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+//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