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 be completely identified using gas chromatography analysis. For the white fractions (e.g. Naphtha), only a limited number of components can be completely identified using PINA, PIONA, or Detailed HC GC analyzers. To overcome this shortcoming, petroleum refiners resort to define (characterize) petroleum and its fractions using global (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 N and metals content are increasing in importance. The price of petroleum fractions is influence by other properties in addition. API Gravity (API) A measure of crude oil density/specific gravity. API = 141.5 11.5 sp.gr. (4.1) sp.gr. = (60 F ) oil (60 F ) water (4.) Light crude has lower density (sp.gr.) API is higher. (heavy crude) 10 < API < 50 (light crude) (or less) (or more) Crudes with high API gravity produce more distillates (more valuable) than residue (less valuable) and are usually low in sulfur. 4-1
Copyrights 001 015, Dr. Tareq Albahri, Chem. Eng. Dept., Kuwait University Distillation Range (Curve) The boiling point range is an alternate method to represent 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 the most desirable processing sequence to obtain the required products. - Can be used to determine whether the crude is suitable for asphalt or lube oil manufacture. Types of distillation curves 1. TBP (True Boiling Point) distillation curve.. ASTM (D86/D1160) distillation curve.. EFV (Equilibrium Flash Vaporization) curve. TBP is the most useful. Gases Crude Oil Residue LPG Naphtha Kerosene Diesel Gas Oil - However, no standard test exists for measuring the TBP. - Most common TBP test is Hempel analysis & D-85. (Neither specifies # of stages or reflex ratio used). - Trend is toward 15/5 distillation (D-89) 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. It is used to characterize petroleum products. Kuwait refineries use ASTM curve (which is then converted to TBP curve). EFV curve is used to predict the vapor-liquid phase separation of petroleum products such as when predicting the amount of vapor and liquid resulting from a flash drum. 4-
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Copyrights 001 015, Dr. Tareq Albahri, Chem. Eng. Dept., Kuwait University Example 4-1: Estimate the true boiling point (TBP) distillation curve of the petroleum fraction having the following ASTM D86 distillation temperatures: a. Using graphical method b. Using numerical method (equations) Solution: Vol % T (ºF) IBP 80 5% - 10% 400 0% 40 50% 48 70% 460 90% 485 490 95% - FBP 50 57.5 Recovery = - Using Fig. A1.1 from the API technical Data Book on Petroleum Refining (1) Correct the ASTM D86 distillation temperatures above 475 ºF for cracking using Hadden equation. Log D = 1.587 + 0.0047 T, where, T & D are in ºF For the 90% temp Log D = 1.587 + 0.0047 (485) = 0.707 D = 5 ºF The corrected 90% temp becomes = 485 + 5 = 490 ºF For the FBP Log D = 1.587 + 0.0047 (50) = 0.876 D = 7.5 ºF The corrected 90% temp becomes = 50 + 7.5 = 57.5 ºF () Find the atmospheric TBP mid (50%) temperature using the mid temp of the ASTM D86 and the lower part of Figure A1.1. 48 ºF on the x-axis T= ºF on the y-axis [correction] Therefore, the TBP 50% T =ASTM 50% T + T = 48+ = 440 ºF () Find the temp for each segment of the TBP curve using upper part of Figure A1.1. Segment of Curve (V%) 0 to 10 10 to 0 0 to 50 50 to 70 70 to 90 90 to 100 ASTM T (ºF) [from table above] 0 0 18 0 7.5 TBP T (ºF) [Figure A1.1] 40 9 1 40 41 TBP (ºF) Vol % 70 40 = 0 409 9 = 70 440 1 = 409 440 + = 47 47 + 40 = 51 51 + 41 = 556 IBP 10% 0% 440 50% 70% 90% FBP Note: it is possible to convert TBP temperatures to ASTM D86 using Figure A.1 from the API technical data book, which involves a trial and error procedure. 4-4
Alternative method: (Do not correct temperatures for cracking) Conversion of ASTM 50% Point to TBP 50% Point Temperature The following equation can be used to convert an ASTM D86 50% temperature to a TBP 50% temperature: TBP (50%) = 0.87180 x [ASTM D86 (50%)T] 1.058 Where: TBP (50%) = true boiling point distillation temperature at 50 vol% distilled ( ⁰ F); ASTM D86 (50) = observed ASTM D86 distillation temperature at 50 vol% distilled ( ⁰ F). Determination of TBP cut points from ASTM D86 The difference between adjacent TBP can be determined by the following equation: Yi = Ai Xi Bi where: Yi = difference in TBP distillation between two cut points ( ⁰ F); Xi = observed difference in ASTM D-86 distillation between two cut points ( ⁰ F); A, B = constants varying for cut points ranges, shown in the following table: i Cut point range (%) A B Xi ( T) Yi 1 4 5 6 0 10 10 0 0 50 50 70 70 90 90 100 7.401 4.9004.005.58.0419 0.11798 0.6044 0.71644 0.80076 0.8007 0.75497 1.6606 TBP (0) = TBP (50) Y Y Y1 TBP (10) = TBP (50) Y Y TBP (0) = TBP (50) Y TBP (70) = TBP (50) + Y4 TBP (90) = TBP (50) + Y4 + Y5 TBP (100) = TBP (50) + Y4 + Y5 + Y6 Source: T. E. Daubert, Petroleum fraction distillation inter-conversions, Hydrocarbon Process. 7(8) (1994) 75-78. Class work: solve the previous example using this alternative method. 4-5
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering 4-6
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Prof. Tareq Albahri 018 Kuwait University Chemical Engineering Characterization & Classification Characterization Factors Correlate between the yield and the aromaticity & paraffinicity of petroleum oils. Several correlations exist, 1. UOP or Watson characterization factor (Kw) K w = 1 B T (sp.gr.) @60F (4.5) More commonly used. TB is the normal boiling point of the pure compound or the mean average boiling point of a petroleum fraction in ºR [ºF+460]. Higher for lighter components and petroleum fractions e.g. Kw 1.1 (Naphtha), Kw 11.9 (Diesel) For petroleum fractions 8 < Kw < 15 Contain Highly aromatic compounds For crude oil 10.5 < Kw < 1.9 Contain Highly paraffinic compounds Highly naphthenic crude Highly paraffinic crude For pure hydrocarbons Kw = 1 for paraffins. Kw = 1 for HC with equivalent chain and ring weights. Kw = 11 for pure naphthenes. Kw = 10 for pure aromatics.. US Bureau of Mines Correlation Index CI = 87,55 + 47.7 (sp.gr.)@60f 456.8 (4.6) T B 4-8
- 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. Higher CI values high conc. of naphthenes & aromatics. Example 4-: 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 0.6 Using equation (5.) above 1 (97+460) K w = = 1.06 (0.6) Using equation (5.4) above = 87,55 CI + 47.7 (0.6) 456.8 0 (97+460) Table 4-1: Comparison of Kw and CI for pure components Compound TB (ºF) Sp.gr. Kw CI n-hexane -Methylpentane Cyclohexane Benzene Naphthalene 155.7 140.47 177.9 176.18 884 0.664 0.658 0.785 0.8845 1.176 1.8 1.8 11.0 9.7 8.16 0 0.7 51.7 99.8 199. 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. 4-9
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering Characterization of Well Defined Hydrocarbon Mixtures For hydrocarbon mixtures for which the composition is known, the pure component physical properties and acentric factors adequately characterize the system. For more complex mixtures such as petroleum fractions, it neither practical nor possible to analyze the entire mixture to define the concentration of all components. These undefined mixtures are usually characterized by parameters that are derived from the normal inspection tests, an ASTM D86 or D1160 distillation, and the specific gravity of the mixture. Many characterizing parameters have been proposed but very few are generally useful. Among the useful parameters are five different boiling points and the Watson characterization factor Boiling points Each boiling point reduces to the normal BP for pure hydrocarbons and is significant for a different group of correlations. Volumetric average boiling point VABP = n xvitbi i=1 (4.7) xvi = volume fraction of component i. Tbi = normal boiling point of component i (ºF or R). Molal average boiling point MABP = n xt i bi i=1 xi = mole fraction of component i. Tbi = normal boiling point of component i (ºF or R). (4.8) Weight average boiling point WABP = n xwitbi i=1 xwi = weight fraction of component i. Tbi = normal boiling point of component i (ºF or R). (4.9) Cubic average boiling point [R] CABP = n 1/ (4.10) vi bi i=1 x T Tbi = normal boiling point of component i (R only). Mean average Boiling point MeABP = MABP + CABP (4.11) 4-10
Watson characterization factor 1/ (MeABP) (4.1) K = (sp.gr.) w @60 F - 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 is known. n K = (4.1) w xwiki Example 4- Calculate the Watson characterization factor and the correlation index for the following LPG mixture. Vol % C 45 n-c4 50 n-c5 5 i=1 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) Kw C 45 44 0.507-4.75 (416) 14.70 n-c4 50 58 0.584 1 (491) 1.50 n-c5 5 7 0.6 97 (557) 1.0 Average Sp.gr. = (0.45)(0.507)+(0.5)(0.584)+(0.05)(0.6) = 0.55 VABP = (0.45)(416)+(0.5)(491)+(0.05)(557) = 460.7 R MABP = VABP (for ideal gas) C wt% = (0.45 44)/[(0.45 44)+(0.5 58)+(0.05 7)] = 7.8 wt% C4 wt% = (0.5 58)/[(0.45 44)+(0.5 58)+(0.05 7)] = 55. wt% C5 wt% = (0.05 7)/[(0.45 44)+(0.5 58)+(0.05 7)] = 6.9 wt% WABP = (0.78)(416)+(0.55)(491)+(0.069)(557) = 467. ºR 1/ 1/ 1/ CABP = 0.45(416) 0.5(491) 0.05(557) 459.4 R MeABP = (MABP+CABP)/ = (460.7+459.4)/ = 460 R Into equation () above 1 (460) K w = = 1.9 (0.55) or Kw = xwi kwi = 0.78(14.7) + 0.55(1.5) + 0.069(1.0) = 1.9 Into equation (4) above = 87,55 CI + 47.7 (0.55) 456.8 5 (460) 4-11
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering Example 4-4: Watson K for petroleum fractions For the following SR Naphtha, 69.1 ºAPI & ASTM D86 distillation 1 Vol % T (ºF) IBP 9 5% 118 10% 18 0% 164 50% 198 70% 0 90% 6 95% 7 FBP 00 Recovery = 98.8 vol.% Calculate, (1) The volumetric average boiling point, VABP () The weight average boiling point, WABP () 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, VABP 90 Ti i 10 VABP 5 VABP = T10% T 0% T 50% T 70% T 90% = 18+164+198+0+6 = 196.4 5 5 Assuming ASTM curve is linear between 10 & 90% ASTM slope = T90% T10% 618 1.675 9010 80 ºF From Figure B1.1 API data book For MABP T = -17 ºF For CABP T = - ºF For WABP T = +4 ºF For MeABP T = -10 ºF Using the corrections MABP = 196.4 17 = 179.4 ºF CABP = 196.4 = 19.4 ºF WABP = 196.4 + 4 = 00.4 ºF 1 D1160 can t be used 4-1
MeABP = 196.4 10 = 186.4 ºF or = (179.4+19.4)/ = 186.4 ºF The Watson characterization factor, Kw Sp.gr. = 141.5/(69.1+11.5) = 0.705 1/ 1/ (MeABP) (186.4+460) K = 1. w (sp.gr.)@60f (0.705) (mainly saturated) The US bureau of mines correlation index, CI. = 87,55 CI + 47.7 (0.705) 456.8 0.7 (paraffinic) (186.4+460) 4-1
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering 4-14
Coordination Number (Hydrogen Deficiency) or Z Number Cn H n+z As Z-number decreases, density increases. Group Paraffins Olefins Naphthenes Aromatics Z 0 0-6 Examples of Z-values for Aromatic compounds C6H6 Z = -6 C10H8 Z = -1 C14H10 Z = -18 C10H1 Z = -8 C10H10 Z = -10 C4H1 Z = - 6 4-15
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering Properties Estimation of Petroleum Fractions 1. Graphical Methods.. Numerical Methods. Example 4-5: Graphical Method For the SR Naphtha in the previous example, Calculate the MW, aniline point, C/H ratio, the true and pseudocritical temperatures and pressures. Solution: Previously calculated, the mean average boiling point, MeABP = 186.4 ºF Using Figure B.1 from the API technical data book The molecular weight, MW = 96 and the Watson characterization factor KW = 1. (Compare to 1. calculated previously). The aniline point = 15 ºF. C/H ratio = 5.5 MABP = 179.4 ºF Figure 4A1. Tpc = 490 ºF. WABP = 00.4 ºF Figure 4A1. Tc = 510 ºF. MeABP = 186.4 ºF Figure 4B1. Ppc = 440 psia. Tc/Tpc = (510+460)/(490+460) = 970/950 = 1.0 Ppc = 440 & Tc/Tpc = 1.0 Figure 4B1.1 Pc = 50 psia. 4-16
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Prof. Tareq Albahri 018 Kuwait University Chemical Engineering 4-18
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Prof. Tareq Albahri 018 Kuwait University Chemical Engineering 4-0
Numerical Methods for Characterization and Properties Estimation of Petroleum Fuels Introduction Graphical methods have long been replaced by numerical methods with the developments of correlation techniques (software) and advancement in computers (hardware). Using correlations is more preferable due to their simplicity and widespread of (programmable) hand calculators and personal computers. These correlations are currently incorporated in chemical engineering process simulators Liquid Viscosity 1. The absolute (or dynamic) viscosity. Defined as the ratio of shear resistance to the shear velocity gradient. This ratio is constant for Newtonian fluids. Expressed in Pa.s (poise) Commonly used unit is mpa.s (centipoise, cp). The kinematic viscosity. Defined as the ratio between the absolute viscosity and the density. Expressed in mm /s (centistokes, cst) The liquid dynamic viscosities at 100 ºF and 10 ºF are used to characterize (heavy) petroleum fractions. When the viscosities are not known one of the following relations can be used to estimate them. Abbott et al (1971) for example uses the Watson characterization factor (Kw) and API gravity (A) to predict viscosity at 100 and 10 F as follows where logv 100 = 4.971 1.947 K w + 0.1769 K w +.69. 10 4 A 1.1846. 10 K w A + (0.171617K w + 10.994A + 9.5066. 10 A 0.86018 K w A) (A + 50.64 4.781K w ) logv 10 = 0.4664 0.1665 A + 5.1447. 10 4 A 8.48995. 10 K w A + (8.05. 10 K w + 1.4899A + 0.19768 A ) (A + 6.786.696K w ) Kw = Watson characterization factor (Kw > 10) A = gravity in degrees API (A > 0) v100 = viscosity at 100 [mm /s] 0.5 < v100 < 0 mm /s v10 = viscosity at 10 [mm /s] 0. < v10 < 40 mm /s log = common logarithm (base 10) Average error about 0%. Source: M. M. Abbott, T. G. Kaufmann and L. Domash. A correlation for predicting liquid viscosities of petroleum fractions. The Canadian Journal of Chemical Engineering, Volume 49, Issue, pages 79 84, 1971 (DOI: 10.100/cjce.545049014) 4-1
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering Molecular Weight Can be estimated with two different means (5% ave error) 1. From the normal BP and standard specific gravity. Riazi method: for light fractions (sp.gr. < 0.97 & Tb < 840 K). M = 4.965 (Tb 1.6007 S 4.9808 ) [exp(.097.10-4 Tb 7.7871 S +.08476.10 - Tb S)] Lee-Kesler: for heavy petroleum fractions (Tb > 600 K & 60 < MW < 650). M 17.6 9486.4 S T (8.741 5.9917 S ) (1 0.77084 S 0.0058 S ) 0.7465 Tb 7 10.466 (1 0.8088 S 0.06 S ) 0.84 Tb b 1 10 17.54 T b T b where M = Molecular weight [kg/kmol]. Tb = Normal boiling point [K]. S = Standard specific gravity.. From the viscosities at 10 ºF and 100 ºF and the standard specific gravity (ave. error is 10%). M.56 (1.18 S 1.45) (.4758.08 S ) 0.6665 100 10 S Aniline Point (AP) where M = Molecular weight [kg/kmol]. v100 = viscosity at 100 ºF [mm /s]. v10 = viscosity at 10 ºF [mm /s]. S = Standard specific gravity. The equation of Albahri et al. gives the aniline point in ºC, Ri is the refractivity intercept, n is the refractive index, d is liquid density at reference state of 0⁰C and 1 atm. in g/cm, Tb is the boiling point in K, and I is an index in refractive index. Ri = n d/ n = [(1+I)/(1 I)] 0.5 I = 0.84 (1.8Tb) -0.069 SG 0.918 d = 0.9855 (1.8Tb) 0.00016 SG 1.0055 AP (ºC) = 9805.69 (Ri) + 711.85761(SG) + 9778.7069 Pseudo-Critical Constants for Petroleum Fractions. To make use of the principle of corresponding states. Use the method of Lee-Kesler (ave. error 10%) 4-
1. Pseudo-Critical Temperature. (14,410 100,688 S) T c = 189.8 + 450.6 S + T b (0.444 + 0.1174 S) + where, Tc = Pseudo-critical temperature [K]. Tb = Normal boiling point [K]. S = Standard specific gravity.. Pseudo-Critical Pressure lnp c = 5.6895 0.0566 10 T S b (0.469 + 4.1164 + 0.146 S S ) where, +10 7 T b (0.475794 + 11.819 S Pc = Pseudo-critical pressure [bar]. ln = Napierian logarithm Tb = Normal boiling point [K]. S = Standard specific gravity. Acentric Factor for Petroleum Fractions For Tbr < 0.8 4- T b + 1.5015 S ) 10 10 T b (.45055 + 9.901 S ) ω = 7.904 + 0.15 K w 0.007465 K w + 8.59 T br + (1.408 0.0106 K w) T br where ω = acentric factor. Kw = Watson characterization factor. Tbr = reduced boiling point temperature. T br = T b T c For Tr > 0.8 (use Edmister's equation) where T br ω = 7 ( ) (logp 1 T c 1.0057) br Pc = Pseudo-critical pressure [bar]. Tc = Pseudo-critical temperature [K]. log = common logarithm (base 10) Tb = Normal boiling point [K]. T br = T b T c Flash Point temperature (Closed Cup) The API method (average error is about 5 C)
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering T f = where 1 0.041 +.84947 T 10 + 0.00454 ln T 10 Tf = flash point temperature [K] T10 = temperature at the 10% volume distilled point from ASTM D86 [K]. Estimation of the Interfacial Tension σ f = 67.7 (1 9.15 1. T ) cf K w Where σ f = the interfacial tension of a petroleum fraction f at 0 ᵒC [mn/m] Kw = Watson characterization factor Tcf = presudocrirtical temperature of the fraction f [K] Error is about 10% Estimation of carbon to hydrogen weight ratio For hydrocarbons with MW ranging from 70-00 Riazi proposed the following extended correlation, CH = 8.774.10-10 Tb -0.98445 SG -18.75 exp (7.176.10 - Tb + 0.064 SG 7.5.10 - Tb SG) where Tb is the normal boiling point [K] SG is the standard specific gravity. This equation gives an average error of %. Estimation of the solubility of water in oil log 10 x = (0.0016 1 ) (400 y + 1050) T Where y = weight ratio of hydrogen to carbon = 1/(CH) x = solubility of water as mole fraction T = temperature [K] Average error is about 0% This equation also applies to pure substances and mixtures. 4-4
Estimation of the pour point (API equation) TEC = 10.47 S (.971) M (0.61 0.474 S) (0.1 0. S) v100 where S = standard specific gravity M = molecular weight v100 = kinematic viscosity at 100 o F TEC = pour point temperature [kg/kmol] [mm /s] (K) Average error is about 5 o C. The method should not be used for pour points less than 60 C. Thermal Conductivities of Liquids: (ifp1 page 1) λl = 0.17 1.418.10-4 T where T = temperature [K] λl = thermal conductivity of the liquid [W/(m.K)] this equation is simple but can be highly inaccurate Influence of Pressure on the Viscosity of Liquids (Kouzel's method): M log ( P Ps)(5.89E 4Ms Ms 0.181 1.479E 4) Specific Heats for liquid Petroleum Fractions (Lee Kessler 1975) Page 11: Cp l 4.185(0.5 0.055Kw )(0.065 0.1674 S T (1.467 *10 5.508 *10 4 S )) Specific Heat of Petroleum Fraction in the Ideal Gas State H gp C pgp T 1.8T 4.5 A BT CT DT ET FT 4 4.185 B CT DT 4ET 5FT ' 0.846 B 0.5644 0.097 KW 0.950 S 4 10 C.947 1.554 K 0.0554 K C ' W W ' 5.0694 C 6.08 S 5 4-5
Prof. Tareq Albahri 018 Kuwait University Chemical Engineering 10 7 D 1.6946 0.0844 1.8 10 1 1 S 0.885 S 0.7.10 KW KW Where Kw = Watson characterization factor S = standard specific gravity 4 Liquid Enthalpy H L = A 1 T 59. 7 A 59.7 T A T 59.7 A 1 = 10 - A = 10-6 A = -10-9 Where 1171.6 (.7 4.907 ) K 1.0 0.846K ( W 1.0 0.846K ( W 1.817 ) 56.086.65 ) 9.6757 W 1149.84655K W Temperature (T) Specific Gravity (SG) Characterization Factor (Kw) Vapor enthalpy H V H B L B 1 T 0.8T B T 0.64T T 0.51T C O RT C H H 4.507 5.66 C MW RT C 48.46 56.44 9.7K B 95.0 B 10 1 W 4 B 6 5.46 10 146.4 (77.6.77KW ) KW B 01.4 4 S B 9 10 56.487.95B4 1.8 10.0 4 B4 1.0 1.0 KW KW Where S 0.885 S 0.7 10 C S 4-6
HL = Liquid Enthalpy of Petroleum Fractions T = Temperature Tc = Critical Temperature R = Gas Constant MW = Molecular Weight ω = Acentric Factor S = Specific Gravity KW = Watson Characterization Factor (Ho- H)/RTC = Pressure Effect on Enthalpy Hv = Vapor Enthalpy of Petroleum Fractions Exercises 4.1. Repeat Example 4-4 using another method by calculating Kw for pure components to find the same for the mixture. 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.. Estimate the MW, aniline point, and H/C ratio for the crude oil handed out to you. 4-7