Question 1

Equal volumes of a 30°API, 20°API, and 15°API crude are blended to produce a feed to refinery. Estimate the °API of the resulting blend. Clearly state your assumptions and comment on your result. (Hint: Would °API blend linearly?) 15 pts

Answer:
Blend: Equal volumes of 30°API, 20°API, and 15°API

°API = 141.5/SG₆₀^o_F - 131.5

°Theoretically APIs cannot be averaged, but SGs can be, assuming that the volume change upon mixing the three crudes is negligible.

Determine SG of each crude, calculate average SG, and use this value to calculate the blend °API.

30= 141.5/SG1–131.5,   20= 141.5/SG2–131.5, 15=141.5/SG3 – 131.5
SG1 = 0.876             SG2 = 0.933                       SG3= 0.966

SG_avg = AVG (0.876:0.933:0.966)
SG_avg = 0.925

°API_blend = 141.5/0.863 – 131.5
°API_blend = 21.4

Note that the blend °API is not a simple a mean of the component °API’s (AVG 30:20:15) = 21.7.  Although the difference appears to be small, it is important to use the precise °API in several calculations and correlations that are used to predict the quality and yield of refinery products.

Question 2

Watson K factors for n-pentane, n-hexane, 2,2,4 trimethylpentane, cyclohexane, methylcyclohexane, benzene, and xylene to consider the effects of hydrocarbon type and methyl substitution on the characterization factor. Present your results in a table and comment on the trends you observe. (You need to find the necessary data yourself to calculate the characterization factor. Watch the units.) 50 pts

Answer:

Compound	BP°C	BP (R)	SG	Kw
n-pentane	36.1	556.6	.626	13.14
n-hexane	68.7	615.7	.664	12.81
2,2,4 trimethylpentane	99.2	670.2	.692	12.65
Cyclohexane	80.7	636.9	.779	11.04
Methylcyclohexane	101	673.5	.769	11.40
Benzene	80.1	635.9	.877	9.81
Toluene	110.6	691.0	.866	10.21
m-Xylene	139.1	742.4	.860	10.53

Observations:

1. Order of KW: paraffins>naphthenes>aromatics
2. The longer the hydrocarbon chain, the lower the Kw.
3. Methyl substitution on naphthenic, or aromatic rings increases Kw.

Question 3

Determine the UOP K factor for a petroleum fraction with the following properties (25 pts):

°API @60°F: 23.5

ASTM Distillation Data:

Vol%	T, °F
10	672
30	764
50	846
70	947
90	1095

Answer:

°API = 23.5 at 60°F
°API = 141.5/ SG_60°F – 131.5
23.5 = 141.5/ SG_60°F – 131.5
SG_60°F = 0.913

K_UOP or K_W = (T_b)^1/3/ SG_60°F

Find VABP=T_b (R) = (672+764+846+947+1095)/5
VABP= T_b = 864.8°F= 1324.8 R
K_UOP = (1324.8)^1/3/0.913
K_UOP = 11.99

25 pts

Question 4

In one sentence explain what kind of refineries can process crude oils with Watson Characterization Factors that are less than 10, and why? 10 pts

Answer:

Complex refineries are equipped with processing heavy (i.e., aromatic, or aromatic asphaltic) crudes because of their capability of upgrading and changing the chemical composition of heavy crude oils.