13 th Aug Chemical Reaction Engineering CH3010. Home work problems

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Abraham Park
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1 13 th ug 18. Chemical Reaction Engineering CH31. Home work problems 1. Batch reactor, variable volume. Consider a gas phase reaction B, conducted isothermally and at constant pressure in a batch reactor. Initially, the reactor contains 1 moles, 5 moles of B and 1 moles of inert materials. The reaction is second order with a rate constant of.5 lit mol -1 s -1. Initial reactor volume is 1 lit. (a) Determine the time required for 5 % conversion. (b) What would be the time needed for the same conversion if the experiment were conducted in a constant volume batch reactor?. Batch reactor, complex reactions. Consider the following gas phase reaction, conducted isothermally in a constant volume batch reactor. Volume of the reactor is 1 lit. k1 B C k1 k B D Given that mol/lit/s, r BC mol/lit/s, r B r mol/lit/s, determine the concentration of each species after 1 h, if (a) initial mixture contains 5 mol B and 5 mol C (b) initial mixture contains mol each of, B, C, D and inert (I). Plot the concentration of each species as a function of time, till 1 h. 3. CSTR and PFR. Consider the following examples. The reaction is B. It is a second order reaction, and the rate constant k =. lit/mol/min. Feed contains 5 % and 5% inert. The operation is at constant temperature and constant pressure. If the total molar flow rate at the inlet is 1 mol/min and the volumetric flow rate is 1 lit/min, determine the volume of reactor needed to get % conversion in case of a (a) PFR and (b) CSTR

2 dn 1. ns: The design equation is dt V r. (a) Isobaric conditions. We assume ideal gas law applies. NT N NB Ninert PV N RT in standard notations. In this particular case, N N N N N N N N N T B inert T T N N Definition of conversion is x N F F x F. In flow systems, under steady state conditions, V RT V Therefore, NT NT Nx, and since P is a constant we can write. Therefore, N P N V V NT N x NT T T dx N 1 N N x N T N V r V kc Vk k k dt V V V NT N x Therefore, dx N 1 x N k dt V N N x T T. Initial condition, t=, conversion =. time =? for conversion =.5 N N x N N dx k dt T T 1 x V

3 1 x 5 1x 1 5 dx.5 dt 1 x x 1 x 1 x 1 x 1 x dx dx dx.15dt ln 1 x.15t 1 x ln ln t Time t = 45 sec = approx. 6 hr 14 min If it were conducted in constant volume reactor, dn dt V r dx N N N N V r V kc Vk k k dx N k dt dx 1 x dt V V V.5 x 1 V N 1 k dt.5 dt.5t V t 1 x.5 1 x Time t = s= approx. 5 hr 33 min dn V ri dt i 1. ns: The design equation is, where the subscript 'i' can take values,b,c, D and I. Of course, inert material, by definition, does not participate in the reaction. It just dilutes the mixture and keeps each species concentration low, thereby changing the reaction rate.

4 , and r k r k B C r k B! r r r r B 1 1 r r r r 1 1 rc r1 r 1 rd r Since it is a constant volume reactor, we can write the design equations as B d C r k k B C k B D r B k1 k 1 B C k B dt rc r 1 1 k1 k 1BC k B D (i) initial concentrations [] =, [B] = 5 mol/lit, [C] = 5 mol/lit, [D] = (ii) initial concentrations [] = [B] = [C] = [D] =[I] = mol/lit % Matlab program to solve this is given below. % % CRE I. ssignment 3. Example problem with multiple reactions clear all; close all; k1 = 1e-4; % 1/s km1 = 1e-4; % lit/mol/s k = e-4; % lit/mol/s tfinal =36; %seconds

5 C = ; CB = 5; CC = 5; CD = ; Cinit = [C CB CC CD]'; p(1)=k1;p()=km1;p(3)=k; [t,conc] = ode45(@(t,c)rate3(t,c,p),[ 36],Cinit); C = Conc(:,1);% first column CB = Conc(:,); % second column CC = Conc(:,3); %third column CD = Conc(:,4);%fourth column figure;set(gcf,'color','w'); p1 = plot(t,c,'-b*');hold on; p = plot(t,cb,'-rd'); p3 = plot(t,cc,'--k'); p4 = plot(t,cd,'gs'); legend('[]','[b]','[c]','[d]'); title('first part of question '); % variation of the problem. C = ; CB = ; CC = ; CD = ; Cinit = [C CB CC CD]'; [t,conc] = ode45(@(t,c)rate3(t,c,p),[ 36],Cinit);

6 C = Conc(:,1);% first column CB = Conc(:,); % second column CC = Conc(:,3); %third column CD = Conc(:,4);%fourth column figure;set(gcf,'color','w'); p1 = plot(t,c,'-b*');hold on; p = plot(t,cb,'-rd'); p3 = plot(t,cc,'--k'); p4 = plot(t,cd,'gs'); legend('[]','[b]','[c]','[d]'); title('second part - Initial conditions changed'); %% The following is stored in a separate file called rate3.m function dc = rate3(t,c,p) k1 = p(1); km1 = p(); k = p(3); C = C(1); CB = C(); CC = C(3); dc = zeros(4,1); r1 = k1 * C; rm1 = km1 * CB * CC; r = k * C * CB; r = -r1 + rm1 - r; rb = r1-rm1-r;

7 rc = r1-rm1; rd = r; dc(1) = r; dc()=rb; dc(3)=rc; dc(4)=rd; end ---- Results are 3. ns: (a) Second order reaction. PFR

8 df dx Fin r kc kcin 1 x dv dv dx..5 3 dv 1 dv 1 x x.8 3 Volume = 5 lit (b) Second order reaction, CSTR F r V F 5 kcv V 4 in out out 1 V V Volume = 31.5 lit Under isothermal operation, for a second order reaction, a CSTR needs higher volume than a PFR to yield the same conversion.

For a recycle reactor the relationship between the volume and other parameters is given by

For a recycle reactor the relationship between the volume and other parameters is given by 9 ug 7. CRE Tutorial Problem. or a recycle reactor the relationship between the volume and other parameters is given by V R in R r or simple kinetics such as first order reaction (under isothermal conditions),