MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Exergy Balance – Closed Systems
For a closed system:
2
Energy Balance: E2 − E1 = dQ − W [1]
1
2
dQ
Entropy Balance: S2 − S1 = +σ [2]
1
T b
denotes boundary Entropy
production
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 2
dQ
E2 − E1 − To ( S 2 − S1 ) = dQ − To − W − Toσ [3]
1 1
T 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 ) [4]
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Substituting [4] into [3] for E2-E1 gives the closed system
exergy balance:
2
To
Ex2 − Ex1 = 1− dQ − [W − Po (V2 − V1 )] − Toσ [5]
1
Tb
where: Ex2 − Ex1 is the exergy change;
2
To
1− dQ is the exergy transfer accompanying heat
1
Tb
transfer;
[W − Po (V2 − V1 )] is the exergy transfer accompanying work.
2
To
∴ 1− dQ − [W − Po (V2 − V1 )]
1
Tb
is the sum of the exergy transfers.
Toσ is the exergy destruction Ed
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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).
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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 To dV
= 1− Q j − W − Po − Ed [6]
dt j Tj dt
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
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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 To Wmax for a Carnot
Accompanying = 1− Q = Power Cycle
Tb
Heat
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:
Work done = Po (V2 − V1 )
pushing against
surroundings
Thus,
Wmax = W − Po (V2 − V1 )
(net)
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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
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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 m3 1kJ
∆e = 2087.56 + 1.014 x103 2 1.672
kg m kg 103 N ⋅ m
kJ
− ( 293.15 K ) 6.048
kg ⋅ K
∆e = 484 kJ / kg
W
Exergy accompanying work = − Po (υ g − υ f )
m
(since constant P process) = Pυ fg − Po (υ fg )
= ( P − Po )υ fg
=0 (since P = Po)
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Exergy accompanying To Q
heat transfer = 1 −
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
heat transfer = 1 − (2257 kJ / kg )
373.15 K
= 484 kJ / kg
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
ed = 0 (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)
W
Exergy accompanying work = − Po (υ g − υ f )
m
W
To find , perform an energy balance:
m
0 0 0
∆U + ∆KE + ∆PE = Q − W
W
= −(u g − u f ) ug and uf in Table A-2
m
W kJ
= −(2506.5 − 418.94)
m kg
W kJ
= −2087.56
m kg
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Thus,
Exergy accompanying work =
kJ N m3 kJ
−2087.56 − 1.014 x105 2 1.672
kg m 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:
∆e = Exergy transfer with Q – Exergy transfer with work
– Ed
i.e.
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 .
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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 [7]
2
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.
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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 To dV
= (1 − )Q j − (Wcv − Po cv ) + mi e fi − me e fe − Ed [8]
dt j Tj dt i e
dEX cv
the rate of exergy change
dt
mi e fi the flow exergy at inlets
i
me e fe the flow exergy at exits
e
Ed the rate of exergy destruction
To dV
(1 − )Q j − (Wcv − Po cv ) + mi e fi − me e fe
j Tj dt i e
the rate of exergy transfer
where,
i = inlet
e = exit
efi = flow exergy at inlet (evaluated with [7] )
efe = flow exergy at exit (evaluated with [7] )
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Steady State Exergy Rate Balance
dEX cv dVcv
For steady state conditions, =0 , =0
dt dt
Equation [8] becomes:
To
0= (1 − )Q j − Wcv + mi e fi − me e fe − Ed [9]
j Tj i e
or, simply:
0= EX qj − Wcv + EX fi − EX fe − Ed [10]
j i e
For the particular case of one inlet and one outlet denoted as 1
and 2, respectively:
To
0= (1 − )Q j − Wcv + m(e f 1 − e f 2 ) − Ed [11]
j Tj
where,
2 2
v − v2
ef 1 − ef 2 = (h1 − h2 ) − To ( s1 − s2 ) + 1 + g ( z1 − z2 ) [12]
2
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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
Wturb = 540 kJ/kg
400oC Turbine
160 m/s 1 Saturated vapour
2 100oC
100 m/s
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Solution
We use the steady state exergy rate balance ([12] with only one
Q).
To
0 = (1 − )Q − Wcv + m(e f 1 − e f 2 ) − Ed [13]
Tb
Step 1
Solving first for net rate of flow exergy carried in:
2 2
v − v2
ef 1 − ef 2 = (h1 − h2 ) − To ( s1 − s2 ) + 1 + g ( z1 − z2 )
2
Evaluating properties:
Point 1: 30 bar, 400oC
Using Table A4: h1 = 3230.9 kJ/kg
s1 = 6.912 kJ/kgK
Point 2: Sat. vapour, 100oC
Using Table A2: h2 = 2676.1 kJ/kg
s2 = 7.3549 kJ/kgK
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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
+
(2)(1000) kg
e f 1 − e f 2 = 691.84 kJ / kg
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Step 2
Q
Solve for 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
EX q
Solve for the rate of exergy transfer accompanying heat,
m
EX q To Q
= 1−
m Tb m
EX q 298 K
= 1− ( −22.6 kJ / kg )
m 350 K
EX q
= −3.36 kJ / kg
m
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MECH 330: APPLIED THERMODYNAMICS II LECTURE 05
Step 4
W
Evaluate [13], dividing all terms by m and noting that is
m
already given as Wturbine = 540 kJ/kg, to solve for Ed .
To Q W E
0 = (1 − ) − + (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 (78.05%)
Exergy carried out with heat: 3.36 kJ/kg (0.49%)
Rate of Exergy Destruction: 148.48 kJ/kg (21.46%)
691.84 kJ/kg (100%)
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