updated the code

1 files changed, 52 insertions, 39 deletions

diff --git a/249/CH3/EX3.1/3_01.sce b/249/CH3/EX3.1/3_01.sce
index 1bbf163bd..4897377b9 100755
--- a/249/CH3/EX3.1/3_01.sce
+++ b/249/CH3/EX3.1/3_01.sce
@@ -1,39 +1,52 @@
-clear

-clc

-//Given

-t=[0 20 40 60 120 180 300];

-C_A=[10 8 6 5 3 2 1];

-CAo=10;

-//Guessing 1st order kinetics

-//This means log(CAo/C_A) vs t should give a straight line

-for i=1:7

-    k(i)=log(CAo/C_A(i));

-    CA_inv(i)=1/C_A(i);

-end

-//plot(t,k)

-//This doesn't give straight line.

-//Guessing 2nd Order Kinetics so 

-//1/C_A vs t should give a straight line

-//plot(t,CA_inv)

-//Again this doesn't give a straight line

-//Guessing nth order kinetics and using fractional life method with F=80%

-//log Tf=log(0.8^(1-n)-1/(k(n-1)))+(1-n)logCAo

-//plot(t,C_A)

-

-//Picking different values of CAo

-//Time needed for 3 runs,,from graph

-T=[18.5;23;35];

-CAo=[10;5;2];

-for i=1:3

-    CA(i)=0.8*CAo(i);

-    log_Tf(i)=log10(T(i));

-    log_CAo(i)=log10(CAo(i));

-end

-plot(log_CAo,log_Tf)

-xlabel('log CAo');ylabel('log t');

-coeff1=regress(log_CAo,log_Tf);

-n=1-coeff1(2);

-printf("From graph we get slope and intercept for calculating rate eqn")

-k1=((0.8^(1-n))-1)*(10^(1-n))/(18.5*(n-1));

-printf("\n The rate equation is given by %f",k1)

-printf("CA^1.4 mol/litre.sec")

+clear
+clc
+function [coefs]=regress(x,y)
+coefs=[]
+  if (type(x) <> 1)|(type(y)<>1) then error(msprintf(gettext("%s: Wrong type for input arguments: Numerical expected.\n"),"regress")), end
+  lx=length(x)
+  if lx<>length(y) then error(msprintf(gettext("%s: Wrong size for both input arguments: same size expected.\n"),"regress")), end
+  if lx==0 then error(msprintf(gettext("%s: Wrong size for input argument #%d: Must be > %d.\n"),"regress", 1, 0)), end
+  x=matrix(x,lx,1)
+  y=matrix(y,lx,1)
+  xbar=sum(x)/lx
+  ybar=sum(y)/lx
+  coefs(2)=sum((x-xbar).*(y-ybar))/sum((x-xbar).^2)
+  coefs(1)=ybar-coefs(2)*xbar
+endfunction
+//Given
+t=[0 20 40 60 120 180 300];
+C_A=[10 8 6 5 3 2 1];
+CAo=10;
+//Guessing 1st order kinetics
+//This means log(CAo/C_A) vs t should give a straight line
+for i=1:7
+    k(i)=log(CAo/C_A(i));
+    CA_inv(i)=1/C_A(i);
+end
+//plot(t,k)
+//This doesn't give straight line.
+//Guessing 2nd Order Kinetics so 
+//1/C_A vs t should give a straight line
+//plot(t,CA_inv)
+//Again this doesn't give a straight line
+//Guessing nth order kinetics and using fractional life method with F=80%
+//log Tf=log(0.8^(1-n)-1/(k(n-1)))+(1-n)logCAo
+//plot(t,C_A)
+
+//Picking different values of CAo
+//Time needed for 3 runs,,from graph
+T=[18.5;23;35];
+CAo=[10;5;2];
+for i=1:3
+    CA(i)=0.8*CAo(i);
+    log_Tf(i)=log10(T(i));
+    log_CAo(i)=log10(CAo(i));
+end
+plot(log_CAo,log_Tf)
+xlabel('log CAo');ylabel('log t');
+coeff1=regress(log_CAo,log_Tf);
+n=1-coeff1(2);
+printf("From graph we get slope and intercept for calculating rate eqn")
+k1=((0.8^(1-n))-1)*(10^(1-n))/(18.5*(n-1));
+printf("\n The rate equation is given by %f",k1)
+printf("CA^1.4 mol/litre.sec")
\ No newline at end of file