Question 2 of 6: Alkylation, Naphtha Cracking and a Coke-Fired Furnace

Question 2: Alkylation, Naphtha Cracking and a Coke-Fired Furnace (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) Alkylation unit — flow sheet and description (4 marks)

Alkylation reacts light olefins (propylene, butylenes, amylenes) with isobutane over a strong-acid catalyst to make alkylate — a mixture of highly branched, high-octane C₇–C₈ paraffins that is a premium, low-vapour-pressure gasoline blendstock. The exothermic reaction is run cold in an intensely mixed contactor; the acid is then settled out and recycled, the hydrocarbon is neutralised, and the products are fractionated with unreacted isobutane recycled to extinction.

Isobutane is held in large excess (isobutane:olefin ≈ 5–8:1) to suppress olefin polymerisation and maximise alkylate quality.

(b) Alkylation catalysts and feedstock (2 marks)

Catalysts: concentrated sulphuric acid (H₂SO₄) and hydrofluoric acid (HF) are the two standard liquid catalysts; solid-acid catalysts are a newer alternative. Feedstock: isobutane plus light olefins (C₃–C₅: propylene, butylenes, amylenes), typically supplied from the fluid catalytic cracker and coker gas plants.

(c) Naphtha cracking for light olefins

(i) Increasing light-olefin yield. (2 marks) Light olefins (ethylene, propylene) are favoured by high severity: raise the coil outlet temperature (≈ 800–850 °C), shorten the residence time (milliseconds), lower the hydrocarbon partial pressure (more steam dilution), and use a lighter, more paraffinic feed. High temperature with short contact time cracks the paraffins to olefins while quenching quickly enough to prevent the secondary reactions (further cracking, condensation, coking) that destroy the olefins.

(ii) Why steam is introduced. (2 marks) Steam is a diluent that lowers the hydrocarbon partial pressure, which by Le Chatelier shifts the mole-increasing cracking reactions toward olefins and suppresses the bimolecular condensation reactions that form coke. It also reduces coke deposition on the tube walls (partly by the reaction $C + H_2O \rightarrow CO + H_2$), improves selectivity, lowers the residence time, and carries heat into the reaction — which is why these units are called steam crackers.

(d) Coke-fired furnace material balance

Given. Basis 100 kg coke: 80 kg C, 0.5 kg H, 19.5 kg mineral ash. Combustion with 50% excess air; the solid ash formed contains 2% unburned carbon; of the carbon that burns, 95% forms CO₂ and 5% forms CO.

Find. (i) reaction equations; (ii) flue-gas composition; (iii) ash produced; (iv) carbon lost, all per 100 kg coke.

Approach. Fix the solid residue from the ash + unburned-carbon rule; that gives the carbon actually burned. Base the theoretical air on complete combustion of all fuel C and H; apply the 50% excess to get O₂ and N₂ supplied; then close a species balance on the actual products (95% CO₂ / 5% CO split) to get the flue gas.

(i) Reactions.

$$C + O_2 \rightarrow CO_2 \qquad C + \tfrac{1}{2}O_2 \rightarrow CO \qquad H_2 + \tfrac{1}{2}O_2 \rightarrow H_2O.$$

1. Ash produced and carbon lost (parts iii, iv). The solid residue is 98% mineral ash and 2% unburned carbon, and the mineral ash (19.5 kg) all reports to the residue:$$W_{\text{ash}} = \frac{19.5}{0.98} = \boxed{19.90\ \text{kg}},\qquad C_{\text{lost}} = 0.02\,W_{\text{ash}} = \boxed{0.398\ \text{kg}}.$$So 0.398 kg of carbon leaves unburned in the ash per 100 kg coke.
2. Carbon actually burned, and its CO₂/CO split.$$C_{\text{burned}} = 80 - 0.398 = 79.60\ \text{kg} = \frac{79.60}{12.011} = 6.628\ \text{kmol}.$$95% to CO₂ and 5% to CO:$$n_{CO_2} = 0.95(6.628) = 6.297,\qquad n_{CO} = 0.05(6.628) = 0.3314\ \text{kmol}.$$Hydrogen burns fully: $n_{H_2O} = 0.5/2.016 = 0.248$ kmol.
3. Theoretical and supplied air. Theoretical O₂ for complete combustion of all fuel C and H:$$O_{2,\text{theo}} = \frac{80}{12.011} + \tfrac{1}{2}(0.248) = 6.661 + 0.124 = 6.785\ \text{kmol}.$$With 50% excess,$$O_{2,\text{sup}} = 1.50(6.785) = 10.18\ \text{kmol},\qquad N_2 = 3.76(10.18) = 38.28\ \text{kmol}.$$
4. Oxygen consumed and left over. O₂ used $= n_{CO_2} + \tfrac12 n_{CO} + \tfrac12 n_{H_2O} = 6.297 + 0.166 + 0.124 = 6.587$ kmol, so$$O_{2,\text{out}} = 10.18 - 6.587 = 3.59\ \text{kmol}.$$
5. Flue-gas composition (part ii). Summing the wet flue gas, $n_{\text{tot}} = 6.297 + 0.331 + 3.59 + 38.28 + 0.248 = 48.75$ kmol, and dividing each species by the total:$$y_{CO_2}=12.9\%,\; y_{CO}=0.68\%,\; y_{O_2}=7.4\%,\; y_{N_2}=78.5\%,\; y_{H_2O}=0.5\%.$$The dry flue gas (excluding the small water term) is essentially $y_{CO_2}=13.0\%$, $y_{CO}=0.68\%$, $y_{O_2}=7.4\%$, $y_{N_2}=78.9\%$.