23-Chem-B6 Petroleum Refining and Petrochemicals · May 2014
Question 6 of 6: Pour Point, Anti-Knock, Hydrogen Recovery and a Cyclohexane Recycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Closed-book, 3 hours; six “Problem” blocks of equal value (20 marks each), of which five constitute a complete paper (the first five in the answer book are marked). Sub-parts (a),(b),(c)… may be treated independently. Most parts call for concise essay answers; several require calculations with all steps shown. All six problems are solved below.
Reference texts: Gary, Handwerk, Kaiser & Geddes, Petroleum Refining: Technology and Economics (5th ed., CRC Press) — refinery processes and product properties; Fahim, Al-Sahhaf & Elkilani, Fundamentals of Petroleum Refining (Elsevier) — hydrogen production, cracking, treating; Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material balances, recycle, combustion and gas-law calculations; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 6: Pour Point, Anti-Knock, Hydrogen Recovery and a Cyclohexane Recycle (20 marks — equal value)
The pour point of a crude oil (or product) is the lowest temperature, measured under a standard cooling test, at which the oil will still flow (pour). By convention it is reported 3 °C (5 °F) above the temperature at which movement of the sample first stops. It is a low-temperature handling property, governed largely by the wax (paraffin) content: a waxy crude has a high pour point and may need heated lines and tankage.
(b) Reducing knock (3 marks)
Knock is suppressed by raising the octane number of the gasoline — with anti-knock additives and high-octane blend components. Historically the additive was tetraethyl lead (TEL); since lead has been phased out for environmental reasons, refiners use oxygenates (ethanol, formerly MTBE) and blend in high-octane refinery streams (reformate, alkylate, isomerate, aromatics). Additive metallic anti-knocks such as MMT have seen limited use. All work by slowing the pre-flame autoignition chemistry in the end gas so the charge burns from the spark rather than detonating.
(c) Two industrial routes to recover concentrated hydrogen (4 marks)
From a dilute refinery gas of H₂ plus methane and heavier hydrocarbon vapours, the two most common recovery methods are:
Pressure-swing adsorption (PSA). The gas is passed at high pressure over molecular-sieve / activated-carbon beds that preferentially adsorb CH₄, CO₂ and the heavier hydrocarbons, letting very pure hydrogen (99.9%+) pass through. The beds are regenerated by depressurisation (blowdown/purge), so several beds cycle in parallel to give continuous product.
Cryogenic (partial-condensation) separation. The gas is cooled and partially condensed; because hydrogen is far more volatile than methane and the C₂+ hydrocarbons, the hydrocarbons condense out and a hydrogen-rich vapour is recovered, with a distillation/flash to sharpen the split. (A third widely used route is membrane separation, in which small, fast-diffusing H₂ permeates a polymeric membrane preferentially, giving a hydrogen-rich permeate.)
PSA is favoured for high purity, membranes for simplicity and modest purity, and cryogenics where hydrocarbon co-products are also recovered.
(d) Recycle-to-fresh-feed ratio for benzene hydrogenation
Find. The ratio of the recycle stream to the total fresh feed, $R/F_{\text{fresh}}$.
Figure 5 — Benzene-hydrogenation loop: fresh benzene with 20% excess hydrogen mixes with the recycle, passes once through the reactor at 20% conversion, and the separator returns unconverted benzene + hydrogen (40% / 60%) to the mixer while cyclohexane leaves.
Approach. Take a basis of 100 mol fresh benzene, fix the fresh hydrogen from the 20% excess, use the overall conversion to fix the benzene reacted (all in the reactor), then apply the single-pass conversion to the combined (fresh + recycle) reactor-inlet benzene to solve for the recycle.
Fresh feed on a 100-mol benzene basis. Stoichiometric hydrogen is $3\times100=300$ mol; with 20% excess,$$\dot n_{H_2,\text{fresh}} = 1.20(300) = 360\ \text{mol},\qquad F_{\text{fresh}} = 100 + 360 = \boxed{460\ \text{mol}}.$$
Benzene reacted overall. At steady state, reaction occurs only in the reactor, so the benzene converted per pass equals the overall amount reacted:$$n_{\text{reacted}} = 0.90(100) = 90\ \text{mol}\quad(\text{10 mol leaves in product}).$$
Reactor-inlet benzene from the single-pass conversion. The reactor inlet benzene is fresh + recycled, $100 + 0.40R$; a 20% single-pass conversion on it must supply the 90 mol reacted:$$0.20\,(100 + 0.40R) = 90.$$
Solve for the recycle.$$100 + 0.40R = \frac{90}{0.20} = 450 \;\Rightarrow\; 0.40R = 350 \;\Rightarrow\; \boxed{R = 875\ \text{mol}}.$$Check: recycled benzene $=0.40(875)=350$; reactor-inlet benzene $=450$, of which 20% (90 mol) reacts and 360 leaves — the separator returns 350 and passes 10 to product, closing the benzene balance at 90% overall.
