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# laws of thermodynamics_ Lecture 6to9

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Engineering Thermodynamics - generalised approach

Module 2 .. Laws of Thermodynamics ...Lectures 6 -9.

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### laws of thermodynamics_ Lecture 6to9

1. 1. Lecture 2.6The Second Law …contd
2. 2. Entropy is a thermodynamicproperty which decides adiabaticaccessibility of statesIn a reversible process S =dTQd…..understanding entropyT should be absolute temperature !
3. 3. Carnot CyclesTTHTC1 24 3The Concept of AbsoluteTemperature
4. 4. Carnot Cycle…….Process Nature H.T.1-2 Isoth QH2-3 Adia 03-4 Isoth -QC4-1 Adia 0
5. 5. For 1-2-3-4 to be a cycle4312 ssss −=−( ) ( ) 03412 =−+− ssssCHCHCCHHTTQQTQTQ=⇒=− 0orCarnot Cycle…….1 24 3
6. 6. HCHCHHcarnotQQQQQQdonework−=−==∴ 1ηHCTT−=1Carnot Cycle…….Since ηcarnot depends only on the valueof the two temperatures, aboveequation enables us to define anabsolute temperature scale
7. 7. Carnot Cycle…….( ) =−=∴CHCcarnotCHQQTTT η1Present Convention :choose TC =273.16 K for Triple-point of water &determine the ‘value’ of anytemperature from above equationusing ηcarnot as determinedexperimentally
8. 8. MEASUREMENT OF ABSOLUTETEMPERATURE♦Difficulty of building heat engineoperating on Carnot cycle♦Need for practically usable methods♦Constant volume gas thermometer♦Platinum resistance thermometers calibratedagainst easily reproducible states : triple pointsof O2, Hg ; M.P. of Ga, Zn etc.
9. 9. The Concept of AbsoluteTemperature• Achieving Absolute Zerotemperature• It can be shown that the definition ofabsolute temperature (also calledThermodynamic temperature)implies that it is impossible toachieve temperatures below absolutezero.
10. 10. DEFINING ABSOLUTE ZEROTEMPERATUREIf a system undergoes a reversibleisothermal process between tworeversible adiabatics, without heattransfer, the temperature at whichthis process takes place is calledabsolute zero.
11. 11. ENTROPY CHANGE IN ANIRREVERSIBLE PROCESSEntropy is a Property ……..?RI12(s2 - s1)I = (s2 – s1)R
12. 12. Entropy Increase due to Friction12mgmg∆hENTROPY CHANGE IN ANIRREVERSIBLE PROCESS
13. 13. …..due to Friction12mgmg∆h.12Thmgss ∆=− 12ENTROPY CHANGE IN ANIRREVERSIBLE PROCESS∆Qin= mg ∆h∆Win= -mg ∆h
14. 14. …..due to short-circuiting of a batteryCI2 I1Qe1Qe2BAEQeI2΄( )avgeeavgavgavgTQQVTQss2112−==−ENTROPY CHANGE IN ANIRREVERSIBLE PROCESS)( 12 eeavgeon QQVVdQW −== ∫
15. 15. ENTROPY CHANGE IN ANIRREVERSIBLE PROCESSOther examples• Free expansion of a compressiblefluid• Heat transfer between tworeservoirs having finite temperaturedifference
16. 16. Lecture 2.7The Second Law…..Corollaries
17. 17. …...Recap• Caratheodory’s formulation of 2ndLaw• Reversible & Irreversible processes• Concept of Entropy• Concept of Absolute Temperature• Entropy change in an irreversibleprocess
18. 18. r : ∆ E = ∆Qr + ∆Wrri12x2x1Work done in an Irreversible Processi : ∆ E = ∆Qi + ∆Wi∆Wi = ∆ Wr + ∆Qr - ∆QiIs ∆Qr - ∆ Qi > < = 0?
19. 19. Work done in an Irreversible Process* Assume 1 - 2 in close proximity so that thetemp T doesn’t change much during theprocess• ∆ S is same in both the processes• ∆Qr = T ∆ S∴ ∆Qr - ∆ Qi= T ∆ S - ∆Qi
20. 20. Work done in an Irreversible ProcessConsider the 3 possibilitiesT ∆ S - ∆Qi > 0T ∆S - ∆Qi = 0orWhich ofthese two iscorrect ?Process I isirreversible & thiseq. is valid only forreversibleprocess.XT ∆ S - ∆Qi < 0
21. 21. Since above discussion is general, itshould also apply to the special caseof an adiabatic process, i.e. ∆Qi = 0For this special case, theseinequalities giveT ∆ S > 0 or T ∆ S < 0Which is correct ?Work done in an Irreversible Process
22. 22. Since in an irreversible adiabaticprocess the entropy can only increase∴T ∆ S > 0Therefore for any irreversible processWork done in an Irreversible ProcessT ∆ S - ∆Qi > 0 ∆ S > ∆Qi/T∆ S = ∆Qr/TRecall, for a reversible process
23. 23. 0>∆−∆ iQST 0>∆−∆ ir QQri WW ∆>∆( ) ( )ri WW ∆−<∆−Work output in anirreversible processis smallerWork input inirreversibleprocess is greaterWork done in an Irreversible Process
24. 24. Work done in an Irreversible ProcessConclusion : Starting from a giveninitial state, to reach the same finalstate work input required is larger inan irreversible process.In a work producing cycle Wirr < Wrev
25. 25. Work done in an Irreversible ProcessCondition under which this result isderived?Pts 1-2 close to each other, T ≈ T1 ≈ T2= temp of thermal reservoirIt is possible to have work output in anirreversible process > that in a reversibleprocess between the same end states
26. 26. Work done in an Irreversible ProcesssT1f abc. 2db′Reversible paths1 - a - 21- b′ - b - 21 - c′ - c - 2
27. 27. Work done in an Irreversible ProcessTa > Tb > TC Qa > Qb > QC• E2 - E1 = Qa+Wona=Qb+Won,b=QC+Won,c∴Won,a < Won,b < Won,cWby,a > Wby,b > Wby,csince E2 < E1Won is -ve1-f-2 irreversible Wby,f < Wby,aBut Wby,f can be > Wby,c !
28. 28. BASIC EQ. OF THERMODYNAMICSTWdETQdSδδ −≥≥TdxfdE ii  ″″Σ+≥Combining the expressions forentropy change in reversibleand irreversible processes
29. 29. ″″Σ+≥ ii dxfdETdS{ ″−=′ii dxdxIf′″Σ−≥ ii dxfdETdSBASIC EQ. OF THERMODYNAMICSFor a reversible process ii ff ′=′′
30. 30. For a reversible process′′Σ−= ii dxfdETdSBasic equation of thermodynamicsapplicable to all processes !!BASIC EQ. OF THERMODYNAMICS
31. 31. Lecture 2.8The Second Law…..Corollaries
32. 32. SECOND LAW FOR CYCLICPROCESSESFor all cycles0=∑ dScyclefor reversible cycle R1 – R2TQdSTQdScyclecycleδδ∑=∑⇒=ITQdScyclecycle−−−−−=∑⇒=∑∴ 00δR2R112I
33. 33. SECOND LAW FOR CYCLICPROCESSESFor irreversible cycle I – R2∑∑ >⇒>II TQdSTQdSδδProcess R2 ∑∑ =⇒=22 RR TQdSTqdSδδProcess I
34. 34. SECOND LAW FOR CYCLICPROCESSES∑∑∑∑ >+=∴cycleRIcycle TQdSdSdSδ2IITQcycle−−−−−−<∑ 0δFor irreversible cycle I – R2
35. 35. SECOND LAW FOR CYCLICPROCESSESCombine I & II to get 0≤∑cycle TQδApplying to power cycles0≤−CChhTQTQorhChCChChTTQQTTQQ≥≤ ;Clausius Inequality
36. 36. SECOND LAW FOR CYCLICPROCESSEShChCTTQQ−≤−=∴ 11η⇒ (η<1) :Kelvin Planck Statement of 2ndLawSimilarly derive Clausius Statement of2ndLaw by considering a Reversed CC.Carnot Cycle (Reversible Engine)is the most efficient inconversion of heat to Work.
37. 37. Isotherm t1Isotherm t21′2′1txRev. ad.2EQUIVALENCE OF CARATHEDORY’SFORMULATION with Kelvin-PlanckStatement
38. 38. Caratheodory: 2’inaccessibleadiabatically from 1Suppose this is not true i.e. 2’ isthen adiabatically accessible from1 (say process 1-2’)EQUIVALENCE OF CARATHEDORY’SFORMULATION with Kelvin-PlanckStatement
39. 39. EQUIVALENCE OF CARATHEDORY’SFORMULATION WITH Kelvin-PlanckStatementConsiderCycle 1-2’-1’-1THIS VIOLATES K-P STATEMENT1′2′1t2 Heat absorbed fromonly one reservoir at t1& equal amount ofwork is produced
40. 40. Equivalence of Caratheodory’sformulation with Clausius Statementt12′′ t21′′2′21tx ≡ VCaratheodory : 1-2′ can’t be anadiabatic processNB 2′ has been solocated thatQ1-1″ = Q2‘-2 ″{both +ve}
41. 41. Assuming Caratheodory’s axiom is incorrect1-2’ could be an adiabatic process.Then consider cycle 1- 2- 2" -1"– 1It has two heattransfers, vizeQeQνν−→+→−1"1"22( ) 0"221"1 =+Σ − QQEquivalence of Caratheodory’sformulation with Clausius Statementt12′′ t21′′2′21t
42. 42. Qnet = 0 = Wnet ⇒ Heat |Q22" | hasbeen absorbed at low temp. t2 &delivered to high temp. t1, withoutany work inputVIOLATION OF CLAUSIUSSTATEMENTEquivalence of Caratheodory’sformulation with Clausius Statement
43. 43. Lecture 2.9The Second Law …..Corollaries &Applications
44. 44. ENTROPY MAX PRINCIPLEfor any process0≥+=⇒≥σσdwheredTdQdSTdQdS≡ entropygen.In the absence of thermalinteraction 0≥= σddsPrinciple of Increase of entropyσσ +=+=TQdtdTQdtdS..In rate terms:
45. 45. ENTROPY MAX PRINCIPLE34 1Object 2 Preventingthermal int.Preventingwork int.Composite system is isolatedThe entropy reaches maximum possiblevalue at the equilibrium
46. 46. ENTROPY MAX PRINCIPLETrend to Equilibrium••EquilibriumFinal StateSInitialTime
47. 47. Examples of Second LawAnalysisA heat engine operates in a cycle between twothermal reservoirs at 300K and 900K and produces100kW power. Find the heat rejection andentropy generation when the engine is internallyreversible but receives heat and rejects heat at800K and 400k respectively
48. 48. Th = 900 KENGINETc = 300 K800K400K100 kWFirst Law inrate terms:WQdtdWdtdQdtdE  +=+=For this cycle:0= ΣQ -100Qh-Qc =100QhQcSince the engine is internallyreversible : Qh/800 -Qc/400=0Qh=200kWQc =100kW
49. 49. Second Law inrate terms:dtdTQdtdS σ+= ∑.For this cycle : 〉−−=300900Ch QQdtdσThis gives entropy generation rate as:-[ 200/900 - 100/300 ] = 0.111 kW / KExample 1….
50. 50. EXAMPLE 2A simple steam power cycle producing100 MW of power receives heat at900K in the boiler and rejects heat at320K. The condensate pumpconsumes 50kW of power, and theboiler consumes 54 tonnes/hour ofcoal. Assuming that the combustion of1Kg of coal releases 20MJ of heatdetermine the thermal efficiency andentropy generation in the cycle.
51. 51. EXAMPLE 2……BoilerPumpTurbineWpWtbyQoutQin
52. 52. EXAMPLE 2 ………….First Law (for System withindashed line boundaries)byToutin WWQQdtdEin , −+−= ΡUnder steady stateconditions, 0=dtdEinWWQQ byToutin Ρ−=−⇒ ,HereMWSMJQin300/203600100054=××=MWQout200100300=−=%3.33300100==η
53. 53. EXAMPLE 2 ………….Second Law ( for system within dashedline boundaries)σ+Σ=TQdtdSUnder steady stateconditions, 0=dtdSΣ−=⇒TQσHereKMWKMW/2917.320200900300=−−=σ
54. 54. EXAMPLE 3The gear box of a machine isoperating under steady-stateconditions. The input shaft receives25 kW from a prime mover andtransmits 22kW to the output shaft,the rest being lost due to friction etc.The gear box surface is at anaverage temperature of 500C andloses heat to the surroundings at300C. Estimate the rate of entropyproduction inside the gear box.
55. 55. GearBoxCTb050=CT 00 30=outQkW22kW25EXAMPLE 3
56. 56. EXAMPLE 3Since the gear box is operating in steadystate,First Law KWQQ outout 322250 =⇒−−= Second Law KWTQTQ out/289.932310003=×==−= ∑σEntropy production outside the box?