Fahim M.A., Sahhaf T.A., Elkilani A.S. Fundamentals of Petroleum Refining

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482–538



C (900–1000



F). The coke in the spent catalyst, produced in the

cracking reaction, is burned off in the regenerator by introducing excess air,

which is used to ensure the efficient combustion of coke. The produced flue

gas exits at the top of the regenerator. The hot flue gas contains carbon

dioxide, carbon monoxide water and excess air at the regenerator tempera-

ture. These gases are sent to the power recovery unit to produce super-

heated steam. The operation of the FCC remains in a steady state as long as a

heat balance exists between the heat produced in the regenerator and the

heat consumed in the reactor.

In both the reactor and the regenerator, hydrocyclones are installed to

catch any solid particles carried out in the overheated stream. The product

gases from the reactor are sent to the fractionator which produces light gases,

heavy gasoline (main product), light cycle gas oil (LCO), heavy cycle gas oil

(HCO) and decant slurry. The light gases are sent to the gas concentration

unit where flue gas, propane, butane, LPG and light gasoline are produced.

The operating conditions are usually adjusted to produce the maximum

amount of gasoline from the VGO as shown in Table 8.4 (Parkash, 2003).

The decant slurry from the bottom of the fractionator is a mixture of

heavy aromatics and fine catalyst particles. These particles are formed during

the circulation of the catalyst and are carried out to the fractionator. The

filtered decant can be used as an aromatic solvent or recycled back to the

riser with the HCO.

Expander

Cyclone

Regenerator

Air

Feed

Steam

Catalyst

Reactor

Recycle

Heavy cycle gas oOil

Decant slurry

Fractionator

Light cycle gas oil

Heavy gasoline

Water

Light gasoline

Fuel gas + H

Cut

Gas concentration unit

Hot flue gas

Superheated

Steam

Water

Stripper

steam

Figure 8.7 Fluid c atalytic cracking process f low sheet

Fluidised Catalytic Cracking 211

8.10. Modes of Fluidisation in FCC unit

The modes of fluidisation are used to operate the different parts of

the FCC unit (Tominaga and Tameki, 1997). The regenerator is run at the

higher velocity of the stable turbulent fluidisation. The line connecting

the regenerator to the riser is carried out at the ‘bubbling fluidisation’ zone.

The riser is operated at the ‘pneumatic transport’ zone, where the catalyst

and products are carried out from the riser. In the stripper section, where

steam is injected, the mode is ‘bubbling fluidisation’. The regenerated

catalyst is transported back in the left line in a ‘pneumatic transport’

mode. The fluidisation modes are summarised in Table 8.5.

8.11. FCC Yield Correlations

The yields of the products involved in fluid cracking are obtained by

the regression of plant data compiled by Maples (1993) using a zeolite

catalyst. The correlations given in Table 8.6 require target conversion

(LV%), feed API and sulphur in the feed. Conversion is defined as the

percentage of the oil fed that has been cracked into lighter fractions than

gasoline and lighter products:

CONV% ¼

volume of oil feed volume of cycle stock

volume of oil feed



100 ð8:12Þ

Table 8.4 Reactor and regenerator operating condition for max

gasoline production (Parkash, 2003)

Vari able Value

Reactor Feed Rate, MBPSD 40

Feed Temperature,



F 446

Catalyst/Oil Ratio 5.4

Catalyst Circulation Rate, tons/min 21.7

Catalyst Makeup Rate, tons/day 2.5

Riser Outlet Temperature,



F 991

Dispersion Steam, wt% feed 0.9

Stripping Steam, tons/ton catalyst 0.0213

Reactor Pressure, psig 30

Regenerator Pressure, psig 33

Regenerator Temperature,



F 1341

Flue Gas Temperature,



F 1355

212 Chapter 8

Table 8.5 Modes of ﬂuidisation in FCC

Location in FCC Mode of fluidisation

Regenerator Turbulent ﬂuidisation: to attain uniform

burning temperature in bed.

Line for catalyst transport

from regenerator to riser

Bubbling ﬂuidisation

Riser Pneumatic transport: Catalyst and products are

carried out from riser. Plug ﬂow has a few

seconds of residence time.

Stripper Bubbling ﬂuidisation: Steam is injected in the

stripper to vaporise and recover heavy oil

and reduce coke formation.

Lift line from regenerator to

reactor

Pneumatic transport

Table 8.6 FCC yield correlations

Products Correlation

Coke wt% 0.05356  CONV – 0.18598  API + 5.966975

LCO LV% 0.0047  CONV

– 0.8564  CONV + 53.576

Gases wt% 0.0552  CONV + 0.597

Gasoline LV% 0.7754  CONV – 0.7778

LV% 0.0007  CONV

+ 0.0047  CONV + 1.40 524

LV% 0.0002  CON V

+ 0.019  CONV + 0.0476

LV% 0.0993  CONV – 0.1556

LV% 0.0436  CONV – 0.8714

LV% 0.0003  CONV

+ 0.0633  CONV + 0.01 43

HCO 100 – CONV – (LCO LV%)

Wt% S in Gases 3.9678  (wt% S in feed) + 0.2238

Wt% S in LCO 1.04994  (wt% S in feed) + 0.00013

Wt% S in

HCO

1.88525  (wt% S in feed) + 0.0135

S in Coke

wt% S in feed – wt% S in gases – wt% S LCO – wt% S HCO

Gasoline API 0.19028  CONV + 0.02772  (Gasoline LV%) + 64.08

LCO API 0.34661  CONV + 1.725715  (Feed API)

Assuming no sulphur in gasoline

Fluidised Catalytic Cracking 213

The recycle stock is the portion of the feedstock which is not cracked to

fractions lighter than gasoline. For example, for 75% conversion, the cycle

stock is 25%.

Example E8.1

A feed of 20,000 BPD of AGO (650–850



F) having an API of 24 and a sulphur

content of 0.2 wt%, is mixed with another of feed of 15,000 BPD of VGO

(850–1050



F) that has an API of 15 and a sulphur content of 0.35 wt%. They

are used as a feed to FCC unit. Use the FCC correlations to ﬁnd the material

balance around the reactor unit. Assume a conversion of 75 LV%. Figure E8.1

shows the reactor input and output streams.

Solution:

AGO ¼ 20000 (bbl/day)  318.6 (lb/bbl)  (1 day/24 h) ¼ 265,000 lb/h

VGO ¼ 15000 (bbl/day)  338 (lb/bbl)  (1 day/24 h) ¼ 211250 lb/h

S in AGO ¼ 265000  0.2/100 ¼ 530 lb/h

S in VGO ¼ 211250  0.35/100 ¼ 739 lb/h

S in feed ¼ 1269/476250  100 ¼ 0.266%

Conversion ¼ ((Vol. of feed – Vol. of cycle stock)/Vol. of feed)  100 ¼ 75%

Cycle stock ¼ unconverted portion below gasoline ¼ (LCG + HCGO) ¼ 25

API of mixture:

Total feed ¼ 476,250 lb/h

SG for AGO ¼ 0.9099 and SG for VGO ¼ 0.9659

Then SG for mixed feed ¼

20000

20000 þ 15000

ð0:9099Þþ

15000

20000 þ 15000

ð0:9659Þ

¼ 0:9339

Products

AGO + VGO + Steam

Spent

Catalyst

Regenerated

Catalyst

Steam

Gases

/iC

Gasoline

LCO

Figure E8.1 Reactor input and output streams

214 Chapter 8

This gives feed API ¼ 20.02

A summary of the feed mixture is shown in Table E8.1.1.

Using the above data and the yield correlations from Table 8.6 , the material

balance and product properties can be calculated as shown in Table E8.1.2

Table E8.1.2 Yields and properties of products

lb/h

Coke wt% 0.05356  (75) – 0.18598  (20.02)

+ 5.966975

6.3 30,004

LCO LV% 0.0047  (75)

– 0.8564  (75) +

53.576

15.8 81,337

Gases wt% 0.0552  (75) – 0.597 4.7 22,574

Gasoline LV% 0.7754  (75) – 0.7778 57.4 226,816

LV% 0.0007  (75)

+ 0.0047  (75) +

1.40524

5.7 16,375

LV% 0.0002  (75)

+ 0.019  (75) +

0.0476

2.6 7735

LV% 0.0993  (75) – 0.1556 7.3 22,356

LV% 0.0436  (75) – 0.8714 2.4 6230

LV% 0.0003  (75)

+ 0.0633  (75) +

0.0143

6.4 16,987

HGO wt% 100 – 75 – 15.8 9.7 46,027

SinH

S wt% 3.9678  (0.266) + 0.2238 1.28 289

S in LCO wt% 1.04994  (0.266) + 0.00013 0.278 226

S in HCO wt% 1.88525  (0.266) + 0.0135 0.515 237

S in Coke wt% (1269 – 289 – 226 – 237)/29,813 1.734 517

Gasoline API 0.1 9028  (75) + 0.02772  (59.1)

+ 64.08

51.4

LCO API 0.34661  (75) + 1.725715 

(20.02)

8.5

Table E8.1.1 Feed properties

Stream BPD API lb/h wt% S lb S/h

AGO 20,000 24 265,000 0.2 530

VGO 15,000 15 211,250 0.35 739

20.02 476,250 1269

Fluidised Catalytic Cracking 215

8.12. Material and Energy Balances

The material and energy balance around the reactor and regenerator

can be calculated by defining the input and output streams.

8.12.1. Material Balance

8.12.1.1. Reactor Material Balance

The input and output streams to the reactor (Figure 8.8) are:

Reactor input:



Oil feed (VGO) to the riser: F (BPD) or m

(lb/h)



Injection steam: S

(lb/h)



Regenerated catalyst: m

cat

(lb/h)

Reactor output:



Masses of products m

, as calculated from FCC yield correlations.

These correlations require some feed properties such as: API, sulphur

content and degree of severity expressed as conversion.



Spent catalyst circulation rate m

scat

(lb/h)



Steam present in cracked products, S

out

(lb/h)

Thus, a material balance around the reactor is

þ S

þ m

cat

i¼1

þ S

out

þ m

scat

ð8:13Þ

Flue gases

Products

Feed oil

, Q

Air

air

cat

Catalyst

Exothermic

reactions

Endothermic

reactions

Regenerator

Reg

Reactor

scat

∑

i =1

∑

i =1

Steam

in,

s,in

out,

s,out

Figure 8.8 Input and output streams for reactor and regene rator in FCC unit

216 Chapter 8

where p is the total number of vapour products and assuming S

does

not conden se a nd is present in the exiting vapou r pr oducts at the same

rate (S

¼ S

out

). m

is the mass of each product that can be calculated

using the FCC correlations. The produced coke is pres ent in spent

catalyst. Thus

coke

¼ m

scat

 m

cat

Equation (8.13) can be rewritten as follows:

i¼1

þ m

coke

ð8:14Þ

8.12.1.2. Regenerator Material Balance

Regenerator input:



Spent catalyst circulation rate m

scat

(lb/h)



Air for coke burning m

air

(lb/h)

Regenerator output:



Flue gases n

(lb/h)



Regenerated catalyst m

cat

(lb/h)

Thus, the material balance around the regenerator produces:

air

þ m

scat

i¼1

þ m

cat

ð8:15Þ

where n

is the mass of each gas produced from the coke burning which may

contain CO

, CO, H

O, SO

and O

(from excess air).

8.12.2. Energy Balance

8.12.2.1. Reactor Heat Balance

Heat input:



Heat of feed oil Q

(Btu/h) at inlet feed temperature (T

)



Heat of steam injected Q

(Btu/h) at T



Heat of regenerated catalyst Q

cat

(Btu/h) at regenerator outlet temper-

ature (T

Reg

)

Heat output:



Heat in vapour products, Q

(Btu/h) at reactor outlet temperature

)



Heat of spent catalyst Q

scat

(Btu/h) at T

Fluidised Catalytic Cracking 217



Heat of exit steam Q

s,out

(Btu/h) at T

Then the energy balance can be expressed as

p;f

ðT

T

Þþm

ðDH

Þþm

cat

P;cat

ðT

Reg

T

ÞþS

ðT

T

¼ðT

T

P;i

þm

scat

P;scat

ðT

T

ÞþS

out

ðT

T

ð8:16Þ

Since, m

coke

¼ m

scat

– m

cat

and S

¼ S

out

, then equation (8.16) becomes

p;f

ðT

T

Þþm

ðDH

Þþm

cat

P;cat

ðT

Reg

T

ÞþS

ðT

T

¼ðT

T

P;i

þm

coke

P;coke

ðT

T

ð8:17Þ

8.12.2.2. Regenerator Heat Balance

Heat input:



Heat of spent catalyst Q

scat

(Btu/h) at T



Heat of input air for coke burning Q

air

(Btu/h) at T

air



Heat of coke combustion q

coke

(Btu/h)

Heat output:



Heat of flue gas Q

(Btu/h) at T

Reg



Heat of regenerated catalyst Q

cat

(Btu/h) at T

Reg

Thus the heat balance around the regenerator can expressed as

air

p;air

ðT

air

 T

Þþm

coke

P;coke

ðT

 T

Þþq

coke

¼ðT

Reg

 T

P;gi

þ m

cat

P;cat

ðT

Reg

 T

ð8:18Þ

Example E8.2

Find the catalyst circulation rate for the FCC unit used in example E8.1 by

carrying out a heat balance around the regenerator. A sketch of the stream ﬂows

and temperatures around the regenerator is presented in Fig. E8.2.1

Data:

Heat of combustion of coke, DH

¼0.393  10

kJ/kmol,

P,cat

¼ 1.11 kJ/kg.K,

P,air

¼ 29.6 kJ/kmol.K, C

P,N

¼ 32.6 kJ/kmol.K, C

P,CO

¼ 46.9 kJ/kmol.K,

P,coke

¼ 21.1 kJ/kg.K

Solution:

Coke produced ¼ 29,813 lb/h ¼ 13,551 kg/h (from example E8.1)

Coke ¼ 13,551 kg/h ¼ 1129 kmol/h

coke

¼ 1129 (kmol/h)  0.393  10

(kJ/kmol) ¼ 443.8  10

(kJ/h)

218 Chapter 8

Air required ¼ O

required/0.21

required:

C+O

! CO

¼ O

required ¼ 1129 kmol/h

CO2

¼ 1129 kmol/h

air

¼ air required ¼ 1129/0.21 ¼ 5376 kmol/h

¼ 5379  0.79 = 4247 kmol/h

Thus, the heat balance from equation (8.18) gives

5376 29:6ð200 15Þþm

cat

1:11 ð520 15Þþ13551 21:1 ð520 15Þ

þ443:8 10

¼ð700 15Þ½1129 46:9 þ4247 32:6þm

cat

1: 11 ð690 15Þ

cat

¼ 2.58  10

kg/h ¼ 43.0 t/min ¼ 0.717 t/s

8.13. Kinetic Model for FCC Reactor

The FCC process involves a network of reactions producing a large

number of components. Therefore, lumping models can be used to describe

the reaction system in terms of the feed and a defined number of products.

Three, four and up to over ten lumps can be used theoretically. A three

lump model (Figure 8.9) assumes that VGO produces two products: gaso-

line and a combined product of gas and coke, where y refers to the mass

fraction of each lump and k is the reaction rate constant.

A more realistic model is the four-lump model (Figure 8.10) in which

VGO produces gasoline, gas and coke.

Assuming that the VGO cracking rate is second order and gasoline rate is

first order. The three and four lump models in Figures 8.9 and 8.10,

respectively, are shown below. In these models the catalyst deactivation

rate f is considered equal for all reactions.

Regenerator

, m

CO2

Reg

= 700 ⬚C

air

= 200 ⬚C

scat

+ m

coke

scat

= T

= 520 ⬚C

cat

Reg

= 690 ⬚C

Figure E8.2.1 Regenerator

Fluidised Catalytic Cracking 219

Three-lump model:

for VGO

¼k

f  k

f ¼k

þ k

ðÞy

f ð8:19Þ

for gasoline

¼ k

f  k

f ¼ k

 k



f ð8:20Þ

for gas + coke

¼ k

þ k



f ð8:21Þ

Four-lump model:

for VGO

¼k

f  k

f ð8:22Þ

for gasoline

¼ k

f  k

f ¼ k

 k

þ k



ð8:23Þ

for gas

¼ k

þ k



f ð8:24Þ

VGO (y

) Gasoline (y

)

Gas + Coke

)

Figure 8.9 Three-lump model

VGO (y

)

Gasoline (y

)

Gas (y

)

Coke (y

)

Figure 8.10 Four-lump model

220 Chapter 8