1 files changed, 44 insertions, 33 deletions
diff --git a/249/CH18/EX18.3/18_03.sce b/249/CH18/EX18.3/18_03.sce
index a281ee41e..6080fb0bd 100755
--- a/249/CH18/EX18.3/18_03.sce
+++ b/249/CH18/EX18.3/18_03.sce
@@ -1,33 +1,44 @@
-clear
-clc
-CAo=0.1;//mol/litre
-FAo=2;//mol/hr
-eA=3;
-CA=[0.074;0.06;0.044;0.029];//mol/litre
-W=[0.02;0.04;0.08;0.16];//kg
-//Gussing 1st order,plug flow rxn
-//(1+eA)*log(1/(1-XA))-eA*XA=k*(CAo*W/FAo)
-for i=1:4
-XA(i)=(CAo-CA(i))/(CAo+eA*CA(i));
-y(i)=(1+eA)*log(1/(1-XA(i)))-eA*XA(i);
-x(i)=CAo*W(i)/FAo;
-W_by_FAo(i)=W(i)/FAo;
-CAout_by_CAo(i)=CA(i)/CAo;
-XA1(i)=(1-CAout_by_CAo(i))/(1+eA*CAout_by_CAo(i));
-end
-plot(x,y)
-coeff3=regress(x,y);
-xlabel('CAoW/FAo'),ylabel('4ln(1/1-XA)-3XA')
-k=coeff3(2);
-printf("\n Part a,using integral method of analysis")
-printf("\n The rate of reaction(mol/litre) is %f",k)
-printf("CA")
-//Part b
-//plotting W_by_FAo vs XA1,the calculating rA=dXA/d(W/FAo) for last 3 points,we get
-rA=[5.62;4.13;2.715];
-coeff2=regress(CA(2:4),rA);
-printf("\n Part b,using differential method of analysis")
-printf("\n The rate of reaction(mol/litre) is %f",coeff2(2))
-printf("CA")
-
-
+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
+CAo=0.1;//mol/litre
+FAo=2;//mol/hr
+eA=3;
+CA=[0.074;0.06;0.044;0.029];//mol/litre
+W=[0.02;0.04;0.08;0.16];//kg
+//Gussing 1st order,plug flow rxn
+//(1+eA)*log(1/(1-XA))-eA*XA=k*(CAo*W/FAo)
+for i=1:4
+XA(i)=(CAo-CA(i))/(CAo+eA*CA(i));
+y(i)=(1+eA)*log(1/(1-XA(i)))-eA*XA(i);
+x(i)=CAo*W(i)/FAo;
+W_by_FAo(i)=W(i)/FAo;
+CAout_by_CAo(i)=CA(i)/CAo;
+XA1(i)=(1-CAout_by_CAo(i))/(1+eA*CAout_by_CAo(i));
+end
+plot(x,y)
+coeff3=regress(x,y);
+xlabel('CAoW/FAo'),ylabel('4ln(1/1-XA)-3XA')
+k=coeff3(2);
+printf("\n Part a,using integral method of analysis")
+printf("\n The rate of reaction(mol/litre) is %f",k)
+printf("CA")
+//Part b
+//plotting W_by_FAo vs XA1,the calculating rA=dXA/d(W/FAo) for last 3 points,we get
+rA=[5.62;4.13;2.715];
+coeff2=regress(CA(2:4),rA);
+printf("\n Part b,using differential method of analysis")
+printf("\n The rate of reaction(mol/litre) is %f",coeff2(2))
+printf("CA")
+\ No newline at end of file