Saturday, April 21, 2012

10.8

The following system of equations is designed to determine concentrations (the c's in g/m^3) in a series of coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day),

15c1 - 3c2 - c3 = 3300
-3c1 +18c2 - 6c3 = 1200
-4c1 - c2 + 12c3 = 2400

(a) Determine the matrix inverse
(b) Use the inverse matrix to determine a solution(c) Determine how much the rate of mass input to reactor 3 must be increased to include a 10 g/m^3 rise in the concentration of reactor 1.(d)How much will the concentration in reactor 3 be reduced if the rate of mass input to reactors 1 and 2 is reduced by 700 and 350 g/day, respectively?


Matlab code:

%(a) Determine the matrix inverse
A = [15 -3 -1; -3 18 -6; -4 -1 12]

inversematrix = inv(A)

%(b) Use the inverse matrix to determine a solution

B = [3300; 1200; 2400]

x = inversematrix * B

OriginalB = A * x

%(c) Determine how much the rate of mass input to reactor 
%3 must be increased to incduce a 10 g/m^3 rise in the concentration of
%reactor 1.
[x(1)+10; x(2); x(3)]

OriginalBplus10 = A * [x(1)+10; x(2); x(3)]
%(d)How much will the concentration in reactor 3 be reduced if the rate of
%mass input to reactors 1 and 2 is reduced by 700 and 350 g/day,
%respectively?
Bnew = [2600; 850; 2400]

newx = inversematrix * Bnew

newB = A * newx

diffx = x - newx


Results
(A)

A =

    15    -3    -1
    -3    18    -6
    -4    -1    12


inversematrix =

   0.072538860103627   0.012780656303972   0.012435233160622
   0.020725388601036   0.060794473229706   0.032124352331606
   0.025906735751295   0.009326424870466   0.090155440414508

(B)
B =

        3300
        1200
        2400


x =

   1.0e+02 *

   2.845595854922280
   2.184455958549223
   3.130569948186529


OriginalB =

   1.0e+03 *

   3.299999999999999
   1.200000000000001
   2.400000000000000


ans =

   1.0e+02 *

   2.945595854922280
   2.184455958549223
   3.130569948186529

(C)
OriginalBplus10 =

   1.0e+03 *

   3.450000000000000
   1.170000000000001
   2.360000000000000

(D)
Bnew =

        2600
         850
        2400


newx =

   1.0e+02 *

   2.293091537132988
   1.826597582037997
   2.916580310880829


newB =

   1.0e+03 *

   2.600000000000000
   0.850000000000001
   2.400000000000000


diffx =

  55.250431778929169
  35.785837651122620
  21.398963730569960





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