Question 4 of 6: Octane Numbers, Flash Point and a Butadiene Recycle Reactor

Question 4: Octane Numbers, Flash Point and a Butadiene Recycle Reactor (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) MON and RON

(i) Difference between the methods. (4 marks) Both octane numbers are measured on the same standardised single-cylinder CFR knock-test engine, but under different severities. The Research method (RON) runs at low speed (600 rpm) with modest inlet-air temperature and fixed spark — mild conditions. The Motor method (MON) runs at higher speed (900 rpm) with a pre-heated mixture (≈ 149 °C) and spark advance — harsher conditions that stress the fuel more. Because the Motor test is more severe, MON is lower than RON for the same fuel; the gap, RON − MON, is the fuel’s sensitivity.

(ii) What they represent. (4 marks) An octane number is the volume percent of iso-octane (2,2,4-trimethylpentane, ON = 100) in a blend with n-heptane (ON = 0) that knocks identically to the test fuel. RON represents anti-knock performance at low speed and mild acceleration (city driving); MON represents performance at high speed and heavy load (highway, hard pull). The pump “anti-knock index” posted in North America is their average, AKI = (RON + MON)/2.

(b) Flash point (2 marks)

The flash point is the lowest temperature at which a fuel gives off vapour in sufficient concentration to form an ignitable mixture with the air immediately above its surface, so that a momentary flash occurs when a small flame or ignition source is applied. It is a volatility-based safety property governing storage and handling classification (it is not sustained combustion — that is the higher fire point).

(c) Butadiene by catalytic dehydrogenation with recycle

Given. Reaction $C_4H_{10} \rightarrow C_4H_6 + 2H_2$. Product stream: 65 mol/hr H₂, 15 mol/hr C₄H₁₀, $n$ mol/hr C₄H₆. Recycle: 20 mol/hr at 20% C₄H₁₀ (4 mol/hr) and 80% C₄H₆ (16 mol/hr), returned to the mixer. Fresh feed is pure butane $F$.

Find. (i) fresh butane feed $F$; (ii) product butadiene $n$; (iii) single-pass conversion of butane.

Approach. Hydrogen appears only from reaction, so the product H₂ fixes the moles reacted. A butane balance around the reactor (inlet = fresh + recycle butane; outlet = product + recycle butane) then gives the feed, and the butadiene balance gives $n$.

1. Extent of reaction from the hydrogen. Each butane converted makes 2 H₂, and all H₂ leaves in the product:$$x = \tfrac{1}{2}\,\dot n_{H_2} = \tfrac{1}{2}(65) = \boxed{32.5\ \text{mol/hr butane reacted}}.$$
2. (i) Fresh butane feed. Butane balance around the reactor: (fresh $F$ + recycle 4) enters, $x$ reacts, and (product 15 + recycle 4) leaves:$$(F + 4) - x = 15 + 4 \;\Rightarrow\; F = 19 + 32.5 - 4 = \boxed{47.5\ \text{mol/hr}}.$$
3. (ii) Product butadiene. C₄H₆ balance around the reactor: 16 enters with the recycle, $x$ is produced, and ($n$ product + 16 recycle) leaves:$$16 + x = n + 16 \;\Rightarrow\; n = x = \boxed{32.5\ \text{mol/hr}}.$$
4. (iii) Single-pass conversion. Butane fed to the reactor is $F + 4 = 51.5$ mol/hr, of which 32.5 reacts:$$X_{sp} = \frac{x}{F+4} = \frac{32.5}{51.5} = \boxed{0.631\ (63.1\%)}.$$Overall check: fresh butane 47.5 in, 15 out unreacted ⇒ 32.5 consumed overall = product butadiene, and $2(32.5)=65$ mol H₂ — consistent.