Hydrocarbon Processing - July 2021 - 68

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R = R + 0.03324 RJ - R× J⎦⎤ + 0.00085 O2
(
⎣⎡
RONBlend= R + 0.03324 RS - R× S)
where:
R = 91( )
(
X1
XT
S= 91−81)
+ 89.5(
X1
XT
( 97−96)
RS= 91−81)
(
( 97−96)
(
)
X2
XT
+ 97(
+ 89.5−85.5)
X3
XT
X1
XT
X3
XT
+ 93−92)
(
(
)
X3
XT
X2
XT
X4
XT
+ 89.5−85.5)89.5
+ 93−92)
(
X1 ≥ 0, X2 ≥ 0, X3 ≥ 0 and X4
X4
XT
≥ 0
An additional constraint is shown in Eq. 57:
X1
+ X2 + X3 + X4 < 15,000 barrels
(56)
(57)
The constraint equation for the blend RVP can be obtained
by calculating the RVP index from Eq. 58:
BIRVPi
where:
BIRVPi
= RVPi
1.25
(58)
is the RVP blending index for component i and RVPi
is the RVP of component i in psi. Using the index, the RVP of a
blend is (Eqs. 59-61):
n
BIRVPi ,Blend= xvi
i =1
12.11.25 X1
XT
∑ BIRVPi
+ 15.51.25 X2
XT
51.61.25 X4
XT
or
22.56
X1
XT
138.3
+ 30.755
X4
XT
X2
XT
− 22.33
+ 7.105
XT
XT
=0
X3
XT
+
(61)
+ 4.81.25 X3
XT
− 12.01.25 XT
XT
=0
O3 =
+
(60)
O2 =
x12N1+ x22N2+ x32N3+ x42N4
x12+ x22+ x32+ x42
x13N1+ x23N2+ x33N3+ x43N4
x13+ x23+ x33+ x43
(63)
(64)
Develop a linear programming model to determine how
much of each base fuel should be blended into gasoline and
how much should be sold " as is " to maximize profit. Solve the
model using numerical data in TABLE 4 and below:
Maximize y = P1
(x11
Component
LSR gasoline
Hydrocracker gasoline
FCC gasoline
n-Butane
Blend
TABLE 3. Components properties for the blend
Barrels RVP, psi
X1
X2
X3
X4
X5
12.1
15.5
4.8
51.6
12
68 JULY 2021 | HydrocarbonProcessing.com
+ P3 (x13
+ (S2
RON
91
89.5
97
93
Max
MON
81
85.5
96
92
+ (S4
(42C3
/42) (42C2
- x31
/42) (42C4
Subject to ∑ xij
+ x23 + x33 + x43) + (S1
)
- x32
≥ 42Dj
∑ xij Ni x11N1 + x21N2 + x31N3 + x41N4 ≥ R1 (x11
x12N1 + x22N2 + x32N3 + x42N4 ≥ R2 (x12
x13N1 + x23N2 + x33N3 + x43N4 ≥ R3 (x13
where xij
+ x22
+ x23
+ x21
+ x32
+ x33
+ x42
+ x43
= gallons of base fuel i used in gasoline j.
+ x31
)
)
+ x41
)
- x33
- x41
- x42
+ x21 + x31 + x41) + P2
- x21 - x22 - x23) + (S3
)
- x43
(x12
/42) (42C1
+ x22
- x11
/42)
+ x32
- x12
- x13
+ x42
)
)
(59)
O1 =
X2
XT
+
+
(54)
(55)
+ 93(
)
X4
XT
(53)
⎢⎣
⎡( )- O2
⎤
⎥⎦
(51)
(52)
Using the solver in an Excel spreadsheet to maximize RON,
the objective function of Eq. 52 with the above constraints
(Eqs. 56, 57 and 60), the results are:
RON = 96.54 and X1
X4 = 1,741 bbl.
= 0, X2 = 0, X3 = 13,259 and
If a constraint for FCC gasoline (6,000 barrels) is added, the
results for the maximum blend storage would be:
RON= 93.23 and X1
X4 = 145 bbl.
Case study.9,10
An oil refinery produces a stream of hydro,
C2, C3 and C4
carbon-based fuel from each of four different processing units.
The processing units have capacities of C1
bpd,
respectively. A portion of each base fuel is fed to a central blending
station where the base fuels are mixed into three grades of
gasoline. The remainder of each base fuel is sold " as is " at the
refinery. Storage facilities are available at the blending station
for temporary storage of the blended gasolines, if required.
Let N1
, N2, N3 and N4
base fuels, and let S1
their direct sale. Let O1
, S2, S3 and S4
, O2 and O3
be the octane ratings of the respective
be the profit derived from
be the octane numbers of
, D2 and D3
. Let R1, R2 and
be the profit per gallon
the three grades of gasoline that must be blended on any given
day to satisfy customer demands D1
R3 be the minimum octane requirements of the three grades of
blended gasoline, and let P1
, P2 and P3
that is realized from the sale of each grade of blended gasoline.
Finally, let xij
be the quantity of the ith base fuel used to blend
the jth gasoline.
The octane number of each gasoline can be expressed as
the weighted average of the octane numbers of the constituent
base fuels. The weighting factors are the fractions of the base
fuels in each gasoline. Therefore (Eqs. 62-64):
x11N1+ x21N2+ x31N3+ x41N4
x11+ x21+ x31+ x41
(62)
= 0, X2 = 8,855, X3 = 6,000 and
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