Estimating petroleum-fraction properties from correlations: 23-Chem-B6 Petroleum Refining and Petrochemicals, December 2018, Q5

Question 5 — Estimating petroleum-fraction properties from correlations (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $M=300\ \text{kg/kmol}$, $\mathrm{SG}=0.90$ at 60 °F.

Find. API, $K$, mean average boiling point (MeABP), standard density, pseudo-critical $T_c$/$P_c$, liquid heat capacity at 100 °C and viscosity at 80 °C.

Check — correlation basis. This is a "use charts/correlations" estimation, so the method is what is marked. Boiling point, $T_c$ and $P_c$ use the Riazi–Daubert (1980) two-parameter correlations $\theta=a\,T_B^{\,b}\,\mathrm{SG}^{\,c}$ (English units, $T_B$ in °R); the liquid heat capacity uses the API/Watson–Nelson correlation and the viscosity the Abbott (API) kinematic-viscosity correlations with an ASTM D341 (Walther) temperature interpolation. Chart readings would give values within a few percent of these; parts (f) and (g) are the most method-sensitive.

(a) API gravity. Directly from the definition, $$^{\circ}\text{API}=\frac{141.5}{\mathrm{SG}}-131.5=\frac{141.5}{0.90}-131.5=\boxed{25.7}.$$
(c) Mean average boiling point (needed first, for $K$ and the criticals). Invert the Riazi–Daubert molecular-weight correlation $M=4.5673\times10^{-5}\,T_B^{2.1962}\,\mathrm{SG}^{-1.0164}$ ($T_B$ in °R) for $T_B$: $$T_B=\left[\frac{M}{4.5673\times10^{-5}\,\mathrm{SG}^{-1.0164}}\right]^{1/2.1962}=1211\ ^{\circ}\text{R}.$$ Converting, $T_B=1211-459.67=751\ ^{\circ}\text{F}=\boxed{399\ ^{\circ}\text{C}}$ (672.6 K) — a heavy gas-oil fraction, consistent with $M=300$.
(b) Watson characterization factor. $$K=\frac{(T_B)^{1/3}}{\mathrm{SG}}=\frac{(1211)^{1/3}}{0.90}=\frac{10.66}{0.90}=\boxed{11.8}.$$ A value near 11.8 indicates an intermediate paraffinic–naphthenic stock.
(d) Density at standard conditions. With the density of water at 60 °F taken as 999.0 kg/m³, $$\rho=\mathrm{SG}\times\rho_{\text{water}}=0.90\times999.0=\boxed{899\ \text{kg/m}^3}.$$
(e) Pseudo-critical temperature and pressure (Riazi–Daubert). $$T_c=24.2787\,T_B^{0.58848}\,\mathrm{SG}^{0.3596}=1524\ ^{\circ}\text{R}=\boxed{847\ \text{K}},$$ $$P_c=3.12281\times10^{9}\,T_B^{-2.3125}\,\mathrm{SG}^{2.3201}=181.5\ \text{psia}=\boxed{1251\ \text{kPa}} \;(\approx 1.25\ \text{MPa}).$$
(f) Liquid heat capacity at 100 °C (212 °F). Using the API/Watson–Nelson correlation with $t$ in °F, $$c_p=\bigl(0.355+1.28\times10^{-3}\,{}^{\circ}\text{API}\bigr)+\bigl(0.503+1.17\times10^{-3}\,{}^{\circ}\text{API}\bigr)\times10^{-3}\,t,$$ $$c_p=0.388+0.113=0.501\ \tfrac{\text{BTU}}{\text{lb}\cdot^{\circ}\text{F}}=\boxed{2.10\ \tfrac{\text{kJ}}{\text{kg}\cdot^{\circ}\text{C}}}.$$ (The simpler Watson–Nelson form $c_p=(0.388+4.5\times10^{-4}t)/\sqrt{\mathrm{SG}}$ gives 2.13 kJ/(kg·°C), confirming the estimate.)
(g) Absolute viscosity at 80 °C. The Abbott (API) correlations, evaluated at $^{\circ}\text{API}=25.7$ and $K=11.8$, give kinematic viscosities $\nu_{100^{\circ}\text{F}}=24.7\ \text{cSt}$ and $\nu_{210^{\circ}\text{F}}=4.74\ \text{cSt}$. Fitting the ASTM D341 (Walther) line $\log\log(\nu+0.7)=A-B\log T$ (T in K) through these two points and evaluating at 80 °C (353 K) gives $\nu_{80^{\circ}\text{C}}=7.0\ \text{cSt}$. With the temperature-corrected liquid density $\rho_{80^{\circ}\text{C}}\approx859\ \text{kg/m}^3$, $$\mu=\nu\,\rho=7.0\ \text{cSt}\times0.859\ \tfrac{\text{g}}{\text{cm}^3}=\boxed{6.0\ \text{cP}}.$$

Property	Estimate
(a) API gravity	25.7
(b) Watson characterization factor $K$	11.8
(c) Mean average boiling point	399 °C (1211 °R)
(d) Density at standard conditions	899 kg/m³
(e) Pseudo-critical $T_c$ / $P_c$	847 K / 1251 kPa
(f) Liquid heat capacity at 100 °C	2.10 kJ/(kg·°C)
(g) Absolute viscosity at 80 °C	≈ 6.0 cP

Concept & Theory

Topic. Predicting petroleum-fraction physical properties from two easily measured inputs (molecular weight and specific gravity).

Key relations. $^{\circ}\text{API}=141.5/\mathrm{SG}-131.5$; Watson $K=(T_B)^{1/3}/\mathrm{SG}$; Riazi–Daubert $\theta=a\,T_B^{\,b}\,\mathrm{SG}^{\,c}$ for $M$, $T_c$, $P_c$; API liquid $c_p$ correlation; Abbott kinematic-viscosity correlations with the ASTM D341 (Walther) viscosity–temperature line; $\mu=\nu\rho$.

Why this works. Petroleum fractions are complex mixtures, but their properties correlate tightly with just the boiling point and specific gravity (equivalently $M$, SG and $K$), because these capture average molecular size and hydrogen content. The Riazi–Daubert forms are fitted to hundreds of fractions, so inverting the $M$ correlation for $T_B$ then chaining to $T_c$, $P_c$, $c_p$ and $\mu$ is standard practice.

Common pitfall. Mixing unit systems — the Riazi–Daubert constants require $T_B$ in °R (not K or °F), and Watson $K$ uses $T_B$ in °R. Viscosity must be interpolated on the double-log Walther scale, never linearly in temperature, and remember to convert kinematic (cSt) to absolute (cP) with the density at the same temperature.

Source. M. R. Riazi, Characterization and Properties of Petroleum Fractions (Riazi–Daubert, Abbott, ASTM D341); Fahim et al. ch. 3–4; API Technical Data Book.

Question 5 of 5: Estimating petroleum-fraction properties from correlations

Question 5 — Estimating petroleum-fraction properties from correlations (10 marks)