energy enter the room by heat for each joule of work done.

Second Law Clausius Form It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work. Or energy does not transfer spontaneously by heat from a cold object to a hot object. Section 22.2

The Limits of Efficiency A perfectly reversible engine must use only two types of processes: 1. Frictionless mechanical interactions with no heat transfer (Q = 0) 2. Thermal interactions in which heat is transferred in an isothermal process (ΔE th = 0). Any engine that uses only these two types of processes is called a Carnot engine. A Carnot engine is a perfectly reversible engine; it has the maximum possible thermal efficiency and, if operated as a refrigerator, the maximum possible coefficient of performance.

2 nd Law: Carnot s Theorem No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc. 1796 1832 French engineer

The Carnot cycle is an ideal-gas cycle that consists of the two adiabatic processes (Q = 0) and the two isothermal processes (ΔE th = 0) shown. These are the two types of processes allowed in a perfectly reversible gas engine. Section 22.4

Carnot Cycle, A to B A B is an isothermal expansion. The gas is placed in contact with the high temperature reservoir, T h. The gas absorbs heat Q h. The gas does work W AB in raising the piston. Section 22.4

Carnot Cycle, B to C B C is an adiabatic expansion. The base of the cylinder is replaced by a thermally nonconducting wall. No energy enters or leaves the system by heat. The temperature falls from T h to T c. The gas does work W BC. Section 22.4

Carnot Cycle, C to D The gas is placed in thermal contact with the cold temperature reservoir. C D is an isothermal compression. The gas expels energy Q c. Work W CD is done on the gas. Section 22.4

Carnot Cycle, D to A D A is an adiabatic compression. The base is replaced by a thermally nonconducting wall. So no heat is exchanged with the surroundings. The temperature of the gas increases from T c to T h. The work done on the gas is W DA. Section 22.4

Carnot Engine Carnot Cycle A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible. This sets an upper limit on the efficiencies of all other engines Q T T = and ec = 1 Q T T c c c h h h Temperatures must be in Kelvins

(a) (b) Carnot Cycle Problem An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250 C, and the isothermal compression takes place at 50.0 C. The gas takes in 1 200 J of energy from the hot reservoir during the isothermal expansion. Find the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle.

Carnot Cycle in Reverse Theoretically, a Carnot-cycle heat engine can run in reverse This would constitute the most effective heat pump available This would determine the maximum possible COPs for a given combination of hot and cold reservoirs

COP, Heating Mode COP is similar to efficiency Q h is typically higher than W Values of COP are generally greater than 1 It is possible for them to be less than 1 We would like the COP to be as high as possible

COP, Cooling Mode In cooling mode, you gain energy from a cold temperature reservoir COP = Q c W A good refrigerator should have a high COP Typical values are 5 or 6

Can this refrigerator be built? W = Q Q H COP = Q c W C COP C Q Tc W T T c = = h c

2nd Law of Thermo Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction. Heat flows from hot to cold. Alternative: Irreversible processes must have an increase in Entropy; Reversible processes have no change in Entropy. Entropy is a measure of disorder in a system

The Limits of Efficiency Everyone knows that heat can produce motion. That it possesses vast motive power no one can doubt, in these days when the steam engine is everywhere so well known.... Notwithstanding the satisfactory condition to which they have been brought today, their theory is very little understood. The question has often been raised whether the motive power of heat is unbounded, or whether the possible improvements in steam engines have an assignable limit. Sadi Carnot

Reversible and Irreversible Processes The Arrow of Time! Play the Movie Backwards!

Entropy

Entropy on a Microscopic Scale We can treat entropy from a microscopic viewpoint through statistical analysis of molecular motions. A connection between entropy and the number of microstates (W) for a given macrostate is S = k B ln W The more microstates that correspond to a given macrostate, the greater the entropy of that macrostate. This shows that entropy is a measure of disorder. Section 22.8

Free Expansion Consider an adiabatic free expansion. This process is irreversible since the gas would not spontaneously crowd into half the volume after filling the entire volume. The change in entropy is greater than zero for a irreversible processes. Section 22.7

S in a Free Expansion Consider an adiabatic free expansion. This process is irreversible since the gas would not spontaneously crowd into half the volume after filling the entire volume. Q = 0 but we need to find Q r Choose an isothermal, reversible expansion in which the gas pushes slowly against the piston while energy enters from a reservoir to keep T constant. dq T f r S = = i S = nr 1 T V ln f Vi i f dq r Since V f > V i, S is positive This indicates that both the entropy and the disorder of the gas increase as a result of the irreversible adiabatic expansion. Section 22.7

Entropy and Heat The original formulation of entropy dealt with the transfer of energy by heat in a reversible process. Let dq r be the amount of energy transferred by heat when a system follows a reversible path. The change in entropy, ds is ds = dq r T The change in entropy depends only on the endpoints and is independent of the actual path followed. The entropy change for an irreversible process can be determined by calculating the change in entropy for a reversible process that connects the same initial and final points. Section 22.6

dq T f r S = = i 1 T i f dq r Entropy increases when: Temperature increases Q flows into the system Volume increases Pressure decreases??

Heat Death of the Universe Ultimately, the entropy of the Universe should reach a maximum value. At this value, the Universe will be in a state of uniform temperature and density. All physical, chemical, and biological processes will cease. The state of perfect disorder implies that no energy is available for doing work. This state is called the heat death of the Universe.

Big Bang Cosmogenesis

The Stellar Age Stars Rule for a Trillion Years

In about 5 billion years, Our Sun will swell into a cool Red Giant, engulfing Mercury, Venus and possibly Earth!

The Degenerate Age After about 100 trillion, years, the stars are dead. By 10 37 years, the material in the "Local Galaxy" consists of isolated stellar remnants and black holes. Everything is cold and dark.

The Black Hole Age All the stars turn into black holes. After about 10 100 years, all the black holes are gone.

The Dark Era The remaining black holes evaporate: first the small ones, and then the supermassive black holes. All matter that used to make up the stars and galaxies has now degenerated into photons and leptons.

Boltzmann was subject to rapid alternation of depressed moods with elevated, expansive or irritable moods, likely the symptoms of undiagnosed bipolar disorder. On September 5, 1906, while on a summer vacation in Duino, near Trieste, Boltzmann hanged himself during an attack of depression. He is buried in the Viennese Zentralfriedhof; his tombstone bears the inscription.