Availability. Second Law Analysis of Systems. Reading Problems , 10.65, 10.66, , 10.75, 10.81, 10.

Transcription

1 Availability Readg Problems , 10.65, 10.66, , 10.75, 10.81, Second Law Analysis of Systems AVAILABILITY: the theoretical maximum amount of reversible 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. What do we mean by work potential of a system? Notice that: η = Ẇ Q = 1 Therefore Ẇ = Q ( 1 T ) 0 This term represents the work potential (availability) of a given TER with respect to the surroundgs (dead state) at. 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.. Availability is a composite property - sce its value depends upon an external datum 3. Availability of a system is 0 at its dead state when T = and P = P Unless otherwise stated, assume the dead state to be: P 0 = 1 atm = 5 C 1

2 5. The maximum work is obtaed through a reversible process to the dead state. REV ERSIBLE W ORK } {{ } W rev = USEF } UL{{ W ORK} W useful + IRREV ERSIBILIT Y } {{ } I Control Mass Analysis we know W rev = W useful + I but as shown the figure, the actual work of the process is divided to two components W actual = W useful + W sur

3 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 ) To fd W actual, from the 1st law E 1 Q W actual = E Q = E 1 E W actual From the nd law S gen = S system + S 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 S gen = S S 1 + E 1 E W actual which leads to W actual = (E 1 E ) + (S S 1 ) S gen or by reversg the order of S and S 1 W actual = (E 1 E ) (S 1 S ) S gen But we also know that W useful = W actual W sur 3

4 therefore W useful = (E 1 E ) (S 1 S ) + P 0 (V 1 V ) S gen } {{ } W sur and W rev = W useful + I = W actual W sur + I where I = S gen Therefore W rev = (E 1 E ) (S 1 S ) + P 0 (V 1 V ) Defe 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 The availability destroyed is I = W rev W useful = S gen = S gen This can be referred to as: irreversibilities, availability destruction or loss of availability. 4

5 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 ṁs + Q + Ṡ gen () 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 Ṡ gen = ṁ s + h + KE + P E] ( Ṡ gen ) (3) Ẇ actual is the actual work put of the turbe. The specific flow availability, ψ, is given as ψ = (s s 0 ) + (h h 0 ) + ( (v ) (v 0 0 ) ) + g(z z 0 0 ) (4) 5

6 For a steady state, steady flow process where we assume KE=PE=0 Ẇ rev = (ṁψ) (ṁψ) (5) I = Ẇ rev Ẇ actual = Ṡ gen = Ṡ gen (6) ψ = (h h 0 ) (s s 0 ) (7) The Exergy Balance Equation From the 1st law de cv = Ẇ Ẇ Q 0 + Q 1 Q + ṁ(e + P v)] ṁ(e + P v)] (1) From the nd law ds cv = ṁs Q 0 + Q 1 T 1 ṁs + Q T + Ṡ gen () 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 + P v s)] 6

7 ṁ(e + P v s)] + Q Q T Q 1 Q 1 T 1 Ṡ gen (3) We can rewrite Eq. (3) a generalized form by troducg the defitions of X and ψ. dφ = P 0 dv CV + Ẇ + ṁψ + Q Ẇ + ṁψ + Q ( 1 T )] 0 ( 1 T )] 0 I where I = Ṡ gen = 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 = flow exergy Ẇ useful = ( ) dv CV (Ẇ Ẇ } {{ ) P } 0 Ẇ actual } {{ } W sur (v ) (v 0 )] + g(z z 0 ) 7

8 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 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 8

9 PROBLEM STATEMENT: kg of air a piston-cylder device is expanded reversibly and isothermally from 700 kp a and 50 C to a pressure of 15 kp a. Durg the process, heat is added from a thermal energy reservoir at 50 C. Assume the dead state conditions to be = 5 C and P 0 = kp a. a) Determe the amount of work transfer kj] to the piston and the amount of heat transfer kj] from the source. b) Determe the availability transfer kj] due to work and heat. c) Determe the crease availability kj] of the air the cylder. d) Physically, how do you expla that, although the ternal energy of the air did not change, its availability did? 9