2 files changed, 89 insertions, 65 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
diff --git a/249/CH3/EX3.2/3_02.sce b/249/CH3/EX3.2/3_02.sce
index a2a1ef1b6..bb129725c 100755
--- a/249/CH3/EX3.2/3_02.sce
+++ b/249/CH3/EX3.2/3_02.sce
@@ -1,26 +1,37 @@
-clear
-clc
-CA=[10;8;6;5;3;2;1];//mol/litre
-T=[0;20;40;60;120;180;300];//sec
-//plot(T,CA)
-//xlabel('Time(sec)');ylabel('CA(mol/litre)');
-//From graph y=-dCA/dt at different points are
-y=[-0.1333;-0.1031;-0.0658;-0.0410;-0.0238;-0.0108;-0.0065];
-//Guessing nth rate order
-//rA=kCA^n
-//log(-dCA/dt)=logk+nlogCA
-for i=1:7
-log_y(i)=log10(y(i));
-log_CA(i)=log10(CA(i));
-end
-plot(log_CA,log_y)
-xlabel('logCA');ylabel('log(-dCA/dt)')
-coeff1=regress(log_CA,log_y);
-n=coeff1(2);
-k=-10^(coeff1(1));
-printf("\n After doing linear regression,the slope and intercept of the graph is %f , %f",coeff(2),coeff(1))
-printf("\n The rate equation is therefore given by %f",k)
-printf("CA^1.375 mol/litre.sec")
-disp('The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept')
-
-
+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
+CA=[10;8;6;5;3;2;1];//mol/litre
+T=[0;20;40;60;120;180;300];//sec
+//plot(T,CA)
+//xlabel('Time(sec)');ylabel('CA(mol/litre)');
+//From graph y=-dCA/dt at different points are
+y=[-0.1333;-0.1031;-0.0658;-0.0410;-0.0238;-0.0108;-0.0065];
+//Guessing nth rate order
+//rA=kCA^n
+//log(-dCA/dt)=logk+nlogCA
+for i=1:7
+log_y(i)=log10(y(i));
+log_CA(i)=log10(CA(i));
+end
+plot(log_CA,log_y)
+xlabel('logCA');ylabel('log(-dCA/dt)')
+coeff1=regress(log_CA,log_y);
+n=coeff1(2);
+k=-10^(coeff1(1));
+printf("\n After doing linear regression,the slope and intercept of the graph is %f , %f",coeff(2),coeff(1))
+printf("\n The rate equation is therefore given by %f",k)
+printf("CA^1.375 mol/litre.sec")
+disp('The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept')
+\ No newline at end of file