Exam 2013-2015, questions and answers PDF

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A24687 Calculators may be used in this examination but must not be used to store text. Calculators with the ability to store text should have their memories deleted prior to the start of the examination Special Requirements: Graph paper availableSchool of Chemical EngineeringB Chemical Engineering, ...

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A24687

Calculators may be used in this examination but must not be used to store text. Calculators with the ability to store text should have their memories deleted prior to the start of the examination Special Requirements: Graph paper available

School of Chemical Engineering B.Eng Chemical Engineering, 3rd Year M.Eng Chemical Engineering, 3rd Year M.Eng Chemical Engineering with Industrial Experience, 3rd Year

04 21233 Petrochemical Engineering Summer Examinations 2013 Time Allowed: 3 hours

Answer THREE questions out of four in Section A, and ONE question out of two in Section B. All questions carry equal marks. The distribution of marks within each question is given as a percentage of the total marks for the question. ANSWER EACH QUESTION IN A SEPARATE ANSWER BOOK.

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SECTION A: Answer three questions out of four in this section. 1. (a)

Petroleum is essentially a complex mixture predominantly composed of hydrocarbons containing small quantities of other elements. The hydrocarbons are classified into three general types.

(i)

What are these 3 types? The three types are: paraffins, naphthenes and aromatics. [8%]

(ii)

Identify the type that each of compounds 1 to 9 belongs to, and justify your selection in terms of the characteristic molecular features of each type. 1 paraffin; 2 paraffin; 3 aromatic; 4 naphthene; 5 aromatic; 6 naphthene; 7 aromatic; 8 naphthene; 9 paraffin [12%]

(iii)

There exists a fourth type; that is formed during processing (not in crude). What is it called? Is it stable? The fourth type is ‘olefins’; they are unstable; i.e. they easily change structure and degrade into some undesirable intermediates [8%]

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1. (b)

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Equation 1 represents a certain petrochemical reaction:

(i)

Identify the reaction that is represented by Equation (1). The reaction is de-hydrocyclisation [12%]

(ii)

What is the refining process in which, this reaction is likely to take place? Dehydrocyclisation takes place in the reforming process [12%]

(iii)

For such process could we use n – hexane instead of n – heptanes as feedstock? Why? No we should not use n-hexane because it would produce benzene, which is under strict environment legislations. [16%]

1. (c) Equation 2 represents the reactions of another petroleum refining process:

(i)

Name the process. The process is catalytic hydrocracking [8%]

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(ii)

Name the reactions that take place in it. The reactions are catalytic cracking and hydrogenation [8%]

(iii)

Which fractions of the crude distillate undergo this process? Heavy gas oil undergoes hydrocracking [8%]

(iv)

Are these reactions exothermic or endothermic? Catalytic cracking is endothermic while hydrogenation is exothermic [8%]

2. (a)

(v)

How is the temperature level controlled in this process? As hydrogenation is highly exothermic the overall heated generated from the system could cause hot spots. Temperature is therefore kept under control by using cold hydrogen streams (for hydrogenation) [8%]

2. (b)

Fluid catalytic cracking (FCC) one of the major processes in petroleum refining, to produce a spectrum of fuels from crude oil distillates. (i)

Sketch a conceptual process flow diagram for the FCC process

. [12%]

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(ii)

(iii)

Name two types of catalysts employed for FCC. Any two out of the following three: 1. Acid treated alumina silicate 2. Amorphous synthetic silica alumina combination 3. Crystalline synthetic silica alumina catalyst called zeolite. [8%] What is the main difference between the “old” and “new” designs of FCC reactors? What instigated the new design? Old – large Reactor and small Riser (in comparison), where regenerated catalyst meets oil feedstock before they enter the reactor. In the new design the Riser is much larger, where most of the reaction occurs; the reactor’s role becomes mainly for separating spent catalyst from products and oil. This was instigated by the high activity of new catalysts (crystalline), causing the reaction to start at the moment of contact between feedstock and catalyst in the riser, whose role has consequently become much more prominent. [12%]

(iv)

Why is FCC favoured to thermal cracking? Because thermal cracking caused heat damage to feedstock, resulting in coke laydown and undesired intermediates. [8%]

Figure 2 is an Exxon Mobil catalytic hydroprocessing process flow diagram

Lean Solvent

STREAM ɸ Unit A

REACTORS Unit B Rich Solvent

H2 Make-up

FEED

STREAM Δ

Unit C

HEATER High Temp Separator

Low Temp Separator

Figure 2 – Exxon Mobil catalytic hydroprocessing (i)

Which fractions of the crude distillate constitute the ‘feed’? The ARC (atmospheric resid crude) and VRC (vacuum resid crude) Page 5

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[10%] (ii)

What chemical changes does hydroprocessing entail? How it will ‘upgrade’ the feed streams. Chemical changes involve (1) raising the H/C ratio in the feedstock, (2) breaking down long chains of HC, (3) lowering the boiling point of product. It upgrades the feedstock by making it more ‘amenable’ for further processing by FCC or Reforming. [10%]

(iii)

What is the main difference between hydroprocessing and hydrotreating? Hydrprocessing involves ‘cracking’ long HC chains and adding H, resulting in lowering the boiling point of products and upgrading their quality, while hydrotreating involves only adding H without cracking. [10%]

(iv)

Identify units (A), (B), (C), streams (ɸ), (Δ). Unit A: Guard reactor; Unit B: H2S removal; unit C: product fractionation, stream ɸ: recycled hydrogen; stream Δ, oil reactor exit [10%]

(v)

What type of reactors does the Exxon Mobil design employ? Fixed bed reactors [10%]

3. (a)

Figure 3 shows the Continuous catalyst regeneration (CCR Platforming™) catalytic reformer.

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(i)

List (using simple equations or sketches), the main reactions that take place during reforming.

[16%] (ii)

Based on Figure 3, describe the reforming process, starting from the feed (Charge), till (and including) separation. Feedstock is preheated via heat exchangers (as heat economy system) by cooling the final product exiting from the bottom reactor. After heating, feed is brought in contact with the freshly regenerated catalyst at the top reactor. The feedstock-catalyst combination flow down the reactor at a residence time designed to allow reforming reactions to take place. After every reactor the reaction stream is reheated and sent back to the successive reactor. The final product stream is separated from the catalyst, where the latter is taken for regeneration. The product stream is cooled down in a heat exchange system (CFE) before separation in a topping column for the various products [12%]

(iii)

Why does the reactor diameter get progressively larger as the process moves downwards in the continuous system? As the process goes on and the catalyst progressively ages, more residence time is required to keep the severity of the process to the right level. [12%]

(iv)

Why is the reactants stream sent to ‘reheat’ after each reactor? The reaction stream is reheated in order to supply the energy required to Page 7

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sustain high conversion reaction [10%] 3.(b)

4.

Crude Distillation is the first step of petroleum refining, after obtaining crude oil from underground. (i)

Describe briefly the functions of the crude distillation column. This is the operation where crude oil is boiled and fractionated to a large number of major components, withdrawn according to their respective boiling point ranges. Each ‘cut’ is further processed by a series of operations on route to a series of petroleum-based fuels. [20%]

(ii)

Outline five of the product streams, which exit the distillation column. C1 – C4 gases; light straight run (LSR) gasoline; heavy straight run (HSR) gasoline (naphtha); light gas oil (LGO); heavy gas oil (HGO); atmospheric resid crude (ARC); vacuum resid crude (VRC) [12%]

(iii)

Briefly explain the pretreatment that crude oil undergoes, before entering the distillation column. Talk about de-salting [18%]

(a)

Draw a simplified flow diagram and briefly describe a typical visbreaking. [40%]

(b)

Describe different types of cokes that can be produced [30%]

(c)

State the main differences between delayed coking and flexi-coking [30%]

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SECTION B: Answer one question out of two in this section. 5. (a)

In Marcotte Field in Kansas Refinery, the reformer unit processes 500 kg of Naphtha of the composition shown in Table 5.1. Table 5.1 – Composition of the feedstock.

Compound Weight percent (%) Toluene 28 Xylene 35 Heptane 31.5 1,2,3 Trimethyl benzene (TMB) 4.6 Dimethyl-cyclohexane [DMCC6] 0.5 Cycloheptane (CC7) 0.4 Based on data supplied in Tables 5.2 and 5.3, calculate the composition of the product. Table 5.2 – Naphthenes’ conversion to products.

Feed (Naphthenes) Methylcyclohexane (MCC6) Cyclohexane (CC6) Dimethylcyclopentane (DMCC5) Dimethylcyclohexane (DMCC6) Cycloheptane (CC7) Methylcyclopentane (MCC7)

Conversion

Products

0.98 0.98 0.98 0.98 0.98 0.98

Toluene Benzene Toluene Xylene Toluene Xylene

Table 5.3 – Paraffins’ conversion to products.

Feed (Paraffins) Hexane [C6H14] Heptane [C7H16] Octane [C8H18] Nonane [C9H20] Decane [C10H22]

Conversion 0.05 0.10 0.25 0.45 0.45

Products Benzene Toluene Xylene 1,2,3 - trimethylbenzene 1,2,3,4 - tetramethylbenzene

Solution: Material balance is conducted on the system and the result is shown as below: 500 mols of input Naphtha Component CC7[C7H14] DMCC6[C8H16] Heptane[C7H16] 1,3,5-TMB SUMS

% Comp 20 40 38 2

toluene 23.4

mol comp toluene 100 98 200 190 19 10 500 117

% IN PRODUCTS xylene CC7 DMCC6 39.2 0.4 0.8

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PRODUCTS CC7 DMCC6 2 196 4

xylene

Heptane

TMB

171 196

2

Heptane TMB 34.2 2

4

171

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5.(b)

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A feed of 100,000 BPD (Barrels per Day) of AGO (Atmospheric Gas Oil) (650 850°F) having an API of 22 and a sulphur content of 0.23 wt%, is mixed with another feed of 30,000 BPD of VGO (Vacuum Gas Oil) (850 - 1050°F) that has an API of 12 and a sulphur content of 0.0.38 wt%. They are used as a feed to FCC unit. Assume a conversion of 80 LV%. Use correlations from Table 5.5 when required. (i) (ii) (iii) (iv)

PRODUCT

Determine the weight % sulphur in the feedstock. Determine the API of the feedstock. Determine the weight % of sulphur in the products (gas, HCO and LCO). Determine the weight % Gasoline in the product.

CORRELATIONS

Coke wt%

0.05356 x C - 0.18598 x API + 5.966975

LCO LV%

0.0047 x C2 – 0.8564 x C + 53.376

Gases wt%

0.0552 x C + 0.597

Gasoline LV%

0.7754 x C -0.7778

iC4 LV%

0.0007 x C2 + 0.0047 x C + 1.40524

nC4 LV%

0.0002 xC2 + 0.019 x C + 0.0476

C4= LV%

0.0993 x C – 0.1556

PRODUCT

CORRELATIONS

C3 LV%

0.0436 x C – 0.8714

C3= LV%

0.0003 x C2 + 0.0633 x C + 0.0143

HCO

100 – C – [LCO LV%]

Wt% S in gases

3.9678 x (wt% S in feed) + 0.2238

Wt% S in LCO

1.04994x (wt% S in feed) + 0.00013

Wt% S in HCO

1.88525 x (wt% S in feed) + 0.0135

S in coke

wt% S in feed – wt% S in gases – wt% S HSO

Gasoline API

-0.19028 x C + 0.02772 x (gasoline LV%) + 64.08

LCO API

-0.34661 x C + 1.725715 x (feed API)

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Solution: Given: 100,000 BPD of AGO API = 22 S wt% = 0.23

30,000 BPD of VGO API = 12 S wt% = 0.38

(i) Using API and table 5.4 find the densities for AGO and VGO, then find the weight of each stream from which, you can determine wt% in the feedstock. For AGO, API = 22, density = 322.8 lb/bbl For VGO, API = 12, density = 345.3 lb/bbl Therefore: Mass of AGO = 100,000 x 322.8/24 = 1345000 lb/hr Mass of VGO = 30,000 x 345.3/24 = 431625 lb/hr Total mass = 1776625 lb/hr S in AGO = 0.23/100 x 1345000 = 3093.3 S in VGO = 0.38/100 x 431625 = 1602.2 Wt% S in feedstock = [[3093.3 + 1602.2]/1776625] x 100% = 0.264 wt% (ii) For API of Feedstock: From Table 5.4: SG for AGO = 0.922, SG for VGO = 0.986 therefore API for feedstock is found from weighted average SG Avg SG = 0.922 x [100,000/130,000] + 0.986 x [30,000/130,000] = 0.937 From Table 5.4, API of feedstock = 19.7 (iii) Apply the S correlations in Table 5.5: Wt% S in gases = 1.27, Wt% in LCO = 0.277, Wt% in HCO = 0.511 (iv) % Gasoline in product (LV%) = 61.25 6. (a) For the mixture shown in Table 6.1, estimate the amount of n-butane that should be added for a 11psi RVP Component

BPCD RVP

n-Butane

52

LSR gasoline

2000

11.1

Reformate

7000

2.8

Alkylate

3000

4.6

FCC Gasoline

8000

4.4

Table 6.1 [50%]

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6. (b) For the problem shown in Table 6.2, set up a linear programming problem to estimate the amounts needed to produce gasoline blend of maximum octane number and 12 psi RVP. The maximum blend capacity us 15000 barrels.

Component LSR gasoline Hydrocracker gasoline FCC gasoline n-Butane

RVPpsi RON 12.1 91 15.5 89.5 4.8 51.6

97 93

Table 6.2 [50%]

END OF PAPER

Table 5.4 Density Conversion Table

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Calculators may be used in this examination but must not be used to store text. Calculators with the ability to store text should have their memories deleted prior to the start of the examination Special Requirements: Graph paper available

School of Chemical Engineering B.Eng Chemical Engineering, 3rd Year M.Eng Chemical Engineering, 3rd Year M.Eng Chemical Engineering with Industrial Experience, 3rd Year

04 21233 Petrochemical Engineering Summer Examinations 2014 Time Allowed: 2 hours

Answer ALL Questions All questions carry equal marks. The distribution of marks within each question is given as a percentage of the total marks for the question. ANSWER EACH QUESTION IN A SEPARATE ANSWER BOOK.

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Figure 1 is a block diagram of a typical petroleum refinery, including the main processes:

STABILISER SECTION

TYPICAL REFINERY PRODUCTS LPG & GAS

STRAIGHT RUN GASOLINE

VAPOUR RECOVERY

FUEL GAS

LPG

ISOMERISATION

HYDRO TREATING

NAPHTHA

REFORMING

HYDRO TREATING

MIDDLE DISTILLATES

HEAVY ATM GAS OIL

REFORMATE

CATALYTIC CRACKING

GASOLINE

VACUUM TOWER SECTION

ALKYLATION VAC GAS OIL AROMATICS REDUCTION LUBE BASE STOCKS

FRACTIONATION BOTTOMS

HYDRO CRACKING

GASOLINE, NAPHTHA & MIDDLE DISTILLATES

TREATING AND BLENDING

ATMOSPHERIC TOWER SECTION

REGULAR GASOLINE

CRUDE UNIT

1. (a)

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PREMIUM GASOLINE

AVIATION FUELS DIESELS HEATING OILS LUBE OILS GREASES

ASPHALTS

DEWAXING

LUBE OILS WAXES

PROPANE DEASPHALTER VISBRAKER

GASOLINE, NAPHTHA & MIDDLE DISTILLATES

INDUSTRIAL FUELS FUEL OILS

FUEL OIL ASPHALT

DELAYED COKER

COKE

GASOLINE, NAPHTHA & MIDDLE DISTILLATES

Answer the following questions: (i) Identify the units where the following takes place: (1) Hydro-de-cyclisation of distillates for products of higher octane numbers: Reforming (2) Production of oils with reduced size molecules: FCC (3) Separation of high boiling fractions (> 300°C): Vacuum distillation [10%] (ii) Describe how composition of the vacuum resid crude (VRC), is found. Use sketches to aid your explanation [20%] See attached sheet (iii) Name the catalysts, which are suitable for: (1) Reforming: Platinum on Alumina (2) Catalytic Cracking: amorphous zeolites, crystalline synthetic zeolites. [15%]

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1. (b)

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In the crude distillation process “Reflux” is a design and operating aspect. Maximum reflux and fractionation are obtained by removing all heat at the top of the tower. However, this is not practised in refineries. (i)

Why is this option not adopted in refineries? [15%] Maximum reflux and fractionation are obtained by removing all heat at the top of the tower. This results in an inverted cone-type liquid loading, which requires a very large diameter at the top of the tower.

(ii)

Explain briefly the alternative arrangement that is used. [20%] To reduce the top diameter of the tower and even the liquid loading over the length of the tower intermediate heat-removal streams are used to generate reflux below the side stream removal points. To accomplish this, liquid is removed from the tower cooled by a heat exchanger, and returned to the tower or, alternatively, a portion of the cooled side stream may be returned to the tower.

(iii)

Explain briefly how energy efficiency and distribution can benefit from this arrangement. [20%] The cold stream returned to the tower condenses more of the vapours coming up the tower and thereby increases the reflux below that point. The energy efficiency of the distillation operation is also improved by using pump-around reflux. If sufficient reflux was produced in the overhead condenser to provide for all side stream draw offs as well as the required reflux, all of the heat energy would be exchanged at the bubble-point temperature of the overhead stream. By using pump-around reflux at lower points in the column, the heat transfer temperatures are higher, and a higher fraction of the heat energy can be recovered by preheating the feed.

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2 (a)

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