1S(hot reservoir) = - |QH|/ THS(cold reservoir) = + |QC|/ TCS(engine) = 0(cyclic process)S(surroundings) = 0(no heat transfer to surroundings)S(total) = - |QH|/ TH + |QC|/ TC Useful or! can only "e done if S(total) 0 | QH| / TH | QC | / TC2An ideal engine e.g. Carnot CycleS = 0 | QH / TH | = | QC / TC |e = W / QHW = |QH| - |QC|e = (|QH| - |QC| / QH)= # - |QC| / |QH|e = # $ TC / TH% #This is the absolute max efficiency of a heat engineSadi Carnot (1824)renchman!"T#$ All heat (QH % QC) exchanges occurred isothermally in calculating the efficiency e = 1 TC/ TH &CARNOT CYCLE'eat engine (ith the maximum )ossible efficiency consistent (ith 2nd la(.All thermal )rocesses in the cycle must be re*ersible+ all heat transfer must occur isothermally because con*ersion of (or, to heat is irre*ersible.-hen the tem)erature of the (or,ing substances changes. there must be /ero heat exchange 0 adiabatic )rocess.Carnot cycle 0 consists of t(o re*ersible isotherms and t(o re*ersible adiabats 44312QHQCreleased tosurroundingsVPCarnot Cycleadia"atic isother&alsDiagram not to scale, adiabats are much steeer than sho!nW!"T#$ All heat (QH % QC) exchanges occurred isothermally in calculating the efficiency e = 1 TC / TH 12 &Adiabaticcom)ression1 22sothermalcom)ressionQCQH& 42sothermalex)ansion4 1Adiabaticex)ansionAll energy exchanges are re*ersible 0 there are no non+reco*erable energy lossesV

43121 and 2$ 3cold4& and 4$ 3hot456roof$ e = # - |QC| / |QH| = # - TC / THIsothermal changeQ = n " T ln(Vf /Vi)1 2$ 2sothermal com)ression heat QC re'ected to sin! at constant te&perature TC( |QC | = n " TC ln(V) / V#)* 4$ 2sothermal ex)ansion heat QH supplied fro& source at constant te&perature|QH | = n " TH ln(V+ / V*) e = # - |QC| / |QH| = # $ TC ln(V) / V#) / TH ln(V+ / V*) (#7. 1)V

43121 and 2$ 3cold4& and 4$ 3hot4Adiabatic change Q = 0TiVi-# = T#V#-# 82 &$Adiabatic ex)ansione = # - TC ln(V# / V)) / TH ln(V+ / V*)2n )ractice. the Carnot cycle cannot be used for a heat engine because the slo)es of the adiabatic and isothermal lines are *ery similar and the net (or, out)ut (area enclosed by V diagram) is too small to o*ercome friction % other losses in a real engine.#fficiency of Carnot engine+ max )ossible efficiency for a heat engine o)erating bet(een TC and TH# #) ) * *T V TV =# ## # + +TV T V =4 1$Adiabatic ex)ansion# ) * +T T T T = =# ## +) *# +) *V VV VV VV V = = #CHTeT= (#7. 1)8Carnot Engine99.19.29.&9.49.19.59.89.89.:19 299 499 599 899 1999 1299 1499 1599 1899 2999Temperature TH(oC)efficiencyThe strength % hardness of metalsdecreases ra)idly abo*e 819 oC TC = ), oCCCarnotH#TeT= 3;e+author4 isobaric com)ression122nta,e1Com)ression 26o(er haust 41&54312PoQHQCreleased tosurroundingsVPOtto Cycleadia"atic isother&alsWcooling of exhaust gases2?!2T2"!fuel combustion)o(er stro,ecom)ression stro,einta,e stro,e exhaust stro,eV# = r V) r = co&pression ratioV2### er = 14% &$ inlet stro"e *olume increases as )iston mo*es do(n creating a )artial *acuum to aid air>fuel entering cylinder *ia the o)en inlet *al*e.54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als11& 2$ compression stro"e inlet *al*e closes )iston mo*es u) com)ressing the air>fuel mixture a#ia$atically.54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als15' ($ ignition 0 s)ar, )lug fires igniting mixture + constant *olume combustion.54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als18( )$ e*pansion or po+er stro"e 0 heated gas ex)ands adiabatically as the )iston is )ushed do(n doing (or, (V&a- = r V&in).Com)ression ratio. r54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als18) & start of E*haust stro"e 0 outlet *al*e o)ens and mixture ex)elled at constant *olume then& %54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als1:& %$ E*haust stro"e 0 )iston mo*es u) )roducing a com)ression at constant )ressure. Po (atmos)heric )ressure).54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"atic isother&als29Otto cycle ! stan#ar# petrol engine () stro"e)2deali/ed model of the thermodynamic )rocesses in a ty)ical car engine.or com)ression ratio. r @ 8 and A 1.4 (air)TH ()ea,) @ 1899 BCTC (base) @ 19 BC e @ 15C (ideal engine) e @ &1C (real engine).#fficiency increases (ith larger r engine o)erates at higher tem)eratures )re0ignition ,noc,ing sound and engine can be damaged."ctane rating 0 measure of anti+,noc,ing 0 )remium )etrol r @ 12.2n )ractice. the same air does not enter the engine again. but since an e7ui*alent amount of air does enter. (e may consider the )rocess as cyclic. ### er = 21Com)arison of theoretical and actual )D diagrams for the four+stro,e "tto Cycle engine. V22,iesel Cycle1 to 1$inta,e stro,eisobaric ex)ansion1 to 2$com)ression stro,e abiabatic com)ression2 to &$ignitionisobaric heating& to 4$)o(er stro,e adiabatic ex)ansion4 to 1$exhaust stro,e (start) isochoric cooling1 to 1$ exhaust stro,e (finish) isobaric com)ression54312PoV)V#QHQCreleased tosurroundingsVPadia"atic isother&als;udolf =iesel2&54312PoV)V#QHQCreleased tosurroundingsVP Diesel Cycleadia"atic isother&alspoer stro!ecooling of e-haust gasescom)ression stro,efuel ignition2421,IE-EL CYCLE!o fuel in the cylinder at beginning of com)ression stro,e. At the end of the adiabatic com)ression high tem)eratures are reached and then fuel is inEected fast enough to ,ee) the )ressure constant. The inEected fuel because of the high tem)eratures ignites s)ontaneously (ithout the need for s)ar, )lugs. =iesel engines o)erate at higher tem)eratures than )etrol engines. hence more efficient.or r @ 18 and A 1.4 (air) TH ()ea,) @ 2999 BCTC (base) @ 19 BC e @ 58C (ideal engine)real efficiency @ 49 C6etrol engine@ 15C (ideal engine)real efficiency @ &1 C25,iesel enginesF'ea*ier (higher com)ression ratios). lo(er )o(er to (eight ratio.F'arder to start.FGore efficient than )etrol engines (higher com)ression ratios).F!o )re+ignition of fuel since no fuel in cylinder during most of the com)ressionFThey need no carburettor or ignition system. but the fuel+inEection systemre7uires ex)ensi*e high+)recision machining.FHse chea)er fuels less refined hea*y oils 0 fuel does need to be *a)ori/ed in carburettor. fuel less *olatile hence safer from fire or ex)losion.F =iesel cycle 0 can control amount of inEected fuel )er cycle 0 less fuel used atlo( s)eeds.28htt)$>>)eo)le.bath.ac.u,>ccsshb>12cyl>These engines (ere designed )rimarily for *ery large container shi)s. 286roblemor the theoretical "tto cycle. calculate$(a) max cycle tem)erature(b) (or, )er ,ilogram of fuel(c) #fficiency(d) Gax efficiency Carnot engine (or,ing bet(een same tem)eratures#ngine characteristics$cp A 1.991 ,I.,g+1.J+1 c. A 9.818 ,I.,g+1.J+1com)ression ratio A 8$12nlet conditionsA :8.1 ,6a and T A 19 oC'eat su))lied A :19 ,I.,g+12:SolutionIdentify / Setupcp A 1.991 ,I.,g+1.J+1

c. A 9.818 ,I.,g+1.J+1V#/ V) A 8

1 A :8.119& 6aT1 A 19 oC A &2& JQH A :19 ,I.,g+1 TH A T* A K JW A K ,I.,g+1e A K eCarnot A KAdiabatic change Q = 0T V -# = constant = cp / c. = #(00, / 0(/#0 = #(+Qp = m cp TQ. = m c. T W = |QH| - |QC|e = W / |QH|eCarnot = # $ TC / TH54312PoV)V#QHQCreleased tosurroundingsVP Otto Cycleadia"aticisother&als&9Adiabatic com)ressionExecuteT) / T# = (V#/V))-#T) = T#(V# / V))-# = (*)*)(0)0(+ 1 = /+) 1Constant *olume heatingQH = m c. (T* $ T))T* = (QH/m)/c. +T) = (2,0/0(/#0 + /+)) 1 = 2065 K = 1792 oCAdiabatic ex)ansionT+ / T* = (V* /V+)-#T+ = T*(#/0)0(+ = 022 1Constant *olume heat reEectionQC = m c. (T+ $ T*) QC/m = (0(/#0)(022 - *)*) !3(!g-# = +#+ !3(!g-#W = |QH| - |QC| = (2,0 $ +#+) !3(!g-# =536 k!k"#1 e = W / |QH| = ,*4 / 2,0 = 0(,4 (real engine % 0(+)eCarnot = # $ TC / TH = # $ *)* / )04/ = 0!$4&154312PoV) V#QHQCreleased tosurroundingsVPDiesel Cycleadia"atic isother&alsCalculate the abo*e 7uantities for the diesel cycle (ith a com)ression ratio A 29&2Adiabatic com)ressionV# / V) = )0T) / T# = (V#/V))-# T) = T#(V# / V))-# = (*)*)()0)0(+ 1 = #0/# 12sobaric heatingQH = m c

(T*-T)) QH/m = c

(T* $T))T* = T) + (#/c

)(QH/m) = #0/# +(#/#(00,)(2,0) 1 = )0#4 12sobaric heating > Adiabatic ex)ansionV) / T) = V* / T* V* / V) = T* / T) V+ / V) = )0V) = V+ / )0V* / V+ = (#/)0)(T*/T)) = (#/)0)()0#4/#0/#) = 0(02+#T*V*-# = T+V+-#T+ = T* (V*/V+)-# = ()0#4)(0(02+#)0(+ = /0* 1 2sochoric coolingQC = m c$ (T+ $ T#)QC/m = c$ (T+ $ T#) = (0(/#0)(/0* $ *)*) 1 = **0(* !3(!g-#W = QH $ QC = (2,0 $ **0) !3 = 4)0 !3(!g-#e = W / QH = 4)0 / 2,0 = 0(4,e = #- QC/QH = # $ **0/2,0 = 0(4, The (or, out)ut and efficiency are considerably higher than for "tto Cycle&&E*ample Consider t(o engines. the details of (hich are gi*en in the follo(ing diagrams. or both engines. calculate the heat flo( to the cold reser*oir and the changes in entro)y of the hot reser*oir. cold reser*oir and engine. -hich engine *iolates the Second La(K -hat is the efficiency of the (or,ing engineK&4Solutionirst La($ % = Qnet $ W #ngine$ cyclic )rocess % A 9 Qnet = W |QH| - |QC| |QC | = |QH| - W#ngine 1$ MQC| A 1999 + 299 A 899 I#ngine 2$ MQC| A 1999 + &99 A 899 I&1S(total) = - |QH|/ TH + |QC|/ TC#ngine 1$ S = (- )(, + )(/) 3(1-# = + 0()3(1-# 5 0 Second La( *alidated#ngine 2$ S = (- )(, + )(*) 3(1-#= - 0() 3(1-# % 0 Second La( not *alidated #ngine 1 is the (or,ing engineefficiency. e A ((or, out > energy in)ut) 199A (299 > 1999)(199) A 29 C&5Semester 1. 2998 #xamination 7uestion (1 mar,s)A hybrid )etrol+engine car has a higher efficiency than a )etrol+only car because it reco*ers some of the energy that (ould normally be lost as heat to the surrounding en*ironment during brea,ing.(a)2f the efficiency of a ty)ical )etrol+only car engine is 29C. (hat efficiency could be achie*ed if the amount of heat loss during brea,ing is hal*edK(b)2s it )ossible to reco*er all the energy lost as heat during bra,ing and con*ert it into mechanical energyK #x)lain your ans(er.&8SolutionIdentify / Setupefficiency #H C CH H HW Q Q QeQ Q Q= = = Second La( of Thermodynamics 199C of heat can not be transformed into mechanical energy e N 1Execute(a)# 0() 0(00C CH HQ QeQ Q= = =;educe heat loss by half0(+0 0(4CHQeQ= =(b) -ould re7uire MQCM A 9. this (ould be a *iolation of the Second La(# # 0 0C CCH HQ Qe QQ Q= < > >&8-hat is a heat )um) KOetter buy this 7uic,$ 199 C efficiency&:WQCQHe.aporatorgas absorbs heat from interior of frig.coldhotcompressorheats gas by com)ressioncon#ensergas to li7uid (high )ressure)e*pansion .aluera)id ex)ansion$ li7uid to gas. sudden large dro) in tem) (@&1 oC)|6H| = |6c| + |7|49The compressor com)resses the gas (e.g. ammonia). The com)ressed gas heats u) as it is )ressuri/ed (orange). The gas re)resents the (or,ing substance eg ammonia and the com)ressor dri*en by an electric motor does (or, W.The con#enser coils at the bac, of the refrigerator let the hot ammonia gas dissi)ate its heat QH. The ammonia gas condenses into ammonia li7uid (dar, blue) at high )ressure gas (gas li7uid). The high+)ressure ammonia li7uid flo(s through the e*pansion .al.e The li7uid ammonia immediately boils and *a)ori/es (light blue). its tem)erature dro))ing to about 0&1 BC by the ex)ansion. This ma,es the inside of the refrigerator coldby absor)tion of heat QC .The cold ammonia gas enters the compressor and the cycle re)eats. 41Refrigeration Cycle'eat engine o)erating in re*erse 0 it ta,es heat from a cold )lace and gi*es it off at a (armer )lace. this re7uires a net in)ut of (or,. |QH| = |QC| + |W|Oest refrigerator 0 one that remo*es the greatest amount of heat MQCM from inside the refrigerator for the least ex)enditure of (or, MWM coe%%icient o% &er%or'ance8 & (higher & value8 "etter the refrigerator)CH CCQ Q&W Q Q= =& A (hat (e (ant > (hat (e )ay for& A extraction of max heat from cold reser*oir > least amount of (or,42

VQHQC74&(e%ri"erator7alls of roo&TCW ' (QCQHQC= QH - 7refrigerator9nside refrigerator|6H| = |6c| + |7|44)ir Con*itioner7alls of roo&TCW ' (QCQHQH= W ) QC:utside|6H| = |6c| + |7|41Heat +,'&7alls of roo&TCW ' (QCQHQH= W ) QC :utside|6H| = |6c| + |7|456'