Question 1 of 5: Petroleum-refining terminology and API blending

Question 1: Petroleum-refining terminology and API blending (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.

(a) Definitions

(i) Asphaltenes. The heaviest, most polar, polynuclear-aromatic fraction of a crude oil, defined operationally as the fraction that is insoluble in a light n-paraffin (n-heptane or n-pentane, ASTM D6560/D3279) but soluble in aromatic solvents such as toluene. They carry most of the crude's metals (Ni, V), sulphur and nitrogen, and are the principal precursors of coke, sediment and fouling.

(ii) Petroleum coke. The solid, carbon-rich residue (typically >90 wt% carbon) left when a heavy residuum is subjected to severe thermal cracking in a coker (delayed or fluid coking). Fuel-grade (“sponge”) coke is burned as a solid fuel; high-purity calcined needle coke is used to make graphite anodes and electrodes.

(iii) Watson characterization factor (UOP K factor). A dimensionless index of a fraction's paraffinicity, $K_W=\dfrac{T_B^{1/3}}{\text{SG}}$, where $T_B$ is the mean average boiling point in degrees Rankine and SG is the 60/60 °F specific gravity. $K_W\approx 12.5$ indicates a highly paraffinic stock, $\approx 11$ naphthenic, and $\approx 10$ highly aromatic.

(iv) Reid vapour pressure (RVP). The absolute vapour pressure of a volatile product measured at 100 °F (37.8 °C) in a standardized bomb at a 4:1 vapour-to-liquid ratio (ASTM D323), reported in psi (or kPa). It is the standard proxy for a gasoline's front-end volatility and evaporative / vapour-lock behaviour.

(v) Olefins. Unsaturated hydrocarbons containing at least one carbon–carbon double bond ($\text{C=C}$), i.e. alkenes such as ethylene and propylene. They are not present in crude but are formed by cracking; they raise octane yet are chemically reactive and promote gum formation and storage instability.

(vi) Fire point. The lowest temperature at which the vapour above a heated sample ignites and sustains combustion for at least 5 s when a test flame is applied (ASTM D92, open cup). It lies a few degrees above the flash point (which is only a momentary flash) and is a fire-safety property.

(vii) Total Acid Number (TAN). The mass of potassium hydroxide, in milligrams, required to neutralize the acidic constituents in 1 g of oil (ASTM D664). It quantifies naphthenic-acid content and is the standard index of a crude's corrosivity toward refinery metallurgy.

(viii) Visbreaking. A mild thermal-cracking process (roughly 450–500 °C at 3–10 bar, in a soaker drum or a coil furnace) applied to atmospheric or vacuum residue purely to break the viscosity of the residue. Light cracking of the long paraffinic side chains lowers the residue's viscosity and pour point by an order of magnitude, so far less valuable light cutter stock is needed to blend it to a heavy fuel-oil specification; a small yield of gas and cracked naphtha/gas oil is produced as a by-product. Severity is limited by the onset of coke formation and by fuel-oil storage stability.

(ix) API gravity. An inverse density scale for petroleum defined by the American Petroleum Institute, ${}^{\circ}\text{API}=\dfrac{141.5}{\text{SG}(60/60\,{}^{\circ}\text{F})}-131.5$. Water is 10 °API; lighter (less dense) oils have higher °API.

(b) Density of the blended oil

Given.

Find. The mixture density in the two units the question asks for, (i) lb/ft³ and (ii) lb/US gallon. (SI density in kg/m³ is carried as the working intermediate.)

Approach. Convert each API gravity to a real density, then — because volumes are additive — take the volume-weighted mean density and convert its units.

1. Convert API gravity to specific gravity. Using $\text{SG}=\dfrac{141.5}{131.5+{}^{\circ}\text{API}}$: $$\text{SG}_1=\frac{141.5}{161.5}=0.8762,\qquad \text{SG}_2=\frac{141.5}{146.5}=0.9659.$$
2. Convert to density. With $\rho=\text{SG}\times\rho_w$ and $\rho_w=999.0\ \text{kg/m}^3$: $$\rho_1=875.3\ \text{kg/m}^3,\qquad \rho_2=964.9\ \text{kg/m}^3.$$
3. Volume-weighted mixture density. Mass is conserved and, with no volume change, volume is additive, so $$\rho_{\text{mix}}=\frac{V_1\rho_1+V_2\rho_2}{V_1+V_2} =\frac{1000(875.3)+5000(964.9)}{6000}=950.0\ \text{kg/m}^3.$$ (Note the fuel oil dominates 5:1, so the blend sits close to the heavier component.)
4. (i) Convert to lb/ft³. $1\ \text{lb/ft}^3=16.0185\ \text{kg/m}^3$, so $$\rho_{\text{mix}}=\frac{950.0}{16.0185}=\boxed{59.3\ \text{lb/ft}^3}.$$
5. (ii) Convert to lb/US gallon. $1\ \text{lb/US gal}=119.826\ \text{kg/m}^3$, so $$\rho_{\text{mix}}=\frac{950.0}{119.826}=\boxed{7.93\ \text{lb/US gal}}.$$