new_ooi chel gee_005093 %28red2&3%29 (1)

8/2/2019 New_OOI CHEL GEE_005093 %28RED2&3%29 (1)

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THE UNIVERSITY OF NOTTINGHAM MALAYSIA

H83 RED Coursework

2010/2011Section 2 & 3 AnswerOoi Chel Gee (005093)

12/15/2010


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Reactor Design Coursework 2 & 3 OOI CHEL GEE (005093)

Question 2(a)

At 73% conversion, weight of catalyst needed is 0.0425 g cat per tube.

If the rate law is r A = kP A 2/3 P B 4/3 with k = 0.1K mol atm -2 g-1 cat s -1, it will increase the required weight of catalyst.Derived:

FAO x = k P A1/3 PB2/3 New:

FAO x( )* = k* P A2/3 PB4/3

= 0.1k (P A1/3 PB2/3 ) 2 (at gas phase, P=CRT)

= 0.1[k (P A1/3 PB2/3 ) (P A1/3 PB2/3 )]

= 0.1/k . [F AO x (dX/dW)] 2

( )* = 0.1 F A0/k . (dX/dW)2

(F A0=2; k = 0.0167)= 12 (dX/dW) 2

If the rate law is r A = kP A 2/3 P B 4/3 with k = 0.1K mol atm -2 g-1 cat s -1, it requires more catalyst.Function of dX/dW is smaller than 1, thus (dX/dW) 2 becomes even smaller. The multiplication of

integer will not change the answer significantly but also remain the answer to be smaller than 1.

Therefore, more catalyst is needed to reach the desired conversion. The conclusion is the weight of

catalyst increases.

Question 2(b)CA = 0.0015 mol/m 3 CB = 158.8778 mol/m 3 CC = 0.0002 mol/m 3

XA = 0.9768XB = 0.3889


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Question 3

Figure 3: Temperature profile along PFR

Figure 4: Flowrate profile along PFR


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Temperature profile along the reactor:The reaction is an exothermic reaction. Therefore along the reactor, the temperature has increased until asteady temperature around 0.6 dm 3 to 0.7 dm 3. The temperature 740K that reached is where the reactionreaches steady state and its maximum. Moreover, the reaction should be complete at that segment.Conclusively, volume of reactor 0.7 dm 3 is sufficient for the system reaction.

Flowrate profile along the reactor:Complete reaction is also shown to be done at 0.7 dm 3. A is fully utilised to produce B and C. The productionof B is more than C as stoichoimetry amount of A needed to produce B is lesser than that of C.

Change of reaction heat:Figure 4: Temperature Profile after Changing H

Figure 5: Flowrate Profile after Changing H

The reaction has shifted from exothermic reaction to endothermic reaction. The temperature of reaction has therefore reduced. In addition, productions of both B and C also reduce as bothreactions require heat output then input, somehow, the production of C is far lesser than that of Bas its reaction heat output is far more than B. Moreover, the first reaction is the only reactionundergoes thus C is only produced in very small amount. The reaction has to be complete and A canonly be fully converted with reactor with volume of 5000dm 3. It is too big and not cost effective forthe production of the desired product.


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Question 2a

A) Information given:

Customised data:i=0.3; j=1.2; K=0.0167; W(atm)=11; X=10.0; y(mm)=26; Z=0.41.

Reaction: 0.3A + 1.2B -> CA + 4B -> 10/3 CaA + bB -> cC@ T = 300C ; P o = W= 11 atm-r A = kP A1/3 PB2/3 ; k=0.0167mol/atm.g cat.sFAO = 100 x X = 100 x 10 = 1000 kmol/hr

Equipment:Total numbers of tubes = X x 50 = 500Db = y = 26mm = 0.026m

C = 1950 kg/m3

Inflow fluid characteristics: = Z = 0.41 0 = 6.4223 kg/m 3 (Perrys handbook) f = 0.1068 kg/m.hr (Perrys handbook) MwA = 30 g/mol; Mw B = 20 g/mol

B) Calculation:

Substances Initial Change RemainA F AO -XF AO FAO-XF AO FAO (1-X)B F BO -(b/a)XF AO FBO-(b/a)XF AO FAO ( -4X)C 0 +(c/a)XF AO (c/a)XF AO FAO (10/3)X Total F TO+(-1-b/a+c/a)X F TO+X

Let =F BO /F AO , =Change in total number of mole per mole of A reacted.Assume reactor is fed by stoichiometry mixture, thus, =b/a ; = -1-b/a+c/a

Molar flow rate for each tube:

FAtube = ; FBtube = b/a x F Atube

i) Design equation

FAO = -r A

ii) Rate Law

-r A = kP A1/3 PB2/3 (at gas phase, P=CRT)

= k(C ART) 1/3 (CBRT) 2/3 = kRT.C A1/3 .CB2/3


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iii) StoichiometryFor gas phase:

v = v o(1+X) (isothermal: T=T o; y=P/P o)= v o(1+X)/y

CA = F A /v (F A = F AO (1-X))

= ( )( ) = ( )( )

CB = FB /v

= ( )( ) = ( )( ) = ( )( )

iv) CombiningFAO x = kRT.C A1/3 .CB2/3

= kRT. ( ( )( ) ) 1/3 .( ( )( )) 2/3= (1/4) 2/3 . kRT ( )( ) = k( )( )

Where k= (1/ )2/3 . kRTCAO = (1/4) 2/3 . kPAO (P AO = C AORT)Hence:

= ( )( )

_____________________________________________(a)

Reactant A at the entrance into reactoryAO = F AO / F TO

= F AO / (F AO + F BO)= F AO / (F AO + (b/a)F AO)

= y AO . PAO = y AO .PO

Therefore, kis obtained.

v) Pressure drop

= (1+ )__________________________________________________(b)

= 2 (1 ) = (1 )


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Therefore, m AO = F AO x MwA ; m BO = F BO x MwB Total mass flowrate: m TO = m AO + m BO

Superficial velocity, G = m TO / A C

O can be obtained so as . vi) Combining with mole balance and considering the pressure drop equation, solve the

two differential equation (a) and (b)

Matlab codeM file 1: RedQuestion2a Calculation and graph plotting% Matlab coursework 2010/2011 % Question2aclc clear all

% Given value: i=0.3; j=1.2; K=0.0167; W=11.0; X=10.0; Y=26; Z=0.41; T=300; FaoTotal=100*X; Po=W*101.325; %kPa Db=Y/1000; Dp= 0.005; denc=1950; void=Z;

conv=0.73; k=0.0167; Mra=30; Mrb=20;

a=i; b=j; c=1; disp([num2str(a) 'A +' num2str(b) 'B ->' num2str(c) 'C' ]) disp([ '-rA=kPa^(1/3)Pb^(2/3)' ])

%Molar flowrate Tubes=X*50;Fao=FaoTotal/Tubes %mol/hr %Mole balance sita=j/i; yao=1/(1+sita); esp=yao*(-a/a-b/a+c/a) Pao=yao*Po; %kPa k1=k*Pao/101.325/(4^(2/3))*3600 %mol/g cat.hr disp([ '-rA=' num2str(k1) '(1-X)y/(1+sigma*X)' ]) %Superficial velocity mao=Fao*Mra/1000; %kg/hr mbo=sita*Fao*Mrb/1000; %kg/hr mTo=mao+mbo; %kg/hr Ac=pi()/4*(Db)^2; %m2; G=mTo/Ac; %kg/m2.hr %beta %at 300C, 11atm: den0=6.4223 ; %kg/m3 vis0=0.1068; %kg/m.h gc=12960000; %kg/h2.m.Pa beta=G*(1-void)/den0/gc/Dp/(void)^3*(150*(1-void)*vis0/Dp+1.75*G); %Pa/m %alpha alpha=2*beta/Ac/denc/(1-void)/Po/1000/1000 %1/g cat

%key value of k1,Fao,alpha,esp obtained into funtion2a.m


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wspan = [0 0.0425]; z0=[0;1]; [w,z] = ode15s(@function2a,wspan,z0); x=z(:,1); y=z(:,2); plot (w,x,w,y); xlabel( 'W,weight(g)' ); ylabel( 'X,y' );

M file 2: Funtion2a ode solver% "function2a" % %Inputs are: % w= weight % z= [z(1);z(2)]=[x;y] % Outputs is: % zdot=[k1*(1-z(1))/(1+esp*z(1))*z(2)/Fao;-alpha/2/(z(2))*(1+esp*(z(1)))] % z(1)=X % z(2)=y % zdot(1)=dX/dW % zdot(2)=dy/dW

function zdot=function2a(w,z)

k1=52.489; Fao=2; esp=-0.3333; alpha=6.5490e-008;

zdot(1,:)=(k1*(1-z(1))/(1+esp*z(1))*z(2)/Fao); zdot(2,:)=(-alpha/2/(z(2))*(1+esp*(z(1))));

Figure 1: Overall graph of X,Y versus weight of catalyst


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Figure 2: Overall graph of X,Y versus weight of catalyst (zoomed in)

At 73% conversion, weight of catalyst needed is 0.0425 g cat.

If the rate law is r A = kP A 2/3 P B 4/3 with k = 0.1K mol atm -2 g-1 cat s -1, it will increase the required weight of catalyst.Derived:

FAO x = k P A1/3 PB2/3 New:

FAO x( )* = k* P A2/3 PB4/3

= 0.1k (P A1/3 PB2/3 ) 2 (at gas phase, P=CRT)

= 0.1[k (P A1/3 PB2/3 ) (P A1/3 PB2/3 )]

= 0.1/k . [F AO x (dX/dW)] 2

( )* = 0.1 F A0/k . (dX/dW) 2 (F A0=2; k = 0.0167)

= 12 (dX/dW) 2

Function of dX/dW is smaller than 1, thus (dX/dW) 2 becomes even smaller. The multiplication will

not change the answer significantly but also remain the answer to be smaller than 1. Therefore,

more catalyst is needed to reach the desired conversion. The conclusion is the weight of catalyst

increases.


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Question 2b


Customised data:i=1.2; j=1.05; K=0.0167; X%=20.0; Y%=80; Z(atm)=40

Reaction: (in CSTR )1.2 A + B -> C + D ..(1) 1.05 C + B - > D + E .(2) @ T = 1500K; P o = Z= 40 atmrC = k 1 CACB0.5 ; k 1 =5 (m 3 mol -1 ) 0.5 hr -1 rE = k 2 CCCB0.5 ; k 2 =3 (m 3 mol -1 ) 0.5 hr -1

Equipment:

Volume, V = 200 ft 3 = 5.663369 m 3

Inflow fluid characteristics:Volumetric flowrate, v 0 = 0.1 ft/s = 10.1941 m 3 /hrFeed: 20% A, 80% B

B) Calculation:

(i) Mole balanceFor CSTR, assume steady state and no accumulation, thus mole balance: F io Fi + r i V =0

A FAO FA = -r A V CAO v0CA v0= -r A V CAOCA = -r A B FBO FB = -r B V CBO v0CB v0= -r B V CBOCB = -r B C FC FCO = r C V CC v0CC0 v0= r C V CCCC0 = r B

(ii) Relative ratesReaction (1): .= = = A: r1A = -1.2r 1C = -1.2 k 1CACB0.5

B: r1B = -r 1C = - k 1CACB0.5

D: r1D = r 1C = k 1CACB0.5

Reaction (2): .= = = C: r2C = -1.05r 2E = -1.05 k 2CCCB0.5 B: r2B = -r 2E = - k 2CCCB0.5 D: r2D = r 2E = k 2CCCB0.5

(iii) Net rate of reactionA: rA = r 1A = -1.2 k 1CACB0.5


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B: rB = r 1B + r 2B = - k 1CACB0.5 - k 2CCCB0.5 C: rC = r 1C + r 2C = k 1CACB0.5 -1.05 k 2CCCB0.5 D: rD = r 1D + r 2D = k 1CACB0.5 + k 2CCCB0.5 E: r E = r 2E = k 2CCCB0.5

(iv) Evaluating parametersCTO =

CAO = y AO CTO (isothermal and isobaric reaction)CBO = y BO CTO CCO = 0

Let = v0/V(v) Combine mole balances, net rate and concentrations

Combining Solving using Matlab

A yAO.CTO CA = 1.2 k 1CACB0.5 . CAO CA - 1.2 k 1CACB0.5 . = 0B yBO. CTO CB = (k 1CACB0.5 + k 2CCCB0.5) CBO CB - (k 1CACB0.5 + k 2CCCB0.5) =0 C CC = (k 1CACB0.5 -1.05 k 2CCCB0.5) - CC + (k 1CACB0.5 -1.05 k 2CCCB0.5) =0

(vi) ConversionXA = = XB = =

Matlab codeM file 1: RedQuestion2b Calculation% Matlab coursework 2010/2011 % Question2bclc clear all

%Given value: i=1.2; j=1.05; X=0.2; Y=0.8; Z=40; T0=1500; %KP0=Z; %atm R=8.205746e-5; %m3.atm/K/mol yao=X; ybo=Y; v0=10.1941; %m3/hrV=5.663369; %m3 k1=5; %(m3mol-1)^0.5 x hr-1 k2=3; %(m3mol-1)^0.5 x hr-1 a1=i; b1=1; c1=1; d1=1;c2=j;b2=1;d2=1;e2=1;

tau=V/v0; %hr CT0=P0/R/T0; %mol/m3 Cao=yao*CT0;


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Cbo=ybo*CT0;

syms Ca Cb Cc ; f1=Cao-Ca-(1.2*k1*Ca*Cb^0.5)*tau; f2=Cbo-Cb-(k1*Ca*Cb^0.5+k2*Cc*Cb^0.5)*tau; f3=-Cc+(k1*Ca*Cb^0.5-1.05*k2*Cc*Cb^0.5)*tau;

[Ca,Cb,Cc]=solve(f1,f2,f3); Ca=double(Ca) %mol/m3 Cb=double(Cb) %mol/m3 Cc=double(Cc) %mol/m3

Xa=(Cao-Ca)/Cao Xb=(Cbo-Cb)/Cbo

The calculation has given several answers for each parameter, upon filtering, the appropriateanswer is listed as below:CA = 0.0015 mol/m 3 CB = 158.8778 mol/m 3 CC = 0.0002 mol/m 3

XA = 0.9768XB = 0.3889


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Question 3


Customised data:i=1.2; j=1.0; K=1; X(C) = 150

Reaction: (in PFR )1.2 A -> 1.0 B ..(1) 1.0 A -> C .(2) @ T0 = X= 150 C ;r1A = -k 1A CA ; k 1A =10 exp * s -1 ; = 4000Kr2A = -k 2A CA2 ; k 2B =0.1 exp * s -1 ; = 8000K

= 20 000 (1) = 0 000 (2)

Inflow fluid characteristics:FA0 = 200 mol/s; C A0 = 0.2 mol/dm 3 ; T A = 90C CPA = 90 J/mol.K; C PB = 90 J/mol.K ; C PC = 180 J/mol.KU = 3000 J/m 3 .s.K

B) Calculation:

(i) Temperature profile along reactorEnergy Balance:

+ = At steady state no accumulation of energy.No shaft work.Therefore,

+ = 0

( ) + ( ) =0 CPi is constant:

+ ( ) =0

Design equation:

= Thus:


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( ) + ( ) =0 Arranging:

( ) = ( )+ ( )

Area is constant, hence:

( ) = ( ( )+ ( )) However, A is not given, thus equation (a) is used instead of (b).

= ( ) ( )+ ( )( + + ) (ii) Flowrate profile along reactor

For PFR, = :

= = =

(iii) Evaluating parameters

Rate Laws

Reaction (1): .= B: r1B = -1/1.2 x r 1A = -1/1.2 k 1ACA

Reaction (2): = A: r2C = -r 2A = -k 2ACA2

Net Rate

r A = r 1A + r 2A = - k 1A CA - k 2A CA2 r B = -r 1B = 1/1.2 k 1ACA r C = -r 2C = k 2ACA2

Concentrations

Isobaric and non-isothermal reaction (gas phase):

CA = () CB =

CC =

FT = + +


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Matlab codeM file 1: Function3% "function3" % Inputs are: % v= volume %

% Outputs is: % z(1)=Fa, z(2)=Fb, z(3)=Fc, z(4)=T; % zdot(1)=dFa/dv;zdot(2)=dFb/dv;zdot(3)=dFc/dv;zdot(4)=dT/dv;

function zdot=function3(v,z)

FT=z(1)+z(2)+z(3); k1a=10*exp(4000*(1/300-1/z(4))); %1/s k2a=0.1*exp(8000*(1/300-1/z(4))); %1/s CT0= 0.2; % mol/dm3 T0=150+273.15; Ta=90+273.15; %deg C Ca=CT0*(z(1)/FT)*(T0/z(4)); %J/mol/K Cb=CT0*(z(2)/FT)*(T0/z(4)); %J/mol/K Cc=CT0*(z(3)/FT)*(T0/z(4)); %J/mol/K U=3000/1000; %J/dm3/s/K r1a=-k1a*Ca; r2a=-k2a*Ca^2; %Reaction Rate: zdot(1,:)=r1a+r2a; zdot(2,:)=-1/1.2*r1a; zdot(3,:)=-r2a; zdot(4,:)=(20000*(-(r1a))+60000*(-(r2a))+U*(Ta-z(4)))/(z(1)*90+z(2)*90+z(3)*180);

M file 2: RedQuestion3 (Temperature)

% Matlab coursework 2010/2011 % Question3 (Temperature Profile) clc clear all

%Temperature Profile vspan = [0 1]; tspan=[200;0;0;423.15]; [v,z] = ode45(@function3,vspan,tspan); T=z(:,4);plot(v,T);title( 'Temperature profile' ); xlabel( 'Volume,v(dm3)' ); ylabel( 'Temperature,T(K)' ); M file 3: RedQuestion3 (Flowrate) % Matlab coursework 2010/2011

% Question3 (Temperature Profile) clc clear all

%Flowrate Profile vspan = [0 1]; fspan=[200;0;0;423.15]; [v,z] = ode45(@function3,vspan,fspan); F=z(:,1:3) plot(v,F); title( 'Flowrate profile' ); xlabel( 'Volume,v(dm3)' ); ylabel( 'Flowrate,F (mol/s)' );


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Figure 3: Temperature profile along PFR

Figure 4: Flowrate profile along PFR


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After changing the reaction rate:

= 20 000 = 0 000 Figure 4: Temperature Profile after Changing H

Figure 5: Flowrate Profile after Changing H

Reference:

Pratap, Rudra (2006), Getting Started with Matlab 7: A Quick Introduction for Scientists andEngineers. New York: Oxford. p.144-147.

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