Greetings!

By now, you are all relatively familiar with mass balance, and with Excel.   So you should be well prepared to do this week’s lab.

 

Your assignment is Problem #23 in the Mass Balance chapter.  It is found on Page 128 of your text.  You will need to use Excel to solve four equations simultaneously.  We covered this topic in class on Thursday; the method for doing this is found in your textbook in Appendix C Excel Applications, under the topic entitled “Simultaneous Linear Equations by Matrix Inversion”.  This topic begins on page 216.  When you are done, please save your spreadsheet under the name created by merging your initials plus the letters SLE (for simultaneous linear equations).  So for example if your name is Dan Schuyler, your spreadsheet would be named DSSLE.

 

Below, I discuss two topics that may be of use to you in doing this spreadsheet:  first, a brief and highly non-technical description of what’s going on with those matricies; and second, a guide to the setup of the problem, in terms of what you put in and what will come out.  The mechanics of how to do this is in your textbook, so I will not be discussing it here.

 

FIRST, matricies.

In our previous mass balance problems, when we set up our work, we always had some “trick” that helped us with the number crunch:  such as, one or two variables could be solved for directly, then those numbers could be plugged into another equation to solve for everything else.  Your authors did this for a reason: mathematically, the computation for such problems is much easier.  So you learned the concepts of mass balance without the rigor of a lot of math.

 

Well now, in real life you can’t count on a trick to solve problems like these.  Very frequently if you have, say, four unknowns and you can generate four equations, you need to have some good techniques for solving those equations SIMULTANEOUSLY  rather than one at a time.  When confronted with such a situation, most  students will deny that they really need to do this and then spend a huge amount of time playing around looking for a trick.  This, in general, will not be effective, and we have some perfectly good math that will allow you to solve these systems without any tricks at all.  This body of math is known as Linear Algebra.

 

Linear algebra is a lot like, but not exactly like, single variable algebra.  It has its own rules, but there are rules and just like in other bodies of math, if you apply those rules the problems will work out.

 

In terms of solving 4 equations for 4 unknowns, as in Problem 22, it works like this:

 

You know that in regular algebra if you have an equation

 

                                                A x X = B

If you know what A is and what B is you can find out what X is by multiplying B by the multiplicative inverse (i.e., reciprocal) of A.  What I mean is:

 

                                                X = B x (1/A)

The reason this works is because A x (1/A) = 1. In other words, 1/A is the inverse of A because A x (1/A)  is equal to one.  This is referred to multiplicative identity. This causes X to be the product of  B and the inverse of A

Similarly in a system of equations, such as:

 

.265A + .215B + .501C + 0D = .25T

.517A + .220B + .201C + 0D = .25T

.205A + .512B + .188C + 0D = .25T

.013A + .053B + .110C + 1D = .25T

 

 

 

 

 

we can think of this as being the product of two matricies:

|.265   .    215          .501         0|              |A|                          |.25|

|.517        .220         .201         0|             |B|                          |.25|

|.205        .512         .188         0|X          |C|           =             |.25|

|.013        .053         .110         1|             |D|                          |.25|

[            matrixA                 ] X               [X]           =           [B]

Since matricies A and B are composed of numbers, matrix X, which is composed of variables, is what we want to find.

So if we could find a matrix such that this matrix multiplied by A would give us an identity (the matrix equivalent to the number 1) then we could multiply it by the matrix B and we would get the matrix X.  Wow!

 

Well, all this stuff is pretty hard to do.  So excel does it for you!

 

Excel uses two functions MINVERSE (stands for matrix inverse) and MMULT (stands for matrix multiplication).  First you find the MINVERSE of matrix A; then you multiply this inverse matrix by the matrix B.  The product of these two is the matrix X, which is a nice column corresponding to all the variables you solved for.

 

So, this describes a little bit about the process.  Below is a description of the problem setup for your lab assignment this week.

 

For this problem, if you think about it, if you didn’t know some limiting fact such as you have 22 kg of pure zinc (part a) or you have 10 kg of Alloy B (part b) there would be an infinite number of solutions to these equations.  To visualize this, just think of one set of numbers that would work, then mentally double them.  Triple them.  Multiply them by 1000.  See? They all work.

 

Therefore, you can set up a general matrix, one that would have infinite solutions, then set a limiting factor, such as how much zinc you have.  So first I will talk about the general solution.  A very important definition that you must be careful not to confuse is this:  THE ALLOYS (A,B,C,D)  are the mixtures.  THE ELEMENTS (Cu, Sn, Pb, and Zn) are the components of the alloys and of the final product, and these are the substances that have to have equal mass in the final product.

Matrix A is just as described above.  It is comprised of the MASS FRACTIONS of each element (Cu, Sn, Pb, and Zn.

 

Matrix B is also just as described above.  It is comprised of the MASS FRACTIONS of each element in the total; here, those fractions are all .25 because we are told in the problem that they are equal (i.e., each element contributes the same mass to the finished product).

 

Matrix X is a little trickier to understand.  It is the mass fractions of Alloys A, B, C, and D in the final product.  It is NOT the mass fractions of the elements, which are the numbers actually called for.

 

BUT, if you can find the mass fractions of A, B, C, and D, you can multiply them by the amount of each element in each alloy.  Then, if you add these amount up for each element, you will come out with the mass of each element in the final product, and if you’ve done your work right, they will all be equal!

 

So the process is one of setting up the matrix for the general solution, and calculating the mass fraction of each alloy in the total.  Then using your limiting component, calculate the mass of each alloy in the final product, then multiply the mass of each alloy by the mass fraction of each element, giving you the mass of each element in each alloy.  Finally, add up the masses of each element contributed by each alloy and this will give you the total mass of the element in the final product.  \

 

So, good luck.  The crunch on this is not too hard, but understanding what the problem calls for is pretty tough.

Astrid