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Page 1
Before we look at more complicated problems, we need to have a
systematic procedure for analyzing interconnected units. We need
particularly to have some way to determine which units in a plant can
be (and should be) analyzed first. Then we want to establish an order
for further analysis. The modules and units considered in the last
sections could all be analyzed completely. In many systems, we will
find that only a few units can be treated in that way. A system
(module, unit or plant) that can be analyzed completely to determine
all pertinent flow rates and reaction rates is said to be completely
determined. Such a system is also said to have a Degree of Freedom of
0.
The degree of freedom for a unit is a measure of how many
specifications we can (or need to) make so that all unknown flows and
reaction rates can be found. The larger the degree of freedom for a
unit, the less we know about it. Since flow and reaction rates are
the unknowns while balances, given compositions, flows, flow ratios
and conversions and splitter restrictions all reduce the uncertainty,
the differences are what we call the degree of freedom.
Just as each unit in a plant has unknowns (flow and reaction rates)
that we may try to solve for, the overall plant has a similar set of
unknowns. In fact all we need to do is look strictly at those flow
streams that enter and leave the plant and forget the streams that
lead from one unit to another. We may have to lump reaction rates if
the same reaction occurs in more than one unit because the overall
system does not care where a reaction takes place. We also can find
much information that is useless in our analysis of the overall
system. Only if the information specifies something about the plant
feed and product streams can it be used to reduce the degree of
freedom for the overall system.
An entirely different problem is connected with analyzing the
"process". In this case we look at all the unknowns in the plant.
This includes the flows of all compounds in all streams and the rates
of all reactions in all units. This time the same reaction is counted
as many times as it occurs. In addition all information is
usable.
The degree of freedom for the overall system and for the process are
difficult to find unless you are careful about the meaning of each
term in them. The following table shows a precise way to find the
degree of freedom for each unit, the overall system and the
process.
Unit 1 Unit 2 ..... Overall Process
Flow Variables FV1 FV2 FVO FVP
Reaction Variables R1 R2 RO RP
Balances B1 B2 BO BP
Given Compositions CP1 CP2 CPO CPP
Given Flows F1 F2 FO FP
Given Flow Ratios FR1 FR2 FRO FRP
Given Conversions CV1 CV2 CVO CVP
Splitter Restrictions S1 S2 SO SP
----- ----- ----- -----
Net Degree of Freedom °F1 °F2 °FO °FP
Specific definitions of the entries in the table are given next. Some of these are obvious, others can be quite confusing unless you are careful to think precisely about what they represent.
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FVk= |
the sum of the number of compounds that are involved in the balance equations in all the streams entering or leaving unit k. |
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FVO= |
the sum of the number of compounds that are involved in the balance equations in all the streams entering or leaving the overall system. |
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FVP= |
the sum of the number of compounds that are involved in the balance equations in all the streams anywhere in the system including those entering or leaving the overall system as well as those connecting the units. |
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Rk= |
the number of independent reactions occurring inside unit k. |
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RO= |
the number of independent reactions occurring inside the overall system. Do not count each reaction more than once even if it occurs in more than one unit. |
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RP= |
sum (Rk) |
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Bk= |
the number of compounds that are found in the flow streams in or out of unit k or take part in the reactions in the unit. |
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Bk= |
the number of compounds that are found in the flow streams in or out of the overall system or take part in the reactions inside the system. |
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BP= |
sum(Bk) |
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CPk= |
the number of compositions given for any flow stream into or out of unit k. The maximum number that is counted in any one stream is one less than the number of compounds in that stream. |
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CPO= |
the number of compositions given for any flow stream into or out of the overall system. |
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CPP= |
the number of compositions given for any flow stream anywhere in the system. |
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Fk= |
the number of flow rates given for the streams in or out of unit k. |
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FO= |
the number of flow rates given for the streams in or out of the overall system. |
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FP= |
the number of flow rates given for all the streams anywhere in the system. |
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FRk= |
the number of flow ratios of streams into or out of unit k. Note that both streams in the ratio must be involved with unit k to be counted. |
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FRO= |
the number of flow ratios of streams into or out of the overall system. Note that both streams in the ratio must be ones that enter or leave the overall system to be counted. |
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FRP= |
the number of flow ratios given for streams anywhere in the system. |
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CVk= |
the number of conversions specified for reactants fed to unit k. Do not count a conversion of 100% if the reactant that is the basis of the conversion does not appear in any exit stream from the unit. |
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CVO= |
the number of conversions specified for reactants fed to the overall system. Do not count a conversion of 100% if the reactant that is the basis of the conversion does not appear in any exit stream from the system. |
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CVP= |
sum(CVk) |
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Sk= |
0 unless the unit is a stream splitter. If it is a splitter, then it is (number streams leaving - 1)(number of compounds in the stream -1). |
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SP= |
sum(Sk) |
°Fk = FVk +
Rk - (Bk + CPk + Fk +
FRk + CVk + Sk)
°FO = FVO + RO - (BO
+ CPO + FO + FRO + CVO +
SO)
°FP = FP + RP - (BP
+ CPP + FP + FRP + CVP +
SP)
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Unit Name: |
Unit 1 |
Unit 2 |
Unit 3 |
Unit 4 |
Unit 5 |
Unit 6 |
Over-all |
Process |
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Three procedures will be demonstrated in this section of the notes
for simulating the steady state behavior of an entire plant. Each
procedure uses the module functions shown in the previous section of
the notes. The three methods are:
Each method has its advantages and disadvantages. The modules may
be readily combined to simulate almost any steady state chemical
system. As soon as we encounter systems with a large number of units,
we find that it is tedious to keep track of the order of the units
and the information that needs to be given as we proceed through the
system. It also becomes tedious if we find that a trial and error
solution must be used to find some of the flow or reaction variables.
Thus we need to adopt one of the last two procedures to be efficient.
If we have only a few systems to analyze, a specific program written
to simulate each system may be the best approach. This becomes less
practical as the number of systems to be analyzed grows. Then we will
find general purpose drivers will be very useful.
Two general purpose programs will be demonstrated in this
chapter:
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Function |
Used For |
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1 |
drive |
Driver function for combining modules. (in ~ceng301/matlab/massmods) |
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2 |
vweg2 |
Vector root finding module. (in ~ceng301/matlab/misc) |
Multi-unit systems can be simulated by combining several of the
modular programs. To illustrate this, let's do Example Problem
3.13 in the text. Let the streams be numbered as in Figure
3.7 in the text. The chlorination reactor has two inlet streams
and two outlet streams, but react can handle only one inlet and one
outlet stream. Therefore, streams 1 and 2 must be mixed; then this
mixture can be reacted. Afterwards the outlet from react can be
separated using sep.
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The new system will have 6 streams. The seven components are:
Component Name Formula
1 Benzene C6H6
2 Chlorobenzene C6H5Cl
3 Dichlorobenzene C6H4Cl2
4 Trichlorobenzene C6H3Cl3
5 Tetrachlorobenzene C6H2Cl4
6 Chlorine Cl2
7 Hydrogen Chloride HCl
Using start301 to set the compounds and reactions,
select New Session, followed by Mass Balances Only. We
then wish to use the CENG 301 database since all the compounds
we are going to use can be found in the database.
>> start301 Copyright 1999 Rice University All rights reserved Welcome to CENG301's start301!! You have three choices, Please read them carefully Then click on the appropriate choice in the menu bar 1: Click (1) to start a new session Input the name of your new file: chloros The output file name is: chloros Input the number of compounds: 7 The number of compounds is: 7 Enter the name of compound # 1: C6H6 Enter the name of compound # 2: C6H5Cl Enter the name of compound # 3: C6H4Cl2 Enter the name of compound # 4: C6H3Cl3 Enter the name of compound # 5: C6H2Cl4 Enter the name of compound # 6: Cl2 Enter the name of compound # 7: HCl Enter the number of reactions: 4 Enter the coefficients for each compound in the same order that the compounds are listed. Coefficients for reactants should be Negative, and coefficients for products should be positive Enter the coefficients for each compound in reaction # 1 C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl -1 1 0 0 0 -1 1 Enter the coefficients for each compound in the same order that the compounds are listed. Coefficients for reactants should be Negative, and coefficients for products should be positive Enter the coefficients for each compound in reaction # 2 C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl 0 -1 1 0 0 -1 1 Enter the coefficients for each compound in the same order that the compounds are listed. Coefficients for reactants should be Negative, and coefficients for products should be positive Enter the coefficients for each compound in reaction # 3 C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl 0 0 -1 1 0 -1 1 Enter the coefficients for each compound in the same order that the compounds are listed. Coefficients for reactants should be Negative, and coefficients for products should be positive Enter the coefficients for each compound in reaction # 4 C6H6 C6H5Cl C6H4Cl2 C6H3Cl3 C6H2Cl4 Cl2 HCl 0 0 0 -1 1 -1 1 Do you want to check to see if the coefficients are correct? type "y" for yes or simply press enter to move on: Your reactions are currently as follows: 1) C6H6 + Cl2 --> C6H5Cl + HCl 2) C6H5Cl + Cl2 --> C6H4Cl2 + HCl 3) C6H4Cl2 + Cl2 --> C6H3Cl3 + HCl 4) C6H3Cl3 + Cl2 --> C6H2Cl4 + HCl Type "y" to change an equation, or press "enter" to continue: Here are your compounds' formulae and names: No. Formula Name ---------------------------------------- 1 C6H6 benzene 2 C6H5Cl chlorobenzene 3 C6H4Cl2 m-dichlorobenzene 4 C6H3Cl3 trichlorobenzene 5 C6H2Cl4 tetrachlorobenzene 6 Cl2 chlorine 7 HCl hydrogen chloride Here are your reactions: ---------------------------------------- 1) C6H6 + Cl2 --> C6H5Cl + HCl 2) C6H5Cl + Cl2 --> C6H4Cl2 + HCl 3) C6H4Cl2 + Cl2 --> C6H3Cl3 + HCl 4) C6H3Cl3 + Cl2 --> C6H2Cl4 + HCl Enter the number of streams: 6 The variables for your compounds have now been created, you may continue, or come back later and reload the same data.
The flow of Benzene in stream 1 is 1000 and of Chlorine in stream
2 is 3600. These may be set by:
>> ns(1,1)=1000; >> ns(2,6)=3600;
Mix streams 1 and 2 to get stream 5.
>> mix([1 2],5)
and we see:
>> showm([1 2],5,10,1)
Compound Inlet | Outlet
Stream 1 2 Total | 5
benzene 1000.0 0.0 1000.0 | 1000.0
chlorine 0.0 3600.0 3600.0 | 3600.0
Total 1000.0 3600.0 4600.0 | 4600.0
Then, stream 5 will be the inlet to the reactor. From material
balances the rates of the reactions can be found (as in the text) to
be:
990 920 800 50
React stream 5 to get stream 6 and display the result:
>> react([990 920 800 50],5,6)
>> showm(5,6,10,1)
Compound Inlet | Outlet
Stream 5 | 6
benzene 1000.0 | 10.0
chlorobenzene 0.0 | 70.0
m-dichlorobenzen 0.0 | 120.0
C6H3Cl3 0.0 | 750.0
C6H2Cl4 0.0 | 50.0
chlorine 3600.0 | 840.0
hydrogen chlorid 0.0 | 2760.0
Total 4600.0 | 4600.0
Now, stream 6 must be separated into streams 3 and 4. All of the
chlorine and HCl goes into stream 3 and the rest goes to stream
4:
>> sep([0 0 0 0 0 1 1],6,[3 4])
>> showm(6,3:4,10,1)
Compound Inlet | Outlet
Stream 6 | 3 4 Total
benzene 10.0 | 0.0 10.0 10.0
chlorobenzene 70.0 | 0.0 70.0 70.0
m-dichlorobenzen 120.0 | 0.0 120.0 120.0
C6H3Cl3 750.0 | 0.0 750.0 750.0
C6H2Cl4 50.0 | 0.0 50.0 50.0
chlorine 840.0 | 840.0 0.0 840.0
hydrogen chloride 2760.0 | 2760.0 0.0 2760.0
Total 4600.0 | 3600.0 1000.0 4600.0
Let's check to see that stream 4 has the compositions given in the
problem.
>> 100*ns(4,:)/sum(ns(4,:))
ans =
1 7 12 75 5 0 0
It does. The flowrate of primary product,
C6H3Cl3 , is seen in stream 4 to be
750.