Fahim M.A., Sahhaf T.A., Elkilani A.S. Fundamentals of Petroleum Refining
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(a) Calculate the RON and MON for the blend using equation (9.22)
and (9.23), then compare with the reported experimental value.
(b) Calculate the RON of the blend using the blending index method
(equation 9.20) and compare with the reported experimental
value.
Solution:
(a) All terms in equations (9.22) and (9.23) are calcula ted and listed in
Table E9.11.2.
RON
Blend
¼ 89:38 þ 0:03324ð941:74 89:38 9:91Þ
þ 0:00085ð1852:38 28: 11
2
Þ¼92:13
MON
Blend
¼ 79:47 þ 0:04285ð824:67 79:47 9:91Þ
þ 0:00066ð1852:38 28: 11
2
Þ
0:00632ðð1821:11 34:44
2
Þ=100Þ
2
¼ 81:49
It can be observed from the above calculations that the RON without correc-
tion ð
R Þ is 89.38 and with olefin and aromatic corrections, it is 92.13.
The laboratory tested RON for the above blend was 91.8 (Twu and Coon,
1998). Therefore, the correction using the olefin and aromatic contents adjusts
the calculations to be closer to the experimental value with a difference of
0.36%.
(b) The blending index method can be used to calculate the RON of the above
blend.
Component BI
RON
x
vi
Reformate 68.16 0.41
Thermally cracked 54.53 0.04
Catalytically cracked 63.71 0.32
Polymerised 67.17 0.09
LSR 51.52 0.14
BI
RON;Blend
¼ 0:41ð68:16Þþ0:04ð54:53Þþ0:32ð63:71Þ
þ 0:09ð67:17Þþ0:14ð 51 :52Þ¼63:77
Then RON
Blend
¼ 92.69 with a difference of 0.97%.
252 Chapter 9
Table E9.11.2 Summary of calculations
Component
R M S RS MS A O A
2
O
2
Reformate 40.10 35.67 4.43 433.06 385.24 25.87 0.57 1632.46 0.80
Thermally cracked 2.82 2.60 0.21 14.92 13.80 0.39 1.30 3.84 42.25
Catalytically cracked 29.63 25.66 3.97 367.44 318.23 7.65 17.06 182.79 909.08
Polymerised 8.71 7.41 1.31 126.32 107.40 0.00 9.00 0.00 900.00
LSR 8.12 8.12 0.00 0.00 0.00 0.53 0.18 2.02 0.24
Sum 89.38 79.47 9.91 941.74 824.67 34.44 28.11 1821.11 1852.38
Product Blending 253
9.10. Linear Programming (LP) For Blending
LP is the most widely applied method for optimising many diverse
applications, including refineries and chemical plants. The application
of LP has been successfully applied for selecting the best set of variables
when a large number of interrelated choices exist. A typical example is in
a large oil refinery in which the stream flow rates are very large, and a
small improvement per unit of product is multiplied by a very large number.
This is done to obtain a significant increase in profit for the refinery.
Optimisation means the action of finding the best solution within
the given constraints and flexibilities. LP is a mathematical technique for
finding the maximum value of some equation subject to stated linear
constraints. The linear in LP stands for the algebraic aspect, which means
that all the constraints and objective functions are linear and satisfy two
fundamental properties, proportionality and additive. The programming in
LP means planning. The implementation of LP involves the development
of an integrated LP model representing the refinery operations with all
constraints and flexibilities and then solving it to determine the optimum
plan.
Refinery optimisation using an LP model has been proven to bring
economic gains higher than unit-specific simulation models or advance
process control techniques. Once all the data is configured, the model is
updated with the variable data. The required variable data includes the
following:
Crude oil or any other raw material prices with minimum and maximum
availability
Selling prices with minimum and maximum demands for the refinery
products
Available process unit capacities
Available inventory stocks with minimum and maximum storage limits
Quality specifications, etc.
The following methodology describes how LP is applied to the blending
problem. For N independent variables, the objective function (P) that
should be maximised is
P ¼ a
1
X
1
þ a
2
X
2
þþa
N
X
N
ð9:24Þ
The objective function is subjected to primary constraint equations
such as
X
1
0; X
2
0; ...; X
N
0 ð9:25Þ
254 Chapter 9
and
a
i1
X
1
þ a
i2
X
2
þþa
iN
X
N
b
i
a
j1
X
1
þ a
j2
X
2
þþa
jN
X
N
b
j
a
k1
X
1
þ a
k2
X
2
þþa
kN
X
N
¼ b
k
ð9:26Þ
can be solved utilising the Excel solver.
The objective function (P) is usually the octane number of the blend that
needs to be maximised. Sometimes the objective function can be minimis-
ing cost or setting the RVP to a certain value. The primary constraint
variables (X
i
to X
N
) are the volume of each cut for a blend which must be
greater than zero. Other constraint can be the capacity of the tank in which
the summation of X
i
s will not exceed. Additional constraints can be any
of the blend properties discussed in this chapter. Therefore, a
i1
to a
kN
are
the properties or their indices for components 1 to N and b
i
to b
k
are the
targeted blending properties or their indices.
In most blending cases, two or more properties are required in order to be
adjusted by the addition of some modifiers to the blend. For example,
oxygenates are added to enhance gasoline octane number and n-butane is
added to adjust the RVP. Determining the quantities of the additives becomes
more difficult in non-linear property cases. Determining the amounts of these
blends is based on solving two (or more) equations with two (or more)
unknowns depending on the property equations as discussed above.
Example E9.12
A blend is performed of 3000 barrels of LSR gasoline and 2000 barrels of
hydrocracker gasoline. Calculate the amount of FCC gasoline and n-butane
that must be added to the blend to produce a gasoline of 12 psi RVP and 0.525
cSt viscosity. The properties of all the components are shown in Table E9.12.1.
Solution:
Calculate the RVP index and viscosity index using equations (9.2) and (9.17),
respectively. The results are listed in Table E9.12.2
Table E9.12.1 Components amounts and properties for the blend
Component Barrels RVP psi Viscosity cSt
LSR gasoline 3000 12.1 0.2886
Hydrocracker gasoline 2000 15.5 0.4140
FCC gasoline X 4.8 0.8607
n-Butane Y 51.6 0.3500
Product Blending 255
RVP blend is considered as the objective function
12
1:25
ð5000 þ X þ Y Þð3000Þð22:56Þð2000Þð30:76Þ
ðXÞð 7:11ÞðY Þð138:3Þ¼0
ðE9:12:1Þ
Viscosity index of the blend is
BI
visi
¼
log
10
ð0:525Þ
3 þ log
10
ð0:525Þ
¼0:1028
Viscosity blend is considered as the first constraint equation
0:1028ð5000 þX þYÞþð3000 Þð0:21936Þþð2000Þð0:14635Þ
þðXÞð0:0222ÞþðYÞð0:1792Þ¼0
ðE9:12:2Þ
Other constraints are added such as X > 0 and Y > 0 (E9.12.3)
Solving using the Excel solver with the above objective function and constraints
gives
X = 6025 barrels
Y = 640 barrels
This example can be solved by simultaneous solving of equations (E9.12.1) and
(E9.12.2) algebraically.
Example E9.13
A blend is performed for LSR gasoline, hydrocracker gasoline, FCC gasoline
and n-butane. Calculate the amount of each component that must be added to
produce gasoline blend of maximum octane number and 12 psi RVP. The
properties of each cut are shown in Table E9.13. The olefin contents in each cut
can be assumed to be zero.
Table E9.12.2 RVP and viscosity indices
Component BI
RV P
BI
vis
LSR gasoline 22.56 –0.21936
Hydrocracker gasoline 30.76 –0.14635
FCC gasoline 7.11 – 0.0222
n-Butane 138.3 –0.1792
Table E9.13 Components properties for the blend
Component Bar rels RVPpsi RON MON
LSR gasoline X
1
12.1 91 81
Hydrocracker gasoline X
2
15.5 89.5 85.5
FCC gasoline X
3
4.8 97 96
n-Butane X
4
51.6 93 92
Blend X
T
12 Max
256 Chapter 9
The maximum blend storage capacity is 15,000 barrels.
Redo the calculation if the available amount of FCC gasoline is 6000 barrels.
Solution:
The objective fun ction is maximising octane number.
Since olefin contents are zero, equation (9.22) becomes
RON
Blend
¼ R þ 0:03324ðRS R SÞðE9:13:1Þ
where
R ¼
X
1
X
T
ð91Þþ
X
2
X
T
ð89:5Þþ
X
3
X
T
ð97Þþ
X
4
X
T
ð93Þ
S ¼
X
1
X
T
ð91 81Þþ
X
2
X
T
ð89:5 85:5Þþ
X
3
X
T
ð97 96Þþ
X
4
X
T
ð93 92Þ
RS ¼
X
1
X
T
91ð91 81Þþ
X
2
X
T
89:5ð89:5 85:5Þþ
X
3
X
T
97ð97 96Þ
þ
X
4
X
T
93ð93 92Þ
The following are four constraints:
X
1
0; X
2
0; X
3
0; and X
4
0 ðE9:13:2Þ
An additional constraint is
X
1
þ X
2
þ X
3
þ X
4
< 15; 000 barrels ðE9:13:3Þ
The constraint equation for the RVP of the blend can be obtained by calculating
the RVP index from equation (9.2)
22:56X
1
þ 30:755X
2
þ 7:105X
3
þ 138:3X
4
22:33X
T
¼ 0 ðE9:13:4Þ
Using the solver in Excel with the above constrains (equations E9.13.2, E9.13.3,
and E9.13.4) and maximising the RON value from equation (E9.13.1), the
results are RON equal 96.1 and
X
1
= 500, X
2
= 440, X
3
= 12,457 and X
4
= 1603 barrels
If the constraint for the amount of FCC gasoline is added, the results will be a
RON of 95.7 and
X
1
= 500, X
2
= 440, X
3
= 6000 and X
4
= 755 barrels
Example E9.14
The following gasoline cuts (Table E9.14.1) are blended to get a RON of 89
and a RVP of 10.2 psi. Calculate X
1
and X
2
. Assume that the olefin contents are
negligible.
Product Blending 257
Solution:
The volume fractions and RVP indices for each component are calculated and
listed in Table E9.14.2.
The first equation for the RVP blend is the objective function
ð10:2
1:25
Þð47603 þ X
1
þ X
2
Þ5735ð25:9Þ14749ð2:7Þ20117ð6:4Þ
814ð24:4Þ4117ð6:7Þ2071ð14:9Þ138X
1
15:6X
2
¼ 0
ðE9:14:1Þ
then becomes
18:22ð47603 þ X
1
þ X
2
Þ394608 138X
1
15:6X
2
¼ 0
The second equation for octane number is considered as the first constraint
RON
Blend
¼ R þ 0:03324ðRS R SÞ¼89 ðE9 :14:2Þ
Table E9.14.1 Components properties for the blend
Component Barrels RVP (psi) RON MON
n-Butane X
1
51.6 93 92
Isomer gasoline 5735 13.5 83 81.1
Reformer gasoline 14,749 2.2 98.5 86.0
FCC gasoline 20,117 4.4 92.3 86.8
Hydrocracker
gasoline
814 12.9 82.8 82.4
Alkylate 4117 4.6 97.3 95.9
Polymer 2071 8.7 96.9 84
MTBE X
2
9.0 118 101
Blend 47,603 + X
1
+ X
2
10.2 89
Table E9.14.2 RVP indices and volume fractions
Component x
vi
Barrels BI
RV P
n-Butane X
1
/(47,603 + X
1
+ X
2
) X
1
138
Isomer gasoline 5735/(47,603 + X
1
+ X
2
) 5735 25.9
Reformer gasoline 14,749/(47,603 + X
1
+ X
2
) 14,749 2.7
FCC gasoline 20,117/(47,603 + X
1
+ X
2
) 20,117 6.4
Hydrocracker
gasoline
814/(47,603 + X
1
+ X
2
) 814 24.4
Alkylate 4117/(47,603 + X
1
+ X
2
) 4117 6.7
Polymer 2071/( 47,603 + X
1
+ X
2
) 2071 14.9
MTBE X
2
/(47,603 + X
1
+ X
2
) X
2
15.6
Blend 47,603 + X
1
+ X
2
47,603 +
X
1
+ X
2
18.22
258 Chapter 9
where,
Rð47603 þ X
1
þ X
2
Þ¼3; 927; 638 þ 93 X
1
þ 118 X
2
Sð47603 þ X
1
þ X
2
Þ¼181; 332:9 þ X
1
þ 17 X
2
RSð47603 þ X
1
þ X
2
Þ¼16; 518; 930 þ 93 X
1
þ 2006 X
2
The other constraints are X
1
> 0 and X
2
> 0
Using the solver in Excel with the above constraints, the results are
X
1
= 3927 barrels
X
2
= 3984 barrels
Questions and Problems
9.1. Calculate the number of barrels of n-butane that have to be added to a
mixture of 1250 barrels of HSR gasoline, 750 barrels of LSR gasoline,
and 620 barrels of C
þ
5
FCC gasoline to produce a 9 psi RVP. What are
the RON and MON of the blend?
Component Barrels
RV P
(psi) RON MON
Olefin
(%)
Aromatics
(%)
HSR gasoline 1250 1 62.3 58.7 0 7
LSR gasoline 750 11.1 66.4 61.6 0 5
C
5
+
FCC 620 13.1 91 80.7 30.6 33.1
n-Butane 51.6 93 92 0 0
9.2. What is the viscosity of a blend of 2000 bbl of oil with a viscosity of
75.5 cSt at 130
F, 3000 bbl with 225 cSt at 130
F and 5000 bbl with
6500 cSt at 130
F?
9.3. What is the flash point of a mixture of 2000 barrels of oil with a flash
point of 100
F, 3000 barrels with a flash point of 80
F and 5000 barrels
with a 150
F flash point?
9.4. Calculate the pour point of the following mixture:
Component Barrels Pour point (
F)
A 5200 10
B 3000 50
C 6500 65
D 3250 45
Product Blending 259
9.5. Calculate the number of barrels of n-butane that have to be added to a
mixture of 3000 barrels of LSR gasoline, 2000 barrels of hydrocrakate
gasoline, and 6000 barrels of FCC gasoline to produce a 12 psi RVP
blend. What is the viscosity of the blend?
Component Barrels RVP (psi) Viscosity (c St)
n-Butane X
B
51.6 0.35
LSR gasoline 3000 12.1 0.2886
Hydrocrakate gasoline 2000 15.5 0.414
FCC gasoline 6000 4.8 0.8607
9.6. A blend is performed of 3000 barrels of LSR gasoline and 3000 barrels
of hydrocrakate gasoline. Calculate the amount of FCC gasoline and n-
butane that must be added to the mixture to produce a 12 psi RVP
blend with a viscosity of 0.525 cSt.
Component Barrels RVP (psi) Viscosity cSt
n-Butane X 51.6 0.35
LSR gasoline 3000 12.1 0.2886
Hydrocrakate gasoline 3000 15.5 0.414
FCC gasoline Y 4.8 0.8607
9.7. A blend is performed for LSR gasoline, hydrocracker gasoline, FCC
gasoline and n-butane. Calculate the amount of each component that
must be added to produce gasoline blend of maximum octane number
and 10 psi RVP using the Linear Programming (LP) technique. The
properties of each cut are shown in the table below. The olefin contents
in each cut can be assumed to be zero. The maximum blend storage
capacity is 10,000 barrels.
Component Barrels RVP (psi) RON MON
LSR gasoline X
1
12.1 91 81
Hydrocracker gasoline X
2
15.5 89.5 85.5
FCC gasoline X
3
4.8 97 96
n-Butane X
4
51.6 93 92
Blend X
T
10 Max
260 Chapter 9
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Product Blending 261