Question 5 of 6: Pour Point, Thermal-Cracking Reactions and Ethanol Distillation

Question 5: Pour Point, Thermal-Cracking Reactions and Ethanol Distillation (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) Pour point

The pour point of a crude oil (or fuel) is the lowest temperature at which the oil will still flow, or "pour," under the conditions of a standard test. Just below the pour point the paraffin waxes crystallise into a network that gels the oil so that it no longer flows. It is an important cold-handling property governing pipelining, pumping and storage of waxy crudes in cold climates.

(b) Thermal-cracking reactions illustrated with but-1-ene

Thermal cracking proceeds by a free-radical chain mechanism with three types of step. Initiation is homolytic scission of a C–C bond to form two radicals; for but-1-ene the weakest bond is the allylic $CH_3\!-\!CH_2$ bond (the C–C bond $\beta$ to the double bond), whose scission gives a methyl and a resonance-stabilised allyl radical: $CH_3CH_2CH{=}CH_2 \rightarrow CH_3\!\cdot + \cdot CH_2CH{=}CH_2$. Propagation combines hydrogen abstraction, radical addition and $\beta$-scission (cleavage of the bond $\beta$ to the radical centre, releasing an olefin and a new radical). A methyl radical abstracts an allylic H to give methane and a methylallyl radical, $CH_3\!\cdot + C_4H_8 \rightarrow CH_4 + CH_3\dot{C}HCH{=}CH_2$, which loses an H atom by $\beta$-scission to form 1,3-butadiene, $C_4H_6 + H\!\cdot$. The H atom adds to the terminal carbon of another butene to give the 2-butyl radical, $CH_3CH_2\dot{C}HCH_3$, whose $\beta$-scission yields propylene and a methyl radical, $C_3H_6 + CH_3\!\cdot$ (a 1-butyl radical would instead split into ethylene and an ethyl radical, $C_2H_4 + C_2H_5\!\cdot$). The chain therefore cracks butene to lighter olefins (ethylene, propylene) and methane. Overall this also dehydrogenates part of the butene (net: butene → 1,3-butadiene $C_4H_6 + H_2$) and, at high severity, polymerisation/condensation of unsaturated fragments toward heavy tar and coke. Termination is radical recombination, e.g. $2\,CH_3\!\cdot \rightarrow C_2H_6$. The net effect is conversion of the feed into lighter olefins and paraffins plus a heavier residue.

(c) Temperature and pressure ranges for thermal cracking

Liquid-phase thermal cracking (visbreaking, delayed coking, older thermal cracking) operates typically at about 450–540 °C (roughly 850–1000 °F) and pressures from a few bar up to ~20–70 bar. High-severity gas-phase steam cracking to produce olefins runs hotter and leaner, about 750–900 °C at near-atmospheric to a few bar with steam dilution. In short, temperatures rise and pressures fall as the target shifts from residue upgrading toward light-olefin production.

(d) Alcohol lost in the bottoms

Given. Feed $F=5000$ kg at 5% ethanol / 95% water; distillate $D=\tfrac{1}{5}F$ at 75% ethanol / 25% water; the balance leaves as bottoms $W$.

Find. The ethanol (alcohol) leaving in the bottoms.

Approach. Use an overall mass balance for the bottoms rate and an ethanol component balance for the alcohol split; the alcohol lost in the bottoms is simply the ethanol in the feed minus the ethanol in the distillate.

1. Overall balance → bottoms rate. With $D=\tfrac15(5000)=1000$ kg, $$W = F - D = 5000 - 1000 = \boxed{4000\ \text{kg}}.$$
2. Ethanol entering. The feed carries $$m_{E,F} = 0.05(5000) = 250\ \text{kg ethanol}.$$
3. Ethanol component balance. The alcohol lost in the bottoms is what enters minus what leaves overhead: $$m_{E,W} = m_{E,F} - m_{E,D} = 250 - 0.75\,D.$$
4. Evaluate and check feasibility. With the figure's $D=1000$ kg, $m_{E,D}=0.75(1000)=750$ kg, which already exceeds the 250 kg of ethanol available in the feed. The component balance then gives $$m_{E,W} = 250 - 750 = -500\ \text{kg} < 0,$$ which is physically impossible — a distillate cannot carry more ethanol than the feed supplies. The three printed figure values (5% feed, 75% distillate, $D=\tfrac15F$) are therefore mutually inconsistent: at 75% ethanol the largest attainable distillate is $250/0.75 = 333.3$ kg, one fifteenth of the feed rather than one fifth. The answer is therefore given as the balance itself, with the inconsistency stated, as the exam's note 1 invites.