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-rw-r--r--1040/CH1/EX1.4.a/Chapter1_Ex4_a.sce69
-rw-r--r--1040/CH1/EX1.4.a/Chapter1_Ex4_a_Output.txt13
-rw-r--r--1040/CH1/EX1.4.a/Chapter1_Ex4_a_Result.pdfbin0 -> 9755 bytes
-rw-r--r--1040/CH1/EX1.4.b/Chapter1_Ex4_b.sce67
-rw-r--r--1040/CH1/EX1.4.b/Chapter1_Ex4_b_Output.txt13
-rw-r--r--1040/CH1/EX1.4.b/Chapter1_Ex4_b_Result.pdfbin0 -> 9254 bytes
-rw-r--r--1040/CH1/EX1.4/Chapter1_Ex4.sce118
-rw-r--r--1040/CH1/EX1.5/Chapter1_Ex5.sce72
-rw-r--r--1040/CH1/EX1.5/Chapter1_Ex5_Result1.pdfbin0 -> 12476 bytes
-rw-r--r--1040/CH1/EX1.5/Chapter1_Ex5_Result2.pdfbin0 -> 12488 bytes
10 files changed, 352 insertions, 0 deletions
diff --git a/1040/CH1/EX1.4.a/Chapter1_Ex4_a.sce b/1040/CH1/EX1.4.a/Chapter1_Ex4_a.sce
new file mode 100644
index 000000000..7dfb3c1c6
--- /dev/null
+++ b/1040/CH1/EX1.4.a/Chapter1_Ex4_a.sce
@@ -0,0 +1,69 @@
+//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436.
+//Chapter-1 Ex1.4.a Pg No. 23
+//Title: Activation energy from packed bed data - I Order Reaction
+//=========================================================================================================
+clear
+clc
+clf
+//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
+//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)))
+ u(i)=(1/(T_avg(i)+273.15));
+end
+v=(log(k_1));
+plot(u.*1000,v,'o');
+xlabel("1000/T (K^-1)");
+ylabel("ln k_1");
+xtitle("ln k_1 vs 1000/T" );
+// Least square regression to obtain activation energy and pre-exponential factor
+i=length(u);
+X=[u ones(i,1) ];
+result= X\v;
+k_0=exp(result(2,1));
+E=(-R)*(result(1,1));
+
+
+
+//OUTPUT
+//Console Output
+mprintf('========================================================================================\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',E );
+
+//File Output
+fid= mopen('E:\Chapter1-Ex4-a-Output.txt','w');
+mfprintf(fid,'========================================================================================\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',E );
+mclose(fid);
+//=================================================================END OF PROGRAM========================================
+//Disclaimer: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
diff --git a/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Output.txt b/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Output.txt
new file mode 100644
index 000000000..31a8d8816
--- /dev/null
+++ b/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Output.txt
@@ -0,0 +1,13 @@
+========================================================================================
+L T x T_average (7+x)ave k_1
+(ft) (°C) (°C)
+========================================================================================
+1.000000 338.000000 0.061538 334.000000 7.030769 0.446548
+2.000000 348.000000 0.138462 343.000000 7.100000 0.607207
+3.000000 361.000000 0.238462 354.500000 7.188462 0.886905
+4.000000 380.000000 0.384615 370.500000 7.311538 1.558039
+5.000000 415.000000 0.653846 397.500000 7.519231 4.326296
+6.000000 447.000000 0.900000 431.000000 7.776923 9.656708
+9.000000 458.000000 0.984615 452.500000 7.942308 4.955476
+
+The activation energy from the slope =21.393545 kcal/mol \ No newline at end of file
diff --git a/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Result.pdf b/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Result.pdf
new file mode 100644
index 000000000..e7a3d1dd4
--- /dev/null
+++ b/1040/CH1/EX1.4.a/Chapter1_Ex4_a_Result.pdf
Binary files differ
diff --git a/1040/CH1/EX1.4.b/Chapter1_Ex4_b.sce b/1040/CH1/EX1.4.b/Chapter1_Ex4_b.sce
new file mode 100644
index 000000000..c89a78537
--- /dev/null
+++ b/1040/CH1/EX1.4.b/Chapter1_Ex4_b.sce
@@ -0,0 +1,67 @@
+//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436.
+//Chapter-1 Ex1.4.b Pg No. 23
+//Title: Activation energy from packed bed data - II Order Reaction
+//=========================================================================================================
+clear
+clc
+clf
+//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
+//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_2(i)=((P_molavg(i))/delta_L(i))*((x(i+1)-x(i))/((1-x(i+1))*(1-x(i))))
+ u(i)=(1/(T_avg(i)+273.15));
+end
+v=(log(k_2));
+plot(u.*1000,v,'o');
+xlabel("1000/T (K^-1)");
+ylabel("ln k_2");
+xtitle("ln k_2 vs 1000/T ");
+i=length(u);
+X=[u ones(i,1) ];
+result= X\v;
+k_0=exp(result(2,1));
+E=(-R)*(result(1,1));
+
+//OUTPUT
+//Console Output
+mprintf('========================================================================================\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',E );
+
+//File Output
+fid= mopen('.\Chapter1-Ex4-b-Output.txt','w');
+mfprintf(fid,'========================================================================================\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',E );
+mclose(fid);
+
+//============================================================END OF PROGRAM===========================================
+//Disclaimer: Least Square method is used to find the slope and intercept in this example.
+// Hence the values differ from the graphically obtained values of slope and intercept in the textbook.
+// 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
+
diff --git a/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Output.txt b/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Output.txt
new file mode 100644
index 000000000..712170a8a
--- /dev/null
+++ b/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Output.txt
@@ -0,0 +1,13 @@
+========================================================================================
+L T x T_average (7+x)ave k_2
+(ft) (°C) (°C)
+========================================================================================
+1.000000 338.000000 0.061538 334.000000 7.030769 0.461034
+2.000000 348.000000 0.138462 343.000000 7.100000 0.675498
+3.000000 361.000000 0.238462 354.500000 7.188462 1.095644
+4.000000 380.000000 0.384615 370.500000 7.311538 2.280240
+5.000000 415.000000 0.653846 397.500000 7.519231 9.503472
+6.000000 447.000000 0.900000 431.000000 7.776923 55.302564
+9.000000 458.000000 0.984615 452.500000 7.942308 145.608974
+
+The activation energy from the slope =43.147788 kcal/mol \ No newline at end of file
diff --git a/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Result.pdf b/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Result.pdf
new file mode 100644
index 000000000..5d6c4a023
--- /dev/null
+++ b/1040/CH1/EX1.4.b/Chapter1_Ex4_b_Result.pdf
Binary files differ
diff --git a/1040/CH1/EX1.4/Chapter1_Ex4.sce b/1040/CH1/EX1.4/Chapter1_Ex4.sce
new file mode 100644
index 000000000..c4f83141c
--- /dev/null
+++ b/1040/CH1/EX1.4/Chapter1_Ex4.sce
@@ -0,0 +1,118 @@
+//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
+
+
diff --git a/1040/CH1/EX1.5/Chapter1_Ex5.sce b/1040/CH1/EX1.5/Chapter1_Ex5.sce
new file mode 100644
index 000000000..1ff042cb0
--- /dev/null
+++ b/1040/CH1/EX1.5/Chapter1_Ex5.sce
@@ -0,0 +1,72 @@
+//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436.
+//Chapter-1 Ex1.5 Pg No. 29
+//Title: Methods to determine km and vm
+//========================================================================================
+clear
+clc
+clf
+//INPUT
+S=[2;5;10;15]*10^(-3);//Concentration of substrate [HCO3]
+r_reciprocal=[95;45;29;25]*10^(3);//Reciprocal rates (L-sec/mol)
+
+//CALCULATION
+//Plot 1 refer equation 1.24 Pg No.29
+x1=(S).^(-1);
+y1=r_reciprocal;
+scf(0)
+plot(x1,y1*10^(-3),'RED');
+xlabel("1/[S]");
+ylabel("(1/r)*10^-3");
+xtitle("1/r versus 1/S");
+p=length(x1);
+X_1=[x1 ones(p,1)];
+R1=X_1\y1;
+slope(1)=R1(1,1);
+intercept(1)=R1(2,1);
+v_m(1)=(1/(intercept(1)));//Maximum Reaction Rate(mol/L-sec)
+k_m(1)=slope(1)*v_m(1);//Michaelis-Menton constant
+
+//Plot 2 refer equation 1.25 Pg No.29
+x2=S;
+y2=S.*r_reciprocal;
+scf(1)
+plot(x2*10^(3),y2);
+xlabel("(S)*10^3");
+ylabel("(S)/r");
+xtitle("(S)/r versus (S)");
+q=length(x2);
+X_2=[x2 ones(q,1)];
+R2=X_2\y2;
+slope(2)=R2(1,1);
+intercept(2)=R2(2,1);
+v_m(2)=1/(slope(2));//Maximum Reaction Rate (mol/L-sec)
+k_m(2)=intercept(2)/(slope(2));//Michaelis-Menton constant
+
+
+//OUTPUT
+mprintf('\n======================================================================================');
+mprintf('\n \t\tMethod_1\tMethod_2');
+mprintf('\n======================================================================================');
+i=1
+ mprintf('\n Slope \t%f\t%f',slope(i),slope(i+1));
+ mprintf('\n Intercept \t%f\t%f',intercept(i),intercept(i+1));
+ mprintf('\n Km (M) \t%f\t%f',k_m(i),k_m(i+1));
+ mprintf('\n Vm(mol/L-sec) %f\t%f',v_m(i),v_m(i+1));
+
+//FILE OUTPUT
+fid= mopen('.\Chapter1-Ex5-Output.txt','w');
+mfprintf(fid,'\n======================================================================================');
+mfprintf(fid,'\n \t\tMethod_1\tMethod_2');
+mfprintf(fid,'\n======================================================================================');
+i=1
+ mfprintf(fid,'\n Slope \t%f\t%f',slope(i),slope(i+1));
+ mfprintf(fid,'\n Intercept \t%f\t%f',intercept(i),intercept(i+1));
+ mfprintf(fid,'\n Km (M) \t%f\t%f',k_m(i),k_m(i+1));
+ mfprintf(fid,'\n Vm(mol/L-sec) %f\t%f',v_m(i),v_m(i+1));
+mclose(fid);
+
+//========================================================================END OF PROGRAM=================================
+//Disclaimer: Least Square method is used to find the slope and intercept in this example.
+// Hence the values differ from the graphically obtained values of slope and intercept in the textbook.
+
+
diff --git a/1040/CH1/EX1.5/Chapter1_Ex5_Result1.pdf b/1040/CH1/EX1.5/Chapter1_Ex5_Result1.pdf
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index 000000000..51eb6702c
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+++ b/1040/CH1/EX1.5/Chapter1_Ex5_Result1.pdf
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diff --git a/1040/CH1/EX1.5/Chapter1_Ex5_Result2.pdf b/1040/CH1/EX1.5/Chapter1_Ex5_Result2.pdf
new file mode 100644
index 000000000..82cd15558
--- /dev/null
+++ b/1040/CH1/EX1.5/Chapter1_Ex5_Result2.pdf
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generated by cgit (git 2.34.1) at 2024-12-19 12:42:25 +0000