Question 4 of 6: Gasoline properties, API gravity, and a blended-oil density

Question 4: Gasoline properties, API gravity, and a blended-oil density (20 marks — equal value)

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) Three gasoline properties (3 + 3 marks)

(i) Meaning.

• Reid vapour pressure (RVP) — the vapour pressure of the gasoline measured at 37.8 °C (100 °F) in a standard bomb (ASTM D323); an index of front-end volatility (light-end content).
• Boiling range — the ASTM D86 distillation curve from initial to final boiling point (with the 10/50/90% temperatures) that describes how the fuel vaporises across its cut.
• Antiknock characteristics — the fuel’s resistance to autoignition/knock, rated by octane number (Research RON, Motor MON, and pump $(R+M)/2$).

(ii) Why important.

• RVP must be high enough for cold starting yet low enough to avoid vapour lock and evaporative (VOC) emissions; it is therefore capped seasonally by regulation.
• Boiling range governs driveability: the front end sets cold-start and warm-up, the mid-range sets acceleration and smooth running, and the tail affects combustion completeness, deposits and oil dilution.
• Antiknock rating prevents destructive engine knock, which lets engines use higher compression ratios for greater power and thermal efficiency; inadequate octane forces spark retard and power loss.

(b) Definition of API gravity (4 marks)

API gravity is a standardised measure of the density of a petroleum liquid relative to water, defined from the specific gravity measured at 60 °F/60 °F:$$^\circ\!API = \frac{141.5}{SG_{60/60}} - 131.5, \qquad\Leftrightarrow\qquad SG_{60/60} = \frac{141.5}{131.5 + {}^\circ\!API}.$$It is a dimensionless scale (degrees API) on which water is exactly $10^\circ$API; higher API means lower density (lighter oil), and the relation is exactly invertible with specific gravity as shown.

(c) Density of the blended oils

Given. $V_1 = 8000$ bbl of $26^\circ$API gas oil blended with $V_2 = 20{,}000$ bbl of $16^\circ$API fuel oil; volumes additive; water at 60 °F has $\rho_w = 0.999$ g/cm³.

Find. The mixture density in (i) lb/US gal and (ii) lb/ft³.

Approach. Convert each API to specific gravity, volume-average the specific gravities (volumes additive), multiply by the water density to get the actual mixture density, then convert the units.

1. Specific gravities from API. $$SG_1 = \frac{141.5}{131.5+26} = 0.8984, \qquad SG_2 = \frac{141.5}{131.5+16} = 0.9593.$$
2. Volume-averaged mixture specific gravity. With additive volumes, the mass-weighted mean reduces to a volume-weighted mean of the specific gravities:$$SG_{mix} = \frac{V_1 SG_1 + V_2 SG_2}{V_1 + V_2} = \frac{8000(0.8984)+20{,}000(0.9593)}{28{,}000} = \boxed{0.9419}.$$
3. Actual mixture density. Specific gravity is referenced to water at 60 °F:$$\rho_{mix} = SG_{mix}\,\rho_w = 0.9419 \times 0.999 = 0.9410\ \text{g/cm}^3.$$
4. (i) Convert to lb per US gallon. $$\rho_{mix} = 0.9410\,\tfrac{\text{g}}{\text{cm}^3}\times \frac{3785.41\ \text{cm}^3/\text{gal}}{453.592\ \text{g/lb}} = \boxed{7.85\ \text{lb/US gal}}.$$
5. (ii) Convert to lb per ft³. $$\rho_{mix} = 0.9410\,\tfrac{\text{g}}{\text{cm}^3}\times \frac{28{,}316.8\ \text{cm}^3/\text{ft}^3}{453.592\ \text{g/lb}} = \boxed{58.7\ \text{lb/ft}^3}.$$