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Home , Otto cycle, Stirling cycle

Camille Bélanger-ChampagneLuis Anchordoqui LehmanMcGill College University City University of New York

ThermodynamicsCharged Particle and StatisticalCorrelations Mechanics in Minimum Bias Events at ATLAS

• ● Physics changes motivation in cyclic processes Maxwell● Minbias relations event and track selection III • th Dissipative as supply 11 SeptemberFebruary 2014 26 2012,• ● Azimuthal correlation results Entropy of ideal WNPPC 2012 • ● Forward-Backward correlation results • Third Law of Thermodynamics

Saturday, September 13, 14 1 @V @V dV = dP + dT (7) @P @T ✓ ◆T ✓ ◆P W = PdV (17) Z (39) CP = CV + nR

@P @V C = C + T (42) P V @T @T ✓ ◆V ✓ ◆P

nR nR C = C = (50) V 1 P 1 Q = TdS (79)

Saturday, September 13, 14 2 ENTROPY

Consequence of secondOverview law is existence of entropy:

a @ thermodynamic equilibrium whose differential is given by

Q = TdS (79)

S being a state function ☛ does not change in any reversible cyclic process:

Q =0 T I

Since Q is extensive ☛ so is S

Units of entropy are [S]=J/K

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 3 CHANGE OF ENTROPY IN IRREVERSIBLE PROCESSES In non equilibrium statesOverview ☛ thermodynamic entropy is undefined If initial and final states of are in equilibrium entropy of these states is defined so ☛ entropy change S = S S 12 2 1 We can find always reversible process connecting 1 and 2: equivalent reversible process Both processes can be joined into irreversible cyclic process for which Clausius inequality applies and takes the form 2 Q 1 Q + reversible 0 (80) T T  Z1 Z2 Since reversible integral is related to change of entropy 2 2 Qreversible Q (81) =S12 1 T 1 T Z Q Z dS T (82) C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 4 CHANGE OF ENTROPY ON ISOLATED SYSTEMS If system is isolated ☛OverviewQ = W =0 Still ☛ dS 0 due to irreversible processes inside systems e.g. relaxation to equilibrium Consider that consists of two subsystems each of them at internal equilibrium but there is no equilibrium between subsystems Changes in reservoir are:

QHOT Q2 (S)HOT = = < 0 T T2 T =Z T2

QCOLD 1 Q1 (S)COLD = = Q = > 0 T T1 T1 T =Z T1

because hot reservoir loses heat Q 2 to and cold reservoir gains heat from engine Q1 C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 5 ENTROPY CHANGES IN CYCLIC PROCESSES Overview Then for the whole system follows

Q1 Q2 (S) = + 0 (83) TOTAL T T ✓ 1 2 ◆

According to second law of thermodynamics heat flows from hot to cold body

T1

Reversible cycle ☛ net change of total entropy of the engine + reservoir is zero

Irreversible cycle ☛ increase of total entropy due to spontaneous processes

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 6 MAXWELL RELATION Inserting (17) and (79) into first law of thermodynamics mainOverview thermodynamic identity dU = TdS PdV (84) (84) is a differential of as a function of two variables

@U @U Correspondingly ☛ T = P = (85) @S @V !V !S

As second mixed derivative does not depend on order of differentiation

@2U @2U = @S@V @V@S (86) it follows that @T @P = @V @S !S !V (87)

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 7 DISSIPATIVE WORK AS HEAT SUPPLY Overview Q = Qreversible + Qdissipative Qreversible = TdS and (88) W = Wreversible + Wdissipative Wreversible = PdV

Then using the first law (25), (84), and (88) we have

Q W = TdS PdV = Q W (89) reversible reversible

It follows that ☛ Q W =0 dissipative dissipative

or equivalently ☛ Q W = TdS (90) dissipative

This shows that dissipative work is equivalent to heat supply

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 8 DEPENDENCE OF ENTROPY AND INTERNAL ENERGY 1 P OverviewdS = dU + dV (91) T T @U @U Use dU = dT + dV @T @V ✓ ◆V ✓ ◆T to obtain differential of entropy as a function of T and V 1 @U 1 @U dS = dT + + P dV (92) T @T T @V !V " !T # Use uniqueness of mixed second derivatives of S to obtain a relation involving derivatives of U @S 1 @U C (92) implies = = V (93) @T T @T T !V !V and @S 1 @U = + P @V T @V (94) !T " !T #

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 9 INTERNAL ENERGY OF Uniqueness of secondOverview mixed derivative of S ( T,V ) is expressed as @ 1 @U @ 1 @U = + P @V T @T @T T @V (95) " !V #T ( " !T #)V Performing differentiation 1 @2U 1 @U 1 @2U 1 @P = + P + + (96) T @V@T T 2 @V T @T@V T @T " ◆T # !V

Taking into account uniqueness of mixed second derivative of U ( T,V ) @U @P = T P @V @T (97) ◆T !V that is (41) For the ideal gas from this formula and follows

(@U/@V )T =0 that is U = U(T )

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 10 MORE ON Substituting (97) intoOverview (94) @S @P = (98) @V @T !T !V

Maxwell relation (98) allows derivation of (97) in a shorter way

From (84) it follows that

@U @S = T P (99) @V @V !T !T

Substituting in (98) yields (97)

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 11 THERMODYNAMIC COEFFICIENTS FOR ADIABATIC PROCESSES Adiabatic Overview1 @V  = (100) S V @P !S To express  S through experimentally measurable quantities consider @S @S dS = dT + dV (101) @T @V !V !T

Setting dS =0 and inserting (93) and (98) C @P 0= V dT + dV (102) T @T !V Combining (102) with (7)

@V @V T @P dV = dP dV (103) @P @T CV @T !T !P !V

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 12 EQUIVALENTLY...

@OverviewV T @P @V 1+ dV = dP (104) @T CV @T @P " !P !V # !T @V dV = dP (105) can be simplified with help of (42) to @P (105) !T T Adiabatic compressibility ☛ S = (106)

Since C P >C V for all substances  S <  T is universally valid

Adiabatic compression is accompanied by temperature increase

that is described by thermodynamic coefficient (@T/@P )S Combining (102) and (105)

C @P C @V V dT = V dP T @T CP @P (107) !V !T

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 13 THIS YIELDS... @T T @P @V (108) Overview= dP CP @T @P !S !V !T With help of (13) this simplifies to @T T @V VT = = (109) dP CP @T CP !S !P For ideal gas T =1 and this formula gives @T T nR V (110) = = dP CP P CP !S Adiabat equation of ideal gas in the form of (53) can be rewritten as 1 1/ T = AP A = const (111) Differentiating this equation

@T 1 1/ 1 T 1 V = A 1 P = 1 = 1 dP P nR (112) !S ! ! !

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 14 ENTROPY OF IDEAL GAS Energy of ideal gas depends on temperature only U = U(T ) For perfect gas ☛ CV Overview= const and (93) can be integrated over T (113) S(T,V )=CV ln T + f(V )

Using (98) @S P nR = = @V T V (114) !T Inserting S ( T,V ) into (114) ☛ df/dV = nR/V Integration yields ☛ f = nR ln V + S 0 Then with help of (50) one obtains 1 S = CV ln T + nR ln V + S0 = CV ln (TV )+S0

This formula defines entropy up to an arbitrary constant S0 1 In of a perfect gas TV = const and entropy does not change

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 15 THIRD LAW OF THERMODYNAMICS Analyzing experimental data Walther Nernst concluded that In the limit T 0 entropyOverview becomes constant independent of other! thermodynamic parameters such as volume and @S @S = =0 (117) @V @P !T 0 !T 0 ! ! Since in thermodynamics entropy is defined up to a constant Planck has suggested to define S(T 0) = 0 (118) !

Explanation of these results is possible only within statistical physics It turns out that statistically defined entropy always satisfies (117) whereas (118) holds for most substances

Some materials have a degenerate ground state and in this case entropy tends to a finite constant at T 0 !

C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 16 CONSEQUENCES OF THE THIRD LAW

Overview T C Integrating (93) one obtains ☛ S = V dT + S T 0 (119) Z0 If C V is finite at T 0 entropy logarithmically diverges ! contradicting third law ☛ C (T 0) = 0 V ! (120) Same condition for C P can be proved in a similar way Note that divergence of entropy of ideal gas (115) at T 0 ! only proves that concept of ideal gas breaks down at low temperatures where become liquid and solid From (98) ☛ pressure thermal coefficient vanishes at absolute zero @P =0 @T ! ? (121) V ?T 0 Next class we will see that ?thermal! expansion coefficient also vanishes: ? @V ? =0 @T ! ? (122) P ?T 0 ? ! C.Luis B.-Champagne Anchordoqui ? 2 Saturday, September 13, 14 ? 17 2.6. HEAT AND THE SECONDTHE LAWSTIRLING OF THERMODYNAMICS CYCLE 39 Consists of two isotherms and two isochores

Figure 2.15: A consists of two isotherms (blue) and two isochores (green). Isothermal ideal gas equation of state ☛ d ( pV )=0

AB: This isothermal expansion is the power . Assuming ν moles of ideal gas throughout, we have pV = νRTP 2 =Vp1V1=,henceP V and P V = P V A 1 B 2 V2 D 1 C 2 νRT2 V2 WAB = dV = νRT2 ln . (2.89) V V1 PB V! PC V1" # 1= = Since AB is an isotherm, we have E = E ,andfrom∆E =0we conclude Q = W . ApA B PD AB V2 AB AB dV =0 W =0 Saturday,BC: Isochoric September cooling. 13, 14 Since we have BC .Theenergychangeisgivenby 18 νR(T T ) ∆E = E E = 1 − 2 , (2.90) BC C − B γ 1 −

which is negative. Since WBC =0,wehaveQBC = ∆EBC. CD: Isothermal compression. Clearly

V1 νRT V W = dV 1 = νRT ln 2 . (2.91) CD V − 1 V " 1 # V!2

Since CD is an isotherm, we have EC = ED,andfrom∆ECD =0we conclude QCD = WCD.

DA: Isochoric heating. Since dV =0we have WDA =0.Theenergychangeisgivenby νR(T T ) ∆E = E E = 2 − 1 , (2.92) DA A − D γ 1 −

which is positive, and opposite to ∆EBC.SinceWDA =0,wehaveQDA = ∆EDA.

We now add up all the work contributions to obtain

W = WAB + WBC + WCD + WDA V (2.93) = νR(T T )ln 2 . 2 − 1 V " 1 # THE STIRLING CYCLE II

AB : This isothermal expansion is the power stroke

For n moles of ideal gas throughout V2 nRT V PV = nRT = P V W = dV 2 = nRT ln 2 2 1 1 ) AB V 2 V ZV1 ✓ 1 ◆ Since AB is isotherm ☛ U A = U B ☛ U AB =0 ☛ QAB = WAB

BC : isochoric cooling dV =0 W =0 ) BC T1 U = U U = C dT = C (T T ) < 0 BC C B V V 1 2 ZT2

Since W =0 Q =U BC ) BC BC

Saturday, September 13, 14 19 THE STIRLING CYCLE III CD : Isothermal compression V1 nRT V W = dV 1 = nRT ln 2 CD V 1 V ZV2 ✓ 1 ◆ Since CD is isotherm ☛ UC = UD ☛UCD =0☛ QCD = WCD

DA : Isochoric heating dV =0 W =0 ) DA U = U U = C (T T ) > 0 DA A D V 2 1 Since W =0 Q =U DA ) DA DA We now add up all work contributions to obtain V2 W = WAB + WBC + WCD + WDA = nR(T2 T1)ln V1 W T T ✓ ◆ Cycle efficiency ☛ ⌘ = = 2 1 Q + Q T + C (T T )/[nR ln(V /V )] AB DA 2 V 2 1 B A

Because of the additional positive term on the denominator efficiency is smaller than the efficiency of the Saturday, September 13, 14 20 THE OTTO AND DIESEL CYCLES 40 CHAPTER 2. THERMODYNAMICS is rough approximation to physics of engine

Consists of two adiabats and two isochores

Q =0

Q =0

Assuming Figureideal 2.16: gas An Ottoalong cycle adiabats consists of two ☛ adiabatsd( (darkpV red )=0)andtwoisochores(green). TheP cycleV efficiency= isP onceV again and PDV = PC V A 1 B 2 W T 1 2 η = =1 1 . (2.94) Q − T P P AB V 2 B = C = 1 2.6.6 The Otto and DieselP cyclesA PD V2 Saturday, September 13, 14 ✓ ◆ 21 The Otto cycle is a rough approximation to the physics of a gasoline engine. It consists of two adiabats and two isochores, and is depicted in Fig. 2.16. Assuming an ideal gas, along the adiabats we have d(pV γ)=0.Thus,

γ γ γ γ pA V1 = pB V2 ,pD V1 = pC V2 , (2.95) which says p p V γ B = C = 1 . (2.96) p p V A D ! 2 "

AB: Adiabatic expansion, the power stroke. The heat transferisQAB =0,sofromtheFirstLawwehaveWAB =

∆EAB = EA EB,thus − − γ 1 p V p V p V V − W = A 1 − B 2 = A 1 1 1 . (2.97) AB γ 1 γ 1 − V − − # ! 2 " $ γ γ Note that this result can also be obtained from the adiabatic equation of state pV = pAV1 :

V V 2 2 γ 1 − γ γ pAV1 V1 W = pdV = p V dV V − = 1 . (2.98) AB A 1 γ 1 − V # ! 2 " $ V%1 V%1 −

BC: Isochoric cooling (exhaust); dV =0hence WBC =0.TheheatQBC absorbed is then V Q = E E = 2 (p p ) . (2.99) BC C − B γ 1 C − B − In a realistic engine, this is the stage in which the old burnedgasisejectedandnewgasisinserted. THE OTTO AND DIESEL CYCLES II

AB : Adiabatic expansion (power stroke) Q =0 W = U = U U AB ) AB AB A B 1 P V P V P V V W = A 1 B 2 = A 1 1 1 AB 1 1 V " ✓ 2 ◆ # Equivalently ☛ using adiabatic equation of state

V2 V2 1 PAV1 V1 W = PdV = P V dV v = 1 AB A 1 1 V ZV1 ZV1 " ✓ 2 ◆ # BC : Isochoric cooling (exhaust) dV =0 WBC =0 ) V2 heat absorbed ☛ Q = U U = C (T T )= (P P ) BC C B V 2 1 1 C B In realistic engine this is stage in which old burned gas is ejected and new gas is inserted

Saturday, September 13, 14 22 THE OTTO AND DIESEL CYCLES III

CD : adiabatic compression Q =0 W = U U CD ) CD C D 1 P V P V P V V W = C 2 D 1 = D 1 1 1 CD 1 1 V " ✓ 2 ◆ # DA : isochoric heating (combustion of the gas) dV =0 W =0 ) DA heat abosrbed by gas V Q = U U = 1 (P P ) DA A D 1 A D total work done per cycle 1 (P P )V V W = W + W + W + W = A D 1 1 1 AB BC CD DA 1 V " ✓ 2 ◆ # efficiency 1 W V ⌘ =1 1 ⌘ Q V DA ✓ 2 ◆ Saturday, September 13, 14 23 THE OTTO AND DIESEL CYCLES IV

Ratio r = V 2 /V 1 ☛

Otto cycle becomes more efficient simply by increasing compression ratio

Problem with this scheme is that

if fuel mixture becomes too hot ☛ it will spontaneously ‘preignite‘ and pressure will jump up before point D in cycle is reached

Diesel engine avoids preignition by compressing air only and then later spraying fuel into cylinder when air temperature is sufficient for fuel ignition

Rate at which fuel is injected is adjusted so that the ignition process takes place at constant pressure

Diesel engine ☛ step DA is isobararic

Saturday, September 13, 14 24 2.6. HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS 43 JOULE- Consists of two adiabats and two isobars

Q =0 Q =0

AlongFigure adiabats 2.18: A Joule-Brayton we have cycle consists of two adiabats (dark red) and two isobars (light blue). P V = P V and P2VB = P1VC CD: Isobaric2 compressionA at1 p =Dp1.

V 1 D 1 γ−1 VD VC P2 p1 − WCD = dV p1 ==p1 (VD =VC)= p2 (VB VA) (2.114) − − − p2 V! VA VB P1 " # C ✓ ◆ Saturday, September 13, 14 p (V V ) 25 ∆E = E E = 1 D − C (2.115) CD D − C γ 1 − 1 γ−1 γ p p − Q = ∆E + W = 2 (V V ) 1 . (2.116) CD CD CD −γ 1 B − A p − " 2 #

DA: Adiabatic expansion; Q =0and W = E E .Theworkdonebythegasis DA DA D − A p V p V p V p V W = 1 D − 2 A = 2 A 1 1 D DA γ 1 −γ 1 − p · V − − " 2 A # 1 γ−1 (2.117) p V p − = 2 A 1 1 . −γ 1 − p − $ " 2 # %

The total work done per cycle is then

W = WAB + WBC + WCD + WDA 1 γ−1 γ p (V V ) p − (2.118) = 2 B − A 1 1 γ 1 − p − $ " 2 # % and the efficiency is defined to be

1 γ−1 W p − η =1 1 . (2.119) ≡ Q − p AB " 2 # JOULE-BRAYTON CYCLE II AB : Isobaric expansion

VB W = P dV = P (V V ) AB 2 2 B A ZVA P (V V ) U = U U = 2 B A AB B A 1 P (V V ) Q =U + W = 2 B A AB AB AB 1

BC : Adiabatic expansion Q =0 W = U U BC ) BC B C work done by the gas is:

1 1 P V P V P V P V P V P W = 2 B 1 C = 2 B 1 1 C = 2 B 1 1 BC 1 1 p V 1 P ✓ 2 B ◆ " ✓ 2 ◆ # Saturday, September 13, 14 26 JOULE-BRAYTON CYCLE III

CD : Isobaric compression

1 V 1 D P W = P dV = P (V V )= P (V V ) 1 CD 1 1 D C 2 B A P ZVC ✓ 2 ◆ P (V V ) U = U U = 1 D C CD D C 1 1 1 P P Q =U + W = 2 (V V ) 1 CD CD CD 1 B A P ✓ 2 ◆ DA : Adiabatic compression Q =0 W = U U DA ) DA D A

1 1 P V P V P V P V P V P W = 1 D 2 A = 2 A 1 1 D = 2 A 1 1 DA 1 1 P V 1 P ✓ 2 A ◆ " ✓ 2 ◆ # Saturday, September 13, 14 27 JOULE-BRAYTON CYCLE IV

Total work done per cycle:

1 1 P (V V ) P W = W + W + W + W = 2 B A 1 1 AB BC CD DA 1 P " ✓ 2 ◆ #

1 1 W P1 Cycle efficiency ☛ ⌘ =1 ⌘ Q P AB ✓ 2 ◆

Saturday, September 13, 14 28