Reading Problems , 8-34, 8-50, 8-53, , 8-93, 8-103, 8-118, 8-137

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1 Availability Readg Problems , 8-34, 8-50, 8-53, , 8-93, 8-103, 8-118, Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of work that can be obtaed from a system at a given state P 1 and T 1 when teractg with a reference atmosphere at the constant pressure and temperature P 0 and. describes the work potential of a given system. also referred to as exergy. The followg observations can be made ab availability: 1. Availability is a property - sce any quantity that is fixed when the state is fixed is a property. For a system at state 1 and specified values of the atmosphere of and P 0,the maximum useful work that can be produced is fixed.. Availability is a composite property - sce its value depends upon an external datum - the temperature and pressure of the dead state. 3. Availability of a system is 0 at its dead state when T = and P = P 0. It is not possible for the system to teract with the reference atmosphere at the dead state. The system is said to be thermodynamic equilibrium with its surroundgs. 4. Unless otherwise stated, assume the dead state to be: P 0 = 1atm = 5 C 1

2 5. The maximum work is obtaed through a reversible process to the dead state. REV ERSIBLE W ORK }{{} W rev = USEFUL }{{ WORK} W usef ul + IRREV ERSIBILIT Y }{{} I Control Mass Analysis we know W rev = W usef ul + I but as shown the figure, the actual work of the process is divided to two components W actual = W usef ul + W sur where W sur is the part of the work done agast the surroundgs to displace the ambient air W sur = P 0 (V V 1 )= P 0 (V 1 V )

3 this is unavoidable this is not useful work. Nothg is gaed by pushg the atmosphere away. To fd W actual, from the 1st law E 1 Q W actual = E Q = E 1 E W actual From the nd law P s = ΔS system +ΔS sur 0 = S S 1 + Q But from the 1st law balance we know Q = E 1 E W actual and when we combe this with the nd law P s = S S 1 + E 1 E W actual which leads to W actual =(E 1 E )+ (S S 1 ) P s or by reversg the order of S and S 1 W actual =(E 1 E ) (S 1 S ) P s But we also know that W usef ul = W actual W sur 3

4 therefore and W usef ul =(E 1 E ) (S 1 S )+P 0 (V 1 V ) P s }{{} W sur W rev = W usef ul + I = W actual W sur + I where I = P S Therefore W rev =(E 1 E ) (S 1 S )+P 0 (V 1 V ) In summary W actual = (E 1 E ) (S 1 S ) P s W usef ul = (E 1 E ) (S 1 S )+P 0 (V 1 V ) P s X =Φ=W rev = (E 1 E ) (S 1 S )+P 0 (V 1 V ) Defe X =Φ = CONT ROL MASS AV AILABILIT Y = W rev ( gog to the dead state) = (E E 0 ) (S S 0 )+P 0 (V V 0 ) where the specific availability is defed as φ = Φ m 4

5 What is the availability gog from one state to another? The reversible work is W rev =(Φ 1 Φ 0 ) (Φ Φ 0 )=Φ 1 Φ but we also know Φ 1 = (E 1 E 0 ) (S 1 S 0 )+P 0 (V 1 V 0 ) Φ = (E E 0 ) (S S 0 )+P 0 (V V 0 ) Φ 1 Φ = (E 1 E ) (S 1 S )+P 0 (V 1 V ) The availability destroyed is X des = I = W rev W usef ul = P s = S gen This can be referred to as: irreversibilities, availability destruction or loss of availability. Control Volume Analysis Consider a steady state, steady flow (SS-SF) process From the 1st law de cv 0 = Ẇ actual Q + [ṁ(h + (v ) + gz) ] [ṁ(h + (v ) + gz) ] (1) From the nd law ds cv 0 = ṁs + Q 0 T TER ṁs + Q + Ṗ s () 5

6 Combg (1) and () through the Q term, leads to the actual work put of the turbe, given as Ẇ actual = [ ṁ (h + (v ) )] [ + gz s ṁ (h + (v ) )] + gz s Ṗ S = ṁ [ Δs +Δh +ΔKE +ΔPE] ( Ṗ s ) (3) Ẇ actual is the actual work put of the turbe. The specific flow availability, ψ,isgivenas ψ = (s s 0 )+(h h 0 )+ ( (v ) (v 0 0 ) ) + g(z z 0 0 ) (4) For a steady state, steady flow process where we assume KE=PE=0 Ẇ rev = (ṁψ) (ṁψ) (5) Ẋ des = I = Ẇ rev Ẇ actual = Ṗ s = Ṡ gen (6) ψ = (h h 0 ) (s s 0 ) (7) 6

7 The General Exergy Equation From the 1st law de cv = Ẇ Ẇ Q 0 + Q 1 Q +[ṁ(e + Pv)] [ṁ(e + Pv)] (1) From the nd law ds cv = ṁs Q 0 + Q 1 T 1 ṁs + Q T + Ṗ s () Multiply () by and subtract from (1) to elimate Q 0, which leads to the generalized exergy equation d (E S) CV = Ẇ Ẇ +[ṁ(e + Pv s)] [ṁ(e + Pv s)] + Q 1 T Q 0 1 T 1 Q T Q 0 Ṗ S (3) T 7

8 We can rewrite Eq. (3) a generalized form by troducg the defitions of Φ and ψ. dφ = P 0 dv CV [ + [ Ẇ + ṁψ + Q Ẇ + ṁψ + Q ( 1 T )] 0 T TER ( 1 T )] 0 T TER I where I = Ẋ des = Ṗ s = exergy destruction rate Φ = [(E E 0 )+P 0 (V V 0 ) (S S 0 )] = non flow exergy ψ = (h h 0 ) (s s 0 )+ 1 [ (v ) (v 0 )] + g(z z 0 ) = f low exergy Ẇ usef ul = ( ) dv CV (Ẇ Ẇ }{{ ) P } 0 Ẇ actual }{{} W sur Aside: The left side of the above equation is dφ = de + P 0dV ds but the left side of Eq. (3) does not conta the term P 0 dv. Therefore, we must add a term P 0 dv to both the left and right side of the above equation order to preserve a balance. Hence the term P 0 dv appears on the right side of the above equation to preserve this balance, while the left side is dφ = (de + P 0dV ds) 8

9 Efficiency and Effectiveness 1. First law efficiency (thermal efficiency) η = Carnot cycle net work put gross heat put = W net Q η = Q H Q L Q H =1 T L T H. Second Law Efficiency (effectiveness) η nd = net work put maximum reversible work = net work put availability Turbe η nd = Ẇ/ṁ ψ e ψ i Compressor η nd = ψ e ψ i Ẇ/ṁ Heat Source η nd = Ẇ/ṁ [ Q/ṁ 1 T ] 0 T TER 3. Isentropic efficiency (process efficiency) (a) adiabatic turbe efficiency η T = work of actual adiabatic expansion work of reversible adiabatic expansion = W act W S (b) adiabatic compressor efficiency η C = work of reversible adiabatic compression work of actual adiabatic compression = W S W act 9

Lecture 34: Exergy Analysis- Concept

ME 200 Thermodynamics I Lecture 34: Exergy Analysis- Concept Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 200240, P. R. China Email : liyo@sjtu.edu.cn