Non-energy Refineries in Petroleum Processing

Abstract

Not all refineries are aimed at producing fuels. This chapter focuses on two significant types of non-energy facilities: (1) lube oil refineries and (2) petrochemical refineries. In the lube oil discussion, we explore the nature of lubes and the various process routes used to make them. Routes include conventional deasphalting and extraction, hydroprocessing, Fisher-Tropsch synthesis, and oligomerization. Related to lubes, we explore the production of asphalt, with a process discussion and an example design. In petrochemicals, an aromatics complex is described, along with the processes used to make the typical BTX streams.

David S. J. Jones: deceased.

Steven A. Treese has retired from Phillips 66.

Appendix: Sizing a Bitumen Oxidizer

Design Specification

Product required 816 BPSD of 25-pen asphalt

Crude source Pennington – West African offshore (for assay data, see Figs. 21, 22, and 23)

Fig. 21

The TBP and gravity curves for Pennington crude

Fig. 22

Plot of softening point versus penetration and penetration versus specific gravity for Pennington crude

Fig. 23

Plot of viscosity versus penetration for asphalt

Vacuum unit feed = +650 °F on crude =>75 vol%

Sizing of the oxidizing reactor is required.

Step 1: Predict Yield of Asphalt

Patel’s equation shall be used for this purpose.

Trial 1

Assume the initial boiling point of asphalt is 1,015 °F which equates to a cut of 95 vol% on crude.

Vol% cut point on crude = 0.75 × 0.95 = 71 vol%.

Using Fig. 8 and data from the TBP curve (Fig. 21):

$$ \begin{array}{l}\mathrm{Slope}\ \mathrm{A} = \frac{1,015-400}{95-31}=9.6\\ {}\mathrm{Slope}\ \mathrm{B} = \frac{1,015-625}{95-71}=16.25\\ {}\mathrm{Slope}\ \mathrm{C} = \frac{625-400}{71-31}=5.6.\end{array} $$

Conradson carbon content of +750 °F residue = 4.6 wt% (see Fig. 24).

Fig. 24

Asphalt yields versus carbon residue % weight on crude

$$ \begin{array}{l}\mathrm{Watson} K\;\mathrm{factor}\ \mathrm{at}\ 750\ {}^{\circ}\mathrm{F} = 11.2\\ {} \mathrm{Exponent}\ n = 1.2\\ {} F=\frac{9.6\times {(4.6)}^{1.2}}{\left(16.25/5.6\right)\times \left(11.2-10.4\right)}\\ {} =25.8.\end{array} $$

From Fig. 9, percent asphalt yield = 4.3 (assumed was 5.0 vol%).

Trial 2

Assume the boiling point of asphalt is 950 °F which equates to a cut of 93.5 vol% on crude or a yield of 6.5 vol%.

Calculated F factor is 34.6 correlating to a yield of 7.6 vol%.

Trial 3

Assume the boiling point of asphalt is 850 °F which equates to a cut of 90 vol% on crude or a yield of 10 vol%.

Calculated F factor is 42.0 correlating to a yield of 10.0 % which is that assumed and is accepted as the yield of 100-pen asphalt from this crude.

Step 2: Calculate Throughput of 100 Penetrations and Air to Oxidizer

From Fig. 25, the ratio of 100-pen asphalt to a 25-pen asphalt is read off at 0.89 for a characterization factor of 11.2. Then:

Fig. 25

Fraction of yield of 100 pen asphalts

For 816 BPSD of 25-pen crude, we require a throughput of 816/0.89 = 917 BPSD of 100-pen asphalt.

For a continuous process, the recommended air rate is 0.3–1.0 scfm per BPSD. A rate of 0.55 will be used here.

Then air rate will be 0.55 × 917 = 504.4 scfm.

Step 3: Calculate Residence Time and Capacity of Fresh Feed

Required final penetration = 25 mm at 77 °F, 100 g/s.

Equivalent softening point = 131 °F (Fig. 22).

Initial pen = 100 and equivalent softening point is 119 °F.

Change in softening point = 12 °F approximate residence time = $\frac{131-119}{12}$ = 1 h.

Capacity of fresh feed: Cuft of feed per hour = $\frac{917\times 42}{24\times 7.48}$ = 214.9

$$ \begin{array}{l}\mathrm{Asphalt}\ \mathrm{S}\mathrm{G}\ @60\ {}^{\circ}\mathrm{F} = 0.966\\ {} @500\ {}^{\circ}\mathrm{F} = 0.815.\end{array} $$

The volume of fresh feed at 500 °F = 254.7 cuft/h. Add 15 % as contingency to 292.9 cuft/h.

Step 4: Reactor Material Balance

Material balance over the reactor is as follows:

$$ \begin{array}{l}\mathrm{F}\mathrm{eed}\ \mathrm{of}\ 100\ \mathrm{pen}\ \mathrm{in} = \frac{917\mathrm{BPSD}\times 42}{24}\\ {} =1,605\;\mathrm{gal}/\mathrm{h}\\ {} \mathrm{S}\mathrm{G}@60{}^{\circ}\mathrm{F} = 0.966 = 8.044\;1\mathrm{b}/\mathrm{gal}\\ {} 1\mathrm{b}/\mathrm{h} = 12,908\end{array} $$

$$ \begin{array}{l}\mathrm{Product}\ 25\ \mathrm{pen}\ \mathrm{asphalt}\ \mathrm{out} = \frac{816\times 42}{24}\\ {} =1,428\;\mathrm{gal}/\mathrm{h}\end{array} $$

$$ \begin{array}{l}\mathrm{S}\mathrm{G}\ @\ 60{}^{\circ}\mathrm{F} = 1.006 = 8.33\;1\mathrm{b}/\mathrm{gal}\\ {} = 11,895\;1\mathrm{b}/\mathrm{h}\ \left( = 0.92\ \mathrm{wt}\%\ \mathrm{on}\ \mathrm{feed}\right)\end{array} $$

$$ \begin{array}{l}\mathrm{Air}\ \mathrm{into}\ \mathrm{the}\ \mathrm{oxidizer} = 504.4\;\mathrm{scft}/ \min = \frac{504.4\times 60\times 29}{378}\\ {} = 2,322\;1\mathrm{b}/\mathrm{h}.\end{array} $$

Approximately 90 % of the oxygen in the air reacts with the asphalt:

$$ \begin{array}{l}\mathrm{lb}/\mathrm{h}\ \mathrm{of}\ \mathrm{oxgen}\ \mathrm{reacted} = 2,322\times 0.232\times 0.9\\ {} = 485\;1\mathrm{b}/\mathrm{h}.\end{array} $$

Approximately 20 % of oxygen reacted will be in the asphalt product = 97 lb/h:

$$ \begin{array}{l}\mathrm{Unreacted}\ \mathrm{oxygen}\ \mathrm{leaves}\ \mathrm{unit}\ \mathrm{in}\ \mathrm{the}\ \mathrm{overhead}\ \mathrm{vapor} = \left(2,322 \times 0.232\right) - 485\\ {} =54\;\mathrm{lb}/\mathrm{h}.\end{array} $$

Nitrogen in overhead vapor = 2,322 − 539 = 1,783 lb/h

Hydrocarbon vapor in overhead vapor:

$$ \begin{array}{l}=\mathrm{Feed} + \mathrm{air}-\left(\mathrm{product} + \mathrm{oxygen}\ \mathrm{in}\ \mathrm{product} + \mathrm{nitrogen} + \mathrm{unreacted}\ \mathrm{oxygen}\right)\\ {}=12,908 + 2,322 - \left(11,895 + 97 + 1,783 + 54\right)\\ {}=1,401\;\mathrm{lb}/\mathrm{h}\end{array} $$

$$ \begin{array}{l}\mathrm{Total}\ \mathrm{overhead}\ \mathrm{vapor} = 1,401\;\mathrm{lb}/\mathrm{h}\ \mathrm{hydrocarbons} + 1,783\;\mathrm{lb}/\mathrm{h}\ {\mathrm{N}}_2 + 54\;\mathrm{lb}/\mathrm{h}\ {\mathrm{O}}_2\\ {} =3,238\;\mathrm{lb}/\mathrm{h}.\end{array} $$

Step 5: Heat Balance Over the Reactor and Amount of Cooling Stream

It is intended to maintain a reactor temperature of 500 °F by recycling cold run-down product.

Feed temperature to the oxidizer shall be 550 °F. Heat of reaction in Btu/lb = 1.3 × diff in softening point.

The amount of recycle cooled product as reactor coolant is obtained from the following heat balance where xlb/h is the recycle.

Stream	V or L	°API	°F	lb/h	Btu/lb	MMBtu/h
In
Feed	L	15	550	12,908	254	3.279
Air	V		86	2,322	21	0.049
Recycle	L	10	338	x	148	238x
Ht. of reaction	–	–	–	–	–	0.201a
Total in				15,230 + x		3.529 + 148x
Out
Product	L	10	500	11,895	238	2.831
Dis. O2	V	–		97	–	Negl.
Recycle	L	10	500	x	238	238x
O/head vap.	V	–	300	3,238	124	0.402
Total out						3.233 + 238x

1. $\begin{array}{l}x = \frac{3,529,000-3,233,000}{238-148}\\ {} = 3,289\;1\mathrm{b}/\mathrm{h}\end{array}$
2. The volume of recycle at 500 °F (SG 0.872) = 7.26 lb/gal = 453 gal/h or 60 cuft/h
3. ^aHeat of reaction as Btu/lb of asphalt feed is calculated as 1.3 × the difference in softening point. In this case, it is 12,908 × (12 × 1.3) = 201,000 Btu/h

Step 6.0: Reactor Sizing

The reactor size shall be based on the hot volume of feed + contingency and the hot volume of recycle over the calculated residence time of 1 h. The volume of feed and recycle shall be based on the reactor temperature (500 °F).

The volume of fresh feed and contingency is 292.9 cuft. The recycle is 60 cuft and the total sizing volume is 352.9 cuft.

This amount shall occupy 65 % of the oxidizer’s total volume. This allows for the disengaging of vapor leaving the vessel.

The total volume of the oxidizer is $\frac{352.9}{0.65}$ = 543 cuft.

The ratio length to diameter shall be 4.5.

$$ \mathrm{Thus} \frac{4.5\pi {D}^3}{4}=543\;\mathrm{cuft} $$

and

D = 5.36 f. say 5.5 f. and L = 4.5 × 5.5 = 24.8 f. T-T.

Check height of liquid to NLL = $\frac{352.9}{\mathrm{X}\ \mathrm{sect}\ \mathrm{area}}$ = 14.8 ft = 60 % of total vessel height which is acceptable.