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23-Chem-B6 Petroleum Refining and Petrochemicals · December 2014

Question 1 of 6: Petroleum-fraction properties, TAN, flash point, and a binary distillation balance

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, December 2014 — 04-Chem-B6. Closed-book; non-communicating calculator permitted. Five of six equally-weighted problems constitute a complete paper; all six are solved here. Parts (a)–(g) of each problem are independent. Most parts are essay-format; some require calculations.

Reference texts: Gary, Handwerk, Kaiser & Geddes, Petroleum Refining: Technology and Economics (5th ed., CRC Press) — refinery processes, product properties and treating; Fahim, Al-Sahhaf & Elkilani, Fundamentals of Petroleum Refining (Elsevier) — hydrogen production, coking, gas treating; Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material balances, combustion and recycle calculations; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the ASTM test-method standards (D323, D86, D93, D97).

Question 1: Petroleum-fraction properties, TAN, flash point, and a binary distillation balance (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) Four key physical properties of petroleum fractions (4 marks)

(Sulphur content, pour/cloud point, flash point, molecular weight and the Watson characterisation factor $K$ are other commonly reported properties.)

(b) Meaning of the total acid number (TAN) (3 marks)

The total acid number is the mass of potassium hydroxide, in milligrams, required to neutralise all the acidic constituents in one gram of oil, determined by titration (ASTM D664). Reported in mg KOH/g, it measures the organic-acid content of a crude — chiefly naphthenic acids. A high TAN (> 0.5–1.0 mg KOH/g) signals a corrosive (“acid”) crude that attacks carbon-steel piping and hot distillation internals, so it drives metallurgy selection and blending limits.

(c) Definition of flash point (3 marks)

The flash point of a fuel is the lowest temperature, at a specified barometric pressure, at which the vapour above the liquid forms a mixture with air that momentarily ignites (“flashes”) when a small test flame is applied, under a standardised apparatus such as the Pensky–Martens closed cup (ASTM D93). It is the temperature at which the vapour concentration just reaches the lower flammability limit. It is a flash, not sustained burning (that higher temperature is the fire point), and it is the primary index for the safe storage, handling and transport classification of a fuel.

(d) Binary distillation material balance

Given. Feed $F = 1000$ kg/h of a C1/C2 mixture with C1 = 30 wt% (so 300 kg/h C1, 700 kg/h C2). The overhead (distillate) is 80 wt% C2, and 70% of the C1 fed leaves in the bottoms.

QuantityValue
Feed F1000 kg/h
C1 in feed (30%)300 kg/h
C2 in feed (70%)700 kg/h
Overhead C2 fraction0.80 (so 0.20 C1)
C1 fraction to bottoms0.70 of feed C1

Find. (i) the overhead flow rate $D$; (ii) the mass flow rates of C1 and C2 in the bottom stream $B$.

Approach. Split the C1 fed between overhead and bottoms using the 70% rule, then use the overhead C1 fraction (20%) to size $D$; close overall and component balances for the bottoms.

  1. Split the C1 between products. Of the 300 kg/h C1 fed, 70% leaves in the bottoms:$$\dot m_{C1,B} = 0.70(300) = 210\ \text{kg/h}, \qquad \dot m_{C1,D} = 300 - 210 = 90\ \text{kg/h}.$$
  2. Size the overhead from its C1 content. The overhead is 80 wt% C2, hence 20 wt% C1, and it carries 90 kg/h of C1:$$0.20\,D = \dot m_{C1,D} = 90 \;\Rightarrow\; D = \boxed{450\ \text{kg/h}}.$$Its C2 content is $\dot m_{C2,D} = 0.80(450) = 360$ kg/h.
  3. Overall balance for the bottoms rate. $$B = F - D = 1000 - 450 = 550\ \text{kg/h}.$$
  4. C2 in the bottoms by component balance. $$\dot m_{C2,B} = \dot m_{C2,F} - \dot m_{C2,D} = 700 - 360 = 340\ \text{kg/h}.$$Check: $210 + 340 = 550$ kg/h $= B$ — the bottoms closes.
QuantityResult
(i) Overhead (distillate) rate D450 kg/h (90 C1 + 360 C2)
(ii) C1 in bottoms210 kg/h
(ii) C2 in bottoms340 kg/h
Bottoms total B550 kg/h
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