Question 4 of 6: Cloud Point, Hydrogen Production, API Gravity and a Blend Density

Question 4: Cloud Point, Hydrogen Production, API Gravity and a Blend Density (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) Cloud point (2 marks)

The cloud point is the temperature at which, on cooling a petroleum product (typically a diesel or a distillate fuel/oil) under standard conditions, the first wax crystals appear and the clear liquid turns hazy/cloudy. It marks the onset of paraffin-wax precipitation and is a low-temperature-operability specification (waxing can plug filters and lines); it is a few degrees above the pour point.

(b) Two hydrogen-production processes (4 marks)

• Steam methane reforming (SMR): natural gas (or refinery off-gas) reacts with steam over a nickel catalyst at 700–900 °C, $CH_4 + H_2O \rightleftharpoons CO + 3H_2$, followed by the water–gas shift $CO + H_2O \rightleftharpoons CO_2 + H_2$; the H₂ is purified by pressure-swing adsorption. This is the dominant on-purpose route.
• Catalytic (naphtha) reforming — by-product hydrogen: in raising gasoline octane, dehydrogenation of naphthenes to aromatics ($C_6H_{12} \rightarrow C_6H_6 + 3H_2$) and dehydrocyclisation release large quantities of hydrogen, which supplies much of the refinery’s hydrotreating demand. (Partial oxidation/gasification of heavy residue is a third route where cheap gas is unavailable.)

(c) API gravity (2 marks)

API gravity is an inverse density scale defined by the American Petroleum Institute for petroleum liquids at 60 °F:

$$^\circ API = \frac{141.5}{SG} - 131.5 \qquad\Longleftrightarrow\qquad SG = \frac{141.5}{^\circ API + 131.5},$$

where $SG$ is the specific gravity (60/60 °F) relative to water. Water is 10° API; lighter (less dense) oils have higher API. It is a linear, convenient field measure of oil density and, loosely, of quality/value.

(d) Density of the blend

Given. 10,000 bbl of 22° API gas oil blended with 20,000 bbl of 11° API fuel oil, volumes additive; 1 bbl = 42 US gal; water density at 60 °F = 0.999 g/cm³.

Find. Mixture density in lb/US-gal and lb/ft³.

Approach. Convert each API to specific gravity, then blend by volume (never by API degrees — API is not additive). Because volumes are additive and both use the same 60 °F water reference, the mixture SG is the volume-weighted average of the component SGs; convert to the requested units with the 0.999 g/cm³ water density.

1. Component specific gravities. $$SG_1 = \frac{141.5}{22 + 131.5} = 0.9218,\qquad SG_2 = \frac{141.5}{11 + 131.5} = 0.9930.$$
2. Volume-weighted blend SG. With $V_1 = 420{,}000$ and $V_2 = 840{,}000$ gal,$$SG_{mix} = \frac{V_1 SG_1 + V_2 SG_2}{V_1 + V_2} = \frac{420{,}000(0.9218) + 840{,}000(0.9930)}{1{,}260{,}000} = \boxed{0.9693}.$$
3. Absolute density. $\rho_{mix} = SG_{mix}\times 0.999 = 0.9683$ g/cm³.
4. (i) lb per US gallon. Using 1 US gal $= 3785.4$ cm³ and 1 lb $= 453.59$ g,$$\rho_{mix} = 0.9683\times\frac{3785.4}{453.59} = \boxed{8.08\ \text{lb/US gal}}.$$
5. (ii) lb per ft³. Using 1 ft³ $= 28{,}317$ cm³,$$\rho_{mix} = 0.9683\times\frac{28{,}317}{453.59} = \boxed{60.4\ \text{lb/ft}^3}.$$