Question 2 of 6: Acid-gas removal, pour point, and percent excess air

Question 2: Acid-gas removal, pour point, and percent excess air (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) Three removal processes for each acid gas (6 marks)

Removal of hydrogen sulphide (H₂S):

• Regenerative amine absorption — the sour gas is contacted counter-currently with an aqueous alkanolamine (MEA, DEA or the H₂S-selective MDEA); the amine chemically binds the acid gas and is stripped (regenerated) with steam, releasing concentrated H₂S to a Claus sulphur-recovery unit.
• Physical-solvent absorption — at high partial pressure the H₂S dissolves physically in a solvent such as Selexol (glycol ethers) or Rectisol (chilled methanol) and is released by simple pressure let-down, economical for high-pressure, high-acid gases.
• Solid-bed / liquid-redox scavenging — for small loads, H₂S is chemisorbed on iron-oxide (“iron sponge”) or on zinc-oxide/molecular-sieve beds, or converted directly to elemental sulphur in a liquid-redox process (LO-CAT, Stretford).

Removal of carbon dioxide (CO₂):

• Chemical (amine) absorption — MEA or promoted MDEA solutions absorb CO₂ and are thermally regenerated; the workhorse for bulk CO₂ removal from natural and refinery gas.
• Hot potassium carbonate — the Benfield / Catacarb process absorbs CO₂ in hot activated K₂CO₃ solution at elevated pressure and regenerates by pressure let-down, favoured at high CO₂ partial pressures.
• Physical separation — physical-solvent absorption (Selexol, Rectisol, Fluor), or non-absorptive routes such as membrane permeation and pressure-swing adsorption (PSA) on molecular sieves.

Because amine and physical solvents co-absorb both acid gases, real gas plants often remove H₂S and CO₂ together and separate them downstream; selective MDEA is used when H₂S must be taken out while slipping some CO₂.

(b) Meaning of pour point (4 marks)

The pour point is the lowest temperature, in 3 °C increments, at which a cooled oil still flows (pours) under the standardised gravity-flow test (ASTM D97) — conventionally 3 °C above the temperature at which movement ceases. It marks the onset of wax crystallisation / gelling that immobilises the oil, so it indicates the paraffin (wax) content and sets the minimum temperature for pumping, pipelining and cold-weather handling of the crude. Units: °C (or °F).

(c) Percent excess air for propane combustion

Given. $\dot n_{C_3H_8} = 100$ mol/h propane and $\dot n_{air} = 3000$ mol/h air (21 mol% O₂) fed to a combustor. Complete combustion is the reference for the theoretical air.

Find. The percent excess air.

Approach. Write the stoichiometric combustion reaction to get the theoretical O₂, compare it with the O₂ actually supplied in the 3000 mol/h of air, and express the surplus as a percentage of the theoretical requirement.

1. Stoichiometric (theoretical) O₂. Complete combustion of propane is$$C_3H_8 + 5\,O_2 \rightarrow 3\,CO_2 + 4\,H_2O,$$so the theoretical oxygen is $\dot n_{O_2,\text{theo}} = 5 \times 100 = 500$ mol/h.
2. Oxygen actually supplied. Air is 21 mol% O₂:$$\dot n_{O_2,\text{sup}} = 0.21 \times 3000 = 630\ \text{mol/h}.$$
3. Percent excess air. Excess air is referenced to the theoretical requirement:$$\%\,\text{excess} = \frac{\dot n_{O_2,\text{sup}} - \dot n_{O_2,\text{theo}}}{\dot n_{O_2,\text{theo}}}\times100 = \frac{630 - 500}{500}\times100 = \boxed{26\%}.$$(The percentage is the same whether written on O₂ or on total air, since the O₂-to-air ratio is fixed.)