Repository navigation
Expand file tree
/
Copy pathprocess.py
More file actions
201 lines (181 loc) · 5.4 KB
/
Copy pathprocess.py
File metadata and controls
201 lines (181 loc) · 5.4 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
"""
## GAMSSOURCE: https://www.gams.com/latest/gamslib_ml/libhtml/gamslib_process.html
## LICENSETYPE: Demo
## MODELTYPE: NLP
## KEYWORDS: nonlinear programming, alkylation process, chemical engineering
Alkylation Process Optimization (PROCESS)
Optimization of a alkylation process.
Bracken, J, and McCormick, G P, Chapter 4. In Selected Applications
of Nonlinear Programming. John Wiley and Sons, New York, 1968.
"""
from __future__ import annotations
from gamspy import Container, Equation, Model, Problem, Sense, Variable
from gamspy.math import sqr
def main():
m = Container()
# Variables
olefin = Variable(
m, name="olefin", type="positive", description="olefin feed (bpd)"
)
isor = Variable(
m, name="isor", type="positive", description="isobutane recycle (bpd)"
)
acid = Variable(
m,
name="acid",
type="positive",
description="acid addition rate (1000lb per day)",
)
alkylate = Variable(
m, name="alkylate", type="positive", description="alkylate yield (bpd)"
)
isom = Variable(
m, name="isom", type="positive", description="isobutane makeup (bpd)"
)
strength = Variable(
m,
name="strength",
type="positive",
description="acid strength (weight pct)",
)
octane = Variable(
m, name="octane", type="positive", description="motor octane number"
)
ratio = Variable(
m,
name="ratio",
type="positive",
description="isobutane makeup to olefin ratio",
)
dilute = Variable(
m, name="dilute", type="positive", description="acid dilution factor"
)
f4 = Variable(
m, name="f4", type="positive", description="f-4 performance number"
)
profit = Variable(m, name="profit")
rangey = Variable(m, name="rangey")
rangem = Variable(m, name="rangem")
ranged = Variable(m, name="ranged")
rangef = Variable(m, name="rangef")
# Equations
yield1 = Equation(
m, name="yield1", description="alkylate yield definition"
)
rngyield = Equation(
m, name="rngyield", description="ranged alkylate yield definition"
)
makeup = Equation(
m, name="makeup", description="isobutane makeup definition"
)
sdef = Equation(m, name="sdef", description="acid strength definition")
motor = Equation(m, name="motor", description="motor octane number")
rngmotor = Equation(
m, name="rngmotor", description="ranged motor octane number"
)
drat = Equation(m, name="drat", description="isobutane to olefin ratio")
ddil = Equation(m, name="ddil", description="dilution definition")
rngddil = Equation(
m, name="rngddil", description="ranged dilution definition"
)
df4 = Equation(m, name="df4", description="f-4 definition")
rngdf4 = Equation(m, name="rngdf4", description="ranged f-4 definition")
dprofit = Equation(m, name="dprofit", description="profit definition")
yield1[...] = alkylate == olefin * (
1.12 + 0.13167 * ratio - 0.00667 * sqr(ratio)
)
makeup[...] = alkylate == olefin + isom - 0.22 * alkylate
sdef[...] = acid == alkylate * dilute * strength / (98 - strength) / 1000
motor[...] = octane == 86.35 + 1.098 * ratio - 0.038 * sqr(
ratio
) - 0.325 * (89 - strength)
drat[...] = ratio == (isor + isom) / olefin
ddil[...] = dilute == 35.82 - 0.222 * f4
df4[...] = f4 == -133 + 3 * octane
dprofit[...] = (
profit
== 0.063 * alkylate * octane
- 5.04 * olefin
- 0.035 * isor
- 10 * acid
- 3.36 * isom
)
rngyield[...] = rangey * alkylate == olefin * (
1.12 + 0.13167 * ratio - 0.00667 * sqr(ratio)
)
rngmotor[...] = rangem * octane == 86.35 + 1.098 * ratio - 0.038 * sqr(
ratio
) - 0.325 * (89 - strength)
rngddil[...] = ranged * dilute == 35.82 - 0.222 * f4
rngdf4[...] = rangef * f4 == -133 + 3 * octane
# Define Models
process = Model(
m,
name="process",
equations=[yield1, makeup, sdef, motor, drat, ddil, df4, dprofit],
problem=Problem.NLP,
sense=Sense.MAX,
objective=profit,
)
rproc = Model(
m,
name="rproc",
equations=[
rngyield,
makeup,
sdef,
rngmotor,
drat,
rngddil,
rngdf4,
dprofit,
],
problem=Problem.NLP,
sense=Sense.MAX,
objective=profit,
)
rangey.lo = 0.9
rangey.up = 1.1
rangey.l = 1
rangem.lo = 0.9
rangem.up = 1.1
rangem.l = 1
ranged.lo = 0.9
ranged.up = 1.1
ranged.l = 1
rangef.lo = 0.9
rangef.up = 1.1
rangef.l = 1
strength.lo = 85
strength.up = 93
octane.lo = 90
octane.up = 95
ratio.lo = 3
ratio.up = 12
dilute.lo = 1.2
dilute.up = 4
f4.lo = 145
f4.up = 162
olefin.lo = 10
olefin.up = 2000
isor.up = 16000
acid.up = 120
alkylate.up = 5000
isom.up = 2000
olefin.l = 1745
isor.l = 12000
acid.l = 110
alkylate.l = 3048
isom.l = 1974
strength.l = 89.2
octane.l = 92.8
ratio.l = 8
dilute.l = 3.6
f4.l = 145
profit.l = 872
process.solve()
print(f"Profit in model 'process': {profit.records.level[0]:.2f}")
rproc.solve()
print(f"Profit in model 'rproc': {profit.records.level[0]:.2f}")
if __name__ == "__main__":
main()