《化工热力学》课程授课教案（讲义）Chapter 5 The Second Law of Thermodynamics

Chapter 5 The Second Law of Thermodynamics Introduction 5.1 Statements of the Second Law 5.2 Heat Engines 5.4 Entropy 5.5 Entropy Change of an Ideal as 5.6 Mathematical Statement of the Second Law 5.7 Entropy Balance for Open Systems 5.8 Calculation of Ideal Work 5.9 Lost Work Introduction The First Law of thermodynamics Conservation and transformation of energy Can the process which accords with the First Law occur spontaneousty? :A process which can occur automatically without any outside effect eous process exist mposes no restriction on the process direction The Second Law of thermodynamics to describe the direction.conditions and bounds of the spontaneous process 5.1 Statements of the Second Law Statement /(Kelvin-Planck Statement):No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system. 不可能从单一热源取热，并使之完全转变为有用功而不产生其它影响 Statement 2 (Clausius t)No is possible which consists solely in the transfer of heat from one temperature level to a higher one 不可能将热从低温物体传至高温物体而不引起其它变化 两种表述完全等效：违反一种表述，必违反另一种表述： It is impossible by a eyclic process to convert the heat absorbed by a system mpletely into work done by the Se:perpetual-motion machine of the kind can'tbe buil 5.2 Heat Engines Heat engine is a device or machine that roduce work from heat in a cyclic pr The First Law therefore reduces to w=lo-loc The thermal efficiency Offering water

Chapter 5 The Second Law of Thermodynamics Introduction 5.1 Statements of the Second Law 5.2 Heat Engines 5.4 Entropy 5.5 Entropy Change of an Ideal Gas 5.6 Mathematical Statement of the Second Law 5.7 Entropy Balance for Open Systems 5.8 Calculation of Ideal Work 5.9 Lost Work Introduction The First Law of thermodynamics Conservation and transformation of energy Can the process which accords with the First Law occur spontaneously? spontaneous process: A process which can occur automatically without any outside effect In nature, the spontaneous process exist direction The First Law of thermodynamics reflects the observation that energy is conserved, but it imposes no restriction on the process direction The Second Law of thermodynamics to describe the direction, conditions and bounds of the spontaneous process 5.1 Statements of the Second Law Statement 1 (Kelvin－Planck Statement):No apparatus can operate in such a way that its only effect is to convert heat absorbed by a system completely into work done by the system. 不可能从单一热源取热，并使之完全转变为有用功而不产生其它影响 Statement 2 (Clausius statement):No process is possible which consists solely in the transfer of heat from one temperature level to a higher one. 不可能将热从低温物体传至高温物体而不引起其它变化 两种表述完全等效! 违反一种表述,必违反另一种表述! Statement 1a: It is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system. Statement 1b: perpetual-motion machine of the second kind can’t be built 5.2 Heat Engines Heat engine is a device or machine that produce work from heat in a cyclic process For example: a steam power plant The First Law therefore reduces to: The thermal efficiency condenser Q c boiler turbin e Generato r W Q H Offering water W Q Q = − H C

net work output n heat absorbed m_-1- -jeu eul 2h :thermal efficiency of a steam power plant is about40% Thinking(5-1) What is the upper limit of the thermal efficiency of a heat engine Carnot heat engine Carnot engine: ismade up of four steps Step 1-4:Reversible adiabatic expansion process Step4-3:Reversible isothermal evaporation process T Step3-2:Reversible adiabatic compression process Step2-1:Reversible isothermal condensing process TS diagram# Carnot engine is an engine operating a Camot cycle Carnot's Theorem For two given heat reservoirs no engine can have a thermal TH efficiency higher than that of a Carnot engine. Carnot efficiency -1-T Tu ◆W The thermal efficiency of a Carnot engine depends only on the temperature levels and not upon the working substance of the 5.4Entropy Te For Carnot Cycle =11 1C l2ml .2c-0 T I For micro Carnot Cycle Te 4-0 S'is the total (rather than s.0-or0n=7☒ molar)entropy of the system The discussion of entropy:(1)There exists a property called entropy S.which is an intrinsic able coo swhich characterize the 1T 存在一种性质称为嫡，它是系统性质，在函数上与表现系统特征的可测量的量相联系，对于

For example: thermal efficiency of a steam power plant is about 40% Thinking(5-1) What is the upper limit of the thermal efficiency of a heat engine ? Carnot heat engine Carnot engine: Carnot cycle is made up of four steps: Step 1-4：Reversible adiabatic expansion process Step 4-3：Reversible isothermal evaporation process Step 3-2：Reversible adiabatic compression process Step 2-1：Reversible isothermal condensing process TS diagram# Carnot engine is an engine operating a Carnot cycle Carnot’s Theorem For two given heat reservoirs no engine can have a thermal efficiency higher than that of a Carnot engine. Carnot efficiency The thermal efficiency of a Carnot engine depends only on the temperature levels and not upon the working substance of the engine. 5.4 Entropy For Carnot Cycle For micro Carnot Cycle Definition of entropy The discussion of entropy: (1) There exists a property called entropy S, which is an intrinsic property of a system, functionally related to the measurable coordinates which characterize the system. For a reversible process, changes in this property are given by equation as follow: 存在一种性质称为熵，它是系统性质，在函数上与表现系统特征的可测量的量相联系，对于 net work output heat absorbed   H C C H H H 1 W Q Q Q Q Q Q  −  = = − 2 4 3 1 T S TH Tc TH TC Rc QH QC W 1 C H H W T Q T  − = = − 1 1 c C H H Q T Q T  = − = − c H C H Q Q T T = 0 H C H C Q Q T T + = 0 H C H C dQ dQ T T + = 0 rev dQ T =  t rev dQ dS T = or t rev S dQ TdS = t is the total (rather than molar) entropy of the system t rev dQ dS T =

可逆过程，该性质的变化量可用下式求出 (2)For any system undergoing a finite reversible process is 4s-∫№ (3)For a system undergoes an irreversible process between two equilibrium states.the entropy change of the ystem is evaluated byqution followed to a arbitrarily chosen reversibl proces that accomplishes the same change of state as the actual process For example: 5.5 entropy change of an ideal gas For a process of an ideal gas Initial state Final state PoTo △S=? PT For one mole or a unit mass of ideal gas undergoing a mechanically reversible process in a closed system du =de-pdv H=U+pV dS=Cg T-R T P as-cd-In P A general equation for the calculation of entropy change of an ideal gas 5.6 Mathematical statement of the Second Law roceeds in such a that the tol entropy change with it is positive,value ofe being attained only by a reversible process.No process is possible for which the total entropy decreases. △Saa≥2d theorem of entropy increasing The entropy of an isolated system during a process always increase or,in the limiting case of a reversible process,remains constant.孤立系统的熵只能增大，或者不变，绝不能减小

可逆过程，该性质的变化量可用下式求出 (2) For any system undergoing a finite reversible process is (3) For a system undergoes an irreversible process between two equilibrium states, the entropy change of the system is evaluated by equation followed to an arbitrarily chosen reversible process that accomplishes the same change of state as the actual process For example: 5.5 entropy change of an ideal gas For a process of an ideal gas For one mole or a unit mass of ideal gas undergoing a mechanically reversible process in a closed system 5.6 Mathematical statement of the Second Law This mathematical statement of the second law affirms that every process proceeds in such a direction that the total entropy change associated with it is positive, the limiting value of zero being attained only by a reversible process. No process is possible for which the total entropy decreases. theorem of entropy increasing The entropy of an isolated system during a process always increase or, in the limiting case of a reversible process, remains constant. 孤立系统的熵只能增大，或者不变，绝不能减小。 Initial state P0 T0 Final state P T Δ ？ S = ig P dT dP dS C R T P = − 0 0 ln ig T P T S dT P C R R T P = −  Δ A general equation for the calculation of entropy change of an ideal gas rev dU dQ pdV = − H U pV = + rev dQ dS T = t rev dQ S T = Δ  t rev ACB dQ S T = Δ  t rev ADB dQ S T = Δ  V P D B A C Q dS T  0  total ΔS 

Thinking(5-2) 第二类永动机？ 如果三峡水电站用降温法发电，使水温降低5C,发电能力可提高11.7倍。 设水位差为180米 重力势能转化为电能 E=mgh =1800m(J] mkg水降低5C放热： 2_21000m=117 E1800m Q=cm△1=21000mlJ] Thinking(5-3) 若工质从同一初态，分别经可逆和不可逆过程，到达同一终态，己知两过程热源相同，问传 热量是否相同？ △S≥∫一可递过程 >:不可逆过程 相同初终态，△s相同 热源T相同 80R>60R Q+AU+W 阳 Thinking(5-4) 若工质从同一初态出发，从相同热源吸收相同热量，问末态箱可逆与不可逆谁大？ AS2 ：可逆过程 相阿路层格短而明不思 相同初态相同S2R>S2R

Thinking(5-2) 第二类永动机??? 如果三峡水电站用降温法发电，使水温降低 5C，发电能力可提高 11.7 倍。 设水位差为 180 米 重力势能转化为电能： Thinking(5-3) 若工质从同一初态，分别经可逆和不可逆过程，到达同一终态，已知两过程热源相同，问传 热量是否相同？ 相同初终态，s 相同 热源 T 相同 Thinking(5-4) 若工质从同一初态出发，从相同热源吸收相同热量，问末态熵可逆与不可逆谁大？ E mgh m J = =1800 [ ] mkg水降低5C放热： Q cm t m J =  = 21000 [ ] 21000 11.7 1800 Q m E m = = 相同热量，热源T相同 Q S T     =：可逆过程 >：不可逆过程 IR R    S S 相同初态s1相同 2,IR 2,R S S  Q S T     =：可逆过程 >：不可逆过程 Q Q R IR    Q U W =  + 相同 W W R IR 

5.7 Entropy Balance for Open System 5.7.1 Entropy Balance for Open Systems ∑s (mS) For a control volume(CV)inunit time rroundings (1)Entropy changes within the CV d(mS)/dt (2)Entropy changes in the streams flowing in and out of the CVA( (3)Entropy changes in heat transfer of the CV with surroundings (entropy change in the surroundings) ds /dt (4)Entropy generationS Time rate of Time rate of Net rate of changes of Total rate of changes of changes of entropy entropy of flowing entropy in the in surroundings streams entropy gen tion 物流熵的净 控制体内箱随时间 环境熵随时间的变化 变化率 三熵的总产率 的变化率 率 )d( @-20 d。d山 This equation is the general rate form of the entropy balance,applicable at any instant △)a=∑iS-∑成S, T 4(d。+dmsa-ygQ d ?

( )fs i i j j i j  = − mS m S m S   , t j sur j j dS Q dt T = − , t j sur j j dS Q dt T = − dSsur Q dt T = − sur dQ dS T = − dQ dS T = , ( ) ( ) 0 CV j fs G j j d mS Q mS S dt T  + − =   G S 5.7 Entropy Balance for Open System 5.7.1 Entropy Balance for Open Systems For a control volume (CV) in unit time (1)Entropy changes within the CV (2) Entropy changes in the streams flowing in and out of the CV (3) Entropy changes in heat transfer of the CV with surroundings (entropy change in the surroundings) (4) Entropy generation This equation is the general rate form of the entropy balance, applicable at any instant Time rate of changes of entropy in the CV Time rate of changes of entropy in surroundings Total rate of entropy generation Net rate of changes of entropy of flowing + + streams = mSi i i j j j m S Qj ( ) mS cv Surroundings GS ( ) / cv d mS dt ( )  mS fs G S / t sur dS dt 物流熵的净 变化率 控制体内熵随时间 的变化率 环境熵随时间的变化 率 + = 熵的总产率 + ( ) ( ) 0 t CV sur fs G d mS dS mS S dt dt  + + =  ？ ？ ? ？ ？

for irreversible processes for reversible processes of irreversibility implying:(1)the process is reversible within theCV (1)Internal irreversibility (2)Heat transfer between the CV and its (2)external thermal irreversibility surroundings is reversible A+0mL-zg-号之0 dt T For a steady-state flow process in the CV m=constant d(mS)=0△s-∑g=S。≥0 S=constant Foronend-∑g=号之0 Thinking(5-5)Ifthe CV is a closed system A+msL-Σg=号≥02 dt T 成=成=0△成)6=0 dmsa-∑g-{≥0 d山 T Thinking(5-6)Other form of entropy balance in unit time -空-%足0 dt ndr tme dr∑is,-∑is+dmsa-dΣ号=$ ∑6mS,-∑6mS+dms-∑2=。 T Entropy balance in differential form m5-工n5+aMm.空2-a Entropy balance in Chinese textbook ∑mS-∑mS+∫2+a,=AS

0 G S  , ( ) ( ) 0 CV j fs G j j d mS Q mS S dt T  + − =   ( ) 0 CV d mS dt = , 0 j G j j Q S S T  − =   , ( ) 0 j fs G j j Q mS S T  − =   , ( ) ( ) 0 CV j fs G j j d mS Q mS S dt T  + − =   , ( ) 0 CV j G j j d mS Q S dt T − =   , ( ) 0 CV j j j i i G j i j j d mS Q m S m S S dt T    − + − =  , ( ) j j j i i CV G j i j j Q dt m S dt m S d mS dt S dt T    − + − = , ( ) j j j i i CV G j i j j Q m S m S d mS dS T       − + − = , ( ) j j j i i CV G j i j j Q m S m S mS S T     − +  − =   i i j j G sys i j Q m S m S S S T   − + +  =   for irreversible processes for reversible processes two sources of irreversibility implying: (1) the process is reversible within the CV (1) Internal irreversibility (2) Heat transfer between the CV and its (2) external thermal irreversibility surroundings is reversible For a steady-state flow process in the CV For one entrance and exit Thinking(5-5) If the CV is a closed system Thinking(5-6) Other form of entropy balance in unit time in dt time Entropy balance in differential form Entropy balance in integration form in Chinese textbook 0 G S = m = constant S = constant ？ 0 m m i j = = ( ) 0  = mS fs

Thinking(5-7刀 Σms-ΣnS+Aw-2-a For a closed system ms-2到2=Aa For a steady-flow process Σ叫5-Ym-2=as ∑m,S,-∑m,S=AS。 For a adiabatic and steady-flow process Thinking(5-8) 1.如果系统经历某一变化过程后，嫡值不变，该过程一定是可逆绝热的 2.如果系统经历某一绝热变化过程后，熵值不变，该过程一定是可逆的 3.在自发过程中，系统的熵变恒大于零 2-2aaa2%-a战 4.某封闭系统经历一可逆过程，系统所做的功和排出的热量分别为100和45，问系 统的熵变： A为正B为负C不能判断 5.某封闭系统经历一不可逆过程，系统所做的功和排出的热量分别为100k)和45，问 系续的培变： A为正B为负 C不能判断 6。某流体在稳流装置内经历一个不可逆绝热过程，向外做的功为24,问流体流入、流出 系统的熵变： A为正B为负C不能判断 7.某流体在稳流装置内经历一个不可逆过程，加给装置的功25从，从此装置带走的热量（即 流体吸热)为10k,问流体流入、流出系统的熵变： A为正B为负 C不能判断 5.8 Ideal Work In any steady-state flow process requiring work,there is an absolute minimum amount which must be expected the desired change of stae In,there maximum mount which may be accomplished as the result of a given change of a state 系统在变化过程中，由于途径不同，所产生（或消耗）的功不一样。理想功就是系统的状 态变化以完全可逆方式完成，理论上产生最大功或者消耗最小功。因此理想功是一个理想的 极限值，可作为实际功的比较标准

, ( ) j j j i i CV G j i j j Q m S m S mS S T     − +  − =   , ( ) j CV G j j Q mS S T   − =   , j j j i i G j i j j Q m S m S S T     − − =   j j i i G j i   m S m S S − =  , ( ) ( ) 0 CV j fs G j j d mS Q mS S dt T  + − =   , ( ) j j j i i CV G j i j j Q m S m S mS S T     − +  − =   Thinking(5-7) For a closed system For a steady-flow process For a adiabatic and steady-flow process Thinking(5-8) 1. 如果系统经历某一变化过程后，熵值不变，该过程一定是可逆绝热的 2. 如果系统经历某一绝热变化过程后，熵值不变，该过程一定是可逆的 3. 在自发过程中，系统的熵变恒大于零 4. 某封闭系统经历一可逆过程，系统所做的功和排出的热量分别为100 kJ 和 45 kJ， 问系 统的熵变： A 为正 B 为负 C 不能判断 5. 某封闭系统经历一不可逆过程，系统所做的功和排出的热量分别为100 kJ 和 45 kJ， 问 系统的熵变： A 为正 B 为负 C 不能判断 6. 某流体在稳流装置内经历一个不可逆绝热过程，向外做的功为 24 kJ，问流体流入、流出 系统的熵变： A 为正 B 为负 C 不能判断 7. 某流体在稳流装置内经历一个不可逆过程，加给装置的功 25 kJ，从此装置带走的热量（即 流体吸热）为10 kJ，问流体流入、流出系统的熵变： A 为正 B 为负 C 不能判断 5.8 Ideal Work In any steady-state flow process requiring work, there is an absolute minimum amount which must be expected to accomplish the desired change of state ; In a process producing work, there is an absolute maximum amount which may be accomplished as the result of a given change of a state. 系统在变化过程中，由于途径不同，所产生(或消耗)的功不一样。理想功就是系统的状 态变化以完全可逆方式完成，理论上产生最大功或者消耗最小功。因此理想功是一个理想的 极限值, 可作为实际功的比较标准

完全可逆指的是不仅系统内的所有变化是完全可逆的，而且系统和环境之间的能量交换 (如传热)过程也可逆的。 For a steady-flow,irreversible process at the uniform surroundings temperature A)6+ -9-0 d =Tn△(m) Energy balance neglect the kinetic-and potential-energy terms Wa=△(Hms-Tn△(Sms Expressed as ratesA(Hm)-TA(Sm) On a unit-mass basis W =AH-TAS 结论 (①)理想功决定与体系的始、终状态和环境的状态，与过程无关： (②)体系发生状态变化的每一个实际过程都有其对应的理想功。 Thermal efficiency for process re g work 成a>0 n,(work required)= or process producing work Wdoa >W Wioa <0 n≤1 rk prodced）=a. 例6-2求298K,0.1013MPa的水变成273K,同压力下冰的过程的理想功。设环境温度分别 为1)298K:(2)248K。 解：忽略压力的影响。查得有关数据

( ) 0 fs Q mS T  − = ( ) Q T mS =  fs , ( ) ( ) 0 CV j fs G j j d mS Q mS S dt T  + − =   2 ( ) 2 s fs u H gz m Q W rev      + + = +         2 ( ) 2 ideal fs fs u W H gz m T Sm      =  + + −          ( ) ( ) W Hm T Sm ideal fs fs =  −  ( ) ( ) W Hm T Sm ideal fs fs =  −  W H T S ideal =  −  W W ideal s  0 Wideal  ( ) ideal t s W work required W  = W W ideal s  0 Wideal  ( ) t ideals W work produced W  =  1 完全可逆指的是不仅系统内的所有变化是完全可逆的，而且系统和环境之间的能量交换 （如传热）过程也可逆的。 For a steady-flow, irreversible process at the uniform surroundings temperature Energy balance neglect the kinetic- and potential-energy terms Expressed as rates On a unit-mass basis 结论： ⑴ 理想功决定与体系的始、终状态和环境的状态，与过程无关； ⑵ 体系发生状态变化的每一个实际过程都有其对应的理想功。 Thermal efficiency for process requiring work for process producing work 例6-2 求298K，0.1013MPa的水变成273K，同压力下冰的过程的理想功。设环境温度分别 为(1)298K；(2)248K。 解：忽略压力的影响。查得有关数据 0

状态■温度K格kg)熵kgK) H2O(1) 298 104.8 0.3666 H00273-0.02 ￥0 273K下冰的熔化焓变：3347kJ/kg 则对于273K下的冰：H-0.02-334.7=-334.72 S=0-334.71273 122 ()取严1kg,环境温度为298K,高于冰点时 Wm=△H-T.AS=(-334.9-104.8)-298(-12265-0.3666)=35.04kJ1kg 若使水变成冰，需用冰机，理论上应消耗的最小功为35.04kJkg (2)取m=1kg,环境温度为248K,低于冰点时 Wa=△H-T.△S=(-334.9-104.8)-248(-1.2265-0.3666)=-44.61kJ1kmol 当环境温度低于冰点时，水变成冰，不仅不需要消耗外功，而且理论上可以回收的最大 功为44.61kJ/kg 理想功不仅与系统的始、终态有关，而且与环境温度有关 5.9 Lost work Work that is wasted as the result of irreversibilities in a process is called lost work WL.and is defined as the difference between the actual work of a process and the ideal work for the process. Win =WV-Wakn or Wioe =W,-Waindl 系统在相同的状态变化过程中，不可逆过程的实际功与完全可逆过程的理想功之差称为 损失功。 for steady-state flow process W=△ H+ 2+&品 -0 T△(Sm- W-A H +8 -TA()o S=A(Sm)s →T,5。=T,A(Sm-⑨ On a unit-mass basis Wion =TAS-0 Wios =TaSG W.ETS m≥0→3≥0 for a completely reversible process >For the irreversible process

W W W lost s ideal = − W W W lost s ideal = − 0 Wlost  0 G S  W T S lost G   状态 温度/K 焓/(kJ/kg) 熵/(kJ/(kg·K)) H2O(l) 298 104.8 0.3666 H2O(l) 273 -0.02 ≈0 273K下冰的熔化焓变：334.7 kJ/kg 则对于273K下的冰：H=－0.02－334.7=－334.72 S=0－334.7 / 273=－1.226 (1) 取 m=1 kg， 环境温度为298K，高于冰点时 若使水变成冰，需用冰机，理论上应消耗的最小功为35.04kJ/kg (2) 取 m=1 kg，环境温度为248K，低于冰点时 当环境温度低于冰点时，水变成冰，不仅不需要消耗外功，而且理论上可以回收的最大 功为44.61kJ/kg 理想功不仅与系统的始、终态有关，而且与环境温度有关 5.9 Lost work Work that is wasted as the result of irreversibilities in a process is called lost work, WL , and is defined as the difference between the actual work of a process and the ideal work for the process. 系统在相同的状态变化过程中，不可逆过程的实际功与完全可逆过程的理想功之差称为 损失功。 for steady-state flow process at the uniform surroundings T W H T S kJ kg ideal =  −  = − − − − − =  ( 334.9 104.8 298 1.2265 0.3666 35.04 / ) ( ) W H T S kJ kmol ideal =  −  = − − − − − = −  ( 334.9 104.8 248 1.2265 0.3666 44.61 / ) ( ) or 2 2 s fs u W H gz m Q     =  + + −         ( ) W T Sm Q lost fs =  −  2 ( ) 2 ideal fs fs u W H gz m T Sm      =  + + −          ( ) T S T Sm Q   G fs =  − ( ) G fs Q S Sm T =  − W T S lost G   W T S lost G   W T S Q lost  =  − On a unit-mass basis = for a completely reversible process ＞For the irreversible process

The engineering significance of this result is The greater the irreversibility of a process.the greater the rate of entropy production and the greater the amount of energy that becomes unavailable for work Thus every irreversibility carries with it a price 实际过程总是有损失功的，过程的不可逆程度越大，熵增越大，损失功也越大。损失的功转 化为热，使系统作功本领下降，因此，不可逆过程都是有代价的。 例5-3用1.57MPa,484℃的过热蒸汽推动透平机作功，并在0.0687MPa下排出。此透平机 既不可逆也不绝热，实际输出的轴功相当干可逆绝热功的85%。另有少量热散入293K的环 境，损失热为7.12kkg。求此过程的理想功、损失功和热力学效率。 思路： Wia =AHAS) AHH2-H AS=S2-S △H=Q4m 可逆给热过程 @085 △H'=WR AS=∑m-∑mS+∫2+AS △H'=H-H △H'=Q+W ④ 保 1.57MPa,484℃时的蒸汽H1=3437.5kJgS1=7.5035kJ/kgK 若蒸汽按绝热可逆膨胀，则是等熵过程，当膨胀至0.0687MPa S'2=S1=7.5035kJ/kgK) 查过热水蒸汽表 0.035MPa 0.07MP 0.0687MP H 饱和蒸汽 263147715326600747972658974885 100Ym 268447860426800753412680275462 P=0.0687MP H kJ/kg S kJ/(kg-K) 7.5035-7.4885H,-2658.9 75462-7488526802-26589 2658.97.4885 H2' 7.5035 2680.21 7.5462 ;=2664.4kJ/kg Wn=H5-H1=2664.4-3437.5=-773.1kJ/kg

2680.2 2658.9 2658.9 7.5462 7.4885 7.5035 7.4885 2 −  − = − − H The engineering significance of this result is: The greater the irreversibility of a process, the greater the rate of entropy production and the greater the amount of energy that becomes unavailable for work. Thus every irreversibility carries with it a price 实际过程总是有损失功的，过程的不可逆程度越大，熵增越大，损失功也越大。损失的功转 化为热，使系统作功本领下降，因此，不可逆过程都是有代价的。 例5-3 用1.57MPa，484℃的过热蒸汽推动透平机作功，并在0.0687MPa下排出。此透平机 既不可逆也不绝热，实际输出的轴功相当干可逆绝热功的85％。另有少量热散入293K的环 境，损失热为7.12kJ/kg。求此过程的理想功、损失功和热力学效率。 思路： 解： 1.57MPa， 484℃ 时的蒸汽 H1=3437.5kJ/kg S1=7.5035kJ/(kg·K) 若蒸汽按绝热可逆膨胀，则是等熵过程，当膨胀至0.0687MPa S´2=S1=7.5035kJ/(kg·K) 查过热水蒸汽表 0.035MPa H S 0.07MPa H S 0.0687MPa H S 饱和蒸汽 2631.4 7.7153 2660.0 7.4797 2658.9 7.4885 100 ℃ 2684.4 7.8604 2680.0 7.5341 2680.2 7.5462 P = 0.0687 MPa H kJ/kg S kJ/(kg·K) 2658.9 7.4885 H2´ 7.5035 2680.2 7.5462 H WR  = H Q WR  = +  H H H 2 1  = −   1 2 S S =  sys i i i i gen Q S m S m S S T   = − + +     in out 可逆绝热过程 W H T S ideal =  −   = − H H H 2 1 2 1  = − S S S 0.85 s R W W =  = + H Q Ws T1 P1 P1 Ws WR = H2  − H1 = 2664.4 − 3437.5 = −773.1kJ / kg H2  = 2664.4kJ / kg