Report

The Second Law of Thermodynamics Consider the following process: mgh A rock spontaneously rises by lowering its temperature such that mCpDT = mgh so that DU = 0. Since energy is conserved, this type of process is not forbidden by the first law, but we know this never happens! There is a natural flow of things or direction for spontaneous processes to occur. We also know that various forms of work can be completely converted into heat, e.g., rubbing of two rocks together in a heat reservoir such that they undergo no temperature change so that W = Q and DU = 0. In general work of any kind can be done on a system in contact with a reservoir giving rise to a flow of heat without altering the state of a system, W = Q. Work can be converted entirely in 1 to heat by a suitable dissipative process. Heat Engines - The conversion of heat into work In order to convert heat into work we require a machine that will consume heat and produce work. The machine itself must not suffer any permanent change; it must play a passive role in that following the process it must return to its initial state. The machine must pass through a cycle. def. Thermal efficiency, work out W . heat in Q1 Applying the first law to the operation of the machine or engine, W = Q1Q2 where Q2 corresponds to any heat rejected from the engine, Q Q2 Q Q1 1 1 2 Q1 Q1 Heat engine system Q2 W 2 Carnot Cycle The Carnot cycle is a reversible cycle operating between two temperatures. A B: Isothermal expansion adsorbing heat Q1. B C: Adiabatic expansion decreasing T from T1 to T2. C D: Isothermal compression rejecting heat Q2. D A: Adiabatic compression increasing T from T2 to T1. *Note that if the cycle is operated in reverse refrigerator. p T1 A Q1 W Carnot cycle for a gas W B T1 D Q2 Q1 C T2 V T2 Q2 Q2 Since all the steps are reversible DU = 0, W = Q1 - Q2 and 1 Q1 3 The 2nd Law Kelvin Statement - No process is possible whose sole result is the complete conversion of heat into work. This addresses the efficiency of conversion. Clausius Statement - No process is possible whose sole result is the transfer of heat from a colder to a hotter body. Spontaneity of processes and the irreversibility of nature. Kelvin Statement Clausius Statement Carnot’s Theorem: No engine operating between two given reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. 4 Proof of Carnot’s Theorem T1 QC1 WC = QC1 - QC2 QH1 C H WH = QH1 - QH2 QH2 QC2 Assume the existence of a Hypothetical engine such that, H C , or T2 WH WC . QH 1 QC1 Since the Carnot engine is reversible we can drive it backwards using the mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so that in one cycle it uses exactly as much work as H produces. WC WH and therefore QC1 QH 1. 5 Proof of Carnot’s Theorem T1 QC1 WC = QC1 - QC2 QH1 C H Composite Engine WH = QH1 - QH2 QH2 QC2 T2 This is a violation of the Clausius statement of the second law! Now consider C and H as a Composite Engine. This composite engine produces no net work but simply extracts heat from a cold reservoir and delivers an amount of heat, QC1 QH1 0 to a hotter reservoir. Carnot’s Theorem: any engine reversible 6 Corollary: All reversible engines operating between the same temperature reservoirs are equally efficient. Thus the efficiency of any reversible engine operating between the same reservoirs are equally efficient. For any reversible engine, Q1 f T1 ,T2 . Q2 Qn+2 Cn+2 Tn+2 Wn+2 Qn+1 Tn+1 Wn+1 Cn+1 Qn Cn Tn Consider a series of Carnot Engines Q Cn 2 : n 2 f Tn 2 ,Tn1 Qn1 Qn1 Cn1 : f Tn1 ,Tn Qn Qn Cn : f Tn ,Tn1 Qn1 Wn Qn-1 Tn-1 7 Qn+2 Cn+2 Tn+2 Wn+2 Qn+1 Tn+1 Wn+1 Cn+1 Qn Tn Cn Wn Qn-1 Therefore: Tn-1 For the composite engine, Q Composite : n 2 f Tn 2 ,Tn1 Qn1 Then, f Tn2 ,Tn1 f Tn2 ,Tn1 f Tn1 ,Tn f Tn ,Tn1 This can only be true if the f’s factorize such that f Tn 2 ,Tn1 f Tn 2 Tn 2 f Tn1 Tn1 Qn 2 Tn 2 The ratio of the temperatures of the reservoirs is equal Qn1 Tn1 to the ratios of heat exchanged by a reversible engine operating between the same reservoirs. 8 According to Carnot’s theorem and its corollary we can make the following statements: reversible Q2 Qr 2 1 1 Q1 Qr1 Q2 Qr 2 T2 Q1 Qr1 T1 Q2 T2 Therefore . Taking the heat entering the system as positive, we can say Q1 T1 Q1 Q2 Q Q 0 and generalizing this result, 0 or 0. T1 T2 T T For any closed cycle, , where the equality necessarily holds for a reversible cycle. 9 Entropy def: We can now define a new variable, the entropy S, by the relation dS = Qrev T for an infinitesimal reversible change. This definition holds for reversible changes only. For a finite reversible change of state, the change in entropy is given by, 2 S 2 S1 1 Qrev. T . 10 Entropy in Irreversible Processes Since entropy is a state function, the change in entropy accompanying a state change must always be the same regardless of how the state change occurs. Only when the state change occurs reversibly is the entropy change related to the heat transfer by the equation Q DS . T B X Consider an irreversible change AB. Construct any reversible path R thus forming an irreversible cycle ABRA. For the irreversible cycle the Clausius theorem says, R A x Determination of the change in entropy for an irreversible change 11 B Taking the integral in two parts, Airrev . B i.e., Airrev . Q B But Arev . T Q T SB S A , Arev . Airrev . dS Q T dS 0 . T T A Brev . Q T 0 . by definition of entropy. Thus B or Q B Q Q T SB S A , Thus we have this general result for a differential irreversible change. For a thermally isolated system Q = 0 and we have the general result known as the law of increase of entropy. 12 Some Interesting Examples Isothermal expansion of an Ideal Gas dU Q W 0 dU TdS pdV 0 Adiabatic Free expansion of an Ideal Gas (Joule expansion - no Q or W exchanged with surroundings) Combined 1st & 2nd Law W RT ln V2 / V1 RT ln p1 / p2 DS system R ln V2 / V1 DS surr DS system ; DStotal 0 This is an irreversible process, but we can always use the combined Law and integrate from the initial to final state by a convenient reversible path. DS system R ln V2 / V1 DS surr 0; DStotal DS system 0 13 Isothermal dissipation of Work P1, V1 Diathermal Walls Electrical work is dissipated isothermally by heat flow into a reservoir. There is no entropy change of the system because it’s thermodynamic coordinates do not change. The reservoir adsorbs Q = W units of heat at temperature T so its entropy change is DS resevoir DStotal Q / T W / T Reservoir (T) Adiabatic dissipation of Work Composite system P, T1 dS system Adiabatic Walls DQ Any Reversible Path T Electrical work is dissipated in a thermally isolated system maintained at fixed pressure. The T of the system increases irreversibly. The coordinates of the system change from P,T1 to P,T2 . The entropy change can be calculated by Q C p dT T2 C p dT T1 T dS system T2 C p ln 0. T1 14 Examples An inventor claims to have developed a power cycle capable of delivering a net work output of 410 kJ for an energy input by heat transfer of 1000 kJ. The system undergoing the cycle receives heat transfer from hot gases at T = 500 K and discharges energy by heat transfer to the atmosphere at T = 300 K. Evaluate this claim. The thermal efficiency is Wcycle QH 410 kJ 0.41 1000 kJ The maximum efficiency of any power cycle is 1 QC T 300 K 1 C 1 0.40 QH TH 500 K No good! 15 Different Forms of the Combined 1st and 2nd Law Using the definition of entropy dU TdS pdV Since enthalpy is defined as, H = U + pV dH TdS Vdp Rearranging these equations and writing them on a unit mass basis, Tds du pdv Tds dh vdp We can use these forms to determine the entropy change of an ideal gas subjected to changes in p, v, T. du p dv T T dh v ds dp T T ds We already know that for an ideal gas, du cv dT and dh c p dT and pv RT dT dv R T v dT dp ds c p R T p ds cv 16 Entropy Production B B X Airrev . B R Airrev . x Determination of the change in entropy for an irreversible change Recall the determination of entropy change for an irreversible process. Airrev . Q T T Arev . Q T B Q T . SB S A , It is convenient to define a quantity such that A B Q A Brev . Q T B Airrev . Q T A Brev . Q T Def: is necessarily a positive 0 quantity called entropy production. 17 Entropy Production Rewriting the expression Q B Airrev . T Q A T Brev . , The 2 terms on the RHS of the equation are path dependant. we obtain, B Airrev . Q T S A S B Q B S A SB Airrev . B SB S A entropy change Airrev . T Q T entropy transfer If the end states are fixed the entropy change on the left hand side of this equation can be determined. The 1st term on the RHS of the equation is the entropy transfer associated with heat transfer. The direction or sign of the entropy transfer is the same as heat transfer. The 2nd term is the entropy production term. entropy production 0 irreversibilities present = 0 no irreversibilities present : 18 Entropy Production can not be less than zero By contrast the change in the entropy of the system can be positive, negative or zero: 0 S B S A : = 0 < 0 Here Qj/Tj is the amount of entropy transferred to the portion of the boundary at temperature Tj . On a time rate basis for a closed system Qj dS dt j Tj The entropy balance can be expressed in various forms. If heat transfer takes place along several locations on the boundary of the system where the temperatures do not vary with position or time, SB S A j Qj Tj 19 Examples Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly. The water undergoes a process to the corresponding saturated vapor during which the piston moves freely in the cylinder. If the change of state is brought about by heating the water as it undergoes an internally reversible process at constant pressure and temperature determine the work and the heat transfer per unit of mass in kJ/kg At constant pressure the work is simply, W p vg v f 170 kJ/kg Table A-2 m 20 Examples Since the process is reversible and occurs at constant temperature g Q TdS mT sg s f f Q T sg s f 2257 kJ/kg m This could also have been calculated our old way ug u f Q W m m Q u g u f p vg v f hg h f m 21 Examples The figure shows a system receiving heat Q from a reservoir. By definition the reservoir is free of irreversibilities, but the system is not, fluid friction, etc. Let’s determine the entropy change of the system and that of the reservoir. For the system, S2 S1 Q Tb For the reservoir DS res Qres Q 0 Tb Tb 22 Examples Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly. The water undergoes a process to the corresponding saturated vapor during which the piston moves freely in the cylinder. There is no heat transfer with the surroundings. If the change in state is brought about by the action of a paddle wheel, determine the net work per unit mass in kJ/kg and the amount of entropy produced per unit mass in kJ/kg-K. As the volume of the system increases during this process, there is an energy transfer by work from the system during the expansion as well as an energy transfer by work to the system done by the paddle wheel. The net work is evaluate from the change in internal energy. 23 Examples From the 1st Law, DU = -W. On a unit mass basis we have, W ug u f 2087.56 kJ/kg m The minus sign indicates that the work input by stirring is greater in magnitude than the work done by the water as it expands. The entropy produced is evaluated by applying the entropy balance, 2 Q DS 0 1 T On a unit mass basis, m sg s f 6.048 kJ/kg-K 24 Entropy Diagrams isentrop T isenthalp isochor isobar isotherm S 25 Entropy Diagrams CCW – refrigeration cycle T CW – power cycle S Carnot cycle on a T – S diagram. Area representation of heat transfer for an internally reversible process of a closed system. 26 Thermodynamic Potentials Combined 1st and 2nd Law Enthalpy is a function of S and p H H S , p dU = TdS - pdV U U S ,V H H dp dH dS S p p S U U dU dS dV S V V S H H , T V S p p S U T S V U p - , V S T V p S S p T p - V S S V Potential Function in terms of S and p, Enthalpy Potential Function in terms of T and p, Gibbs Free Energy Lengendre Transform subtract a -d(pV) term from dU Lengendre Transform add a -d(TS) term to dH d ( H-TS ) = Vdp - SdT where G = (H-TS) is the Gibbs Free Energy and G = G (p, T) dU + d(pV) = TdS - pdV + d(pV) d(U + pV) = TdS + Vdp where H = (U + pV) is the Enthalpy and H = H (S, p) G G dp dG dT T p p T G G , S ,V T p p T S V p T T p 27 Potential Function in terms of T and V, Four Fundamental Thermodynamic Potentials Helmholtz Free Energy Lengendre Transform subtract a -d(TS) term from dU d(U-TS) = -pdV - SdT = dA where A = A(V, T) is the Helmholtz Free Energy A A dA dV dT V T T V A A S p S , p , T V V T V T T V The Maxwell relations are useful in that the relate quantities that are difficult or impossible to measure to quantities that can be measured. dU = TdS - pdV dH = TdS + Vdp dG = Vdp - SdT dA = -pdV - SdT The appropriate thermodynamic potential to use is determined by the constraints imposed on the system. For example, since entropy is hard to control (adiabatic conditions are difficult to impose) G and A are more useful. Also in the case of solids p is a lot easier to control than V so G is the most useful of all potentials for solids. 28 Some important bits of information For a mechanically isolated system kept at constant temperature and volume the A = A(V, T) never increases. Equilibrium is determined by the state of minimum A and defined by the condition, dA = 0. For a mechanically isolated system kept at constant temperature and pressure the G = G(p, T) never increases. Equilibrium is determined by the state of minimum G and defined by the condition, dG = 0. Consider a system maintained at constant p. Then DG SdT T2 T2 T1 T1 DS S T dT C p T d ln T T2 S T2 S T1 C p T d ln T T1 T2 T2 DG G T2 G T1 T2 T1 S T1 dT C p T d ln T T T1 1 29