updated the code

1 files changed, 116 insertions, 118 deletions

diff --git a/1040/CH1/EX1.4/Ex1_4.sce b/1040/CH1/EX1.4/Ex1_4.sce
index c4f83141c..57de822b4 100644
--- a/1040/CH1/EX1.4/Ex1_4.sce
+++ b/1040/CH1/EX1.4/Ex1_4.sce
@@ -1,118 +1,116 @@
-//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436. 

-//Chapter-1 Ex1.4  Pg No. 23

-//Title: Activation energy from packed bed data

-//=========================================================================================================

-clear

-clc

-clf

-// COMMON INPUT

-L= [0 1 2 3 4 5 6 9];//Bed length in feet(ft)

-T=[330 338 348 361 380 415 447 458 ] //Temperature Corresponding the bed length given (°C) 

-R=1.98587E-3;//Gas constant (kcal/mol K)

-

-//CALCLATION (Ex1.4.a)

-//Basis is 1mol of feed A(Furfural) X moles reacted to form Furfuran and CO 

-x=(T-330)./130;//Conversion based on fractional temperature rise

-n=length (T);//6 moles of steam per mole of Furfural is used to decrease temperature rise in the bed

-P_mol=x+7;//Total No. of moles in product stream

-for i=1:(n-1)

-    T_avg(i)= (T(i)+T(i+1))/2

-    P_molavg(i)= (P_mol(i)+P_mol(i+1))/2

-    delta_L(i)=L(i+1)-L(i)

-    k_1(i)=((P_molavg(i))/delta_L(i))*log((1-x(i))/(1-x(i+1)))

-    u1(i)=(1/(T_avg(i)+273.15));

-end

-v1=(log(k_1));

-i=length(u1);

-X1=[u1 ones(i,1) ];

-result1= X1\v1;

-k_1_dash=exp(result1(2,1));

-E1=(-R)*(result1(1,1));

-

-//OUTPUT (Ex1.4.a)

-//Console Output

-mprintf('\n OUTPUT Ex1.4.a');

-mprintf('\n========================================================================================\n')

-mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')

-mprintf('\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')

-mprintf('\n========================================================================================')

-for i=1:n-1

-mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))

-mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))

-end

-mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E1 );

-//=====================================================================================================

-

-

-//Title: II Order Reaction 

-//=========================================================================================================

-//CALCULATION (Ex 1.4.b)

-for i=1:(n-1)

-    T_avg(i)= (T(i)+T(i+1))/2

-    P_molavg(i)= (P_mol(i)+P_mol(i+1))/2

-    delta_L(i)=L(i+1)-L(i)

-    k_2(i)=((P_molavg(i))/delta_L(i))*((x(i+1)-x(i))/((1-x(i+1))*(1-x(i))))

-    u2(i)=(1/(T_avg(i)+273.15));

-end

-v2=(log(k_2));

-plot(u1.*1000,v1,'o',u2.*1000,v2,'*');

-xlabel("1000/T (K^-1)");

-ylabel("ln k_1 or ln k_2");

-xtitle("ln k vs 1000/T ");

-legend('ln k_1','ln k_2');

-j=length(u2);

-X2=[u2 ones(j,1) ];

-result2= X2\v2;

-k_2_dash=exp(result2(2,1));

-E2=(-R)*(result2(1,1));

-

-//OUTPUT (Ex 1.4.b)

-mprintf('\n OUTPUT Ex1.4.b');

-mprintf('\n========================================================================================\n')

-mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')

-mprintf('\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')

-mprintf('\n========================================================================================')

-for i=1:n-1

-mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))

-mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))

-end

-mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E2 );

-

-//FILE OUTPUT

-fid= mopen('.\Chapter1-Ex4-Output.txt','w');

-mfprintf(fid,'\n OUTPUT Ex1.4.a');

-mfprintf(fid,'\n========================================================================================\n')

-mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')

-mfprintf(fid,'\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')

-mfprintf(fid,'\n========================================================================================')

-for i=1:n-1

-mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))

-mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))

-end

-mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E1 );

-mfprintf(fid,'\n\n========================================================================================\n')

-mfprintf(fid,'\n OUTPUT Ex1.4.b');

-mfprintf(fid,'\n========================================================================================\n')

-mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')

-mfprintf(fid,'\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')

-mfprintf(fid,'\n========================================================================================')

-for i=1:n-1

-mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))

-mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))

-end

-mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E2 );

-mclose(all);

-

-//============================================================END OF PROGRAM===========================================

-//Disclaimer (Ex1.4.a):The last value of tavg and k_1  corresponding to L=9 in Table 1.6 (Pg No. 25)of the textbook is a misprint.

-// The value should be 452.5 and 4.955476 respectively instead of 455 and 18.2 as printed in the textbook.

-//Hence there is a change in the activation energy obtained from the code 

-// The answer obtained is 21.3935 kcal/mol instead of 27 kcal/mol as reported in the textbook.

-//Figure 1.8 is a plot between ln k_1 vs 1000/T instead of k_1 vs 1000/T as stated in the solution of Ex1.4.a

-//=========================================================================================================

-//Disclaimer (Ex1.4.b): There is a discrepancy between the computed value of activation energy and value reported in textbook 

-// Error could have been on similar lines as reported for example Ex.1.4.a 

-// Further, intermeidate values for Ex.1.4.b is not available/ reported in textbook and hence could not be compared. 

-//Figure 1.8 is a plot between ln k_2 vs 1000/T instead of k_2 vs 1000/T as stated in the solution of Ex1.4.b

-

-

+//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436. 
+//Chapter-1 Ex1.4  Pg No. 23
+//Title: Activation energy from packed bed data
+//=========================================================================================================
+clear
+clc
+clf
+// COMMON INPUT
+L= [0 1 2 3 4 5 6 9];//Bed length in feet(ft)
+T=[330 338 348 361 380 415 447 458 ] //Temperature Corresponding the bed length given (°C) 
+R=1.98587E-3;//Gas constant (kcal/mol K)
+
+//CALCLATION (Ex1.4.a)
+//Basis is 1mol of feed A(Furfural) X moles reacted to form Furfuran and CO 
+x=(T-330)./130;//Conversion based on fractional temperature rise
+n=length (T);//6 moles of steam per mole of Furfural is used to decrease temperature rise in the bed
+P_mol=x+7;//Total No. of moles in product stream
+for i=1:(n-1)
+    T_avg(i)= (T(i)+T(i+1))/2
+    P_molavg(i)= (P_mol(i)+P_mol(i+1))/2
+    delta_L(i)=L(i+1)-L(i)
+    k_1(i)=((P_molavg(i))/delta_L(i))*log((1-x(i))/(1-x(i+1)))
+    u1(i)=(1/(T_avg(i)+273.15));
+end
+v1=(log(k_1));
+i=length(u1);
+X1=[u1 ones(i,1) ];
+result1= X1\v1;
+k_1_dash=exp(result1(2,1));
+E1=(-R)*(result1(1,1));
+
+//OUTPUT (Ex1.4.a)
+//Console Output
+mprintf('\n OUTPUT Ex1.4.a');
+mprintf('\n========================================================================================\n')
+mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')
+mprintf('\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')
+mprintf('\n========================================================================================')
+for i=1:n-1
+mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))
+mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))
+end
+mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E1 );
+//=====================================================================================================
+
+
+//Title: II Order Reaction 
+//=========================================================================================================
+//CALCULATION (Ex 1.4.b)
+for i=1:(n-1)
+    T_avg(i)= (T(i)+T(i+1))/2
+    P_molavg(i)= (P_mol(i)+P_mol(i+1))/2
+    delta_L(i)=L(i+1)-L(i)
+    k_2(i)=((P_molavg(i))/delta_L(i))*((x(i+1)-x(i))/((1-x(i+1))*(1-x(i))))
+    u2(i)=(1/(T_avg(i)+273.15));
+end
+v2=(log(k_2));
+plot(u1.*1000,v1,'o',u2.*1000,v2,'*');
+xlabel("1000/T (K^-1)");
+ylabel("ln k_1 or ln k_2");
+xtitle("ln k vs 1000/T ");
+legend('ln k_1','ln k_2');
+j=length(u2);
+X2=[u2 ones(j,1) ];
+result2= X2\v2;
+k_2_dash=exp(result2(2,1));
+E2=(-R)*(result2(1,1));
+
+//OUTPUT (Ex 1.4.b)
+mprintf('\n OUTPUT Ex1.4.b');
+mprintf('\n========================================================================================\n')
+mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')
+mprintf('\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')
+mprintf('\n========================================================================================')
+for i=1:n-1
+mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))
+mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))
+end
+mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E2 );
+
+//FILE OUTPUT
+fid= mopen('.\Chapter1-Ex4-Output.txt','w');
+mfprintf(fid,'\n OUTPUT Ex1.4.a');
+mfprintf(fid,'\n========================================================================================\n')
+mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')
+mfprintf(fid,'\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')
+mfprintf(fid,'\n========================================================================================')
+for i=1:n-1
+mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))
+mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))
+end
+mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E1 );
+mfprintf(fid,'\n\n========================================================================================\n')
+mfprintf(fid,'\n OUTPUT Ex1.4.b');
+mfprintf(fid,'\n========================================================================================\n')
+mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')
+mfprintf(fid,'\n(ft) \t \t (°C) \t\t   \t\t (°C) \t  ')
+mfprintf(fid,'\n========================================================================================')
+for i=1:n-1
+mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))
+mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))
+end
+mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E2 );
+close;
+
+//============================================================END OF PROGRAM===========================================
+//Disclaimer (Ex1.4.a):The last value of tavg and k_1  corresponding to L=9 in Table 1.6 (Pg No. 25)of the textbook is a misprint.
+// The value should be 452.5 and 4.955476 respectively instead of 455 and 18.2 as printed in the textbook.
+//Hence there is a change in the activation energy obtained from the code 
+// The answer obtained is 21.3935 kcal/mol instead of 27 kcal/mol as reported in the textbook.
+//Figure 1.8 is a plot between ln k_1 vs 1000/T instead of k_1 vs 1000/T as stated in the solution of Ex1.4.a
+//=========================================================================================================
+//Disclaimer (Ex1.4.b): There is a discrepancy between the computed value of activation energy and value reported in textbook 
+// Error could have been on similar lines as reported for example Ex.1.4.a 
+// Further, intermeidate values for Ex.1.4.b is not available/ reported in textbook and hence could not be compared. 
+//Figure 1.8 is a plot between ln k_2 vs 1000/T instead of k_2 vs 1000/T as stated in the solution of Ex1.4.b
\ No newline at end of file