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The first element balance program: **el1**, is an
interactive version in which all data is input in reply to prompts.
It requires that all flows be given in molar units. Typing
**el1** will set the program into motion as shown in
analyzing *Problem 4.2* in the text with a basis chosen of 100
mols of feed gas:

wsname% el1 ____________________________________________________ el1 asks for the number of mols of each compound IN and OUT of a System and then its Formula. A Table then gives the molar flows for all compounds and a second Table gives the element Flows. ____________________________________________________ Give the number of mols IN and OUT then the FORMULA for COMPOUND 1 or stop to STOP..8 0 CH4<-- Flow in, Flow out, FormulaGive the number of mols IN and OUT then the FORMULA for COMPOUND 2 or stop to STOP.7.6 0 CO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 3 or stop to STOP.35.2 0 COGive the number of mols IN and OUT then the FORMULA for COMPOUND 4 or stop to STOP.51.9 0 H2Give the number of mols IN and OUT then the FORMULA for COMPOUND 5 or stop to STOP.4 4 N2Give the number of mols IN and OUT then the FORMULA for COMPOUND 6 or stop to STOP.0.5 0 O2Give the number of mols IN and OUT then the FORMULA for COMPOUND 7 or stop to STOP.stopCompound Mols in Mols out CH4 0.8000 0.0000 CO2 7.6000 0.0000 CO 35.2000 0.0000 H2 51.9000 0.0000 N2 4.0000 4.0000 O2 0.5000 0.0000 Total 100.0000 4.0000 Element Atoms IN Atoms OUT Atoms Lost C 43.6000 0.0000 43.6000 H 107.0000 0.0000 107.0000 O 51.4000 0.0000 51.4000 N 8.0000 8.0000 0.0000

We did not include water in this list given to the program and put in
zeroes for all the terms we did not know. Now let's copy the flows
shown in the display to an editor where we can add symbols for all
unknowns:

COMPOUND MOLS IN MOLS OUT CH4 0.8000 CH4 CO2 7.6000 .438D CO 35.2000 .002D D=CH4+.44D+H2+4 H2 51.9000 H2 N2 4.0000 4.0000 O2 0.5000 0.0000 H2O 0.0000 W Total 100.0000 4 + CH4 + .44D + H2 + W Element Atoms IN Atoms OUT Atoms Lost C 43.6000 CH4+.44D 43.6 - CH4 - .44D H 107.0000 4CH4+2H2+2W 107 - 4CH4-2H2-2W O 51.4000 .878D+W 51.4 - .878D - W N 8.0000 8.0000 0.000

The Atoms lost should be zero for all elements. Only nitrogen
satifies this condition in our first execution of
**el1**. We have three simultaneous equations to
satisify to make the other balances zero. The relation shown for the
total amount of dry gas give a fourth relation among the four
unknowns. Here are the four equations:

CH_{4}+ .440D = 43.6<-- C balance4CH_{4}+ 2H_{2}+ 2W = 107<-- H balance.878D + W = 51.4<-- O balance-CH_{4}+ .560D - H_{2}= 4.0<-- Dry gas relation

Here is a MATLAB session to solve these:

>>a=[ 1 0.44 0 0 4 0 2 2 0 0.878 0 1 -1 0.56 -1 0]a = 1.0000 0.4400 0 0 4.0000 0 2.0000 2.0000 0 0.8780 0 1.0000 -1.0000 0.5600 -1.0000 0 >>b=[43.6 107 51.4 4];>>(a\b')'ans = 21.8322 49.4723 1.8723 7.9633

or: CH_{4}=21.8322, D=49.4723, H_{2}=1.8723, and
W=7.9633. We can test these values in the **el1**
program to find:

wsname%el1____________________________________________________ el1 asks for the number of mols of each compound IN and OUT of a System and then its Formula. A Table then gives the molar flows for all compounds and a second Table gives the element Flows. ____________________________________________________ Give the number of mols IN and OUT then the FORMULA for COMPOUND 1 or stop to STOP.0.8 21.8322 CH4Give the number of mols IN and OUT then the FORMULA for COMPOUND 2 or stop to STOP.7.6 21.6689 CO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 3 or stop to STOP.35.2 .0989 COGive the number of mols IN and OUT then the FORMULA for COMPOUND 4 or stop to STOP.51.9 1.8723 H2Give the number of mols IN and OUT then the FORMULA for COMPOUND 5 or stop to STOP.0 7.9633 H2OGive the number of mols IN and OUT then the FORMULA for COMPOUND 6 or stop to STOP.4 4 N2Give the number of mols IN and OUT then the FORMULA for COMPOUND 7 or stop to STOP.0.5 0 O2Give the number of mols IN and OUT then the FORMULA for COMPOUND 8 or stop to STOP.stopCompound Mols in Mols out CH4 0.8000 21.8322 CO2 7.6000 21.6689 CO 35.2000 0.0989 H2 51.9000 1.8723 H2O 0.0000 7.9633 N2 4.0000 4.0000 O2 0.5000 0.0000 Total 100.0000 57.4356 Element Atoms IN Atoms OUT Atoms Lost C 43.6000 43.6000 0.0000 H 107.0000 107.0000 0.0000 O 51.4000 51.4000 0.0000 N 8.0000 8.0000 0.0000

There are two difficulties with the first element balance program.
One difficulty is the requirement that you type in all data each time
the program is executed. This may be fine for small problems, but it
can be very cumbersome in analyzing programs with many feeds and
products or that require many executions to find a satisfactory
solution. A second difficulty is that all flows must be entered in
molar units. We will eliminate the second of these in the program
**el2**, and then at least alleviate the first problem
in **el3** by allowng the user to store all data in a
file where it can be edited.

The program **el2** helps the user in problems that
involve flows given in both mass and molar units. We will demonstrate
the program by using it on *Problem 4.15* in the text with a
basis of 100 lb of the fuel oil:

wsname%el2_____________________________________________________ el2 asks for the number of mols of each compound IN and OUT of a System and then its Formula. When you finish giving molar flows, you can then give Mass flows; again IN, then OUT and finally Formula. A Table then gives the molar and mass flows for all compounds as well as their molecular weights. Finally a Table of element Flows is given. _____________________________________________________ Give the number of mols IN and OUT then the FORMULA for COMPOUND 1 or stop to give Mass flows.0 0 N2Give the number of mols IN and OUT then the FORMULA for COMPOUND 2 or stop to give Mass flows.0 0 O2Give the number of mols IN and OUT then the FORMULA for COMPOUND 3 or stop to give Mass flows.0 0 SO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 4 or stop to give Mass flows.0 0 CO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 5 or stop to give Mass flows.0 0 H2OGive the number of mols IN and OUT then the FORMULA for COMPOUND 6 or stop to give Mass flows.stopGive the mass IN and OUT then the FORMULA for COMPOUND 6 or stop to STOP.84 0 CGive the mass IN and OUT then the FORMULA for COMPOUND 7 or stop to STOP.11.4 0 HGive the mass IN and OUT then the FORMULA for COMPOUND 8 or stop to STOP.1.4 0 NGive the mass IN and OUT then the FORMULA for COMPOUND 9 or stop to STOP.3.2 0 SGive the mass IN and OUT then the FORMULA for COMPOUND 10 or stop to STOP.stopCompound Mol Wt Mols in Mols out Mass in Mass out N2 28.0134 0.0000 0.0000 0.0000 0.0000 O2 31.9988 0.0000 0.0000 0.0000 0.0000 SO2 64.0588 0.0000 0.0000 0.0000 0.0000 CO2 44.0098 0.0000 0.0000 0.0000 0.0000 H2O 18.0152 0.0000 0.0000 0.0000 0.0000 C 12.0110 6.9936 0.0000 84.0000 0.0000 H 1.0079 11.3106 0.0000 11.4000 0.0000 N 14.0067 0.1000 0.0000 1.4000 0.0000 S 32.0600 0.0998 0.0000 3.2000 0.0000 TOTAL 18.5040 0.0000 100.0000 0.0000 Element Atoms IN Atoms OUT Atoms Lost C 6.9936 0.0000 6.9936 H 11.3106 0.0000 11.3106 N 0.1000 0.0000 0.1000 S 0.0998 0.0000 0.0998 O 0.0000 0.0000 0.0000

Again 0's were put in where we had unknowns so that we can edit the
result to show our element balances. Before doing so, we can see that
6.9936 + 0.25 11.3106 + 0.0998 mols of O_{2} will be needed
in the theoretical air. Thus 47.2431 mols of air would be required
for the answer to part a). For part b) our edited table looks
like:

Compound Mol Wt Mols in Mols out N2 28.0134 0.79A N2 O2 31.9988 0.21A O2 SO2 64.0588 0.0000 SO2 CO2 44.0098 0.0000 CO2 H2O 18.0152 0.0000 W C 12.0110 6.9936 0.0000 H 1.0079 11.3106 0.0000 N 14.0067 0.1000 0.0000 S 32.0600 0.0998 0.0000 TOTAL 18.5040 0.0000 Element Atoms IN Atoms OUT Atoms Lost C 6.9936 CO2 6.9936 - CO2 H 11.3106 2W 11.3106 - 2W N 0.1 + 1.58A 2N2 0.1 + 1.58A - 2N2 S 0.0998 SO2 0.0998 - SO2 O .42A 2O2+2SO2+W+2CO2 .42A - 2O2+2SO2+W+2CO2

From the SO_{2} composition in the dry gas:

64.06*SO_{2}= .004(64.06S*O_{2}+ 28.01*N_{2}+ 32.00*O_{2}+ 44.01*CO_{2})

Thus, using three element balances:

SO_{2}=.0998, W=5.6553, and CO_{2}=6.9936.

from the remaining two element balances and the SO_{2}
composition relation, we have the three equations in three
unknowns:

28.01*N_{2}+ 32.0*O_{2}= 1284.1<-- SO2.00*N_{2}composition_{2}- 1.58*A = 0.1<-- N balance- 2.0*O_{2}+ 0.42*A = 19.84<-- O balance

Solving (with MATLAB):

>> a=[28.01 32 0 2 0 -1.58 0 -2.002 .42]a = 28.0100 32.0000 0 2.0000 0 -1.5800 0 -2.0020 0.4200 >>(a\[1284.1 .1 19.84]')'ans = 43.8713 1.7270 55.4701

gave: N_{2}=43.8713, O_{2}=1.7270 and A=55.4701. With
these in our flow table we see:

wsname%el2_____________________________________________________ el2 asks for the number of mols of each compound IN and OUT of a System and then its Formula. When you finish giving molar flows, you can then give Mass flows; again IN, then OUT and finally Formula. A Table then gives the molar and mass flows for all compounds as well as their molecular weights. Finally a Table of element Flows is given. _____________________________________________________ Give the number of mols IN and OUT then the FORMULA for COMPOUND 1 or stop to give Mass flows.43.8214 43.8713 N2Give the number of mols IN and OUT then the FORMULA for COMPOUND 2 or stop to give Mass flows.11.6487 1.727 O2Give the number of mols IN and OUT then the FORMULA for COMPOUND 3 or stop to give Mass flows.0 .0998 SO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 4 or stop to give Mass flows.0 6.9936 CO2Give the number of mols IN and OUT then the FORMULA for COMPOUND 5 or stop to give Mass flows.0 5.6553 H2OGive the number of mols IN and OUT then the FORMULA for COMPOUND 6 or stop to give Mass flows.stopGive the mass IN and OUT then the FORMULA for COMPOUND 6 or stop to STOP.84 0 CGive the mass IN and OUT then the FORMULA for COMPOUND 7 or stop to STOP.11.4 0 HGive the mass IN and OUT then the FORMULA for COMPOUND 8 or stop to STOP.1.4 0 NGive the mass IN and OUT then the FORMULA for COMPOUND 9 or stop to STOP.3.2 0 SGive the mass IN and OUT then the FORMULA for COMPOUND 10 or stop to STOP.stopCompound Mol Wt Mols in Mols out Mass in Mass out N2 28.0134 43.8214 43.8713 1227.5864 1228.9842 O2 31.9988 11.6487 1.7270 372.7444 55.2619 SO2 64.0588 0.0000 0.0998 0.0000 6.3931 CO2 44.0098 0.0000 6.9936 0.0000 307.7869 H2O 18.0152 0.0000 5.6553 0.0000 101.8814 C 12.0110 6.9936 0.0000 84.0000 0.0000 H 1.0079 11.3106 0.0000 11.4000 0.0000 N 14.0067 0.1000 0.0000 1.4000 0.0000 S 32.0600 0.0998 0.0000 3.2000 0.0000 TOTAL 73.9741 58.3470 1700.3307 1700.3076 Element Atoms IN Atoms OUT Atoms Lost N 87.7428 87.7426 0.0002 O 23.2974 23.2961 0.0013 S 0.0998 0.0998 0.0000 C 6.9936 6.9936 0.0000 H 11.3106 11.3106 0.0000

The % excess air was then: 100(55.4701-47.2431)/47.2431= 17.41.

The program **el3** allows the same operations as in
**el2**, but with input data stored in a file. Here is a
session to illustrate its use. First we establish a new file for the
input data:

wsname%el3_____________________________________________________ el3 performs just like el2, but it uses a file to store all data. If you need to set up a new file, the program will give you a template to use. The data stored in the file includes: the number of mols of each compound IN and OUT of a System and its Formula. When you finish with molar flows, you can then give Mass flows; again IN, then OUT and finally Formula. A Table then gives the molar and mass flows for all compounds as well as their molecular weights. Finally a Table of element Flows is given. _____________________________________________________ Give the name of your file or NEW to set up a new file.newGive the name to be used for your file:tst4.6Give the number of compounds that you want to give molar flows for:3Give the number of compounds that you want to give mass flows for:2Now you should edit your file: tst4.6 to put into it the compound Formula and flow rates. When you have completed that and SAVED the file, hit a RETURN.

Now the file *tst4.6* has in it:

The number of compounds with molar flows is = 3 For Compound Number: 1 The Formula is = The Molar Flow Rate IN is = The Molar Flow Rate OUT is = ******************************* For Compound Number: 2 The Formula is = The Molar Flow Rate IN is = The Molar Flow Rate OUT is = ******************************* For Compound Number: 3 The Formula is = The Molar Flow Rate IN is = The Molar Flow Rate OUT is = ******************************* The number of compounds with mass flows is = 2 For Compound Number: 4 The Formula is = The Mass Flow Rate IN is = The Mass Flow Rate OUT is = ******************************* For Compound Number: 5 The Formula is = The Mass Flow Rate IN is = The Mass Flow Rate OUT is = *******************************

We edit the numbers in *tst4.6* to make the file look
like:

The number of compounds with molar flows is = 3 For Compound Number: 1 The Formula is = Na2CO3 The Molar Flow Rate IN is = 1 The Molar Flow Rate OUT is = 0.25 ******************************* For Compound Number: 2 The Formula is = H2O The Molar Flow Rate IN is = 10 The Molar Flow Rate OUT is = 8 ******************************* For Compound Number: 3 The Formula is = CO2 The Molar Flow Rate IN is = 5 The Molar Flow Rate OUT is = 6 ******************************* The number of compounds with mass flows is = 2 For Compound Number: 4 The Formula is = C The Mass Flow Rate IN is = 15 The Mass Flow Rate OUT is = 18 ******************************* For Compound Number: 5 The Formula is = NaOH The Mass Flow Rate IN is = 45 The Mass Flow Rate OUT is = 38 *******************************

Then we continue the execution of **el3** by
*saving* the data file and *then* hitting the RETURN
key in response to:

When you have completed that and SAVED the file, hit a RETURN.

For our simulation we find:

Compound Mol Wt Mols in Mols out Mass in Mass out Na2CO3 105.9887 1.0000 0.2500 105.9887 26.4972 H2O 18.0152 10.0000 8.0000 180.1520 144.1216 CO2 44.0098 5.0000 6.0000 220.0490 264.0588 C 12.0110 1.2489 1.4986 15.0000 18.0000 NaOH 39.9971 1.1251 0.9501 45.0000 38.0000 TOTAL 18.3739 16.6987 566.1898 490.6776 Element Atoms IN Atoms OUT Atoms Lost Na 3.1251 1.4501 1.6750 C 7.2489 7.7486 -0.4998 O 24.1251 21.7001 2.4250 H 21.1251 16.9501 4.1750

In order to eliminate the imbalances in the elements, we need to
revise the input file, but that simply requires editing the file to
change the parts that need revision. This is much simpler and faster
than answering all the questions in the interactive programs
**el1** and **el2**.

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Last modified August 22, 1997.