Fahim M.A., Sahhaf T.A., Elkilani A.S. Fundamentals of Petroleum Refining

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FP;Blend

i¼1

FPi

¼ 356:55

Blend

¼ exp ð0: 91725Þln

356:06

51708



0:5

þ 2:6287

¼ 117:36



The calculated ﬂash point is less than 130



F therefore, component D need to be

added.

The ﬂash point index for component D is

FPi

¼ 51708 exp½ðlnð220Þ2:6287Þ

=ð0:91725Þ ¼ 12 :41

The blend ﬂash point index is

FPi

¼ 51708 exp½ðlnð130Þ2:6287Þ

=ð0:91725Þ ¼ 218:94

If W

is the mass ﬂow rate for component D, then

29,120 (321.36) + 46,410 (731.07) + 65,520 (106.47) +

12.41 W

¼ (141,050 + W

) 218.94

¼ 93,843 lb/hr

The volumetric ﬂow rate of component D is 6785 BPD.

Using the other ﬂash point index correlation (equation 9.5) gives a blend ﬂash

point of 119.6



F which is comparable with the one calculated above.

Table E9.3.2 Summary of calculations

Component x

FPi

A 0.207 321.36

B 0.329 731.07

C 0.465 106.47

9.4. Pour Point Blending

The pour point is the lowest temperature at which oil can be stored

and still capable of flowing or pouring, when it is cooled without stirring

under standard cooling conditions. Pour point is not an additive property

and pour point blending indices are used, which blend linearly on a

volume basis. The pour point of a blend is determined using the follow-

ing equation:

PP;Blend

i¼1

PPi

ð9:7Þ

242 Chapter 9

where x

is the volume fraction of component i, and BI

PPi

is the pour point

index of component i that can be determined from the following correla-

tion (Hu and Burns, 1970):

PPi

¼ 3; 262; 000 ðPP

=1000Þ

12:5

ð9:8Þ

where PP

is the pour point of component i,in

The pour point of the product, PP

Blend

, is then evaluated using the

reverse form of equation (9.8).

Another relation to estimate the pour point blending index is the same

formula for the flash point reported by Riazi (2005) with different value of

the exponent:

FPi

¼ PP

1=x

ð9:9Þ

is the pour point temperature of component i, in K, and the best value of

x is 0.08.

Example E9.4

What is the pour point of the following blend?

Component BPD Pour Point (



Catalytic cracked gas oil 2000 –15

Straight run gas oil 3000 –3

Light vacuum gas oil 5000 42

Heavy vacuum gas oil 1000 45

Solution:

The given pou r point temperatures are converted to Kelvin then the pour point

indices are calculated for each component and listed in Table E9.4. The blend

pour point is calculated from the blend index shown in Table E9.4:

Blend

¼ð1697:2=3; 262; 000Þ

1=12:5



 1000 ¼ 546



Blend

¼ 86



F ¼ 30



Table E9.4 Summary of calculations

Component x

ppi

BI

ppi

Catalytic cracked gas oil 0.1818 227.3 41.32

Straight run gas oil 0.2727 461.0 125.7

Light vacuum gas oil 0.4545 2748 1249

Heavy vacuum gas oil 0.0909 3093 281.2

Sum 1 1697.2

Product Blending 243

9.5. Cloud Point Blending

Cloud point is the lowest temperature at which oil becomes cloudy

and the first particles of wax crystals are observed as the oil is cooled

gradually under standard conditions. Cloud point is not an additive property

and cloud point blending indices are used, which blend linearly on a volume

basis using the following:

CP;Blend

i¼1

CPi

ð9:10Þ

where x

is the volume fraction of component i, and BI

CPi

is the cloud

point blending index of component i that can be determined from the

following correlation (Hu and Burns, 1970; Riazi, 2005):

CPi

¼ CP

1=x

ð9:11Þ

is the cloud point temperature of component i, in K and the value of

x is 0.05.

Example E9.5

Calculate the cloud point of the following blend

Component BPD Cloud point (



A 1000 –5

B 2000 5

C 3000 10

Solution:

The given cloud point temperatures are converted to Kelvin then the cloud

point indices are calculated for each component and listed in Table E9.5.1.The

blend cloud point is calculated from the summation as

Blend

¼ (1.8098  10

)

0.05

¼ 258.75 K ¼14.25



C ¼ 6.35



Table E9.5.1 Summary of calculations

Component x

CP (K) BI

cpi

10

BI

cpi

10

A 0.167 252.4 1.101 0.1839

B 0.333 258.0 1.701 0.5664

C 0.500 260.8 2.119 1.0595

Sum 1.0 1.8098

244 Chapter 9

9.6. Aniline Point Blending

The aniline point indicates the degree of aromaticity of a petroleum

fraction. It is the minimum temperature at which equal volumes of the

aniline and the oil are completely miscible. Aniline point is not an additive

property, and aniline point blending indices are used, which blend linearly

on a volume basis. The aniline point of a blend is determined using the

following formula:

AP;Blend

i¼1

APi

ð9:12Þ

where x

is the volume fraction of component i, and BI

APi

is the aniline

point index of component i that can be determined from the following

correlation (Baird, 1981):

APi

¼ 1:124½expð0:00657AP

Þ ð9:13Þ

is the aniline point of component i,in



Example E9.6

What is the aniline point of the following blend?

Component BPD Aniline point (



Light diesel 4000 71.0

Kerosene 3000 60.7

Light cycle gas oil 3000 36.8

Solution:

The volume fractions and aniline point indices are calculated and listed in

Table E9.6.

AP,Blend

¼ 0.4(1.792) þ 0.3(1.675) þ 0.3(1.432) ¼ 1.6468

Blend

¼ lnð1:6468=1:124Þ=0:00657 ¼ 58: 17



Table E9.6 Summary of calculations

Component x

APi

Light diesel 0.4 1.792

Kerosene 0.3 1.675

Light cycle ga s oil 0.3 1.432

Sum 1.0

Product Blending 245

9.7. Smoke Point Blending

The smoke point is the maximum flame height in millimetre at which

the oil burns without smoking when tested at standard specified conditions.

The smoke point of a blend is determined using the following formula

(Jenkis and Walsh, 1968):

Blend

¼255:26 þ2:04 AP

Blend

 240:8lnðSG

Blend

þ 7727ðSG

Blend

=AP

Blend

ð9:14Þ

where SP

Blend

is the blend smoke point in mm,AP

Blend

and SG

Blend

are the

aniline point and specific gravity of the blend, respectively. Aniline point for

the blend is calculated as discussed in section 9.6. Specific gravity is an

additive property and can be blended linearly on a volume basis. The

specific gravity of a blend is estimated using the mixing rule:

Blend

i¼1

ð9:15Þ

where x

is the volume fraction of component i, and SG

is the specific

gravity of component i. It must be stated here that, API is not an additive

property, and it does not blend linearly. Therefore, API is converted to

specific gravity, which can be blended linearly.

Example E9.7

What is the smoke point of the following blend?

Component BPD

Aniline point

(



C) SG

Kerosene A 4000 50 0.75

Kerosene B 3000 60 0.8

Kerosene C 3000 55 0.85

Solution:

The volume fractions and aniline point indices are calculated and listed in

Table E9.7.

Blend

¼ 0.4(0.75) + 0.3(0.8) + 0.3(0.85) ¼ 0.805

AP,Blend

¼ 0.4 (1.561) + 0.3(1.667) + 0.3(1.613) ¼ 1.6296, gives AP

Blend

56.54



Blend

¼255.26 + 2.04(56.54)  240.8 ln(0.805) + 7727(0.805/56.54) ¼

22.33 mm

246 Chapter 9

9.8. Viscosity Blending

Viscosity is not an additive property; therefore, viscosity blending

indices are used to determine the viscosity of the blended products. A

number of correlations and tables are available for evaluating the viscosity

indices. The viscosity index of the blended product is determined using the

following equation:

vis;Blend

i¼1

visi

ð9:16Þ

where x

is the volume fraction of component i, and BI

visi

is the viscosity

index of component i that can be determined using the following correla-

tion (Baird, 1989):

visi

log

3 þ log

ð9:17Þ

where n

is the viscosity of component i, in cSt.

The viscosity of the blended product is then calculated using

Blend

¼ 10

½3BI

vis;Blend

=ð1BI

vis;Blend

Þ

ð9:18Þ

Example E9.8

Estimate the viscosity of a blend of 2000 barrels of Fraction-1 with a viscosity of

75 cSt, 3000 barrels of Fraction-2 with 100 cSt, and 4000 barrels of Fraction-3

with 200 cSt all at 130



Solution:

The viscosity index is calculated from equation (9.17).

As shown in Table E9.8, the blend viscosity index is calculated

Table E9.7 Summary of calculations

Component x

APi

Light diesel 0.4 1.561

Kerosene 0.3 1.667

Light cycle gas oil 0.3 1.613

Sum 1

Product Blending 247

The blend viscosity is calculated from the blend index as

Blend

¼ 10

½30:4117=ð10:4117Þ

¼ 125:73 cSt

9.9. Gasoline Octane Number Blending

The octane number is a characteristic of spark engine fuels such as

gasoline. Octane number is a measure of a fuel’s tendency to knock in a test

engine compared to other fuels. The octane number posted octane number

(PON), commercially used for gasoline (referred to as the road octane

number), is the average of its research octane number (RON) and motor

octane number (MON). The difference between RON and MON is

known as fuel sensitivity (S).

There are a number of additives, such as oxygenated ethers or alcohols,

that can enhance a gasoline octane number. Tetra-ethyl lead (TEL) which

was used for enhancing octane number, has now been phased out and

replaced by the oxygenates listed in Table 9.2 with their octane numbers.

RONs for selected hydrocarbons are listed in Appendix D.

If the octane number of a blend is calculated by the linear addition of an

octane number for each component, the following equation can be

obtained:

Blend

i¼1

ð9:19Þ

where x

is the volume fraction of component i, and ON

is the octane

number of component i.

Many alternative methods have been proposed for estimating the octane

number of gasoline blends since the simple mixing rule needs minor

corrections. One correction method that uses the octane number index

has been reported by Riazi (2005). The following octane index correlations

depend on the octane number range as follows:

Table E9.8 Summary of calculations

Component Quantity (BPD) x

(cSt) BI

visi

BI

visi

Fraction-1 2000 0.222 75 0.385 0.0854

Fraction-2 3000 0.333 100 0.400 0.1332

Fraction-3 4000 0.445 200 0.434 0.1931

Sum 9000 1.0 0.4117

248 Chapter 9

For 11  ON  76

ONi

36:01 þ38:33ðON=100Þ99:8ðON=100Þ

þ341:3ðON=100Þ

507:02ðON=100Þ

þ268:64 ð ON=100Þ



ð9:20aÞ

For 76  ON  103

ONi

¼299 :5 þ1272ðON=100Þ1552:9ðON=100Þ

þ651ðON=100Þ

ð9:20bÞ

For 103  ON  106

ONi

¼ 2206:3  4313:64ðON=100Þþ2178:57ðON=100Þ

ð9:20cÞ

The octane number index for a blend can be determined using the follow-

ing equation:

ON;Blend

i¼1

ONi

ð9:21Þ

where x

is the volume fraction of component i, and BI

ONi

is the octane

number index of component i that can be determined from equations

(9.20a, 9.20b and 9.20c).

Example E9.9

For a b lend of alkylate C

, coker gasoline, and FCC gasoline. Calculate the

RON by

(a) Linear mixing of octane numbers.

(b) Linear mixing of octane number indices.

Properties and quantities of the components are given in Table E9.9.

Table 9.2 Octane numbers of some oxygenates (alcohols and ethers) (Guibet, 1995)

Compound RON MON

Methanol 125–135 100–105

Ethanol 120–130 98–103

Methyl-tertiary-butyl ether (MTBE) 113–117 95–101

Ethyl-tertiary-butyl ether (ETBE) 118–122 100–102

Tertiary-butyl alcohol (TBA) 105–110 95–100

Tertiary-amyl-methyl ether (TAME) 110–114 96–100

Product Blending 249

Solution:

(a) Using equation (9.19) for simple mixing rule, the RON for the blend is

calculated as

RON

Blend

6000 þ4000 þ5000

6000ð97:3Þþ4000ð67:2Þþ5000ð83:2Þ½¼84:57

(b) Using the index method

RON;Alkylate

¼299:5 þ 1272ð0:973Þ1552:9ð0:973Þ

þ 651ð0 :973Þ

¼ 67:66

RON;coker

36:01 þ 38:33ð0:672Þ99:8ð0:672Þ

þ 341:3ð0 :672Þ

507:02ð0:672Þ

þ 268: 64ð0:672Þ



¼ 53:65

RON;FCC

¼299:5 þ 1272ð0:832Þ1552:9ð0:832Þ

þ 651ð0 :832Þ

¼ 58:78

RON;Blend

¼ 0:4ð67:66Þþ0:267ð53:65Þþ0:333ð58:78Þ¼60:98

To solve for RON

Blend

, equation (9.20b) is solved for RON, when I

RON

equals

60.98. The result ed RON

Blend

is 88.09.

Example E9.10

It is required to produce gasoline with a RON of 95 from the blending

components in Table E9.9 in the previous example. Calculate the amount of

the oxygenate MTBE with a RON of 115 to be added in order to adjust the

blend RON to 95.

Solution:

Using the index method

RON;Blend

¼299:5 þ 1272ð0:95Þ1552:9ð0:95Þ

þ 651ð0:95Þ

¼ 65:56

RON;MTBE

¼ 2206:3  4313:64ð1:15Þþ2178:57ð1:15Þ

¼ 126:77

The RON blending index of alkylate, coker gasoline and FCC gasoline are

calculated in Example E9.9. The blend equation (9.21) is then written as

Table E9.9 Quantities and RON of the blending components

Component Quantity (BP D) RON

Alkylate C

6000 97.3

Coker gasoline 4000 67.2

FCC gasoline 5000 83.2

250 Chapter 9

65:56ð15; 000 þ V

MTBE

Þ¼6000ð67:66Þþ4000ð53:65Þþ5000ð58:78 Þ

þ 126:77V

MTBE

Solving for V

MTBE

= 1126 BPD

Depending on the availability of olefin and aromatic contents in the

blended components, the octane number of the blend can be calculated

using the linear mixing rule method with a correction (Healy et al., 1959;

Maples, 1993)

RON

Blend

¼ R þ 0:03324 RS  R  SðÞþ0:00085 O

 O



ð9:22Þ

MON

Blend

¼ M þ 0:04285ðMS  M  SÞ

þ 0:00066 ð

 O

Þ0:00632

 A

100

ð9:23Þ

where the terms represent volumetric average values of given properties of

components as follows:

R = Research octane number (RON)

M = Motor octane number (MON)

S = Sensitivity (RON-MON)

RS = RON  Sensitivity

MS = MON  Sensitivity

O = Volume percent olefins

A = Volume percent aromatics

Example E9.11

Twu and Coon (1998) reported RON of 91.8 for a blend with the composition

given in Table E9.11.1.

Table E9.11.1 Composition, RON, MON, oleﬁn and aromatic contents of the blending

components (Twu and Coon, 1998)

Component x

RO N M O N

Olefin

content

(%)

Aromatic

content

(%)

Reformate 0.41 97.8 87 1.4 63.1

Thermally cracked 0.04 70.4 65.1 32.5 9.8

Catalytically

cracked

0.32 92.6 80.2 53.3 23.9

Polymerised 0.09 96.8 82.3 100 0

LSR 0.14 58 58 1.3 3.8

Product Blending 251