Crude Oil Characterization

8/9/2019 Crude Oil Characterization

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Petroleum Refining – Chapter 4: Characterization

4-1

Chapter 4 Characterization & properties estimation

of crude oil and petroleum products

Introduction

There is no analytical technique available to determine

(either quantitatively or quantitatively) all the tens of

thousands of chemical species in petroleum and its

fractions.

Only the low-boiling components C1-C5 can becompletely identified using gas chromatography analysis.

For the white fractions (e.g. Naphtha), only a limited

number of components can be completely identifiedusing PINA, PIONA, or Detailed HC GC analyzers

(Table 2.3 – 2.6).

To overcome this shortcoming, petroleum refiners resortto define (characterize) petroleum and its fractions usingglobal (bulk) properties.

This traditional way of oil characterization, though old, is

still being used today.

This type of characterization is used as basis for

assigning a price for crude oils and petroleum fractions

and in design of petroleum processes.

Sulfur content & API gravity have the greatest influence on the value of

crude oil, although N2 and metals content are increasing in importance.

The price of petroleum fractions is influence by other properties in

addition


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Copyrights © 2001 – 2012, Dr. Tareq Albahri, Chem. Eng. Dept., Kuwait University

4-2

API Gravity (API)

A measure of crude oil density/specific gravity.

141.5API = 131.5sp.gr.

(4.1)

oil

water

(60 )

sp.gr. =(60 )

F

F

(4.2)

Light crude have lower density (sp.gr.)→ API is higher.

10 < API < 50(or less) (or more)

High API crude produces more distillates (valuable) thanit does residue (less valuable).

Distillation Range (Curve)

The boiling point range is an alternate method torepresent the composition of petroleum and its products.

It gives an indication of the quantities and qualities of the

various products present in crude oil (i.e. naphtha,kerosene, diesel, gas oil, residue, etc).

- Can be used to determine themost desirable processing

sequence to obtain therequired products.

- Can be used to determinewhether the crude is

suitable for asphalt orlube oil manufacture.

CrudeOil

GasesLPG

Naphtha

Kerosene

Diesel

Residue


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4-3


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4-4

Types of distillation curves

1. TBP (True Boiling Point) distillation curve.

2. ASTM (D86/D1160) distillation curve.3. EFV (Equilibrium Flash Vaporization)


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4-5

TBP is the most useful.

- However, no standard test exists for measuring the

TBP.

- Most common TBP test is Hempel & D-285.

(Neither specifies # of stages or reflex ratio used).

- Trend is toward 15/5 distillation (D-2892) to stand

for TBP (assumed to be the same as TBP).

- An estimate of the composition of the butane and

lighter components is frequently added to the low- boiling end or the IBP of the TBP curve to

compensate for the loss during distillation.

ASTM is more common because it is simple to

determine in the laboratory. Kuwait refineries use ASTM

curve (Which is then converted to TBP curve).


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4-6

Example 4.1:

Estimate the true boiling point (TBP) distillation curve of the petroleum fraction having the following ASTM D86

distillation temperatures:

Vol % T (ºF)IBP -

5% -10% 400

30% 42050% 438

70% 46090% 485 → 490

95% -FBP -

Recovery = -

Solution:

Using Fig. 3A1.1 from the API technical Data Book

(1) Correct the ASTM D86 distillation temperatures above

475 ºF for cracking using Hadden equation.

Log D = – 1.587 + 0.00473 T

where, T & D are in ºF

For the 90% temp. → Log D = – 1.587 + 0.00473 (485) =

0.707

D = 5 ºF

The corrected 90% temp. becomes = 485 + 5 = 490 ºF


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4-7

(2) Find the atmospheric TBP mid (50%) temperature using

the mid temp of the ASTM D86 and the lower part ofFigure 3A1.1.

438 ºF on the x-axis → ∆T= 2 ºF on the y-axis

[correction]

Therefore, the TBP 50% T =ASTM 50% T + ∆T = 438+2= 440 ºF

(3) Find the temp for each segment of the TBP curve using

upper part of Figure 3A1.1.

Segment of

Curve (Vol

%)

ASTM ∆T (ºF)

[from table above]

TBP ∆T (ºF)

[Figure 3A1.1]

TBP (ºF) Vol %

0 to 10

10 to 3030 to 50

2018

3931

409 – 39 = 370440 – 31 = 409

IBP

10%30%

440 50%

50 to 70

70 to 90

90 to 100

22

30

33

40

440 + 33 = 473

473 + 40 = 513

70%

90%

FBP

Note: it is possible to convert TBP temperatures to ASTMD86 using Figure 3A2.1 from the API technical data

book, which involves a trial and error procedure.


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Copyright © 2001-2014 Dr. Tareq Albahri, Chemical Engineering Dept., Kuwait University, All rights reserved

4-8


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4-10

Crude oil analysis are reported in 2 parts,

1. D-86, atmospheric, (IBP up to 527 ºF end point 760

mmHg)

2. D-1160, vacuum, (392 to 572 ºF end point at 40 mmHg)

equivalent to 580 to 790 ºF at 1 atm.

- Vacuum Distillation is necessary to prevent excessive

pot ( وع ) temperatures, which cause cracking of the

crude oil.

- To combine the two results in one curve, the distillation

temperatures at 40 mmHg reported in the analysis must

be corrected to 760 mmHg pressure using Figure 3.6

from text.

The 572 ºF end point at 40 mmHg corresponds to 790 ºF at 760

mmHg.


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4-11

Figure 4.1: Boiling point distribution curve

Estimation of the TBP curve above 790 ºF can be obtained by

1. Extrapolating of probability graph (Figure 3.7 in text) to

1100ºF or higher FBP.


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4-12

Figure 4.2: Crude oil TBP Distillation Curve Probability Chart.


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4-13

Example 4.2: construction of the TBP & API Curves

Draw the TBP and API curves for the following crude assay


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4-14

Solution:

1. Results in the table are in TBP not ASTM so there is no need

for conversion.

2. On a graph paper, plot the sum percent (column 4) as your x-

axis and the cut temperature (column 2) as your y-axis untilyou reach the temperature 527 ºF (this is the atmospheric

distillation data).

3. To plot the vacuum distillation data on the same graph you

would need to correct the vacuum distillation temperatures(at 40 mmHg) to atmospheric pressure (i.e. convert from

D1160 to D86 test pressure) using the following equation,

T b (1atm) = 126.14 + 1.169 T b (40 mmHg)

then continue plotting the curve as above.

4. Plot all the data in step 1 and 2 above on the probabilitygraph extending the straight line to the FBP. Read the valuesof x (sum percent) and y (temperature) above 572 ºF and list

them in a table. Use the new values to continue the rest of

your distillation curve.

5. Plot on the same graph the API (column 6) versus cut mid%

calculated as follows;

Mid% cut 1 = 0.8/2 = 0.4 → API = 78.8

Mid% cut 2 = 0.8 + 1.0/2 = 1.3 → API = 75.1 Mid% cut 3 = 1.8 + 3.0/2 = 3.3 → API = 63.7

Mid% cut 4 = 4.8 + 3.4/2 = 6.5 → API = 55.9 etc.


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4-15

2. Better results are obtained when using the probability densityfunction shown below instead of Figure 4.2.

1/B

o

o

A 1T-T = ln

B 1-xT

(4.3)

Where x is the volume fraction distilled, T is the distillation

temperature, and To, A and B are constants determined byregression using the solver function in Microsoft Excel or any

curve fitting program. Once the constant are determined, the

function may be used in the above equation to construct therest of the boiling point diagram.

When applied to the crude in example 4.2 the following

values are obtained for the constantsTo = 30.788

A = 4986.5

B = 2.5131

Using these constants into the above equation, the followingTBP curve is predicted.

Hints: when applying the same to your own crude oil, use the

above values as initial guess. Also, add an absolute to theabove equation to avoid getting errors

1/B

o

o

A 1T-T = ln

B 1-xT

abs (4.4)


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4-16

vol % T (ºF) x T calculated abs diff

0.8 122 0.008 122 123.43632 1

1.8 167 0.018 167 158.976516 8

4.8 212 0.048 212 221.339965 9

8.2 257 0.082 257 268.288313 11

11.3 302 0.113 302 302.429928 0

15.2 347 0.152 347 339.126056 8

20.1 392 0.201 392 379.358772 13

26.9 437 0.269 437 428.88685 8

34.9 482 0.349 482 482.000019 0

53.1 584 0.531 584 596.320067 12

60.9 637 0.609 637 646.905867 10

67.1 690 0.671 690 689.704546 0

72.8 742 0.728 742 732.454459 10

79.7 795 0.797 795 791.401438 3

85 846 0.85 846 845.847377

90 911 0.9 911 911.151517

95 1008 0.95 1008 1008.33677

98 1118 0.98 1118 1117.85398

99.9 1394 0.999 1394 1393.84728

0

200

400

600

800

1000

1200

1400

1600

0 20 40 60 80 100

Vol. %

TBPTemperature(F)


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4-17

Characterization & Classification

Characterization Factors

Correlate between the yield and the aromaticity & paraffinicity of petroleum oils.

Several correlations exist,

1. UOP or Watson characterization factor (K w )

B

13

w@60 F

T =K (sp.gr.)

(4.5)

More commonly used.

TB is the normal boiling point of the pure compound orthe mean average boiling point of a petroleum fraction

in ºR [ºF+460].

Higher for lighter components and petroleum fractions

e.g. K w ≈ 12.1 (Naphtha), K w ≈ 11.9 (Diesel)

For petroleum fractions 10 < K w < 15

↑ ↑

contain containhighly highly

aromatic paraffinic

compounds compounds


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4-18

For crude oil 10.5 < K w < 12.9↑ ↑

highly highly

naphthenic paraffiniccrude crude

For pure hydrocarbons

K w = 13 for paraffins.

K w = 12 for HC with equivalent chain and ringweights.

K w = 11 for pure naphthenes.

K w = 10 for pure aromatics.

2. US Bureau of Mines “Correlation Index”

B

@60 F= + 473.7 (sp.gr.) 456.8CIT

87,552 (4.6)

- TB & sp.gr. as above.

- Useful in evaluating individual fractions from oils (e.g.

naphtha, kerosene, diesel, etc.).

- The CI scale starts with 0 for straight-chain paraffins

and 100 for benzene.

0 → highly paraffinic

100 → highly naphthenic/aromatic

- CI values are not quantitative.

Low CI values→ high conc. of paraffins in the fraction.


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4-19

Higher CI values → high conc. of naphthenes &aromatics.

Example 4.3: Watson K for pure components

Calculate the Watson characterization factor and correlation index for n-

pentane.

Solution:For n-pentane, the boiling point is 97 ºF and sp.gr. is 0.63

Using equation (5.3) above

13

w

(97+460) = = 13.06K (0.63)

Using equation (5.4) above

= + 473.7 (0.63) 456.8 0CI87,552

(97+460)

Table 5.4: Comparison of K w and CI for pure components

Compound TB (ºF) Sp.gr. K w CI

n-Hexane

2-Methylpentane

CyclohexaneBenzene

Naphthalene

155.72

140.47

177.29176.18

884

0.664

0.658

0.78350.8845

1.176

12.8

12.8

11.09.7

8.16

0

0.7

51.799.8

199.2


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Classification of crude oil

- Paraffinic base

- Naphthenic base- Asphalt base

- Mixed base- Aromatic base (up to 80% in the far east)

These classifications convey the nature of the products to be

expected and the processing necessary.

US Bureau of Mines Classification

Based on the properties of the residue left from non-destructive atmospheric and vacuum distillation tests.

Fraction (1) from 482-527 ºF atm. distillation (760 mmHg)1.

Fraction (2) from 527-572 ºF vac. distillation (40 mmHg)2.

The gravity of these two fractions, obtained from Figure 3.8,

is used to classify crude as shown in table below.

ClassificationKey fractions, ºAPI

Fraction (1) Fraction (2)

ParaffinParaffin, Intermediate

Intermediate, Paraffin

Intermediate

Intermediate, Naphthene Naphthene, Intermediate

Naphthene

4040

33 – 40

33 – 40

33 – 4033

< 33

3020 – 30

30

20 – 30

2020 – 30

< 20

1&2 Both from TBP curve


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4-21

Crude Suitable for Asphalt Manufacture:

There are certain characteristics of crude oils that indicate if

they are possible sources of Asphalt.

Not 100% accurate as experimental tests.

A crude oil is usually suitable for asphalt manufacture if itmeats the three following criteria:

(1) The crude oil gravity < 35 ºAPI

(2) Contains a residue (MeABP=750ºF) with a Watson

characterization factor < 11.8.

(3) If (KW)550ºF – (KW)750ºF < 0.15

If > 0.15, the residue may contain too much

wax to meet most asphalt specifications.


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4-22

Example 4.4:

Classify the following crude using the US bureau of mines classification

then specify whether it is suitable for asphalt manufacture. (See chapter 3)

Solution:

Figure. TBP and gravity-mid percent curves

Fraction (1) from 482-527 ºF atm. distillation (760 mmHg).

Fraction (2) from 527-572 ºF vac. distillation (40 mmHg)2.

The gravity of these two fractions, obtained from above figure, is used to

classify crude as shown in table below.

Fraction 1 → 33 ºAPI

Fraction 2 → 22 ºAPI

2 Equivalent to 740-790 ºF atm. distillation (760 mmHg).

0

200

400

600

800

1000

1200

0 20 40 60 80 100

Percent Distilled

Temperature(F

0

10

20

30

40

50

60

70

80

90

APIGravit


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4-23

From the table above, this crude would be classified as → Naphthene,

Intermediate

For the crude oil to be suitable for asphalt manufacture it should satisfy the

following

(1) API < 35 (specified in the data sheet for crude oil)

(2) (KW)550ºF – (KW)750ºF < 0.15

at 750 ºF → API = 25 → SG = 0.904

1/3

w

(750+460)K = 11.79

(0.904)

at 550 ºF → API = 32 → SG = 0.865

1/3

w

(550+460)K = 11.6

(0.865)

∆K w = 11.79 – 11.6 = 0.19 (> 0.15)

0

200

400

600

800

1000

1200

0 20 40 60 80 100

Percent Distilled

Temperature(F

0

10

20

30

40

50

60

70

80

90

APIGra

vit


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4-24

(3) Residue (MeABP = 750 ºF) (Kw < 11.8).

Calculate as above, only for the 790 ºF+ residue

The TBP curve for resid is first reconstructed as follows;

IBP is the final T reached by vac. distill. = 790 ºF

10 % T is 0.1(100-79) + 79 = 81.1%, T = 800 ºF

30 % T is 0.3(100-79) + 79 = 85.3%, T = 820 ºF

50 % T is 0.5(100-79) + 79 = 89.5%, T = 890 ºF

70 % T is 0.7(100-79) + 79 = 93.7%, T = 930 ºF

90 % T is 0.9(100-79) + 79 = 97.9%, T = 1030 ºF

FBP remains the same as before at about 1100 ºF

as an alternative, using computer program

(petrochem toolkit) API = 19 → MeABP = 860 ºF

Kw = 11.67 Ok

0

200

400

600

800

1000

1200

0 20 40 60 80 100

Percent Distilled

Temperature(F

0

10

20

30

40

50

60

70

80

90

APIGravit


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Molal average boiling point

n

i bii=1

MABP =x T (4.8)

xi = mole fraction of component i.

T bi = normal boiling point of component i (ºF or R).

Weight average boiling point

n

wi bii=1

WABP = x T (4.9)

xwi = weight fraction of component i.

T bi = normal boiling point of component i (ºF or R).

Cubic average boiling point [R]

3

1/3

bi

n

vii=1

CABP = x T (4.10)

T bi = normal boiling point of component i (R only).

Mean average Boiling point

MABP + CABPMeABP =

2

(4.11)

Watson characterization factor

1/3

@60 Fw

(MeABP)K =

(sp.gr.) (4.12)


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- The Watson K is an approximate index of paraffinicity,with high values corresponding to high degrees of

saturation.

For multicomponent mixtures for which the composition isknown.

n

iwi

i=1

K = x K w (4.13)

Example 4.5:

Calculate the Watson characterization factor and the correlation index for

the following LPG mixture.

Vol %

C3 45

n-C4 50n-C5 5

Solution:

LPG is at high pressure, but the laboratory analysis for gas composition

are done at atmospheric pressure. This is close to ideal conditions and the

volume % is equal to the mole %.

mole % MW sp.gr. Boiling Point, ºF (ºR)C3 45 44 0.507 - 43.75 (416)

n-C4 50 58 0.584 31 (491)

n-C5 5 72 0.63 97 (557)

Average Sp.gr. = (0.45)(0.507)+(0.5)(0.584)+(0.05)(0.63) = 0.552

VABP = (0.45)(416)+(0.5)(491)+(0.05)(557) = 460.7 R

MABP = VABP (for ideal gas)


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C3 wt% = (0.45*44)/[(0.45*44)+(0.5*58)+(0.05*72)] = 37.8 wt%

C4 wt% = (0.5*58)/[(0.45*44)+(0.5*58)+(0.05*72)] = 55.3 wt%

C5 wt% = (0.05*72)/[(0.45*44)+(0.5*58)+(0.05*72)] = 6.9 wt%

WABP = (0.378)(416)+(0.553)(491)+(0.069)(557) = 467.3 ºR

31/3 1/3 1/3

CABP = 459.40.45(416) 0.5(491) 0.05(557)

R

MeABP = (MABP+CABP)/2 = (460.7+459.4)/2 = 460 R

Into equation (3) above1

3

w

(460) = = 13.9K (0.552)

orn

iwi

i=1

K = x K w = 0.378(14.7) + 0.553(13.5) + 0.069(13.03) = 13.92

Into equation (4) above

= + 473.7 (0.552) 456.8 5CI

87,552

(460)

Example 4.6: Watson K for petroleum fractions

For the following SR Naphtha,

69.1 ºAPI & ASTM D86 distillation3

Vol % T (ºF)

IBP 925% 11810% 128

30% 164

50% 198

70% 23090% 262

95% 272

3 D1160 can’t be used


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4-29

FBP 300Recovery = 98.8 vol.%

Calculate,

(1) The volumetric average boiling point, VABP(2) The weight average boiling point, WABP

(3) The molal average boiling point, MABP

(4) The cubic average boiling point, CABP

(5) The mean average boiling point, MeABP

(6) The Watson characterization factor, Kw

(7) The US bureau of mines correlation index, CI.

Solution:

No correction for temperature is required (since T < 475 ºF).

The volumetric average boiling point, VABP90

i

10

T

VABP5

i

10% 30% 50% 70% 90%

5

T T T T TVABP =

128+164+198+230+262 = = 196.45

ºF

Assuming ASTM curve is linear between 10 & 90%

90% 10% 262 128T TASTM slope = 1.67590 10 80

From Figure 2B1.1 API data book

For MABP → ∆T = -17 ºFFor CABP → ∆T = -3 ºF

For WABP → ∆T = +4 ºF For MeABP → ∆T = -10 ºF


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Using the corrections

MABP = 196.4 – 17 = 179.4 ºFCABP = 196.4 – 3 = 193.4 ºF

WABP = 196.4 + 4 = 200.4 ºFMeABP = 196.4 – 10 = 186.4 ºF

or = (179.4+193.4)/2 = 186.4 ºF

The Watson characterization factor, Kw

Sp.gr. = 141.5/(69.1+131.5) = 0.705

1/3 1/3

@60 Fw

(MeABP) (186.4+460)K = 12.3

(sp.gr.) (0.705) (mainly saturated)

The US bureau of mines correlation index, CI.

= + 473.7 (0.705) 456.8 0.7CI87,552

(186.4+460)

(paraffinic)


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Coordination Number (Hydrogen Deficiency) or “Z” Number

Cn H 2n+Z

As Z-number decreases, density increases.

Group Z

Paraffins

Olefins

NaphthenesAromatics

2

0

0-6

Examples of Z-values for Aromatic compounds

C6H6

Z = -6

C10H8

Z = -12

C14H10

Z = -18

C10H12Z = -8

C10H10Z = -10

C24H12Z = - 36


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Properties Estimation of Petroleum Fractions

1. Graphical Methods.

2. Correlations.

Example 4.7: Graphical Method

For the following SR Naphtha; 69.1 ºAPI & ASTM D86 distillationVol % T (ºF)

IBP 92

5% 118

10% 128

30% 164

50% 198

70% 230

90% 26295% 272

FBP 300

Recovery = 98.8 vol.%

Calculate the MW, aniline point, H/C ratio, the true and pseudocritical

temperatures and pressures.

Solution:

Previously calculated, the mean average boiling point,

MeABP = 186.4 ºF

Using Figure 2B2.1 from the API technical data book

The molecular weight, MW = 96

and the Watson characterization factor K W = 12.2(Compare to 12.3 calculated previously).


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The aniline point = 125 ºF.

H/C ratio = 5.5

MABP = 179.4 ºF → Figure 4A1.2 → T pc = 490 ºF.

WABP = 200.4 ºF → Figure 4A1.2 → Tc = 510 ºF.

MeABP = 186.4 ºF → Figure 4B1.2 → P pc = 440 psia.

Tc/T pc = (510+460)/(490+460) = 970/950 = 1.02

P pc = 440 & Tc/T pc = 1.02 → Figure 4B1.1 → Pc = 520 psia.


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Liquid Viscosity

1. The absolute (or dynamic) viscosity.

Defined as the ratio of shear resistance to the shear velocitygradient.

This ratio is constant for Newtonian fluids.

Expressed in Pa.s (poise)

Commonly used unit is mPa.s (centipoise, cP)

2. The kinematic viscosity.

Defined as the ratio between the absolute viscosity and thedensity.

Expressed in mm2/s (centistokes, cSt)

The liquid dynamic viscosities at 100 ºF and 210 ºF are used

to characterize (heavy) petroleum fractions.

Viscosities can be estimated by the relation of Abbott et al.

2

100log 4.39371 1.94733 0.12769

W W K K

4 2 2

3.2629.10 1.18246.10W

A K A


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4-40

2 2(8.0325.10 1.24899 0.19768 )

( 26.786 2.6296 )

W

W

K A A

A K

(4.14)

210

4 2 3

2 2

log 0.463634 0.166532

5.13447.10 8.48995.10

(8.0325.10 1.24899 0.19768 )

( 26.786 2.6296 )

W

W

W

A

A K A

K A A

A K

(4.15)

whereK w = Watson characterization factor

A = API gravity

v100 = viscosity at 100 ºF [mm2/s]

v210 = viscosity at 210 ºF [mm2/s]log = common logarithm (base 10)

notes:

Should not be used if K w < 10 and A < 0.

Recommended for the following range;

0.5


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4-41

1. From the normal BP and standard specific gravity.

Riazi method: for light fractions

(sp.gr. < 0.97 & T b < 840 K).

M = 42.965 (T b1.26007 S4.98308)

[exp(2.097.10-4

T b – 7.78712 S + 2.08476.10-3

T b S)]

Lee-Kesler: for heavy petroleum fractions

(T b > 600 K & 60 < MW < 650).

7

2

12

2

3

12272.6 9486.4 (8.3741 5.9917 )

10 222.466 (1 0.77084 0.02058 ) 0.7465

10 17.3354 (1 0.80882 0.02226 ) 0.32284

b

b b

b b

M S T S

S S T T

S S T T

Where

M = Molecular weight [kg/kmol].

T b = Normal boiling point [K].

S = Standard specific gravity.


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2. From the viscosities at 210 ºF and 100 ºF and the standardspecific gravity (ave. error is 10%).

(1.1228 1.2435) (3.4758 3.038 ) 0.6665

100 210223.56 S S

M S

where

M = Molecular weight [kg/kmol].

v100 = viscosity at 100 ºF [mm2/s].

v210 = viscosity at 210 ºF [mm2/s].


Pseudo-Critical Constants for Petroleum Fractions.

To make use of the principle of corresponding states.

Use the method of Lee-Kesler (ave. error 10%)

1. Pseudo-Critical Temperature.

189.8 450.6 (0.4244 0.1174 )

(14,410 100,688 )

C b

b

T S T S

S

T

where,


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4-43

Tc = Pseudo-critical temperature [K].



2. Pseudo-Critical Pressure

3

2

7 2

2

10 3

2

0.0566ln 5.68925

4.12164 0.213426 10 0.436392

11.819 1.53015 10 4.75794

9.901 10 2.45055

C

b

b

b

P S

T S S

T

S S

T S

where,

Pc = Pseudo-critical pressure [bar].

ln = Napierian logarithm




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4-44

Acentric Factor for Petroleum Fractions.

For T r < 0.8

2

7.904 0.1352 0.007465

1.408 0.1063 8.359

W W

W

br

br

K K

K T

T

b

br

c

T

T T

where

ω = acentric factor.

T br = reduced boiling point temperature.

K w = Watson characterization factor.

For T r > 0.8 (use Edmister's equation)

3

(log 1.0057)7 1 c x

P x

where b

c

T x

T

and

Pc = Pseudo-critical pressure [bar].

Tc = Pseudo-critical temperature [K].


45/51


4-45

log = common logarithm (base 10)


Specific Heat of Petroleum Fraction in the Ideal Gas State

5432325.2 FT ET DT CT BT A H gp

432 5432185.4 FT ET DT CT BC pgp

'8.1 T T

S K B

W

2846.029502.002972.035644.0

'24

05543.05524.19247.22

10C K K C

W W

S C

0694.50283.6

'

0844.06946.13

10 7

D

2

410.7.0885.0

1011

8.12

S S

K K W W

Kw = Watson characterization factor


46/51


4-46

S = standard specific gravity

Flash Point

The API method (error 5°C)

T T

f

ln 1010

0034254.0

84947.2

02421.0

1

T10 = temperature at the 10% volume distilled point from ASTM

D86 [k].

Liquid EnthalpyHL = A1 7.259T A2 22 7.259T A3 33 7.259T

A1= 10-3

W

W

K K

4653582.1149)907.24722.23(26.1171

A2= 10-6

817.13086.56)82463.00.1( W K

A3= -10-9

3653.26757.9)82463.00.1( W K

Temperature (T)

Specific Gravity (SG)

Characterization Factor (Kw)


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4-47

Vapor enthalpy

266.5507.4 512.0

64.0 8.0

33

3

22

21

C

O

C

C

C C LV

RT

H H

MW

RT

T T B

T T BT T B H H

S

BW

B K 46.248

02.2953

1041

72.2944.356

S

BW W

B K K 46.25342.3016102 4)772.262.77(24.146

4

95.2487.569

103

B B

2

4

4 107.0885.00.10

0.10.18.12

S S

K K W W

B

HL Liquid Enthalpy of Petroleum Fractions

T Temperature

Tc Critical Temperature

R Gas Constant

MW Molecular Weight

ω Acentric Factor

S Specific Gravity

K W Watson Characterization Factor

(Ho- H)/RTC Pressure Effect on Enthalpy


48/51


4-48

Hv Vapor Enthalpy of Petroleum Fractions

Calculation of Density by the Lee and Kesler Method

RT z

PM

V T

V V c

Vri

Di

Vri

Ci

Vri

Bi

Tr

Vri zi

T

d d D

T

c

T

ccC

T

b

T

b

T

bb B

ww

ww Z Z Z Z

Z Z w Z Z

l

l

rir

ri

ii

ri

ii

r

i

ii

r

i

r

i

ii

r

i

r

i

r

i

ii

23

224

52

21

3

321

3

4

2

321

12

1121

121

)exp()(

1Pr

)(

)('''5138.2

Pressure for Saturation

exp[ ( , )]m m s c r m

P P f T

0 0( , ) ln ln

mr m r m r T P P

6

0

6.09648ln 5.92714 1.28862ln 0.169347

m m

m

r r r

r

P T T

T


49/51


4-49

1

615.6875ln 15.2518 13.4721ln 0.43577r m m

m

r r

r

P T T T

Pressure Correction for Density

1 ln1

s

s

B P C

B P

0.0861488 0.0344483m

C

4

1

a 1k

cm k

k

B P

1/ 3

1mr

T

Estimation of the pour point (page 172):

)333.031.0(

100

)474.0612.0(971.247.130 S S

EC v M S T

Estimation of the I nterfacial Tension of Petroleum fractions:

(page 167)

Kw

Tcf f

232.1]

15.2931[7.673

Thermal Conductivities of L iquids: (page 132)

T E l

4418.117.0

Influence of Pressure on the Viscosity of Liquids (Kouzel's

method):


50/51


4-50

)4479.14829.5)((log 181.0 E Ms E Ps P Ms

M

Specific Heats for liquid Petroleum Fractions (Lee Kessler

1975) Page 121:

))410*508.5310*467.1(16734.03065.0)(055.035.0(185.4 S T S Kwl

Cp


51/51


Exercises

4.1. Repeat example 4.5 using another method by calculating K w for

pure components to find the same for the mixture.

4.2. Classify the crude oil handed out to you using the US bureau of

mines classification.

4.3. Is your crude oil suitable for asphalt or lube oil manufacture?

4.4. Draw the TBP and API curves for the crude oil assay handed out to

you. Use the probability density function to construct the heavier

portion of your TBP curve.

4.5. Estimate the MW, aniline point, and H/C ratio for the crude oil

handed out to you.

Documents

Crude Oil Characterization