Alkylation, Naphtha Cracking and a Coke Furnace: 23-Chem-B6 Petroleum Refining and Petrochemicals, May 2015, Q2

Question 2: Alkylation, Naphtha Cracking and a Coke 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 combines a light olefin (propylene, butylenes or amylenes) with isobutane over a strong-acid catalyst to make a high-octane, low-vapour-pressure branched paraffin (“alkylate”) for the gasoline pool. The olefin and a large excess of isobutane contact the liquid acid in a cooled reactor; the two-phase effluent is settled to recover and recycle the acid; the hydrocarbon is neutralised (caustic/water wash) and fractionated to recycle unreacted isobutane, reject n-butane/propane, and draw the alkylate product.

Figure 2 — Acid-catalysed alkylation unit: olefin + isobutane react over liquid acid; the settler recycles acid, the wash neutralises entrained acid, and the fractionator recycles isobutane and draws the alkylate.

(b) Catalysts and feedstock (3 marks)

The two industrial catalysts are concentrated sulphuric acid (H₂SO₄, ~90–98%) and hydrofluoric acid (HF). The feed is an olefin stream — chiefly butylenes, with propylene and amylenes — reacted with a large excess of isobutane (typically an isobutane:olefin ratio of 5:1 to 15:1 to suppress olefin polymerisation and favour the desired mono-alkylation).

(c) Naphtha cracking for light olefins (2 + 2 marks)

(i) Increasing the light-olefin yield. Steam (thermal) cracking of naphtha to ethylene/propylene is favoured by high temperature (short, hot coil, ~800–850 °C outlet), low hydrocarbon partial pressure (more steam dilution), short residence time (milliseconds, with rapid quench), and a lighter/more paraffinic feed. High severity and low partial pressure shift the radical chemistry toward light olefins and away from heavier condensation products.

(ii) Why steam is added to the tubes. Steam is a diluent that lowers the hydrocarbon partial pressure, which thermodynamically and kinetically favours the mole-increasing cracking-to-olefins reactions; it also reduces residence time, suppresses coke deposition on the tube walls (and helps gasify coke via C + H₂O), and provides sensible heat — allowing high conversion without excessive coking.

(d) Petroleum-coke furnace combustion

Given. Per 100 kg coke: 75 kg C, 2 kg H, 23 kg ash (inert). 40% excess air; the ash formed carries 5% unburned carbon; of the carbon that burns, 80% → CO₂ and 20% → CO. Air: 1 mol O₂ : 3.76 mol N₂.

Quantity (per 100 kg coke)	Value
Carbon	75 kg
Hydrogen	2 kg (= 1 kmol H₂)
Inert ash	23 kg
Excess air	40%
Unburned C in ash	5% of the ash
CO₂ : CO of burned C	80 : 20

Find. (i) reactions; (ii) flue-gas composition; (iii) ash per 100 kg; (iv) carbon lost per 100 kg.

(i) Reactions

$$C + O_2 \rightarrow CO_2 \qquad C + \tfrac12 O_2 \rightarrow CO \qquad H_2 + \tfrac12 O_2 \rightarrow H_2O.$$

Figure 3 — Coke furnace: 100 kg coke and 40%-excess air in; flue gas (CO₂/CO/O₂/N₂/H₂O) and ash (with 5% unburned carbon) out.

Approach. Fix the ash and unburned carbon from the 5%-in-ash rule, split the burned carbon 80/20 into CO₂/CO, base the theoretical O₂ on the fuel as fired (all C→CO₂, all H→H₂O), apply 40% excess for the supplied air and its N₂, and close the flue gas by species balances.

(iii) Ash and (iv) carbon lost. The inert ash (23 kg) is 95% of the ash formed:$$m_{ash} = \frac{23}{0.95} = \boxed{24.21\ \text{kg}},\qquad C_{lost} = 24.21 - 23 = \boxed{1.21\ \text{kg unburned carbon}}.$$
Carbon that burns, split into CO₂/CO. Burned C $= 75 - 1.21 = 73.79$ kg $= 6.149$ kmol. Then$$n_{CO_2} = 0.80(6.149) = 4.919,\qquad n_{CO} = 0.20(6.149) = 1.230\ \text{kmol}.$$
Hydrogen. $2$ kg H $= 1$ kmol H₂ $\rightarrow$ $n_{H_2O} = 1.0$ kmol.
Theoretical and supplied O₂. On the fuel as fired (all C to CO₂, all H to H₂O):$$O_2^{theo} = \frac{75}{12} + \tfrac12(1) = 6.75\ \text{kmol};\quad O_2^{sup} = 1.40(6.75) = 9.45\ \text{kmol},\ \ N_2 = 3.76(9.45) = 35.53\ \text{kmol}.$$
O₂ in the flue gas. O₂ actually consumed $= n_{CO_2} + \tfrac12 n_{CO} + \tfrac12 n_{H_2O} = 4.919 + 0.615 + 0.5 = 6.034$ kmol, so$$O_2^{flue} = 9.45 - 6.034 = 3.416\ \text{kmol}.$$
(ii) Flue-gas composition. Total wet flue $= 4.919 + 1.230 + 3.416 + 35.53 + 1.0 = 46.10$ kmol, giving the mole fractions below.$$\boxed{\text{CO}_2\ 10.7\%,\ \ \text{CO}\ 2.7\%,\ \ \text{O}_2\ 7.4\%,\ \ \text{N}_2\ 77.1\%,\ \ \text{H}_2\text{O}\ 2.2\%.}$$

Quantity	Result
(iii) Ash produced	24.2 kg / 100 kg coke
(iv) Carbon lost (unburned)	1.21 kg / 100 kg coke
(ii) Flue gas (wet mol%)	CO₂ = 10.7%
	CO = 2.7%
	O₂ = 7.4%
	N₂ = 77.1%
	H₂O = 2.2%

Concept & Theory

Topic. Combustion stoichiometry with excess air, incomplete combustion (CO) and a solid-ash tie.

Key relations. Theoretical O₂ = O₂ to burn all C→CO₂ and H→H₂O; excess air multiplies the theoretical O₂ and its N₂ (3.76:1); C, H, O species balances close the flue gas.

Why this works. The 5%-unburned-carbon rule fixes the ash and the carbon that actually burns; the 80/20 split apportions that carbon between CO₂ and CO. Nitrogen is inert and the natural tie for the supplied air, and O₂ in the flue is simply supplied minus consumed.

Common pitfall. Forgetting that CO consumes only ½ O₂ per carbon (so incomplete combustion leaves more excess O₂), or basing theoretical air on the burned carbon rather than the fuel as fed — state the convention explicitly.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §9.5 (combustion, excess air, wet/dry flue gas).

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