Second Law of Thermodynamics and Entropy

Transcription

1 5 Scond Law of hrmodynamics and Entropy 5.. Limitations of first law of thrmodynamics and introduction to scond law. 5.. Prformanc of hat ngins and rvrsd hat ngins Rvrsibl procsss Statmnts of scond law of thrmodynamics Clausius statmnt Klvin-planck statmnt Equivalnc of clausius statmnt to th klvin Planck statmnt Prptual motion machin of th scond kind hrmodynamic tmpratur Clausius inquality Carnot cycl Carnot s thorm Corollary of Carnot s thorm. 5.. Efficincy of th rvrsibl hat ngin. 5.. Entropy ntroduction Entropy A proprty of a systm Chang of ntropy in a rvrsibl procss Entropy and irrvrsibility Chang in ntropy of th univrs mpratur Entropy diagram Charactristics of ntropy Entropy changs for a closd systm Gnral cas for chang of ntropy of a gas Hating a gas at constant volum Hating a gas at constant prssur sothrmal procss Adiabatic procss Polytropic procss Approximation for hat absorbd Entropy changs for an opn systm h third law of thrmodynamics Highlights Objctiv yp Qustions hortical Qustions Unsolvd Exampls. 5.. LMAONS O RS LAW O HERMODYNAMCS AND NRODUCON O SECOND LAW t has bn obsrvd that nrgy can flow from a systm in th form of hat or work. h first law of thrmodynamics sts no limit to th amount of th total nrgy of a systm which can b causd to flow out as work. A limit is imposd, howvr, as a rsult of th principl nunciatd in th scond law of thrmodynamics which stats that hat will flow naturally from on nrgy rsrvoir to anothr at a lowr tmpratur, but not in opposit dirction without assistanc. his is vry important bcaus a hat ngin oprats btwn two nrgy rsrvoirs at diffrnt tmpraturs. urthr th first law of thrmodynamics stablishs quivalnc btwn th quantity of hat usd and th mchanical work but dos not spcify th conditions undr which convrsion of hat into work is possibl, nithr th dirction in which hat transfr can tak plac. his gap has bn bridgd by th scond law of thrmodynamics. 5.. PERORMANCE O HEA ENGNES AND REVERSED HEA ENGNES Rfr ig. 5. (a). A hat ngin is usd to produc th maximum work transfr from a givn positiv hat transfr. h masur of succss is calld th thrmal fficincy of th ngin and is dfind by th ratio : hrmal fficincy, η th W Q...(5.) whr, W Nt work transfr from th ngin, and Q Hat transfr to ngin. or a rvrsd hat ngin [ig. 5. (b)] acting as a rfrigrator whn th purpos is to achiv th maximum hat transfr from th cold rsrvoir, th masur of succss is calld th co-fficint of prformanc (C.O.P.). t is dfind by th ratio : 7

2 8 ENGNEERNG HERMODYNAMCS Co-fficint of prformanc, (C.O.P.) rf. Q...(5.) W whr, Q Hat transfr from cold rsrvoir, and W h nt work transfr to th rfrigrator. or a rvrsd hat ngin [ig. 5. (b)] acting as a hat pump, th masur of succss is again calld th co-fficint of prformanc. t is dfind by th ratio : Co-fficint of prformanc, (C.O.P.) hat pump Q W whr, Q Hat transfr to hot rsrvoir, and W Nt work transfr to th hat pump....(5.3) Hot rsrvoir Hot rsrvoir Q Q Q + W Hat ngin W (Q Q ) Hat pump or rfrigrator W Q Q Cold rsrvoir Cold rsrvoir (a) Hat ngin (b) Hat pump or rfrigrator ig. 5. n all th abov thr cass application of th first law givs th rlation Q Q W, and this can b usd to rwrit th xprssions for thrmal fficincy and co-fficint of prformanc solly in trms of th hat transfrs. Q Q η th Q..(5.4) (C.O.P.) Q rf Q Q...(5.5) Q (C.O.P.) hat pump..(5.6) Q Q t may b sn that η th is always lss than unity and (C.O.P.) hat pump is always gratr than unity REVERSBLE PROCESSES A rvrsibl procss should fulfill th following conditions :. h procss should not involv friction of any kind.. Hat transfr should not tak plac with finit tmpratur diffrnc. /M-thrm/th5-.pm5

3 SECOND LAW O HERMODYNAMCS AND ENROPY 9 3. h nrgy transfr as hat and work during th forward procss should b idntically qual to nrgy transfr as hat and work during th rvrsal of th procss. 4. hr should b no fr or unrstrictd xpansion. 5. hr should b no mixing of th fluids. 6. h procss must procd in a sris of quilibrium stats. Som xampls of idal rvrsibl procsss ar : (i) rictionlss adiabatic xpansion or comprssion ; (ii) rictionlss isothrmal xpansion or comprssion ; (iii) Condnsation and boiling of liquids. Som xampls of irrvrsibl procsss ar : (i) Combustion procss ; (ii) Mixing of two fluids ; (iii) All procsss involving friction ; (iv) low of lctric currnt through a rsistanc ; (v) Hat flow from a highr tmpratur to lowr tmpratur. Rvrsibl procsss ar prfrrd bcaus th dvics which produc work such as ngins and turbins, rvrsibl procss of th working fluid dlivrs mor work than th corrsponding irrvrsibl procsss. Also in cas of fans, comprssors, rfrigrators and pumps lss powr input is rquird whn rvrsibl procsss ar usd in plac of corrsponding irrvrsibl ons. n thrmodynamic analysis concpt of rvrsibility, though hypothtical, is vry important bcaus a rvrsibl procss is th most fficint procss. Only rvrsibl procsss can b truly rprsntd on proprty diagrams. hrmodynamic rvrsibility can only b approachd but can nvr b achivd. hus th main task of th nginr is to dsign th systm which will volv approximat rvrsibl procsss SAEMENS O SECOND LAW O HERMODYNAMCS h scond law of thrmodynamics has bn nunciatd mticulously by Clausius, Klvin and Planck in slightly diffrnt words although both statmnts ar basically idntical. Each statmnt is basd on an irrvrsibl procss. h first considrs transformation of hat btwn two thrmal rsrvoirs whil th scond considrs th transformation of hat into work Clausius Statmnt t is impossibl for a slf acting machin working in a cyclic procss unaidd by any xtrnal agncy, to convy hat from a body at a lowr tmpratur to a body at a highr tmpratur. n othr words, hat of, itslf, cannot flow from a coldr to a hottr body Klvin-Planck Statmnt t is impossibl to construct an ngin, which whil oprating in a cycl producs no othr ffct xcpt to xtract hat from a singl rsrvoir and do quivalnt amount of work. Although th Clausius and Klvin-Planck statmnts appar to b diffrnt, thy ar rally quivalnt in th sns that a violation of ithr statmnt implis violation of othr Equivalnc of Clausius Statmnt to th Klvin-Planck Statmnt Rfr ig. 5.. Considr a highr tmpratur rsrvoir and low tmpratur rsrvoir. ig. 5. shows a hat pump which rquirs no work and transfrs an amount of Q from a low tmpratur to a highr tmpratur rsrvoir (in violation of th Clausius statmnt). Lt an amount of hat Q (gratr than Q ) b transfrrd from high tmpratur rsrvoir to hat ngin which dvolops a nt work, W Q Q and rjcts Q to th low tmpratur rsrvoir. Sinc thr is no hat intraction with th low tmpratur, it can b liminatd. h combind systm /M-thrm/th5-.pm5

4 30 ENGNEERNG HERMODYNAMCS of th hat ngin and hat pump acts thn lik a hat ngin xchanging hat with a singl rsrvoir, which is th violation of th Klvin-Planck statmnt. High tmp. rsrvoir, Q Q Systm boundary Hat pump Hat ngin W Q Q Q Q Low tmp. rsrvoir, ig. 5.. Equivalnc of Clausius statmnt to Klvin-Planck statmnt PERPEUAL MOON MACHNE O HE SECOND KND A machin which voilats th first law of thrmodynamics is calld th prptual motion machin of th first kind (PMM). Such a machin crats its own nrgy from nothing and dos not xist. hrmal rsrvoir Q Prptual motion machin W Q ig Prptual motion machin of scond kind (PMM). Without violating th first law, a machin can b imagind which would continuously absorb hat from a singl thrmal rsrvoir and would convrt this hat compltly into work. h fficincy of such a machin would b 00 pr cnt. his machin is calld th prptual motion machin of th scond kind (PMM). ig. 5.3 shows th prptual motion machin of th scond kind. A machin of this kind will vidntly violats th scond law of thrmodynamics. /M-thrm/th5-.pm5

5 SECOND LAW O HERMODYNAMCS AND ENROPY HERMODYNAMC EMPERAURE ak th cas of rvrsibl hat ngin oprating btwn two rsrvoirs. ts thrmal fficincy is givn by th qn. (5.4), η th Q Q Q Q Q h tmpratur of a rsrvoir rmains uniform and fixd irrspctiv of hat transfr. his mans that rsrvoir has only on proprty dfining its stat and th hat transfr from a rsrvoir is som function of that proprty, tmpratur. hus Q φ (K), whr K is th tmpratur of rsrvoir. h choic of th function is univrsally accptd to b such that th rlation, Q K φ ( ) Q φ( K) bcoms Q Q...(5.7) whr and ar th thrmodynamic tmpraturs of th rsrvoirs. Zro thrmodynamic tmpratur (that tmpratur to which tnds, as th hat transfr Q tnds to zro) has nvr bn attaind and on form of third law of thrmodynamics is th statmnt : h tmpratur of a systm cannot b rducd to zro in a finit numbr of procsss. Aftr stablishing th concpt of a zro thrmodynamic tmpratur, a rfrnc rsrvoir is chosn and assignd a numrical valu of tmpratur. Any othr thrmodynamic tmpratur may now b dfind in trms of rfrnc valu and th hat transfrs that would occur with rvrsibl ngin, Q rf....(5.8) Q rf. h dtrmination of thrmodynamic tmpratur cannot b mad in this way as it is not possibl to build a rvrsibl ngin. mpraturs ar dtrmind by th application of thrmodynamic rlations to othr masurmnts. h S unit of thrmodynamic tmpratur is th klvin (K). h rlation btwn thrmodynamic tmpratur and clsius scal, which is in common us is : hrmodynamic tmpratur Clsius tmpratur h klvin unit of thrmodynamic tmpratur is th fraction of thrmodynamic 73.5 tmpratur of ripl point of watr CLAUSUS NEQUALY Whn a rvrsibl ngin uss mor than two rsrvoirs th third or highr numbrd rsrvoirs will not b qual in tmpratur to th original two. Considration of xprssion for fficincy of th ngin indicats that for maximum fficincy, all th hat transfr should tak plac at maximum or minimum rsrvoir tmpraturs. Any intrmdiat rsrvoir usd will, thrfor, lowr th fficincy of th hat ngin. Practical ngin cycls oftn involv continuous changs of tmpratur during hat transfr. A rlationship among procsss in which ths sort of changs occur is ncssary. h idal approach to a cycl in which tmpratur continually changs is to considr th systm to b in communication with a larg numbr of rsrvoirs in procssion. Each rsrvoir is considrd to hav a tmpratur diffring by a small amount from th prvious on. n such a modl it is possibl to imagin that ach rsrvoir is rplacd by a rvrsibl hat ngin in communication with standard rsrvoirs at sam tmpratur 0. ig. 5.4 shows on xampl to this substitution. /M-thrm/th5-.pm5

6 3 ENGNEERNG HERMODYNAMCS + δ 0 δw R δq δw Original systm boundary δq Rvrsibl hat ngin Nw systm boundary (a) (b) ig h clausius inquality. h systm to which th hat transfr is ffctd is nithr concrnd with th sourc of nrgy it rcivs nor with th mthod of transfr, sav that it must b rvrsibl. Associatd with th small hat transfr dq to th original systm is a small work transfr dw and for this systm th first law givs ( δq δw) 0...(5.9) cycl Now considr th ngin rplacing th rsrvoirs and apply th scond law to th nw systm in ig. 5.4 (b). f th nw systm is not a prptual motion machin of scond kind, no positiv work transfr is possibl with a singl rsrvoir. hrfor, ( δw δw R ) 0 But by th dfinition of thrmodynamic tmpratur in quation (5.8) δw R δq0 δq 0 δq δq and by combination of qns. (5.9), (5.0) and (5.) δq 0 K J 0 but 0 0 and thrfor ; cycl cycl δq K J 0 cycl his is known as Clausius inquality. Lt us now considr th cas of a rvrsibl ngin for which cycl δq K J 0,...(5.0)...(5.)..(5.) /M-thrm/th5-.pm5

7 SECOND LAW O HERMODYNAMCS AND ENROPY 33 rvrs th ngin and for th rvrsibl hat pump obtaind it is possibl to dvlop th xprssion, H G δq K J 0 cycl h ngativ sign indicats that th hat transfrs hav all rvrsd in dirction whn th ngin was rvrsd. his mans that for th sam machin w hav two rlations which ar only satisfid if in th rvrsibl cas, cycl δq K J 0...(5.3) or a rvrsibl cas, as th numbr of rsrvoirs usd tnds to infinity, th limiting valu of th summation will b cycl δq K J n words, th Clausius inquality may b xprssd as follows : Whn a systm prforms a rvrsibl cycl, thn but whn th cycl is not rvrsibl 5.8. CARNO CYCLE cycl cycl δq K J δq K J < 0 0, 0. h cycl was first suggstd by a rnch nginr Sadi Carnot in 84 which works on rvrsibl cycl and is known as Carnot cycl. Any fluid may b usd to oprat th Carnot cycl (ig. 5.5) which is prformd in an ngin cylindr th had of which is supposd altrnativly to b prfct conductor or a prfct insulator of a hat. Hat is causd to flow into th cylindr by th application of high tmpratur nrgy sourc to th cylindr had during xpansion, and to flow from th cylindr by th application of a lowr tmpratur nrgy sourc to th had during comprssion. Hat sourc at Cylindr Sink at (a) Piston /M-thrm/th5-.pm5

8 34 ENGNEERNG HERMODYNAMCS (b) Carnot ngin cycl ig. 5.5 (c) Carnot hat pump cycl h assumptions mad for dscribing th working of th Carnot ngin ar as follows : (i) h piston moving in a cylindr dos not dvlop any friction during motion. (ii) h walls of piston and cylindr ar considrd as prfct insulators of hat. (iii) h cylindr had is so arrangd that it can b a prfct hat conductor or prfct hat insulator. (iv) h transfr of hat dos not affct th tmpratur of sourc or sink. (v) Working mdium is a prfct gas and has constant spcific hat. (vi) Comprssion and xpansion ar rvrsibl. ollowing ar th four stags of Carnot cycl : Stag. (Procss -). Hot nrgy sourc is applid. Hat Q is takn in whilst th fluid xpands isothrmally and rvrsibly at constant high tmpratur. Stag. (Procss -3). h cylindr bcoms a prfct insulator so that no hat flow taks plac. h fluid xpands adiabatically and rvrsibly whilst tmpratur falls from to. Stag 3. (Procss 3-4). Cold nrgy sourc is applid. Hat Q flows from th fluid whilst it is comprssd isothrmally and rvrsibly at constant lowr tmpratur. Stag 4. (Procss 4-). Cylindr had bcoms a prfct insulator so that no hat flow occurs. h comprssion is continud adiabatically and rvrsibly during which tmpratur is raisd from to. h work dlivrd from th systm during th cycl is rprsntd by th nclosd ara of th cycl. Again for a closd cycl, according to first law of th thrmodynamics th work obtaind is qual to th diffrnc btwn th hat supplid by th sourc (Q ) and th hat rjctd to th sink (Q ). W Q Q Work don Q Q Also, thrmal fficincy, η th Hat supplid by th sourc Q /M-thrm/th5-.pm5

9 SECOND LAW O HERMODYNAMCS AND ENROPY 35 Q Q L NM O P Q Q Q mcp Q m cp P whr, m mass of fluid.p Such an ngin sinc it consists ntirly of rvrsibl procsss, can oprat in th rvrs dirction so that it follows th cycl shown in ig. 5.5 (b) and oprats as a hat pump. Q is bing takn in at th lowr tmpratur during th isothrmal xpansion (procss 4-3) and hat Q is bing rjctd at th uppr tmpratur (procss -). Work W will b ndd to driv th pump. Again, th nclosd ara rprsnts this work which is xactly qual to that flowing from it whn usd as ngin. h Carnot cycl cannot b prformd in practic bcaus of th following rasons :. t is imposibl to prform a frictionlss procss.. t is impossibl to transfr th hat without tmpratur potntial. 3. sothrmal procss can b achivd only if th piston movs vry slowly to allow hat transfr so that th tmpratur rmains contant. Adiabatic procss can b achivd only if th piston movs as fast as possibl so that th hat transfr is ngligibl du to vry short tim availabl. h isothrmal and adiabatic procsss tak plac during th sam strok thrfor th piston has to mov vry slowly for part of th strok and it has to mov vry fast during rmaining strok. his variation of motion of th piston during th sam strok is not possibl CARNO S HEOREM t stats that of all ngins oprating btwn a givn constant tmpratur sourc and a givn constant tmpratur sink, non has a highr fficincy than a rvrsibl ngin. Rfr ig ig wo cyclic hat ngins HE A and HE B oprating btwn th sam sourc and sink, of which HE B is rvrsibl. HE A and HE B ar th two ngins oprating btwn th givn sourc at tmpratur and th givn sink at tmpratur. /M-thrm/th5-.pm5

10 36 ENGNEERNG HERMODYNAMCS Lt HE A b any hat ngin and HE B b any rvrsibl hat ngin. W hav to prov that fficincy of HE B is mor than that of HE A. Lt us assum that η A > η B. Lt th rats of working of th ngins b such that Q A Q B Q Sinc η A > η B WA WB > QA QB W A > W B Now, lt HE B b rvrsd. Sinc HE B is a rvrsibl hat ngin, th magnituds of hat and work transfr quantitis will rmain th sam, but thir dirctions will b rvrsd, as shown in ig Sinc W A > W B, som part of W A (qual to W B ) may b fd to driv th rvrsd hat ngin H B. Sinc Q A Q B Q, th hat dischargd by H B may b supplid to HE A. h sourc may, thrfor, b liminatd (ig. 5.8). h nt rsult is that HE A and H B togthr constitut a hat ngin which, oprating in a cycl producs nt work W A W B whil xchanging hat with a singl rsrvoir at. his violats th Klvin-Planck statmnt of th scond law. Hnc th assumption that η A > η B is wrong. ig HE B is rvrsd. ig HE A and H B togthr violat th Klvin-Planck statmnt. η B η A. /M-thrm/th5-.pm5

11 SECOND LAW O HERMODYNAMCS AND ENROPY COROLLARY O CARNO S HEOREM h fficincy of all rvrsibl hat ngins oprating btwn th sam tmpratur lvls is th sam. Rfr ig Lt both th hat ngins HE A and HE B b rvrsibl. Lt us assum η A > η B. Similar to th procdur outlind in th Articl 5.9, if HE B is rvrsd to run say, as a hat pump using som part of th work output (W A ) of ngin HE A, w s that th combind systm of hat pump HE B and ngin HE A, bcoms a PMM. So η A cannot b gratr than η B. Similary, if w assum η B > η A and rvrs th ngin HE A, w obsrv that η B cannot b gratr than η A η A η B. Sinc th fficincis of all rvrsibl ngins oprating btwn th sam hat rsrvoirs ar th sam, th fficincy of a rvrsibl ngin is indpndnt of th natur or amount of th working substanc undrgoing th cycl. 5.. ECENCY O HE REVERSBLE HEA ENGNE h fficincy of a rvrsibl hat ngin in which hat is rcivd solly at is found to b η rv. η max or η rv. Q Q rv. rom th abov xprssion, it may b notd that as dcrass and incrass, th fficincy of th rvrsibl cycl incrass. Sinc η is always lss than unity, is always gratr than zro and + v. h C.O.P. of a rfrigrator is givn by Q (C.O.P.) rf. Q Q Q Q or a rvrsibl rfrigrator, using Q Q (C.O.P.) rv. [(C.O.P.) rf. ] rv....(5.4) Similarly, for a rvrsibl hat pump [(C.O.P.) hat pump ] rv....(5.5) Exampl 5.. A hat ngin rcivs hat at th rat of 500 kj/min and givs an output of 8. kw. Dtrmin : (i) h thrmal fficincy ; (ii) h rat of hat rjction. /M-thrm/th5-.pm5

12 38 ENGNEERNG HERMODYNAMCS Solution. Hat rcivd by th hat ngin, Q 500 kj/min kj/s 60 Work output, W 8. kw 8. kj/s. Sourc Q 500 kj/min (i) hrmal fficincy, η th W Q % 5 Hnc, thrmal fficincy 3.8%. (Ans.) (ii) Rat of hat rjction, Q Q W kj/s Hnc, th rat of hat rjction 6.8 kj/s. (Ans.) HE Sink Q W 8. kw HE Hat ngin +Exampl 5.. During a procss a systm rcivs 30 kj of hat from a rsrvoir and dos 60 kj of work. s it possibl to rach initial stat by an adiabatic procss? Solution. Hat rcivd by th systm 30 kj Work don 60 kj p ig. 5.9 B A ig. 5.0 Procss - : By first law of thrmodynamics, Q (U U ) + W 30 (U U ) + 60 (U U ) 30 kj. Procss - : By first law of thrmodynamics, Q (U U ) + W W W 30 kj. hus 30 kj work has to b don on th systm to rstor it to original stat, by adiabatic procss. V /M-thrm/th5-.pm5

13 SECOND LAW O HERMODYNAMCS AND ENROPY 39 Exampl 5.3. ind th co-fficint of prformanc and hat transfr rat in th condnsr of a rfrigrator in kj/h which has a rfrigration capacity of 000 kj/h whn powr input is 0.75 kw. Solution. Rfr ig. 5.. Rfrigration capacity, Q 000 kj/h Powr input, W 0.75 kw ( kj/h) Co-fficint of prformanc, C.O.P. : Hat transfr rat : Hat absorbd at lowr tmpratur (C.O.P.) rfrigrator Work input C.O.P. Q W Hnc C.O.P (Ans.) Hnc transfr rat in condnsr Q According to th first law Q Q + W kj/h Hnc, hat transfr rat 4700 kj/h. (Ans.) Condnsr ( ) HE Evaporator ( ) Exampl 5.4. A domstic food rfrigrator maintains a tmpratur of C. h ambint air tmpratur is 35 C. f hat laks into th frzr at th continuous rat of kj/s dtrmin th last powr ncssary to pump this hat out continuously. Solution. rzr tmpratur, K Ambint air tmpratur, K Rat of hat lakag into th frzr kj/s Last powr rquird to pump th hat : h rfrigrator cycl rmovs hat from th frzr at th sam rat at which hat laks into it (ig. 5.). or minimum powr rquirmnt Q Q Q Q kj/s 6 W Q Q kj/s 0.36 kw Hnc, last powr rquird to pump th hat continuously 0.36 kw. (Ans.) 308 K Ambint air HE Q Q ig. 5. Q Q 6 K rzr Q kj/s ig. 5. Exampl 5.5. A hous rquirs 0 5 kj/h for hating in wintr. Hat pump is usd to absorb hat from cold air outsid in wintr and snd hat to th hous. Work rquird to oprat th hat pump is kj/h. Dtrmin : (i) Hat abstractd from outsid ; (ii) Co-fficint of prformanc. W W /M-thrm/th5-.pm5

14 40 ENGNEERNG HERMODYNAMCS Solution. (i) Hat rquirmnt of th hous, Q (or hat rjctd) 0 5 kj/h Work rquird to oprat th hat pump, W kj/h Now, Q W + Q whr Q is th hat abstractd from outsid Q hus Q kj/h Hnc, hat abstractd from outsid kj/h. (Ans.) (ii) (C.O.P.) Q hat pump Q Q Hnc, co-fficint of prformanc (Ans.) Not. f th hat rquirmnts of th hous wr th sam but this amount of hat had to b abstractd from th hous and rjctd out, i.., cooling of th hous in summr, w hav Q (C.O.P.) rfrigrator Q Q Q W hus th sam dvic has two valus of C.O.P. dpnding upon th objctiv. Exampl 5.6. What is th highst possibl thortical fficincy of a hat ngin oprating with a hot rsrvoir of furnac gass at 00 C whn th cooling watr availabl is at 5 C? Solution. mpratur of furnac gass, K mpratur of cooling watr, K Now, η max ( η carnot ) or 87.8%. (Ans.) Not. t should b notd that a systm in practic oprating btwn similar tmpraturs (.g., a stam gnrating plant) would hav a thrmal fficincy of about 30%. h discrpncy is du to irrvrsibility in th actual plant, and also bcaus of dviations from th idal Carnot cycl mad for various practical rasons. Exampl 5.7. A Carnot cycl oprats btwn sourc and sink tmpraturs of 50 C and 5 C. f th systm rcivs 90 kj from th sourc, find : (i) Efficincy of th systm ; (ii) h nt work transfr ; (iii) Hat rjctd to sink. Solution. mpratur of sourc, K mpratur of sink, K Hat rcivd by th systm, Q 90 kj (i) η carnot %. (Ans.) /M-thrm/th5-.pm5

15 SECOND LAW O HERMODYNAMCS AND ENROPY 4 W (ii) h nt work transfr, W η carnot Q Q η carnot Q kj. (Ans.) (iii) Hat rjctd to th sink, Q Q W [Q W Q Q ] kj. (Ans.) Exampl 5.8. An invntor claims that his ngin has th following spcifications : mpratur limits C and 5 C Powr dvlopd ul burnd pr hour Hating valu of th ful Stat whthr his claim is valid or not kw kg kj/kg Solution. mpratur of sourc, K mpratur of sink, K W know that th thrmal fficincy of Carnot cycl is th maximum btwn th spcifid tmpratur limits. Now, η carnot or 70.86% 03 h actual thrmal fficincy claimd, Work don η thrmal Hat supplid or 9.9% Sinc η thrmal > η carnot, thrfor claim of th invntor is not valid (or possibl). L N M (Ans.) Exampl 5.9. A cyclic hat ngin oprats btwn a sourc tmpratur of 000 C and a sink tmpratur of 40 C. ind th last rat of hat rjction pr kw nt output of th ngin? Solution. mpratur of sourc, K mpratur of sink, K Last rat of hat rjction pr kw nt output : or a rvrsibl hat ngin, th rat of hat rjction will b minimum (ig. 5.3) Now Now η max η rv. W Q nt η max Q W nt kw Q Q W nt kw Hnc, th last rat of hat rjction 0.36 kw. (Ans.) ig. 5.3 O Q P /M-thrm/th5-.pm5

16 4 ENGNEERNG HERMODYNAMCS Exampl 5.0. A fish frzing plant rquirs 40 tons of rfrigration. h frzing tmpratur is 35 C whil th ambint tmpratur is 30 C. f th prformanc of th plant is 0% of th thortical rvrsd Carnot cycl working within th sam tmpratur limits, calculat th powr rquird. Givn : ton of rfrigration 0 kj/min. Solution. Cooling rquird 40 tons kj/min Ambint tmpratur, K rzing tmpratur, K Prformanc of plant 0% of th thortical rvrsd Carnot cycl (C.O.P.) rfrigrator Actual C.O.P Now work ndd to produc cooling of 40 tons is calculatd as follows : Cooling rqd. (C.O.P.) actual Work ndd or W W 073. kj/min 9.5 kj/s 9.5 kw Hnc, powr rquird 9.5 kw. (Ans.) Exampl 5.. Sourc can supply nrgy at th rat of 000 kj/min at 30 C. A scond sourc can supply nrgy at th rat of 0000 kj/min at 70 C. Which sourc ( or ) would you choos to supply nrgy to an idal rvrsibl hat ngin that is to produc larg amount of powr if th tmpratur of th surroundings is 35 C? Solution. Sourc : Rat of supply of nrgy 000 kj/min mpratur, K. Sourc : Rat of supply of nrgy 0000 kj/min mpratur, K mpratur of th surroundings, 35 C K Lt th Carnot ngin b working in th two cass with th two sourc tmpraturs and th singl sink tmpratur. h fficincy of th cycl will b givn by : and η η h work dlivrd in th two cass is givn by or 48.06% 0.0 or 0.% W kj/min W kj/min. hus, choos sourc. (Ans.) Not. h sourc is slctd vn though fficincy in this cas is lowr, bcaus th critrion for slction is th largr output. /M-thrm/th5-.pm5

17 SECOND LAW O HERMODYNAMCS AND ENROPY 43 +Exampl 5.. A rvrsibl hat ngin oprats btwn two rsrvoirs at tmpraturs 700 C and 50 C. h ngin drivs a rvrsibl rfrigrator which oprats btwn rsrvoirs at tmpraturs of 50 C and 5 C. h hat transfr to th ngin is 500 kj and th nt work output of th combind ngin rfrigrator plant is 400 kj. (i) Dtrmin th hat transfr to th rfrigrant and th nt hat transfr to th rsrvoir at 50 C ; (ii) Rconsidr (i) givn that th fficincy of th hat ngin and th C.O.P. of th rfrigrator ar ach 45 pr cnt of thir maximum possibl valus. Solution. Rfr ig ig. 5.4 mpratur, K mpratur, K mpratur, K h hat transfr to th hat ngin, Q 500 kj h ntwork output of th combind ngin rfrigrator plant, W W W 400 kj. (i) Maximum fficincy of th hat ngin cycl is givn by Again, η max W Q W kj 3 48 (C.O.P.) max Also, C.O.P. Q 4 W /M-thrm/th5-.pm5

18 44 ENGNEERNG HERMODYNAMCS Sinc, W W W 400 kj W W W kj Q kj Q 3 Q 4 + W kj Q Q W kj. Hat rjction to th 50 C rsrvoir Q + Q kj. (Ans.) (ii) Efficincy of actual hat ngin cycl, η 0.45 η max W η Q kj W kj C.O.P. of th actual rfrigrator cycl, C.O.P. Q 4 W Q kj. (Ans.) Q kj Q kj Hat rjctd to 50 C rsrvoir Q + Q kj. (Ans.) +Exampl 5.3. (i) A rvrsibl hat pump is usd to maintain a tmpratur of 0 C in a rfrigrator whn it rjcts th hat to th surroundings at 5 C. f th hat rmoval rat from th rfrigrator is 440 kj/min, dtrmin th C.O.P. of th machin and work input rquird. (ii) f th rquird input to run th pump is dvlopd by a rvrsibl ngin which rcivs hat at 380 C and rjcts hat to atmosphr, thn dtrmin th ovrall C.O.P. of th systm. Solution. Rfr ig. 5.5 (a). (i) mpratur, K mpratur, K Sourc 5 C Sourc 380 C Sourc 0 C Q Q 3 Q W Hat pump Hat ngin W Hat pump Q Q 4 Q Sink 0 C (a) Singl systm Sink (Atmosphr) 5 C (b) Combind systm ig. 5.5 /M-thrm/th5-.pm5

19 SECOND LAW O HERMODYNAMCS AND ENROPY 45 Hat rmoval rat from th rfrigrator, Q 440 kj/min 4 kj/s Now, co-fficint of prformanc, for rvrsibl hat pump, 98 C.O.P..9. (Ans.) ( 98 73) 73 (C.O.P.) rf Now, 0.9 Q 4 W W W. kw i.., Work input rquird. kw. (Ans.) Q Q + W kj/s (ii) Rfr ig. 5.5 (b). h ovrall C.O.P. is givn by, Hat rmovd from th rfrigrator C.O.P. Hat supplid from th sourc Q Q3 or th rvrsibl ngin, w can writ Q3 Q4 3 4 or Q4 W Q or Q4 +. Q 4 ( ) ( ) or Q 4. Q (i) or 98(Q 4 +.) 653 Q 4 or Q 4 (653 98) or Q kj/s ( ) Q 3 Q 4 + W kj/s Substituting this valu in qn. (i), w gt 4 C.O.P (Ans.) f th purpos of th systm is to supply th hat to th sink at 5 C, thn Ovrall C.O.P. Q + Q (Ans.) Q Exampl 5.4. An ic plant working on a rvrsd Carnot cycl hat pump producs 5 tonns of ic pr day. h ic is formd from watr at 0 C and th formd ic is maintaind at 0 C. h hat is rjctd to th atmosphr at 5 C. h hat pump usd to run th ic plant is /M-thrm/th5-.pm5

20 46 ENGNEERNG HERMODYNAMCS coupld to a Carnot ngin which absorbs hat from a sourc which is maintaind at 0 C by burning liquid ful of kj/kg calorific valu and rjcts th hat to th atmosphr. Dtrmin : (i) Powr dvlopd by th ngin ; (ii) ul consumd pr hour. ak nthalpy of fusion of ic kj/kg. Solution. (i) ig. 5.6 shows th arrangmnt of th systm. Amount of ic producd pr day 5 tonns. (0 + 73) 493 K (5 + 73) 98 K Q Q p Hat ngin W Hat pump Q Q p (5 + 73) 98 K (0 + 73) 73 K ig. 5.6 h amount of hat rmovd by th hat pump, Q p kj/min C.O.P. of th hat pump Qp 73 W W Q p kj/min his work must b dvlopd by th Carnot ngin, W kj/s 5.3 kw hus powr dvlopd by th ngin 5.3 kw. (Ans.) (ii) h fficincy of Carnot ngin is givn by η carnot W Q W 5. 3 Q kj/s Q ( pr hour) kj Quantity of ful consumd/hour kg/h (Ans.) /M-thrm/th5-.pm5

21 SECOND LAW O HERMODYNAMCS AND ENROPY 47 Exampl 5.5. wo Carnot ngins work in sris btwn th sourc and sink tmpraturs of 550 K and 350 K. f both ngins dvlop qual powr dtrmin th intrmdiat tmpratur. Solution. ig. 5.7 shows th arrangmnt of th systm. mpratur of th sourc, 550 K mpratur of th sink, K ntrmdiat tmpratur, : h fficincis of th ngins HE and HE ar givn by η W Q W Q W...(i) η W Q 3 W Q3 W...(ii) rom qn. (i), w gt W (Q + W) W L NM W O P Q Q P Q W Q rom qn. (ii), w gt W Q 3...(iii) ig (iv) Now from qns. (iii) and (iv), w gt K Hnc intrmdiat tmpratur 450 K. (Ans.) Exampl 5.6. A Carnot hat ngin draws hat from a rsrvoir at tmpratur and rjcts hat to anothr rsrvoir at tmpratur 3. h Carnot forward cycl ngin drivs a Carnot rvrsd cycl ngin or Carnot rfrigrator which absorbs hat from rsrvoir at tmpratur and rjcts hat to a rsrvoir at tmpratur 3. f th high tmpratur 600 K and low tmpratur 300 K, dtrmin : (i) h tmpratur 3 such that hat supplid to ngin Q is qual to th hat absorbd by rfrigrator Q. (ii) h fficincy of Carnot ngin and C.O.P. of Carnot rfrigrator. Solution. Rfr ig mpratur, 600 K mpratur, 300 K /M-thrm/th5-.pm5

22 48 ENGNEERNG HERMODYNAMCS (300 K) (600 K) Q Q Carnot rfrig. W Q Q carnot Carnot ngin Q Q 3 ig. 5.8 Efficincy of Carnot ngin, η carnot ngin Q Q 3 Q Work of Carnot ngin Hat supplid to th Carnot ngin W Q or W Q carnot 3 Q Also C.O.P. (carnot rfrigrator) Q Q 3 Hat absorbd Q W W carnot or W carnot Q (i) mpratur, 3 : rom qns. (i) and (ii), w gt Q 3 3 Q Q Q carnot Q or Q 300 or Hnc, tmpratur, K. (Ans.) ( 3 300) or K carnot...(i)...(ii) /M-thrm/th5-.pm5

23 SECOND LAW O HERMODYNAMCS AND ENROPY 49 (ii) Efficincy of Carnot ngin, η carnot ngin C.O.P. rfrigrator %. (Ans.) 3. (Ans.) Exampl 5.7. A hat pump working on a rvrsd carnot cycl taks in nrgy from a rsrvoir maintaind at 5 C and dlivrs it to anothr rsrvoir whr tmpratur is 77 C. h hat pump drivs powr for its opration from a rvrsibl ngin oprating within th highr and lowr tmpraturs of 077 C and 77 C. or 00 kj/kg of nrgy supplid to rsrvoir at 77 C, stimat th nrgy takn from th rsrvoir at 077 C. (U.P.S.C., 994) Solution. Givn : K ; K ; K ; Enrgy takn from th rvrvoir at 077 C, Q : K Q 4 Q Pump Engin 350 K Q Q 3 78 K 3 ig. 5.9 or rvrsibl ngin, η Q Q Q...(i) or Q Q or Q Q Q4 4 or rvrsibl hat pump, C.O.P....(ii) Q4 Q3 4 3 Sinc work for running th pump is bing supplid by th ngin Q Q Q 4 Q 3 Q Q4 ( ) ( 4 3) [rom (i) and (ii)] 4 /M-thrm/th5-.pm5

24 50 ENGNEERNG HERMODYNAMCS Q Q or Q 4 and 4 Q Q Q Q 350 Q Q Q 4 + Q ( ) Q 00 Q kj. (Ans.) CLAUSUS NEQUALY +Exampl kj/s of hat is supplid at a constant fixd tmpratur of 90 C to a hat ngin. h hat rjction taks plac at 8.5 C. h following rsults wr obtaind : (i) 5 kj/s ar rjctd. (ii) 50 kj/s ar rjctd. (iii) 75 kj/s ar rjctd. Classify which of th rsult rport a rvrsibl cycl or irrvrsibl cycl or impossibl rsults. Solution. Hat supplid at 90 C 300 kj/s Hat rjctd at 8.5 C : (i) 5 kj/s, (ii) 50 kj/s, (iii) 75 kj/s. Applying Clausius inquality to th cycl or procss, w hav : δq (i) cycl < 0. Cycl is irrvrsibl. (Ans.) (ii) δq cycl Cycl is rvrsibl. (Ans.) (iii) δq cycl > 0. his cycl is impossibl by scond law of thrmodynamics, i.., Clausius inquality. (Ans.) Exampl 5.9. A stam powr plant oprats btwn boilr tmpratur of 60 C and condnsr tmpratur of 50 C. Watr ntrs th boilr as saturatd liquid and stam lavs th boilr as saturatd vapour. Vrify th Clausius inquality for th cycl. Givn : Enthalpy of watr ntring boilr 687 kj/kg. Enthalpy of stam laving boilr 760 kj/kg Condnsr prssur N/m. /M-thrm/th5-.pm5

25 SECOND LAW O HERMODYNAMCS AND ENROPY 5 and Solution. Boilr tmpratur, K Condnsr tmpratur, K rom stam tabls : Enthalpy of watr ntring boilr, h f 687 kj/kg Enthalpy of stam laving boilr, h 760 kj/kg Condnsr prssur N/m Boilr prssur N/m...(corrsponding to 60 C) Enthalpy of vapour laving th turbin, h 3 60 kj/kg (assuming isntropic xpansion) Enthalpy of watr laving th condnsr, h f4 09 kj/kg Now Q boilr, Q h h f kj/kg Q condnsr, Q h f4 h kj/kg δq Q Q K J cycl.5 kj/kg K < Provd. +Exampl 5.0. n a powr plant cycl, th tmpratur rang is 64 C to 5 C, th uppr tmpratur bing maintaind in th boilr whr hat is rcivd and th lowr tmpratur bing maintaind in th condnsr whr hat is rjctd. All othr procsss in th stady flow cycl ar adiabatic. h spcific nthalpis at various points ar givn in ig Vrify th Clausius nquality. ig. 5.0 Solution. mpratur maintaind in boilr, K mpratur maintaind in condnsr, K /M-thrm/th5-.pm5

26 5 ENGNEERNG HERMODYNAMCS Hat transfrrd in th boilr pr kg of fluid, Q h h kj/kg Hat transfrrd out at th condnsr pr kg of fluid, Q h 4 h kj/kg Sinc thr is no transfr of hat at any othr point, w hav pr kg δq Q Q K J cycl kj/kg K < 0. h Clausius nquality is provd. h stady flow cycl is obviously irrvrsibl. f th cycl is rvrsibl btwn th sam tmpratur limits and th hat supplid at highr tmpratur is sam, th hat rjctd can b calculatd as follows : i.., η rvrsibl or 5.86% Hat rjctd pr kg is givn by Q ( 0.586) Q ( 0.586) kj/kg δq cycl δq Q Q addd rjctd 0 sourc cycl sink hus Clausius Equality sign for a rvrsibl ngin is vrifid. 5.. ENROPY 5... ntroduction n hat ngin thory, th trm ntropy plays a vital rol and lads to important rsults which by othr mthods can b obtaind much mor laboriously. t may b notd that all hat is not qually valuabl for convrting into work. Hat that is supplid to a substanc at high tmpratur has a gratr possibility of convrsion into work than hat supplid to a substanc at a lowr tmpratur. Entropy is a function of a quantity of hat which shows th possibility of convrsion of that hat into work. h incras in ntropy is small whn hat is addd at a high tmpratur and is gratr whn hat addition is mad at a lowr tmpratur. hus for maximum ntropy, thr is minimum availability for convrsion into work and for minimum ntropy thr is maximum availability for convrsion into work Entropy a Proprty of a Systm Rfr ig. 5.. Lt us considr a systm undrgoing a rvrsibl procss from stat to stat along path L and thn from stat to th original stat along path M. Applying th Clausius thorm to this rvrsibl cyclic procss, w hav z δq 0 R (whr th subscript dsignats a rvrsibl cycl) /M-thrm/th5-3.pm5

27 SECOND LAW O HERMODYNAMCS AND ENROPY 53 Hnc whn th systm passs through th cycl -L--M-, w hav z z δq δq (5.6) ( L) ( M) Now considr anothr rvrsibl cycl in which th systm changs from stat to stat along path L, but rturns from stat to th original stat along a diffrnt path N. or this rvrsibl cyclic procss, w hav z z δq δq (5.7) ( L) ( N) or ig. 5.. Rvrsibl cyclic procss btwn two fixd nd stats. Subtracting quation (5.7) from quation (5.6), w hav z z δq δq 0 ( M) ( N) z z ( M) δq ( N) δq As no rstriction is imposd on paths L and M, xcpt that thy must b rvrsibl, th quantity δq is a function of th initial and final stats of th systm and is indpndnt of th path of th procss. Hnc it rprsnts a proprty of th systm. his proprty is known as th ntropy Chang of Entropy in a Rvrsibl Procss Rfr ig. 5.. Lt S Entropy at th initial stat, and S Entropy at th final stat. hn, th chang in ntropy of a systm, as it undrgos a chang from stat to, bcoms δq S S K J...(5.8) z Lastly, if th two quilibrium stats and ar infinitsimal nar to ach othr, th intgral sign may b omittd and S S bcoms qual to ds. R /M-thrm/th5-3.pm5

28 54 ENGNEERNG HERMODYNAMCS Hnc quation (5.8) may b writtn as K J...(5.9) ds δq R whr ds is an xact diffrntial. hus, from quation (5.9), w find that th chang of ntropy in a rvrsibl procss is qual to δq. his is th mathmatical formulation of th scond law of thrmodynamics. Equation (5.9) indicats that whn an inxact diffrntial δq is dividd by an intgrating factor during a rvrsibl procss, it bcoms an xact diffrntial. h third law of thrmodynamics stats Whn a systm is at zro absolut tmpratur, th ntropy of systm is zro. t is clar from th abov law that th absolut valu of ntropy corrsponding to a givn δq stat of th systm could b dtrmind by intgrating R and th givn stat. Zro ntropy, howvr, mans th absnc of all molcular, atomic, lctronic and nuclar disordrs. As it is not practicabl to gt data at zro absolut tmpratur, th chang in ntropy is calculatd ithr btwn two known stats or by slcting som convnint point at which th ntropy is givn an arbitrary valu of zro. or stam, th rfrnc point at which th ntropy is givn an arbitrary valu of zro is 0 C and for rfrigrants lik ammonia, ron-, carbon dioxid tc. th rfrnc point is 40 C, at which th ntropy it takn as zro. hus, in practic w can dtrmin th chang in ntropy and not th absolut valu of ntropy. K J btwn th stat at absolut zro 5.3. ENROPY AND RREVERSBLY W know that chang in ntropy in a rvrsibl procss is qual to us now find th chang in ntropy in an irrvrsibl procss. δq K J (qn. 5.9). Lt R p L M ig. 5.. Entropy chang for an irrvrsibl procss. Considr a closd systm undrgoing a chang from stat to stat by a rvrsibl procss -L- and rturns from stat to th initial stat by an irrvrsibl procss -M- as shown in ig. 5. on th thrmodynamic coordinats, prssur and volum. V /M-thrm/th5-3.pm5

29 SECOND LAW O HERMODYNAMCS AND ENROPY 55 Sinc ntropy is a thrmodynamic proprty, w can writ z z z R ( L) ( M ) ds ( ds) + ( ds) 0...(5.0) (Subscript rprsnts th irrvrsibl procss). Now for a rvrsibl procss, from qn. (5.9), w hav z δq ( ds) R ( L) ( L) H G K J z z R ds R in qn. (5.0), w gt ( L) K J z + R Substituting th valu of ( ) z δq ( ds) ( L) ( M )...(5.) 0...(5.) Again, sinc in qn. (5.0) th procsss -L- and -M- togthr form an irrvrsibl cycl, applying Clausius quality to this xprssion, w gt z z δq δq δq ( L) ( M ) H G K J + H G K J Now subtracting qn. (5.3) from qn. (5.), w gt z z ( ds) ( M) ( M) δq > H G K J R z < 0...(5.3) which for infinitsimal changs in stats can b writtn as z Q ( ds) > δ K J...(5.4) Eqn. (5.4) stats that th chang in ntropy in an irrvrsibl procss is gratr than Combining qns. (5.3) and (5.4), w can writ th quation in th gnral form as δq. ds δq...(5.5) whr quality sign stands for th rvrsibl procss and inquality sign stands for th irrvrsibl procss. t may b notd hr that th ffct of irrvrsibility is always to incras th ntropy of th systm. Lt us now considr an isolatd systm. W know that in an isolatd systm, mattr, work or hat cannot cross th boundary of th systm. Hnc according to first law of thrmodynamics, th intrnal nrgy of th systm will rmain constant. Sinc for an isolatd systm, δq 0, from qn. (5.5), w gt (ds) isolatd 0...(5.6) Eqn. (5.6) stats that th ntropy of an isolatd systm ithr incrass or rmains constant. his is a corollary of th scond law. t xplains th principl of incras in ntropy CHANGE N ENROPY O HE UNVERSE W know that th ntropy of an isolatd systm ithr incras or rmains constant, i.., (ds) isolatd _ 0 /M-thrm/th5-3.pm5

30 56 ENGNEERNG HERMODYNAMCS By including any systm and its surrounding within a singl boundary, as shown in ig. 5.3, an isolatd systm can b formd. h combination of th systm and th surroundings within a singl boundary is somtims calld th Univrs. Hnc, applying th principl of incras in ntropy, w gt (ds) univrs _ 0 whr (ds) univrs (ds) systm + (ds) surroundings. Systm tmpratur Boundary of th univrs Surrounding tmpratur 0 ig Entropy chang of univrs. n th combind closd systm considr that a quantity of hat δq is transfrrd from th systm at tmpratur to th surroundings at tmpratur 0. Applying qn. (5.4) to this procss, w can writ (ds) systm > δq ( v sign indicats that hat is transfrrd from th systm). Similarly, sinc an amount of hat δq is absorbd by th surroundings, for a rvrsibl procss, w can writ (ds) surroundings δq 0 Hnc, th total chang in ntropy for th combind systm (ds) systm + (ds) surroundings δq δ Q + or (ds) univrs dq + 0 h sam rsult can b obtaind in th cas of an opn systm. or both closd and opn systms, w can writ (ds) univrs 0...(5.7) Eqn. (5.7) stats that th procss involving th intraction of a systm and th surroundings taks plac only if th nt ntropy of th combind systm incrass or in th limit rmains constant. Sinc all natural procsss ar irrvrsibl, th ntropy is incrasing continually. 0 /M-thrm/th5-3.pm5

31 SECOND LAW O HERMODYNAMCS AND ENROPY 57 h ntropy attains its maximum valu whn th systm rachs a stabl quilibrium stat from a non-quilibrium stat. his is th stat of maximum disordr and is on of maximum thrmodynamic probability EMPERAURE-ENROPY DAGRAM f ntropy is plottd-horizontally and absolut tmpratur vrtically th diagram so obtaind is calld tmpratur-ntropy (-s) diagram. Such a diagram is shown in ig f working fluid rcivs a small amount of hat dq in an lmntary portion ab of an opration AB whn tmpratur is, and if dq is rprsntd by th shadd ara of which is th man ordinat, th width of th figur must b dq. his is calld incrmnt of ntropy and is dnotd by ds. h total hat rcivd by th opration will b givn by th ara undr th curv AB and (S B S A ) will b corrsponding incras of ntropy. (mp.) B a b A S A ds S B S (Entropy) ig mpratur-ntropy diagram. rom abov w conclud that : Hat chang ( Q) Entropy chang, ds Absolut tmpratur ( ). Entropy may also b dfind as th thrmal proprty of a substanc which rmains constant whn substanc is xpandd or comprssd adiabatically in a cylindr. Not. s stands for spcific ntropy whras S mans total ntropy (i.., S ms) CHARACERSCS O ENROPY h charactristics of ntropy in a summarisd form ar givn blow :. t incrass whn hat is supplid irrspctiv of th fact whthr tmpratur changs or not.. t dcras whn hat is rmovd whthr tmpratur changs or not. /M-thrm/th5-3.pm5

32 58 ENGNEERNG HERMODYNAMCS 3. t rmains unchangd in all adiabatic frictionlss procsss. 4. t incrass if tmpratur of hat is lowrd without work bing don as in a throttling procss ENROPY CHANGES OR A CLOSED SYSEM Gnral Cas for Chang of Entropy of a Gas Lt kg of gas at a prssur p, volum v, absolut tmpratur and ntropy s, b hatd such that its final prssur, volum, absolut tmpratur and ntropy ar p, v, and s rspctivly. hn by law of consrvation of nrgy, dq du + dw whr, dq Small chang of hat, du Small intrnal nrgy, and dw Small chang of work don (pdv). Now dq c v d + pdv Dividing both sids by, w gt dq cd v pdv + But dq ds and as pv R p R v Hnc cd v ds + R dv v ntgrating both sids, w gt s d v dv ds cv + R s v or or z z z (s s ) c v log v + R log v v his xprssion can b rproducd in th following way : According to th gas quation, w hav pv pv p v p v Substituting th valu of in qn. (5.8), w gt s s c v log p p v + R log v v v...(5.8) c v log p p + c v log v v + R log v v /M-thrm/th5-3.pm5

33 SECOND LAW O HERMODYNAMCS AND ENROPY 59 c v log p p c v log p p + (c v + R) log v v + c p log v v s s c v log p + c p log p v v Again, from gas quation, pv pv or v v p p...(5.9) Putting th valu of v v in qn. (5.8), w gt (s s ) c v log + R log p p c v log + R log p p + R log (c v + R) log R log p p c p log R log p p s s c p log R log p....(5.30) p Hating a Gas at Constant Volum Rfr ig Lt kg of gas b hatd at constant volum and lt th chang in ntropy and absolut tmpratur b from s to s and to rspctivly. hn Q c v ( ) Diffrntiating to find small incrmnt of hat dq corrsponding to small ris in tmpratur d. dq c v d Dividing both sids by, w gt dq c v. d or or ds c v. d ntgrating both sids, w gt s d ds cv s z z s s c v log s v Const....(5.3) ig s diagram : Constant volum procss Q s s /M-thrm/th5-3.pm5

34 60 ENGNEERNG HERMODYNAMCS Hating a Gas at Constant Prssur Rfr ig Lt kg of gas b hatd at constant prssur, so that its absolut tmpratur changs from to and ntropy s to s. v const. p const. Q s s s d. ig s diagram : Constant prssur procss. hn, Q c p ( ). Diffrntiating to find small incras in hat, dq of this gas whn th tmpratur ris is dq c p. d Dividing both sids by, w gt dq c p. d or ntgrating both sids, w gt zs s ds c p. d ds c z p s s c p log...(5.3) sothrmal Procss An isothrmal xpansion - at constant tmpratur is shown in ig Entropy changs from s to s whn gas absorbs hat during xpansion. h hat takn by th gas is givn by th ara undr th lin - which also rprsnts th work don during xpansion. n othr words, Q W. s But Q ds ( s s ) z s d and W p v log v v R log v v pr kg of gas [Q p v R ] /M-thrm/th5-3.pm5

35 SECOND LAW O HERMODYNAMCS AND ENROPY 6 ig s diagram : sothrmal procss. (s s ) R log v v or s s R log v. [Q v ]...(5.33) Adiabatic Procss (Rvrsibl) During an adiabatic procss as hat is nithr supplid nor rjctd, dq 0 dq or d 0 or ds 0...(5.34) ig s diagram : Adiabatic procss. his shows that thr is no chang in ntropy and hnc it is known as isntropic procss. ig. 5.8 rprsnts an adiabatic procss. t is a vrtical lin (-) and thrfor ara undr this lin is nil ; hnc hat supplid or rjctd and ntropy chang is zro. /M-thrm/th5-3.pm5

36 6 ENGNEERNG HERMODYNAMCS Polytropic Procss Rfr ig h xprssion for ntropy chang in polytropic procss (pv n constant) can b obtaind from qn. (5.8) i.., s s c v log + R log v v or or or or Also Again, as ig s diagram : Polytropic procss. p v n p v n p p pv p p rom (i) and (ii), w gt v v v v n v v n H G K J...(i) pv v v n v v v v Substituting th valu of v v...(ii) n in qn. (5.8), w gt s s c v log + R log n cv log + R n log /M-thrm/th5-3.pm5

37 SECOND LAW O HERMODYNAMCS AND ENROPY 63 c v log c v log c v log cv γ n c v c v. s s c v L N M n n R n (c p c v ) (γ. c v c v ) O Q P n γ + ) n n γ n γ log log log log log n log log n L n γ cvnm ( n ) pr kg of gas ( ) ( ) O QP [Q R c p c v ] log [Q c p γ. c v ] pr kg of gas...(5.35) Approximation or Hat Absorbd h curv LM shown in th ig is obtaind by hating kg of gas from initial stat L to final stat M. Lt tmpratur during hating incrass from to. hn hat absorbd by th gas will b givn by th ara (shown shadd) undr curv LM. M L Q s s s ig As th curv on -s diagram which rprsnts th hating of th gas, usually has vry slight curvatur, it can b assumd a straight lin for a small tmpratur rang. hn, Hat absorbd Ara undr th curv LM (s s ) + K J...(5.36) /M-thrm/th5-3.pm5

38 64 ENGNEERNG HERMODYNAMCS n othr words, hat absorbd approximatly quals th product of chang of ntropy and man absolut tmpratur. abl 5.. Summary of ormula S. No. Procss Chang of ntropy (pr kg). Gnral cas (i) c v log + R log v v (in trms of and v) (ii) c v log p p (iii) c p log. Constant volum c v log 3. Constant prssur c p log 4. sothrmal R log v v 5. Adiabatic Zro 6. Polytropic c v n γ n + c v log v v R log p p log (in trms of p and v) (in trms of and p) 5.8. ENROPY CHANGES OR AN OPEN SYSEM n an opn systm, as compard with closd systm, thr is additional chang of ntropy du to th mass crossing th boundaris of th systm. h nt chang of ntropy of a systm du to mass transport is qual to th diffrnc btwn th product of th mass and its spcific ntropy at th inlt and at th outlt of th systm. hrfor, th total chang of ntropy of th systm during a small intrval is givn by ds dq + Σs i. dm Σs. dm 0 i 0 0 whr, 0 mpratur of th surroundings, s i Spcific ntropy at th inlt, s 0 Spcific ntropy at th outlt, dm i Mass ntring th systm, and dm 0 Mass laving th systm. (Subscripts i and 0 rfr to inlt and outlt conditions) h abov quation in gnral form can b writtn as ds dq + s. dm 0...(5.37) n quation (5.37) ntropy flow into th systm is considrd positiv and ntropy out-flow is considrd ngativ. h quality sign is applicabl to rvrsibl procss in which th hat intractions and mass transport to and from th systm is accomplishd rvrsibly. h inquality sign is applicabl to irrvrsibl procsss. /M-thrm/th5-3.pm5

39 SECOND LAW O HERMODYNAMCS AND ENROPY 65 f quation (5.37) is dividd by dt, thn it bcoms a rat quation and is writtn as ds dq. + dt 0 s. dm...(5.38) dt dt ds n a stady-stat, stady flow procss, th rat of chang of ntropy of th systm H dtk bcoms zro. or whr Q. dq dt and &m dm dt. O 0 dq dt + s. dm dt Q. + Σs. &m 0...(5.39) 0 or adiabatic stady flow procss, Q. 0 s. m& 0...(5.40) f th procss is rvrsibl adiabatic, thn s. m& 0...(5.4) 5.9. HE HRD LAW O HERMODYNAMCS l h third law of thrmodynamics is statd as follow : h ntropy of all prfct crystallin solids is zro at absolut zro tmpratur. l h third law of thrmodynamics, oftn rfrrd to as Nrnst Law, provids th basis for th calculation of absolut ntropis of substancs. According to this law, if th ntropy is zro at 0, th absolut ntropy s ab. of a substanc at any tmpratur and prssur p is xprssd by th xprssion. z z z s f d h sf f g d hfg sab cps + + d cpf + cpg 0 s s g g...(5.4) whr s f sf sat... for fusion, f g fg sat...for vaporisation c ps, c pf, c pg Constant prssur spcific hats for solids, liquids and gas, h sf, h fg Latnt hats of fusion and vaporisation. hus by putting s 0 at 0, on may intgrat zro klvin and standard stat of 78.5 K and atm., and find th ntropy diffrnc. l urthr, it can b shown that th ntropy of a crystallin substanc at 0 is not a function of prssur, viz., s p 0 0 /M-thrm/th5-3.pm5