Dr. Belal Al Zaitone Zuriqat - kausfalhuthali.kau.edu.sa/GetFile.aspx?id=238735&Lng... · Chapter 6 MATERIAL BALANCES FOR MULTI-UNIT SYS. 2 Chapter 6: Recycle, Bypass and Purge 1

King Abdulaziz University

Faculty of Engineering

Department of Chemical and

Materials Engineering

Dr. Belal Al Zaitone Zuriqat

eMail: [email protected] http://balzaitone.kau.edu.sa Office: Building No. 40, Room No. 24G58

Ch

E2

01

Introduction to Chemical Engineering

Chapter 6MATERIAL BALANCES FOR MULTI-UNIT SYS.

2

Chapter 6: Recycle, Bypass and Purge

1 Primary Concepts

3 Recycle without Chemical Reaction

4 Recycle with Chemical Reaction

5 Bypass and Purge

6 The Industrial Application of Material Balances

2 Sequential Multi-Unit Systems

3

Chapter 6: Material Balance Involving Multiple Units

Production Plant

4


thyssenkrupp-uhde.deProcess Flow Sheet

5


Process Block Diagram

6


3 5Reactor

4

7

1

2

6

Mixing point Splitting pointm1+m2=m3 m5=m6+m7

x5,A= x6,A= x7,A

x5,B= x6,B= x7,B

A

BA

BA

B

7


3 5Reactor

4

7

1

2

6

Mixing point Splitting pointm1+m2=m3

m3+m4=m5

8


3 5Reactor

4

7

1

2

6

Mixing point Splitting pointm1+m2=m3 m5=m6+m7

m3+m4=m5

m1+m2+m4=m6+m7

9


3 5Reactor

4

7

1

2

6

m1+m2+m4=m5

10


3 5Reactor

4

7

1

2

6

m3+m4=m6+m6

11


Example 6.1: Examine Block diagram given, No reaction takes place. The

system is open and steady state. The arrows designate mass flows. The

composition of each stream is as follows:

(1) Pure A

(2) Pure B

(3) A and B, concentrations known: xA = 0.800, xB = 0.200

(4) Pure C

(5) A, B, and C, concentrations known: xA = 0.571, xB = 0.143, xC = 0.286

(6) Pure D

(7) A and D, concentrations known: xA = 0.714, xD = 0.286

(8) B and C, concentrations known: xB = 0.333, xC = 0.667

Solution:

12


3 5II

4

7

1

2 6

I III8

m1,A=100 kg

m2,B

m3

m3,A

m3,B

x3,A=0.8

x3,B=0.2

m4,C

m5

m5,A

m5,B

m5,C

x5,A=0.571

x5,B=0.143

x5,C=0.286

m6,D

m7

m7,A

m7,D

x7,A=0.714

x7,D=0.286

m8

m8,B

m8,C

x8,B=0.333

x8,C=0.667

13


14


Example 6.2: Acetone is used in the manufacture of many chemicals and

also as a solvent. In its latter role, many restrictions are placed on the

release of acetone vapor to the environment.

You are asked to design an acetone recovery system having the flow sheet

illustrated in Figure shown.

All the concentrations shown of both the gases and liquids are specified in

weight percentage in this special case to make the calculations simpler.

Calculate, mass flow from each process unit in the flow sheet,

if m1,A = 1400 kg/hr.

Solution:

15


3 5

II

4

7

1

2

6

I

m2,W

III

m1= 1400 kg/hr

x1,A=0.95

x1,T=0.03

x1,W=0.02

m3

x3,A=0.995

x3,W=0.005

m4

x4,T=0.19

x4,W=0.81

m7

x7,T=0.99

x7,W=0.01

m6

x6,T=0.04

x6,W=0.96

Absorber column

CondenserDistillatio

n co

lum

n

16

3 5

II

4

7

1

2

6

I

m2,W

III

m1= 1400 kg/hr

x1,A=0.95

x1,T=0.03

x1,W=0.02

m3

x3,A=0.995

x3,W=0.005

m4

x4,T=0.19

x4,W=0.81

m7

x7,T=0.99

x7,W=0.01

m6

x6,T=0.04

x6,W=0.96


17


Example 6.3: Material Balances Involving Multiple Units and Reactions

A Company operates two furnaces, one fired with natural gas and the other

with fuel oil.

The gas furnace uses air while the oil furnace uses an oxidation stream that

analyzes: 02, 20%; N2, 76%; and C02, 4%. The stack gases go up a common

stack.

The reserve of fuel oil was only 60 bbl (oil barrel).

The minimum heating load for the company when translated into the stack

gas output was 6205 mol/hr of dry stack gas.

1) How many hours could the company operate before shutting down if no

additional fuel oil was attainable?

2) How many mol/hr of natural gas were being consumed?

18


3

5

II

4

7

1

2

6

I

n5,H2O

n1(Nat. gas)

y1,CH4=0.96

y1,C2H2=0.02

y1,CO2=0.02

8

n2 (Air)

y2,O2=0.21

y2,N2=0.79

n8(Fuel Oil)

y8,C =0.50

y8,H2=0.47

y8,S=0.03

n7 (Air*)

y7,O2=0.20

y7,N2=0.76

y7,CO2=0.04

n4 =6205 mol/hr)

y4,O2=0.0413

y4,N2=0.8493

y4,CO2=0.1084

y4,SO2=0.001

Gas

Fu

rnace

Oil

Fu

rnace

19


3

5

II

4

7

1

2

6

I

n1(Nat. gas)

y1,CH4=0.96

y1,C2H2=0.02

y1,CO2=0.02

8

n2 (Air)

y2,O2=0.21

y2,N2=0.79

n8(Fuel Oil)

y8,C =0.50

y8,H2=0.47

y8,S =0.03

n7 (Air*)

y7,O2=0.20

y7,N2=0.76

y7,CO2=0.04

n4 =6205 mol/hr)

y4,O2 =0.0413

y4,N2 =0.8493

y4,CO2=0.1084

y4,SO2=0.001

n5,H2O

Gas

Fu

rnace

Oil

Fu

rnace

20


Example 6.4: Analysis of a Sugar Recovery Process

Process block diagram shows a sugar. You are asked to calculate the

compositions of every flow stream, and the fraction of the sugar in

the cane that is recovered?

Solution

Suger = S

Water = W

Pulp = P

15

Screen

4

8

2

3

7

Mill

m8,S=1000 Ib

9

m1

x1,S=0.16

x1,W=0.25

x1,P=0.59

Evap

Cryst

6

m9,w

m6,w

m4

x4,s=

x4,w=

x4,P=0.95

m3

x3,S=0.13

x3,P=0.14

x3,w=0.73

m7

x7,S=0.40

x7,w=0.60

m2

x2,s=

x2,w=

x2,P=0.80

m5

x5,S=

x5,W=0.85

21


1 5Screen

4

8

2

3

7

Mill

m8,S=1000

9

m1

x1,S=0.16

x1,W=0.25

x1,P=0.59

Evap

Cryst

6

m9,w

m6,w

m4

x4,s=

x4,w=

x4,P=0.95

m3

x3,S=0.13

x3,P=0.14

x3,w=0.73

m7

x7,S=0.40

x7,w=0.60

m2

x2,s=

x2,w=

x2,P=0.80

m5

x5,S=

x5,W=0.85

22


• A recycle system is a system that includes one or more recycle

streams.Examples of the application of material recycling in the industry:

• Increase reactant conversion.

• Continuous catalyst regeneration.

• Circulation of a working fluid.

I IIFeed 1

Feed 2

Recycle

• A recycle stream is a stream that is fed back from a down-stream

unit (unit II) to up-stream unit (unit I)

23


I IIFeed 1

Feed 2

Bypass

• A Bypass stream is a stream that Skip one or more process units and

goes directly to down-stream unit (unit III).

III

• A Purge stream is a stream bled off (drained) stream from process to

remove an accumulation of inert that might build up in the recycle

stream.

I IIFeed 1

Feed 2

RecycleS

III

Purge

24


Example 6.5: Continuous Filtration Involving a Recycle StreamThe following diagram is a schematic of a process for the production of flake

NaOH, which is used in households to clear plugged drains in the plumbing.

The fresh feed to the process is 10,000 lb/hr of a 40% aqueous NaOH solution.

The fresh feed is combined with the recycled filtrate from the crystallizer, and fed to

the evaporator where water is removed to produce a 50% NaOH solution, which in

turn is fed to the crystallizer.

The crystallizer produces a filter cake that is 95% NaOH crystals and 5% solution

that itself consists of 45% NaOH. The filtrate contains 45% NaOH.

1. Determine the flow rate of water removed by the evaporator, and the recycle

rate for this process.

2. Assume that the same production rate of NaOH flakes occurs, but the filtrate is

not recycled. What would be the total feed rate of 40% NaOH have to be then?

Assume that the product solution from the evaporator still contains 50% NaOH.

Solution:

25


Evap. Cryst.4

1

3m1=10000 Ib/hr

x1,NaOH=0.40

x1,H2O=0.60

m3,H2O

5

6

2

m6

x6,NaOH=0.45

x6,H2O=0.55

m4

x4,NaOH=0.50

x4,H2O=0.50

m5

x5,NaOH,cake=0.95

x5, Soln =0.05(NaOH +H2O)0.45 wt% NaOH

0.55 wt% H2O

26


Evap. Cryst.4

1

3m1=

x1,NaOH=0.40

x1,H2O=0.60

m2,H2O

5m4

x4,NaOH=0.50

x4,H2O=0.50

m5 =4113 Ib/hr

x5,NaOH,cake=0.95

x5, Soln =0.05(NaOH +H2O)

0.45 wt% NaOH

0.55 wt% H2O

m6

x6,NaOH=0.45

x6,H2O=0.55

6

With Recycle process

m1=10000 Ib/hr produce m5=4113 Ib/hr

Without Recycle stream process

m1=53730 Ib/hr produce m5=4113 Ib/hr

You can notice that in order to produce the same amount of m5 (4113) without recycle

stream process, a lager amount of m1 is required!!!

27

𝑓𝑆𝑃 =𝑛𝐿𝑅,𝑖𝑛 − 𝑛𝐿𝑅,𝑜𝑢𝑡

𝑛𝐿𝑅,𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑜𝑟

× 100%


Recycle with Chemical Reaction : A B• Fractional Conversion on reactor unit:

𝑓 =𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑖𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑟𝑒𝑎𝑐𝑡𝑒𝑑

𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑× 100%

𝑓 =𝑛𝐿𝑅,𝑟𝑒𝑎𝑐𝑡𝑒𝑑𝑛𝐿𝑅,𝑓𝑒𝑒𝑑

× 100%

Single-Pass fractional conversion

Reactor Seperator1

3

n1,A=

100 mol/s

4

5

2

n5,A=900 mol/s

n3

y3,A=0.90

y3,B=0.10

n4,B

=100 mol/s

n2,A=1000 mol/s

𝑓𝑆𝑃 =𝑛2,𝐴 − 𝑛3,𝐴

𝑛2,𝐴× 100%

28


Recycle with Chemical Reaction : A B

• Fractional Conversion on process units (Reactor & Seperator):

Over-All fractional Conversion

Reactor Seperator1

3

n1,A=

100 mol/s

4

5

2

n5,A=900 mol/s

n3

y3,A=0.90

y3,B=0.10

n4,B =

100 mol/s

n2,A=1000 mol/s

𝑓𝑂𝐴 =𝑛1,𝐴 − 0

𝑛1,𝐴× 100%

𝑓𝑂𝐴 =𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 − 𝑛𝐿𝑅,𝑜𝑢𝑡

𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠

× 100%

29


Recycle with Chemical Reaction : A B

𝑓𝑂𝐴 =𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 − 𝑛𝐿𝑅,𝑜𝑢𝑡

𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝒑𝒓𝒐𝒄𝒆𝒔𝒔

× 100%

Over-All fractional Conversion

𝑓𝑂𝐴 =−𝜉 ∙ 𝜐𝐿𝑅


𝜉 =𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛

𝜐𝐴

𝑓𝑆𝑃 =−𝜉 ∙ 𝜐𝐿𝑅

𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑟𝑒𝑎𝑐𝑡𝑜𝑟

Single-Pass fractional conversion

𝑓𝑆𝑃 =𝑛𝐿𝑅,𝑖𝑛 − 𝑛𝐿𝑅,𝑜𝑢𝑡

𝑛𝐿𝑅,𝑖𝑛 𝒓𝒆𝒂𝒄𝒕𝒐𝒓

× 100% 𝜉 ∙ 𝜐𝐴= 𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛

30


Reactor Seperator1

3

n1,A=

100 mol/s

4

5

2

n5,A=900 mol/s

n3

y3,A=0.90

y3,B=0.10

n4,B =

100 mol/sn2,A=1000 mol/s

n2,A=n1,A+n5,A

MB on the mixing point:

nLR,into reactor= nLR,into process+nLR,recycle

𝑓𝑂𝐴 =−𝜉 ∙ 𝜐𝐿𝑅

𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑓𝑆𝑃 =

−𝜉 ∙ 𝜐𝐿𝑅𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑟𝑒𝑎𝑐𝑡𝑜𝑟

𝑓𝑆𝑃𝑓𝑂𝐴

=𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠

𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑟𝑒𝑎𝑐𝑡𝑜𝑟=


𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 + 𝑛𝐿𝑅,𝑅𝑒𝑐𝑦𝑐𝑙𝑒 𝑠𝑡𝑟𝑒𝑎𝑚

A B

31


Example 6.6: Recycle with Chemical Rxn.

Cyclohexane (C6H12) can be made by the reaction of benzene (Bz) (C6H6) with

hydrogen according to the following reaction:

C6H6 + 3H2 C6H12

For the process shown in Figure bellow, determine the ratio of the recycle stream

to the fresh feed stream if the overall conversion of benzene is 95%, and the

single-pass conversion is 20%.

Assume that 20% excess hydrogen is used in the fresh feed, and that the

composition of the recycle stream is 22.74 mol% benzene and 78.26 mol%

hydrogen.

Solution:

32


Reactor Seperator1

3n1,Bz=100 mole

n1,H2

H2 is 20% excess

4

5

2

n5

y5,Bz=0.2274

y5,H2=0.7726

n4

n4,Bz

n4,H2

n4,C6H12

fSP =0.20

fOA=0.95

C6H6 + 3H2 C6H12

33


Example 6.7: Recycle with Chemical Rxn.

Immobilized glucose is used as a catalyst in producing fructose from glucose in a

fixed-bed reactor (water is the solvent). For the system shown, what percent

conversion of glucose results on one pass through the reactor when the ratio of the

exit stream to the recycle stream in mass units is equal to 8.33? The reaction is

C12H22O11 C12H22O11

Glucose Fructose

Solution:

34


Reactor1

3m1=100 kg

x1,G=0.4

x1,W=0.65

2

m5

x5,G

x5,F

x5,W

m4

x5,G

x5,F

x5,W

4

m2

x2,G

x2,F=0.04

x2,W

C12H22O11 C12H22O11

Glucose Fructose

# m4/m5 =8.33

# Recycle stream m5 has

the same composition as

the exist stream, m4

A B

35


Example 6.8: Bypass Calculation

In the feedstock preparation section of a plant manufacturing natural

gasoline, isopentane is removed from butane-free gasoline. What fraction of

the butane-free gasoline is passed through the isopentane tower?

The process is in the steady state and no reaction occurs.

Iso

pe

nta

ne

To

we

r1

3

m1=100 kg

x1,n-pantane= 0.8

x1,I-pentane= 0.2

4

6

2

5

m4,I-pentane=

m5,n-pentane=

m6=100 kg

x6,n-pantane= 0.9

x6,I-pentane= 0.1

m2=

x2,n-pantane= 0.8

x2,I-pentane= 0.2

m3=

x3,n-pantane= 0.8

x3,I-pentane= 0.2

36


37


Example 6.9: Purge Calculation

Considerable interest exists in the conversion of coal into more convenient

liquid products for subsequent production of chemicals. Two of the main

gases that can be generated under suitable conditions from in situ (in the

ground) coal combustion in the presence of steam (as occurs naturally in

the presence of groundwater) are H2 and CO. After cleanup, these two

gases can be combined to yield methanol according to the following

equation:

CO + 2H2 CH3OHNote that some CH4 enters the process, but does not participate in the reaction.

A purge stream is used to maintain the CH4 concentration in the exit from the

separator at no more than 3.2 mol%, and prevent hydrogen buildup as well. The

once-through conversion of the CO in the reactor is 18%.

Compute the moles of recycle, CH3OH, and purge per mole of feed, and also

compute the purge gas composition.

The flow chart illustrates a steady-state process for the production of methanol. All of the

compositions are in mole fractions. The stream flows are in moles.

38


Reactor

Recycle

S

Separator

Purge6

1

3

4

5

2 7

n1

y1,H2=0.673

y1,CO= 0.325

y1,CH4=0.002 n7,CH3OH

n5

y5,H2=

y5,CO=

y5,CH4=

n6

y6,H2=

y6,CO=

y6,CH4=0.032

fSP =0.18CO + 2H2 CH3OH

Documents

Dr. Belal Al Zaitone Zuriqat - kausfalhuthali.kau.edu.sa/GetFile.aspx?id=238735&Lng... · Chapter 6 MATERIAL BALANCES FOR MULTI-UNIT SYS. 2 Chapter 6: Recycle, Bypass and Purge 1