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1、Chapter 20 - Thermodynamics,A PowerPoint Presentation byPaul E. Tippens, Professor of PhysicsSouthern Polytechnic State University, 2019,Chapter 20 - ThermodynamicsA P,THERMODYNAMICS,Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy a
2、nd heat transfer.,Central Heating,THERMODYNAMICSThermodynamics i,Objectives: After finishing this unit, you should be able to:,State and apply the first and second laws of thermodynamics.,Demonstrate your understanding of adiabatic, isochoric, isothermal, and isobaric processes.,Write and apply a re
3、lationship for determining the ideal efficiency of a heat engine.,Write and apply a relationship for determining coefficient of performance for a refrigeratior.,Objectives: After finishing th,A THERMODYNAMIC SYSTEM,A system is a closed environment in which heat transfer can take place. (For example,
4、 the gas, walls, and cylinder of an automobile engine.),Work done on gas or work done by gas,A THERMODYNAMIC SYSTEMA system,INTERNAL ENERGY OF SYSTEM,The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system.,Usually the internal energy
5、consists of the sum of the potential and kinetic energies of the working gas molecules.,INTERNAL ENERGY OF SYSTEMThe i,TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.,HEAT PUT INTO A SYSTEM (Positive),TWO WAYS TO INCREASE THE INTER,WORK DONE BY EXPANDING GAS: W is positive,-UDecrease,TWO WAYS TO DECREA
6、SE THE INTERNAL ENERGY, U.,WORK DONE BY EXPANDING GAS: ,THERMODYNAMIC STATE,The STATE of a thermodynamic system is determined by four factors:,Absolute Pressure P in PascalsTemperature T in KelvinsVolume V in cubic meters,Number of moles, n, of working gas,THERMODYNAMIC STATEThe STATE o,THERMODYNAMI
7、C PROCESS,Increase in Internal Energy, U.,Initial State:P1 V1 T1 n1,Final State:P2 V2 T2 n2,Wout,Work by gas,THERMODYNAMIC PROCESSIncrease,The Reverse Process,Decrease in Internal Energy, U.,Initial State:P1 V1 T1 n1,Final State:P2 V2 T2 n2,The Reverse ProcessDecrease in,THE FIRST LAW OF THERMODYAMI
8、CS:,The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system.,Q = U + W final - initial),Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.,THE FIRST LAW OF THERMODYAMICS,SI
9、GN CONVENTIONS FOR FIRST LAW,Heat Q input is positive,Q = U + W final - initial),Heat OUT is negative,Work BY a gas is positive,Work ON a gas is negative,+Qin,+Wout,U,-Win,-Qout,U,SIGN CONVENTIONS FOR FIRST LAW,APPLICATION OF FIRST LAW OF THERMODYNAMICS,Example 1: In the figure, the gas absorbs 400
10、J of heat and at the same time does 120 J of work on the piston. What is the change in internal energy of the system?,Wout =120 J,APPLICATION OF FIRST LAW OF T,Example 1 (Cont.): Apply First Law,U = +280 J,U = Q - W = (+400 J) - (+120 J) = +280 J,DW is positive: +120 J (Work OUT),Q = U + W,U = Q - W
11、,DQ is positive: +400 J (Heat IN),Example 1 (Cont.): Apply First,Example 1 (Cont.): Apply First Law,U = +280 J,The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J,The increase in internal energy is:,Energy is conserved:,E
12、xample 1 (Cont.): Apply First,FOUR THERMODYNAMIC PROCESSES:,Isochoric Process: V = 0, W = 0 Isobaric Process: P = 0 Isothermal Process: T = 0, U = 0 Adiabatic Process: Q = 0,Q = U + W,FOUR THERMODYNAMIC PROCESSES:,Q = U + W so that Q = U,ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0,0,HEAT IN = I
13、NCREASE IN INTERNAL ENERGYHEAT OUT = DECREASE IN INTERNAL ENERGY,No Work Done,Q = U + W so that,ISOCHORIC EXAMPLE:,Heat input increases P with const. V,400 J heat input increases internal energy by 400 J and zero work is done.,ISOCHORIC EXAMPLE: Heat input,Q = U + W But W = P V,ISOBARIC PROCESS: CON
14、STANT PRESSURE, P = 0,+U,-U,QIN,QOUT,HEAT IN = Wout + INCREASE IN INTERNAL ENERGY,Work Out,Work In,HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY,Q = U + W But W,ISOBARIC EXAMPLE (Constant Pressure):,Heat input increases V with const. P,400 J heat does 120 J of work, increasing the internal energy by
15、 280 J.,400 J,ISOBARIC EXAMPLE (Constant Pre,ISOBARIC WORK,400 J,Work = Area under PV curve,PA = PB,ISOBARIC WORK 400 JWork = Area,ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0,NET HEAT INPUT = WORK OUTPUT,Q = U + W AND Q = W,U = 0,U = 0,QOUT,Work In,Work Out,QIN,WORK INPUT = NET HEAT OUT,ISO
16、THERMAL PROCESS: CONST. TE,ISOTHERMAL EXAMPLE (Constant T):,PAVA = PBVB,Slow compression at constant temperature: - No change in U.,U = T = 0,ISOTHERMAL EXAMPLE (Constant T,ISOTHERMAL EXPANSION (Constant T):,400 J of energy is absorbed by gas as 400 J of work is done on gas. T = U = 0,U = T = 0,Isot
17、hermal Work,ISOTHERMAL EXPANSION (Constant,Q = U + W ; W = -U or U = -W,ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0,Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy,Q = U + W ; W = -U o,ADIABATIC EXAMPLE:,Insulated Walls: Q = 0,Expanding gas does work with zero heat loss.
18、Work = -DU,ADIABATIC EXAMPLE: Insulated W,ADIABATIC EXPANSION:,400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0,Q = 0,ADIABATIC EXPANSION: 400 J of,MOLAR HEAT CAPACITY,Check with your instructor to see if this more thorough treatment of thermodynamic p
19、rocesses is required.,MOLAR HEAT CAPACITYOPTIONAL TR,SPECIFIC HEAT CAPACITY,Remember the definition of specific heat capacity as the heat per unit mass required to change the temperature?,For example, copper: c = 390 J/kgK,SPECIFIC HEAT CAPACITYRemember,MOLAR SPECIFIC HEAT CAPACITY,The “mole” is a b
20、etter reference for gases than is the “kilogram.” Thus the molar specific heat capacity is defined by:,For example, a constant volume of oxygen requires 21.1 J to raise the temperature of one mole by one kelvin degree.,MOLAR SPECIFIC HEAT CAPACITYTh,SPECIFIC HEAT CAPACITYCONSTANT VOLUME,How much hea
21、t is required to raise the temperature of 2 moles of O2 from 0oC to 100oC?,Q = (2 mol)(21.1 J/mol K)(373 K - 273 K),Q = nCv T,Q = +4220 J,SPECIFIC HEAT CAPACITYCONSTAN,SPECIFIC HEAT CAPACITYCONSTANT VOLUME (Cont.),Since the volume has not changed, no work is done. The entire 4220 J goes to increase
22、the internal energy, U.,Q = U = nCv T = 4220 J,U = nCv T,Thus, U is determined by the change of temperature and the specific heat at constant volume.,SPECIFIC HEAT CAPACITYCONSTAN,SPECIFIC HEAT CAPACITYCONSTANT PRESSURE,We have just seen that 4220 J of heat were needed at constant volume. Suppose we
23、 want to also do 1000 J of work at constant pressure?,Q = U + W,Q = 4220 J + J,Q = 5220 J,Cp Cv,Same,SPECIFIC HEAT CAPACITYCONSTAN,HEAT CAPACITY (Cont.),Cp Cv,For constant pressureQ = U + WnCpT = nCvT + P V,U = nCvT,Heat to raise temperature of an ideal gas, U, is the same for any process.,HEAT CAPA
24、CITY (Cont.)Cp CvFo,REMEMBER, FOR ANY PROCESS INVOLVING AN IDEAL GAS:,PV = nRT,U = nCv T,Q = U + W,REMEMBER, FOR ANY PROCESS INVO,Example Problem:,AB: Heated at constant V to 400 K.,A 2-L sample of Oxygen gas has an initial temp-erature and pressure of 200 K and 1 atm. The gas undergoes four process
25、es:,BC: Heated at constant P to 800 K.CD: Cooled at constant V back to 1 atm.DA: Cooled at constant P back to 200 K.,Example Problem:AB: Heated at,PV-DIAGRAM FOR PROBLEM,How many moles of O2 are present?,Consider point A: PV = nRT,PV-DIAGRAM FOR PROBLEMBAPB2 L,PROCESS AB: ISOCHORIC,What is the press
26、ure at point B?,PROCESS AB: ISOCHORICWhat is t,PROCESS AB: Q = U + W,Analyze first law for ISOCHORIC process AB.,W = 0 Q = U = nCv T,U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K),Q = +514 J,W = 0,U = +514 J,PROCESS AB: Q = U + WAnalyz,PROCESS BC: ISOBARIC,What is the volume at point C (& D)?,V C = V
27、D = 4 L,PROCESS BC: ISOBARICWhat is th,FINDING U FOR PROCESS BC.,Process BC is ISOBARIC.,P = 0 U = nCv T,U = (0.122 mol)(21.1 J/mol K)(800 K - 400 K),U = +1028 J,FINDING U FOR PROCESS BC. Pro,FINDING W FOR PROCESS BC.,Work depends on change in V.,P = 0 Work = P V,W = (2 atm)(4 L - 2 L) = 4 atm L = 4
28、05 J,W = +405 J,FINDING W FOR PROCESS BC. Wor,FINDING Q FOR PROCESS BC.,Analyze first law for BC.,Q = U + WQ = +1028 J + 405 JQ = +1433 J,Q = 1433 J,W = +405 J,U = 1028 J,FINDING Q FOR PROCESS BC. Ana,PROCESS CD: ISOCHORIC,What is temperature at point D?,T D = 400 K,PROCESS CD: ISOCHORICWhat is t,PR
29、OCESS CD: Q = U + W,Analyze first law for ISOCHORIC process CD.,W = 0 Q = U = nCv T,U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K),Q = -1028 J,W = 0,U = -1028 J,PROCESS CD: Q = U + WAnalyz,FINDING U FOR PROCESS DA.,Process DA is ISOBARIC.,P = 0 U = nCv T,U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K),U
30、= -514 J,FINDING U FOR PROCESS DA. Pro,FINDING W FOR PROCESS DA.,Work depends on change in V.,P = 0 Work = P V,W = (1 atm)(2 L - 4 L) = -2 atm L = -203 J,W = -203 J,FINDING W FOR PROCESS DA. Wor,FINDING Q FOR PROCESS DA.,Analyze first law for DA.,Q = U + WQ = -514 J - 203 JQ = -717 J,Q = -717 J,W =
31、-203 J,U = -514 J,FINDING Q FOR PROCESS DA. Ana,PROBLEM SUMMARY,PROBLEM SUMMARYDQ = DU + DWFor,NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA,Area = (1 atm)(2 L)Net Work = 2 atm L = 202 J,NET WORK FOR COMPLETE CYCLE IS,ADIABATIC EXAMPLE:,Q = 0,Example 2: A diatomic gas at 300 K and 1 atm is compressed
32、 adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4),ADIABATIC EXAMPLE: Q = 0ABPBV,ADIABATIC (Cont.): FIND PB,Q = 0,PB = 32.4 atm or 3284 kPa,ADIABATIC (Cont.): FIND PB Q,ADIABATIC (Cont.): FIND TB,Q = 0,TB = 810 K,ADIABATIC (Cont.): FIND TB
33、Q,ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND W,Q = 0,W = - U = - nCV T & CV= 21.1 j/mol K,Find n from point A,PV = nRT,ADIABATIC (Cont.): If VA= 96 c,ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND W,n = 0.000325 mol & CV= 21.1 j/mol K,T = 810 - 300 = 510 K,W = - U = - nCV T,W = - 3.5
34、0 J,ADIABATIC (Cont.): If VA= 96 c,Absorbs heat Qhot Performs work WoutRejects heat Qcold,A heat engine is any device which through a cyclic process:,HEAT ENGINES,Absorbs heat Qhot A heat engin,THE SECOND LAW OF THERMODYNAMICS,It is impossible to construct an engine that, operating in a cycle, produ
35、ces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.,Not only can you not win (1st law); you cant even break even (2nd law)!,THE SECOND LAW OF THERMODYNAMI,THE SECOND LAW OF THERMODYNAMICS,THE SECOND LAW OF THERMODYNAMI,EFFICIENCY OF A
36、N ENGINE,The efficiency of a heat engine is the ratio of the net work done W to the heat input QH.,EFFICIENCY OF AN ENGINECold Re,EFFICIENCY EXAMPLE,An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency?,e = 25%,Question: How many joules of work is done?,EFFICIENCY EXAMPLECold
37、 Res. TC,EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine),For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.,EFFICIENCY OF AN IDEAL ENGINE,Example 3: A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the a
38、ctual efficiency is only half of the ideal efficiency, how much work is done during each cycle?,e = 40%,Actual e = 0.5ei = 20%,W = eQH = 0.20 (600 J),Work = 120 J,Example 3: A steam engine abs,REFRIGERATORS,A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from co
39、ld reservoir and depositing heat into hot reservoir.,Win + Qcold = Qhot,WIN = Qhot - Qcold,REFRIGERATORSA refrigerator is,THE SECOND LAW FOR REFRIGERATORS,It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0.,If th
40、is were possible, we could establish perpetual motion!,THE SECOND LAW FOR REFRIGERATO,COEFFICIENT OF PERFORMANCE,The COP (K) of a heat engine is the ratio of the HEAT Qc extracted to the net WORK done W.,For an IDEAL refrigerator:,COEFFICIENT OF PERFORMANCECold,COP EXAMPLE,A Carnot refrigerator oper
41、ates between 500 K and 400 K. It extracts 800 J from a cold reservoir during each cycle. What is C.O.P., W and QH ?,C.O.P. (K) = 4.0,COP EXAMPLEA Carnot refrigerat,COP EXAMPLE (Cont.),Next we will find QH by assuming same K for actual refrigerator (Carnot).,QH = 1000 J,COP EXAMPLE (Cont.)Next we wil
42、,COP EXAMPLE (Cont.),Now, can you say how much work is done in each cycle?,Work = 1000 J - 800 J,Work = 200 J,COP EXAMPLE (Cont.)Now, can yo,Summary,Q = U + W final - initial),The First Law of Thermodynamics: The net heat taken in by a system is equal to the sum of the change in internal energy and
43、the work done by the system.,Isochoric Process: V = 0, W = 0 Isobaric Process: P = 0 Isothermal Process: T = 0, U = 0 Adiabatic Process: Q = 0,SummaryQ = U + W ,Summary (Cont.),U = nCv T,The Molar Specific Heat capacity, C:,Units are:Joules per mole per Kelvin degree,The following are true for ANY p
44、rocess:,Q = U + W,PV = nRT,Summary (Cont.) c = QU = nCv,Summary (Cont.),The Second Law of Thermo: It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work.,Not only can you not win (1st law); you cant even break even (2nd law)!,Summary (Cont.)The Second Law,Summary (Cont.),The efficiency of a heat engine:,The coefficient of performance of a refrigerator:,Summary (Cont.)The efficiency,CONCLUSION: Chapter 20Thermodynamics,CONCLUSION: Chapter 20Thermod,
