1. No category

Utility systems
How to satisfy the energy
requirements of a process
MER François Maréchal
LENI - ISE-STI - EPFL
[email protected]
FM_07/ 2000
Le système énergétique industriel
Environnement
Air
Energie
Transformation de
l'énergie
Matières
Premières
Eau
Supports de
production
Energie
Produits
Sous-produits
Procédés
Traitements des rejets
Rejets et déchets
Air
Eau
Sol
Traitement
Laboratoire d!énergétique industrielle LENI -ISE-STI-EPFL
grid
R˙ r ,yw ,fw w=1
nw
!
grid
grid
grid
i(1 +
+
(C1w yw ) +
(1 + i)nyears − 1
Grand composite curves
i)nyears
w=1
600
600
nw
!
(ICFw y
w=1
ect to the heat
balance of the550 temperature intervals
550
Cold composite curve
Hot composite curve
Hot Utility : 6854 kW
Hot utility
w=1
400
n
!
i=1
350
Self sufficient
"Pocket"
500
450
˙ r+1
Q˙ −
+
R
− R˙ r400=
K 0
i,r
Heat recovery
T(K)
450
fw Q˙ −
w,r +
T(K)
nw
!
500
400
350
cooling utility
300systems
0.2 300 Methodology
for designing integrated utility
Ambient temperature
5
Cold utility : 6948 kW
250
0.2250 0 Methodology
for designing
integrated
utility
systems
ricity consumption:
0
2000
4000
6000
8000
10000 12000
5000 10000 15000 20000 25000 30000
refrigeration
nw
!
w=1
Refrigeration : 1709 kW
Q(kW)
Q(kW)techniques aim at identifying the maximum
The process integration
energy recovery that could
be obtained by counter-current heat exchange between the hot and the cold streams of the proFM_07/ 2000
cess. This technique, based on the assumption of a minimum temperature difference
between
the cold streams (∆T
+ min), allows the calculation of the so-called minimum en−the hot and +
(MER) targetprocess
for the system. The identification of the pinch point, the point
w ergy requirement
grid
where the hot and cold composite curves of the process are the closest, is further used to design the heat recovery heat exchanger network structure. Using the concept of the hot and
cold composite curves, it is possible to compute graphically the MER. Mathematically, the
minimum energy requirement is computed by solving the heat cascade model (1). This model
is a one degree of freedom linear programming problem that computes the the energy required
to balance the energy required by the cold streams with the hot streams for each temperatures
of the problem. The Grand composite curve is the plot of this heat requirement as a function
of the temperature.
− E˙
fw E˙ + E˙
≥0
ricity exportation
Linear programming formulation
−
nw
˙
!
E
min+
Rn +1
grid
+
R
˙
˙
− E˙ process
=0
fw Ew + Egrid −
subject to heat balance of the
temperature
intervals
ηgen
w=1
n
!
Qi,r + Rr+1 − Rr = 0
ence of technology w
i=1
(1)
r
r
Rr ≥ 0
∀r = 1, ..., nr
(2)
∀r = 1, ..., nr + 1
(3)
∀w=1,...,nw
f minw yw ≤
f max
with fn
the number
of specified
process
streams;
w ≤
w yw
yw
∈{0,1}
nr
Rr
the number of temperature intervals;
the energy cascaded from the temperature interval r to the lower temperature intervals in the time period p;
the heat load of the reference level of process stream i in the temperature
interval r; Qir > 0 for hot streams and ≤ 0 for cold streams;
modynamic feasibility
of the heat recovery and utility syste
Q
ir
The heat cascade constraints (2) is the equation system that is solved by the problem table
+
≥ 0, E˙ grid
≥ 0An alternative set of equations (4) may be used to compute the heat cascade. This
method.
formulation has the advantage of involving only one R per equation, each of the equation
being related to one temperature in the temperature list. From the analysis of the pinch point
˙ demonstrated
0, R˙ n
=
0,itR
0 that the list of temperatures (and therefore of equations) may
location
mayrbe≥
FM_07/ 2000
r
r+1
be reduced in this case to the list of inlet temperature conditions of all the streams. By saying
that, we assume that the fluid phase changing streams are divided into streams segments.
∀
Utility definition
Type of utility (e.g. : Hot stream)
inlet conditions (T, P)
(fuel type, combustion , operation)
T
Outlet conditions (T,P)
(environment, operation)
H
3 characteristics :
- H-T diagram (from GCC analysis)
- Cost as a function of flowrate
- Flowrate : to be determined to satisfy the
FM_07/ 2000
Utility integration
T(°C)
Counter current analogy
Hot utility -cold process
220
Heat sink
T
Zone 1
200
T
GCC = cold stream
180
P
160
H
Q
H
Q
140
Self sufficient zones
Exchange process -> process
120
Zone 2
100
D
80
E
60
Ambient T
F
40
Zone 3
GCC = hot stream
0
Heat source
200
600
H(kW)
Counter current analogy
Hot process -cold utility
T
P
T
Q
H
Q
H
FM_07/ 2000
Multiple utilities
Identify the temperature levels
Qhmin
T
(°C)
T1
Q1
Q2
T2
T3
PINCH
H (kW)
FM_07/ 2000
Hot Utility : combustion
O2 (air)
Fuel : CxHyOzSsNn
xC + xO2 + = xCO2
y
y
yH + O2 = H2 O
4
2
nN + nO2 = nNO2
sS + sO2 = sSO2
y
z&
#
x + + n+s"
%
4
2 ( * (1 + ) ) * (0.21O + 0.79N )
a
2
2
%
(
0.21
%
(
$
'
Heat
Flue gases
y
z&
y
z&
#
#
x + + n+s"
x + + n + s"
%
(
%
4
2 * (1 + ) ) * (0.79N ) +
4
2 ( * () ) * (0.21O ) + xCO + y H O + nNO + sSO
a
2
a
2
2
2
2
%
(
%
(
0.21
0.21
2 2
%
(
%
(
$
'
$
'
Combustion
tf
Qfuel = LHV + "25 m˙ fuel cp fuel dT = LHV + m˙ fuel cp fuel (tf # 25)
T
Fuel
Adiabatic temperature of combustion
Useful
heating
value
Lower
Heating
Value
O2 (air)
ta
Qair = "25 m˙ aircpair dT = m˙ aircpair (ta # 25)
Stack temperature Tc
T0=25°C
Tc
˙ fumées cp fumées (tc # 25)
Qpertes = "25 m˙ fuméescp fumées dT = m
Environnement
!
Eléments spéciaux
Conversion du soufre : contrôle cinétique
!
0.95 * sS + 0.95 * sO2 = 0.95* sSO2
" 3$
0.05 * sS + 0.05 * # % sO2 = 0.05* s * SO3
2
0.05 * s * SO3 + 0.05* s * H 2O = 0.05* s * H 2 SO4
Affinités avec l!eau => temperature de condensation + élevée
(+70°C)
Problèmes de corrosion
NOx = NO, N2O, NO2 différentes formes contrôlées par la
cinétique (vitesse) donc par T et conditions de mélange
Imbrûlés = combustible non brulés => cendres et poussières,
perte de pouvoir calorifique.
Fuels containing S
avec
SO3 (ppm)
29.863
=- 27,712 + 4,331 S(%) + O (%)
2
SO3(ppm)
S(%)
O2(%)
la teneur en ppm de SO3 dans les fumées;
le pourcentage massique de soufre dans le combustible;
le pourcentage molaire de O2 dans les fumées;
limite de validité:
Tra
0,5
1
<S(%) <5
<O2(%) <4
[H2O]
Tch2o
!Tc
Tra
P
Temperature de rosée acide
= Tch2o + !Tc
= 273.15+ 100
avec
S(%)
- 3,707 O (%)! + 54,396 LOG(O2(%))
2
4
!SO3(ppm)!
![H2O]!P + 49,7041 ( H O!(%) )0.1173
2
la fraction molaire de H2O dans les fumées;
le point de rosée de la vapeur d'eau dans les fumées;
l'avance à la condensation;
la température de rosée acide des fumées (°K);
la pression dans la cheminée (bar).
FM_07/ 2000
30
Contents
Fuels example
Table 5: Values for some typical fuels
LHV
Tad
Tstack
Air
Cost
CO2
◦
◦
kJ/kg
K
K
kgair /kg e/GJLHV kgCO2 /GJ
CH4
50000 2646.5
374
17.1
55
Natural gas
39680
2270
374
13.9
7.67
55
Light fuel oil
45316
2425
440
14.4
6.9
70
Heavy fuel oil 44500
2423
441
14.3
4.67
71
Coal (lignite)
25450
2111
438
7.29
2.53
81
Wood
18900 2185.43
374
7.9
5
0
Biogas
13358
2077
374
4.63
0
Natural gas composition: 87% Methane, 13% N2
Light fuel oil: C-86.2% mass, H-12.4% mass, S-1.4% mass
Heavy fuel oil: C-86.1% mass, H-11.8% mass, S-2.1% mass
Lignite: C-56.52%, H-5.72%, O-31.89%, N-0.81%, S-0.81%, Ash-4.25%
Wood (wt) composition: C-49.5%, H-6% , O-44.6% → C1 H1,4 O0,7 , CO2 neutral
50%CO2 and 50% CH4 , CO2 neutral
Cost 2004, European market
Air preheating : outlet temperature calculation
When combustion is considered, the air preheating plays the role of a chemical heat pump,
it is used to pump waste heat available below the pinch point and make it available above
the pinch point (by an increase of the adiabatic temperature of combustion). The effect of
air preheating is however limited to the preheating of the stoechiometric air flow. When the
Values for different fuels
LHV
(kJ/kg)
Gaz naturel
Gaz de cokerie
Distrigaz
Mer du Nord
Gaz de charbon
Essence
Vaporizing oil
Diesel
Kérozène
Fuel léger
Fuel lourd
Anthracite
Bitume
Lignite
39680
2780
44945
47900
27270
47798
46105
45867
46924
45316
44500
33220
31520
25450
s
O2
(kg/kg)
13,879
0,696
15,693
16,735
8,557
15,003
15
14,583
14,826
14,454
14,264
11,53
10,21
7,23
Tads0
(K)
2270
1688
2290
2292
2373
2431
2389
2426
2429
2425
2423
1819
1954
2111
Tch
(K)
LHVu
(kJ/kg)
374
346
374
374
376
438
438
439
438
440
441
434
435
438
"LHVu
38163
2685
43225
46047
26246
44665
43023
42838
43850
42303
41507
30256
28929
23488
-3,82%
-3,43%
-3,83%
-3,87%
-3,75%
-6,55%
-6,69%
-6,60%
-6,55%
-6,65%
-6,73%
-8,92%
-8,22%
-7,71%
FM_07/ 2000
CO2 Impact
DPCU
Gaz naturel
Gaz de cokerie
Distrigaz
Mer du Nord
Gaz de charbon
Essence
Vaporizing oil
Diesel
Kérozène
Fuel léger
Fuel lourd
Anthracite
Bitume
Lignite
-3,82%
-3,43%
-3,83%
-3,87%
-3,75%
-6,55%
-6,69%
-6,60%
-6,55%
-6,65%
-6,73%
-8,92%
-8,22%
-7,71%
CO2
(LHV)
55
186
57
55
58
65
69
69
67
70
71
100
86
81
CO2
(LHVu)
57
193
59
58
60
70
74
74
72
75
76
109
94
88
DCO2
3,97%
3,56%
3,98%
4,03%
3,90%
7,01%
7,16%
7,07%
7,01%
7,12%
7,21%
9,80%
8,96%
8,35%
!CO2
(LHV)
0,00%
239%
3,54%
1,26%
5,19%
19,53%
25,77%
25,71%
22,88%
27,08%
29,29%
81,53%
56,84%
48,39%
!CO2
(LHVu)
0,00%
238%
3,55%
1,31%
5,11%
23,02%
29,63%
29,45%
26,47%
30,92%
33,31%
91,69%
64,35%
54,64%
FM_07/ 2000
!
Utilities definitions
Utility cost : C(CHF/s) = cost(CHF/kg) * flow(kg/s)
T(K)
Tad
QMER
Increasing flow
Infeasible
Tstack
Q(kW)
-20
0
20
40
60
Corrected temperature domain
FM_07/ 2000
Useful heating value of a fuel
[
]
LHVu = cp f * T 0 " Tstack : Useful heating value
ad
cp f #
LHV
(T
0
ad
" T0 )
in kW
,T = 25°C
(kg fuel °C) 0
m˙ gc
cp f # cpgc *
m˙ fuel
LHVu
LHV
˙ fuel * LHVu " Q˙ mer
Real losses = m
Intrisic Losses = 1"
FM_07/ 2000
!
Utilities definitions
Different utility heat
loads
For the same MER !!!
Tad2
Tad1
Q2
Q1
T(K)
Utility 1 : C1 = cost 1 * flow 1
Utility 2 : C2 = cost 2 * flow 2
Q(kW)
-20
0
20
40
60
Energy available or excess
FM_07/ 2000
Intégration de la combustion
Effet de la préchauffe
Pour un même débit de combustible :
plus d’énergie au-dessus du pincement
2500
o2=10%
Stoechiometrique
Préchauffe de l’air
Combustion du méthane
Excès d’air : moins d’énergie disponible
2000
1500
T (°C)
1000
T pincement
500
Cheminée
0
-10000
0
10000
20000
30000
Q (kJ/kg)
40000
50000
60000
Condensation
FM_07/ 2000
Cogeneration : gas turbines
Gaz
%
(
'
*
˙
Qmerk
*
max '
m˙ fuel =
'
*
k ' "GTth * LHV fuel %
(*
min
*
* ToT # max T
, Tk + $Tmin /2 f *
)*
stack
' (ToT # T min ) '&
&
)
stack
(
E˙ = "GTe * m˙ fuel * LHV fuel
))
(
!
ToT
Air
T(K)
!
P1
H
P2
"GTe = 30 # 38%
T min
"GTt
stack
!
0
20
40
60
Energy available or excess
= 55 # 47%
"GT = 85%
Q(MW)
-20
gc
S
Inv = 1000 - 2500 CHF/kWe
Size = 50000 - 1000 kWe
FM_07/ 2000
!
Gas Turbine : fixed size
Fuel
m˙ fuel PC + m˙ fuelGT * (1" #GT e )
ToPC = T0 +
m˙ fuel PC
m˙ fuelGT * (1" #GT e )
+
(Tad " T0 )
(ToT " T0 )
ToPC
Q˙ GT = m˙ fuelGT * LHV fuel * "GTth
!
!
!
ToGT
Air
Q˙ = m˙ fuel PC *UHV fuel + m˙ fuel PC * LHV fuel * ("GTth )
T(K)
!
E˙ = "GTe * m˙ fuelGT * LHV fuel
!
P1
H
!
T
!
P2
"GTe = 30 # 38%
min
stack
-20
"GTt
Q(MW)
0
20
40
60
gc
S
= 55 # 47%
"GT = 85%
Energy available or excess
FM_07/ 2000
!
Systèmes de cogénération
ηel
∗
ηel
ηel
1
ηth
ηth
∗
ηth
from the grid
∗
ηel
by combustion
∗
ηth
Marginal efficiency : Electricity/(additional fuel)
marg
ηel
=
E˙
m
˙ f uel − m
˙ th,equiv
f uel
∗
ηel
ηth
∗ ηel
=
=
∗ −η
(1 − ηηth
ηth
∗ )
th
th
Energy savings : equivalent energy consumption
∗
ηth ∗ ηth
+ ηel ∗ ηel∗ − 1
∗
ηth ∗ ηth
+ ηel ∗ ηel∗
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
_________ ______________________________________________ ___Février 2002
Systèmes de cogénération
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
_________ ______________________________________________ ___Février 2002
Systèmes de cogénération
th
el
CO2
50%
35%
55
Energy savings
Fuel
Natural gas
Light Fuel Oil Heavy Fuel Oil Coal (anthracite)
Coal (Lignite)
LHV (GJ/kg)
47900
45316
44500
33220
25450
CO2 (kg/GJLHV)
55.00
70.00
71.00
100.00
81.00
[nuth]
95%
92%
91%
89%
90%
[nuel equiv]
55%
42%
42%
40%
40%
Fuel el\fuel th Natural gas
Light Fuel Oil Heavy Fuel Oil Coal (anthracite)
Coal
Natural gas
13.99%
15.24%
15.67%
16.54%
Light Fuel Oil
26.45%
27.37%
27.68%
28.32%
Heavy Fuel Oil
26.45%
27.37%
27.68%
28.32%
Coal (anthracite)
28.64%
29.50%
29.80%
30.40%
Coal (Lignite)
28.64%
29.50%
29.80%
30.40%
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
(Lignite)
16.10%
28.00%
28.00%
30.10%
30.10%
_________ ______________________________________________ ___Février 2002
Systèmes de cogénération
Economie de CO2
Using the CO 2 production factor of each fuel, the CO 2 production without cogeneration would have been :
$#
'
#
th*
th
el*
el
th
el
m˙ *CO2 = Ffuel
* " CO
+ Ffuel
* " CO
= m˙ fuel * LHV fuel * & th* * " CO
+ *e * " CO
)
2
2
2
2
% #th
# el
(
th
where " CO
2
is the CO 2 production factor in
el
" CO
2
is the CO 2 production factor in
kgCO2
kgCO2
kJLHV for the fuel subsituted in the boiler
kJLHV for the fuel subsituted in the electricity production
relative CO2 reduction is therefore :
m˙
*
CO 2
* m˙ CO2
*
m
˙ CO
2
where " CO2
$ # th
'
#e
th
el
& * * " CO2 + * * " CO2 ) * " CO2
# *th * # *el * " CO2
%#
# el
(
= th
= 1* *
th
el
$ # th
'
#el * # th * " CO
+ # *th * #e * " CO
#e
th
el
2
2
& * * "CO2 + * * " CO2 )
% # th
#el
(
is the CO 2 production factor in
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
kgCO2
kJLHV for the fuel used in the cogeneration system
_________ ______________________________________________ ___Février 2002
Systèmes de cogénération
Economie de CO2
%
(
$
" CO2 # th* * " thCO2 *
th*
th
*
th
'
˙
˙
F fuel * " CO2 # F fuel * " CO2 &
$th
) $ th * " CO2 # $ th * " CO2
el
CO2 =
=
=
E˙
$e
$*th * $e
% kgCO2
(
'
GJ el *)
&
Fuel el\fuel th Natural gas
Light Fuel Oil Heavy Fuel Oil Coal (anthracite)
Coal
Natural gas
13.99%
24.70%
25.69%
39.68%
Light Fuel Oil
36.98%
42.93%
43.50%
51.97%
Heavy Fuel Oil
37.58%
43.42%
43.98%
52.32%
Coal (anthracite)
52.77%
56.19%
56.53%
61.72%
Coal (Lignite)
44.90%
49.50%
49.95%
56.71%
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
(Lignite)
31.25%
46.77%
47.20%
58.49%
52.54%
_________ ______________________________________________ ___Février 2002
Systèmes de cogénération
System cost
•
•
Fuel cost
Electricity
– selling
– or bill reduction
•
•
•
Maintenance
Operating time
Annualized investment
1
˙+ +
˙
cost = (M˙ f uel cf uel − E˙ − c−
grid + E cgrid − Qcq ) · time + Cmaint + I
τ
LABORATOIRE D!’ENERGETIQUE INDUSTRIELLE___
_________ ______________________________________________ ___Février 2002
0.8
Conclusions
59
Cost estimation
˙ e electrical power in [kW]
Aeroderivative gas turbines - W
˙ e ) + 0.8687, W
˙ e ≤ 845 [kW ]
[−]
ηgen = 0.015 · ln(W
˙ e ) + 0.9611, W
˙ e > 845 [kW ]
ηgen = 0.0013 · ln(W
˙ e ) − 0.0684
ηelec .
[−]
ηel = 0.0439 · ln(W
˙ e−0.0587
ηheat
[−]
ηth,echap = 0.838 · W
N Ox
[mg/m3 N ]
80
CO
[mg/m3 N ]
50
˙ e0.8503 , W
˙ e < 50000[kW ]
Turbine
[e]
Cturbine = 1564. · W
˙ 0.7791 , W
˙ e > 50000[kW ]
Cturbine = 2977. · W
e
Lifetime
[hour]
48000
˙
Heavy duty gas turbines - We electrical power in [kW]
˙ e ) + 0.8687, W
˙ e ≤ 845 [kW ]
ηgenerator
[−]
ηgen = 0.015 · ln(W
˙ e ) + 0.9611, W
˙ e > 845 [kW ]
ηgen = 0.0013 · ln(W
˙ e ) + 0.1317
ηelec .
[−]
ηel = 0.0187 · ln(W
−0.0315
˙
ηheat
[−]
ηth,echap = 0.7058 · We
N Ox
[mg/m3 N ]
50
CO
[mg/m3 N ]
50
˙ e0.7338 , W
˙ e < 50000[kW ]
Turbine
[e]
Cturbine = 4786. · W
˙ 0.7791 , W
˙ e > 50000[kW ]
Cturbine = 2977. · W
e
Lifetime
[hour]
55000
˙ e electrical power in [kW], Q
˙ th heat recovery boiler heat load in [kW]
Auxiliaries - W
ηgenerator
Reference cost
[e]
Cref =
Recovery boiler
[e]
Connection Charges
Instrumentation
Civil engineering
Engineering
Others
Maintenance
[e]
[e]
[e]
[e]
[e]
[ctse/kWh]
We
) · Cref
Cboiler = (125.5 − 0.4 · 1000
˙ th [kW ])0.8462
Cboiler = 0.2436 · (Q
˙e
W
Cconn = (0.09318 − 0.00011 · 1000
) · Cref
˙e
W
Cins = (0.0494 − 0.0047 · ln( 1000
)) · Cref
˙e
W
CG.C. = (0.1232 − 0.0005 · 1000 ) · Cref
˙e
W
Ceng = (0.1211 − 0.000735 · 1000
) · Cref
˙e
W
Cdiv = (0.1403 − 0.0142 · ln( 1000
) · Cref
˙ −0.3081
Cmaint = 8.15 · W
Cturbine
˙ e
W
(0.0503·ln( 1000
)+0.3208)
˙
e
Table 14: Aeroderivative and heavy duty gas turbines
FM_07/ 2000
Cogeneration engines
Fuel
Inv = 800 - 4000 CHF/kWe
Size = 10000 - 100 kWe
E˙ = "ENG e * m˙ fuel ENG * LHV fuel
Tcg " 500°C
Q˙ cg = "ENG t * m˙ fuel ENG * LHV fuel
cg
!
!
Tc
T(K)
!
!
Air
V1
H
V2
T min " 120°C
Tcg
stack
Q˙ c = "ENG t * m˙ fuel ENG * LHV fuel
c
Tc " 80 # 90°C
!
!
!
"ENG e = 30%
S
"ENG t = 30%
c
!
-20
Q(MW)
0
20
40
60
Energy available or excess
"ENG t
gc
= 25%
"ENG total = 85 # 90%
FM_07/ 2000
!
Combined heat and power
Example : Rankine cycle
Carnot : E˙ = Q˙ ∗ (1 −
Heat source
T
Vaporisation
Tvap
)
Tcond
Heat source
Turbine
E
Vapo
Condensation
Pompe
Condens
Heat sink W
Heat sink
H
FM_07/ 2000
Steam network and restricted matches
fuel
H
Process 1
Cooling system
C
Process 2
FM_07/ 2000
Combined heat and power
T
Q1+W
Q2
Q˙1 + W˙ + Q˙ 2 = Q˙ Hotu + W˙
MER
W
Q1
!
Q3
W
Q3-W
Q4
Coldu
Q˙ 3 " W + Q˙ 4 = Q˙ MER
" W˙
FM_07/ 2000
!
Combined heat and power
Q+W
TQ
˙ Hotu + W˙ + Q˙
MER
!
W
Q
Coldu
Q˙ MER
+ Q˙
FM_07/ 2000
!
How to integrate mechanical power production ?
Above the pinch point :
1 thermal kW
=
1mechanical kW
T
F10+W
F9-Q
! 0 !!!
Below the pinch point :
1 cold utility kW
=
1 mechanical kW
F10+(Q+W)
F10
Q+W
Across the pinch point:
Energy penalty
W
F7
F8
F8
F8-Q
Q
F6
Q+W
F9
F9
F7
F7
F6
F6
0
W
F5
F5
F4
F4-Q-W
F3
F2
Q+W
F2-W
F3+Q
F2+Q
Q
F1-W
F1
Q
F4+Q
W
F3-Q-W
! 0 !!!
F5+Q
F1+Q
The constraint limits the flowrate
FM_07/ 2000
Choose the appropriate pressure levels
•
Exploit the self sufficient zones to maximise the
mechanical power production : insert rectangles
T(K) flowrate
Max
650
Tv
Q˙
600
115 bar
550
Max
Expansion
!
! Ratio
# T " 273%
P=$
100 &
58 bar
500
35 bar
450
7,7 bar
400
1,0 bar
1,5 bar
Tc
350
300
4
0
5000
10000
15000
20000
Tv
E˙ Carnot = Q˙ * (1" )
Tc
25000
30000
35000
Q(kW)
!
!
FM_07/ 2000
Estimate the mechanical power production
•
Above the pinch point: A is the area of the integrated
rectangles = "Q#"$
$
'
& T " Tc )
A
E˙ = q˙c & T v
)=T
v
& v " (Tv " Tc ))
" (Tv " Tc )
% #c
( #c
• Below the pinch point A is the area of the integrated
rectangles = "Q#"$
$T #T '
A
E˙ = q˙v" c & v c ) = "c
% Tv (
Tv
FM_07/ 2000
Heat pump and refrigeration
Cold utility (cooling water)
T
CONDENSATION
HIGH TEMPERATURE
E˙ k
!
EXPANSION
VALVE
COMPRESSOR
EVAPORATION
LOW TEMPERATURE
E˙ k
Process stream
Q
!
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Heat pump
T Q
˙ Hotu " (Q˙ + W˙ )
MER
!
W
Q+W
Q
Coldu
Q˙ MER
" Q˙
FM_07/ 2000
!
Heat pump
T
Q˙ Hotu " (W˙ )
0.5 Solvingtheenergyconversionproblemusingmathematicalprogramming
37
MER
T
W
!
Above pinch point
Heat supplement from W
Q+W
Q
Q+W
High temperature recompressed stream
fmvrhigh
W
pinch temperature
fmvrhigh
Low temperature stream : fmvrlow
W
Below pinch point
fmvrlow
Q+W
Q
Coldu
Q˙ MER
+ W˙
Q
Heat below the pinch
Figure16:Mechanicalvapourrecompression
FM_07/ 2000
!
Theequality represents thesituationpresentedonfigure 16where thestreamis partly recompressed,theremainingpart beingcooleddownatlowertemperature. It should benotedthat
theheatloadofthehightemperature streamis higherthantheoneofthecolderonebecause
it includesthemechanicalpowerofcompression. Thetargettemperature ofthehightemperature streamis identicalto theoneofthecolderstream. theadiabatic valvebeingconsidered
afterliquid sub-cooling. Theusefulheatoftherecompressedstreamis therefore lowerthan
Definition des niveaux de recompression
700
Grand composite curve
650
600
550
500
T(K)
450
Emec
400
350
300
250
0
10000
20000
30000
40000
50000
60000
Q(kW)
FM_07/ 2000
Integrating MVR
Above the pinch point:
Wmec = hot utility
T
Below the pinch point :
Wmec = hot utility !!!
F10-W
F9+Q
Q+W
Accross the pinch point:
Wmec => energy savings: W+Q
F10
F10-(Q+W)
F9
F9-(Q+W)
W
F7
F6
! 0 !!!
F8
F8+Q
Q
F7-(Q+W)
F7
F6
F6
F5
F5-Q
Q+W
0
F5
W
F4
F4+Q+W
F3
F3+Q+W
F2
F2+W
F1
F1+W
Q+W
F4-Q
W
Q
F3-Q
Q
! 0 !!!
F2-Q
F1-Q
FM_07/ 2000
Refrigeration
T
36
Contents
340
Process & cooling water
refrigeration
320
Qr+W
Ambient T
T(K)
Qc
W
300
280
Qr
260
Coldu
Q˙ c + Q˙ r + W˙ = Q˙ MER
+ W˙
-2000
-1500
-1000
-500
0
500
1000
Q(kW)
Figure 15: Integrated composite curve of a single stage refrigeration cycle
FM_07/ 2000
0.5.4 Heat pumps
When appropriately placed heat pumping systems should drive heat from below to above the
pinch point. Several types of heat pumping systems may be considered (mechanical vapour
recompression, mechanically driven heat pumps or absorption heat pumps). Table 15 give
some useful thermoeconomical values for the evaluation of the industrial heat pumping systems. The optimal integration is determined by first identifying the streams or the temperature
levels for which heat pumping is envisaged. The simulation of the heat pump system using
process modelling tools defines the hot and cold streams characteristics that are added to the
system integration model and the optimal flowrate for each of the levels will be computed by
solving the MILP problem. When mechanical vapour recompression (MVR) is used, a hot
stream at lower temperature (mvrlow ) is replaced by another hot stream (mvrhigh) with a
higher temperature and a mechanical power consumption Wmvr . From the Grand composite
curve analysis, it may happen that only part of the stream will be recompressed. This situation will be represented by considering as variable the fraction used in the two streams (resp.
fmvrlow and fmvrhigh) and by adding the following constraints (35).
!
Refrigeration cycle integration
Single stage cooling
fmvrhigh + fmvrlow ≤ 1
(35)
C/W
T
Three stages cooling
+ condensation
A
Tamb
B
C/W
C
D
2
E
F
1
2
(a)
B
4
3
B'
D
C/W
7
T
Tamb
7
6
5
D'
C
F
1
2
A
3
B
(f)
5
(e)
4
C
D
2
E
F
1
2
1
8
E
8
4
(c)
A
H
(b)
Two stages cooling
T
Tamb
4
Qr
1
3
6
2
1
Qr
(d)
H
FM_07/ 2000
H
Combined heat and power : important rules
Producing mechanical power
Appropriate placement : Never accross the pinch
Above : 1 thermal kW for 1 mechanical kW
Below : 1 kW cold utility becomes 1 mechanical kW
Use self sufficient zones
Using mechanical power
Take heat below the pinch and sent it back above
Use if:
Hot utility at low temperature
Cold utility at high temperature
Small temperature difference
FM_07/ 2000
Utilities
1100
T(K)
1000
900
800
700
Gas turbine ?
Fuels ?
Gas engines ?
Enriched air ?
Preheating ?
Steam ?
Liquid Fuel ?
Natural gas ?
PSA ?
VSA ?
Membrane ?
Cryogenic ?
Temperature ?
Pressure ?
Turbine ?
600
500
400
300
200
Heat pumps ?
Organic Rankine Cycles ?
Water ?
Refrigerant ?
Pressure ?
Air ?
0
2000
4000
8000
Compressor
?
6000
Compressor ?
Absorption ?
refrigeration ?
10000
12000
Q(kW)
FM_07/ 2000
Energy balance of a temperature interval
Integrating heat producers and consumers
Sharing energy by counter current heat exchange
Excess of energy from the upper intervals
Ri+1
T *i+1
Energy for the hot streams in interval i
fj qji
fj qji
Energy for the cold streams in interval i
T*
i
Ri
Energy to the lower intervals
For each utility stream
fj
Unknown flowrate->
Use YES/NO ? ->
: Continuous variable
y j :Interger variable 1/0
FM_07/ 2000
Optimalisation
Minimum cost of energy requirement (MCER)
Mixed Integer Linear Programming (MILP)
min
Ri, f j, yj
Submit to :
nu
Cost =& C1jyj+ C2jf j
j=1
Heat balance
nc
Ri+1+ & fj qji
j =1
Utility
f min,j yj ' f j
(1)
Hot stream j in interval i
Cold stream j in interval i
n
f
- & f j qji - Ri =0
j =1
(2)
' f max,j yj
2nd principle
Ri! 0; R1=0; Rni+1=0
yj % {0,1}
ideal HEN model
FM_07/ 2000
Combined mechanical power production
Linear constraints
Mechanical power consumption
nu
& w w fw ( w + Wel - Wp ! 0
w=1
Process requirement
(3.1)
Electricity import
Production from the utilities(W w > 0 )
- Export of electricity
Electricity export
nu
& w w fw ( w + Wel - Welv - Wp = 0
w=1
(3.2)
- Operating cost
buy
nu
Cost = & (C1w y w + C2w fw ) + Cel Wel - Celv Welv
Welv w=1
sell
FM_07/ 2000
Use of integer variables
1 integer variable for each flowrate
yj = 0 : the utility j is not used
yj = 1 : the utility j is used
f min,j yj ' f j ' fmax,j yj
if yj =0 => f min,j 0 ' f j ' f max,j 0 j is not selected
=> 0 ' f j ' 0 => f j =0
j is selected
if yj =1 => fmin,j ' f j ' f max,j
Linear constraint
Linear cost :
Costj = C1j yj + C2j f j
if yj =0=> Costj = C1j 0 + C2j 0 =0
if yj =1=> Costj = C1j + C2j f j
j is not selected
j is selected
FM_07/ 2000
(1)
MILP formulation
Operating cost
Fixed maintenance
Subject to
Electricity consumption
Investment
Electricity production
Technology selection
Feasibility
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Targeting the optimal integration : model
• MILP formulation
Gas turbine g : hot stream from ToT to Tstack
unknown
Fuel
Electricity
Part load efficiency
Operating cost
Investments
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Linear cost function : MILP formulation
Cost
C2*f
C2 =
Cost(f max ) − Cost(f min )
f max − f min
C1 = Cost(f min ) − C2f min
C1*y
f
fmin
fmax
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Combustion model
Post combustion
O2 balance
Post comb
Heat above Trad
Air
Fuels
Air
Enriched air
Heat from Trad to stack
Air
Fuels
Enriched air
Preheating
Fuels
Enriched air
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Air preheating : chemical heat pumping
2500
Qcnv
o2=10%
Stoechiometrique
Qrad
Préchauffe de l’air
2000
Qprh
Qprh
1500
Trad
T (°C)
1000
500
0
-10000
0
10000
20000
30000
40000
50000
60000
Q (kJ/kg)
FM_07/ 2000
Gas Turbine : fixed size
T(K)
Qpc
Trad
Fuel
ToPC
!
ToGT
!
Air
P1
H
P2
T
!
"GTe = 30 # 38%
min
stack
-20
"GTt
Q(MW)
0
20
40
60
gc
S
= 55 # 47%
"GT = 85%
FM_07/ 2000
!
Outlet temperature calculation
Stream
Add linear constraints
Compute the temperature a posteriori
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
FM_08/2002
Outlet temperature calculation
500
450
Qi+2
Qi+1 fi+1 To
400
T(K)
fi
Ti+2
Ti+1
Ti
350
300
250
0
200
Laboratory of Industrial EnergySystems
LENI -ISE-STI-EPFL
400
600
Q(kW)
800
1000
FM_08/2002
the integer variable used to select the segment i
yi
Figure
17:17:
Integrated
composite
:boiler,refrigeration
refrigerationand
andcooling
cooling
system
Figure
Integrated
compositecurve
curveofofthe
theheat
heat pump
pump system
system :boiler,
system
The linearization by segments is also applicable for performances indicators of technologies like power and efficiency of a gas turbine. The linear formulation may in this case be
extended to account for piece-wice linearized of non linear functions.
0.5.5 Handling
Handlingnon
nonlinear
linearcost
cost functions
functions
0.5.5
0.5.6 Using range
the exergy losses as anisobjective
function
In the
problem
formulation,linear
linearcost
costare
are needed.
needed. When
covered
In the
problem
formulation,
When the
thewhole
whole rangeofofsizes
sizes is
covered
model,
a piece-wiselinearisation
linearisation technique
technique may
used
totorepresent
non
linear
cost
to the linear
nature
of the problem,
the use
of the energy
cost as an objective function may
by by
thethe
model,
a piece-wise
may be
beDue
used
represent
non
linear
cost
Sdifficulties
b [26]. When the cost of fuel and electricity is such that the electrical
reveals
function.
The
genericinvestment
investmentcost
costfunction
function C(S)
C(S) =
∗∗some
((Sref
bebeapproximated
S ) bwill
function.
The
generic
=C
Cref
)
will
approximated
ref
efficiency ofSaref
cogeneration unit is attractive without the use of heat (i.e. when the electrical
by a set of segments (figure 18) defined by the following set
of constraints
(36) in the linear
W˙ (36) in Cthe(e/kJ)
by•a setpiecewize
of segments (figure
18) defined by the following set
of ofconstraints
linear
linearisation
efficiency
the unit ηel = LHV
˙ is greater than C (e/kJ )) there is an economical interest
optimisation problem.
to produce electricity even without cogeneration). In this case, the linear programming prooptimisation problem.
Non linear cost function
LHV
el
e
el
cedure leads to a situation where the cogeneration unit is used at its maximum. This situation
nsegments
!
nsegments
C(S max ) − C(S max )
C(S max ) − C(S max )
not occur when the investment
cost are properly iconsidered
or when the cost of
i−1
imax usually does
i−1
max
max
max
max
max
Si−1
) ∗of yenergy
∗Sp
C= !
{(C(Smax
) −tomax
C(S
) i }Neverthe) − C(S
)the∗different
i + areC(S
i−1 ) −C(Si−1
i
i−1
i efficiency.
forms
coherent
with respect
the electrical
max
max
max
max
S
−
S
S
−
S
∗
S
)
∗
y
+
∗Spi }
C=
{(C(S
)
−
i−1
imax
i−1
i max
i−1 price of thei different forms
i−1
max
max
i=1
less, the relative
of energy−
will
influence the technology selection
S
−
S
S
S
i−1
i
i−1
i
i=1
nsegments
!
nsegments
S=
Spi
!
S=
Spi
i=1
max
max
yi ∗ Si=1
i−1 ≤ Spi ≤ yi ∗ Si
max
max
yi y∗i S∈i−1
≤ Spi ≤ yi ∗ Si
{0, 1}
yi ∈ {0, 1}
16000
∀i = 1, ..., nsegments
∀i = 1, ..., nsegments
14000
(36)
12000
with
C(S)
cost
ofinthe
equipment
of size S
Simax the
Theinstalled
maximum
size
segment
i
Spi
the size of the equipment in segment i
yi
the integer variable used to select the segment i
Cost
0.5 Solving the energy conversion problem using mathematical programming
10000
with C(S) the installed cost of the equipment of size S
(36)
39
8000
6000
4000
max
S
max
S
i
The linearization by segments is also applicable for performances2000indicators ofi!1technologies like power and efficiency of a gas turbine. The linear formulation may in this case be
extended to account for piece-wice linearized of non linear functions. 0
Sp
0
100
i
200
300
400
500
Size
Laboratory of Industrial EnergySystems
0.5.6LENIUsing
the exergy losses as an objective function
-ISE-STI-EPFL
FM_08/2002
Figure 18: piecewize linearisation of cost function (exponent 0.75)
Due to the linear nature of the problem, the use of the energy cost as an objective function may
reveals some difficulties [26]. When the cost of fuel and electricity is such that the electrical
efficiency of a cogeneration unit is attractive without the use of heat (i.e. when the electrical
(e/kJ)
W˙el
efficiency of the unit ηel = LHV
is greater than CCLHV
)) there is an economical interest
˙
el (e/kJe
to produce electricity even without cogeneration). In this case, the linear programming procedure leads to a situation where the cogeneration unit is used at its maximum. This situation
usually does not occur when the investment cost are properly considered or when the cost of
the different forms of energy are coherent with respect to the electrical efficiency. Neverthe: balanced
hot will
andinfluence
coldthecomposite
curves
less, the relative Results
price of the different
forms of energy
technology selection
T(K)
1100
Multiple pinch points
optimal use of the cheapest utility
16000
1000
14000
900
combustion
800
12000
Cost
700
10000
600
exothermal reactor
8000
500
steam consumption
6000
steam production
400
4000
300
water cooling
2000
200
0
0
max
Air
S cooling
i!1
max
S
i
5000
0
10000 15000 20000
Q(kW)
Spi 200
100
300
25000
400
30000
35000
500
Size
Figure 18: piecewize linearisation of cost function (exponent 0.75)
FM_07/ 2000
Balanced Grand composite curve
T(K)
1100
1000
Understand the results ?
900
800
700
600
500
400
300
200
0
1000
2000
3000
4000
Q(kW)
5000
6000
7000
FM_07/ 2000
Evaluate : the Integrated Composite Curves
Hot and cold streams
Sub-set B : complement
n k n Bw
nB
r=k w=1
i=1
RB k = R ref + & ( & f w q wr + & Q ir )
Sub-set A
n k n Aw
nA
r=k w=1
i=1
RA k = R ref - & ( & f w q wr + & Q ir ) - R n k +1
T
n k n Bw
RBkp = 0 => Rref = - & ( &
r=kp w=1
nB
f w q wr +& Qir )
i=1
Q
FM_07/ 2000
8000
ICC for utility system integration
T(K)
1100
Process
Utility system
1000
Furnace
900
800
700
600
500
400
air cooling
300
200
-2000
0
2000
4000
6000
Q(kW)
water
8000
10000
fridge
12000
14000
FM_07/ 2000
ICC for the integration of the fumes
T(K)
1100
Other systems
Furnace
1000
900
800
700
600
500
400
300
200
-1000
0
1000
2000
3000
Q(kW)
4000
5000
6000
7000
Excess of heat in the fumes
FM_07/ 2000
ICC for refrigeration cycle integration
50
T(K)
340
Process & cooling water
refrigeration
320
"process"
"Fridge system"
1100
1000
T(K)
900
800
300
280
700
600
260
500
-2000
-1500
-1000
-500
0
500
Q(kW)
400
Figure 13: Integrated composite curve of a single stage refrigeration cycle
300
200
-1000
Contents
0
1000
2000 3000 4000 5000
Q(kW)
6000 7000
8000
FM_07/ 2000
ICC of the steam network
T(K)
1100
Other systems
Steam network
1000
900
800
Energy supplement
700
600
500
400
Steam production
Steam cons.
300
200
-1000 0
1000 2000 3000 4000 5000 6000 7000 8000
Q(kW)
Mechanical production
FM_07/ 2000
1000
Heat exchanger placement audit
Systematic drawing of the ICC of the existing heat exchangers
Heat exchange through the pinch point
700
Penality
650
MER
Existing Heat Exchanger
600
550
Rest of the process
cf remaining problem
T(K)
500
450
400
350
300
250
-2000
0
2000
4000
6000
Q(kW)
8000
10000
12000
FM_07/ 2000
Heat exchanger placement audit
Heat exchanger below the pinch point
600
T(K)
MER
550
500
Penalty of heat exchanger
450
400
Heat exchanger
Overall process
350
Remaining
300
250
-1000
Q(kW)
-500
0
500
1000
1500
2000
2500
3000
FM_07/ 2000
Integration of refrigeration cycles
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Content
•
•
•
Problem statement : goals and strategy
Modelling the refrigeration cycle integration
Application to the olefins plant refrigeration cycle
– Targeting the minimum energy requirement of the plant
•
Reducing the refrigeration requirements
– Options for improving the energy efficiency of the refrigeration system
•
Conclusions
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
R&D context and goals
•
•
In the frame of a European R&D project
Goal of the R&D project
– Develop methodology and tools to optimise the integration of energy
technologies in industrial processes
– Refrigeration systems is one of the energy technologies considered
•
Methodology
– Method development for energy technologies integration
•
targeting method for preliminary design in grassroot and process retrofit cases
– Validation with real cases studies
•
Real case study :olefin plant
– Results of a data reconciliation and on-line optimisation software
•
Prof Pierucci , Politecnicdi Milano
– Interface with energy integration software
•
LASSC, University of Liege
– Evaluation of feasibility
•
LASSC & Politecnico
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Problem statement : the context
•
The olefin plant
Furnace 1
Cold Box
H2, C1, C2
110 - 313 K
Furnace 2
300 - 750 K
Furnace 3
Quench section
C3 - C4
separation
Compression
C3 - C4
298 - 1100 K
Fuel Oil
280 - 380 K
288 - 350 K
Furnace 9
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
C5 - C6+
separation
C5 - C6+
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
The energy requirement
•
Compute the composite curves
120 hot and cold streams
From real process data (model)
Heat Requirement
1600
1400
Cracking section
1200
T(K)
1000
800
Quench
600
400
Cooling
200
0
Compression
C3-C4 and C5-C6 separations
Ambient temperature
0
Cold box
Refrigeration Requirement
20000
40000
60000
80000
100000 120000 140000
Q(kW)
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
The refrigeration system
•
•
•
•
•
Multi components
Multi pressure levels
Methane : 1 level
Ethylene : 3 levels
Propylene : 4 levels
Cycle T evaporator (K)
METH1
136
ETH1
171,6
ETH2
199,25
ETH3
212,51
PROP1
233
PROP2
248
PROP3
277
PROP4
291
Complex systems
1
PROPC
PROPC_1
2
Condenser or under cooling
Evaporator
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Integrated process and refrigeration system
1
PROPC
Propylene condenser : 319 K
300
T(K)
ETHC
Prop4
250
ETHC
291 K
Prop3
Prop2
243K
2
277K
Proc_1 : intermediate condenser
248K
Ethc
233K
Prop1
Eth3
212K
200
199K Eth2
180KMethc
279 K
365 K
171K
366 K
METHI
457 kW
Eth1
302 K
Condenser
150
Evaporator
METHP
826 kW
METHC
2016 kW
137K
139 K
136 K
136 K
Meth1
Q(kW)
159 K
METH1
879 kW
100
0
5000
10000 15000
20000 25000
30000
35000 40000
45000
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Goal and strategy
•
Goal : develop a preliminary retrofit step model to target :
–
Optimal integration of process and refrigeration system
–
Optimal flow-rates in the refrigeration system
–
Interactions between refrigeration cycles
–
The minimum mechanical power consumption
–
And to identify the major energy penalty and improvement routes
•
–
•
PROPC_1
Process and Heat exchanger network modification
Quick evaluation of process changes impact (support to the engineer creativity)
Method proposed : EMO : Effect Modelling and Optimisation
–
Based on process data and thermodynamic based models
–
Heat exchanger network modelling using the heat cascade constraints
•
–
HEN modifications allowed
Combined heat (fridge) and power
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
50000
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
EMO for one refrigeration effect : reference flow = flow in the condenser
Condenser : hot stream
c,r
r
r
qcond
= (hout
_ comp " hout _ cond )
Tc
h0
h1
)4
Compression : mechanical power
w
c,r
n sc, r
h2
)3.(1- )4)
c ,r
'%n s "1
')
r
c,r
= & $ ((1" # c ,r (Pt ))* * (hout
_ comp " hsat " vap (Pn cs ,r )))
'( t =1
'+
c, r
% s #1
(
" c,r (Psc ,r )* ' $ (1# " c,r (Pt ))*
& t =1
)
(1-)4)
)2.(1-)3).(1-)4)
s"1
%
)
r
c,r
+ ,# (Ps )* &$ ((1" # c,r (Pt ))* * (hout
_ comp " hsat "vap (Ps )))
s=1
( t =1
+
c ,r
h3
(1-)3).(1-)4)
wPROP_1= (1-!1).(1-!2).(1-!3).(h4 -h0) + !3(1-!1).(1-!2).(h3 -h0)
+ !2(1-!1). (h2-h0) + !1(h1 -h0)
PROP1
Evaporation : cold stream
c,r
eva
q
h4
(1- )2).(1-)3).(1- )4)
r
out _ cond
= (h
c ,r
sat _ vap
"h
Te
c, r
%' n s
)'
c ,r
c ,r
(Pn cs ,r )) = &$ ((1" # c,r (Pt ))* * (hsat
_ liq (Pn sc,r ) " hsat _ vap (Pn sc, r ))
(' t =1
+'
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute
of Technology (Lausanne-CH)
!
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Propylene cycle model
Heat load in the condenser
PROPC
PROPC
Wmec 4
)4
Wmec 4
r
cond
Q
)4
c,r
= " f c ,r * qcond
c =1
Wmec 3
Wmec 3
n rc
) .(1-) )
3
4
(1-) 4)
PROPC
Mechanical power
Wmec 2
PROP 3
1
(1-) 4)
PROPC
r
(1-) 3).(1- )4 )
PROP 2
PROP 4
c =1
PROPC
Wmec 4
PROPC_1
)4
Wmec 2
PROPC_1
Wmec 3
2
PROP 2_1
) .(1-) )
3
4
PROPC_1
Wmec 2
(1-) 4)
)2 .(1-) 3).(1-) 4)
)2
Wmec 2
(1-)3).(1- ) 4)
Wmec 1
Wmec 1
Condenser or under cooling
Evaporator
(1-) 2).(1- )3).(1- ) 4)
(1-) 2)
PROP 1_1
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
n rc
W = " f c,r * w c,r
Wmec 4
PROP 1
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Optimisation model
•
Goal : to compute the optimal flow-rate in each effect
nw
300
Minimise " ( ywC1w + f wC2w ) + Cel * ELi # Celo * ELo
R k ,y w ,f w
250
subject to :
heat balance of the temperature interval k
nw
"f
(P1)
w =1
nr
n cr
nr
200
n cr
n
,r
c,r
+ " " f c,r qccond,k
# " " f c,r qeva,k
+ " Qik + Rk +1 # Rk = 0
q
w wk
w =1
r=1 c =1
r=1 c =1
i=1
$k = 1,...,n k
150
34941 kW
r
nc
Wcr # " f
c,r
*w
c,r
=0
$r = 1,...,nr
100
30000
35000
40000
45000
50000
c =1
c,r
f min y
55000
60000
65000
70000
Q(kW)
c,r
% f
c,r
c,r
c,r
c,r
% f max y ,
y & {0,1}
nw
Electricity production :
"f
r
c
$c = 1,...,n ,$r = 1,...,nr
nr
w
w =1
Wmec 4=117,63 kW
!4
* ww # "Wc r + ELi # ELo = 0
Wmec 3=20,82 kW
!3.(1-!4)
r =1
Wmec 2=41,41 kW
nw
nr
(1-!4)
!2.(1-!3).(1-!4)
consumption: " f w * ww # " Wc r + ELi ' 0
w =1
Wmec1 =21,4 kW
(1-!3).(1-!4)
r=1
f minw y w % f w % f maxw y w , yw& {0,1}
Rk ' 0 $k = 1,...,nk + 1
$w = 1,...,nw
(1-!2).(1-!3).(1-!4)
PROP
1
R1 = 0, Rn k +1 = 0
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
The integrated system : graphical representation of solution
Propylene condenser : 319 K
Prop4
300
Prop3
Prop2
250
1,6
Proc_1 : intermediate condenser
Eth_C
Flow-rate modifications
Prop1
Eth3
1,4
Eth2
Methc
Eth1
1,2
1
Actual
0,8
Optimised
0,6
0,4
Meth1
0
5000
10000
METH1
15000
TOTAL
ETH3
ETH2
ETH1
PROP2_1
PROP1_1
PROP3
100
PROP4
PROP2
0
PROP1
0,2
20000
25000
30000
Q(kW)
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
35000
40000
45000
50000
75000
80000
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
The first results
Actual
Optimized
Simulation
Wmec
Wmec
Wmec
kW
kW
kW
PROP1
17297
12685
-
PROP2
5314
4281
-
PROP3
6598
4759
-
PROP4
1489
377
-
PROP1_1
1520
2029
-
PROP2_1
0
0
32218
24131
ETH1
919
705
-
ETH2
312
463
-
ETH3
1378
655
Total ethylene
2609
1823
2001
746
655
655
0
0
35573
26609
Total propylene
METH1
Off streams contribution
TOTAL
25%
26284
-
28940
8%
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Sensitivity to refrigeration requirements saving
Reducing the refrigeration requirement of 1 unit
reduces the mechanical requirement of :
Cycles
Meth
Eth1
Eth2
Eth3
Prop1
Prop2
Total
135 K
0.833
0.483
0.056
0.014
0.830
0.064
2.281
170 K
0.070
0.621
0.005
0.001
0.906
0.064
1.667
Energy saving temperature
197 K
210 K
0
0
0
0
0.525
0
0
0.484
0.835
0.810
0.059
0.057
1.419
1.351
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
231 K
0
0
0
0
0.750
0
0.750
246 K
0
0
0
0
0.526
0.199
0.725
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Ethane valorisation
T=270 K
P=20 bar
452
Sep
T=205 K
P=3 bar
METH
T=298 K and + ETH1
P=3 bar
ETH2
ETH3
To furnace PROP1
Reheater
PROP2
PROP3
PROP4
Total
C2H6VLV
Evaporator
with C2H6 recov
100%
100%
57%
54%
96%
97%
94.50%
860 kW
Total saving
860 kW (FRG) => 1500 kW (MEC)
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Methane valorisation
132 K
1.35 b
147 K
Preheating
633 kW
347
173 K
6.7 bar
400 K
1.35 b
To furnace
Cooling recovery
Total saving
+ 633 kW mec produced by the turbine
+ 1414 kW mec saved in the refrigeration system
+ 2047 kW mec
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Hydrogen valorisation
Other usage ?
!
if not re-expand
298 K
10bar
130 K
10 bar
100 K
Preheating
320 kW
347
Cooling recovery
173 K
32bar
Total saving
+ 320 kW mec produced by the turbine
+ 764 kW mec saved in the refrigeration system
+ 1084 kW mec
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
After process modifications
Actual
Optimized
Simulation
Off streams
Wmec
Wmec
Wmec
kW
kW
kW
Wmec
kW
PROP1
17297
12685
-
10924
PROP2
5314
4281
-
4114
PROP3
6598
4759
-
4589
PROP4
1489
377
-
357
PROP1_1
1520
2029
-
2061
PROP2_1
Total propylene
0
0
-
24131
ETH1
919
705
-
281
ETH2
312
463
-
248
ETH3
1378
655
-
343
Total ethylene
2609
1823
2001
872
746
655
655
0
0
0
-
-827
METH1
Off streams contribution
TOTAL
35573
26284
0
32218
26609
28940
17%
38%
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
22045
22090
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Interest of the proposed targeting method
•
•
Suitable for retrofit and grassroots problems
Based on the existing process performances
– Offers an overall view of the energy integration including the utility system
•
Identify and quantify energy savings potentials (heat and power)
–
–
–
–
•
Optimal flow-rates in the cycle
Incentive for cycles improvements in terms of costs of energy (and investment)
Graphical representation => better knowledge of the system interactions
Quick evaluation of the process modifications with an overall system view
Study the valorisation of “off-streams”
– Quick evaluation (no heat exchanger network model needed)
– Sensitivity analysis
•
First step of process improvement : Target optimal flowrates
– Use heat load distribution to identify heat exchanger network modifications
– Estimate the investment and the profitability of solutions
– More precise simulation needed
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
A tool for optimal synthesis of industrial refrigeration systems :
Application to an olefins plant
Outcome for the olefin plant test case
•
The refrigeration system integration has been included in an overall
approach
–
Hot and cold utilities
–
Steam network integration also important
•
•
Targeted energy savings for the refrigeration system
–
–
–
Up to 25 % of refrigeration power consumption
New flow-rates targeted
Heat exchanger network modifications
•
•
After heat load distribution and heat exchanger network modelling
Valorise “off-streams”
–
–
–
•
gas turbine, partial oxidation gas turbine,…
Exergy valorisation
Expansions + refrigeration effects
12 % of additional energy savings
HEN evaluation : preliminary step
–
–
Structure modifications
Requires further optimization to estimate the investment
Laboratory of Industrial Energy Systems - LENI - Swiss Federal Institute of Technology (Lausanne-CH)
Integration of Organic Rankine Cycles
Why integrating ORC systems ?
Transform low grade energy into mechanical power
– Geothermal
• hot source up to 150°C, size : 300 kW to 2.5 MW
– Renewables
• Biogas, Biomass burning, solar energy (hybrid systems)
• Up to 300°C efficiency up to 20%
• Sizes from 5kW (scroll) to 1MW (turbo-expander)
– Use in industrial processes
• Valorise low quality process / utilities waste heat
• Typical pinch point location = 100°C to 150°C
2nd International Symposium on
process integration, Halifax (CA) 2001
What are the questions ?
• Process requirement
* Composite curves
H
H
C
H
Tin
773.00
373.00
523.00
493.00
Tout
393.00
373.00
523.00
493.00
•
•
Heat load
100.00
120.00
100.00
100.00
Grand composite curves
800
• Evaporation pressure
• Condensing pressure
• Superheating temperature
Grand composite curve
750
700
– What are the technologies ?
– Which process/cycle configuration ?
650
600
• Multiple cycles ?
• Process/cycle integration ?
550
500
– What are the competing technologies ?
450
• Heat pumps ?
• Steam Rankine cycle ?
• District heating ?
400
350
300
0
50
100
150
200
Q(kW)
2nd International Symposium on
process integration, Halifax (CA) 2001
Exergy Grand Composite Curve
0.7
0.6
0.5
Carnot factor (-)
T(K)
Should I integrate an ORC to my process ?
If yes
– What is(are) the fluid(s) ?
– What are the best operating conditions ?
0.4
0.3
0.2
0.1
0
0
50
100
150
200
Q(kW)
2nd International Symposium on
process integration, Halifax (CA) 2001
250
300
STEP1 : Identify the temperature levels
Minimum heat transfer
exergy losses
Utilities = qk at Tk cst
Limit the number of levels
!
qk " 0
2nd International Symposium on
process integration, Halifax (CA) 2001
STEP1 : Results of the levels identification
•
Temperature levels analysis
800
system
Steam level
750
• Combinatorial
• Use of bounds
• Distinction between close
levels
Combustion is integrated
700
650
600
– High precision not required
Steam levels ?
550
T(K)
Difficulties
– Integer variables
• DTmin assumption
500
– “Intelligent” results evaluation
algorithm
450
Heat pump ?
400
ORC ?
350
300
0
50
100
150
200
Q(kW)
2nd International Symposium on
process integration, Halifax (CA) 2001
STEP 2 : selection
ICC of the ORC with refrigerant BZ
tb=353K
•
Benzene
For each couple : ((Tvap,Qvap),(Tcnd,Qcnd))
ORC cycle for temperatures between 486.0 K and 298.1 Next T = 463.5(K)
Heat load :167.2 kW(vaporisation: 116.340 kW - preheating: 50.914 kW)
800
ORC
Frame
Wmec
process
– For each fluid in the list
700
• Compute Pvapr&Pcndr
600
• Compute wmecr,qvapr, qliqr => flowr
! intersection with requirement curve
"
T(K)
• Compute Tsuperheatingr
– Vapour fraction > 0.85
– Wmecr*flowr increases
500
Organic Rankine cycle :OZ
Fluid– < Max temperature
:BZ
tb=353K
Benzene
Mechanical power : 31.9451
kW
Heat load
:
213.22
kW (eff : 15.0%)
Superheating temp : 486.0 (K) and 17.41 (bar)
Vaporisation temp : 486.0 (K)
Condensation temp : 353.2 (K) and 1.00 (bar) vapf 100.0
(%)
Reference Flowrate: 0.495328E-02 kmol/s
Inlet volumic flow: 0.862047E-02 m3/s (VR= 19.08)
400
300
0
50
100
150
200
Q(kW)
– Select the n bests
2nd International Symposium on
process integration, Halifax (CA) 2001
STEP 2 : Results
ORC requirement between
357.77
(K) and
298.15
(K)
45
40
• List of possible fluids
– Heat transfer coefficient => Dtmin/2
• Vaporisation temperature
– Compute the pressure
• Superheating temperature
• Condensation temperature
• Flowrate reference
• Efficiency
35
25
20
15
10
PP2
BZ
tb=415K perfluorodecaline_c10f18
CA
2nd International Symposium on
process integration, Halifax (CA) 2001
tb=383K Toluene
360
TOL
340
tb=349.15K c7f14
tb=353K
Benzene
320
tb=330K c6f14
300
Tboil(K)
PP1
280
tb=298.98K pentafluoropropane
R123PP50 tb=301K
c2H1cl2f3
tb=302K
c5f12
Tb=282.05K DichloroFluoro-methane
R160
Tb=285.35K Chloro-ethane
260
69.15K BromoChloroDifluoro-Methane
R600
Tb=273.1K N-butane
276.65K DichloroTetrafluoro-ethane
R764
Tb=263.14K
Sulfur dioxide
Tb=263.9K
Chlorodifluoro-ethane
240
R134A
0
220
tb=247. c2h2f4
5
Tb=232.35K ChloroDifluoro-methane
WTOT(kW)
30
STEP 3 : Cycles calculation
• For each cycle and its reference flowrate
ORC composite curves
440
Ts
Qbr (151.3 kW)
Cost_boilerr=a*(Qbr)b
=> C1br + xr * C2br
420
400
1)Type selection
scroll -turbo => eff,nstages
2) Compute Wr (22.46 kW)
3) Cost_turbiner=at*(Wr)bt
T(K)
380
Tv
360
=> C1br + xr * C2br
340
Linearising cost estimation functions
90
320
Tc
300
cost estimate
linearised
80
70
60
50
280
0
50
100
Wpr (0.8 kW)
Cost_pumpr
40
Qcr (134.71 kW)
150
200
Cost_condr=ac*(Qcr)bc
Q(kW)
=> C1cr + xr * C2cr
250
30
300
20
10
0
0
50
100
150
200
2nd International Symposium on
process integration, Halifax (CA) 2001
Integration of an ORC
FROM the hot source : Qh
Grand composite curve of an ORC
500
bz
Mech. power
480
• The composite has to “fit”
the process
Grand Composite curve
460
T(K)
440
• Qh “Free” waste energy
= W/Qh = 17.2 %
= 23.72/137.7
(
420
400
Liquid preheating
By vapor desuperheating
380
360
-20
0
W
20
40
60
Q(kW)
80
100
120
To the cold source
140
Heat load saving
-> Greater production
*+9%
2nd International Symposium on
process integration, Halifax (CA) 2001
STEP 3 : MILP superstructure model for selection
nw
Minimise " ( ywCo1w + f wCo2w ) + Cel * ELi # Celo * ELo
R k ,y w ,f w
(P2)
Objective function = total cost
!
Including competing
!
technologies
w =1
nr % yr * C1w + W * C2w
)/
1 , nw
'( r
r
r
r ) + ( yrr * C1c r + Qcr *C2c r ) '
+ * . " ( y wC1w + f wC2 w ) + " &
*1
$ .- w=1
r =1 (
'+ ( yrr *C1br + Qbr * C2br ) + ( yrr * C1pr + Wpr *C2pr )+'10
subject to :
heat balance of the temperature interval k
nw
"f
w =1
nr
n
r=1
i=1
q + " (qc k,r # qbk,r ) * f r + " Qik + Rk +1 # Rk = 0
Heat cascade constraint
2k = 1,...,nk
w wk
nk
Qc r # " (qc k,r ) * f r = 0
2r = 1,...,nr
k =1
nk
Qbr # " (qbk ,r ) * f r = 0
2r = 1,...,nr
k =1
ORC sizing
4
%
,
3 isr r
/)
'
4
.5 Pv 8 1#4 r
1'
Wr # &3isr * r R *Tsr * .7 r :
# 11* * f r = 0
1# 4 r
Pc
6
9
'
j
.10'
(
+
v *(Pv # Pc )
Wpr # r
* fr = 0
3 pr
2r = 1,...,nr
2r = 1,...,nr
f rmin yrr ; f r ; f rmax yrr
ORC selection
2r = 1,...,nr
nw
nr
ELo
Electricity production : " f w * ww + " (W r # Wpr ) + 3 i * ELi #
=0
3o
w =1
r=1
nw
nr
w =1
r=1
Electricity balances
Electricityconsumption : " f w * ww + " (Wr # Wpr ) + 3 i * ELi < 0
f minw y w ; f w ; f maxw y w , yw= {0,1}
Rk < 0 2k = 1,...,nk + 1
Other technologies selection
2w = 1,...,n w
R1 = 0,Rn k +1 = 0
2nd International Symposium on
process integration, Halifax (CA) 2001
Linearising cost estimation functions
90
cost estimate
linearised
Cost
80
70
60
50
40
30
20
10
0
0
fmin
fmax
50
100
150
2nd International Symposium on
process integration, Halifax (CA) 2001
Size
200
ORC level analysis
500
system
ORC levels
450
Profile 1
T(K)
400
350
Profile 2
300
0
50
100
150
200
250
300
Q(kW)
2nd International Symposium on
process integration, Halifax (CA) 2001
STEP 3 : Difficulties/Limitations
•
•
•
Solving integer programming
– Consistent bounds set !
Linear model
– linearised cost
– Estimated A
• Flowrate estimation
– reference
• + DTmin assumption
• Integer cuts
– Multiple solutions
– Multi objective (linear)
Multiple alternatives
– Multi criteria selection
• Break even
•
Heat Exchanger Network
– Structure unknown
– Cost not considered
• List of streams complete
– Automated or Pinch design
method
– Complete system !
2nd International Symposium on
process integration, Halifax (CA) 2001
Result of selection : report & evaluation
• Solution evaluation
ORC integrated composite
1200
1100
1000
900
T(K)
800
MER : Hot =
36.8 kW Tpinch = 255 C
Cold = 256.8 kW
Fuel :
41.76 kW (-> 88%)
curve (cycle cycle integration)
Air preheating :
0.39 kW (-> 51C)
Cooling water
:
213.13 kW
Others
(ORC)
-------------------------------------Mech. power
Heat(kW)
Wmec(kW)
ORC 1 c6h6 :
157.6
21.6
ORC 2 r134a:
237.0
22.4
-------------------------------------Total :
44.0 (17%)
700
600
500
C6H6
400
R134a
300
200
-50
0
50
100
150
200
250
300
Q(kW)
2nd International Symposium on
process integration, Halifax (CA) 2001
Conclusion
•
ORC allow to valorise industrial process waste heat
– Efficiency of 7 to 17 % can be reached (according to process)
– At the price of the system investment
•
3 steps method for ORC integration into industrial processes
– Minimum exergy losses for temperature levels
– Thermodynamic based algorithm for fluid and T
– MILP for ORC integration wi a holistic approach
• Represent the competition with other technologies
• Multi criteria evaluation -Multi-solution generation
•
"
Preliminary step before
(super-)structures thermo-economic optimisation
– Time varying demand
– Non linear optimisation required
2nd International Symposium on
process integration, Halifax (CA) 2001
Integration of steam network
0.7
Example
51
Site scale integration
T
T
T
Total site representation
Process 1
Process 2
Energy saving through process/process integration
ignoring pockets
with pockets
FM_07/ 2000
Process 2
0.7
Example
51
Process 2
Process 1
site scale integration
Energy saving through process/process integration
ignoring pockets
T
T
T
with pockets
Total site representation
Process 2
Process 1
Process 2
Process
Energy
saving1through process/process integration
ignoring pockets
f lu
eg
ase
s
FM_07/ 2000
with pockets
fuel
H
Role of the steam network
Process 2
fuel
C
Cooling system
f lu
eg
ase
s
Process 1
H
Process 2
Process 1
4: Total site integration and steam network
C
Cooling system
Process 1
Process 2
FM_07/ 2000
Figure 14: Total site integration and steam network
Steam network model
Constant T
preheating ?
Superheating ?
What about using a gas turbine ?
What about CHP ?
From boiler
HP
T
Source profile
MP
Sink profile
LP
Q
FM_07/ 2000
Steam network
fuel
H
Process 1
Cooling system
C
Process 2
FM_07/ 2000
Integrated Combined Heat and
Power
Balanced Grand Composite Curve
T(K)
1000
Choose the pressure levels
from Tv and Tc
Estimate the CHP potential
900
using Carnot cycle approximation
HT diagram of steam production ?
800
700
600
W =Ar ea *
80b
VHP
1
Tv
( - ( T v - Tc)
c
500
257
kJ
400
27.5b
266
kJ
LP
3.3b
HP
148
kJ
MP
90
kJ
8.5b
VMP
14.5b
300
200
100
0
Q(kJ)
W(kJ
)
400
0
2000
4000
6000
HT diagram of condensation ?
FM_07/ 2000
From the list of steam levels
Generate a steam network superstructure
With fixed steam qualities
Pi, Ti => hi
1
Effects
• Hot streams
• Cold streams
• Mechanical power
Modelling
Heat and mass balance
Optimisation
Minimum Cost Target
Df14
A1
L1 DT12
DT13 Dc14
A2
2
L2 DT
Dc24
23
A3
3
4
S1
Df24
S2
Df34
Dc34
S3
Ac4
Sc4
Vapor
Condensate
FM_07/ 2000
Effects in the steam network
j-1
wij = min{ (hi - hj),
! wk(k+1) }
k=i
1
wk(k+1) = ! {hk - his(Pk+1,s(Pk,hk))}
where
hk = h(k-1) -w(k-1)k if k = i+1,...,nv
and
hk = hi
if k=i
Df14
A1
L1 DT12
DT13 Dc14
A2
2
L2 DT23
Dc24
A3
3
Pi, Ti => hi
Hot stream
700
600
T(K)
De-superheating
550
S3
500
Vaporising
450
350
Preheating
300
Sc4
250
-2000
Condensing
450
Superheating
600
S2
Df34
Dc34
Cold stream
650
Ac4
4
500
700
Df24
400
650
550
S1
T(K)
Mechanical power
0
2000
4000 6000
Q(kW)
8000
10000
400
350
Undercooling
300
250
-2000
0
2000
4000 6000
Q(kW)
8000
10000
12000
FM_07/ 2000
The steam network model
Steam header model
nco
j-1
rj-1 Lj-1 + Aj + ! kij * DTij + ! Dfkj
i=1
k=1
nv
nco
- ! DTji - ! Dcjk - Sj - Lj = 0
i=j+1
k=1
Where :
hi - wij - hkj
kij =
hj - hkj
hj-1 - hkj
rj-1 = (1+
)
hj - hkj
1
4
DT13 Dc14
L1 DT12
A2
2
3
Df14
A1
L2 DT23
S1
Dc24
Pi, Ti => hi
Df24
S2
Df34
A3
Dc34
S3
Ac4
Sc4
Condensate header model
nv
nco
k-1
Drk-1+Ak + !Dcjk + ! Dfuk - !Dfki
j=1
u=k+1
i=1
nv nv
nv-1
- Sk - Drk - ! !{(kij - 1)*DTij} - !{(rj - 1) * Lj}= 0
i=1 j=i+1
j=1
FM_07/ 2000
12000
Targeting the optimal integration
Minimise
nw
! ( C1w yw + C2w fw ) + Celin Welin - Celex Welex +
w=1
nco+nv
nco+nv
! (C1ai yai + C2ai Ai) + ! (C1si ysi - C2si Si)
i=1
i=1
Using Rk,yw,fw, Ai, yai, ysi, Si, Li, DTij, Dcij, ycij, Dfij, yfij
Subject to
nw
nco nv
nv nco
n
!fw qwk + ! !Dcij qcijk + ! !Dfij qfijk + ! Qik +Rk+1- Rk = 0
w=1
i=1 j=1
i=1 j=1
i=1
k=1,...,nk
fminw yw " fw " fmaxw , fmincij ycij " Dcij " fmaxcij ycij , fminfij yfij " Dfij " fmaxfij yfij
yw {0,1}
Rk # 0
k=1,...,nk+1
R1 = 0, Rnk+1 = 0
nv-1 nv
nco nv+nco
1
DTij wij - !
!
m Dfki *wpki
a
i=1 j=i+1
k=1 i=1
nv-1 nv
nco nv+nco
1
Export ! !
DTij wij - !
!
m Dfki *wpki
a
i=1 j=i+1
k=1 i=1
Import
!
!
+
nw
! fw ww + Welin - Welex =0
w=1
nw
+
! fw ww + Welin # 0
w=1
+ Model of steam network
FM_07/ 2000
Steam network integration
Integrated composite curves of the steam network
1100
T(K)
1000
900
800
700
600
90b
500
30b
11b
6b
400
300
200
-1000
0
1000
2000
127 kW
3000
Q(kW)
4000
5000
6000
7000
436 kW
FM_07/ 2000
Restricted matches : our goals
• Targeting the Minimum Cost Energy Requirement
taking into account the restricted matches
– Energy penalty of the constraints ?
– Cost of the energy penalty ?
– Solve site scale problems
– Find technological solutions to minimise the cost
penalty of the restricted matches
• choice of the heat transfer fluids
• Before the HEN synthesis task
– Start HEN synthesis with the complete list of streams
FM_07/ 2000
The heat cascade a LP formulation
mini m i s e
Rk
Rnk+1
Heat balance of a temperature interval
nc
nh
Rk+1 + ! fi qik - Rk - ! fj qjk = 0
j=1
i=1
Rk " 0
! ( R*k,Tk)
k=1,...nk
k=1,...nk+1
the MER heat cascade
( G r and composite curve)
FM_07/ 2000
The constraints
Accepted Matches
Accepted Matches
New variables
Heat balance of a temperature interval
nc
nh
Rk+1 + ! fi qik - Rk - ! fj qjk = 0
k=1,...nk
j=1
i=1
Heat balance of a hot stream i in a restricted match
ncai
k=1,...,ki , i =1,...,nh
! Qijk + Rik - fi qik - Rik+1 = 0
j=1
Heat cascade of the cold stream j
nh
k=1,...,nk, j=1,...,nc
! Qijk " fj qjk
i=1
Overall heat cascade
nh
k=1,...,nk
! Rik " Rk
i=1
Rik ! 0 k=1,..., ki, i =1,...,nh ;
Rik+1 = 0 when k ! ki ;
Qijk ! 0 k=1,..., nk, i =1,...,nh,
(1)
(R1)
(R2)
(R3)
j=1,...,nc
FM_07/ 2000
Compute the energy penalty
• Target the Minimum Cost Energy Requirement
without constraints
• Optimal heat cascade : R*k
• Fix the flowrates
• Add the restricted matches constraints
• Solve the problem (P1)
minimise
Rnk+1
Rik , Qijk , Rk
R nk+1 - R* nk+1 is the energy penalty
• if penalty is acceptable goto HEN Synthesis
• if not choose the heat transfer fluid
FM_07/ 2000
Choose the heat transfer fluid
nh
Rok =& Rik
i=1
Restricted matches cascade
Energy
MER
penalty
4807
kJ
6810 kJ
Grand composite curve
800
700
R hot
5250 kJ
600
Heat sink
nh
&
Rik = Rk
i=1
500
Process pinch point
400
Heat source
300
200
-6000
-4000
Rcold
15600kJ 2000
-2000
4000
6000
8000
10000
12000
FM_07/ 2000
Heat transfer fluid characteristics
Process : Hot streams
- - - Heat transfer fluid : cold stream
- - - Heat transfer fluid hot stream
Process : Cold streams
Conditions to be satisfied by the heat transfer fluid
1)
2)
3)
All the Rk must be positive (definition of the MER);
Rnk+1 = R *nk+1 (no energy penalty is due to the use of the
intermediate stream);
Rok = Rk for all k =1,...,n
k (the intermediate fluids solve the
restricted matches constraints)
FM_07/ 2000
Choose the heat transfer fluid
Above the pinch point
900
T(K)
Step 1
Step 2
Process GCC
MER
4807 kJ
RMC
Rhot
5250 kJ
800
Step 3
New GCC
MER
10057 kJ
700
Rhot
5250 kJ
600
Heat from hot stream
500
400
Heat to cold stream
300
200
Add cold
stream
Cold stream
0
2000 4000
0
2000 4000
Hot stream
Q(kJ)
0
2000 4000
6000
8000
FM_07/ 2000
Choose the heat transfer fluid
Hot streams
Cold stream s
540
T(K)
520
Rcold
1560 kJ
Rhot
5250 kJ
27.5 b
500
480
12
460
b
440
6
420
b
400
380
1.5
b
Energy penalty if
1.5 b steam is used
360
-2000
-1000
0
100 0
200 0
Q(kJ)
300 0
400 0
500 0
600 0
Place the heat transfer fluid between Red and Blue lines
FM_07/ 2000
Steam network and restricted matches
fuel
H
Process 1
Cooling system
C
Process 2
FM_07/ 2000
Target the minimum cost of energy requirement
• Add the heat transfer fluids
– Hot and cold streams
– Unknown flowrate
• Use the MILP model
– Heat cascade
– Mechanical power balance
– EMO models
• Energy technologies
• Heat transfer fluids or steam network
– Restricted matches constraints
• Objective function : Minimum Cost of Energy
Fuels - Electricity - CHP - Investments
FM_07/ 2000
Targeting the Minimum Cost Energy Requirement
Minimise
nw
! ( C1w yw + C2w fw ) + Celin Welin - Celex Welex
w=1
Rk,yw,fw
subject to
nw
n
!fw qwk + ! Qik +Rk+1- Rk = 0
w=1
i=1
k=1,...,nk
fminw yw " fw " fmaxw yw
w=1,...,nw
yw {0,1}
Rk # 0
R1 = 0, Rnk+1 = 0
k=1,...,nk+1
Import
Export
nw
! fw ww + Welin - Welex=0
w=1
nw
! fw ww + Welin # 0
w=1
FM_07/ 2000
Evaluate the results
Energy supplement
Integrated composite curves
1100
T(K)
1000
Others
Steam network
900
800
Heat transfer
700
600
Mechanical power
500
400
300
200
1000
Penalty
2000
3000
Q(kW)
4000
5000
6000
FM_07/ 2000
Example
System with constraints
MER :
4143,0
kW
With steam network:
4560,0
kW
CHP :
417,0
kWe (! = 100 %)
System with constraints
MER :
Energy penalty :
10017,0
kW
5874,0
kW
(142 % MER)
System with constraints and steam network
Steam network without modif.
CHP :
6557,0
755,0
Energy penalty :
1659,0
Adapted steam network
4658,0
CHP :
Energy penalty
418,0
97,0
kW
kWe (! = 31 %)
kW (40 % MER)
kW
kWe
(! = 81 %)
kW (2,3 % MER)
FM_07/ 2000