Chapter 4 Total Entropy Cannot Decrease 2012/5/6

Transcription

1 Chapter 4 Total Entropy Cannot Decreae

2 "Ice melting" - a claic example of entropy increaing It happen all the time! Nothing you can do About it. Spontaneou and irreverible. It i approaching to an equilibrium tate (liquid water at room temp) I there a natural law Involved for thi pontaneou phenomenon?

3 Needed for a new thermodynamic function The equation of ma and energy conervation are not ufficient to olve all the thermodynamic energy flow problem in which we might be intereted. What i needed i a balance equation for an additional tate variable. 3

4 INSTRUCTIONAL OBJECTIVES FOR CHAPTER 4 The goal of thi chapter are for the tudent to Be able to ue the difference form of the pure component entropy balance in problem olving Be able to calculate the entropy change between two tate of an ideal ga Be able to calculate the entropy change of a real fluid uing thermodynamic propertie chart and table 4

5 4. ENTROPY: A NEW CONCEPT All pontaneou procee that occur in an iolated contant-volume ytem reult in the evolution of the ytem to a tate of equilibrium. The eral balance equation (Eq..-4) i for any extenive variable of a cloed, iolated, contantvolume ytem dθ Rate of change of Rate at which θ i erated = = dt θ in the ytem within the yem t (4.-) 5

6 Alternatively, we can write Eq. 4.- a dθ = & θ (4.-) dt where θ, i the rate of internal eration of the yetunpecified tate variable θ. Now, if the ytem under conideration were in a true time-invariant equilibrium tate, d θ /dt = 0 (ince, by definition of a time-invariant tate, no tate variable can change with time). Thu & 0 at equilibrium tate (4.-3) θ = θ & time or time 6

7 Equation 4.- and 4.-3 ugget a way of quantifying the qualitative obervation of the unidirectional evolution of an iolated ytem to an equilibrium tate. In particular, uppoe we could identify a thermodynamic variable θ whoe rate of internal eration θ, wa poitive, except at equilibrium, where θ =0. For thi variable dθ > 0 away from equilibrium dt dθ = 0 (or θ = contant) at equilibrium dt (4.-4) 7

8 Definition of entropy (S ) The entropy (denoted by the ymbol S) i a tate function. () In a ytem in which there are flow of both heat by conduction (Q) and work [ W and P(dV/dt)] acro the ytem boundarie, the conductive heat flow, but not the work flow, caue a change in the entropy of the ytem: thi rate of entropy change i Q/T, where T i the abolute thermodynamic temperature of the ytem at the point of the heat flow. Q ΔS; Work ΔS () If, in addition, there are ma flow acro the ytem boundarie, the total entropy of the ytem will alo change due to the convective ma flow. That i, each element of ma entering or leaving the ytem carrie with it entropy (a well a internal energy, enthalpy, etc.). 8

9 Entropy balance equation ds dt K k= K ˆ Q& = MS & k k + + S& T k= MS & ˆ k k = (4.-5a) net rate of entropy flow due to the flow of ma into and out of the ytem( Sˆ k entropy per unit ma) Q& = rate of entropy flow due to the flow of heat acro the ytem T boundary S & = rate of internal eration of entropy within the ytem 9

10 For a cloed ytem M& k = 0 ds Q& = + S & dt T For adiabatic proce, Q& = 0 (4.-5b) In General : S& 0, ds dt = 0 at equilibrium (4.-5c) 0

11 An illutration for S & ye(energy) no(energy) Fig 4.- Sytem A and B are free to interchange energy, but the compoite (A+ B) i Iolated from the environment. A compoite of two ubytem, A and B. (a) well inulated (b) heat tranfer that occur i a flow of heat from the high-temperature ubytem A to the low-temperature ubytem B, (c) the ubytem were well-mixed, the temperature of each ubytem i uniform, but varying with time

12 (4.-5b) (4.-5b) (4.-5b) (a) If ytem = ubytem A or B; ds A Q& A TA T B = = h dt TA TA dsb Q& B TA T B = =+ h dt TB TB Q& = Q& = h A ( T T ) B A B S& =0, Q& A 0 (4.-6a) (4.-6b) where h i the heat tranfer coefficient. (b) If ytem = ubytem A + B; S& 0, Q& = 0 ds = S& dt A (4.-7) ds dsa dsb TA TB TA TB S& = = + = h + h dt dt dt TA TB ( ) ( Δ ) hta TB h T = = 0 TT TT A B A B

13 (4.-5a) 3

14 Difference form of entropy balance Integrating Eq 4.-5a over the time t to t S S M ks t kdt dt S t k T t ) t Q& = & + + t S = S& dt total entropy eration = 0 t (4.-9) 4

15 Two implification of Eq If the entropy per unit ma of each tream entering and leaving the ytem i contant in time (even though the flow rate may be vary) t t & & t t k k k k k k k k k t Δ M = k ) ) ) M S dt = S M dt = ΔM S t M& dt k in (4.-9) If the temperature i contant at the location where the heat flow occur t t Q& t Q dt = Qdt & = T T t T 5

16 (4.-9) 6

17 Illutration 4.- Clauiu Statement of the Second Law The econd-law tatement of Rudolf Clauiu (8-888) i that it i not poible to contruct a device that operate in a cycle and whoe ole effect i to tranfer heat from a colder body to a hotter body. Show from the axiom (S 0) that the proce below i impoible, o that the Clauiu tatement of the econd law i conitent with what ha been preented in thi book. It i impoible 7

18 Solution of Illutration 4.- The energy balance over one complete cycle, o that the device i in the ame tate at the beginning and at the end of the proce, i U U = 0 = Q + Q + W = Q + Q f i ince there i no work produced or aborbed in the device. Therefore, the energy balance yield Q = -Q. Thu the energy balance tell u that the proce may be poible a long a the heat flow in and out balance, that i, If energy i conerved. The entropy balance over one complete cycle i 0 Q Q Sf Si = = + + S T T Q Q Q Q T T S = = + = Q = Q T > T, Q < 0, S > 0 T T T T T T TT Only If Q were negative, o that the heat flow could be from high temperature to low temperature (or equivalently T > T ), the proce would be poible. 8

19 Illutration 4.- Kelvin-Planck Statement of the Second Law An alternative tatement of the econd law, due to Lord Kelvin (84-907) and Max Planck ( ), i that it i not poible to contruct a device operating in a cycle that reult in no effect other than the production of work by tranferring heat from a ingle body. A chematic diagram of a Kelvin-Planck device i hown below. Show from our axiom (S 0) that thi proce i indeed impoible. 9

20 Solution of Illutration 4.- (4.-9) The energy balance for a complete cycle: U U = 0 = Q+ W f i W = Q The entropy balance for a complete cycle: Q Sf Si = 0 = + S T Q S = T Q mut be negative for S > 0 0

21 Remark on the lt and nd law of thermodynamic Q converted to W (00%) i not poible (econd law); W converted to Q (00%) i alway happen (firt law). Q converted to W (not 00%) i poible W i mechanical energy, high level of energy, can be ued to produce ueful work; Q i thermal energy The firt law et ome contraint; The econd law i definitive in that it etablihe that it i impoible to contruct device that would operate in the manner propoed.

22 4. THE ENTROPY BALANCE AND REVERSIBILITY S ΔT S = 0, dt 0, approach reverible proce i.e. initial tate reverible proce final tate. where initial tate = final tate only for S& = 0, the reverible proce On the other hand, Irreverible proce S& g en 0, initial tate final tate a c b Initial tate(a) = Final tate(c)

23 3 The deignation reverible arie from the following obervation. Conider the change in tale of a eral ytem open to the flow of ma, heat, and work, between two equal time interval, 0 to t and t to t where t = t. The ma, energy, and entropy balance for thi ytem are, from Eq..-4, 3.-6, and 4.-9, ˆ ˆ ˆ ˆ dt S dt S dt T Q S M dt T Q S M S S dt Q W dt dv P H M dt Q W dt dv P H M U U dt M dt M M M t t t t t k k k t k k k t t k k k t k k k k t t k k t k = = + = & & & & & & & & & & & & & & Final tate Initial tate = a c b

24 Now, uppoe that all of the ma, heat, and work flow are jut revered between t and t from what they had been between 0 and t, o that M& dt = M& dt Qdt & = Qdt & t t t t 0 k t k 0 ˆ ˆ Q& Q& M& H dt = M& H dt dt = dt T t T M& Sˆ dt = M& Sˆ dt W& dt = W& dt t t t t 0 k k t k k 0 t t t t 0 k k t k k 0 t t dv t dv P dt = P dt 0 dt t dt In thi cae the equation reduce to M = M0 U = U 0 S = S0 + S 0 dt+ S t dt For S& t & = 0, the ytem in equilibrium tate at any time of the proce, i.e., the reverible proce, S = S0. For S& > 0, S > S, irreverible proce. 0 t & t a c b Initial tate = Final tate 4

25 Characteritic of a reverible proce The reverible proce ) proceed with infiniteimal gradient (dt, dp, dμ) within the ytem, ) occur lowly on the time cale of macrocopic relaxation time. 3) change of tate in real ytem can be approximated a being reverible if there i no appreciable internal heat flow or vicou diipation. 4) Example: ) Expanion and compreion that occur uniformly throughout a fluid, or ) Well-deigned turbine, compreor, and nozzle for which vicou diipation and internal heat flow are unimportant, can erally be conidered to occur reveribly (i.e., S = 0). The irreverible proce ) proceed with finite gradient (ΔT, Δ P, Δ μ) within the ytem, ) occur quickly on the time cale of macrocopic relaxation time. 3) Change of tate in real ytem can be approximated a being irreverible if there i appreciable internal heat flow or vicou diipation, friction. 5

26 6

27 We are intereting in the following W?( ><, ) ( irreverible) = W( reverible)?( ><, ) Q( irreveribl e ) = Q( reverible) 7

28 (I) Cloed ytem 8

29 Work and Reveribility of Proce Sytem ====(perform work) ===>>> urrounding, work (reverible proce) = maximum work produced For example, a well-deigned team turbine in a power plant produce the maximum amount of electricity. Surrounding ====(perform work) ===>>> ytem, work (reverible proce) = minimum work required For example, a well-deigned ga compreor need the minimum amount of electricity. The calculation of the work baed on the reverible proce provide the upper (or lower) bound to the true value. 9

30 Work: reverible/irreverible proce (3.-4a) Conider the energy and entropy balance for a cloed, iothermal, contant-volume ytem: U = U+ Q+ W and Q S = S+ + S T Q= T S S TS ( ) ( ) ( ) ( ) ( U TS ) ( U TS ) TS W = U U Q= U U T S S + TS rev S = + = A A + TS Reverible work at contant N, V, and T W = A A rev rev S ( ) ( ) rev For irreverible, iothermal, contant-volume proce between the ame final and initial tate W = A A + TS = W + TS Q= T S S TS Q (4.-3) No flow energy dv = 0 (4.-4) (4.-5) ( 4.-7) = T S S ; Q= Q TS (4--9a) 30

31 Conider the energy and entropy balance for a cloed, iothermal, contant-preure ytem: U = U+ Q+ W ( PV PV ) = U+ Q+ W P( V V) (4.-3) and Q S = S+ + S T Q= T S S TS ( ) ( ) ( ) ( ) ( ) ( ) ( U PV TS ) ( U PV TS ) TS (4.-4) W = U U Q+ PV V = U U T S S + TS + PV V = = G G + TS Reverible work at contant N, P, and T rev WS = G G For irreverible, iothermal, contant-preure proce between the ame final and initial tate rev W = G G + TS = W + TS ( ) ( ) Q= T S S TS rev Q = T S S ; S (4.-8) (4.-7) (4--9b) rev Q= Q TS (4--9a) 3

32 For either contant N, V, T or contant N, P, T proce between the ame initial and final tate W = W + TS rev rev Q= Q TSgn e rev W + Q= W + Q rev Will be applied for Turbine and Compreor For irreverible proce, the workability of the ytem decreae in the form of Q; however, the total energy of the ytem doe not change. Thu, the conervation of the total energy i till held. W =====> Q, workability decreae 3

33 Good Reverible No-good Irreverible W Q W Q Should be applied for Turbine and Compreor 33

34 (II) Open ytem with only one ma flow rate Thermo: Fall

35 The rate-of-change form of the ma, energy, and entropy balance for a ytem in which the kinetic and potential energy term are unimportant, there i only one ma flow tream, and the ma and heat flow occur at the common temperature T: dm = M & dt du ˆ dv du ; ˆ dm dv = MH & + Q& P + W& = H + Q& P + W& dt dt dt dt dt ds ˆ Q& ds ˆ dm = MS & + + S& ; T = TS + Q& + TS& dt T dt dt For a differential time interval dt, Q= Q& dt = heat flow into the ytem in the time interval dt W = W& dt = haft work into the ytem in the time interval dt S = S& dt = entropy erated in the ytem in the time interval dt du = HdM ˆ + Q PdV + W (4.-0a) TdS = TSdM ˆ + Q + TS (4.-0b) Q = TdS TS TSd ˆ M du = HdM ˆ + TdS TS TSdM ˆ PdV + W ( ˆ ˆ) = TdS TS PdV + W + H TS dm (4.-a) = TdS TS PdV + W + GdM ˆ (4.-b) 35

36 Q = TdS TS TSdM ˆ (4.-a) ( ) irrev du = TdS TS + PdV + W + GdM ˆ For a reverible proce, S = 0 rev Q = TdS TSdM ˆ rev ( ) ˆ ( ) (4.-b) (4.-a) rev du = TdS + PdV + W + GdM ; PdV + W = du TdS Gd ˆ M (4.-b) For the ame initial and final tate, Q irrev irrev rev Q = Q TS rev ( ) ( ) irrev ( PdV W ) ( PdV + W ) irrev = Td ˆ = rev S TS TSdM Q TS From (4.- b) and(4. b): rev = W + irrev PdV + W = PdV + W TS + = + TS W TS Same reult a thoe obtained for cloed ytem. rev 36

37 Differential entropy change for an open ytem VS energy change for cloe ytem ( ) ( ) irrev du = TdS TS + PdV + W + GdM ˆ (4.-b) For reverible proce rev du = TdS + PdV + W + GdM ˆ For W = 0 du = TdS PdV + GdM ˆ (4.-3a) For a cloed ytem, dm = 0 du = TdS PdV du = TdS PdV (4.-3b) Recalled EQ (3.-9b) for a cloed ytem and reverible proce at no haft work, no KE, no PE. 37

38 4.3 HEAT, WORK, ENGINES, AND ENTROPY Gaoline car engine Steam Turbine Figure (a) Schematic diagram of a imple heat engine. (b) Schematic diagram of a fluid flow engine. 38

39 A imple heat engine: The open teady tate engine for a complete cycle (3.-6) (4.-9) 0 = Q + Q + W (4.3-) Q Q S 0 = + + (4.3-) T T where Q Qdt an t = T Q = Q T S T t ( ) t & d = & ( / ) Work done by the engine W W P dv dt dt T T T T W = Q TS = Q TS (4.3-3) T T Maximum work done by the engine Engine Efficiency Fraction of heat upplied that i converted to work = = t T = W = Q + Q = Q Q T S W T T = Q T = W = Q T T T (4.3-5) (4.3-4) 39

40 Reverible/Irreverible proce efficiency Efficiency of a heat engine: (a) Reverible heat engine η reverible heat engine = T T T (b) Irreverible proce T T ηactural heat engine T due to deign and operating limitation, heat loe, and friction of the proce. 40

41 Limitation of the reverible heat engine Efficiency of reverible heat engine: η reverible heat engine (a) limη.0, however, T can not approach to the infinity T = reverible heat engine due to the material of the heater T T T (b) limη.0, however, T can not approach to abolute zero T 0 reverible heat engine temperature. The T i the temperature of the cooling water or the temperature of the lake (or ea water) Therefore, η.0 i impoible; Q can not be completely reverible heat engine converted into W, the econd law of thermodynamic. 4

42 Ideal Carnot Heat Engine Aborbed Q at T Releaed Q at T Figure 4.3- The Carnot cycle, (a) Schematic diagram of a Carnot engine. (b) The Carnot cycle on a preure-volume plot. (c) The Carnot cycle on a temperature-entropy plot. (d) Heat flow into the cycle going from point a to b. (e) Heat flow into the cycle going from point c to point d. (f) Net work flow. 4

43 The Carnot Heat Engine: Four reverible tep 43

44 The Carnot Heat Engine For "one complete cycle" : 0 = Q + Q + W (4.3-7) Q Q 0 = + T T b c d a W = PdV PdV PdV PdV Va Vb Vc Vd Q T = T V V V V Q T Work done by the engine = W = Q + Q = Q Q T ( 4.3-8) T T T W = Q = Q (4.3-9) T T W T T Carnot cycle Efficiency = = Q T 44

45 Flow Fluid Engine Which ide i the inlet, which one i the outlet? Figure Sketch of a team turbine 45

46 Fluid Flow Engine The teady-tate ma, energy, and entropy balance A molar bai 0 = N& + N& or N& = N& (4.3-0) 0 = H H N& + Q& + W& ( ) Q& 0 = ( S S) N& + + S& T Q& = T S S N& TS& ( ) ( ) & ( ) ( ) N& ( H TS ) ( H TS ) TS& (4.3-) (4.3-) W& = H H N& Q = H H N& + T S S N& + TS& = + (4.3-3) W& = N& G G + TS& for T = T = T ( iothermal) 46

47 Fluid Flow Engine: reverible For reverible engine at iothermal proce (4.3-3) (4.3-) rev W = N& G G rev Q& = T S S N& ( ) I it poible in the real world? 47

48 Fluid Flow Engine: adiabatic/reverible (4.3-) (4.3-3) (4.3-) For adiabatic flow engine W& = H H N& 0 ( ) ( ) & ( ) = S S N S& S& = S S N& (4.3-4) (4.3-5) For adiabatic and reverible proce W& = H H N& S = S ( ) 48

49 Reverible work: PV-work and W Total work can be expreed a W = W + W For a reverible proce, one ha W rev rev PV = W rv e PV rev rev For a reverible PV -work W = where P and V are propertie of the ytem. We know PV + W PdV rev PV W W = PdV rev Could that be W = VdP???????????? 49

50 Reverible Shaft work ( rev W ) related to property change S Figure Device with fluid, heat and work flow. 50

51 Sytem: volume element with length ΔL open, teady-tate flow ytem dn = 0 = ( N& ) ( N& ) dt L L+ΔL du = 0 = ( NH & ) ( NH & ) + q& Δ L+ w& L L L ΔL dt +Δ (4.3-6a) (4.3-6b) ds q = 0 = ( NS & ) ( NS & & ) + Δ L+ & σ L L L ΔL dt +Δ T (4.3-6c) &,, & σ are the rate of heat,, and entropy eration q w& per unit length of the device haft work 5

52 ( ) ( ) N& N& L L L dn& +Δ lim = = 0 or N& = contant ΔL 0 ΔL dl ( H) ( H) L L L dh N& +Δ lim = N& = q& + w& ΔL 0 ΔL dl ( S) ( S) L+ΔL L ds q N& lim = N& & = + & L 0 L dl T σ Δ Δ For a reverible proce, & σ = 0 ds q ds N& & = ; q& = TN& dl T dl dh dh ds dh ds w& = N& q& = N& TN& = N& T dl dl dl dl dl (4.3-7a) (4.3-7 b) (4.3-7c) 5

53 dh = TdS+ VdP; dh TdS = VdP dh ds dp w& = N& T = NV & dl dl dl dp W& = w & dl = N& V dl = N& VdP dl rev W& W = = VdP= f ( tate of the ytem ) (4.3-8) N Proved! Reverible haft work (W) related to property change can be calculated by (4.3-8), which H i not equal to H and i alo not neceary an adiabatic proce (i.e. Q might not be zero). 53

54 W& W = = VdP N (4.3-8) (a) For undergoe an iothermal PRESSURE change an ideal ga RT P W = VdP = dp = RT ln P P (4.3-9) (b) For an expanion or compreion for with polytropic proce gae γ γ PV = PV = contant γ ( PV PV ) W = γ γ = 0, for an iobaric proce γ =, for an iothermal proce γ =, for a contant-volume (iochoric) proce C γ = C * P * V, for a contant-entropy(ientropic) proce in an ideal ga of contant heat capacity (4.3-0) (4.3-) It i eay to prove for ideal ga in textbook! 54

55 ILLUSTRATION 4.3- Converion of Radiant Energy to Mechanical or Electrical Energy Show that a olar or photovoltaic cell that convert olar energy to mechanical or electrical energy mut emit ome of the energy of the incident radiation a heat. The teady-tate energy balance on a olar cell aborbing radiation but not releaing any heat 0 = Q& R + W& ; or W& = Q& R the entropy balance Q& R Q& R 0 = + S& ; S& = T T R R S& 0, mut be Q& 0 and radiant energy mut be releaed R rather than aborbed. 55

56 ILLUSTRATION 4.3- Maximum Converion of Solar Energy to Mechanical or Electrical Energy Baed on analyi of the frequency ditribution of radiation from the un, it can be conidered to be emitting radiant energy with a Stefan-Boltzmann ditribution at a temperature of 6000 K. Etimate the maximum efficiency with which thi radiant energy can be converted to electrical (or mechanical) energy uing olar cell (commonly called photovoltaic cell). For thi analyi, aume that the olar cell i operating in teady tate and i receiving radiant energy, that it urface temperature i 300 K, and that it i loing heat by conduction to the environment. 56

57 The energy balance on thi olar cell i 0 = Q& + Q& + W& = Q& R + Q& + W& W& = Q& + Q& R and the entropy balance i & & 0 = + + QR Q S & TR T For maximum converion efficiency, S& = 0 Q& Q& = T R R T Q& R W& = Q& R + Q& = Q& R T = Q& T R T R T W& T = Q& T R R T = 6000 K, T = 300K R 300 Carnot efficiency = = R Doe it poible? 57

58 4.4 ENTROPY CHANGES OF MATTER (I) For an ideal ga du = C dt and P = RT / V * V du = TdS PdV * P CV R ds = du + dv = dt + dv T T T V Fo r C ( ) (, ) (, ) * V f T S T V S T V C * = V + T T dt T T V S( T V ) S( T V ) C R R V V (4.-3b) (4.4-) dv V (T, V) *,, = V ln + ln (4. 4-) T V 58

59 * T RT P * T ( ) ( ) P S T, P S T, P = CV ln + Rln = ( CV + R) ln Rln T RT P T P T P,, = ln ln T P ( ) ( ) ( P ) * S T P S T P C R ( T, P) (4.4-3) * PV / R V * V * P S( P, V ) S( P, V) = CV ln ( ) + Rln = CV + R ln + CV ln P V / R V V P V P = + ( ) S( P V) ( CP) S P V ( PV, ) * *,, ln C ln V (4.4-4) V P What i the ideal ga (Cp/Cv) at contant S and T? 59

60 (II) For liquid and olid, V contant, dv 0 P ds = du + dv du T T T and CV CP du = CVdT CPdT dt ds = du = CP T T Entropy change for a olid or liquid dt T S( T) S( T) = CP T T (4.4-5) (4.4-6) 60

61 ILLUSTRATION 4.4- Calculation of Entropy Generation for a Proce Compute the entropy erated on mixing kg of team at bar and 00 o C (tate ) with kg of team at bar and 300 o C (tate ). (3.-6) (4.-9) Solution: the mixing of kg of team Conidering to be, M f ( M+ M) = 0 or M f = ( M+ M) = kg The energy balance MUˆ MUˆ MUˆ = PMVˆ MVˆ MVˆ ( ) f f f f Since the preure i contant, the Uˆ and PVˆ can be combined to give M ˆ ˆ ˆ fh f MH MH = 0 kj o Hˆ f = [ ] = ; Tf = 50 C ( at br a ) kg M ˆ ˆ ˆ fsf MS MS= S kj S = ( ) = K a cloed ytem No flow energy Checking T-S diagram 6

62 ILLUSTRATION 4.4- Compute the entropy erated by the flow of kg/ of team at 400 bar and 500 o C undergoing a Joule-Thomon expanion to bar. Adiabatic, open and teady tate ytem 4.-5a, Q=0 Solution: ( ) ˆ ( ) Hˆ 500 C,400bar = H? C, bar, T = 4. C o o o ds = MS & ˆ ˆ ˆ + MS + S& = 0 dt S& = M& S S = ( ˆ ˆ ) ( ) kj =.40 K Checking team table Checking H-S Note: The J-T expanion i an irreverible proce! 6

63 4.5 APPLICATIONS OF THE ENTROPY BALANCE Solve PROBLEMS uing both the energy and entropy balance equation 63

64 Recalled ILLUSTRATION Example of a Thermodynamic Problem That Cannot Be Solved with Only the Ma and Energy Balance A compreor i a ga pumping device that take in ga at low preure and dicharge it at a higher preure. Since thi proce occur quickly compared with heat tranfer, it i uually aumed to be adiabatic; that i, there i no heat tranfer to or from the ga during it compreion. Auming that the inlet to the compreor i air [which we will take to be an ideal ga with C P * = 9.3 J/(mol K)] at bar and 90 K and that the dicharge i at a preure of 0 bar, etimate the temperature of the exit ga and the rate at which work i done on the ga (i.e., the power requirement) for a ga flow of.5 mol/. 64

65 ILLUSTRATION 4.5- Solved Illutration 3.4-4, uing the additional entropy balance equation: From the teady-tate ma balance N& = N& = N& From the teady-tate energy balance W& = NC & T T * P ( a) ( ) ( b) The entropy balance 0 = S S N & + S & c ( ) ( ) 65

66 The entropy balance 0 = + & c ( S S ) N & S ( ) To obtain an etimate of the exit temperature and the power requirement, we aume that the compreor i well deigned and operate reverible that i S& = 0 d S = S (e) ( ) For an ideal ga, the entropy difference i * T P S( P, T) S( P, T) = CP ln Rln = 0 T P T T T P = P * P * RC P * P ( T ) ( ) ( f ) RC P 0 o = T = 90 = 557.4K = 84. C P W = C T = = J/mol ; mol J W & = N & W = = 9.59 kj mol 66

67 Solving Illutration Uing (a) and (b) (a) ILLUSTRATION 4.5- An Alternative Way to Solve Illutration Sometime it i poible to olve a thermodynamic problem everal way, baed on different choice of the ytem. To ee thi, we conider Illutration 3.4-5, which wa concerned with the partial evacuation of a compreed ga cylinder into an evacuated cylinder of equal volume. Suppoe we chooe for the ytem of interet only that portion of the content of the firt cylinder that remain in the cylinder when the preure (0 bar) have equalized (ee Fig. 4.5-, where the thermodynamic ytem of interet i within the dahed line). Note that with thi choice the ytem i cloed, but of changing volume. Furthermore, ince the ga on one ide of the imaginary boundary ha preciely the ame temperature a the ga at the other ide, we can aume there i no heat tranfer acro the boundary, o that the ytem i adiabatic. 67

68 Imaginary cylinder before and after evacuation under adiabatic and reverible condition Figure 4.5- The dahed line encloe a ytem coniting of ga initially in the firt cylinder that remain in that cylinder at the end of the proce. 68

69 The ga in the cylinder i undergoing a uniform expanion o there will be no preure velocity, or temperature gradient in the cylinder. Therefore, we can aume that the change taking place in the ytem occur reveribly. Ma balance N = N (a) Energy balance f i f f i i f N U N U PdV = f f i i = NS Entropy balance N S (c) * P C f R f P = i i P f i = S From eq (a) and (c) S (d) The proce i ientropic. f f f f i i * T P S( P, T ) S( P, T ) = CP ln Rln 0 i i = T P T T Thi i preciely the reult obtained in Eq. f of IIutration uing the energy balance on the open ytem coniting of the total content of cylinder. The remainder of the problem can now be olved in exactly the ame manner ued in Illutration (Chap.3 gave T=64.3K, T=55.6K) V i V (b) 69

70 One might expect that a imilar relation would hold for the ytem hown in cylinder. Thi i not the cae, however, ince the ga entering cylinder i not necearily at the ame temperature a the ga already there (hydrodynamic will enure that the preure are the ame). Therefore, temperature gradient will exit within the cylinder, and our ytem, the partial content of cylinder, will not be adiabatic. Conequently, Eq. e will not apply. Can not be choen ince it i not adiabatic! T i changed x Figure 4.5- Incorrect ytem choice for ga contained in cylinder. 70

71 Recalled ILLUSTRATION Ue of Ma and Energy Balance to Solve an Ideal Ga Problem A ga cylinder of m 3 volume containing nitro initially at a preure of 40 bar and a temperature of 00 K i connected to another cylinder of m 3 volume that i evacuated. A valve between the two cylinder i opened until the preure in the cylinder equalize. Find the final temperature and preure in each cylinder if there i no heat flow into or out of the cylinder or between the ga and the cylinder. You may aume that the ga i ideal with a contant-preure heat capacity of 9.3 J/(mol K). (Chap.3 gave T=64.3K, T=55.6K after the proce) T=64.3K T=55.6K 7

72 (b) Illutration Solved Illutration Uing the Entropy Balance Intead of the Energy Balance A ga cylinder of m 3 volume containing nitro initially at a preure of 40 bar and a temperature of 00 K i connected to another cylinder of m 3 volume, which i evacuated. A valve between the two cylinder i opened only until the preure in both cylinder equalize and then i cloed. Find the final temperature and preure in each cylinder if there i no heat flow into or out of the cylinder' or between the ga and the cylinder wall. The propertie of nitro ga are given in the following Fig

73 Figure Preure-enthalpy diagram for nitro Thermo: Fall

74 Solution: Except for the fact that nitro i i f = S now being conidered to be a real the problem here i the ame a in Illutration In fact, Eq. h-n in the comment to that illutration apply here. The additional equation needed to olve thi problem i obtained in the ame manner a in Illutration Thu, for the cylinder initially filled we have S For an ideal ga of contant heat capacity ( which i not the cae here) thi reduce to * P f f P T i = i P T lited by Eq C R For real nitro ga, Eq.(O) require that the initial and final tate in cylinder be connected by a line of contant entropy in Fig (O) V f f f f i f P = P (H); N + N = N (I) ; N = V f f f f ( ) ( ) f f f f = = i i f f f f i f = + = cyl f V V T, P (L); V V T, P (M) NU N U N U (N); S S (o) f cyl (J); N = (K) f V With eight equation Eq. (H-O) and eight unknown, thi problem can beolved, though the olution i a trial-and-error proce. V rather than an ideal ga, f f f f f f f f,,,,,,, V V T T P P N N 74

75 In eral, a reaonable firt gue for the preure in the nonideal ga problem i the ideal ga olution, which wa f f P = P = 0 bar Initial condition( P= 40bar = 4 MPa and T = 00 K) in Fig yield Hˆ ˆ i kj/kg and V m /kg i 3 i V m 3 V ˆ i ˆi ˆ f f S = S T 3 f V m ˆ i 3 V f i f = =.9 M = = = kg 0.36m /kg From, 65 K ˆ = M = = = 44.6 kg 0.04 m /kg M M M Vˆ f f f f 3, H 30 kj/kg and V 0.04 m /kg 8 kg, which inplie 3 m 3 = = m /kg 8.9kg = = ˆ = 40 K and Hˆ = 39 kj/kg f 3 Locating P 0bar MPa and V m /kg, T f ˆ Recalled: Chap. 3 gave T= 64.3K T= 55.6K May you obtain accurate T and T value uing NITROGEN TABLE DATA? 75

76 Figure Preure-enthalpy diagram for nitro. H=337 V=0.036 i f f Thermo: Fall Hf=30 Hf= 39

77 Finally, we mut check whether the energy balance i atified for the condition computec here baed on the final preure we have aumed. To do thi we mut firt compute the intenal energie of the initial and final tate a follow: 3 i i i i kj m 5 Pa J kj kj = = = 3 Uˆ Hˆ PVˆ bar kg kg bar m Pa 000 J kg ˆ f kj ˆ f kj Similarly, U = 65. and U = 3.8 kg kg The energy balance i ˆ i f ˆ f f MU = MU + M ˆ f U = kj.57 0 kj Thu, to the accuracy of our calculation, the energy balance can be conidered to be atified and the problem olved. Chap. 4 gave T= 40K for real ga 77

78 Illutration Illutration Continued, Showing That Entropy I a State Function Show that the entropy S i a tate function by computing ΔS for each of the three path of Illutration SOLUTION Since the piton-and-cylinder device i frictionle, each of the expanion procee will be reverible. Thu, the entropy balance for the ga within the piton and cylinder reduce to ds Q& Q = ; Δ S = dt T T Path A i. Iothermal compreion. QA J/mol J Since T i contant, Δ SA = = = 9.4 T 98.5 K mol K ii. Iobaric heating. * * dt d S Q& CP dt Q& = CP, = = dt dt T T dt Δ S = C B * P T J 98.5 mol K ln = 38ln = 4.83 T J Δ S =Δ SA +Δ S B = = 5.69 mol K 78

79 Path B T J i. Iobaric heating. Δ SA = C ln = 38ln = mol K * P T B ii. Iothermal compreion. Since T i contant, S B 9.4 Δ S =Δ S +Δ S = = 5.69 Path C A * 3 P T B γ i. Compreion with PV = contant Q Δ S A = = 0 T ii. Iobaric heating T J Δ S B = C ln = 38ln = mol K ΔS = J 5.69 mol K J mol Q 097. J/mol J Δ = = = T K mol K K 79

80 State function and path function Path Q J/mol W J/mol ΔU J/mol ΔS J/(mol K) A B C

81 ILLUSTRATION Showing That the Entropy Reache a Maximum at Equilibrium in a Cloed, Iolated Sytem. Figure how a well-inulated box of volume 6 m 3 divided into two equal volume. The left-hand cell i initially filled with air at 00 o C and bar, and the right-hand cell i initially evacuated. The valve connecting the two cell will be opened o that ga will lowly pa from cell to cell. The wall connecting the two cell conduct heat ufficiently well that the temperature of the ga in the two cell will alway be the ame. Plot on the ame graph () the preure in the econd tank veru the preure in the firt tank, and () the change in the total entropy of the ytem veru the preure in tank. At thee temperature and preure, air can be conidered to be an ideal ga of contant heat capacity. 00 C and bar 00oC and 0 bar Figure A well-inulated box divided into two equal compartment 8

82 Solution For thi ytem Total ma = N = N + N Total energy = U = U + U Total entropy = S = S + S = N S + N S From the ideal ga equation of tate 3 i PV bar 3 m N = = = 93.4 mol = kmol 3 RT -5 bar m K mol K Now ince U = contant, T = T and for the ideal ga U i a function of temperature only, o we conclude that T = T = 00 C at all time. Thi reult greatly implifie the computation. Suppoe that the preure in cell i decreaed from bar to.9 bar by tranferring ome ga from cell to cell. Since the temperature in cell i contant, we have, from the ideal ga law, N = 0.95N i i and by ma conervation, N = 0.05 N. Applying the ideal ga relation, we obtain P =0. bar. For any element of ga, we have, from Eq , f i * T P P S S = CP ln Rln = Rln T P P f f f i i i 8

83 ince temperature i contant. Therefore, to compute the change in entropy of the ytem, we viualize the proce of tranferring 0.05 N mole of ga from cell to cell a having two effect: i i. To decreae the preure of the 0.95 N mole of ga remaining in cell from "total ytem" bar to. 9bar. To decreae the preure of the 0.05 N mole of ga that have been tranferred from cell to cell from "total ytem" bar to 0. br a ( ) ( ) S S = 0.95NRln.9 / 0.05NRln 0./ f i i i Δ S = 0.95ln ln 0.05 = 0.99 for = = NR i i i ( P.9 bar; P 0. bar) Δ S = 0.9ln ln 0. = 0.35 for P =.8 bar; P = 0. bar NR And SO ON ( ) Recalled Table 4.- (b) & Fig. 4.- Sf-Si = S Figure The ytem Entropy change and the Preure in cell a a Function of the preure In cell. 83

84 ILLUSTRATION Showing That the Energy and Entropy Balance Can Be Ued to Determine Whether a Proce i Poible An engineer claim to have invented a teady-flow device that will take air at 4 bar and 0 o C and eparate it into two tream of equal ma, one at bar and -0 o C and the econd at bar and 60 o C. Furthermore, the inventor tate that hi device operate adiabatically and doe not require (or produce) work. I uch a device poible? [Air can be aumed to be an ideal ga with a contant heat capacity of C p * = 9.3 J/(mol K)]. Cooler Heater 84

85 Solution The three principle of thermodynamic -- () conervation of ma, () conervation of energy, and (3) S& 0 - mut be atified for thi or any other d evice. Thee principle can be ued to tet whether any device can meet the pecification given here. The teady-tate ma balance equation for the open ytem coniting of the device and it content i dn / dt = 0 = N & k = N & + N & + N & 3. Since from the problem tatement, k N& = N& = N&, ma i conerved. The teady-tate energy balance for thi device i 3 du = 0 = = 3 dt = NC & 5 K = 0 N& kh k NH & NH & NH & k ( 93.5 K 53.5 K 333. ) * P o the energy balance i alo atified. 85

86 Finally, the teady-tate entropy balance i ds = 0 = N& ks k+ S& = N& S N& S N& S + S& dt k = N& ( S S ) + ( S S 3) + S& Now uing Eq , we have 3 T P T P S& = N& C R + C R * * P ln ln P ln ln T P T 3 P J N 9.3ln Rln = & =.5N& > K Therefore, we conclude, on the bai of thermodynamic, that it i poible to contruct a device with the pecification claimed by the inventon. Thermodynamic, however, give u no inight into how to deign uch a device. EQ T P S T P S T P = C R T, P ( ) ( ) ( P ) *,, ln ln ( ) T P Solution of Illutration of * C P f R f T P i = i T P 86

87 Two poible device for ILLUSTRATION bar / 60 o C 4 bar / 0 o C bar / -0 o C Doe it require NO WORKS? 87

88 ILLUSTRATION Another Example of Uing the Entropy Balance in Problem Solving A team turbine operate at the following condition: a. Compute the horepower developed by the turbine and the entropy change of the team. b. Suppoe the turbine i replaced with one that i well inulated, o that the heat lo i eliminated, and well deigned, o that the expanion i reverible. If the exit preure and velocity are maintained at the previou value, what are the outlet team temperature and the horepower developed by the turbine? 88

89 SOLUTION The teady-tate ma and energy balance on the turbine and it content (the ytem) yield dm = 0 = M& + M& ; M& = M& = 0000 kg/hr Open teady tate dt d v ˆ v ˆ v U + M + gh =0= M& H+ + M& H + + W& + Q& EQ(f) dt a. From the Mollier diagram of Fig ( or Appendix A.II) Hˆ ˆ = 350 J/g and S = 7.3 J/(g K) v 000 m/min J/kg kg Alo = = 0.56 J/g 60 /min m / 000 g ˆ v H + = J/g ˆ ˆ v Similarly, H ˆ = 805 J/g H + = = 8.8 J/g and S = 7.5 J/(g K) Therefore, from EQ(f ) : J kg g 4 kj J -W& = ( ) g hr kg hr kj -W& 9 J kj hr kj 6 = = = Watt= 553 hp hr 000 J 3600 Alo, Watt= J/ ΔSˆ = Sˆ Sˆ ˆ ˆ = 0.7 J/(g K) S& 0, and S S hp= Watt ( ) Thermo: Fall

90 b. The teady-tate entropy balance for the turbine and it content i ds = 0 = MS & ˆ MS & ˆ + S& dt The turbine operate reveribly o that S& = 0, and Sˆ = Sˆ ; that i, the expanion i ientropic. We now ue Fig. 3.3-, the entropy-enthalpy plot (Mollier deagram) for team, to olve thi problem. The initial team condition ( T=800 K, P=3. 5 MPa) on the chart and follow a line of contant entropy to the exit preure (0.5 MPa), to obtain the enthalpy of the exiting team ( Hˆ = 690 J/g) and it final temperature ( T 373 K). J kg 000 g/kg hr -W& = ( ) ( ) 0000 g hr 000 J/kJ 3600 kj 6 = Watt= 308 hp = Could you compute thi TURBINE EFFICIENCY for cae a? 90

91 57 o C 35,00 kpa 50 kpa >00 o C Figure 3.3- (a) Enthalpy Entropy of Mollier diagram For team S= Contant 7.3 J/gK 9

92 COMMENTS. Here, a before, the kinetic energy term i of negligible compared with the internal energy term.. Notice from the Mollier (H-S) diagram that the turbine exit team i right at the boundary of a two-phae mixture of vapor and liquid. For the olution of thi problem, no difficultie arie if the exit team i a vapor (better), a liquid, or a two-phae vapor-liquid mixture ince our ma. energy, and entropy balance are of eral applicability. In particular, the information required to ue thee balance equation i the internal energy, enthalpy, and entropy per unit ma of each of the flow tream. Provided we have thi information, the balance equation can be ued independent of whether the flow tream conit of ingle or multiple phae, or, in fact, ingle or multiple component. Here the Mollier diagram provide the neceary thermodynamic information, and the olution of thi problem i traightforward. 3. Finally, note that more work i obtained from the turbine by operating it in a reverible and adiabatic manner. 9

93 Homework aignment Prob. 4.4, 4.6, 4.9, 4., 4.4, 4.4,