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## 1.6.1 Summary of Degree of Freedom Table

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.

#### Flow Variables

 FVk= the sum of the number of compounds that are involved in the balance equations in all the streams entering or leaving unit k. 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. 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.

#### Reaction Variables

 Rk= the number of independent reactions occurring inside unit k. 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. RP= sum (Rk)

#### Balances

 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. 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. BP= sum(Bk)

#### Specified Compositions

 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. CPO= the number of compositions given for any flow stream into or out of the overall system. CPP= the number of compositions given for any flow stream anywhere in the system.

#### Specified Flow Rates

 Fk= the number of flow rates given for the streams in or out of unit k. FO= the number of flow rates given for the streams in or out of the overall system. FP= the number of flow rates given for all the streams anywhere in the system.

#### Specified Flow Ratios

 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. 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. FRP= the number of flow ratios given for streams anywhere in the system.

#### Specified Conversions

 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. 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. CVP= sum(CVk)

#### Splitter Restrictions

 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). SP= sum(Sk)

#### Degree of Freedom

°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)

### Degree of Freedom Table

Unit Name:

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Over-all

Process

Flow
Variables

Reaction
Variables

Balances

Given
Compositions

Given Flows

Given Flow
Ratios

Given
Conversions

Splitter
Restrictions

Net °F

This is a typical degree of freedom table for a system with no more than 6 units.

## 1.6.2 Using the Modules to Simulate a Plant

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:

1. Interactive Use of the Modules
2. Using a General Purpose Driver
3. Writing a Program for a Specific Plant

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:

Function

Used For

1

drive

Driver function for combining modules. (in ~ceng301/matlab/massmods)

2

vweg2

Vector root finding module. (in ~ceng301/matlab/misc)

### Interactive Use of the Modules

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.

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

Welcome to CENG301's start301!!
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

----------------------------------------
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.

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