Non-Equilibrium Thermodynamics
for Engineers
”How do we find the optimal process unit?”
Signe Kjelstrup,
Chair of Engineering Thermodynamics
Department of Process and Energy
TU Delft
Lecture no. 7
1
Why is the entropy production
important?
• The work output from the maximum available
w = w max − wlost
• Guy-Stodola’s theorem (1889, 1910):
wlost = T 0( dSirr / dt ) > 0
Lecture no. 7
dS irr / dt =
∫
σ dV
2
Mathematical methods
for constrained optimisation
Control theory
Euler Lagrange optimisation:
dS
L = irr + ∑ λ i Pi
dt
i
Constraint examples:
P = P1
H = σ( z , t ) + ∑ λ i ( z , t ) f i
i
dT
= ....
dz
dp
Momentum balance f p =
= ....
dz
dξ
Mass balance f ξ =
= ....
dz
Extra conditions, i.e. pext = const.
Energy balance fT =
Ta = T0
•
Conservation equations are
included in the objective function
Cf. Course on ”Engeering Fundamentals”
Lectured by prof. J. Gross
•
•
•
•
Local control of conservation equations
Defined control variables give a practical handle
Mathematically robust
An autonomous Hamiltonian is constant along the
path
Lecture no. 7
3
Optimal isothermal expansion (1)
•
Find the external pressure in a
one step process that gives
mininmum lost work, when the
pressure of the system
changes from V1 to V2
V2
w = − ∫ pext dV = − NRT0 (
V1
N moles of ideal gas in a piston
wmax = NRT0 ln
p2
p1
1 1
− )
p2 p1
⎡
p ⎤
1
1
wlost = w − wmax = − NRT0 ⎢ pext ( − ) + ln 2 ⎥
p2 p1
p1 ⎦
⎣
The piston moves with time:
dV (t )
f
=−
p (t ) − p (t ) )
2 ( ext
dt
p
t
(
)
[ ]
Lecture no. 7
4
Optimal isothermal expansion (2)
•
Which pext gives minimum
entropy production, given
the values of p1 and p2?
N moles of ideal gas in a piston
wmax = NRT0 ln
p2
p1
Min
⎛ dSirr
⎡
p2 ⎤ ⎞
1
1
=
−
NR
p
−
+
(
)
ln
⎜⎜
⎢ ext
⎥ ⎟⎟
dt
p
p
p
⎣
2
1
1 ⎦⎠
⎝
Given p1 , p2
Solution:
pext
p1
exp(
( p2 / p1 )
f
)
NRT0
V2
Lecture no. 7
w = − ∫ pext dV = − NRT0 (
V1
1 1
− )
p2 p1
5
Optimal isothermal expansion (3)
Max. work
The pressure variation giving
minimum lost work (stipled line) can be
obtained from control theory
25
20
θ
dSirr
1
⎡ dV (t ) ⎤
= ∫ ( pext (t ) − p (t )) ⎢ −
dt
T
dt ⎥⎦
⎣
0 0
Constraint:
dV (t )
f
=−
p (t ) − p (t ) )
2 ( ext
dt
[ p(t )]
Presssure / bar
General objective function:
15
10
5
0
1.2
1.4
1.6
1.8
2
Volume / 10−3 m3
2.2
2.4
2.6
Solution:
pext
⎛ NRT0 p2 ⎞⎛ p2 ⎞
ln ⎟⎜ ⎟
= p1 ⎜1 +
f
p1 ⎠⎝ p1 ⎠
θ
⎝
⎛p ⎞
p (t ) = p1 ⎜ 2 ⎟
⎝ p1 ⎠
t /θ
t /θ
The driving force is constant along
the optimal path, and so is σ!
Lecture no. 7
6
Optimal expansion (4):
Continuous Expansion of Gases in a Turbine
„Multistage“ gas
turbine – a
realization of the
results derived for
the K-step
expansion case?
Lecture no. 7
7
Optimal heat exchange (1)
•
Find the temperature profile T(z)
that gives minimum entropy
production, when a given amount of
heat is transferred from the hot fluid
By fixing Th ,in and Th ,out we fix the heat transferred
Constraints
Fixed heat transferred means fixed:
Th ,in and Th ,out
The energy balance must be obeyed:
FC p dTh ( z ) = J q' ( z )Δydz
Lecture no. 7
8
Optimal heat exchange (3):
Is work is obtainable by heat exchange?
dq0
dq0
dq
=
T0 Tc ( z )
dq = J q' ( z )Δydz
Carnot machine
⎡
T0 ⎤ '
dw = ηC dq = Δy ⎢1 −
⎥ J q dz
⎣ Tc ( z ) ⎦
dq
J
⎡
T0 ⎤ '
w = Δy ∫ ⎢1 −
dz
⎥ J q dz = Fout H out − Fin H in − ΔyT0 ∫
Tc ( z ) ⎦
T ( z)
0 ⎣
0 c
L
L
Lecture no. 7
dw
'
q
T c (z)
Cold fluid
T h (z)
Hot fluid
dz
Entropy production from transfer
9
to cold side
Optimal heat exchange (2):
The entropy production
( x, z)
J q' ( x, z)
'
q
By fixing Th ,in and Th ,out we fix the heat transferred
'
q
J ( x, z)
( z)
d
1
dx T ( x, z)
J ( z)
y
( x, z)dx
yJ q' ( z)
0
1
1
Th ( z) Tc ( z)
⎛1⎞
J q' = lqq Δ ⎜ ⎟
⎝T ⎠
L
L
−1
dSirr
' 2
= Δy ∫ σ ( z )dz = Δy ∫ ( lqq ) ⎡⎣ J q ⎤⎦ dz
dt
0
0
Lecture no. 7
10
Optimal heat exchange (5): Solution
•
dq0
Carnot machine
dw
•
dq
T c (z)
Cold fluid
T h (z)
Hot fluid
Exact solution: Constant
entropy production (EoEP)
Approximate solution:
Constant thermal force (EoF)
450
dz
T , EoEP
h
Tc, EoEP
440
T , EoF
The entropy balance for the hot fluid
h
430
Temperature / K
L
J q'
dSirr
= Fout Sout − Fin Sin − Δy ∫
dz
dt
T
z
(
)
0 c
The entropy production for heat exchange
δ
dSirr
= Δy ∫
dt
0
410
400
L
390
0
0
380
0
∫ σ ( x, z )dzdx =Δy ∫ σ ( z )dz
2
Lecture no. 7
c
420
L
⎡ 1⎤
= Δy ∫ lqq (Th ( z )) ⎢ Δ ⎥ dz
⎣ T⎦
0
L
T , EoF
2
4
6
Position / m
8
10
11
Reasons to minimize the entropy production
• We obtain a realistic target for the efficiency:
The most energy efficient operation for the real system
• We find a zero on a yardstick that measures lost work
• We can find rules for process design:
Rules of thumb, energy efficient design
• A turbine with equipartition of forces
• Heat exchange with equipartition of forces
Lecture no. 7
12
Energy efficient design means that:
1. The path of minimum entropy production has
been used, given the boundary conditions
2. This operating path has constant entropy
production if the system has sufficient degrees of
freedom
3. Constant driving forces seems to be a good
approximation to a state with constant entropy
production
Lecture no. 7
13