Question 1 of 6: Polymerization, Crude Distillation and a Refinery-Boiler Combustion Balance

Question 1: Polymerization, Crude Distillation and a Refinery-Boiler Combustion Balance (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) Polymerization in the petroleum industry (2 marks)

In refining, polymerization is a conversion process that combines two or more small, gaseous light olefins — chiefly propylene and butylenes recovered from cracking (FCC) off-gas — into a larger, liquid hydrocarbon that boils in the gasoline range (“polymer gasoline” or “poly-gasoline”). Unlike plastics manufacture, the reaction is deliberately limited to dimerization/trimerization so the product stays in the motor-fuel boiling range; the branched-olefin product has a high octane number, making it a valuable gasoline blendstock.

(b) Thermal, sulphuric-acid and phosphoric-acid polymerization (4 marks)

• Thermal polymerization. Non-catalytic; relies on high temperature (~500–600 °C) and high pressure (up to ~70 bar) to drive a free-radical combination of olefins. It is unselective (also cracks feed), gives lower-quality product, and is now largely obsolete, displaced by the milder catalytic routes.
• Sulphuric-acid polymerization. Uses liquid sulphuric acid (~65% “cold-acid” process) as a catalyst at low temperature (~30–40 °C). The acid protonates the olefin to a carbocation that adds to a second olefin; it is selective for combining specific olefins (e.g. isobutylene) and yields high-octane polymer gasoline, but requires acid handling and neutralization.
• Phosphoric-acid polymerization. The dominant industrial catalytic route, using solid phosphoric acid (SPA) supported on kieselguhr (or liquid phosphoric acid) at ~180–230 °C and 30–80 bar. Propylene/butylene feed is dimerized/trimerized to gasoline-range olefins; it is milder and more selective than thermal polymerization and can be tuned to make specific dimers (e.g. propylene → nonene, or cumene precursors).

(c) Why distillation separates crude oil (4 marks)

Crude oil is a mixture of thousands of hydrocarbons that differ mainly in boiling point (which rises with molecular size). Fractional distillation exploits this difference in volatility. The desalted crude is heated in a fired heater to ~350–400 °C so that the lighter components vaporize, and the partly vaporized stream is fed near the base of a tall atmospheric distillation column fitted with trays (or packing). As the hot vapour rises it progressively cools; at each tray a component condenses when the local temperature falls to its boiling point, while lighter material continues upward. Side-draws withdraw the successive boiling-range cuts — light/heavy naphtha, kerosene, light and heavy gas oils — with the undistilled residue leaving the bottom (and routed to vacuum distillation).

Why it works: repeated vaporization and condensation on each tray (reflux) continuously enriches the more volatile species in the rising vapour and the less volatile species in the descending liquid, so a single column resolves the crude into fractions defined by boiling range rather than into pure compounds — which is exactly what fuel products are.

(d) Refinery-boiler combustion balance

Given. A fuel gas is burned in a refinery boiler with 150% excess air. Of the propane fed, 80% burns to CO₂, 15% burns to CO, and 5% leaves unburned.

Find. The molar composition of the flue (stack) gas.

Approach. Take 100 mol of feed gas as the basis, size the air from the theoretical-oxygen requirement (crediting the O₂ already in the feed), then tally each product species from the specified split of the propane.

1. Theoretical oxygen for complete combustion. Complete combustion is $C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O$, so the stoichiometric O₂ to burn all 75 mol propane is $5(75)=375$ mol. The feed already supplies 25 mol O₂, hence the oxygen that must come from air is$$O_{2,\text{theo (air)}} = 375 - 25 = 350\ \text{mol}.$$
2. Air actually supplied (150% excess). Excess air is reckoned on the air-borne theoretical oxygen, so the air delivers $2.5\times$ the theoretical amount:$$O_{2,\text{air}} = 2.5(350) = 875\ \text{mol}, \qquad N_2 = 875\times\tfrac{79}{21} = 3291.7\ \text{mol}.$$Total oxygen available to react $= 875 + 25 = 900$ mol.
3. Split the propane and burn it. Of 75 mol propane: 60 mol → CO₂, 11.25 mol → CO, 3.75 mol unburned.
   • $C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$: gives $3(60)=180$ CO₂, $4(60)=240$ H₂O, using $5(60)=300$ O₂.
   • $C_3H_8+\tfrac72 O_2\rightarrow 3CO+4H_2O$: gives $3(11.25)=33.75$ CO, $4(11.25)=45$ H₂O, using $3.5(11.25)=39.375$ O₂.
   Totals: H₂O $=285$ mol, O₂ consumed $=339.375$ mol.
4. Oxygen remaining and grand total. Excess O₂ leaving $=900-339.375=560.625$ mol. Adding N₂ (3291.7) and the 3.75 mol unburned propane:$$n_{\text{flue}} = 180+33.75+285+560.625+3291.7+3.75 = 4354.8\ \text{mol}.$$
5. Wet stack-gas composition. Dividing each species by the total gives the mole fractions boxed below.$$\boxed{y_{CO_2}=4.13\%,\; y_{CO}=0.78\%,\; y_{H_2O}=6.54\%,\; y_{O_2}=12.87\%,\; y_{N_2}=75.59\%,\; y_{C_3H_8}=0.09\%}$$