Question 1 of 5: Gasoline properties; binary distillation split

Question 1 — Gasoline properties; binary distillation split (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) Gasoline property definitions

(i) Boiling range. The temperature interval between the initial boiling point (IBP) and the end (final) boiling point (EP/FBP) of the gasoline as measured by a standard distillation test (ASTM D86). It is the span over which the fuel vaporises — a typical motor gasoline runs from roughly 30 °C (IBP) to about 200 °C (EP) — and its shape controls cold-start behaviour, warm-up, driveability and the tendency to form deposits.

(ii) Reid vapour pressure (RVP). The absolute vapour pressure of the gasoline measured at 100 °F (37.8 °C) in the standard ASTM D323 bomb at a 4:1 vapour/liquid ratio, reported in psi (or kPa). It is a front-end volatility specification that governs starting, hot-fuel handling / vapour-lock tendency and evaporative emissions, and is trimmed seasonally by butane blending (higher RVP in winter).

(iii) Antiknock characteristics. The resistance of the gasoline to spontaneous autoignition (“knock” or detonation) ahead of the spark flame front in a spark-ignition engine. It is quantified by octane number — Research (RON) and Motor (MON) octane, and the pump “(R+M)/2” antiknock index — and is raised by branched paraffins, aromatics, and oxygenates or antiknock additives.

(iv) Fire point. The lowest temperature at which the vapour above the gasoline, once ignited, continues to burn for at least 5 seconds (as opposed to the flash point, where the vapour merely flashes momentarily). It lies a few degrees above the flash point and is a measure of the fire hazard of the fuel during handling and storage.

(b) Binary A/B distillation column

Given. Feed $F=200\ \text{mol/h}$ of A+B, $z_A=0.60$ (so 40% B). Distillate is 90% A ($x_{A,D}=0.90$); bottoms is 85% B, i.e. $x_{A,W}=0.15$.

Find. (i) the distillate (D) and bottoms (W) flow rates; (ii) a labelled schematic giving the A/B composition of every stream.

Approach. Two unknowns (D and W) are fixed by the overall total balance and a single component (A) balance around the column.

1. Overall and component-A balances. With $F=D+W$ and $z_AF=x_{A,D}D+x_{A,W}W$, $$200=D+W,\qquad 0.60(200)=0.90\,D+0.15\,W.$$
2. Solve for D. Substituting $W=200-D$ into the A balance, $$120=0.90D+0.15(200-D)=0.75D+30\ \Rightarrow\ 0.75D=90,$$ $$D=\boxed{120\ \text{mol/h}},\qquad W=200-120=\boxed{80\ \text{mol/h}}.$$
3. Check the A balance. A leaving $=0.90(120)+0.15(80)=108+12=120\ \text{mol/h}$, which equals A entering $=0.60(200)=120\ \text{mol/h}$ — the balance closes exactly.