Question 3 of 6: Characterisation of Petroleum Fractions, Flash Point and Mass of Gas in a Tank

Question 3: Characterisation of Petroleum Fractions, Flash Point and Mass of Gas in a Tank (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) Modes of characterisation of petroleum fractions

(i) Main modes. Petroleum fractions are characterised by (1) boiling behaviour — the true-boiling-point (TBP) curve and the standard ASTM D86 / D1160 distillation, giving initial and final boiling points and the boiling range; (2) density / gravity — specific gravity and API gravity; (3) bulk composition — the PONA/PIONA analysis (paraffins, iso-paraffins, olefins, naphthenes, aromatics) and sulphur/nitrogen content; and (4) correlating indices such as the Watson (UOP) characterisation factor $K = \dfrac{(T_B)^{1/3}}{SG}$ with $T_B$ the mean average boiling point in °R, which classifies a fraction as paraffinic ($K\approx12.5$–13), naphthenic (~11) or aromatic (~10).

(ii) Advantages and limitations. Distillation data are simple, reproducible and directly tied to product cut-points, but they say nothing about chemical type. Density/API is fast and correlates with many properties, yet two very different fractions can share a gravity. PONA gives real molecular insight but is slow and needs sophisticated analysis (chromatography, mass spectrometry). The characterisation factor conveniently bundles boiling point and gravity into one number for property estimation, but it is an empirical correlation and loses accuracy for wide or highly aromatic cuts. In practice several modes are combined.

(b) Flash point

The flash point of a fuel is the lowest temperature at which the liquid gives off enough vapour to form an ignitable (flammable) vapour–air mixture just above its surface, so that a momentary "flash" occurs when a small flame or spark is applied — but at which sustained combustion is not yet supported. It is a key safety property for storage, handling and classification of fuels (measured by closed-cup, e.g. Pensky–Martens, or open-cup methods).

(c) Mass of air in the cylindrical tank

Given. A horizontal cylindrical tank, inner diameter $D=3$ m and length $L=7$ m, full of air ($M=29$ g/mol) at $T=80$ °C and $P=2500$ kPa absolute.

Find. The mass of air in the full tank.

Approach. Compute the cylinder volume, use the ideal-gas law to get the moles of air, then multiply by the molar mass.

1. Tank volume. For a cylinder, $V = \dfrac{\pi D^2}{4}L$: $$V = \frac{\pi (3)^2}{4}(7) = \boxed{49.48\ \text{m}^3}.$$
2. Moles of air (ideal gas). With $P=2.5\times10^{6}$ Pa, $R=8.314\ \text{J/mol}\cdot\text{K}$, $T=353.15$ K: $$n = \frac{PV}{RT} = \frac{(2.5\times10^{6})(49.48)}{(8.314)(353.15)} = 4.213\times10^{4}\ \text{mol}.$$
3. Mass of air. Multiply by the molar mass $M=0.029$ kg/mol: $$m = nM = (4.213\times10^{4})(0.029) = \boxed{1222\ \text{kg}}.$$