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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));
rAt2p=1+kp*(1-sqrt(Tr2p));
rAt3p=1+kp*(1-sqrt(Tr3p));
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);
P2p=PengRobinson2(Rp,T2p,Vp,a2p,bp);
P3p=PengRobinson3(Rp,T3p,Vp,a3p,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));
gamma{3,2}=num2str(gpolar(2));
gamma{4,2}=num2str(gpolar(3));
gamma{5,2}=num2str(gpolar(4));
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{3,3}=num2str(ginduced(2));
gamma{4,3}=num2str(ginduced(3));
gamma{5,3}=num2str(ginduced(4));
gamma{2,4}=num2str(gdisperse(1));
gamma{3,4}=num2str(gdisperse(2));
gamma{4,4}=num2str(gdisperse(3));
gamma{5,4}=num2str(gdisperse(4));
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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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Quiz 2
%problem 1a;
%Data and Constants;
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);
%Data and Constants;
Tb=449.5455; %K;
Pb=7.0909; %bar;
Na=6.022e23; %particles/mol;
Gc=83.1451e-6; %gas constant in bar m3/
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;
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;
factorb=quadl(@(rb)((1-exp(-4.*ekTm.*
((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']);
disp(['B(mixt) of C2H6-C3H8: ',num2str(Bm
(2)),' m3/mol']);
disp(['B(mixt) of C3H8-C3H8: ',num2str(Bm
(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;
Na=6.022e23; %particles/mol;
Gc=83.1451e-6; %gas constant in bar m3/
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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