Question 2 of 6: Visbreaking, API Gravity and Flue-Gas Composition

Question 2: Visbreaking, API Gravity and Flue-Gas Composition (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) Visbreaking

(i) Meaning. Visbreaking (viscosity breaking) is a mild, once-through thermal-cracking process applied to heavy vacuum residue or atmospheric residue. Its purpose is to lower the viscosity and pour point of the residue by cracking a modest fraction of it, so that less light cutter stock is needed to blend the residue to fuel-oil specification, while simultaneously yielding some lighter gas-oil and gasoline-range material.

(ii) Typical conditions. Coil (furnace) visbreaking runs at about 450–500 °C with short residence times at 0.5–2 MPa; soaker visbreaking uses a slightly lower coil temperature (~430–450 °C) with a longer residence time in a soaking drum. Conversion is deliberately limited (typically 5–15% to gas + naphtha) to stay short of coke formation and to keep the product stable.

(iii) Principal reactions. The chemistry is free-radical thermal cracking: scission of long paraffinic side chains from aromatic and naphthenic cores, breaking of C–C bonds in paraffins to give smaller paraffins and olefins, and partial dehydrogenation. These reactions reduce the average molecular weight (hence the viscosity). Over-cracking must be avoided because continued condensation/polymerisation of the aromatic cores leads to asphaltene growth and coke.

(b) Heavier crude and API gravity

API gravity is defined as $ \text{°API} = \dfrac{141.5}{SG_{60/60}} - 131.5 $, where $SG$ is the specific gravity at 60 °F. A heavier crude has a larger density and therefore a larger specific gravity; because $SG$ appears in the denominator, a larger $SG$ gives a lower API gravity. Thus a heavier crude oil has a lower API gravity (light crudes are typically > 31.1 °API, heavy crudes < 22.3 °API). Water ($SG=1$) corresponds to 10 °API.

(c) Mole percent to weight percent

Given. A flue gas of 14.6% CO₂, 9.2% H₂O, 73.4% N₂ and 2.8% O₂ on a molar basis.

Find. The composition of the flue gas in weight percent.

Approach. Take a basis of 100 mol of gas; the mole percents become moles directly. Multiply each by its molar mass to get mass, sum, and divide each mass by the total.

1. Mass of each species (basis 100 mol). Mass $= n_i M_i$: $$m_{CO_2}=14.6(44.01)=642.5,\quad m_{H_2O}=9.2(18.02)=165.8,$$ $$m_{N_2}=73.4(28.02)=2056.7,\quad m_{O_2}=2.8(32.00)=89.6\ \text{g}.$$
2. Total mass. $$m_{\text{tot}} = 642.5+165.8+2056.7+89.6 = \boxed{2954.6\ \text{g per 100 mol}}.$$
3. Weight fractions. Divide each mass by the total: $$w_{CO_2}=\frac{642.5}{2954.6}=21.75\%,\quad w_{H_2O}=\frac{165.8}{2954.6}=5.61\%,$$ $$w_{N_2}=\frac{2056.7}{2954.6}=69.61\%,\quad w_{O_2}=\frac{89.6}{2954.6}=3.03\%.$$ The four sum to 100.0%.