cad lab-ii

8/13/2019 CAD LAB-II

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SOLUTION FOR THE GIVEN COLLECTION OF

TEN NUMERICAL PROBLEMS IN

CHEMICAL ENGINEERING

COMPUTER AIDEd DESIGN LAB-II

SUBMITTED BY

SUSANTA SETHI

ROLL NO. : 110CH0109

UNDER THE GUIDANCE OF

DR. ARVIND KUMAR

ASST. PROF. DEPARTMENT OF CHEMICAL ENGINEERING


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PROBLEM NO. 1 :

The ideal gas law can represent the pressure-volume-temperature (PVT) relationship of gases only atlow (near atmospheric) pressures. For higher pressures more complex equations of state should be used.

The calculation of the molar volume and the compressibility factor using complex equations of statetypically requires a numerical solution when the pressure and temperature are specified. The van derWaals equation of state is given by;

SOLUTION:C++ PROGRAMMING CODE:

//bisection method for finding the root of a function//

//roll no. 110CH0109//

#include

#include

#include

using namespace std;


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int main()

{

double v,v0,b,R,T,p,a,Tc,Pc;

double Tr,Pr,Z;

coutp;

coutT;

coutR;

coutTr;

coutPr;

v0=(R*T)/p;

Tc=T/Tr;

Pc=p/Pr;

cout


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cout


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enter the gas constant value in atm-lit./gmol-K :0.08

enter the given reduced temperature :1

enter the given reduced pressuren :1

the critical temperature was found to be :450

the critical pressure was found to be :56

the calculated constant a is :10.2727

the calculated constant b is :0.0824263

the calculated molar volume is :0.546188

the compressibility factor is given by :0.828297

PROBLEM-2


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C++ PROGRAMMING CODE:-

/************* material balance using Gauss elimination for solving linear equations *************/

#include

#include

#include


int main()

{

int n,i,j,k,temp;

float a[10][10],c,d[10]={0};

float D1,D2,B1,B2;

float D,B;

coutn;

cout


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cin>>a[i][j];

}

for(i=n-1;i>0;i--) // partial pivoting//

{

if(a[i-1][0]


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for(i=k;i


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d[i]=(a[i][n]-c)/a[i][i];

}

//******** RESULT DISPLAY *********//

for(i=0;i


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equation: 1 0.07 0.18 0.15 0.24 10.50

equation: 2 0.04 0.24 0.10 0.65 17.50

equation: 3 0.54 0.42 0.54 0.10 28

equation: 4 0.35 0.16 0.21 0.01 14

the stream D1 in kmol is : 26.25

the stream B1 in kmol is : 17.5

the stream D2 in kmol is: 8.75

the stream B2 in kmol is : 17.5

the stream D flow rate is : 43.75

the stream B flow rate is : 26.25

PROBLEM-3


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C++ PROGRAMMING CODE :

/************* calculation of terminal velocity of faliing coal particles *************/

#include

#include

#include


int main()

{

float Vt,d,dp,g,Dp,u,CD,Rep,V,R;

coutDp;

coutdp;

coutd;

coutu;

coutg;

Rep = ((Dp*Dp*Dp*g*d)* (dp-d))/(18*u*u);

cout


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if (Rep


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getch();

return 0;

}

RESULT:

Enter the coal particle dia Dp in meter unit : 0.000298

Enter the coal particle density in kg/m3 unit : 1800

Enter the fluid density in kg/m3 unit : 994.6

Enter the viscosity of fluid in kg/m.s unit : 0.0008931

Enter the acceleration due to gravity in m/s2 unit : 9.81

The calculated Reynolds No. is : 14.4846

The calculated drag coefficient value is : 3.16378

The calculated terminal velocity in m/s unit is : 0.0315857

PROBLEM NO. 4:


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SOLUTION :

POLYMATH SIMULATION CODE

Model:Pressure = a0 + a1*Temperature + a2*Temperature^2 + a3*Temperature^3 + a4*Temperature^4

Variable Value 95 confidence

a0 24.67876 0.7872334

a1 1.606196 0.0544632

a2 0.0360443 0.0010089

a3 0.0004131 4.005E-05

a4 3.963E-06 4.514E-07

GeneralDegree of polynomial = 4Number of observations = 10Statistics

R^2 0.9999963

R^2adj 0.9999934

Rmsd 0.1410532

Variance 0.3979203

Source data points and calculated data points

Temperature Pressure Pressure calc Delta Pressure

1 -36.7 1 1.047712 -0.04771252 -19.6 5 4.518359 0.481641

3 -11.5 10 10.41537 -0.4153728

4 -2.6 20 20.73923 -0.7392276

5 7.6 40 39.16234 0.8376636


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6 15.4 60 59.69417 0.305826

7 26.1 100 100.3384 -0.3384191

8 42.2 200 200.2647 -0.2647227

9 60.6 400 399.7679 0.232079310 80.1 760 760.0518 -0.0517552

4(a)

Showing 4th degree polynomial curve fitting

4(b)

POLYMATH Results for Regression of Clausius-Clapeyron Equation.


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4(c)

POLYMATH Results for Nonlinear Regression of Antoine Equation.

PROBLEM NO . 5 :


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SOLUTION:


f (CX) =(CX*CY )-(KC2*CB*CC) CX(0) =10 #nonlinear equaiton in terms of CX f (CZ) =CZ-(KC3*CA*CX) CZ(0) =10 #nonlinear equation in terms of CZ f (CD) =(CC*CD)-(KC1*CA*CB) CD(0) =10 #nonlinear equation in terms of CD CB =CB0-CD-CY # linear equation in terms of CB CA =CA0-CD-CZ # linear equation in terms of CA CY =CX+CZ # linear equation in terms of CYCC=CD-CY # KC1=1.06KC2=2.63KC3=5CA0=1.5CB0=1.5#for condition 1 that is CD=CX=CZ=0#for condition 2 that is CD=CX=CZ=1#for condition 3 that is CD=CX=CZ=10RESULT:Condition : 1CD=CX=CZ=0

Calculated values of NLE variables

Variable Value f(x) Initial Guess

1 CD 0.7053344 3.577E-13 0

2 CX 0.1777924 3.588E-13 0

3 CZ 0.3739766 -2.287E-13 0

Variable Value

1 CA 0.420689

2 CA0 1.5

3 CB 0.2428966

4 CB0 1.5


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5 CC 0.1535654

6 CY 0.551769

7 KC1 1.06

8 KC2 2.639 KC3 5.

Nonlinear equations

1 f(CX) = (CX*CY)-(KC2*CB*CC) = 0

nonlinear equaiton in terms of CX

2 f(CZ) = CZ-(KC3*CA*CX) = 0

nonlinear equation in terms of CZ

3 f(CD) = (CC*CD)-(KC1*CA*CB) = 0

nonlinear equation in terms of CD

Explicit equations

1 CB0 = 1.5

2 CA0 = 1.5

3 CY = CX+CZ

linear equation in terms of CY

4 CC = CD-CY

5 KC1 = 1.06

6 KC2 = 2.63

7 KC3 = 5

8 CA = CA0-CD-CZ

linear equation in terms of CA

9 CB = CB0-CD-CY

linear equation in terms of CB

Condition : 2CD=CX=CZ=1Calculated values of NLE variables


1 CD 0.0555561 6.523E-08 1.

2 CX 0.5972196 8.249E-08 1.


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3 CZ 1.082074 -2.602E-08 1.

Variable Value

1 CA 0.3623704

2 CA0 1.5

3 CB -0.2348492

4 CB0 1.5

5 CC -1.623737

6 CY 1.679293

7 KC1 1.06

8 KC2 2.63

9 KC3 5.

Nonlinear equations

1 f(CX) = (CX*CY)-(KC2*CB*CC) = 0


2 f(CZ) = CZ-(KC3*CA*CX) = 0


3 f(CD) = (CC*CD)-(KC1*CA*CB) = 0nonlinear equation in terms of CD

Explicit equations

1 CB0 = 1.5

2 CA0 = 1.5

3 CY = CX+CZ


4 CC = CD-CY

5 KC1 = 1.06

6 KC2 = 2.63

7 KC3 = 5

8 CA = CA0-CD-CZ



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9 CB = CB0-CD-CY


Condition 3:CD=CX=CZ=10Calculated values of NLE variables


1 CD 1.070104 1.574E-09 10.

2 CX -0.3227156 8.159E-09 10.

3 CZ 1.130534 3.941E-09 10.

Variable Value

1 CA -0.7006376

2 CA0 1.53 CB -0.377922

4 CB0 1.5

5 CC 0.2622862

6 CY 0.8078179

7 KC1 1.06

8 KC2 2.63

9 KC3 5.Nonlinear equations

1 f(CX) = (CX*CY)-(KC2*CB*CC) = 0


2 f(CZ) = CZ-(KC3*CA*CX) = 0


3 f(CD) = (CC*CD)-(KC1*CA*CB) = 0

nonlinear equation in terms of CD

Explicit equations

1 CB0 = 1.5

2 CA0 = 1.5

3 CY = CX+CZ



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4 CC = CD-CY

5 KC1 = 1.06

6 KC2 = 2.63

7 KC3 = 58 CA = CA0-CD-CZ


9 CB = CB0-CD-CY


General Settings

Total number of equations 12

Number of implicit equations 3

Number of explicit equations 9

Elapsed time 0.0000 sec

Solution method SAFENEWT

Max iterations 150

Tolerance F 0.0000001

Tolerance X 0.0000001

Tolerance min 0.0000001

PROBLEM 6:


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SOLUTION :


d(T3) / d(t) =(W*Cp*(T2-T3)+UA*(Ts-T3))/(M*Cp) #temperature distribution in tank-III T3(0) =20 d(T2) / d(t) =(W*Cp*(T1-T2)+UA*(Ts-T2))/(M*Cp) #Temperature distribution in tank-II T2(0) =20 d(T1) / d(t) =(W*Cp*(T0-T1)+UA*(Ts-T1))/(M*Cp) #Temperature distribution in tank-I

T1(0) =20#data given:W=100 # in kg per minuteCp=2.0 T0=20UA=10Ts=250M=1000#given intial condition:

t(0)=0#given final value:t(f )=200

RESULT:Calculated values of DEQ variables

Variable Initial value Minimal value Maximal value Final value

1 Cp 2. 2. 2. 2.

2 M 1000. 1000. 1000. 1000.

3 t 0 0 200. 200.

4 T0 20. 20. 20. 20.

5 T1 20. 20. 30.95238 30.95238

6 T2 20. 20. 41.38322 41.38322

7 T3 20. 20. 51.31735 51.31735


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8 Ts 250. 250. 250. 250.

9 UA 10. 10. 10. 10.

10 W 100. 100. 100. 100.

Differential equations

1 d(T3)/d(t) = (W*Cp*(T2-T3)+UA*(Ts-T3))/(M*Cp)

temperature distribution in tank-III


Temperature distribution in tank-II


Temperature distribution in tank-I

Explicit equations

1 W = 100

2 Cp = 2.0

3 T0 = 20

4 UA = 10

5 Ts = 250

6 M = 1000

PROBLEM 7:


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SOLUTION :


d( y1) / d(z) =k *CA1/DAB #variation of y1 w.r.t. z y1(0) = -130.30 d(CA1) / d(z) = y1 #variation of CA1 w.r.t. z CA1(0) =0.2 d( y ) / d(z) =k *CA/DAB #variation of y w.r.t. z y (0) = -130 d(CA) / d(z) = y #variation of CA w.r.t. z CA(0) =0.2 err = y -0 err1 = y1-0 derr =(err1-err)/(0.0001* y0) CAanal =0.2*cosh ( L*(k /DAB)^0.5*(1-z/L) )/(cosh (L*(k /DAB)^0.5)) # variation of CA #initial given data :

k =0.001DAB=1.2E-9 z(0)=0 L=0.001delta=0.0001 y0=-130#final value:z(f )=0.001

Calculated values of DEQ variables Variable Initial value Minimal value Maximal value Final value

1 CA 0.2 0.1404279 0.2 0.1404606

2 CA1 0.2 0.1400944 0.2 0.1401171

3 CAanal 0.2 0.1382726 0.2 0.1382726


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4 DAB 1.2E-09 1.2E-09 1.2E-09 1.2E-09

5 delta 0.0001 0.0001 0.0001 0.0001

6 derr 23.07692 23.07692 33.37887 33.37887

7 err -130. -130. 2.764383 2.7643838 err1 -130.3 -130.3 2.330458 2.330458

9 k 0.001 0.001 0.001 0.001

10 L 0.001 0.001 0.001 0.001

11 y -130. -130. 2.764383 2.764383

12 y0 -130. -130. -130. -130.

13 y1 -130.3 -130.3 2.330458 2.330458

14 z 0 0 0.001 0.001


1 d(y1)/d(z) = k*CA1/DAB

variation of y1 w.r.t. z

2 d(CA1)/d(z) = y1

variation of CA1 w.r.t. z

3 d(y)/d(z) = k*CA/DAB

variation of y w.r.t. z

4 d(CA)/d(z) = y

variation of CA w.r.t. z

Explicit equations

1 err = y-0

2 err1 = y1-0

3 y0 = -130

4 L = 0.001

5 k = 0.0016 DAB = 1.2E-9

7 CAanal = 0.2*cosh( L*(k/DAB)^0.5*(1-z/L) )/(cosh(L*(k/DAB)^0.5))

variation of CA

8 delta = 0.0001


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9 derr = (err1-err)/(0.0001*y0)

PROBLEM 8:

SOLUTION:

POLYMATH SIMULATION CODE #f(Tbp)=xA*PA+xB*PB-760*1.2 #xA=0.6 #PA=10^(6.90565-1211.033/(Tbp+220.79))

#PB=10^(6.95464-1344.8/(219.482+Tbp)) #xB=1-xA #yA=xA*PA/(760*1.2) #yB=xB*PB/(760*1.2) #Search Range: #Tbp(min)=60 #Tbp(max)=120


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d(L) /d(x2) =L/((k2*x2)-x2) d(T)/d(x2)=Kc*errT0=95.59Kc=0.5e6

k2=10^(6.95464 -1344 .8/(T+219.482))/(760*1.2) x1=1-x2 k1=10^(6.90565 -1211 .033/(T+220.79))/(760*1.2) err=(1-k1*x1-k2*x2) #Initial Conditions: x2(0)=0.4L(0)=100T(0)=95.5851#Final Value: x2(f )=0.8RESULTS:Calculated values of DEQ variables


1 err -3.646E-07 -3.646E-07 7.747E-05 7.747E-05

2 k1 1.311644 1.311644 1.856602 1.856602

3 k2 0.5325348 0.5325348 0.7857526 0.7857526

4 Kc 5.0E+05 5.0E+05 5.0E+05 5.0E+05

5 L 100. 14.04555 100. 14.045556 T 95.5851 95.5851 108.5693 108.5693

7 T0 95.59 95.59 95.59 95.59

8 x1 0.6 0.2 0.6 0.2

9 x2 0.4 0.4 0.8 0.8


1 d(L)/d(x2) = L/((k2*x2)-x2)

2 d(T)/d(x2) = Kc*errExplicit equations

1 T0 = 95.59

2 Kc = 0.5e6

3 k2 = 10^(6.95464-1344.8/(T+219.482))/(760*1.2)

4 x1 = 1-x2


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5 k1 = 10^(6.90565-1211.033/(T+220.79))/(760*1.2)

6 err = (1-k1*x1-k2*x2)

PROBLEM 9:

SOLUTION:


d(T)/d(W)=(.8*(Ta-T)+rA*delH)/(CPA*FA0)


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d( y )/d(W)=-0.015*(1-.5*x)*(T/450)/(2* y ) d(x)/d(W)=-rA/FA0 CA=.271*(1-x)*(450/T)/(1-.5*x)* y CC=.271*.5*x*(450/T)/(1-.5*x)* y

Ta=500delH=-40000CPA=40FA0=5k =.5*exp ((41800 /8.314)*(1/450-1/T)) Kc=25000 *exp (delH/8.314*(1/450-1/T)) rA=-k *(CA 2-CC/Kc) #Initial Conditions: W(0)=0x(0)=0T(0)=450 y (0)=1W(f )=20Calculated values of DEQ variables


1 CA 0.271 0.0350883 0.271 0.0350883

2 CC 0 0 0.0510418 0.0462521

3 CPA 40. 40. 40. 40.

4 delH -4.0E+04 -4.0E+04 -4.0E+04 -4.0E+04

5 FA0 5. 5. 5. 5.

6 k 0.5 0.5 475.9749 448.6209

7 Kc 2.5E+04 35.28512 2.5E+04 37.34129

8 rA -0.0367205 -0.9690852 0.0033395 0.0033395

9 T 450. 450. 1165.41 1149.637

10 Ta 500. 500. 500. 500.

11 W 0 0 20. 20.

12 x 0 0 0.7271791 0.7249973

13 y 1. 0.766806 1. 0.766806


1 d(T)/d(W) = (.8*(Ta-T)+rA*delH)/(CPA*FA0)


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2 d(y)/d(W) = -0.015*(1-.5*x)*(T/450)/(2*y)

3 d(x)/d(W) = -rA/FA0

Explicit equations

1 CA = .271*(1-x)*(450/T)/(1-.5*x)*y

2 CC = .271*.5*x*(450/T)/(1-.5*x)*y

3 Ta = 500

4 delH = -40000

5 CPA = 40

6 FA0 = 5

7 k = .5*exp((41800/8.314)*(1/450-1/T))

8 Kc = 25000*exp(delH/8.314*(1/450-1/T))

9 rA = -k*(CA^2-CC/Kc)

(a) The requested plot for part (a) is shown in Figure given below ,where there is a rapid increase inconversion and temperature within the reactor at approximately the midpoint of the catalyst bed. Thebed pressure drop is enhanced by the increased temperature and reduced pressure even though thenumber of moles is decreasing.

(b) This rapid increase is due to the exothermic reaction rapidly accelerating due to the increasingtemperature even though the reactant concentration falling. Equilibrium is rapidly achieved afterthis hot spot is achieved with the temperature and conversion only reducing slightly due to the exter-nal heat transfer which tends to slightly cool the reactor as the reacting mixture continues toward thereactor exit.


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(c) The concentration profiles shown in Figure given below, reflect the net effects of reaction rate andchanges in temperature and pressure within the reactor.

Concentration Profiles in Catalytic Reactor

PROBLEM 10:


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SOLUTION :

POLYMATH SIMULATION CODE #Equations: d(T)/d(t)=(WC*(Ti-T)+q)/rhoVCp d(T0)/d(t)=(T-T0-(taud/2)*dTdt)*2/taud d(Tm)/d(t)=(T0-Tm)/taum d(errsum)/d(t)=Tr-Tm WC=500rhoVCp=4000

taud=1taum=5Tr=80Kc=0tauI=2step=if (t


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7 tauI = 2

8 step = if (t


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10 (c) Closed Loop Performance for Kc = 500The increase of a factor of 10 in the proportional gainfrom the baseline case gives the unstable result plotted in Figure PM-(11). This is clearly an undesirableresult.

Closed Loop Response to Step Down in Inlet Feed Temperature att = 10 min for Kc = 500.

10 (d) Closed Loop Performance for Only Proportional ControlThe removal of the integral controlaction gives the stable result plotted in Figure PM-(12). Note that there is offset from the set point. whenthe system returns to steady state operation. This is always the case for only proportional control,and the use of integral control allows the offset to be eliminated.

Closed Loop Response for only Proportional Control.


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(e) Closed Loop Performance with Limits on qThere are many times in control when limitsmust be established. In this example, the limits on q can be achieved by a POLYMATH if... then... else... statement which can be utilized as shown below:

POLYMATH CODE #Equations: d(T)/d(t)=(WC*(Ti-T)+qlim)/rhoVCp d(T0)/d(t)=(T-T0-(taud/2)*dTdt)*2/taud d(Tm)/d(t)=(T0-Tm)/taum d(errsum)/d(t)=Tr-Tm WC=500Ti=60rhoVCp=4000taud=1taum=5Kc=5000tauI=2step=if (t


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6 rhoVCp 4000. 4000. 4000. 4000.

7 step 0 0 1. 1.

8 T 80. 80. 97.47784 89.08812

9 t 0 0 200. 200.10 T0 80. 80. 98.4204 89.0909

11 taud 1. 1. 1. 1.

12 tauI 2. 2. 2. 2.

13 taum 5. 5. 5. 5.

14 Ti 60. 60. 60. 60.

15 Tm 80. 80. 92.13203 89.09319

16 Tr 80. 80. 90. 90.

17 WC 500. 500. 500. 500.


1 d(T)/d(t) = (WC*(Ti-T)+qlim)/rhoVCp

2 d(T0)/d(t) = (T-T0-(taud/2)*dTdt)*2/taud

3 d(Tm)/d(t) = (T0-Tm)/taum

4 d(errsum)/d(t) = Tr-Tm

Explicit equations

1 WC = 500

2 Ti = 60

3 rhoVCp = 4000

4 taud = 1

5 taum = 5

6 Kc = 5000

7 tauI = 2

8 step = if (t


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I)

Closed Loop Response for only Proportional Control.II)

Closed Loop Response for only Proportional Control.

cad lab-ii

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