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King Abdulaziz University
Faculty of Engineering
Department of Chemical and
Materials Engineering
Dr. Belal Al Zaitone Zuriqat
eMail: balzaitone@kau.edu.sa 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
Chapter 6: Material Balance Involving Multiple Units
thyssenkrupp-uhde.deProcess Flow Sheet
5
Chapter 6: Material Balance Involving Multiple Units
Process Block Diagram
6
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
3 5Reactor
4
7
1
2
6
Mixing point Splitting pointm1+m2=m3
m3+m4=m5
8
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
3 5Reactor
4
7
1
2
6
m1+m2+m4=m5
10
Chapter 6: Material Balance Involving Multiple Units
3 5Reactor
4
7
1
2
6
m3+m4=m6+m6
11
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
14
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
17
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Material Balance Involving Multiple Units
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
Chapter 6: Recycle, Bypass and Purge
• 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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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%
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
Recycle with Chemical Reaction : A B
𝑓𝑂𝐴 =𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 − 𝑛𝐿𝑅,𝑜𝑢𝑡
𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝒑𝒓𝒐𝒄𝒆𝒔𝒔
× 100%
Over-All fractional Conversion
𝑓𝑂𝐴 =−𝜉 ∙ 𝜐𝐿𝑅
𝑛𝐿𝑅,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠
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
Chapter 6: Recycle, Bypass and Purge
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,𝑓𝑒𝑒𝑑 𝑖𝑛𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 + 𝑛𝐿𝑅,𝑅𝑒𝑐𝑦𝑐𝑙𝑒 𝑠𝑡𝑟𝑒𝑎𝑚
A B
31
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
37
Chapter 6: Recycle, Bypass and Purge
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
Chapter 6: Recycle, Bypass and Purge
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
Recommended