Question 3 of 6: Coking processes, API of heavy crude, and an acetylene plant

Question 3: Coking processes, API of heavy crude, and an acetylene plant (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) Delayed coking and fluid coking (4 + 4 marks)

(i) Main characteristics.

• Delayed coking is a semi-batch thermal process. Vacuum residue is heated rapidly in a fired heater to about 485–505 °C and discharged into large insulated coke drums, where the cracking is “delayed” until the residue sits in the drum and lays down solid coke while cracked vapours (gas, naphtha, gas oils) rise to a fractionator. Drums operate in cycles — one fills while a full one is steamed, cooled and hydraulically cut — and produce lump coke (sponge, and, from aromatic feeds, premium needle coke).
• Fluid coking is a continuous process using a fluidised bed of hot coke particles. Feed is sprayed onto the circulating hot coke; it cracks and deposits a thin coke layer on the particles, which circulate to a separate burner/heater that burns part of the coke to supply the reaction heat and returns hot particles to the reactor. It yields fine, hard fluid coke powder; Flexicoking extends it by gasifying most of the coke.

(ii) How they differ. Delayed coking is cyclic/semi-batch in coke drums with heat supplied once in the fired heater, whereas fluid coking is continuous with heat supplied by circulating hot coke burned in a separate vessel. Delayed coking makes lump coke (sponge/needle) and is mechanically simpler; fluid coking makes fine powder coke, runs at a lower coke yield with higher liquid yield, handles heavier/dirtier feeds, and requires no drum-cutting downtime but is more complex. Thus the choice trades product-coke value and simplicity (delayed) against continuity and liquid yield (fluid).

(b) API gravity of a heavier crude (2 marks)

A heavier crude has a lower API gravity. API gravity is defined inversely to density, $^\circ\!API = 141.5/SG - 131.5$, so as the specific gravity rises the API number falls (water, $SG=1$, is $10^\circ$API). Light crudes exceed about $31^\circ$API, while heavy crudes fall below about $22^\circ$API; a heavier, denser crude therefore sits at the low end of the API scale.

(c) Acetylene by partial oxidation of methane

Given (basis 100 lbmol dry gas leaving the condenser; adopted analysis).

Find. (i) the O₂:CH₄ molar feed ratio; (ii) the pounds of water condensed per 100 lbmol dry gas; (iii) the carbon-based overall yield of C₂H₂.

Approach. Read the three reaction extents directly from the dry-gas species (CO₂→Eq.1, CO→Eq.2, C₂H₂→Eq.3), then close carbon, oxygen and hydrogen balances to get the CH₄ and O₂ fed, the water made, and the carbon yield.

1. Identify the reaction extents. Each product tags one reaction: $\xi_1 = n_{CO_2} = 8$, $\xi_2 = n_{CO} = 25$, $\xi_3 = n_{C_2H_2} = 10$ (lbmol). The 30 lbmol H₂ check: $3\xi_3 = 30$ ✓, and the H₂O made is $2\xi_1 + 2\xi_2$.
2. Methane fed (carbon balance). Methane consumed by the three reactions plus the unreacted methane leaving:$$n_{CH_4} = (\xi_1 + \xi_2 + 2\xi_3) + n_{CH_4,\text{un}} = (8 + 25 + 20) + 27 = 80\ \text{lbmol}.$$
3. Oxygen fed (oxygen balance). Only Eqs. 1 and 2 consume O₂:$$n_{O_2} = 2\xi_1 + 1.5\,\xi_2 = 2(8) + 1.5(25) = 16 + 37.5 = 53.5\ \text{lbmol}.$$
4. (i) O₂:CH₄ feed ratio. $$\frac{n_{O_2}}{n_{CH_4}} = \frac{53.5}{80} = \boxed{0.669}.$$
5. (ii) Water removed by the condenser. The condensed water is what reactions 1 and 2 make:$$n_{H_2O} = 2\xi_1 + 2\xi_2 = 2(8) + 2(25) = 66\ \text{lbmol},$$$$m_{H_2O} = 66 \times 18.02 = \boxed{1189\ \text{lb per 100 lbmol dry gas}}.$$
6. (iii) Carbon-based yield of acetylene. The natural gas carbon in equals the methane fed (1 C each), $=80$ lbmol C; the carbon appearing as product acetylene is $2\xi_3 = 20$ lbmol C:$$Y_{C_2H_2} = \frac{2\xi_3}{n_{CH_4}} = \frac{20}{80} = \boxed{25\%}.$$