compiled learning tasks

8/4/2019 Compiled Learning Tasks

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BIO176/A1

Learning TasksCOMPILATION

Submitted by:

Gomez, Marineil C.2008102730

Submitted to:Prof. Allan Soriano


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Learning Task 1

Syntax:

%Evaluate this string in the base work-

space

%Evaluate this string in the base work-

space

close all

clear

clc

%Header;disp(['Homework 1']);

disp(['Pressure-Volume Graph of Carbon

Tetrachloride using Peng-Robinson Equa-

tion of State']);

disp(['Gomez, Marineil C.']);

disp(['2008102730']);

%define variables;

syms Vp

syms P1p P2p P3p;

Pcp=45.6;%Pcp in bar;

Tcp=556.4;

%Tcp in K;

wp=0.194;

Rp=8.31451*10^(-5);

%Rp in bar m3/mol K;

Vcp=0.276*10^3;

%Vcp in m3/mol;

Tr1p=0.5;

Tr2p=1.0;Tr3p=1.5;

T1p=Tr1p*Tcp;

T2p=Tr2p*Tcp;

T3p=Tr3p*Tcp;

kp=0.37464+1.54226*wp-0.26992*(wp^2);

rAt1p=1+kp*(1-sqrt(Tr1p));



At1p=rAt1p^2;

At2p=rAt2p^2;

At3p=rAt3p^2;

a1p=0.45724*(Rp^2)*(Tcp^2)*(At1p/Pcp);

a2p=0.45724*(Rp^2)*(Tcp^2)*(At2p/Pcp);

a3p=0.45724*(Rp^2)*(Tcp^2)*(At3p/Pcp);

bp=0.0778*((Rp*Tcp)/Pcp);

Vp=0.01:0.001:300;

%substitute for equation;

P1p=PengRobinson1(Rp,T1p,Vp,a1p,bp);



%Graph of function;

G1p=semilogx(Vp,P1p,'b.');

xlabel('{\it V}, m^3/mol')

ylabel('{\it P}, bar')

title('{\it Pressure} vs{\it Volume} us-

ing Peng Robinson EOS')

text(1e-3,150,'Carbon Tetrachloride

(CCl4)','EdgeColor','green',...

'LineWidth',3,'FontSize',12, ...

'BackgroundColor',[.7 .9 .8])

axis ([0 400 -0.1 3]);

hold on

G2p=semilogx(Vp,P2p,'r.');

hold on

G3p=semilogx(Vp,P3p,'g.');

legend('Tr = 0.5','Tr = 1.0','Tr =

1.5',1)

hold on

Instructions: Plot Pvs Vfor carbon tetrachloride (CCl4) at temperatures Tr= 0.5, 1.0 and 1.5 based

on the Peng-Robinson Equation of State


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M-File:

function [ P1p ] = PengRobinson1(Rp,T1p,Vp,bp,a1p)%PENGROBINSON1 Summary of this function goes here% Detailed explanation goes hereP1p=((Rp*T1p)./(Vp-bp))-(a1p./(Vp.^2+2*Vp.*bp-bp^2));

End

function [ P2p ] = PengRobinson2( Rp,T2p,Vp,bp,a2p)%PENGROBINSON2 Summary of this function goes here% Detailed explanation goes hereP2p=((Rp*T2p)./(Vp-bp))-(a2p./(Vp.^2+2*Vp.*bp-bp^2));

end

function [ P3p ] = PengRobinson3( Rp,T3p,Vp,bp,a3p )%PENGROBINSON3 Summary of this function goes here% Detailed explanation goes hereP3p=((Rp*T3p)./(Vp-bp))-(a3p./(Vp.^2+2*Vp.*bp-bp^2));

end


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File Viewer


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Learning Task 2

Syntax:

%calculations;

%input parameters;

mudCCl4= 0; % dipole moment in debye, ppt

4 slide 60 ;

mudHCl=1.08; % dipole moment in debye.

ppt 4 slide 60 ;

muCCl4=mudCCl4*3.33569e-30; %convert de-

bye;muHCl=mudHCl*3.33569e-30; %convert debye;

alphaCCl4=10.5e-30; %polarizability in

m3;

alphaHCl=2.6e-30; %polarizability in m3;

iVCCl4=11; %ionization potential in eV;

iVHCl=12.8; %ionization potential in eV;

iCCl4=iVCCl4*1.60218e-19; %conv eV to J

iHCl=iVHCl*1.60218e-19; %conv eV to J

Eo=8.85419e-12; % dielectric permitivity

in vacuum in C^2/J-m;

T=373.15; % temperature (100oC) in K;

kboltz=1.3807e-23; % Boltzman constant in

J/K;

r=[5e-10 10e-10 50e-10 100e-10]; % dis-

tances in m;

disp(['dielectric permitivity in vacuum

(Eo) = ', num2str(Eo),' C^2/J-m']);

disp(['Boltzman Constant = ', num2str

(kboltz),' J/K']);

disp(['Temperature = ', num2str(T),'K']);

disp(['distances (r) = [5 10 50 100] Ang-

strom']);

disp(['For CCl4:']);disp([' dipole moment = ', num2str

(mudCCl4),' debye']);

disp([' polarizability = 10.5e-30 m3']);

disp([' first ionization potential = ',

num2str(iVCCl4),' eV']);

disp(['For HCl:']);

disp([' dipole moment = ', num2str

(mudHCl),' debye']);

disp([' polarizability = 2.6e-30 m3']);

disp([' first ionization potential = ',

num2str(iVHCl),' eV']);

%calculation using eq 4-8 for dipole-

dipole interactions;

E=4*pi*Eo;

k=(E^2)*kboltz*T;

x=-(2*(muCCl4^2)*(muHCl^2))/(3*k); %

collation of constants;

gpolar=x./(r.^6); %mean potential energy;

%calculation using eq 4-14 for induced

dipole interactions;

n=(alphaCCl4*(muHCl^2))+(alphaHCl*

(muCCl4^2));

ginduced=(-n/(E^2))./(r.^6);

%calculation using eq 4-18 for dispersion

interactios;

ion=(iHCl*iCCl4)/(iHCl+iCCl4);

y=(-3/2)*((alphaHCl*alphaCCl4)*ion)/

(E^2);

gdisperse=y./(r.^6);

%graphics diplay;

%table output;

gamma=cell(5,4);

gamma{1,1}='Distance';

gamma{1,2}='Polar/Polar';

gamma{1,3}='Polar/Induction';

gamma{1,4}='Dispersion';

gamma{2,1}='5 Angstrom';gamma{3,1}='10 Angstrom';

gamma{4,1}='50 Angstrom';

gamma{5,1}='100 Angstrom';

gamma{2,2}=num2str(gpolar(1));




Instructions: Calculate the intermolecular potentials: polar/polar, polar/induction, dispersion at 100C

and for distances 5, 10, 50 100 A for the pair-molecules CCl 4/HCl


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gamma{2,3}=num2str(ginduced(1));




gamma{2,4}=num2str(gdisperse(1));





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File Viewer


8/17

Learning Task 3

Syntax:

For Lennard-Jones Potential

%input data;

Na=6.02214e23; %Avogadros number in part/

mol;

sLJ=5.711e-8; %in cm;eKLJ=233.28; %in K;

%data from table 5-1 praustnitz 1999;

%calculations;

B1=((2*pi)/3)*Na*(sLJ^3); %from ppt 5

slide 76;

kTeLJ=0.3:0.01:5;

kToeLJ=kTeLJ(iLJ);

eokTLJ=1/kToeLJ;

FactorLJ=3*quadl(@(rLJ)((1-exp(-

4.*eokTLJ.*((rLJ).^(-12)-(rLJ).^(-6)))).*rLJ.^2),0.01,15);

integ1(iLJ)=FactorLJ;

integLJ=integ1(iLJ);

TLJ=kTeLJ.*eKLJ;

BLJ(iLJ)=B1*integLJ;

%Plot of Lennard Jones potential;

plot(TLJ,BLJ,'*b');

xlabel('{\it T},K');

ylabel('{\it BLJ},cm3/mol');

axis tight;

For Kihara Potential

%Data and constants;

Na= 6.022e23; % particles/mol;

as=0.470; % data from table 5-3 praust-

nitz 1999;

sK=4.611e-8; % cm;

eKK=501.89; % K;

%calculations from ppt 5 slide 78-80;

B2=((2*pi)/3)*Na*(sK^3);

kTeK=0.3:0.01:5;

kToeK=kTeK(iK);

eokTK=1/kToeK;

a=as/(1+as);

beg2a=a+0.0001;

FactorK=quadl(@(rK)((1-exp(-4.*eokTK.*

(((1-a)./(rK-a)).^(12)-((1-a)./(rK-a)).^6))).*rK.^2),beg2a,15);

integ2(iK)=3.*(a.^3+FactorK);

integK=integ2(iK);

TK=kTeK*eKK;

BK(iK)=B2*integK;

%Plot of Kihara potential;

plot(TK,BK,'*r');

xlabel('{\it T},K');ylabel('{\it BLJ},cm3/mol');

axis tight;

Instructions: Calculate the second virial coefficient B2 for temperatures kT/ = 0.3 to 5.0 for C3H8

from (a) Lennard-Jones Potential, (b) Kihara Potantial, (cO Pitzer-Tssonopoulos correlation. Plot B2

(cm3/mol) vs Temperature (K) for three different methods in a single figure


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For Pitzer-Tsonopoulos correlation

%Data and Constants;

Tc=369.8; % K;

Pc=42.5; % bar;

w=0.153; % accentric factor;

Gc=83.1451; % bar cm3/mol K;

TPT=0:10:2500;

tkt=TPT(iPT);

Tr=tkt./Tc;

Fo=0.1445-(0.33./Tr)-(0.1385./(Tr.^2))-

(0.0121./(Tr.^3))-(0.000607./(Tr.^8));

F1=0.0637+(0.331./(Tr.^2))-(0.423./

(Tr.^3))-(0.008./(Tr.^8));

factorPT=Fo+(w*F1);

integ3(iPT)=factorPT;

BPRT=integ3(iPT);

BPT(iPT)=BPRT*((Gc*Tc)/Pc);

%Plot of Pitzer-Tsonopoulos Correlation;

plot(TPT,BPT,'*g');

xlabel('{\it T},K');

ylabel('{\it BLJ},cm3/mol');

axis([0,2500,-7000,100]);

For Combined plot

%Plot of the three methods

plot

(TLJ,BLJ,'*b',TK,BK,'*r',TPT,BPT,'*g')

xlabel('{\it T}, K');

ylabel('{\it B2}, cm^3/mol');

legend('Lennard-Jones potential','Kihara

potential','Pitzer-Tsonopoulos poten',4)

axis([0,2550,-7000,1000]);


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File Viewer


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Quiz 2

%problem 1a;


Ta=449.5455; %K;

Pa=7.0909; %bar;

Na=6.022e23; %particles/mol;

Gc=83.1451e-6; %gas constant in bar m3/

mol K;

ya=[0.22 0.78]; %mole fraction compo-

nents;

s=[5.22e-10 5.711e-10]; %in m;

ek=[194.14 233.28]; %in K;

Bo=2*pi*Na;

b2i=[0 0];

for i=1:2;

ekT=ek(i)/Ta;

factora=quadl(@(ra)((1-exp(-4.*ekT.*

((ra).^(-12)-(ra).^(-

6)))).*ra.^2),0.01,15);

BLJi(i)=Bo*s(i)^3*factora;

factora=quadl(@(ra)((1-exp(-4.*ekT.*

((ra).^(-12)-(ra).^(-

6)))).*ra.^2),0.01,15);

BLJi(i)=Bo*s(i)^3*factora;

end

%output answers;

disp(['Answers']);

disp(['B of C2H6: ',num2str(BLJi(1)),'

m3/mol']);

disp(['B of C3H8: ',num2str(BLJi(2)),'

m3/mol']);

%problem 1b;

%legend;

%C2H6(1)-C2H6(1)=j(1);

%C2H6(1)-C3H8(2)=j(2);

%C3H8(2)-C3H8(2)=j(3);


Tb=449.5455; %K;

Pb=7.0909; %bar;



mol K;

yb=[0.22 0.78]; %mole fraction compo-

nents;

s=[5.22e-10 5.711e-10]; %in m;

ek=[194.14 233.28]; %in K;

Bo=2*pi*Na;

sb=[0 0 0];

sb(1)=0.5*(s(1)+s(1));

sb(2)=0.5*(s(1)+s(2));

sb(3)=0.5*(s(2)+s(2));

ekb=[0 0 0];

ekb(1)=(ek(1)*ek(1))^0.5;ekb(2)=(ek(1)*ek(2))^0.5;

ekb(3)=(ek(2)*ek(2))^0.5;

Bm=[0 0 0];

sb;

ekb;

for j=1:3

ekTm=ekb(j)/Tb;

factorb=quadl(@(rb)((1-exp(-4.*ekTm.*

((rb).^(-12)-(rb).^(-

6)))).*rb.^2),0.01,15);

Bm(j)=Bo*sb(j)^3*factorb;

ekTm=ekb(j)/Tb;


((rb).^(-12)-(rb).^(-

6)))).*rb.^2),0.01,15);

Bm(j)=Bo*sb(j)^3*factorb;

ekTm=ekb(j)/Tb;


((rb).^(-12)-(rb).^(-

6)))).*rb.^2),0.01,15);Bm(j)=Bo*sb(j)^3*factorb;

end

Bm;

%Bmix calculations;

Bmix1=(yb(1)^2)*Bm(1);

Bmix2=2*yb(1)*yb(2)*Bm(2);

Bmix3=(yb(2)^2)*Bm(3);


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Bmix=Bmix1+Bmix2+Bmix3;

%output answers;

disp(['Answers']);

disp(['B(mixt) of C2H6-C2H6: ',num2str(Bm

(1)),' m3/mol']);


(2)),' m3/mol']);


(3)),' m3/mol']);

disp(['B(mix) = ',num2str(Bmix),' m3/

mol']);

%Problem 1c;

%using previous data and parameters;

Tc=449.5455; %K;

Pc=7.0909; %bar;



mol K;

yc=[0.22 0.78]; %mole fraction compo-

nents;

%calculations;

lnfc1=(Pc/(Gc*Tc))*(2*((yc(1)*Bm(1))+(yc

(1)*Bm(2)))-Bmix);

lnfc2=(Pc/(Gc*Tc))*(2*((yc(2)*Bm(2))+(yc

(2)*Bm(3)))-Bmix);

fc1=exp(lnfc1);

fc2=exp(lnfc2);

%output answers;

disp(['Answers']);

disp(['Fugacity Coefficient of C2H6:

',num2str(fc1),' ']);

disp(['Fugacity Coefficient of C3H8:

',num2str(fc2),' ']);

% problem 2;

%input data;

%data [T(K) B(m3/gmol)];

data=[330 -9.7707

370 -7.3893

410 -5.6488

450 -4.3256

490 -3.2880

530 -2.4538

570 -1.7695

610 -1.1984

650 -0.71487

690 -0.30045

730 0.058557];

Gc=83.1451e-6;

Bact=data(:,2);

Tact=data(:,1);

unk0=[374 0.1 16]; %[Tc w Pc] in order,

initial guess;

[unkhat resid J]=nlinfit

(Tact,Bact,pct,unk0);

unkhat;

Bcal=pct(unkhat,Tact);

disp(['Critical Parameters']);

disp([' Critical Temperature: ',num2str

(unkhat(1),' K']);

disp([' Critical Pressure: ',num2str

(unkhat(3),' bar ']);

disp([' Accentric Factor: ',num2str

(unkhat(2),' ']);

PCT M-file

function [ Bhat ] = pct(unk,Tact)%PCT Summary of this function goes here% Detailed explanation goes here

Gc=83.1451e-6;

F0=0.1445-(0.330./(Tact./unk(1)))-

(0.1385./(Tact./unk(1)).^2)-(0.0121./

(Tact./unk(1))^3)-(0.000607./(Tact./unk

(1)).^8);F1=0.0637+(0.331./(Tact./unk(1)).^2)-

(0.423./(Tact./unk(1)).^3)-(0.008./

(Tact./unk(1)).^8);Bfunct=F0+unk(2).*F1;Bhat=(Gc*unk(1).*Bfunct)./unk(3);end


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