23-Chem-B6 Petroleum Refining and Petrochemicals · May 2016
Question 3 of 5: Characterization of a Crude Oil Cut
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Open-book, 3 hours; five questions of equal value (10 marks each), and exactly five questions constitute a complete paper — all five printed questions are solved below. Most parts call for concise, qualitative essay answers built on the given flow sheets; Question III is a short characterization calculation.
Reference texts: Gary, Handwerk, Kaiser & Geddes, Petroleum Refining: Technology and Economics (5th ed., CRC Press) — refinery configuration, conversion and treating units, product cuts; Fahim, Al-Sahhaf & Elkilani, Fundamentals of Petroleum Refining (Elsevier) — crude characterization factors, hydrotreating, coking, catalytic cracking; Jones & Pujadó, Handbook of Petroleum Processing (Springer) — unit operating windows; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 3: Characterization of a Crude Oil Cut (10 marks)
Check The printed molecular weight, “300 000 kg/mol,” is a misprint: no petroleum stream has that mass. The intended value is MW = 300 kg/kmol (300 g/mol), which together with an MeABP of 350 °C describes a realistic heavy cut (about a C₂₁ paraffin). All results below use MW = 300.
Approach. With only $M$ and $T_b$ known, back out the specific gravity from a two-parameter Riazi–Daubert molecular-weight correlation, then compute API, the Watson factor and the C/H ratio, and classify the cut from $K_w$.
Convert the boiling point. The Watson factor needs $T_b$ in Rankine and the MW correlation in Kelvin: $T_b = 350\,{}^{\circ}\!\mathrm{C} = 623.15\ \text{K}$, and $T_b(^{\circ}\!\mathrm{R}) = (350\cdot 1.8 + 32) + 459.67 = 1121.7\ ^{\circ}\!\mathrm{R}$.
Specific gravity from the Riazi–Daubert MW correlation. The two-parameter form (Fahim et al., Ch. 2) is
$$M = 1.6607\times10^{-4}\, T_b^{\,2.1962}\, SG^{-1.0164}\qquad(T_b\ \text{in K}).$$
Solving for $SG$ at $M=300$, $T_b=623.15\ \text{K}$:
$$SG = \left(\frac{1.6607\times10^{-4}\,T_b^{\,2.1962}}{M}\right)^{1/1.0164} = \left(\frac{227.8}{300}\right)^{0.9839}.$$
$$\boxed{SG_{60} \approx 0.763}$$
(The same correlation reproduces n-heptane and n-hexadecane to within a few percent, confirming the constant and units.)
API gravity. By definition
$$^{\circ}\!\mathrm{API} = \frac{141.5}{SG} - 131.5 = \frac{141.5}{0.763} - 131.5.$$
$$\boxed{^{\circ}\!\mathrm{API} \approx 54.0}$$
A high API confirms a light, low-density (paraffin-rich) material.
Watson (UOP) characterization factor. With $T_b$ in Rankine,
$$K_w = \frac{(T_b)^{1/3}}{SG} = \frac{(1121.7)^{1/3}}{0.763} = \frac{10.39}{0.763}.$$
$$\boxed{K_w \approx 13.6}$$
Carbon-to-hydrogen weight ratio. Since $K_w\approx13.6$ marks a strongly paraffinic cut, model the average molecule as an n-paraffin $\mathrm{C}_n\mathrm{H}_{2n+2}$ of $M=300$: $12.011\,n + 1.008(2n+2) = 300 \Rightarrow n \approx 21.2$. Then
$$\frac{C}{H} = \frac{12.011\,n}{1.008\,(2n+2)} = \frac{255.2}{44.8}.$$
$$\boxed{\left(\tfrac{C}{H}\right)_{\text{wt}} \approx 5.7}$$
Classify the crude. The Watson factor bands are $K_w\approx12.5\text{–}13$ (paraffinic), $\approx11$ (naphthenic), $\approx10$ (aromatic). Here $K_w\approx13.6$, the API is high (~54), the SG is low (~0.76) and the C/H ratio is low (~5.7) — every indicator points the same way.
$$\boxed{\text{Paraffinic}}$$
