Exergy Balance - Closed Systems: E E DQW DQ S S T [PDF]

Citation preview

MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Exergy Balance – Closed Systems For a closed system: 2



Energy Balance:



E2 − E1 = dQ − W



[1]



1



Entropy Balance:



S2 − S1 =



2



1



dQ T



+σ



[2]



b



denotes boundary production



Entropy



where W and Q represent work and heat transfers between the system and surroundings. Multipling eq. [2] by To and subtracting it from eq. [1] gives, 2



E2 − E1 − To ( S 2 − S1 ) = dQ − To 1



2



1



dQ T



− W − Toσ



[3]



b



Recall that the change in exergy for closed system can be given as:



Ex2 − Ex1 = ( E2 − E1 ) + Po (V2 − V1 ) − To ( S 2 − S1 )



43



[4]



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Substituting [4] into [3] for E2-E1 gives the closed system exergy balance:



Ex2 − Ex1 =



2



1



1−



To dQ − [W − Po (V2 − V1 )] − Toσ Tb



[5]



where: Ex2 − Ex1 is the exergy change; 2



To dQ is the exergy transfer accompanying heat T b 1 transfer; 1−



[W − Po (V2 − V1 )] is the exergy transfer accompanying work.



∴



2



1



1−



To dQ − [W − Po (V2 − V1 )] Tb



is the sum of the exergy transfers.



Toσ is the exergy destruction Ed



44



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



For the exergy transfer accompanying heat transfer: If Tb > To, Q and exergy will be in the same direction (i.e. both +ve or both –ve). If Tb < To, (e.g., for refrigerators and heat pumps) the opposite is true. This is because if, for example, a gas is cooled below To (i.e., with a –ve Q), it is “moved” further from the dead state, To, and entropy is increased (i.e., made +ve).



45



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



This equation can also be expressed in a time dependent form as the closed system exergy rate balance:



dEx = dt



1− j



To dV Q j − W − Po − Ed Tj dt



[6]



where j refers to points on the boundary where heat is •



transferred and Ed accounts for the destruction of exergy due to irreversibilities within the system. That is, •



•



Ed = To σ



46



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



The exergy transfer accompanying heat can be thought of as the maximum work that could be developed by supplying heat Q to a reversible power cycle operating between Tb and To. Note that this is also the Wmax obtainable from a Carnot Power Cycle, i.e. Exergy Transfer Accompanying Heat



= 1−



Wmax for a Carnot To Q = Power Cycle Tb



The exergy transfer accompanying work can be thought as the maximum work that could be obtained were the system interacting with the environment. For example, if a system expands from V1-V2 due to an input of W, the system must do work on the surroundings at Po equal to:



= Po (V2 − V1 )



Work done pushing against surroundings Thus,



Wmax = W − Po (V2 − V1 ) (net)



47



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Example Water in a piston cylinder assembly, initially in a saturated liquid state (at 100oC), undergoes a process in which it ends up as a saturated vapour. For each process noted below: what is ∆e ? what is ed ? What is the exergy transfer accompanying work? What is the exergy transfer accompanying heat? The piston is allowed to move freely. To = 20oC and Po = 1.014 bar. Process a: Change of state via Q addition at constant T, P and without irreversibilities. Process b: Change of state via stirring action (i.e. W input) and adiabatically (i.e. no Q). Solution Assumptions - Closed system - ∆KE , ∆PE = 0



48



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



a) Q Addition



∆e = (u g − u f ) + Po (υ g − υ f ) − To ( s g − s f ) Note: There are no KE or PE terms due to assumptions. f and g properties from Table A-2:



kJ N ∆e = 2087.56 + 1.014 x103 2 kg m



m3 1.672 kg



1kJ 103 N ⋅ m



− ( 293.15 K ) 6.048



∆e = 484 kJ / kg Exergy accompanying work =



W − Po (υ g − υ f ) m



(since constant P process) = Pυ fg − Po (υ fg )



= ( P − Po )υ fg =0



49



(since P = Po)



kJ kg ⋅ K



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Exergy accompanying To Q = 1 − heat transfer T m



Q kJ = T ∆s = T ( sg − s f ) = (373.15 K )(7.3549 − 1.3069) m kg ⋅ K Q = 2257 kJ / kg m Exergy accompanying 293.15 K 1 (2257 kJ / kg ) = − heat transfer 373.15 K = 484 kJ / kg



50



MECH 330: APPLIED THERMODYNAMICS II



ed = 0



LECTURE 05



(since reversible process)



This is consistent with the fact that, for this process, ∆e = exergy accompanying Q. b) Adiabatic W Addition



∆e = 484 kJ / kg (same as a) ) Exergy accompanying Q = 0 (no Q) Exergy accompanying work =



W − Po (υ g − υ f ) m



W , perform an energy balance: m 0 0 0 ∆U + ∆KE + ∆PE = Q − W



To find



W = −(u g − u f ) m



ug and uf in Table A-2



W kJ = −(2506.5 − 418.94) m kg W kJ = −2087.56 m kg



51



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Thus, Exergy accompanying work =



kJ N −2087.56 − 1.014 x105 2 kg m



m3 kJ 1.672 kg 103 N ⋅ m



Exergy accompanying work = −2257 kJ / kg Note: Since value is negative, exergy is transferred to the system due to work. To calculate the exergy destruction, Ed:



i.e.



∆e = Exergy transfer with Q – Exergy transfer with work – Ed Ed = - Exergy transfer with work - ∆e



(since Q = 0)



Ed = −(−2257 kJ / kg ) − 484 kJ / kg Ed = 1773 kJ / kg Realistic since Ed must be greater or equal to 0. Interpretation: A total exergy of ∆e = 2257 kJ / kg is transferred into the system accompanying work. Of this amount 1773 kJ / kg is destroyed (Ed) due to irreversibilities. The net increase in exergy is thus 484 kJ / kg .



52



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Flow Exergy Consider a flow across a boundary of a control volume: Boundary of control volume Fluid Flow



The specific flow exergy for this individual flow is:



v2 e f = (h − ho ) − To ( s − so ) + + gz 2



[7]



h, s, v and z are for the fluid at the entrance or exit for the control volume where the flow crosses the boundary. ho and so are evaluated at the dead state, To, Po.



53



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Exergy Rate Balance for Control Volumes The closed system exergy balance (eq. [5]), and the flow exergy for individual flows [eq. 7] can be used to develop the following equation for the control volume exergy rate balance:



dEX cv = dt dEX cv dt



(1 − j



To dV )Q j − (Wcv − Po cv ) + Tj dt



mi e fi − i



me e fe − Ed [8] e



the rate of exergy change



mi e fi



the flow exergy at inlets



me e fe



the flow exergy at exits



i



e



Ed



the rate of exergy destruction



To dV )Q j − (Wcv − Po cv ) + Tj dt j the rate of exergy transfer (1 −



mi e fi − i



where, i = inlet e = exit efi = flow exergy at inlet efe = flow exergy at exit



(evaluated with [7] ) (evaluated with [7] )



54



me e fe e



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Steady State Exergy Rate Balance For steady state conditions,



dEX cv dVcv =0 , =0 dt dt



Equation [8] becomes:



0=



(1 − j



To )Q j − Wcv + Tj



mi e fi − i



me e fe − Ed



[9]



e



or, simply:



0=



EX qj − Wcv + j



EX fi − i



EX fe − Ed



[10]



e



For the particular case of one inlet and one outlet denoted as 1 and 2, respectively:



0=



(1 − j



where,



ef 1 − ef 2



To )Q j − Wcv + m(e f 1 − e f 2 ) − Ed Tj 2



[11]



2



v − v2 = (h1 − h2 ) − To ( s1 − s2 ) + 1 + g ( z1 − z2 ) 2



55



[12]



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Example Steam enters a turbine at 30 bar, 400oC and 160 m/s, and leaves as saturated vapour at 100oC and100 m/s. wturbine = 540 kJ/kg of steam (at steady state). Q between turbine and surroundings occurs at Tb = 350 K. ∆PE = 0 , To = 25oC and Po = 1 atm. Develop a full accounting of the exergy carried in by the steam (per unit mass of steam).



30 bar 400oC 160 m/s



1



Wturb = 540 kJ/kg



Turbine 2



56



Saturated vapour 100oC 100 m/s



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Solution We use the steady state exergy rate balance ([12] with only one Q).



0 = (1 − Step 1



To )Q − Wcv + m(e f 1 − e f 2 ) − Ed Tb



[13]



Solving first for net rate of flow exergy carried in:



ef 1 − ef 2



2



2



v − v2 = (h1 − h2 ) − To ( s1 − s2 ) + 1 + g ( z1 − z2 ) 2



Evaluating properties: Point 1:



30 bar, 400oC Using Table A4:



Point 2:



h1 = 3230.9 kJ/kg s1 = 6.912 kJ/kgK



Sat. vapour, 100oC h2 = 2676.1 kJ/kg s2 = 7.3549 kJ/kgK



Using Table A2:



57



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Thus, e f 1 − e f 2 = (3230.9 − 2676.1) kJ / kg − 298 K (6.912 − 7.3549) kJ / kg ⋅ K +



(160 m / s ) 2 − (100 m / s ) 2 kJ



kg



(2)(1000)



e f 1 − e f 2 = 691.84 kJ / kg



58



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Step 2 Solve for



Q by applying an energy balance: m 2



2



Q W v − v1 = + (h2 − h1 ) + 2 m m 2



Q kJ kJ (160 m / s ) 2 − (100 m / s )2 kJ = 540 + (2676.1 − 3230.9) + m kg kg (2)(1000) kg



Q = −22.6 kJ / kg m Step 3 Solve for the rate of exergy transfer accompanying heat,



EX q m



EX q m EX q m



= 1−



To Q Tb m



= 1−



298 K ( −22.6 kJ / kg ) 350 K



= −3.36 kJ / kg



59



EX q m



MECH 330: APPLIED THERMODYNAMICS II



LECTURE 05



Step 4 Evaluate [13], dividing all terms by m and noting that already given as Wturbine = 540 kJ/kg, to solve for Ed .



0 = (1 −



W is m



To Q W E ) − + (e f 1 − e f 2 ) − d Tb m m m



Ed kJ kJ kJ = −3.36 − 540 + 691.84 m kg kg kg Ed kJ = 148.48 m kg Exergy Accounting Net rate of Exergy flowing in: 691.74 kJ/kg (100%) Disposition of Exergy Rates: Exergy carried out with work: 540.00 kJ/kg Exergy carried out with heat: 3.36 kJ/kg Rate of Exergy Destruction: 148.48 kJ/kg 691.84 kJ/kg



60



(78.05%) (0.49%) (21.46%) (100%)