CHAPTER V

1. CHAPTER V THE SECOND LAW OF THERMODYNAMICS

2. Thermodynamics : concerned with transformation of energy The laws of thermodynamics : describe the BOUNDS within which these transformations occur. The FIRST law : • Conservation of energy law : “The energy is conserved in any ordinary process” • DOES NOT explain the direction of energy transform. The SECOND law : • Distinguishes between theQUALITYof work & heat. • Explain the DIRECTION or energy transform.

3. There is a difference in QUALITY between heat & work. WORK : is readily transform into other forms of energy. • e.g. potential energy (elevation of a weight)  kinetic energy (acceleration of a mass)  electrical energy (operation of a generator) • Conversion efficiency 100% (e.g. by elimination of friction  dissipative process that transform WORK into HEAT) HEAT : is NOT readily transform completely and continuously into work (or mechanical energy, or electrical energy) • No matter how well the improvements of the devices involved, conversion efficiency of (HEAT  WORK) ≤ 40%. • FLOW of heat ALWAYS takes place from the hotter to the cooler body; never in reverse direction.

4. STATEMENT OF THE SECOND LAW Many general statements may be made for the second law; two of the most common are : • No apparatus can convert heat absorbed COMPLETELY into work  EFFICIENCY. • No process is possible, which consist SOLELY in the transfer of heat from LOWER level to a HIGHER one. H E A T E N G I N E The classical approach to the 2nd law is based on microscopic viewpoints of properties; independent of any knowledge of the structure of matter or behavior of molecules.

5. Example : a STEAM POWER PLANT 2 WS 1 QH 3 QC 4

7. Note : QH = heat absorbed by heat engine (the system) QC = heat discarded by the heat engine Wt = work done by the system (through turbine) Wp = work done to the system (through pump) Ws = nett amount of work, done by the system. lWsl = lWt – Wpl In a cyclic – closed system; according to 1st law of thermodynamics : ΔH = lQHl – lQCl – lWsl = 0 Or, lWsl = lQHl - lQCl

8. Define : thermal efficiency of the engine, as : η =  That means : • If η 100% ; lQCl must 0. • NO ENGINE can do this ; some heat is always rejected to the cold reservoir. The heat engine above (the steam power plant) can be schematically describe as follow :

9. This is the basis of statement 1 of the 2nd law : “No apparatus can convert heat COMPLETELY into work” Or: “It is impossible by a cyclic process to convert the heat absorbed by the system COMPLETELY into work done by the system” If the thermal energy efficiency of 100% is NOT possible, so what is the UPPER LIMIT possible? • The value of η depend on the degree of REVERSIBILITY • Max. value of η  if the operation is COMPLETELY REVERSIBLE

10. Heat engine operating in a COMPLETELY REVERSIBLE manner  CARNOT ENGINE 2 |QH| 1 4 3 |QC| CARNOT CYCLE

11. The CARNOT CYCLE : 1  2 : system maintains contact with hot reservoir at TH, undergoes a REVERSIBLEisothermal process during which |QH| is absorbed. 2  3 : the system undergoes a REVERSIBLEadiabaticprocess that causes its temperature to decrease to TC. 3  4 : system maintains contact with cold reservoir at TC, undergoes a REVERSIBLE isothermalprocess, during which |QC| is rejected. 4  1 : the system undergoes a REVERSIBLE adiabaticprocess that causes the temperature to riseto TH. Refer to figure A at p. V-3 : Engine E operating a Carnot refrigerator C. • Engine E : absorbing heat |QH’| producing work |W| discarding heat |QH’| - |W| • Refrigerator C : absorbing heat |QH| - |W| consuming work |W| discarding heat |QH| Assuming : ηE > ηC Engine refrigerator

12. Then : |QH| > |QH’| The net heat extracted from the cold reservoir : |QH| - |W| - (|QH’| - |W|) = |QH| - |QH’| So the net heat delivered to the hot reservoir is also : |QH| - |QH’|. Thus : the sole result of those engine / refrigerator combination is the transfer of heat from Tc to TH. • This is THE VIOLATION of the 2nd Law!!! it is impossible to have engine with thermal efficiency greaterthan a Carnot engine. Note that : For a Carnot engine : η = η (TH,TC) ≠ η (working path).

13. THERMODYNAMICS TEMPERATURE (IDEAL GAS as WORKING FLUID) The Carnot engine cycle using ideal gas as the working fluid :

14. a  b : adiabatic compression. b  c : isothermal expansion. c  d : adiabatic expansion. d  e : isothermal compression. At isothermal process : ΔU = Q – W with : W = p.dV = dV • Q = W = RT ln V| step b  c : |QH| = R.THln step d  a : |QC| = R.TCln Hence : …(a) 0

15. For adiabatic process : ΔU = Q – W note : ΔU = Cv.dT W = pdV = dV Step a  b : Step c  d : (b)  (a) : η is defined as : η = 1 -  η = 1 - Or …(b)

16. ENTROPY The equation for a Carnot engine : Can be written as : = If the heat is referred to the engine (NOT the reservoirs), QH = + and QC = - Therefore : =  + = 0

17. Consider an arbitrarily reversible cyclic process : the entire cyclic process can be represented by a series of reversible adiabatic – isothermal processes. Each cycle of reversible process has its own pair of isotherms Tc and TH and associated heat, QH and Qc

18. For infinitesimal – reversible process, the heat quantity becomes : dQH and dQc , so that : + = 0 Integration for the entire cycle gives : [Qrev = Q-reversible] The quantity of :  called as ENTROPY change dS = or dQrev = T dS Where : S = total entropy of the system

19. Consider 2 arbitrarily process connecting points A & B (two equilibrium states) on the PV diagram : For each path gives : ΔSc = And ΔSD= For reversible process, the entropy change must be the same. ΔSc = ΔSD = ΔS = SB - SA

20. Question : If the process is IRREVERSIBLE, is the entropy change ΔS = SA – SB still the same? Answer : YES. The ΔS = SA – SB still the same, as ΔS is independent of path. BUT, the ΔS is NOT given by irreversible.

21. Consider a heat reservoir : QQ T or T The entropy change : ΔS =  whether the transfer is ‘reversible’ or ‘irreversible’ If the process is : ADIABATIC & REVERSIBLE dQrev = 0 dS= 0 (no entropy change; the entropy is constant) ISENTROPIC PROCESS then

22. Question : If the process is MECHANICALLY REVERSIBLE but THERMALLY IRREVERSIBLE(the heat transfer is irreversible), is ΔS can still be calculated by ΔS = ?

23. SUMMARY • For a system undergoing REVERSIBLE process : 2. Entropy (S) is a STATE FUNCTION, so ΔS is INDEPENDENT of PATH (whether the process is reversible or irreversible, ΔS MUST be the same), calculated by :

24. ENTROPY CHANGE OF AN IDEAL GAS Recall the 1st law of Thermodynamics (closed system) : dU = dQ + dW For a reversible process : dU = dQrev – pdV …(a) By definition : H = U + PV dH = dU + pdV + Vdp …(b) Combination of (b) & (a) gives : dQrev = dH – VdP For an ideal gas : dH = Cpig.dT V = Hence : Results in : …(c) = dS Relates to property only, independent of process

25. Define : log – mean heat capacity, Cpmsig = Where if Cpig = R[A + BT + CT2 + DT-2] So, Cpmsig = A + B.Tlm + Tam.Tlm[C + ] Tam = ½ (T1 + T2) Tlm = Eq. (c) becomes : ΔS = Cpmsigln

26. Example 5-1 A central power plant, rated at 800000 kW, generates steam at 585 K and discards heat to a river at 295 K. if the thermal efficiency of the plant is 70% of the maximum possible value, how much heat is discarded to the river at rated power?

27. Solution : The maximum possible thermal efficiency is : (if the process follows a Carnot cycle) Where : TH = steam generation temperature = 585 K TC = river temperature = 295 K The actual efficiency is, then : = 0,3470 By definition : Where : |QH| = W + |QC| Hence : |QC| = Heat discarded to the river, |QC| = 1505500 kJ/s

28. Example 5-2 : For an ideal gas with constant Cp undergoing a reversible adiabatic ( isentropic) process, we found that : Shows that this same equation can be derived from enthalpy equation : !!

29. Answer : For isentropic process, = 0 = or, For an ideal gas : Cpig = Cvig + R  1 = By definition :  1 = Putting back into equation, found :

30. Example 5-3 : Methane gas at 550 K and 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming CH4 an ideal gas, what is the final temperature?

31. Solution : For adiabatic reversible process, ΔS = 0 ; hence : But : Cpmsig = f(T)  depends on T2!! Arrange the equation :  T2 = T1exp () For CH4 (see table 4.1 Smith & VanNess for CpigCH4) : Cpmsig= 1,702 + 9,081x10-3.Tlm – 2,164x10-6.Tam.Tlm Where : Tam = ; Tlm= By trial & error, found T2 = 411,34 K

32. Question : Can T2 be calculated directly from : Where is calculated from Tlm& Tam between T1 & T2? (involving trial & error too!) Answer : • Theoretically, NO, it can’t WHY? • Practically, it’s ok.

33. PRINCIPLE OF THE INCREASE OF ENTROPY (MATHEMATICAL STATEMENT of the SECOND LAW) • Consider two heat reservoir : a resource and a sink Resource  hot reservoir Sink  cold reservoir ΔSHHot reservoir ΔSCCold reservoir TH Q TC

34. the entropy decrease of reservoir TH : ΔSH = • The enthalpy increase of reservoir TC : ΔSC= • Total enthalpy change of the system : ΔStotal = ΔSH + ΔSC = + Or, ΔStotal = |Q| Since : TH > Tc ΔStotal> 0 (irreversible heat tr. Process) if THTc ΔStotal 0 (adiabatic reversible)

35. P Qrev reversible II. Consider a cycle process A  B  A • During REVERSIBLE process (B  A) : ΔS Ξ SA – SB = • The whole process : ΔUtot = Qtot - Wtot A Irreversible adiabatic B V = Qrev = Wirr (A  B) + Wrev (B  A)

36. For reversible – compression process : dQrev = - [negative]  SB > SA Since the original IRREVERSIBLE process is adiabatic, the total entropy change of the process : ΔStot = SB – SA > 0 The general equation : ΔStot≥ 0

37. Example 5-4 : A steel casting [Cp = 0,5 kJ/kg.K] weighting 40 kg at 450oC is quenched in 150 kg oil [Cp = 2,5 kJ/kg.K] at 25oC. If there are no heat losses, what is the entropy change of : • The casting • The oil, and • Both, considered together

38. Solution : Final temperature t2 of the oil & steel casting is found by an energy balance : 40 (0,5)(t2-450) + 150(2,5)(t2-25) = 0 Found : t2 = 46,52oC a. Entropy change of the casting L ΔS = = (40)(0,5)ln The oil Steel casting

39. b. Entropy change of the oil ΔS = (150)(2,5) ln c. Total entropy change ΔStotal = -16,33 + 26,13 = 9,80 kJ/K

40. Example 5-5 An inventor claims to have devised a process which takes in only saturated at 100oC, and by a complicated series of steps, makes a continuously available heat at a temperature level of 200oC. He further claims that, for every kg steam taken into the process, 2000 kJ of energy as heat is liberated at the higher temperature level of 200oC. Is the process that the inventor claims is POSSIBLE? Assume, cooling water is available at 0oC in unlimited quantity.

41. Solution : The process can be schematically describes as follows : From steam table : H1 = 2676,0 kJ/kg S1 = 7,3554 kJ/kg.K Basis on the steam table

42. At the apparatus : ΔH = Q – Ws ; where Q = -2000+Qo  0,0 – 2676,0 = -2000 + Qo • Qo = -676,0 kJ Now we examine this results to find out the entropy change : • For the apparatus (steam): ΔSap = 0,000 – 7,3554 = -7,3554 kJ/K • For the hot reservoir (at 200oC) ΔSh = • For the cold reservoir (cooling water at 0oC) ΔSc =

43. Thus : ΔStotal = -7,3554 + 4,2270 + 2,4748 = -0,6536 kJ/K Since (ΔStotal) < 0  the process as described is IMPOSSIBLE Question : For the process to be possible, what is the MAXIMUM amount of heat which can be transferred to the heat reservoir at 200oC? Other conditions remain the same.

44. THE THIRD LAW OF THERMODYNAMICS Postulate : The absolute entropy is zero for all PERFECT crystalline substance at absolute zero temperature. Hence, if the entropy is ZERO, at T = 0K, then the absolute entropy of GAS at temperature T :

45. Where : Tf= fusion temperature ΔHf = latent heat of fusion Tv = vaporization temperature ΔHv = latent heat of vaporization