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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 1: Homogeneous Kinetics"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1.4: Activation_energy_from_packed_bed_data.sce"
]
},
{
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"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436. \n",
"//Chapter-1 Ex1.4 Pg No. 23\n",
"//Title: Activation energy from packed bed data\n",
"//=========================================================================================================\n",
"clear\n",
"clc\n",
"clf\n",
"// COMMON INPUT\n",
"L= [0 1 2 3 4 5 6 9];//Bed length in feet(ft)\n",
"T=[330 338 348 361 380 415 447 458 ] //Temperature Corresponding the bed length given (°C) \n",
"R=1.98587E-3;//Gas constant (kcal/mol K)\n",
"\n",
"//CALCLATION (Ex1.4.a)\n",
"//Basis is 1mol of feed A(Furfural) X moles reacted to form Furfuran and CO \n",
"x=(T-330)./130;//Conversion based on fractional temperature rise\n",
"n=length (T);//6 moles of steam per mole of Furfural is used to decrease temperature rise in the bed\n",
"P_mol=x+7;//Total No. of moles in product stream\n",
"for i=1:(n-1)\n",
" T_avg(i)= (T(i)+T(i+1))/2\n",
" P_molavg(i)= (P_mol(i)+P_mol(i+1))/2\n",
" delta_L(i)=L(i+1)-L(i)\n",
" k_1(i)=((P_molavg(i))/delta_L(i))*log((1-x(i))/(1-x(i+1)))\n",
" u1(i)=(1/(T_avg(i)+273.15));\n",
"end\n",
"v1=(log(k_1));\n",
"i=length(u1);\n",
"X1=[u1 ones(i,1) ];\n",
"result1= X1\v1;\n",
"k_1_dash=exp(result1(2,1));\n",
"E1=(-R)*(result1(1,1));\n",
"\n",
"//OUTPUT (Ex1.4.a)\n",
"//Console Output\n",
"mprintf('\n OUTPUT Ex1.4.a');\n",
"mprintf('\n========================================================================================\n')\n",
"mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')\n",
"mprintf('\n(ft) \t \t (°C) \t\t \t\t (°C) \t ')\n",
"mprintf('\n========================================================================================')\n",
"for i=1:n-1\n",
"mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))\n",
"mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))\n",
"end\n",
"mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E1 );\n",
"//=====================================================================================================\n",
"\n",
"\n",
"//Title: II Order Reaction \n",
"//=========================================================================================================\n",
"//CALCULATION (Ex 1.4.b)\n",
"for i=1:(n-1)\n",
" T_avg(i)= (T(i)+T(i+1))/2\n",
" P_molavg(i)= (P_mol(i)+P_mol(i+1))/2\n",
" delta_L(i)=L(i+1)-L(i)\n",
" k_2(i)=((P_molavg(i))/delta_L(i))*((x(i+1)-x(i))/((1-x(i+1))*(1-x(i))))\n",
" u2(i)=(1/(T_avg(i)+273.15));\n",
"end\n",
"v2=(log(k_2));\n",
"plot(u1.*1000,v1,'o',u2.*1000,v2,'*');\n",
"xlabel('1000/T (K^-1)');\n",
"ylabel('ln k_1 or ln k_2');\n",
"xtitle('ln k vs 1000/T ');\n",
"legend('ln k_1','ln k_2');\n",
"j=length(u2);\n",
"X2=[u2 ones(j,1) ];\n",
"result2= X2\v2;\n",
"k_2_dash=exp(result2(2,1));\n",
"E2=(-R)*(result2(1,1));\n",
"\n",
"//OUTPUT (Ex 1.4.b)\n",
"mprintf('\n OUTPUT Ex1.4.b');\n",
"mprintf('\n========================================================================================\n')\n",
"mprintf('L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')\n",
"mprintf('\n(ft) \t \t (°C) \t\t \t\t (°C) \t ')\n",
"mprintf('\n========================================================================================')\n",
"for i=1:n-1\n",
"mprintf('\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))\n",
"mprintf('\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))\n",
"end\n",
"mprintf('\n\nThe activation energy from the slope =%f kcal/mol',E2 );\n",
"\n",
"//FILE OUTPUT\n",
"fid= mopen('.\Chapter1-Ex4-Output.txt','w');\n",
"mfprintf(fid,'\n OUTPUT Ex1.4.a');\n",
"mfprintf(fid,'\n========================================================================================\n')\n",
"mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_1')\n",
"mfprintf(fid,'\n(ft) \t \t (°C) \t\t \t\t (°C) \t ')\n",
"mfprintf(fid,'\n========================================================================================')\n",
"for i=1:n-1\n",
"mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))\n",
"mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_1(i))\n",
"end\n",
"mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E1 );\n",
"mfprintf(fid,'\n\n========================================================================================\n')\n",
"mfprintf(fid,'\n OUTPUT Ex1.4.b');\n",
"mfprintf(fid,'\n========================================================================================\n')\n",
"mfprintf(fid,'L \t \t T \t\t x \t\t T_average \t(7+x)ave \tk_2')\n",
"mfprintf(fid,'\n(ft) \t \t (°C) \t\t \t\t (°C) \t ')\n",
"mfprintf(fid,'\n========================================================================================')\n",
"for i=1:n-1\n",
"mfprintf(fid,'\n%f \t %f \t %f ',L(i+1),T(i+1),x(i+1))\n",
"mfprintf(fid,'\t %f \t %f \t %f',T_avg(i),P_molavg(i),k_2(i))\n",
"end\n",
"mfprintf(fid,'\n\nThe activation energy from the slope =%f kcal/mol',E2 );\n",
"mclose(all);\n",
"\n",
"//============================================================END OF PROGRAM===========================================\n",
"//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.\n",
"// The value should be 452.5 and 4.955476 respectively instead of 455 and 18.2 as printed in the textbook.\n",
"//Hence there is a change in the activation energy obtained from the code \n",
"// The answer obtained is 21.3935 kcal/mol instead of 27 kcal/mol as reported in the textbook.\n",
"//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\n",
"//=========================================================================================================\n",
"//Disclaimer (Ex1.4.b): There is a discrepancy between the computed value of activation energy and value reported in textbook \n",
"// Error could have been on similar lines as reported for example Ex.1.4.a \n",
"// Further, intermeidate values for Ex.1.4.b is not available/ reported in textbook and hence could not be compared. \n",
"//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\n",
"\n",
""
]
}
,
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1.5: Methods_to_determine_km_and_vm.sce"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"//Harriot P.,2003,Chemical Reactor Design (I-Edition) Marcel Dekker,Inc.,USA,pp 436. \n",
"//Chapter-1 Ex1.5 Pg No. 29\n",
"//Title: Methods to determine km and vm\n",
"//========================================================================================\n",
"clear\n",
"clc\n",
"clf\n",
"//INPUT\n",
"S=[2;5;10;15]*10^(-3);//Concentration of substrate [HCO3]\n",
"r_reciprocal=[95;45;29;25]*10^(3);//Reciprocal rates (L-sec/mol)\n",
"\n",
"//CALCULATION\n",
"//Plot 1 refer equation 1.24 Pg No.29\n",
"x1=(S).^(-1);\n",
"y1=r_reciprocal;\n",
"scf(0)\n",
"plot(x1,y1*10^(-3),'RED');\n",
"xlabel('1/[S]');\n",
"ylabel('(1/r)*10^-3');\n",
"xtitle('1/r versus 1/S');\n",
"p=length(x1);\n",
"X_1=[x1 ones(p,1)];\n",
"R1=X_1\y1;\n",
"slope(1)=R1(1,1);\n",
"intercept(1)=R1(2,1);\n",
"v_m(1)=(1/(intercept(1)));//Maximum Reaction Rate(mol/L-sec)\n",
"k_m(1)=slope(1)*v_m(1);//Michaelis-Menton constant\n",
"\n",
"//Plot 2 refer equation 1.25 Pg No.29\n",
"x2=S;\n",
"y2=S.*r_reciprocal;\n",
"scf(1)\n",
"plot(x2*10^(3),y2);\n",
"xlabel('(S)*10^3');\n",
"ylabel('(S)/r');\n",
"xtitle('(S)/r versus (S)');\n",
"q=length(x2);\n",
"X_2=[x2 ones(q,1)];\n",
"R2=X_2\y2;\n",
"slope(2)=R2(1,1);\n",
"intercept(2)=R2(2,1);\n",
"v_m(2)=1/(slope(2));//Maximum Reaction Rate (mol/L-sec)\n",
"k_m(2)=intercept(2)/(slope(2));//Michaelis-Menton constant\n",
"\n",
"\n",
"//OUTPUT\n",
"mprintf('\n======================================================================================');\n",
"mprintf('\n \t\tMethod_1\tMethod_2');\n",
"mprintf('\n======================================================================================');\n",
"i=1\n",
" mprintf('\n Slope \t%f\t%f',slope(i),slope(i+1));\n",
" mprintf('\n Intercept \t%f\t%f',intercept(i),intercept(i+1));\n",
" mprintf('\n Km (M) \t%f\t%f',k_m(i),k_m(i+1));\n",
" mprintf('\n Vm(mol/L-sec) %f\t%f',v_m(i),v_m(i+1));\n",
"\n",
"//FILE OUTPUT\n",
"fid= mopen('.\Chapter1-Ex5-Output.txt','w');\n",
"mfprintf(fid,'\n======================================================================================');\n",
"mfprintf(fid,'\n \t\tMethod_1\tMethod_2');\n",
"mfprintf(fid,'\n======================================================================================');\n",
"i=1\n",
" mfprintf(fid,'\n Slope \t%f\t%f',slope(i),slope(i+1));\n",
" mfprintf(fid,'\n Intercept \t%f\t%f',intercept(i),intercept(i+1));\n",
" mfprintf(fid,'\n Km (M) \t%f\t%f',k_m(i),k_m(i+1));\n",
" mfprintf(fid,'\n Vm(mol/L-sec) %f\t%f',v_m(i),v_m(i+1));\n",
"mclose(fid);\n",
"\n",
"//========================================================================END OF PROGRAM=================================\n",
"//Disclaimer: Least Square method is used to find the slope and intercept in this example.\n",
"// Hence the values differ from the graphically obtained values of slope and intercept in the textbook.\n",
" \n",
""
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