Gasoline Properties, Solvent Dewaxing and a Reflux Ratio: 23-Chem-B6 Petroleum Refining and Petrochemicals, May 2015, Q5

Question 5: Gasoline Properties, Solvent Dewaxing and a Reflux Ratio (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) Reid vapour pressure, boiling range, antiknock (3 + 3 marks)

(i) Meaning of each.

Reid vapour pressure (RVP): the vapour pressure of the gasoline measured at 100 °F (37.8 °C) in a standard bomb (ASTM D323), reported in psi — a measure of the fuel’s front-end volatility.
Boiling range: the temperature interval over which the gasoline distils (ASTM D86), described by the initial boiling point, the 10%/50%/90% recovery temperatures and the end point — the full volatility profile.
Antiknock characteristics (octane number): the fuel’s resistance to auto-ignition (knock), reported as RON and MON (and their average, the anti-knock index), relative to iso-octane (100) and n-heptane (0).

(ii) Why they matter. RVP must be high enough for cold starting but low enough to avoid vapour lock and evaporative (VOC) emissions — it is set seasonally and by region. The boiling range governs driveability across the whole operating envelope: the light end controls cold start and warm-up, the mid-range controls acceleration and mixture uniformity, and the heavy end controls combustion completeness, deposits and crankcase dilution. Antiknock quality must match the engine’s compression ratio; too low an octane causes knock (power loss, engine damage), so octane sets which grade a vehicle requires and is a primary value driver in the gasoline pool.

(d) Solvent dewaxing (2 marks)

Solvent dewaxing removes high-melting paraffin wax from lubricating-oil base stocks (vacuum distillate and de-asphalted oil) so that the finished lube oil has an acceptably low pour point. The oil is diluted with a solvent (typically a methyl-ethyl-ketone/toluene mixture, or propane), chilled to precipitate the wax as crystals, and the wax is removed by rotary vacuum filtration; the solvent is then recovered and recycled. The by-product slack wax is further de-oiled to make refined paraffin/microcrystalline waxes.

(e) Reflux-to-product ratio for the benzene–toluene column

Given. Feed $F = 5000$ lb/h at 60% benzene (40% toluene); distillate (and top vapour, and reflux) 85% benzene; bottoms 10% benzene (90% toluene); overhead vapour to the condenser stated as $V = 3000$ lb/h. Streams $V$, $D$ and $R$ have identical composition.

Stream	Flow (lb/h)	Benzene
Feed $F$	5000	60%
Distillate $D$ / vapour $V$ / reflux $R$	$V$ stated as 3000	85%
Bottoms $W$	—	10%

Find. the reflux ratio $R/D$, where $V = D + R$.

Figure 5 — Benzene–toluene column: overhead vapour $V$ condenses and splits into reflux $R$ and distillate $D$ (both 85% benzene); the bottoms $W$ leave at 10% benzene.

Approach. Overall column mass and benzene balances fix the distillate $D$ and bottoms $W$; the condenser split then gives the reflux ratio from $V = D + R$.

Overall balances fix $D$ and $W$. With $D + W = F$ and a benzene balance $x_D D + x_W W = x_F F$:$$0.85\,D + 0.10\,W = 0.60(5000),\quad D + W = 5000 \;\Rightarrow\; \boxed{D = 3333\ \text{lb/h},\ \ W = 1667\ \text{lb/h}}.$$(Check benzene: $0.85(3333)+0.10(1667)=3000$ lb/h = feed benzene. ✓)
Reflux ratio from the condenser split. All the overhead vapour condenses and divides into reflux and product, $V = D + R$, so$$\frac{R}{D} = \frac{V - D}{D}.$$

Check: the printed overhead flow $V = 3000$ lb/h is inconsistent with the separation as specified. The feed/product/bottoms compositions rigorously require the distillate to be $D = 3333$ lb/h, but a condenser must satisfy $V = D + R$ with $R \ge 0$, i.e. $V \ge D$. Since $3000 < 3333$, the stated $V$ cannot be the true overhead vapour (it would imply a negative reflux). This inconsistency is in the printed exam itself (the text and the page-6 schematic both give 3000 lb/h), so per exam note 1 the assumption is stated here rather than a number forced. The defensible results are the mass-balance quantities $D = 3333$ lb/h and $W = 1667$ lb/h; the reflux ratio then follows from $R/D=(V-D)/D$ once a physically consistent overhead ($V > 3333$ lb/h) is supplied. In the practice version the numbers are made consistent so a finite reflux ratio results.

Quantity	Result
Distillate $D$ (85% benzene)	3333 lb/h
Bottoms $W$ (10% benzene)	1667 lb/h
Reflux ratio $R/D$	$(V-D)/D$ — indeterminate as printed ($V=3000

Concept & Theory

Topic. Overall column material balances; the condenser mass split and definition of reflux ratio.

Key relations. $D+W=F$; benzene balance $x_D D + x_W W = x_F F$; condenser split $V = D + R$, reflux ratio $R/D = (V-D)/D$.

Why this works. The two overall balances (total and one component) determine the two unknown product flows independently of the internal reflux. The reflux ratio is then purely the way the fixed overhead vapour is divided between return and draw.

Common pitfall. Trying to read $D$ off the stated overhead flow rather than from the balances; and not checking $V \ge D$ — a stated vapour smaller than the required distillate signals inconsistent data, not a valid reflux ratio.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §4.4 (multi-component separations); Gary & Handwerk, Petroleum Refining Ch. 4 (distillation).

Question 5 of 6: Gasoline Properties, Solvent Dewaxing and a Reflux Ratio

Question 5: Gasoline Properties, Solvent Dewaxing and a Reflux Ratio (20 marks — equal value)

(a) Reid vapour pressure, boiling range, antiknock (3 + 3 marks)

(d) Solvent dewaxing (2 marks)

(e) Reflux-to-product ratio for the benzene–toluene column