Question 3 of 6: Crude Refining, Product Specifications and SO₂ Absorption

Question 3: Crude Refining, Product Specifications and SO₂ Absorption (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) Why crude is refined, and how (4 marks)

Crude oil is a complex mixture of thousands of hydrocarbons (plus sulphur, nitrogen, oxygen and metal compounds) with a wide boiling range and, as produced, no direct end use. Refining separates it into fractions with defined, saleable properties (LPG, gasoline, jet/kerosene, diesel, fuel oil, lube base stocks, asphalt, petrochemical feedstocks) and upgrades the low-value heavy ends into more valuable light products while removing contaminants that would poison catalysts or violate fuel specifications.

This is done in three families of process:

• Separation: atmospheric distillation splits the crude by boiling point; vacuum distillation recovers gas oils and lube stocks from the atmospheric residue without thermal cracking.
• Conversion: catalytic cracking (FCC), hydrocracking, coking and visbreaking break heavy molecules into lighter ones; catalytic reforming, alkylation and isomerisation rearrange molecules to raise octane.
• Treating: hydrotreating/desulphurisation and sweetening remove sulphur, nitrogen and metals to meet product and environmental specifications.

(b) Leading specification features by product (6 × 1 mark)

(c) SO₂ absorption in the packed tower

Given. Gas feed 7380 m³/h at 303 K, 1 bar, 14.8 mol% SO₂ (balance inert); exit gas 1 mol% SO₂; water 1 m³/min. Only SO₂ is absorbed; the inert gas passes through unchanged.

Find. (i) SO₂ concentration in the liquid effluent; (ii) the volumetric flow of the exit gas at 0.95 bar, 293 K.

Approach. Convert the inlet gas to a molar flow (ideal gas), use the inert as a tie to fix the exit gas and the SO₂ absorbed, put the absorbed SO₂ into the water for the effluent concentration, and apply the ideal-gas law again for the exit volume at the new T, P.

1. Inlet molar flow and SO₂ split. With $R = 8.314$ kPa·m³/(kmol·K),$$\dot n_{in} = \frac{PV}{RT} = \frac{100(7380)}{8.314(303)} = 293.0\ \text{kmol/h}.$$Then SO₂ in $= 0.148(293.0) = 43.36$ and inert $= 249.6$ kmol/h.
2. Exit gas from the inert tie. Inert is not absorbed, so it is 99% of the exit gas:$$\dot n_{out} = \frac{249.6}{0.99} = 252.1\ \text{kmol/h},\qquad \text{SO}_2\ \text{out} = 0.01(252.1) = 2.52\ \text{kmol/h}.$$SO₂ absorbed $= 43.36 - 2.52 = \boxed{40.84\ \text{kmol/h}}$.
3. (i) Effluent concentration. The water is $60$ m³/h $= 60{,}000$ kg/h. The absorbed SO₂ mass is $40.84(64) = 2614$ kg/h, so$$C_{SO_2} = \frac{2614\ \text{kg/h}}{60\ \text{m}^3/\text{h}} = \boxed{43.6\ \text{kg/m}^3}\ (\approx 4.17\ \text{wt\%}).$$
4. (ii) Exit-gas volume at 0.95 bar, 293 K.$$\dot V_{out} = \frac{\dot n_{out} R T}{P} = \frac{252.1(8.314)(293)}{95} = \boxed{6465\ \text{m}^3/\text{h}}.$$