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diff --git a/611/CH1/EX1.1/Chap1_Ex1_R1.sce b/611/CH1/EX1.1/Chap1_Ex1_R1.sce new file mode 100755 index 000000000..7a62b1f73 --- /dev/null +++ b/611/CH1/EX1.1/Chap1_Ex1_R1.sce @@ -0,0 +1,21 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-1,Example 1,Page 6
+//Title:Weight of payload
+//================================================================================================================
+clear
+clc
+
+//INPUT
+weight=981; //weight of payload in N
+gmoon=1.62;//acceleration due to gravity on the moon in m/s^2
+g=9.81;//acceleration due to gravity on earth
+
+//CALCULATION
+mass=weight/g;//calculation of mass of the payload in kg (calculated as F=m*g)
+weightmoon=mass*gmoon;//calculation of weight of payload on the moon in N
+
+//OUTPUT
+mprintf('\n The weight of payload on the moon= %d N',weightmoon);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH1/EX1.2/Chap1_Ex2.sce b/611/CH1/EX1.2/Chap1_Ex2.sce new file mode 100755 index 000000000..5e1bc8a3e --- /dev/null +++ b/611/CH1/EX1.2/Chap1_Ex2.sce @@ -0,0 +1,21 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-1,Example 2,Page 7
+//Title:Force due to atmospheric air
+//================================================================================================================
+clear
+clc
+
+//INPUT
+l=15;//length of the child's head in cm
+b=12;//breadth of the child's head in cm
+p=101325;//atmospheric pressure in Pa
+
+//CALCULATION
+area=(l*b)/(10^4);//calculation of area of the child's head in m^2
+force=p*area;//calculation of force exerted on the child's head due to atmospheric air in N
+
+//OUTPUT
+mprintf('\n The force exerted on the childs head due to atmospheric air= %f N',force);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH1/EX1.3/Chap1_Ex3.sce b/611/CH1/EX1.3/Chap1_Ex3.sce new file mode 100755 index 000000000..c31291850 --- /dev/null +++ b/611/CH1/EX1.3/Chap1_Ex3.sce @@ -0,0 +1,22 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-1,Example 3,Page 7
+//Title:Pressure drop
+//================================================================================================================
+clear
+clc
+
+//INPUT
+rho_water=1000;//density of water flowing through the pipeline in kg/m^3
+rho_manomtr=1595;//density of manometric fluid (carbon tetrachloride) in kg/m^3
+l=40;//length between the selected sections in cm
+theta=45;//inclination of the manometer in degrees
+g=9.81;//acceleration due to gravity in m/s^2
+
+//CALCULATION
+delp=(l/100)*sin((theta*%pi)/180)*g*(rho_manomtr-rho_water);//calculation of pressure drop between the required sections in Pa
+
+//OUTPUT
+mprintf('\n The pressure drop between the required sections= %f Pa',delp);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH10/EX10.2/Chap10_Ex2_R1.sce b/611/CH10/EX10.2/Chap10_Ex2_R1.sce new file mode 100755 index 000000000..336fdc6ba --- /dev/null +++ b/611/CH10/EX10.2/Chap10_Ex2_R1.sce @@ -0,0 +1,22 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-10,Example 2,Page 369
+//Title:Number of degrees of freedom
+//================================================================================================================
+clear
+clc
+
+//INPUT
+P=2;//number of phases (no unit)
+C=2;//number of components (no unit)
+
+//CALCULATION
+F=C+2-P;//calculation of the number of degrees of freedom using Eq.(10.35) (no unit)
+
+//As the number of degrees of freedom is 2, two intensive properties of the system are to be specified to describe the thermodynamic state of the system.Therefore, the fundamental relation in the Gibbs free energy representation for this system is of the type G=G(T,P,N1,N2)
+
+//OUTPUT
+mprintf("\n The number of degrees of freedom = %d \n",F);
+mprintf("Two intensive properties are required to be specified to describe the thermodynamic state of the system,and \nthe fundamental relation in the Gibbs free energy representation for this system is of the type, G=G(T,P,N1,N2)");
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH10/EX10.3/Chap10_Ex3_R1.sce b/611/CH10/EX10.3/Chap10_Ex3_R1.sce new file mode 100755 index 000000000..4e0e99e8e --- /dev/null +++ b/611/CH10/EX10.3/Chap10_Ex3_R1.sce @@ -0,0 +1,123 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-10,Example 3,Page 370
+//Title:Vapour Pressure of n-octane using the Peng-Robinson equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+R=8.314;//universal gas constant in J/molK
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+
+//CALCULATION
+Pguess=0.215;//taking a guess value of vapour pressure (P) to check the criterion of equilibrium given by Eq.(10.37) in MPa
+//Using the Cardans method to calculate Z_l and Z_v
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=(Pguess*10^6)/(Pc*10^5);//calculation of reduced pressure (no unit)
+S=0.37464+(1.54226*w)-(0.26992*w^2);//calculation of S using Eq.(3.79)
+alpha1=(1+(S*(1-sqrt(Tr))))^2;//calculation of alpha1 using Eq.(3.78)
+a=(0.45724*R^2*Tc^2*alpha1)/(Pc*10^5);//calculation of the Peng-Robinson constant in (m^6 Pa mol^-2) using Eq.(3.76)
+b=(0.07780*R*Tc)/(Pc*10^5);//calculation of the Peng-Robinson constant in m^3/mol using Eq.(3.77)
+A=(a*Pguess*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*Pguess*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1+B;//calculation of alpha for Peng-Robinson equation of state using Table (3.2)
+beeta=A-(2*B)-(3*B^2);//calculation of beeta for Peng-Robinson equation of state using Table (3.2)
+gaamma=-(A*B)+(B^2)+(B^3);//calculation of gaamma for Peng-Robinson equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ Z_l=Z;
+ Z_v=Z;
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ Z_l=min(Z);
+ Z_v=max(Z);
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ Z_l=min(Z);
+ Z_v=max(Z);
+
+ end
+end
+//calculation of fugacity coefficient for the liquid using Eq.(9.48) (no unit)
+phi_l=exp (Z_l-1-log(Z_l-B)-((a/(2*sqrt(2)*b*R*T))*log((Z_l+(B*(1+sqrt(2))))/(Z_l+(B*(1-sqrt(2)))))));
+//calculation of fugacity coefficient for the vapour using Eq.(9.48) (no unit)
+phi_v=exp (Z_v-1-log(Z_v-B)-((a/(2*sqrt(2)*b*R*T))*log((Z_v+(B*(1+sqrt(2))))/(Z_v+(B*(1-sqrt(2)))))));
+fl=Pguess*phi_l;//calculation of the fugacity of the liquid in MPa
+fv=Pguess*phi_v;//calculation of the fugacity of the vapour in MPa
+tolerance=1e-6;//defining the tolerance to compare fl and fv
+if abs(fl-fv)<tolerance then
+ P=Pguess;//the vapour pressure (in MPa) is taken as the guess value as the criterion of equilibrium given by Eq.(10.37) is established
+else
+ Prevised=Pguess*(fl/fv);//calculation of the revised value of P to check for the criterion of equilibrium given by Eq.(10.37) in MPa, if it fails for Pguess
+while abs(fl-fv)>tolerance
+ //Using the Cardans method to calculate Z_l and Z_v
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=(Prevised*10^6)/(Pc*10^5);//calculation of reduced pressure (no unit)
+S=0.37464+(1.54226*w)-(0.26992*w^2);//calculation of S using Eq.(3.79)
+alpha1=(1+(S*(1-sqrt(Tr))))^2;//calculation of alpha1 using Eq.(3.78)
+a=(0.45724*R^2*Tc^2*alpha1)/(Pc*10^5);//calculation of the Peng-Robinson constant in m^6*Pa*mol^-2 using Eq.(3.76)
+b=(0.07780*R*Tc)/(Pc*10^5);//calculation of the Peng-Robinson constant in m^3/mol using Eq.(3.77)
+A=(a*Prevised*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*Prevised*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1+B;//calculation of alpha for Peng-Robinson equation of state using Table (3.2)
+beeta=A-(2*B)-(3*B^2);//calculation of beeta for Peng-Robinson equation of state using Table (3.2)
+gaamma=-(A*B)+(B^2)+(B^3);//calculation of gaamma for Peng-Robinson equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ Z_l=Z;
+ Z_v=Z;
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ Z_l=min(Z);
+ Z_v=max(Z);
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ Z_l=min(Z);
+ Z_v=max(Z);
+
+ end
+end
+//calculation of fugacity coefficient for the liquid using Eq.(9.48) (no unit)
+phi_l=exp (Z_l-1-log(Z_l-B)-((a/(2*sqrt(2)*b*R*T))*log((Z_l+(B*(1+sqrt(2))))/(Z_l+(B*(1-sqrt(2)))))));
+//calculation of fugacity coefficient for the vapour using Eq.(9.48) (no unit)
+phi_v=exp (Z_v-1-log(Z_v-B)-((a/(2*sqrt(2)*b*R*T))*log((Z_v+(B*(1+sqrt(2))))/(Z_v+(B*(1-sqrt(2)))))));
+fl=Prevised*phi_l;//calculation of the fugacity of the liquid in MPa
+fv=Prevised*phi_v;//calculation of the fugacity of the vapour in MPa
+Prevised=Prevised*fl/fv;//updating the value of Prevised for the next iteration
+end
+P=Prevised;//the vapour pressure (in MPa) is taken as the revised value as the criterion of equilibrium given by Eq.(10.37) is established
+end
+
+//OUTPUT
+mprintf("\n The vapour pressure of n-octane at 427.85K = %f MPa\n",P);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH11/EX11.1/Chap11_Ex1.sce b/611/CH11/EX11.1/Chap11_Ex1.sce new file mode 100755 index 000000000..642fa7d41 --- /dev/null +++ b/611/CH11/EX11.1/Chap11_Ex1.sce @@ -0,0 +1,98 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 1,Page 378
+//Title:P-x-y and T-x-y diagram for a Benzene Toluene system
+//================================================================================================================
+clear
+clc
+
+//INPUT
+antoine_const_benzene=[6.87987;1196.760;219.161];//Antoine's constants for Benzene from Table A.7
+antoine_const_toluene=[6.95087;1342.310;219.187];//Antoine's constants for Toluene from Table A.7
+t=95;//temperature at which the P-x-y diagram has to be prepared in degree celsius
+P=101.325;//pressure at which the T-x-y diagram has to be prepared in kPa
+
+//CALCULATION
+//P-x-y diagram:
+//For convenience Benzene is denoted as 1 and Toluene as 2
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+
+P1_s=10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(t+antoine_const_benzene(3,:))));//calculation of saturation pressure of benzene at t in Torr
+P2_s=10^(antoine_const_toluene(1,:)-(antoine_const_toluene(2,:)/(t+antoine_const_toluene(3,:))));//calculation of saturation pressure of toluene at t in Torr
+x1=0:0.1:1;//mole fraction of benzene in the liquid phase (no unit)
+i=1;//iteration parameter
+n=length(x1);//iteration parameter
+while i<n |i==n
+ P_tot(i)=P2_s+((P1_s-P2_s)*x1(:,i));//calculation of the total pressure using Eq.B, Page 379 in Torr
+ y1(i)=(x1(:,i)*P1_s)/(P_tot(i));//calculation of the mole fraction of Benzene in the vapour phase (no unit)
+ i=i+1;
+end
+
+//T-x-y diagram:
+P=760;//converting pressure from kPa to Torr. 760 Torr=101.325 kPa
+//calculation of the saturation temperature of benzene at P in degree celsius
+t1_s=((antoine_const_benzene(2,:))/(antoine_const_benzene(1,:)-log10(P)))-antoine_const_benzene(3,:);
+//calculation of the saturation temperature of toluene at P in degree celsius
+t2_s=((antoine_const_toluene(2,:))/(antoine_const_toluene(1,:)-log10(P)))-antoine_const_toluene(3,:);
+//calculation of the saturation vapour pressures of Benzene (P1s) and Toluene (P2s)
+//At T=t1_s, P=P1s=760.0 Torr, and at T=t2_s, P=P2s=760.0 Torr.
+//X1 is given by X1=(P-P2s)/(P1s-P2s). Therefore at T=t1_s, X1=1.0 and at T=t2_s, X1=0.0
+//As Y1=X1*P1s/P, Y1=1.0, when X1=1.0 and Y1=0.0,when X1=0.0.Therefore x1_initial=y1_initial=0.0(corresponding to t=t2_s) and x1_final=y1_final=1.0(corresponding to t=t1_s) where X1,x1_initial,x1_final are the mole fractions of benzene in the liquid phase (no unit) and Y1,y1_initial,y1_final are the mole fractions of benzene in the vapour phase (no unit).
+x1_initial=0.0;
+y1_initial=0.0;
+x1_final=1.0;
+y1_final=1.0;
+// An intermediate temperature is chosen such that t1_s<T<t2_s, and at different temperatures upto t2_s, the values of P1s, P2s, X1 and X1 are found out
+T=85:5:105;//temepertaures at which P1s, P2s, X1 and Y1 are to be determined in degree celsius. The initial T is taken as 85 degree celsius, such that t1_s<T<t2_s
+k=1;
+l=length(T);
+while k<l|k==l
+P1s(k)=10^((antoine_const_benzene(1,:))-((antoine_const_benzene(2,:))/(T(:,k)+antoine_const_benzene(3,:))));//calculation of saturation pressure of benzene in Torr
+P2s(k)=10^((antoine_const_toluene(1,:))-((antoine_const_toluene(2,:))/(T(:,k)+antoine_const_toluene(3,:))));//caclculation of saturation pressure of toluene in Torr
+ X1(k)=(P-P2s(k))/(P1s(k)-P2s(k));//calculation of mole fraction of Benzene in liquid phase (no unit)
+ Y1(k)=(X1(k)*P1s(k))/P;//calculation of mole fraction of Benzene in vapour phase (no unit)
+ k=k+1;
+end
+//Generating the T-x-y data
+ j=1;
+ while j<l+2|j==l+2
+ if j==1 then
+ temp(j)=t1_s;
+ x1_benzene(j)=x1_final;
+ y1_benzene(j)=y1_final;
+ else if j==l+2 then
+ temp(j)=t2_s;
+ x1_benzene(j)=x1_initial;
+ y1_benzene(j)=y1_initial;
+ else
+ temp(j)=T(j-1);
+ x1_benzene(j)=X1(j-1);
+ y1_benzene(j)=Y1(j-1);
+ end
+ end
+ j=j+1;
+ end
+
+//OUTPUT
+mprintf('P-x-y results \n');
+i=1;
+for i = 1 : n
+ mprintf('x1=%f \t y1=%f\t P=%f Torr \n\n',x1(i),y1(i),P_tot(i));
+ end
+ mprintf('T-x-y results \n t=%f degree celsius\t P1_s=760.0 Torr \t P2_s=(-) Torr \t\t x1=1.0 \t y1=1.0 \n\n',t1_s);
+ k=1;
+ for k= 1:l
+ mprintf('t=%f degree celsius\t P1_s=%f Torr \t P2_s=%f Torr \t x1=%f \t y1=%f \n\n',T(k),P1s(k),P2s(k),X1(k),Y1(k));
+ end
+ mprintf('t=%f degree celsius\t P1_s=(-)Torr \t\t P2_s=760.0 Torr \t x1=0.0 \t y1=0.0 \n',t2_s);
+ f1=scf(1);
+ scf(f1);
+plot(x1,P_tot,y1,P_tot);//generating the P-x-y plot
+ xtitle('P-x-y diagram for benzene-toluene system at 95 degree celsius','x1,y1','P(Torr)');
+ f2=scf(2);
+ scf(f2);
+ plot(x1_benzene,temp,y1_benzene,temp);//generating the T-x-y plot
+ xtitle('T-x-y diagram for benzene-toluene sytem at 760 Torr','x1,y1','t (degree celsius)');
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH11/EX11.1/Pxy_plot.jpg b/611/CH11/EX11.1/Pxy_plot.jpg Binary files differnew file mode 100755 index 000000000..6396983fe --- /dev/null +++ b/611/CH11/EX11.1/Pxy_plot.jpg diff --git a/611/CH11/EX11.1/Txy_plot.jpg b/611/CH11/EX11.1/Txy_plot.jpg Binary files differnew file mode 100755 index 000000000..6c1274e94 --- /dev/null +++ b/611/CH11/EX11.1/Txy_plot.jpg diff --git a/611/CH11/EX11.10/Chap11_Ex10.sce b/611/CH11/EX11.10/Chap11_Ex10.sce new file mode 100755 index 000000000..55e580184 --- /dev/null +++ b/611/CH11/EX11.10/Chap11_Ex10.sce @@ -0,0 +1,60 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 10,Page 401
+//Title: Activity coefficients using the UNIQUAC equation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, ethanol is taken as 1 and benzene as 2
+T=345;//temperature of the mixture in K
+x1=0.8;//mole fraction of ethanol in the liquid phase (no unit)
+
+//Ethanol (CH3CH2OH) has 1 CH3 group, 1 CH2 group and 1 OH group, while Benzene has 6 ACH groups. The group identification and the parameters R_k and Q_k are given below:
+//Molecule: Ethanol : Group identification: Name: CH3, Main No. 1, Sec.No.1, Name: CH2, Main No. 1,Sec.No.2, Name: OH, Main No.5 ,Sec.No.14
+//Molecule: Benzene : Group identification: Name: ACH, Main No. 3, Sec.No. 9
+nu_ki=[1;1;1;6];//number of groups of type: CH3, CH2, OH and ACH respectively (no unit)
+R_k=[0.9011;0.6744;1.0000;0.5313];//Group volume parameter for CH3, CH2, OH and ACH respectively (no unit)
+Q_k=[0.848;0.540;1.200;0.400];//Area parameter for CH3, CH2, OH and ACH respectively (no unit)
+R=8.314;//universal gas constant in J/molK
+u12_u22=-241.2287;//UNIQUAC parameter for the system in J/molK
+u21_u11=2799.5827;//UNIQUAC parameter for the system in J/molK
+z=10;//co-ordination number usually taken as 10 (no unit)
+
+//CALCULATION
+x2=1-x1;//calculation of mole fraction of benzene in liquid phase (no unit)
+r1=(nu_ki(1,:)*R_k(1,:))+(nu_ki(2,:)*R_k(2,:))+(nu_ki(3,:)*R_k(3,:));//calculation of volume parameter using Eq.(11.108) (no unit)
+r2=(nu_ki(4,:)*R_k(4,:));//calculation of volume parameter using Eq.(11.108) (no unit)
+phi1=(x1*r1)/((x1*r1)+(x2*r2));//calculation of volume fraction of component using Eq.(11.101) (no unit)
+phi2=(x2*r2)/((x2*r2)+(x1*r1));//calculation of volume fraction of component using Eq.(11.101) (no unit)
+q1=(nu_ki(1,:)*Q_k(1,:))+(nu_ki(2,:)*Q_k(2,:))+(nu_ki(3,:)*Q_k(3,:))//calculation of surface area parameter using Eq.(11.109) (no unit)
+q2=(nu_ki(4,:)*Q_k(4,:))//calculation of surface area parameter using Eq.(11.109) (no unit)
+theta1=(x1*q1)/((x1*q1)+(x2*q2));//calculation of area fraction of component using Eq.(11.102) (no unit)
+theta2=(x2*q2)/((x1*q1)+(x2*q2));//calculation of area fraction of component using Eq.(11.102) (no unit)
+l1=((z/2)*(r1-q1))-(r1-1);//calculation of l_i using Eq.(11.107) (no unit)
+l2=((z/2)*(r2-q2))-(r2-1);//calculation of l_i using Eq.(11.107) (no unit)
+tau_12=exp(-(u12_u22)/(R*T));//calculation of the adjustable parameter using Eq.(11.103) (no unit)
+tau_21=exp(-(u21_u11)/(R*T));//calculation of the adjustable parameter using Eq.(11.103) (no unit)
+tau_11=1.0;//by convention taken as 1.0
+tau_22=1.0;//by convention taken as 1.0
+//calculation of the combinatorial part of the activity coefficient using Eq.(11.105) (no unit)
+ln_gaamma1_c=log(phi1/x1)+((z/2)*q1*log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));
+//calculation of the combinatorial part of the activity coefficient using Eq.(11.105) (no unit)
+ln_gaamma2_c=log(phi2/x2)+((z/2)*q2*log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));
+//calculation of the residual part of the activity coefficient using Eq.(11.106) (no unit)
+ln_gaamma1_r=q1*(1-log((theta1*tau_11)+(theta2*tau_21))-(((theta1*tau_11)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_12)/((theta1*tau_12)+(theta2*tau_22)))));
+//calculation of the residual part of the activity coefficient using Eq.(11.106) (no unit)
+ln_gaamma2_r=q2*(1-log((theta1*tau_12)+(theta2*tau_22))-(((theta1*tau_21)/((theta1*tau_11)+(theta2*tau_21)))+((theta2*tau_22)/((theta1*tau_12)+(theta2*tau_22)))));
+ln_gaamma1=ln_gaamma1_c+ln_gaamma1_r;//calculation of the ln(activity coefficient) using Eq.(11.104) (no unit)
+ln_gaamma2=ln_gaamma2_c+ln_gaamma2_r;//calculation of the ln(activity coefficient) using Eq.(11.104) (no unit)
+gaamma1=exp(ln_gaamma1);//calculation of the activity coefficient (no unit)
+gaamma2=exp(ln_gaamma2);//calculation of the activity coefficient (no unit)
+
+//OUTPUT
+
+mprintf('The activity coefficients for the system using the UNIQUAC equation are : gamma1=%f \t gamma2=%f \t\n ', gaamma1,gaamma2);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH11/EX11.11/Chap11_Ex11.sce b/611/CH11/EX11.11/Chap11_Ex11.sce new file mode 100755 index 000000000..8a8b2f22f --- /dev/null +++ b/611/CH11/EX11.11/Chap11_Ex11.sce @@ -0,0 +1,91 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 11,Page 405
+//Title: Activity coefficients using the UNIFAC method
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, acetone is taken as 1 and n-pentane as 2
+T=307;//temperature of the mixture in K
+x1=0.3;//mole fraction of acetone in the liquid phase (no unit)
+
+//Acetone (CH3COCH3) has 1 CH3 group and 1 CH3CO group, while n-Pentane (C5H12) has 2 CH3 groups and 3 CH2 groups.
+//The group identification and the parameters R_k and Q_k are given below:
+//Componenet: Acetone : Group identification: Name: CH3, Main No. 1, Sec.No. 1, Name: CH3CO, Main No. 9, Sec.No. 18
+//Component: n-Pentane : Group identification: Name: CH3, Main No. 1, Sec.No. 1, Name: CH2, Main No. 1, Sec.No. 2
+nu_ki=[1;1;2;3];//no.of groups of type : CH3, CH3CO, CH3, CH2 respectively in the two components making up the systen (no unit)
+R_k=[0.9011;1.6724;0.6744];//Group volume parameter for CH3, CH3CO and CH2 respectively (no unit)
+Q_k=[0.848;1.488;0.540];//Area parameter for CH3, CH3CO and CH2 respectively (no unit)
+a_19=476.40;//group interaction parameter for the system in K
+a_91=26.760;//group interaction parameter for the system in K
+z=10;//co-ordination number usually taken as 10 (no unit)
+
+//CALCULATION
+x2=1-x1;//calculation of mole fraction of benzene in liquid phase (no unit)
+r1=(nu_ki(1,:)*R_k(1,:))+(nu_ki(2,:)*R_k(2,:));//calculation of volume parameter using Eq.(11.108) (no unit)
+r2=(nu_ki(3,:)*R_k(1,:))+(nu_ki(4,:)*R_k(3,:));//calculation of volume parameter using Eq.(11.108) (no unit)
+phi1=(x1*r1)/((x1*r1)+(x2*r2));//calculation of volume fraction of component using Eq.(11.101) (no unit)
+phi2=(x2*r2)/((x2*r2)+(x1*r1));//calculation of volume fraction of component using Eq.(11.101) (no unit)
+q1=(nu_ki(1,:)*Q_k(1,:))+(nu_ki(2,:)*Q_k(2,:))//calculation of surface area parameter using Eq.(11.109) (no unit)
+q2=(nu_ki(3,:)*Q_k(1,:))+(nu_ki(4,:)*Q_k(3,:))//calculation of surface area parameter using Eq.(11.109) (no unit)
+theta1=(x1*q1)/((x1*q1)+(x2*q2));//calculation of area fraction of component using Eq.(11.102) (no unit)
+theta2=(x2*q2)/((x1*q1)+(x2*q2));//calculation of area fraction of component using Eq.(11.102) (no unit)
+l1=((z/2)*(r1-q1))-(r1-1);//calculation of l_i using Eq.(11.107) (no unit)
+l2=((z/2)*(r2-q2))-(r2-1);//calculation of l_i using Eq.(11.107) (no unit)
+//calculation of the combinatorial part of the activity coefficient using Eq.(11.105) (no unit)
+ln_gaamma1_c=log(phi1/x1)+((z/2)*q1*log(theta1/phi1))+l1-((phi1/x1)*((x1*l1)+(x2*l2)));
+//calculation of the combinatorial part of the activity coefficient using Eq.(11.105) (no unit)
+ln_gaamma2_c=log(phi2/x2)+((z/2)*q2*log(theta2/phi2))+l2-((phi2/x2)*((x1*l1)+(x2*l2)));
+a_11=0;//by convention taken as 0.0,in K
+a_99=0;//by convention taken as 0.0,in K
+psi_19=exp(-(a_19)/(T));//calculation of psi_mn using Eq.(11.119) (no unit)
+psi_91=exp(-(a_91)/(T));//calculation of psi_mn using Eq.(11.119) (no unit)
+psi_11=1;//as a_11=0, psi_11=1 (no unit)
+psi_99=1;//as a_99=0, psi_99=1 (no unit)
+//calculation of the residual activity coefficient(tau_k) of group k, in a reference solution containing molecules of type i
+x1_1=nu_ki(1,:)/(nu_ki(1,:)+nu_ki(2,:));//calculation of mole fraction of CH3 group (pure acetone(1)) (no unit)
+x1_18=nu_ki(2,:)/(nu_ki(1,:)+nu_ki(2,:));//calculation of mole fraction of CH3CO group (pure acetone(1)) (no unit)
+theta1_1=(Q_k(1,:)*x1_1)/((Q_k(1,:)*x1_1)+(Q_k(2,:)*x1_18));//calculation of surface area fraction of CH3 group (pure acetone) using Eq.(11.118) (no unit)
+theta1_18=(Q_k(2,:)*x1_18)/((Q_k(2,:)*x1_18)+(Q_k(1,:)*x1_1));//calculation of surface area fraction of CH3CO group (pure acetone) using Eq.(11.118) (no unit)
+//calculation of the residual activity coefficient(tau_k(i))of CH3 group,in a reference solution of (pure acetone) using Eq.(11.117)(no unit)
+ln_tau1_1=Q_k(1,:)*(1-log((theta1_1*psi_11)+(theta1_18*psi_91))-(((theta1_1*psi_11)/((theta1_1*psi_11)+(theta1_18*psi_91)))+((theta1_18*psi_19)/((theta1_1*psi_19)+(theta1_18*psi_11)))));
+//calculation of the residual activity coefficient(tau_k(i))of CH3CO group,in a reference solution of (pure acetone) using (Eq.11.117)(no unit)
+ln_tau1_18=Q_k(2,:)*(1-log((theta1_1*psi_19)+(theta1_18*psi_99))-(((theta1_1*psi_91)/((theta1_1*psi_99)+(theta1_18*psi_91)))+((theta1_18*psi_99)/((theta1_1*psi_19)+(theta1_18*psi_99)))));
+x2_1=nu_ki(3,:)/(nu_ki(3,:)+nu_ki(4,:));//calculation of mole fraction of CH3 group (pure n-pentane(2)) (no unit)
+x2_2=nu_ki(4,:)/(nu_ki(3,:)+nu_ki(4,:));//calculation of mole fraction of CH2 group (pure n-pentane(2)) (no unit)
+//As n-pentane contains only one main group (1)
+ln_tau2_1=0;
+ln_tau2_2=0;
+//calculation of group residual activity coefficients for the given mole fraction of acetone in liquid phase (x1)(no unit)
+//calculation of group mole fraction for CH3 group in Acetone and n-pentane using Eq.(11.115)(no unit)
+x_1=((x1*nu_ki(1,:))+(x2*nu_ki(3,:)))/((((x1*nu_ki(1,:))+(x1*nu_ki(2,:))))+((x2*nu_ki(3,:))+(x2*nu_ki(4,:))));
+//calculation of group mole fraction for CH2 group in n-Pentane using Eq.(11.115)(no unit)
+x_2=((x2*nu_ki(4,:)))/((((x1*nu_ki(1,:))+(x1*nu_ki(2,:))))+((x2*nu_ki(3,:))+(x2*nu_ki(4,:))));
+//calculation of group mole fraction for CH3CO group in Acetone using Eq.(11.115)(no unit)
+x_18=((x1*nu_ki(2,:)))/((((x1*nu_ki(1,:))+(x1*nu_ki(2,:))))+((x2*nu_ki(3,:))+(x2*nu_ki(4,:))));
+theta_1=(Q_k(1,:)*x_1)/((Q_k(1,:)*x_1)+(Q_k(2,:)*x_18)+(Q_k(3,:)*x_2));//calculation of surface area fraction of CH3 group (using Eq.11.118)(no unit)
+theta_2=(Q_k(3,:)*x_2)/((Q_k(1,:)*x_1)+(Q_k(2,:)*x_18)+(Q_k(3,:)*x_2));//calculation of surface area fraction of CH2 group (using Eq.11.118)(no unit)
+theta_18=(Q_k(2,:)*x_18)/((Q_k(1,:)*x_1)+(Q_k(2,:)*x_18)+(Q_k(3,:)*x_2));//calculation of surface area fraction of CH3CO group (using Eq.11.118)(no unit)
+//calculation of group residual activity coefficient of CH3 using Eq.(11.117)(no unit)
+ln_tau_1=Q_k(1,:)*(1-log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));
+//calculation of group residual activity coefficient of CH2 using Eq.(11.117)(no unit)
+ln_tau_2=Q_k(3,:)*(1-log((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91))-((((theta_1*psi_11)+(theta_2*psi_11))/((((theta_1*psi_11)+(theta_2*psi_11))+(theta_18*psi_91)))+((theta_18*psi_19)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));
+//calculation of group residual activity coefficient of CH3CO using Eq.(11.117)(no unit)
+ln_tau_18=Q_k(2,:)*(1-log((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_99))-(((((theta_1+theta_2)*psi_91)/((theta_1*psi_11)+(theta_2*psi_11)+(theta_18*psi_91)))+((theta_18*psi_99)/((theta_1*psi_19)+(theta_2*psi_19)+(theta_18*psi_11))))));
+//calculation of the residual contributions to the activity coefficients using Eq.(11.116)(no unit)
+ln_gaamma1_r=(nu_ki(1,:)*(ln_tau_1-ln_tau1_1))+(nu_ki(2,:)*(ln_tau_18-ln_tau1_18));
+ln_gaamma2_r=(nu_ki(3,:)*(ln_tau_1-ln_tau2_1))+(nu_ki(4,:)*(ln_tau_2-ln_tau2_2));
+ln_gaamma1=ln_gaamma1_c+ln_gaamma1_r;//calculation of the ln(activity coefficient) using Eq.(11.104) (no unit)
+ln_gaamma2=ln_gaamma2_c+ln_gaamma2_r;//calculation of the ln(activity coefficient) using Eq.(11.104) (no unit)
+gaamma1=exp(ln_gaamma1);//calculation of the activity coefficient (no unit)
+gaamma2=exp(ln_gaamma2);//calculation of the activity coefficient (no unit)
+
+//OUTPUT
+
+mprintf('The activity coefficients for the system using the UNIFAC method are : gamma1=%f \t gamma2=%f \t\n ', gaamma1,gaamma2);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH11/EX11.2/Chap11_Ex2.sce b/611/CH11/EX11.2/Chap11_Ex2.sce new file mode 100755 index 000000000..2327a19d5 --- /dev/null +++ b/611/CH11/EX11.2/Chap11_Ex2.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 2,Page 384
+//Title:Composition of liquid
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, benzene is taken as 1 and toluene as 2
+T=95;//temperature of the equimolar vapour mixture of benzene and toluene in degree celsius
+y1=0.5;//mole fraction of benzene in vapour phase (no unit)
+y2=0.5;//mole fraction of toluene in vapour phase (no unit)
+P1_s=1176.21;//saturation pressure of benzene at T, taken from Example 11.1 in Torr
+P2_s=477.03;//saturation pressure of toluene at T, taken from Example 11.1 in Torr
+
+//CALCULATION
+P=1/((y1/P1_s)+(y2/P2_s));//calculation of the total pressure using Eq.(11.21) in Torr
+x1=(y1*P)/P1_s;//calculation of mole fraction of benzene in liquid phase using Eq.(11.15)(no unit)
+x2=1-x1;//calculation of mole fraction of toluene in liquid phase using Eq.(11.15)(no unit)
+
+//OUTPUT
+mprintf('The composition of the liquid which is in equilibrium with the equimolar vapour mixture of benzene and toluene at 95 degree celsius is \n mole fraction of benzene in liquid phase (x1)=%f \n mole fraction of toluene in liquid phase (x2)=%f \n',x1,x2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH11/EX11.3/Chap11_Ex3_R1.sce b/611/CH11/EX11.3/Chap11_Ex3_R1.sce new file mode 100755 index 000000000..86a7faeab --- /dev/null +++ b/611/CH11/EX11.3/Chap11_Ex3_R1.sce @@ -0,0 +1,34 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 3,Page 384
+//Title:Bubble temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, benzene is taken as 1 and toluene as 2
+P=760;//pressure of the binary liquid mixture of benzene and toluene in Torr
+x1=0.4047;//mole fraction of benzene in liquid phase (no unit)
+antoine_const_benzene=[6.87987;1196.760;219.161];//Antoine's constants for Benzene from Table A.7
+antoine_const_toluene=[6.95087;1342.310;219.187];//Antoine's constants for Toluene from Table A.7
+
+//CALCULATION
+
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+//Based on the procedure given in the texbook, the criterion of summation of (y_i)=1.0 will be checked, where y_i=(x_i*Pi_s)/P, where Pi_s is the saturation pressure obtained from the Antoine's equation.
+
+x2=1-x1;//mole fraction of toluene in liquid phase (no unit)
+tol=1e-6;//tolerance limit for convergence of the system using fsolve
+tguess=100;//taking a guess value for the bubble temperature to be used in the fsolve function in degree celsius
+function[fn]=solver_func(ti)
+ fn=(((x1/P)*(10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(ti+antoine_const_benzene(3,:))))))+((x2/P)*(10^(antoine_const_toluene(1,:)-(antoine_const_toluene(2,:)/(ti+antoine_const_toluene(3,:)))))))-1.0;//Function defined for solving the system
+endfunction
+[t]=fsolve(tguess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+
+//OUTPUT
+mprintf('The bubble temperature of a binary liquid mixture of benzene and toluene at 760 Torr=%d degree celsius \n',t);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH11/EX11.4/Chap11_Ex4_R1.sce b/611/CH11/EX11.4/Chap11_Ex4_R1.sce new file mode 100755 index 000000000..0519bb95c --- /dev/null +++ b/611/CH11/EX11.4/Chap11_Ex4_R1.sce @@ -0,0 +1,34 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 4,Page 385
+//Title:Dew temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, benzene is taken as 1 and toluene as 2
+P=760;//pressure of the binary liquid mixture of benzene and toluene in Torr
+y1=0.6263;//mole fraction of benzene in vapour phase (no unit)
+antoine_const_benzene=[6.87987;1196.760;219.161];//Antoine's constants for Benzene from Table A.7
+antoine_const_toluene=[6.95087;1342.310;219.187];//Antoine's constants for Toluene from Table A.7
+
+//CALCULATION
+
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+//Based on the procedure given in the texbook, the criterion of summation of (x_i)=1.0 will be checked, where x_i=(y_i*P)/(Pi_s), Pi_s is the saturation pressure obtained from the Antoine's Equation.
+
+y2=1-y1;//mole fraction of toluene in vapour phase (no unit)
+tol=1e-6;//tolerance limit for convergence of the system using fsolve
+tguess=100;//taking a guess value for the dew temperature to be used in the fsolve function in degree celsius
+function[fn]=solver_func(ti)
+ fn=(((y1*P)/(10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(ti+antoine_const_benzene(3,:))))))+((y2*P)/(10^(antoine_const_toluene(1,:)-(antoine_const_toluene(2,:)/(ti+antoine_const_toluene(3,:)))))))-1.0;//Function defined for solving the system
+endfunction
+[t]=fsolve(tguess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+
+//OUTPUT
+mprintf('The dew temperature of a binary vapour mixture of benzene and toluene at 760 Torr=%d degree celsius \n',t);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH11/EX11.5/Chap11_Ex5.sce b/611/CH11/EX11.5/Chap11_Ex5.sce new file mode 100755 index 000000000..86a1e1858 --- /dev/null +++ b/611/CH11/EX11.5/Chap11_Ex5.sce @@ -0,0 +1,54 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 5,Page 386
+//Title:Composition of the liquid and vapor streams leaving the flash unit
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, n-pentane is taken as 1 , n-hexane as 2, and n-heptane as 3
+P=200;//pressure at which the flash vaporizer is maintained in kPa
+T=90;//temperature at which the flash vaporizer is maintained in degree celsius
+zf1=0.3//mole fraction of n-pentane in feed stream (no unit)
+zf2=0.3//mole fraction of n-hexane in feed stream (no unit)
+zf3=0.4//mole fraction of n-heptane in feed stream (no unit)
+antoine_const_pentane=[6.87632;1075.780;233.205];//Antoine's constants for n-pentane from Table A.7
+antoine_const_hexane=[6.91058;1189.640;226.280];//Antoine's constants for n-hexane from Table A.7
+antoine_const_heptane=[6.89386;1264.370;216.640];//Antoine's constants for n-heptane from Table A.7
+
+//CALCULATION
+
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+P1_s=10^(antoine_const_pentane(1,:)-(antoine_const_pentane(2,:)/(T+antoine_const_pentane(3,:))));//calculation of saturation pressure of n-pentane at T in Torr
+P1_s=P1_s*133.322*10^-3;//conversion from Torr to kPa
+P2_s=10^(antoine_const_hexane(1,:)-(antoine_const_hexane(2,:)/(T+antoine_const_hexane(3,:))));//calculation of saturation pressure of n-hexane at T in Torr
+P2_s=P2_s*133.322*10^-3;//conversion from Torr to kPa
+P3_s=10^(antoine_const_heptane(1,:)-(antoine_const_heptane(2,:)/(T+antoine_const_heptane(3,:))));//calculation of saturation pressure of n-heptane at T in Torr
+P3_s=P3_s*133.322*10^-3;//conversion from Torr to kPa
+K1=P1_s/P;//calculation of K factor using Eq.(11.22) (no unit)
+K2=P2_s/P;//calculation of K factor using Eq.(11.22) (no unit)
+K3=P3_s/P;//calculation of K factor using Eq.(11.22) (no unit)
+tol=1e-6;//tolerance limit for convergence of the system using fsolve
+L_F_guess=0.1;//taking a guess value for the L/F ratio, where L is the mole number of liquid stream leaving the unit at T and P, and F is the mole number of feed stream
+function[fn]=solver_func(L_F)
+ fn=((zf1/((L_F)+((1-L_F)*K1)))+(zf2/((L_F)+((1-L_F)*K2)))+(zf3/((L_F)+((1-L_F)*K3))))-1.0;//Function defined for solving the system
+endfunction
+[L_F]=fsolve(L_F_guess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+x1=(zf1/((L_F)+((1-L_F)*K1)));//calculation of mole fraction of n-pentane in liquid stream leaving the unit at T and P (no unit)
+x2=(zf2/((L_F)+((1-L_F)*K2)));//calculation of mole fraction of n-hexane in liquid stream leaving the unit at T and P (no unit)
+x3=(zf3/((L_F)+((1-L_F)*K3)));//calculation of mole fraction of n-heptane in liquid stream leaving the unit at T and P (no unit)
+y1=K1*x1;//calculation of mole fraction of n-pentane in the vapour stream leaving the unit at T and P (no unit)
+y2=K2*x2;//calculation of mole fraction of n-hexane in the vapour stream leaving the unit at T and P (no unit)
+y3=K3*x3;//calculation of mole fraction of n-heptane in the vapour stream leaving the unit at T and P (no unit)
+V_F=1-(L_F);//calculation of the fraction that has vaporized
+
+//OUTPUT
+mprintf('The composition of the liquid leaving the flash unit is : x1=%f \t\t x2=%f \t\t x3=%f\n',x1,x2,x3);
+mprintf('The composition of the vapour leaving the flash unit is : y1=%f \t\t y2=%f \t\t y3=%f\n',y1,y2,y3);
+mprintf('The fraction of feed that has vaporized in the unit=%f \n ',V_F);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH11/EX11.7/Chap11_Ex7.sce b/611/CH11/EX11.7/Chap11_Ex7.sce new file mode 100755 index 000000000..bff86ddf1 --- /dev/null +++ b/611/CH11/EX11.7/Chap11_Ex7.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 7,Page 397
+//Title:Activity coefficients
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, nitromethane is taken as 1 and carbon tetrachloride as 2
+T=45;//temperature of the mixture in degree celsius
+A=2.230;//van laar constant for the system at T (no unit)
+B=1.959;//van laar constant for the system at T (no unit)
+n1=30;//mole percentage of nitromethane in the mixture ( in percentage)
+
+//CALCULATION
+n2=100-n1;//calculation of mole percentage of carbon tetrachloride in the mixture (in percentage)
+x1=n1/100;//calculation of mole fraction of nitromethane in the mixture (no unit)
+x2=1-x1;//calculation of mole fraction of carbon tetrachloride in the mixture(no unit)
+gaamma1=exp (A/(1+((A/B)*(x1/x2)))^2);//calculation of activity coefficient using Eq.(11.82) (no unit)
+gaamma2=exp (B/(1+((B/A)*(x2/x1)))^2);//calculation of activity coefficient using Eq.(11.83) (no unit)
+
+//OUTPUT
+mprintf('The activity coefficients for the system using van laar equation is : gamma1=%f \t gamma2=%f \t\n ', gaamma1,gaamma2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH11/EX11.8/Chap11_Ex8.sce b/611/CH11/EX11.8/Chap11_Ex8.sce new file mode 100755 index 000000000..65c08e3e0 --- /dev/null +++ b/611/CH11/EX11.8/Chap11_Ex8.sce @@ -0,0 +1,38 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 8,Page 397
+//Title:van Laar constants and Activity coefficients
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, ethanol is taken as 1 and benzene as 2
+n_azeo=44.8;//azeotropic composition given as mole percentage
+Tb=68.24;//boiling point of mixture in degree celsius
+P=760;//pressure in Torr
+P1_s=506;//saturation pressure of ethanol at Tb in Torr
+P2_s=517;//saturation pressure of benzene at Tb in Torr
+n1=10;//mole percentage of ethanol in the mixture (in percentage)
+
+//CALCULATION
+//At azeotropic composition, y_i=x_i, therefore gaamma_i=P/Pi_s
+x1=n_azeo/100;//calculation of the mole fraction of ethanol (azeotropic composition) (no unit)
+x2=1-x1;//calculation of the mole fraction of benzene (azeotropic composition) (no unit)
+gaamma1=P/P1_s;//calculation of the activity coefficient at the azeotropic composition (no unit)
+gaamma2=P/P2_s;//calculation of the activity coefficient at the azeotropic composition (no unit)
+A=log(gaamma1)*(1+((x2*log(gaamma2))/(x1*log(gaamma1))))^2;//calculation of the van Laar constant using Eq.(11.84) (no unit)
+B=log(gaamma2)*(1+((x1*log(gaamma1))/(x2*log(gaamma2))))^2;//calculation of the van Laar constant using Eq.(11.85) (no unit)
+x1=n1/100;//calculation of the mole fraction of ethanol (no unit)
+x2=1-x1;//calculation of the mole fraction of benzene (no unit)
+gaamma1=exp (A/(1+((A/B)*(x1/x2)))^2);//calculation of activity coefficient at the given composition using Eq.(11.82) (no unit)
+gaamma2=exp (B/(1+((B/A)*(x2/x1)))^2);//calculation of activity coefficient at the given composition using Eq.(11.83) (no unit)
+
+
+//OUTPUT
+mprintf('The van Laar constants for the system are : A=%f \t B=%f \n',A,B)
+mprintf('The activity coefficients for the system using van laar equation are : gamma1=%f \t gamma2=%f \t\n ', gaamma1,gaamma2);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH11/EX11.9/Chap11_Ex9.sce b/611/CH11/EX11.9/Chap11_Ex9.sce new file mode 100755 index 000000000..065752c19 --- /dev/null +++ b/611/CH11/EX11.9/Chap11_Ex9.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-11,Example 9,Page 399
+//Title: Activity coefficients using the Wilson's parameters
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience, nitromethane is taken as 1 and carbon tetrachloride as 2
+T=45;//temperature of the system in degree celsius
+A_12=0.1156;//Wilson's parameter for the system at T (no unit)
+A_21=0.2879;//Wilson's parameter for the system at T (no unit)
+x1=0.3;//mole fraction of nitromethane in the liquid mixture (no unit)
+
+//CALCULATION
+x2=1-x1;//calculation of the mole fraction of carbon tetrachloride in the liquid mixture (no unit)
+ln_gaamma1=-log(x1+(A_12*x2))+(x2*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));//calculation of ln(activity coefficient) using Eq.(11.89) (no unit)
+gaamma1=exp(ln_gaamma1);//calculation of activity coefficient (no unit)
+ln_gaamma2=-log(x2+(A_21*x1))-(x1*((A_12/(x1+(A_12*x2)))-(A_21/((A_21*x1)+x2))));//calculation of ln(activity coefficient) using Eq.(11.90) (no unit)
+gaamma2=exp(ln_gaamma2);//calculation of activity coefficient (no unit)
+
+//OUTPUT
+
+mprintf('The activity coefficients for the system using Wilsons parameters are : gamma1=%f \t gamma2=%f \t\n ', gaamma1,gaamma2);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH12/EX12.1/Chap12_Ex1_R1.sce b/611/CH12/EX12.1/Chap12_Ex1_R1.sce new file mode 100755 index 000000000..deb490c98 --- /dev/null +++ b/611/CH12/EX12.1/Chap12_Ex1_R1.sce @@ -0,0 +1,64 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 1,Page 423
+//Title: Margules parameters
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience Benzene is taken as 1 and heptane is taken as 2
+T=60;//temperature of the system in degree celsius
+P=[237.60;265.20;317.50;333.00;368.70;387.20];//Pressure data in Torr (from Danneil et al.)
+x1=[0.0870;0.1800;0.4040;0.4790;0.7130;0.9070];//mole fraction of benzene in the liquid phase corresponding to the given pressure (no unit) (from Danneil et al.)
+y1=[0.1870;0.3400;0.5780;0.6420;0.7960;0.9220];//mole fraction of benzene in the vapour phase corresponding to the given pressure (no unit) (from Danneil et al.)
+antoine_const_benzene=[6.87987;1196.760;219.161];//Antoine's constants for Benzene from Table A.7
+antoine_const_heptane=[6.89386;1264.370;216.640];//Antoine's constants for heptane from Table A.7
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+P1_s=10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(T+antoine_const_benzene(3,:))));//calculation of saturation pressure of benzene at T in Torr
+P2_s=10^(antoine_const_heptane(1,:)-(antoine_const_heptane(2,:)/(T+antoine_const_heptane(3,:))));//calculation of saturation pressure of heptane at T in Torr
+l=length(P);//iteration parameter
+i=1;//iteration parameter
+while i<l|i==l
+ gaamma1(i)=(y1(i,:)*P(i,:))/(x1(i,:)*P1_s);//calculation of activity coefficient using the data points (no unit)
+ gaamma2(i)=((1-y1(i,:))*P(i,:))/((1-x1(i,:))*P2_s);//calculation of activity coefficient using the data points (no unit)
+ ln_gaamma1_expt(i)=log(gaamma1(i));
+ ln_gaamma2_expt(i)=log(gaamma2(i));
+ gE_RTx1x2(i)=((x1(i,:)*ln_gaamma1_expt(i))+((1-x1(i,:))*ln_gaamma2_expt(i)))/(x1(i,:)*(1-x1(i,:)));//calculation of gE/RT using Eq.(11.36) (no unit)
+ i=i+1;
+end
+plot(x1,gE_RTx1x2,'o');//Plot of gE/RTx1x2 vs x1 to determine A12 and A21
+ xtitle('Plot of gE/RTx1x2 vs x1','x1','gE/RTx1x2');
+//The three suffix Margules equation is given by gE/RTx1x2=(A21*x1)+(A12*x2), upon plotting gE/RTx1x2 vs x1 , the values of A12 and A21 were found (corresponding to x1=0 and x1=1.0. At x1=0, we get the value of A12 and at x1=1.0 i.e. x2=0, we get A21)
+A21=0.555;//value of A21 obtained from the plot of gE/RTx1x2 vs x1 (no unit)
+A12=0.315;//value of A12 obtained from the plot of gE/RTx1x2 vs x1 (no unit)
+//From the Margules equations (Eq. 11.79 and 11.80), the activity coefficients, Pressure and the mole fraction of benzene in the vapour phase in the vapour phase are recalculated using the experimental values of the mole fraction of benzene in the liquid phase (x1)repoerted earlier and the Margules paramters found above.
+j=1;//iteration parameter
+while j<l|j==l
+ ln_gaamma1(j)=((1-x1(j,:))^2)*(A12+(2*(A21-A12)*x1(j,:)));//calculation of ln(activity coefficient) (no unit)
+ ln_gaamma2(j)=(x1(j,:)^2)*(A21+(2*(A12-A21)*(1-x1(j,:))));//calculation of ln(activity coefficient) (no unit)
+ gaamma1(j)=exp(ln_gaamma1(j));//calculation of the activity coefficient (no unit)
+ gaamma2(j)=exp(ln_gaamma2(j));//calculation of the activity coefficient (no unit)
+ P_calc(j)=(gaamma1(j)*x1(j,:)*P1_s)+(gaamma2(j)*(1-x1(j,:))*P2_s);//pressure recalculated in Torr
+ y1_calc(j)=(gaamma1(j)*x1(j,:)*P1_s)/P(j);//recalculation of mole fraction of benzene in vapour phase (no unit)
+ j=j+1;
+end
+
+//OUTPUT
+mprintf('Data for the plot of gE/RTx1x2 vs x1: \n\n');
+i=1;
+for i=1:l
+ mprintf('P=%f Torr\t x1=%f\t y1=%f \t ln(gamma1)=%f\t\t ln(gamma2)=%f\t\t gE/RTx1x2=%f\n\n',P(i),x1(i),y1(i),ln_gaamma1_expt(i),ln_gaamma2_expt(i),gE_RTx1x2(i));
+end
+mprintf('Results: \n\n');
+i=1;
+for i=1:l
+ mprintf('x1=%f \t gamma1=%f \t gamma2=%f \t P_Exptl.=%f Torr\t P_Calc=%f Torr\t y1_Exptl=%f \t y1_calc=%f \n\n',x1(i),gaamma1(i),gaamma2(i),P(i),P_calc(i),y1(i),y1_calc(i));
+end
+
+//===============================================END OF PROGRAM===================================================
+//DISCLAIMER:ONE OF THE VALUES OF y1(Exptl) GIVEN IN THE ANSWER HAS A TYPO ERROR. THE VALUE AS GIVEN IN THE QUESTION IS 0.7960,WHILE WHAT IS GIVEN IN THE ANSWER IS 0.7920 THIS HAS BEEN CORRECTED IN THIS CODE.
+
+
diff --git a/611/CH12/EX12.1/Plot_12_1.jpeg b/611/CH12/EX12.1/Plot_12_1.jpeg Binary files differnew file mode 100755 index 000000000..c10e08793 --- /dev/null +++ b/611/CH12/EX12.1/Plot_12_1.jpeg diff --git a/611/CH12/EX12.2/Chap12_Ex2_R1.sce b/611/CH12/EX12.2/Chap12_Ex2_R1.sce new file mode 100755 index 000000000..e8a30bf67 --- /dev/null +++ b/611/CH12/EX12.2/Chap12_Ex2_R1.sce @@ -0,0 +1,82 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 2,Page 427
+//Title: van Laar parameters and t-x-y data
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience methanol is taken as 1 and nitromethane is taken as 2
+P=760;//pressure of the system in Torr
+t=[96.90;68.20;65.10;64.50];//temperature of the system in degree celsius (from Nakanishi et al.)
+x1=[0.0150;0.4260;0.7470;0.9140];//mole fraction of methanol in the liquid phase corresponding to the given temperature (no unit) (from Nakanishi et al.)
+y1=[0.1330;0.7470;0.8380;0.9210];//mole fraction of methanol in the vapour phase corresponding to the given temperature (no unit) (from Nakanishi et al.)
+antoine_const_methanol=[8.08097;1582.271;239.726];//Antoine's constants for methanol from Table A.7
+antoine_const_nmethane=[7.28166;1446.937;227.600];//Antoine's constants for nitromethane from Table A.7
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+l=length(t);//iteration parameter
+i=1;
+while i<l|i==l
+//calculation of saturation pressure of methanol at t in Torr
+P1_s(i)=10^(antoine_const_methanol(1,:)-(antoine_const_methanol(2,:)/(t(i,:)+antoine_const_methanol(3,:))));
+//calculation of saturation pressure of nitromethane at t in Torr
+P2_s(i)=10^(antoine_const_nmethane(1,:)-(antoine_const_nmethane(2,:)/(t(i,:)+antoine_const_nmethane(3,:))));
+gaamma1(i)=(y1(i,:)*P)/(x1(i,:)*P1_s(i));//calculation of activity coefficient using the data points (no unit)
+gaamma2(i)=((1-y1(i,:))*P)/((1-x1(i,:))*P2_s(i));//calculation of activity coefficient using the data points (no unit)
+ln_gaamma1(i)=log(gaamma1(i));//calculating the value of ln(activity coefficient) (no unit)
+ln_gaamma2(i)=log(gaamma2(i));//calculating the value of ln(activity coefficient) (no unit)
+gE_RT(i)=((x1(i,:)*ln_gaamma1(i))+((1-x1(i,:))*ln_gaamma2(i)));//calculation of gE/RT using Eq.(11.36) (no unit)
+x1x2_gE_RT(i)=(x1(i,:)*(1-x1(i,:)))/gE_RT(i);//function for plotting againt x1 to determine the van Laar paramters (no unit)
+i=i+1;
+end
+plot(x1,x1x2_gE_RT,'o');//Plot of RTx1x2/gE vs x1 to determine A and B
+ xtitle('Plot of RTx1x2/gE vs x1','x1','RTx1x2/gE');
+ //The values of the intercepts are read from Plot 12.2. From the intercepts ,the value of A and B are determined as given below
+intercept_A=0.6725;//value of 1/A at x1=0, read from the plot (no unit)
+intercept_B=0.710;//value of 1/B at x1=1.0, read from the plot (no unit)
+A=1/intercept_A;//calculation of A (no unit)
+B=1/intercept_B;//calculation of B (no unit)
+//From the van Laar equations and using the van Laar parameters determined above along with the reported values of the mole fraction of methanol in the liquid phase (x1), the activity coefficients and the mole fraction of methanol in the vapour phase (y1) are found out.
+j=1;//iteration parameter
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+while j<l|j==l
+ ln_gaamma1_new(j)=A/(1+((A/B)*(x1(j,:)/(1-x1(j,:)))))^2;//calculation of ln(activity coefficient) (no unit)
+ ln_gaamma2_new(j)=B/(1+((B/A)*((1-x1(j,:))/x1(j,:))))^2;//calculation of ln(activity coefficient) (no unit)
+ gaamma1(j)=exp(ln_gaamma1_new(j));//calculation of the activity coefficient (no unit)
+ gaamma2(j)=exp(ln_gaamma2_new(j));//calculation of the activity coefficient (no unit)
+ tguess(j)=100;//taking a guess value for the temperature to be used in the fsolve function in degree celsius
+function[fn]=solver_func(ti)
+ fn=((gaamma1(j)*(x1(j,:)/P)*(10^(antoine_const_methanol(1,:)-(antoine_const_methanol(2,:)/(ti+antoine_const_methanol(3,:))))))+((gaamma2(j))*((1-x1(j,:))/P)*(10^(antoine_const_nmethane(1,:)-(antoine_const_nmethane(2,:)/(ti+antoine_const_nmethane(3,:)))))))-1.0;//Function defined for solving the system
+endfunction
+[t_calc(j)]=fsolve(tguess(j),solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+j=j+1;
+end
+//Recalculation of the mole fraction of methanol in the vapour phase using the temperature recalculated above and the activity coefficients calculated above
+j=1;//iteration paramter
+while j<l|j==l
+//recalculation of saturation pressure of methanol at t in Torr
+P1_s_calc(j)=10^(antoine_const_methanol(1,:)-(antoine_const_methanol(2,:)/(t_calc(j,:)+antoine_const_methanol(3,:))));
+//recalculation of saturation pressure of nitromethane at t in Torr
+P2_s_calc(j)=10^(antoine_const_nmethane(1,:)-(antoine_const_nmethane(2,:)/(t_calc(j,:)+antoine_const_nmethane(3,:))));
+y1_calc(j)=(gaamma1(j)*x1(j,:)*P1_s_calc(j))/P;//recalculation of the mole fraction of methanol in vapour phase (no unit)
+j=j+1;
+end
+
+//OUTPUT
+mprintf('Data for the plot of RTx1x2/gE vs x1: \n\n');
+i=1;
+for i=1:l
+ mprintf('t=%f degree celsius\t P1_s=%f Torr \t P2_s=%f Torr \t x1=%f\t y1=%f \n ln(gamma1)=%f\t\t ln(gamma2)=%f\t\t RTx1x2/gE=%f\n\n',t(i),P1_s(i),P2_s(i),x1(i),y1(i),ln_gaamma1(i),ln_gaamma2(i),x1x2_gE_RT(i));
+end
+mprintf('Txy data recalculated: \n \n');
+i=1;
+for i=1:l
+ mprintf('x1=%f \n t_exptl=%f degree celsius \t t_calc=%f degree celsius \n y1_exptl=%f \t\t\t y1_calc=%f \n\n ',x1(i),t(i),t_calc(i),y1(i),y1_calc(i));
+end
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH12/EX12.2/Plot12_2.jpg b/611/CH12/EX12.2/Plot12_2.jpg Binary files differnew file mode 100755 index 000000000..3d77c156d --- /dev/null +++ b/611/CH12/EX12.2/Plot12_2.jpg diff --git a/611/CH12/EX12.3/Chap12_Ex3.sce b/611/CH12/EX12.3/Chap12_Ex3.sce new file mode 100755 index 000000000..074ef4db6 --- /dev/null +++ b/611/CH12/EX12.3/Chap12_Ex3.sce @@ -0,0 +1,49 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 3,Page 430
+//Title: P-x-y data using the Margules parameters
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience acetone is taken as 1 and cyclohexane is taken as 2
+T=25;//temperature of the system in degree celsius
+A12=2.0522;//three suffix Margules parameters for the system (no unit)
+A21=1.7201;//three suffix Margules parameters for the system (no unit)
+P=[118.05;207.70;246.35;259.40;261.50;262.00;261.90;258.70;252.00];//Pressure data in Torr (from Tasic et al.)
+//mole fraction of acetone in the liquid phase corresponding to the given pressure (no unit) (from Tasic et al.)
+x1=[0.0115;0.1125;0.3090;0.5760;0.6920;0.7390;0.7575;0.8605;0.9250];
+//mole fraction of acetone in the vapour phase corresponding to the given pressure (no unit) (from Tasic et al.)
+y1=[0.1810;0.5670;0.6550;0.7050;0.7250;0.7390;0.7460;0.8030;0.8580];
+antoine_const_acetone=[7.11714;1210.595;229.664];//Antoine's constants for acetone from Table A.7
+antoine_const_chexane=[6.85146;1206.470;223.136];//Antoine's constants for cyclohexane from Table A.7
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+P1_s=10^(antoine_const_acetone(1,:)-(antoine_const_acetone(2,:)/(T+antoine_const_acetone(3,:))));//calculation of saturation pressure of acetone at T in Torr
+//calculation of saturation pressure of cyclohexane at T in Torr
+P2_s=10^(antoine_const_chexane(1,:)-(antoine_const_chexane(2,:)/(T+antoine_const_chexane(3,:))));
+
+//From the Margules equations (Eq. 11.79 and 11.80), the activity coefficients are found out
+l=length(P);//iteration parameter
+j=1;//iteration parameter
+while j<l|j==l
+ ln_gaamma1(j)=((1-x1(j,:))^2)*(A12+(2*(A21-A12)*x1(j,:)));//calculation of ln(activity coefficient) (no unit)
+ ln_gaamma2(j)=(x1(j,:)^2)*(A21+(2*(A12-A21)*(1-x1(j,:))));//calculation of ln(activity coefficient) (no unit)
+ gaamma1(j)=exp(ln_gaamma1(j));//calculation of the activity coefficient (no unit)
+ gaamma2(j)=exp(ln_gaamma2(j));//calculation of the activity coefficient (no unit)
+ P(j)=(gaamma1(j)*x1(j,:)*P1_s)+(gaamma2(j)*(1-x1(j,:))*P2_s);//calculation of pressure in Torr
+ y1_calc(j)=(gaamma1(j)*x1(j,:)*P1_s)/P(j);//calculation of mole fraction of acetone in vapour phase (no unit)
+ j=j+1;
+end
+
+//OUTPUT
+mprintf('P-x-y data: \n\n');
+i=1;
+mprintf('x1 \t gamma1\t gamma2 \t P (Torr) \t y1 \n');
+for i=1:l
+ mprintf('%0.4f \t %f \t %f \t %f \t %f \n',x1(i),gaamma1(i),gaamma2(i),P(i),y1_calc(i));
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH12/EX12.4/Chap12_Ex4.sce b/611/CH12/EX12.4/Chap12_Ex4.sce new file mode 100755 index 000000000..190123036 --- /dev/null +++ b/611/CH12/EX12.4/Chap12_Ex4.sce @@ -0,0 +1,48 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 4,Page 432
+//Title: P-x-y data using the van Laar model
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience acetone is taken as 1 and cyclohexane is taken as 2
+T=25;//temperature of the system in degree celsius
+A=2.0684;//the van Laar parameters for the system (no unit)
+B=1.7174;//the van Laar parameters for the system (no unit)
+P=[118.05;207.70;246.35;259.40;261.50;262.00;261.90;258.70;252.00];//Pressure data in Torr (from Tasic et al.)
+//mole fraction of acetone in the liquid phase corresponding to the given pressure (no unit) (from Tasic et al.)
+x1=[0.0115;0.1125;0.3090;0.5760;0.6920;0.7390;0.7575;0.8605;0.9250];
+//mole fraction of acetone in the vapour phase corresponding to the given pressure (no unit) (from Tasic et al.)
+y1=[0.1810;0.5670;0.6550;0.7050;0.7250;0.7390;0.7460;0.8030;0.8580];
+antoine_const_acetone=[7.11714;1210.595;229.664];//Antoine's constants for acetone from Table A.7
+antoine_const_chexane=[6.85146;1206.470;223.136];//Antoine's constants for cyclohexane from Table A.7
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+P1_s=10^(antoine_const_acetone(1,:)-(antoine_const_acetone(2,:)/(T+antoine_const_acetone(3,:))));//calculation of saturation pressure of acetone at T in Torr
+//calculation of saturation pressure of cyclohexane at T in Torr
+P2_s=10^(antoine_const_chexane(1,:)-(antoine_const_chexane(2,:)/(T+antoine_const_chexane(3,:))));
+//From the van Laar equations(Eq. 11.82 and 11.83), the activity coefficients are found out
+l=length(P);//iteration parameter
+j=1;//iteration parameter
+while j<l|j==l
+ ln_gaamma1(j)=A/(1+((A*x1(j,:))/(B*(1-x1(j,:)))))^2;//calculation of ln(activity coefficient) (no unit)
+ ln_gaamma2(j)=B/(1+((B*(1-x1(j,:)))/(A*x1(j,:))))^2;//calculation of ln(activity coefficient) (no unit)
+ gaamma1(j)=exp(ln_gaamma1(j));//calculation of the activity coefficient (no unit)
+ gaamma2(j)=exp(ln_gaamma2(j));//calculation of the activity coefficient (no unit)
+ P(j)=(gaamma1(j)*x1(j,:)*P1_s)+(gaamma2(j)*(1-x1(j,:))*P2_s);//calculation of pressure in Torr
+ y1_calc(j)=(gaamma1(j)*x1(j,:)*P1_s)/P(j);//calculation of mole fraction of acetone in vapour phase (no unit)
+ j=j+1;
+end
+
+//OUTPUT
+mprintf('P-x-y data: \n\n');
+i=1;
+mprintf('x1 \t gamma1 \t gamma2 \t P(Torr) \t y1 \n');
+for i=1:l
+ mprintf('%0.4f \t %f \t %f \t %f \t %f \n',x1(i),gaamma1(i),gaamma2(i),P(i),y1_calc(i));
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH12/EX12.5/Chap12_Ex5_R1.sce b/611/CH12/EX12.5/Chap12_Ex5_R1.sce new file mode 100755 index 000000000..bbcd878e4 --- /dev/null +++ b/611/CH12/EX12.5/Chap12_Ex5_R1.sce @@ -0,0 +1,50 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 5,Page 435
+//Title: VLE data using the van Laar model
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience chloroform is taken as 1 and methanol is taken as 2
+P=760;//pressure in Torr at which chloroform and methanol form an azeotrope
+T=53.5;//temperature in degree celsius at which chloroform and methanol form an azeotrope
+x1=0.65;//mole fraction of chloroform in the liquid phase (no unit) (corresponding to azeotropic composition)
+antoine_const_chloroform=[6.95465;1170.966;226.232];//Antoine's constants for acetone from Table A.7
+antoine_const_methanol=[8.08097;1582.271;239.726];//Antoine's constants for acetone from Table A.7
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+x2=1-x1;//calculation of the mole fraction of methanol in the liquid phase (no unit) (corresponding to azeotropic composition)
+//calculation of saturation pressure of chloroform at T in Torr
+P1_s=10^(antoine_const_chloroform(1,:)-(antoine_const_chloroform(2,:)/(T+antoine_const_chloroform(3,:))));
+//calculation of saturation pressure of methanol at T in Torr
+P2_s=10^(antoine_const_methanol(1,:)-(antoine_const_methanol(2,:)/(T+antoine_const_methanol(3,:))));
+//At the azeotropic conditions, the activity coefficients are determined using Eq.(12.15 and 12.16)
+gaamma1=P/P1_s;//calculation of activity coefficient using Eq.(12.15) (no unit)
+gaamma2=P/P2_s;//calculation of activity coefficient using Eq.(12.16) (no unit)
+A=log(gaamma1)*(1+((x2*log(gaamma2))/(x1*log(gaamma1))))^2;//calculation of the van Laar parameter (no unit) using Eq.(11.84)
+B=log(gaamma2)*(1+((x1*log(gaamma1))/(x2*log(gaamma2))))^2;//calculation of the van Laar parameter (no unit) using Eq.(11.85)
+x1=0.1:0.1:0.9;//taking the values of mole fraction of chloroform in the liquid phase to compute the VLE data (no unit)
+l=length(x1);//iteration parameter
+j=1;//iteration parameter
+while j<l|j==l
+ ln_gaamma1(j)=A/(1+((A*x1(:,j))/(B*(1-x1(:,j)))))^2;//calculation of ln(activity coefficient) (no unit)
+ ln_gaamma2(j)=B/(1+((B*(1-x1(:,j)))/(A*x1(:,j))))^2;//calculation of ln(activity coefficient) (no unit)
+ gaamma1(j)=exp(ln_gaamma1(j));//calculation of the activity coefficient (no unit)
+ gaamma2(j)=exp(ln_gaamma2(j));//calculation of the activity coefficient (no unit)
+ P(j)=(gaamma1(j)*x1(:,j)*P1_s)+(gaamma2(j)*(1-x1(:,j))*P2_s);//calculation of pressure in Torr
+ y1(j)=(gaamma1(j)*x1(:,j)*P1_s)/P(j);//calculation of mole fraction of chloroform in vapour phase (no unit)
+ j=j+1;
+end
+
+//OUTPUT
+mprintf('VLE data: \n\n');
+i=1;
+mprintf('x1 \tgamma1 \t\t gamma2 \t P (Torr) \t y1 \n\n');
+for i=1:l
+ mprintf('%0.1f \t %f \t %f \t %f \t %f \n',x1(i),gaamma1(i),gaamma2(i),P(i),y1(i));
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH12/EX12.6/Chap12_Ex6.sce b/611/CH12/EX12.6/Chap12_Ex6.sce new file mode 100755 index 000000000..311dc8d01 --- /dev/null +++ b/611/CH12/EX12.6/Chap12_Ex6.sce @@ -0,0 +1,38 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 6,Page 443
+//Title: Dew pressure and liquid composition
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience ethane is taken as 1 and propane is taken as 2
+y1=0.3;//mole fraction of ethane in the vapour phase (no unit)
+T=30;//temperature in degree celsius
+
+//CALCULATION
+//An assumption for the total Pressure is taken and the K factors are read from Fig.12.6. Using the K factor value, the value of x1 and x2 are computed.If x1 and x2 add up to 1, the assumption of total pressure is correct. Otherwise, the pressure is suitably adjusted
+y2=1-y1;//calculation of the mole fraction of propane in the vapour phase (no unit)
+P_guess=1;//assuming the value of pressure in MPa to compute the K factors
+K1=3.4;//K factor taken from Fig.(12.6) corresponding to T and P_guess (no unit)
+K2=1.1;//K factor taken from Fig.(12.6) corresponding to T and P_guess (no unit)
+x1_calc=y1/K1;//calculation of the mole fraction of ethane in the liquid phase (no unit)
+x2_calc=y2/K2;//calculation of the mole fraction of propane in the liquid phase (no unit)
+tot=x1_calc+x2_calc;//checking if x1 and x2 add upto 1
+if tot==1 then
+ P=P_guess;//if the total of x1 and x2 sum up to 1, then the assumed pressure is the Dew pressure (in MPa)
+ x1=x1_calc;//if the total of x1 and x2 sum up to 1, then the calculated value of x1 is the correct liquid composition of ethane (no unit)
+ x2=x2_calc;//if the total of x1 and x2 sum up to 1, then the calculated value of x2 is the correct liquid composition of propane (no unit)
+else
+ P=1.5;//assuming a higher value of P in MPa to compute the K factors from Fig.(12.6), as in this case, the sum total of x1 and x2 are less than 1
+ K1=2.4;//K factor taken from Fig.(12.6) corresponding to T and P (no unit)
+ K2=0.8;//K factor taken from Fig.(12.6) corresponding to T and P (no unit)
+ x1=y1/K1;//calculation of the mole fraction of ethane in the liquid phase (no unit)
+ x2=y2/K2;//calculation of the mole fraction of propane in the liquid phase (no unit)
+end
+
+//OUTPUT
+mprintf('The Dew pressure and the liquid composition of a binary vapour mixture of ethane and propane was found to be P=%0.2f MPa\t x1=%0.3f\t x2=%0.3f \t',P,x1,x2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH12/EX12.7/Chap12_Ex7_R1.sce b/611/CH12/EX12.7/Chap12_Ex7_R1.sce new file mode 100755 index 000000000..ddb249abe --- /dev/null +++ b/611/CH12/EX12.7/Chap12_Ex7_R1.sce @@ -0,0 +1,38 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 7,Page 443
+//Title: Bubble temperature and vapour composition
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience ethane is taken as 1 and propane is taken as 2
+x1=0.4;//mole fraction of ethane in the liquid phase (no unit)
+P=1.5;//pressure in MPa
+
+//CALCULATION
+//An assumption for the temperature is taken and the K factors are read from Fig.12.6. Using the K factor value, the value of y1 and y2 are computed.If y1 and y2 add up to 1, the assumption of the temperature is correct. Otherwise, the temperature is suitably adjusted
+x2=1-x1;//calculation of the mole fraction of propane in the liquid phase (no unit)
+t_guess=10;//assuming the value of temperature in degree celsius to compute the K factors
+K1=1.8;//K factor taken from Fig.(12.6) corresponding to t_guess and P (no unit)
+K2=0.5;//K factor taken from Fig.(12.6) corresponding to t_guess and P (no unit)
+y1_calc=K1*x1;//calculation of the mole fraction of ethane in the vapour phase (no unit)
+y2_calc=K2*x2;//calculation of the mole fraction of propane in the vapour phase (no unit)
+tot=y1_calc+y2_calc;//checking if y1 and y2 add upto 1
+if tot==1 then
+ t=t_guess;//if the total of y1 and y2 sum up to 1, then the assumed temperature is the bubble temperature (in degree celsius)
+ y1=y1_calc;//if the total of y1 and y2 sum up to 1, then the calculated value of y1 is the correct vapour composition of ethane (no unit)
+ y2=y2_calc;//if the total of y1 and y2 sum up to 1, then the calculated value of y2 is the correct vapour composition of propane (no unit)
+else
+ t=9;//assuming a lower value of t in degree celsius to compute the K factors from Fig.(12.6), as in this case, the sum total of y1 and y2 are greater than 1
+ K1=1.75;//K factor taken from Fig.(12.6) corresponding to t and P (no unit)
+ K2=0.5;//K factor taken from Fig.(12.6) corresponding to t and P (no unit)
+ y1=K1*x1;//calculation of the mole fraction of ethane in the vapour (no unit)
+ y2=K2*x2;//calculation of the mole fraction of propane in the vapour phase (no unit)
+end
+
+//OUTPUT
+mprintf('The bubble temperature and the vapour composition of a binary vapour mixture of ethane and propane was found to be t=%d degree celsius\n y1=%f\t y2=%f\t',t,y1,y2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH12/EX12.8/Chap12_Ex8.sce b/611/CH12/EX12.8/Chap12_Ex8.sce new file mode 100755 index 000000000..78bd41ba3 --- /dev/null +++ b/611/CH12/EX12.8/Chap12_Ex8.sce @@ -0,0 +1,56 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 8,Page 449
+//Title: Thermodynamic consistency
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience 1-Propanol is taken as 1 and chlorobenzene is taken as 2
+P=[350.00;446.00;518.00;574.50;609.00;632.50;665.00;681.50;691.50];//pressure data in Torr, taken from (Ellis et al.)
+x1=[0.0550;0.1290;0.2120;0.3130;0.4300;0.5200;0.6380;0.7490;0.8720];//mole fraction of 1-propanol in the liquid phase, taken from (Ellis et al.) (no unit)
+y1=[0.3500;0.5110;0.5990;0.6500;0.6970;0.7260;0.7590;0.8130;0.8830];//mole fraction of 1-propanol in the vapour phase, taken from (Ellis et al.) (no unit)
+antoine_const_propanol=[8.37895;1788.020;227.438];//Antoine's constants for 1-Propanol from Table A.7
+antoine_const_cbenzene=[7.17294;1549.200;229.260];//Antoine's constants for Chlorobenzene from Table A.7
+T=95;//temperature of the system in degree celsius
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is is Torr and t is in degree celsius
+P1_s=10^(antoine_const_propanol(1,:)-(antoine_const_propanol(2,:)/(T+antoine_const_propanol(3,:))));//calculation of saturation pressure of propanol at T in Torr
+//calculation of saturation pressure of chlorobenzene at T in Torr
+P2_s=10^(antoine_const_cbenzene(1,:)-(antoine_const_cbenzene(2,:)/(T+antoine_const_cbenzene(3,:))));
+l=length(P);//iteration parameter
+i=1;//iteration paramter
+while i<l|i==l
+ gaamma1(i)=(y1(i,:)*P(i,:))/(x1(i,:)*P1_s);//calculation of activity coefficient using Eq.(12.15) (no unit)
+ gaamma2(i)=((1-y1(i,:))*P(i,:))/((1-x1(i,:))*P2_s);//calculation of activity coefficient using Eq.(12.16) (no unit)
+ lngamma1_gamma2(i)=log(gaamma1(i)/gaamma2(i));//calculation of ln(activity coefficient1/activity coefficient 2) (no unit), to check for the consistency
+ i=i+1;
+end
+plot(x1,lngamma1_gamma2);//Plot of ln(gamma1/gamma2) vs x1 to determine A12 and A21
+xtitle('Plot of ln(gamma1/gamma2) vs x1','x1','ln(gamma1/gamma2)');
+//From the figure, the area above the x-axis and the area below the x-axis are determined and the thermodynamic consistency is checked
+area_above=1515;//area above the x-axis from the above plot (no unit)
+area_below=1540;//area below the x-axis (absolute value) from the above plot (no unit)
+consistency_parameter=abs((area_above-area_below)/(area_above+area_below));//calculating the paramter for checking the thermodynamic consistency (no unit)
+
+
+//OUTPUT
+mprintf('Values of ln(gamma1/gamma2): \n\n');
+i=1;
+mprintf('x1 \t gamma1 \t gamma2 \t ln(gamma1/gamma2)\n');
+
+for i=1:l
+ mprintf('%0.4f \t %f \t %f \t %f \n',x1(i),gaamma1(i),gaamma2(i),lngamma1_gamma2(i));
+end
+mprintf('\nThe value of the consistency parameter=%f\n',consistency_parameter);
+//0.02 is taken as the checking paramter for the consistency as prescribed by the author in the book on Page 449
+if consistency_parameter<0.02|consistency_parameter==0.02 then
+ mprintf('The VLE data is thermodynamically consistent');
+else
+ mprintf('The VLE data is not thermodynamically consistent');
+end
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH12/EX12.8/lngamma1_gamma2_vs_x1.jpg b/611/CH12/EX12.8/lngamma1_gamma2_vs_x1.jpg Binary files differnew file mode 100755 index 000000000..7a9b3b19c --- /dev/null +++ b/611/CH12/EX12.8/lngamma1_gamma2_vs_x1.jpg diff --git a/611/CH12/EX12.9/Chap12_Ex9_R1.sce b/611/CH12/EX12.9/Chap12_Ex9_R1.sce new file mode 100755 index 000000000..46b1266d3 --- /dev/null +++ b/611/CH12/EX12.9/Chap12_Ex9_R1.sce @@ -0,0 +1,86 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-12,Example 9,Page 464
+//Title: Temperature-composition diagram
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//For convenience benzene is taken as 1 and water is taken as 2. They form a completely immiscible system
+P=760;//pressure of the system in Torr
+antoine_const_benzene=[6.87987;1196.760;219.161];//Antoine's constants for Benzene from Table A.7
+t=60:5:100;//temperature range in degree celsius
+//saturation pressure of water(in torr)in the temperature range given by t (from steam tables)
+P2_s=[149.40;187.58;233.71;289.13;355.21;433.51;525.84;634.00;760.00];
+x1=0:0.2:1;//mole fraction of benzene in the liquid phase (no unit) (taken in an arbitrary manner)
+
+//CALCULATION
+//The form of the Antoine's equation used is logP=A-(B/(t+C)), where P is in Torr and t is in degree celsius
+//The three phase equilibrium temperature is estimated using the saturation pressure values,such that at the three phase equilibrium temperature,P=P1_s+P2_s=P as given by Eq.(12.57) Torr
+l=length(t);//iteration parameter
+i=1;//iteration parameter
+while i<l|i==l
+//calculation of saturation pressure of benzene at T in Torr
+P1_s(i)=10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(t(:,i)+antoine_const_benzene(3,:))));
+//calculating the total pressure in Torr so as to narrow down the temperature range for estimating the three phase equilibrium temperature
+P_tot(i)=P1_s(i)+P2_s(i,:);
+i=i+1;
+end
+//From the P_tot values calculated above, it is observed that the temperature range in which the three phase equilibrium temperature lies,is in between 65 and 70 degree celsius. Using linear interpolation , the three phase equilibrium temperature is determined in degree celsius
+T=(((t(:,3)-t(:,2))*(760-P_tot(2,:)))/(P_tot(3,:)-P_tot(2,:)))+t(:,2);//linear interpolation to determine the three phase equilibrium temperature in degree celsius
+//calculation of saturation pressure of benzene at the three phase temperature in Torr
+P1_s_three_phase=10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(T+antoine_const_benzene(3,:))));
+P2_s_three_phase=760-P1_s_three_phase;//calculation of the saturation pressure of water at the three phase temperature in Torr
+y1_three_phase=P1_s_three_phase/760;//calculation of the mole fraction of benzene in the vapour phase at the three phase equilibrium point (no unit)
+//redefining the temeprature range in degree celsius for computing the vapour compositions in the two phase regions. (As the three phase equilibrium temperature lies between 65 and 70 degree celsius)
+//The normal boiling point of benzene is given as 80.1 degree celsius (at a pressure of 760 Torr)
+trange1=T:1:T+11;//temperature range for calculating vapour phase composition of benzene in the two phase region given by (L1+V)
+n=length(trange1);//iteration parameter
+i=1;//iteration parameter
+while i<n|i==n
+ if i==1 then
+ y1(i)=y1_three_phase;//calculation of the vapour composition of benzene in the two phase region (L1+V) using Eq.(12.59) (no unit)
+ else
+ P1_s_calc(i)=10^(antoine_const_benzene(1,:)-(antoine_const_benzene(2,:)/(trange1(:,i)+antoine_const_benzene(3,:))));
+ y1(i)=(P1_s_calc(i))/P;//calculation of the vapour composition of benzene in the two phase region (L1+V) using Eq.(12.59) (no unit)
+ end
+ i=i+1;
+end
+trange2=[70;75;80;85;90;95;100];//temperature range for calculating vapour phase composition of benzene in the two phase region given by (L2+V)
+P2_s_range=[233.71;289.13;355.21;433.51;525.84;634.00;760.00];//saturation pressure of water(in torr)in the temperature range given by trange2 (from steam tables)
+p=length(trange2);//iteration parameter
+i=1;//iteration parameter
+//calculation of the vapour composition of benzene in the two phase region (L2+V) using Eq.(12.61) (no unit)
+y_one(i)=y1_three_phase;
+trange2(i)=T;
+i=i+1;
+while i<p|i==p
+ y_one(i)=(P-P2_s_range(i,:))/P;
+ i=i+1;
+end
+i=1;//iteration parameter
+k=length(x1);//iteration parameter
+while i<k|i==k
+ t_3phase(i)=T;//creating a vector for generating the plot at the three phase temperature
+ i=i+1;
+end
+
+//OUTPUT
+//Generating the T-x-y plot for the benzene-water system
+plot(y1,trange1);
+plot(y_one,trange2);
+plot(x1,t_3phase);
+ xtitle('t-x-y diagram for benzene-water sytem at 760 Torr','x1,y1','t (degree celsius)');
+q=length(t);//iteration parameter
+i=1;//iteration parameter
+mprintf('Calculations performed for determining the three phase equilibrium temperature\n');
+mprintf('t(degree celsius) \t P1_s (Torr) \t P2_s (Torr) \t P1_s+P2_s (Torr) \n');
+for i=1:q
+ mprintf('%d \t \t \t %f \t %0.2f \t %f \n',t(i),P1_s(i),P2_s(i),P_tot(i));
+end
+mprintf('The three phase equilibrium temperature=%0.2f degree celsius \n',T);
+mprintf('The vapour phase composition of benzene at the three phase equilibrium point=%0.4f \n',y1_three_phase);
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH12/EX12.9/txy_diagram.jpeg b/611/CH12/EX12.9/txy_diagram.jpeg Binary files differnew file mode 100755 index 000000000..bf24b4583 --- /dev/null +++ b/611/CH12/EX12.9/txy_diagram.jpeg diff --git a/611/CH13/EX13.1/Chap13_Ex1.sce b/611/CH13/EX13.1/Chap13_Ex1.sce new file mode 100755 index 000000000..66b2bece0 --- /dev/null +++ b/611/CH13/EX13.1/Chap13_Ex1.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-13,Example 1,Page 478
+//Title:Depression in freezing point
+//================================================================================================================
+clear
+clc
+
+//INPUT
+weight=10; //weight of NaCl in grams
+volume=1; //volume of water in litres
+weight_water=1000; // weight of water in grams (Weight=Volume*Density, density of water =1g/cc=1g/ml=1000g/l)
+molwt_NaCl=58.5; //molecular weight of NaCl in grams
+molwt_water=18; //molecular weight of water in grams
+hf=6.002; //enthalpy change of fusion in kJ/mol at 0 degree celsius
+P=101.325; //pressure in kPa
+T=273.15; // freezing point temperature of water at the given pressure in K
+R=8.314; //universal gas constant in J/molK;
+
+//CALCULATION
+x2=(weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water));// calculation of mole fraction of solute NaCl (no unit)
+delt=(R*T^2*x2)/(hf*10^3);//calculation of depression in freezing point of water using Eq.(13.14)
+
+//OUTPUT
+mprintf('\n The depression in freezing point of water when 10g of NaCl solute is added = %0.2f K',delt);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH13/EX13.2/Chap13_Ex2_R1.sce b/611/CH13/EX13.2/Chap13_Ex2_R1.sce new file mode 100755 index 000000000..9f7787158 --- /dev/null +++ b/611/CH13/EX13.2/Chap13_Ex2_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-13,Example 2,Page 480
+//Title:Elevation in Boiling Point
+//================================================================================================================
+clear
+clc
+
+//INPUT
+weight=10; //weight of NaCl in grams
+volume=1; //volume of water in litres
+weight_water=1000; // weight of water in grams (Weight=Volume*Density, density of water =1g/cc=1g/ml=1000g/l)
+molwt_NaCl=58.5; //molecular weight of NaCl in grams
+molwt_water=18; //molecular weight of water in grams
+lat_ht=2256.94; //latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)
+P=101.325; //pressure in kPa
+T=373.15; //boiling point temperature of water at the given pressure in K
+R=8.314; //universal gas constant in J/molK
+
+//CALCULATION
+x2=0.0031;//mole fraction of solute NaCl (From Example 13.1)(no unit)
+hv=(lat_ht*molwt_water)/1000; //conversion of latent heat from kJ/kg to kJ/mol
+delt=(R*T^2*x2)/(hv*10^3); //calculation of elevation in boiling point of water using Eq.(13.24)
+
+//OUTPUT
+mprintf('\n The elevation in boiling point of water when 10g of NaCl solute is added = %0.2f K',delt);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH13/EX13.3/Chap13_Ex3_R1.sce b/611/CH13/EX13.3/Chap13_Ex3_R1.sce new file mode 100755 index 000000000..325b8783d --- /dev/null +++ b/611/CH13/EX13.3/Chap13_Ex3_R1.sce @@ -0,0 +1,24 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-13,Example 3,Page 481
+//Title:Osmotic pressure
+//================================================================================================================
+clear
+clc
+
+//INPUT
+weight=10; //weight of NaCl in grams
+weight_water=1000; // weight of water in grams
+molwt_NaCl=58.5; //molecular weight of NaCl in grams
+molwt_water=18; //molecular weight of water in grams
+T=300; //prevailing temperature of water in K
+R=8.314; //universal gas constant in (Pa m^3)/(mol K);
+v=18*10^-6;//molar volume in m^3/mol
+//CALCULATION
+x2=0.0031;//mole fraction of solute NaCl (From Example 13.1)(no unit)
+pi=((R*T*x2)/v)*10^-3; // calulation of osmotic pressure using Eq(13.30)(in kPa)
+
+//OUTPUT
+mprintf('\n The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K = %0.2f kPa',pi);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH13/EX13.4/Chap13_Ex4.sce b/611/CH13/EX13.4/Chap13_Ex4.sce new file mode 100755 index 000000000..922270d48 --- /dev/null +++ b/611/CH13/EX13.4/Chap13_Ex4.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-13,Example 4,Page 483
+//Title:Ideal solubility
+//================================================================================================================
+clear
+clc
+
+//INPUT
+temp=20; // prevailing tempearture in degree celsius
+melt_temp=80.05; // melting point of naphthalene in degree celsius
+hf=18.574; // enthalpy of fusion in kJ/mol
+R=8.314; // universal gas constant in J/molK
+
+//CALCULATION
+t=temp+273.15; // convesion of prevailing temperature to K
+melt_t=melt_temp+273.15; //conversion of melting point of naphtalene to K
+x2=exp(((hf*10^3)/R)*((1/melt_t)-(1/t))); //calculation of ideal solubility using Eq.(13.40)(no unit)
+
+//OUTPUT
+mprintf('\n The ideal solubility of naphthalene at 20 degree celsius= %0.4f',x2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH13/EX13.5/Chap13_Ex5.sce b/611/CH13/EX13.5/Chap13_Ex5.sce new file mode 100755 index 000000000..c2f90edbd --- /dev/null +++ b/611/CH13/EX13.5/Chap13_Ex5.sce @@ -0,0 +1,21 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-13,Example 5,Page 483
+//Title:Solubility of gas
+//================================================================================================================
+clear
+clc
+
+//INPUT
+t=295.43; //prevailing temperature in K
+sat_p=6.05; //Sasturation pressure of carbon dioxide at the prevailing temperature in MPa
+p=0.1; //pressure at which solubility has to be determined in MPa
+
+//CALCULATION
+x2=p/sat_p; //calculation of solubility using Eq.(13.44)(no unit)
+
+//OUTPUT
+mprintf('\n The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa= %0.4f',x2);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH14/EX14.1/Chap14_Ex1_R1.sce b/611/CH14/EX14.1/Chap14_Ex1_R1.sce new file mode 100755 index 000000000..d2e04e283 --- /dev/null +++ b/611/CH14/EX14.1/Chap14_Ex1_R1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 1,Page 489
+//Title: Standard Gibbs free energy change and equilibrium constant
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The water gas shift reaction is given by : CO2(g)+H2(g)--->CO(g)+H2O(g)
+T=298.15;//temperature in K
+del_Gf=[-137.327;-228.600;-394.815;0];//the standard Gibbs free energy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ
+n=[1;1;-1;-1];//stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//calculation of the standard Gibbs free energy of reaction at 298.15K using Eq.(14.1) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:))+(n(4,:)*del_Gf(4,:));
+Ka=exp((-(del_G*10^3))/(R*T));//calculation of the equilibrium constant using Eq.(14.9) (no unit)
+
+//OUTPUT
+mprintf('The standard Gibbs free energy of the water gas shift reaction at 298.15K=%0.3f kJ \n',del_G);
+mprintf('The equilibrium constant of the water gas shift reaction at 298.15K=%0.3e \n',Ka);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH14/EX14.10/Chap14_Ex10.sce b/611/CH14/EX14.10/Chap14_Ex10.sce new file mode 100755 index 000000000..45776ca71 --- /dev/null +++ b/611/CH14/EX14.10/Chap14_Ex10.sce @@ -0,0 +1,80 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 10,Page 500
+//Title: Adiabatic reaction temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T0=298.15;//temperature at the entrance (feed) in K
+P=0.1;//pressure (operating) in MPa
+//The reaction is given by: H2(g)+(1/2)O2(g)--->H20(g)
+n=[1;-1;-0.5];//stoichiometric coefficients of H2O(g),H2(g)and O2(g) respectively (no unit)
+n_r=[1;0.5];//stochiometric coefficients on the reactant side alone for computing the right hand side of Eq.(A)
+m=[0;1;0.5];//inlet mole number of H2O(g),H2(g) and O2(g) respectively
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively)
+a=[28.850;27.012;30.255];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively)
+b=[12.055*10^-3;3.509*10^-3;4.207*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively)
+c=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively)
+e=[1.006*10^5;0.690*10^5;-1.887*10^5];
+del_H=-241.997;//enthalpy of reaction at 298.15K in kJ
+del_G=-228.600;//Gibbs free energy of reaction at 298.15K in kJ
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+mtot=m(1,:)+m(2,:)+m(3,:);//calculation of the total mole number of feed entering (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant
+I=(1/(R*T0))*(((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3)));
+//The conversion is computed by using Eq.(A) and by Eq.(B) and the two are plotted with respect to temperature. The point of intersection gives the adiabatic reaction temeperature and from that the conversion and the composition are determined.Let E_A denote the conversion obtained by using Eq.A and E_B denote the conversion obtained by using Eq.B (no unit)
+//For both the equations, conversion is determined for a temperature range of 2000 to 3800K, by incrementing temperature by 100K every time.
+T=2000:100:3800;//framing the temperature range in K
+l=length(T);//iteration parameter (no unit)
+i=1;//iteration parameter
+tol=1e-4;//tolerance limit for convergence of the system when using fsolve
+while i<l|i==l
+ del_H_T(i)=((del_H0*10^3)+((del_a*T(:,i))+((del_b/2)*T(:,i)^2)+((del_c/3)*T(:,i)^3)+((del_d/4)*T(:,i)^4)-(del_e/T(:,i))))*10^-3;
+del_G_T(i)=((del_H0*10^3)-(del_a*T(:,i)*log(T(:,i)))-((del_b/2)*T(:,i)^2)-((del_c/6)*T(:,i)^3)-((del_d/12)*T(:,i)^4)-((del_e/2)*(1/T(:,i)))-(I*R*T(:,i)))*10^-3;
+ Ka(i)=exp(-(del_G_T(i)*10^3)/(R*T(:,i)));//calculation of the equilibrium constant (no unit)
+ //using Eq.A to determine the conversion(no unit)
+E_A(i)=(1/del_H_T(i)*10^-3)*(-((((n_r(1,:)*a(2,:))+(n_r(2,:)*a(3,:)))*(T(:,i)-T0))+((((n_r(1,:)*b(2,:))+(n_r(2,:)*b(3,:)))/2)*((T(:,i))^2-(T0^2)))+((((n_r(1,:)*c(2,:))+(n_r(2,:)*c(3,:)))/3)*((T(:,i))^3-(T0^3)))+((((n_r(1,:)*d(2,:))+(n_r(2,:)*d(3,:)))/4)*((T(:,i))^4-(T0^4)))+(((n_r(1,:)*a(2,:))+(n_r(2,:)*a(3,:)))*((1/T(:,i))-(1/T0)))));
+Eguess(i)=0.99;//taking a guess value for the conversion (no unit)
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system (Using Eq.B to determine the conversion (no unit))
+fn=((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)))-Ka(i);
+endfunction
+[E_B(i)]=fsolve(Eguess(i),solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+i=i+1
+end
+//plotting the conversions determined above (using Eqs.A and B respectively) against temperature to determine the adiabatic reaction temperature in K
+plot(T,E_A,T,E_B);
+legends(['Equation (A)';'Equation (B)'],[2,3],opt="lr");
+xtitle('Plot of degree of conversion versus adiabatic reaction temperature','T(K)','E');
+//From the above plot, it is determined that the point of intersection occurs around 3440K, which is taken as the reaction temperature, where the conversion=0.68(no unit). Therefore, the conversion at the adiabatic reaction temperature is 0.68
+T_adiabatic=3440;//the adiabatic reaction temperature in K
+E_adiabatic=0.68;//conversion at the adiabatic reaction temperature (no unit)
+//Calculation of the composition of the burned gas (H2,O2 and H2O respectively) at the adiabatic reaction temperature (no unit)
+y_H2=((m(2,:)+(n(2,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic)));
+y_O2=((m(3,:)+(n(3,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic)));
+y_H2O=((m(1,:)+(n(1,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic)));
+
+//OUTPUT
+mprintf('\n The adiabatic reaction temperature=%d K\n',T_adiabatic);
+mprintf('\n The composition of the burned gases is given by: y_H2=%0.4f \t y_O2=%0.4f \t y_H2O=%0.4f \n',y_H2,y_O2,y_H2O);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.10/temp_vs_conversion.jpeg b/611/CH14/EX14.10/temp_vs_conversion.jpeg Binary files differnew file mode 100755 index 000000000..3a232be6a --- /dev/null +++ b/611/CH14/EX14.10/temp_vs_conversion.jpeg diff --git a/611/CH14/EX14.11/Chap14_Ex11.sce b/611/CH14/EX14.11/Chap14_Ex11.sce new file mode 100755 index 000000000..37c1d9af7 --- /dev/null +++ b/611/CH14/EX14.11/Chap14_Ex11.sce @@ -0,0 +1,58 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 11,Page 506
+//Title: Primary reactions
+//================================================================================================================
+clear
+clc
+
+
+//INPUT
+
+//The reactions occuring during steam reformation are given by:
+//CH4(g)+H2O(g)---->CO(g)+3H2(g)
+//CO(g)+H2O(g)----->CO2(g)+H2(g)
+//CH4(g)+2H2O(g)---->CO2(g)+4H2(g)
+//CO2(g)---->CO(g)+(1/2)O2(g)
+//CH4(g)+2O2(g)---->CO2(g)+2H2O(g)
+//CH4(g)+O2(g)---->CO(g)+H2O(g)+H2(g)
+
+//Let CH4=A1,H2O=A2,CO=A3,H2=A4,CO2=A5,O2=A6
+
+stoichio_matrix=[-1 -1 1 3 0 0;0 -1 -1 1 1 0;-1 -2 0 4 1 0;0 0 1 0 -1 0.5;-1 2 0 0 1 -2;-1 1 1 1 0 -1]//Framing the stoichiometric coefficient matrix
+
+
+
+//CALCULATION
+
+r=rank(stoichio_matrix);//Determining the rank of the given matrix(number of independent row/columns),which is indicative of the number of primary reactions
+
+//Performing elementary row operations to obtain diagonal elements as 0 or 1 and all elements below the diagonal as zero
+stoichio_matrix(1,:)=-stoichio_matrix(1,:);
+stoichio_matrix(3,:)=stoichio_matrix(3,:)+stoichio_matrix(1,:);
+stoichio_matrix(5,:)=stoichio_matrix(5,:)+stoichio_matrix(1,:);
+stoichio_matrix(6,:)=stoichio_matrix(6,:)+stoichio_matrix(1,:);
+stoichio_matrix(2,:)=-stoichio_matrix(2,:);
+stoichio_matrix(3,:)=stoichio_matrix(3,:)+stoichio_matrix(2,:);
+stoichio_matrix(5,:)=stoichio_matrix(5,:)-(3*stoichio_matrix(2,:));
+stoichio_matrix(6,:)=stoichio_matrix(6,:)-(2*stoichio_matrix(2,:));
+x=stoichio_matrix(:,3);
+y=stoichio_matrix(:,4);
+stoichio_matrix(:,3)=y;
+stoichio_matrix(:,4)=x;
+stoichio_matrix(5,:)=stoichio_matrix(5,:)+(4*stoichio_matrix(4,:));
+stoichio_matrix(6,:)=stoichio_matrix(6,:)+(2*stoichio_matrix(4,:));
+
+
+
+//OUTPUT
+mprintf('\n The stoichiometric coefficient matrix after performing the elementary row operations=\n');
+disp(stoichio_matrix);
+mprintf('\n The number of primary reactions=%d\n',r);
+mprintf('\n The non zero rows are (1,2,4)\n');
+mprintf('\n The primary reactions are: CH4(g)+H2O(g)--->CO(g)+3H2(g), CO(g)+H2O(g)--->CO2(g)+H2(g), CO2(g)--->CO(g)+(1/2)O2(g)\n');
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH14/EX14.13/Chap14_Ex13.sce b/611/CH14/EX14.13/Chap14_Ex13.sce new file mode 100755 index 000000000..74d205641 --- /dev/null +++ b/611/CH14/EX14.13/Chap14_Ex13.sce @@ -0,0 +1,55 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 13,Page 510
+//Title: Equilibrium composition in a simultaneous reaction
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The simultaneous reactions are given as:
+//A+B--->C+D =>1
+//A+C--->D+E =>2
+Ka1=0.1429;//equilibrium constant of reaction 1 (no unit)
+Ka2=2;//equilibrium constant of reaction 2 (no unit)
+P=1;//pressure in bar
+m=[1;1;0;0;0];//mole number in the feed (for A,B,C,D and E respectively) (equimolar mixture of A and B are present in the feed)
+n1=[-1;-1;1;1;0];//stoichiometric coefficients for reaction 1 (A,B,C D,E respectively) (no unit)
+n2=[-1;0;-1;1;1];//stoichiometric coefficients for reaction 2 (A,B,C D E respectively) (no unit)
+
+//CALCULATION
+del_n1=n1(1,:)+n1(2,:)+n1(3,:)+n1(4,:);//calculation of the total mole number for reaction 1 (no unit)
+del_n2=n2(1,:)+n2(2,:)+n2(3,:)+n2(4,:);//calculation of the total mole number for reaction 2 (no unit)
+//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (for reaction 1) (K_phi=1.0,assuming ideal gas behaviour)
+Ky1=Ka1/(P^del_n1);
+//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (for reaction 2) (K_phi=1.0,assuming ideal gas behaviour)
+Ky2=Ka2/(P^del_n2);
+mtot=m(1,:)+m(2,:)+m(3,:)+m(4,:)+m(5,:);//calculation of the total mole number of feed entering (no unit)
+//To determine the degree of conversion, a trial and error process is used to solve the equations given by Ky1=(y_C*y_D)/(y_A*y_B), and Ky2=(y_D*y_E)/(y_A*y_C) where y_A,y_B,y_C,y_D,y_E are the mole fractions of A,B,C,D and E respectively. Let the equilibrium conversion be denoted as epsilon1 for reaction 1 and epsilon2 for reaction 2 respectively.
+epsilon1_guess=0.3;//taking a guess value for the degree of conversion (reaction1) for the trial and error process (no unit)
+tol=1e-6;//defining the tolerance limit for obtaining the convergence of the system using fsolve
+E_guess=0.1;//taking a guess value for the degree of conversion (reaction2) to be used for solving the set of equations by the inbuilt function fsolve
+function[fn1]=solver_func1(En)
+ //The system of equations to be solved for reaction 1
+ fn1=Ky1-((((m(3,:)+(n1(3,:)*epsilon1_guess)+(n2(3,:)*En))/(mtot))^n1(3,:))*(((m(4,:)+(n1(4,:)*epsilon1_guess)+(n2(4,:)*En))/(mtot))^n1(4,:))*(((m(1,:)+(n1(1,:)*epsilon1_guess)+(n2(1,:)*En))/(mtot))^n1(1,:))*(((m(2,:)+(n1(2,:)*epsilon1_guess)+(n2(2,:)*En))/(mtot))^n1(2,:)));
+endfunction
+[epsilon2]=fsolve(E_guess,solver_func1,tol);//using inbuilt function fsolve for solving the system of equations
+E_guess=0.2;//taking a guess value for the degree of conversion (reaction1) to be used for solving the set of equations by the inbuilt function fsolve
+function[fn2]=solver_func2(Em)
+ //For reaction 2, the degree of conversion(reaction2), determined above is used along with the guess value and the system of equations below are solved
+fn2=Ky2-((((m(4,:)+(n1(4,:)*Em)+(n2(4,:)*epsilon2))/(mtot))^n2(4,:))*(((m(5,:)+(n1(5,:)*Em)+(n2(5,:)*epsilon2))/(mtot))^n2(5,:))*(((m(1,:)+(n1(1,:)*Em)+(n2(1,:)*epsilon2))/(mtot))^n2(1,:))*(((m(3,:)+(n1(3,:)*Em)+(n2(3,:)*epsilon2))/(mtot))^n2(3,:)));
+endfunction
+[epsilon1]=fsolve(E_guess,solver_func2,tol);//using inbuilt function fsolve for solving the system of equations
+//calculation of the equilibrium composition at 1 bar pressure of A,B,C,D and E respectively (no unit)
+y_A=((m(1,:)+(n1(1,:)*epsilon1)+(n2(1,:)*epsilon2)))/(mtot);
+y_B=(m(2,:)+(n1(2,:)*epsilon1)+(n2(2,:)*epsilon2))/(mtot);
+y_C=((m(3,:)+(n1(3,:)*epsilon1)+(n2(3,:)*epsilon2)))/(mtot);
+y_D=(m(4,:)+(n1(4,:)*epsilon1)+(n2(4,:)*epsilon2))/(mtot);
+y_E=((m(5,:)+(n1(5,:)*epsilon1)+(n2(5,:)*epsilon2)))/(mtot);
+
+//OUTPUT
+mprintf('\nThe degree of conversion :epsilon1=%0.1f\t epsilon2=%0.1f \n',epsilon1,epsilon2);
+mprintf('\nThe equilibrium composition at 1 bar pressure for an equimolar mixture of A and B fed to the reactor:\n y_A=%0.2f \t y_B=%0.2f \t y_C=%0.2f \t y_D=%0.2f \t y_E=%0.1f\n',y_A,y_B,y_C,y_D,y_E);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.14/Chap14_Ex14.sce b/611/CH14/EX14.14/Chap14_Ex14.sce new file mode 100755 index 000000000..839705319 --- /dev/null +++ b/611/CH14/EX14.14/Chap14_Ex14.sce @@ -0,0 +1,38 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 14,Page 515
+//Title: Equilibrium concentration
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The reaction is given by :
+//CH3COOH(l)+C2H5OH(l)--->CH3COOC2H5(l)+H2O(l)
+T=100;//temperature in degree celsius
+Kc=2.92;//equilibrium constant (in terms of concentration) at T (no unit)
+v=1;//volume of the aqueous solution in m^3
+m=[0;10;10;5];//feed composition of CH3COOC2H5(l),H20(l),C2H5OH(l),CH3COOH(l) respectively in kmol
+n=[1;1;-1;-1];//stoichiometric coefficient for the reaction (no unit)(CH3COOC2H5(l),H20(l),C2H5OH(l),CH3COOH(l) respectively)
+
+//CALCULATION
+//For convenience, CH3COOH(l) is denoted as A, C2H5OH(l) is denoted as B, CH3COOC2H5(l) is denoted as C and H2O(l) as D
+//Calculation of the extent of the reaction, expressed in concentration units.The inbuilt function fsolve is used for solving the set of equations
+tol=1e-6;//tolerance limit framed for the convergence of the system of equations by using fsolve
+Eguess=1;//taking a guess value for the extent of reaction (no unit)
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system
+ fn=Kc-(((m(1,:)+n(1,:)*Ei)^n(1,:))*((m(2,:)+n(2,:)*Ei)^n(2,:))*((m(3,:)+n(3,:)*Ei)^n(3,:))*((m(4,:)+n(4,:)*Ei)^n(4,:)));
+endfunction
+[E]=fsolve(Eguess,solver_func,tol);//using inbuilt function fsolve for solving the system of equations
+C_A=(m(4,:)+n(4,:)*E);//equilibrium concentration of CH3COOH(l) (no unit)
+C_B=(m(3,:)+n(3,:)*E);//equilibrium concentration of C2H5OH(l) (no unit)
+C_C=(m(1,:)+n(1,:)*E);//equilibrium concentration of CH3COOC2H5(l) (no unit)
+C_D=(m(2,:)+n(2,:)*E);//equilibrium concentration of H2O(l) (no unit)
+
+//OUTPUT
+mprintf('\n The extent of reaction, expressed in concentration units=%0.4f\n',E);
+mprintf('\n The equilibrium concentration:C_A=%0.4f kmol/m^3 \t C_B=%0.4f kmol/m^3 \t C_C=%0.4f kmol/m^3 \t C_D=%0.4f kmol/m^3 \n',C_A,C_B,C_C,C_D);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH14/EX14.15/Chap14_Ex15.sce b/611/CH14/EX14.15/Chap14_Ex15.sce new file mode 100755 index 000000000..f4a33ed6c --- /dev/null +++ b/611/CH14/EX14.15/Chap14_Ex15.sce @@ -0,0 +1,56 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 15,Page 517
+//Title: Decomposition pressure
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The reaction is given by: CaCO3(s)--->CaO(s)+CO2(g)
+T=1200;//temperature in K
+T0=298.15;//reference temperature in K
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K
+//coefficient in the expression for computing the isobaric molar heat capacity (for CaO(s),CO2(g),CaCO3(s) respectively)
+a=[41.84;45.369;82.34];
+//coefficient in the expression for computing the isobaric molar heat capacity (for CaO(s),CO2(g),CaCO3(s) respectively)
+b=[20.25*10^-3;8.688*10^-3;49.75*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity (for CaO(s),CO2(g),CaCO3(s) respectively)
+c=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity (for CaO(s),CO2(g),CaCO3(s) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity (for CaO(s),CO2(g),CaCO3(s) respectively)
+e=[-4.51*10^5;-9.619*10^5;-12.87*10^5];
+del_Gf=[-604.574;-394.815;-1129.515]//Standard Gibbs free energies of formation of (CaO(s),CO2(g),CaCO3(s)) in kJ
+del_Hf=[-635.975;-393.978;-1207.683]//Standard enthalpies of formation of (CaO(s),CO2(g),CaCO3(s))in kJ
+n=[1;1;-1];//stoichiometric coefficients of CaO(s),CO2(g) and CaCO3(s) respectively (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=(n(1,:)*del_Hf(1,:))+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the value of IR (no unit)
+IR=(1/(T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(IR*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+//Now, Ka=(a_CaO*a_CO2)/a_CaCO3. We get a_CaO=1 and a_CaCO3=1, if we choose the pure component solids CaO(s) and CaCO3(s) at 1200K(at T)and 1 bar pressure as the standard states. Then, Ka=a_CO2=(f/f0)_CO2=((phi*y*P)/f0)_CO2. Assume the gas phase (pure CO2) is ideal. Then, phi=1 and y=1. The usual standard state for the gas gives f0=1 bar.Therefore, Ka=P
+y=1;
+phi=1;
+f0=1;
+P=(Ka*f0)/(phi*y);//calculation of the decomposition pressure in bar
+
+//OUTPUT
+mprintf('\n The decomposition pressure,P=%f bar \n',P);
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE TEXTBOOK, GIVES A VALUE OF 2.42 bar FOR THE VALUE OF THE DECOMPOSITION PRESSURE. HOWEVER, THE ACTUAL VALUE IS ONLY 2.38 bar AND NOT 2.42 bar AS PRINTED IN THE TEXTBOOK.
diff --git a/611/CH14/EX14.2/Chap14_Ex2_R1.sce b/611/CH14/EX14.2/Chap14_Ex2_R1.sce new file mode 100755 index 000000000..a45509b6a --- /dev/null +++ b/611/CH14/EX14.2/Chap14_Ex2_R1.sce @@ -0,0 +1,35 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 2,Page 490
+//Title: Standard Gibbs free energy of formation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=298.15;//temperature in K
+P_s=0.16716;//saturation pressure of CH3OH in bar at T
+//The reactions can be written down as:
+//C(s)+2H2(g)+(1/2)O2(g)---->CH3OH(l)--->del_G1
+//CH3OH(l)--->CH3OH(g)--->del_G2
+//Overall: C(s)+2H2(g)+(1/2)O2(g)--->CH3OH(g)--->del_G=del_G2+del_G1
+del_G1=-166.215;//standard Gibbs free energy of formation of CH3OH(l) in kJ
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//Now, the value of del_G2 has to be computed, from which del_G can be determined. The standard state for CH3OH(l) is 1 bar and 298.15K
+//del_G2 is given by, del_G2=RTln(f_v/f_l), where f_v and f_l are the fugacities of the vapour and liquid phases respectively
+//At 1 bar pressure, the vapour is an ideal gas and hence its fugacity is equal to pressure
+f_v=1;//fugacity of the vapour in bar
+f_l=P_s;//fugacity of the liquid is the saturation pressure at T, in bar
+del_G2=R*T*log(f_v/f_l)*10^-3;//calculation of the value of del_G2 in kJ
+del_G=del_G2+del_G1;//calculation of the standard Gibbs free energy of formation of CH3OH(g) in kJ
+
+//OUTPUT
+mprintf('The standard Gibbs free energy of formation of CH3OH(g)=%0.3f kJ \n',del_G);
+
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH14/EX14.3/Chap14_Ex3.sce b/611/CH14/EX14.3/Chap14_Ex3.sce new file mode 100755 index 000000000..d53024e71 --- /dev/null +++ b/611/CH14/EX14.3/Chap14_Ex3.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 3,Page 491
+//Title: Equilibrium constant
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The water gas shift reaction is given by: CO2(g)+H2(g)--->CO(g)+H2O(g)
+T1=298.15;//initial temperature in K
+Ka1=8.685*10^-6;//equilibrium constant for the water-gas shift reaction at T1 (no unit)
+T2=1000;//temperature at which the equilibrium constant has to be determined in K
+R=8.314;//universal gas constant in J/molK
+del_Hf=[-110.532;-241.997;-393.978;0];//the standard enthalpy of formation of CO(g),H2O(g),CO2(g) and H2(g) in kJ
+n=[1;1;-1;-1];//stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)
+
+//CALCULATION
+//It is assumed that del_H is constant in the temperature range T1 and T2
+del_H=(n(1,:)*del_Hf(1,:))+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:))+(n(4,:)*del_Hf(4,:));//calculation of the standard enthalpy of the reaction in kJ
+Ka2=Ka1*exp(((del_H*10^3)/R)*((1/T1)-(1/T2)));//calculation of the equilibrium constant at T2 (no unit)
+
+//OUTPUT
+mprintf('The equilibrium constant for the water gas shift reaction at 1000K=%f \n',Ka2);
+
+
+//===============================================END OF PROGRAM===================================================
+
+
+
+
diff --git a/611/CH14/EX14.4/Chap14_Ex4.sce b/611/CH14/EX14.4/Chap14_Ex4.sce new file mode 100755 index 000000000..260a8a019 --- /dev/null +++ b/611/CH14/EX14.4/Chap14_Ex4.sce @@ -0,0 +1,53 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 4,Page 492
+//Title: Equilibrium constant with enthalpy of reaction varying with temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//The water gas shift reaction is given by: CO2(g)+H2(g)--->CO(g)+H2O(g)
+P=0.1;//pressure in MPa
+T1=298.15;//initial temperature in K
+Ka1=8.685*10^-6;//equilibrium constant for the water-gas shift reaction at T1 (no unit) (from Example 14.1)
+T2=1000;//temperature at which the equilibrium constant is to be found, in K
+del_H=41.449;//standard enthalpy of the reaction at T1 in kJ (from Example 14.3)
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CO(g),H2O(g),CO2(g),H2(g) respectively)
+a=[28.068;28.850;45.369;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CO(g),H2O(g),CO2(g),H2(g) respectively)
+b=[4.631*10^-3;12.055*10^-3;8.688*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CO(g),H2O(g),CO2(g),H2(g) respectively)
+c=[0;0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CO(g),H2O(g),CO2(g),H2(g) respectively)
+d=[0;0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CO(g),H2O(g),CO2(g),H2(g) respectively)
+e=[-0.258*10^5;1.006*10^5;-9.619*10^5;0.690*10^5];
+n=[1;1;-1;-1];//stoichiometric coefficients of CO(g),H2O(g),CO2(g) and H2(g) respectively (no unit)
+R=8.314;//universal gas constant in J/molK
+Ka2_prev=1.0855;//equilibrium constant calculated in Example (14.3) without considering the variation of del_H between T1 and T2 (no unit)
+
+
+//CALCULATION
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:))+(n(4,:)*a(4,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:))+(n(4,:)*b(4,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:))+(n(4,:)*c(4,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:))+(n(4,:)*d(4,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:))+(n(4,:)*e(4,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=(del_H*10^3)-((del_a*T1)+((del_b/2)*T1^2)+((del_c/3)*T1^3)+((del_d/4)*T1^4)-(del_e/T1));
+//Calculation of the integration constant using Eq.(14.22) (no unit)
+I=(log(Ka1))-((1/R)*((-del_H0/T1)+(del_a*log(T1))+((del_b/2)*T1)+((del_c/6)*T1^2)+((del_d/12)*T1^3)+((del_e/(2*T1^2)))));
+//calculation of the equilibrium constant at T2 using Eq.(14.22) (no unit)
+Ka2=exp(((1/R)*((-del_H0/T2)+(del_a*log(T2))+((del_b/2)*T2)+((del_c/6)*T2^2)+((del_d/12)*T2^3)+((del_e/(2*T2^2)))))+I);
+
+//OUTPUT
+mprintf('The equilibrium constant for the water gas shift reaction at 1000K by taking into account the variation of del_H with temperature=%f \n',Ka2);
+mprintf('The equilibrium constant for the water gas shift reaction at 1000K without considering the variation of del_H with temperature as given by Example(14.3)=%0.4f \n',Ka2_prev);
+
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH14/EX14.5/Chap14_Ex5.sce b/611/CH14/EX14.5/Chap14_Ex5.sce new file mode 100755 index 000000000..4139aaf98 --- /dev/null +++ b/611/CH14/EX14.5/Chap14_Ex5.sce @@ -0,0 +1,70 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 5,Page 494
+//Title: Conversion and composition of the equilibrium mixture
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//Industrial methanol is produced by the following reaction:
+//CO(g)+2H2(g)--->CH3OH(g)
+T0=298.15;//standard temperature in K
+T=500;//temperature in K
+P=5;//pressure in bar
+del_Hv=37.988;//enthalpy of vapourization of CH3OH at 298.15K in kJ/mol
+R=8.314;//universal gas constant in J/molK
+del_Gf=[-161.781;-137.327;0]//Standard Gibbs free energies of formation of CH3OH(g) from Example(14.2),CO(g) and H2(g) respectively in kJ
+del_Hf=[-238.648;-110.532;0]//Standard enthalpies of formation of CH3OH(l), CO(g) and H2(g) respectively in kJ
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+a=[18.382;28.068;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+b=[101.564*10^-3;4.631*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+c=[-28.683*10^-6;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+e=[0;-0.258*10^5;0.690*10^5];
+n=[1;-1;-2];//stoichiometric coefficients of CH3OH(g),CO(g) and H2(g) respectively (no unit)
+m=[0;1;2];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively)
+
+//CALCULATION
+del_Hf_CH3OH_g=del_Hf(1,:)+del_Hv;//calculation of the standard enthalpy of formation of CH3OH(g) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=del_Hf_CH3OH_g+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant (no unit)
+I=(1/(R*T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(I*R*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+Ky=Ka/((P)^del_n);//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (K_phi=1.0,assuming ideal gas behaviour)
+mtot=m(1,:)+m(2,:)+m(3,:);//calculation of the total mole number of feed entering (no unit)
+//To determine the degree of conversion, the inbuilt function fsolve is used to solve the equation given by Ky=(y_CH3OH)/(y_CO*y_H2^2), where y_CH3OH,y_CO,y_H2 are the mole fractions of CH3OH,CO,H2 respectively. Let the equilibrium conversion be denoted as E
+E_guess=0.1;//taking a guess value for the degree of conversion,to be used in the inbuilt function fsolve (no unit)
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)));
+endfunction
+[E]=fsolve(E_guess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+//Calculation of the composition of the equilibrium mixture (for CH3OH(g),CO(g),H2(g) respectively)(no unit)
+y_CH3OH=(m(1,:)+(n(1,:)*E))/(mtot+(del_n*E));
+y_CO=(m(2,:)+(n(2,:)*E))/(mtot+(del_n*E));
+y_H2=(m(3,:)+(n(3,:)*E))/(mtot+(del_n*E));
+
+//OUTPUT
+mprintf('The degree of conversion at 500K and 5bar pressure=%0.4f\n',E);
+mprintf('The composition of the equilibrium mixture at 500K and 5bar pressure: y_CH3OH=%0.4f\t y_CO=%0.4f\t y_H2=%0.4f\n',y_CH3OH,y_CO,y_H2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.6/Chap14_Ex6.sce b/611/CH14/EX14.6/Chap14_Ex6.sce new file mode 100755 index 000000000..b456b6c9b --- /dev/null +++ b/611/CH14/EX14.6/Chap14_Ex6.sce @@ -0,0 +1,78 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 6,Page 496
+//Title: Conversion and composition of the equilibrium mixture at 5 and 100 bar Pressures
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//Industrial methanol is produced by the following reaction:
+//CO(g)+2H2(g)--->CH3OH(g)
+T0=298.15;//standard temperature in K
+T=500;//temperature in K
+P1=5;//pressure in bar
+P2=100;//pressure in bar
+del_Hv=37.988;//enthalpy of vapourization of CH3OH at 298.15K in kJ/mol
+R=8.314;//universal gas constant in J/molK
+del_Gf=[-161.781;-137.327;0]//Standard Gibbs free energies of formation of CH3OH(g) from Example(14.2),CO(g) and H2(g) respectively in kJ
+del_Hf=[-238.648;-110.532;0]//Standard enthalpies of formation of CH3OH(l), CO(g) and H2(g) respectively in kJ
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+a=[18.382;28.068;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+b=[101.564*10^-3;4.631*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+c=[-28.683*10^-6;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+e=[0;-0.258*10^5;0.690*10^5];
+n=[1;-1;-2];//stoichiometric coefficients of CH3OH(g),CO(g) and H2(g) respectively (no unit)
+m=[0;1;2];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively)
+
+//CALCULATION
+//From Example 14.5, the conversion and the equilibrium composition has been determined and this is given below:
+E1=0.0506;
+y_CH3OH_1=0.0175;
+y_CO_1=0.3275;
+y_H2_1=0.6550;
+//Calculation of conversion and equilibrium composition for the pressure of 100 bars(P2)
+del_Hf_CH3OH_g=del_Hf(1,:)+del_Hv;//calculation of the standard enthalpy of formation of CH3OH(g) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=del_Hf_CH3OH_g+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant (no unit)
+I=(1/(R*T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(I*R*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+Ky=Ka/((P2)^del_n);//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (K_phi=1.0,assuming ideal gas behaviour)
+mtot=m(1,:)+m(2,:)+m(3,:);//calculation of the total mole number of feed entering (no unit)
+//To determine the degree of conversion, the inbuilt function fsolve is used to solve the equation given by Ky=(y_CH3OH)/(y_CO*y_H2^2), where y_CH3OH,y_CO,y_H2 are the mole fractions of CH3OH,CO,H2 respectively. Let the equilibrium conversion be denoted as E
+E_guess=0.1;//taking a guess value for the degree of conversion,to be used in the inbuilt function fsolve (no unit)
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)));
+endfunction
+[E2]=fsolve(E_guess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+//Calculation of the composition of the equilibrium mixture (for CH3OH(g),CO(g),H2(g) respectively)(no unit)
+y_CH3OH_2=(m(1,:)+(n(1,:)*E2))/(mtot+(del_n*E2));
+y_CO_2=(m(2,:)+(n(2,:)*E2))/(mtot+(del_n*E2));
+y_H2_2=(m(3,:)+(n(3,:)*E2))/(mtot+(del_n*E2));
+
+//OUTPUT
+mprintf('The degree of conversion at 500K and 5bar pressure=%0.4f\n',E1);
+mprintf('The composition of the equilibrium mixture at 500K and 5bar pressure: y_CH3OH=%0.4f\t y_CO=%0.4f\t y_H2=%0.4f\n',y_CH3OH_1,y_CO_1,y_H2_1);
+mprintf('The degree of conversion at 500K and 100bar pressure=%0.3f\n',E2);
+mprintf('The composition of the equilibrium mixture at 500K and 100bar pressure: y_CH3OH=%0.4f\t y_CO=%0.4f\t y_H2=%f\n',y_CH3OH_2,y_CO_2,y_H2_2);
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.7/Chap14_Ex7.sce b/611/CH14/EX14.7/Chap14_Ex7.sce new file mode 100755 index 000000000..a3739c0df --- /dev/null +++ b/611/CH14/EX14.7/Chap14_Ex7.sce @@ -0,0 +1,70 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 7,Page 497
+//Title: Conversion and composition of the equilibrium mixture with inerts
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//Industrial methanol is produced by the following reaction:
+//CO(g)+2H2(g)--->CH3OH(g)
+T0=298.15;//standard temperature in K
+T=500;//temperature in K
+P=5;//pressure in bar
+del_Hv=37.988;//enthalpy of vapourization of CH3OH at 298.15K in kJ/mol
+R=8.314;//universal gas constant in J/molK
+del_Gf=[-161.781;-137.327;0]//Standard Gibbs free energies of formation of CH3OH(g) from Example(14.2),CO(g) and H2(g) respectively in kJ
+del_Hf=[-238.648;-110.532;0]//Standard enthalpies of formation of CH3OH(l), CO(g) and H2(g) respectively in kJ
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+a=[18.382;28.068;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+b=[101.564*10^-3;4.631*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+c=[-28.683*10^-6;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+e=[0;-0.258*10^5;0.690*10^5];
+n=[1;-1;-2];//stoichiometric coefficients of CH3OH(g),CO(g),H2(g) respectively (no unit)
+//The inert is denoted as A
+m=[0;1;2;5];//mole number in feed (for CH3OH(g),CO(g),H2(g),A(g) respectively)
+
+//CALCULATION
+del_Hf_CH3OH_g=del_Hf(1,:)+del_Hv;//calculation of the standard enthalpy of formation of CH3OH(g) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=del_Hf_CH3OH_g+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant (no unit)
+I=(1/(R*T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(I*R*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+Ky=Ka/((P)^del_n);//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (K_phi=1.0,assuming ideal gas behaviour)
+mtot=m(1,:)+m(2,:)+m(3,:)+m(4,:);//calculation of the total mole number of feed entering (no unit)
+//To determine the degree of conversion, the inbuilt function fsolve is used to solve the equation given by Ky=(y_CH3OH)/(y_CO*y_H2^2), where y_CH3OH,y_CO,y_H2 are the mole fractions of CH3OH,CO,H2 respectively. Let the equilibrium conversion be denoted as E
+E_guess=0.1;//taking a guess value for the degree of conversion,to be used in the inbuilt function fsolve (no unit)
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)));
+endfunction
+[E]=fsolve(E_guess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+//Calculation of the composition of the equilibrium mixture (for CH3OH(g),CO(g),H2(g),A(g) respectively)(no unit)
+y_CH3OH=(m(1,:)+(n(1,:)*E))/(mtot+(del_n*E));
+y_CO=(m(2,:)+(n(2,:)*E))/(mtot+(del_n*E));
+y_H2=(m(3,:)+(n(3,:)*E))/(mtot+(del_n*E));
+y_A=m(4,:)/(mtot+(del_n*E));
+//OUTPUT
+mprintf('The degree of conversion at 500K and 5bar pressure=%0.5f\n',E);
+mprintf('The composition of the equilibrium mixture at 500K and 5bar pressure: y_CH3OH=%0.5f\t y_CO=%0.5f\t y_H2=%0.5f\t y_A=%0.4f \n',y_CH3OH,y_CO,y_H2,y_A);
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.8/Chap14_Ex8.sce b/611/CH14/EX14.8/Chap14_Ex8.sce new file mode 100755 index 000000000..d21b569c2 --- /dev/null +++ b/611/CH14/EX14.8/Chap14_Ex8.sce @@ -0,0 +1,89 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 8,Page 498
+//Title: Degree of conversion for different feed conditions
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//Industrial methanol is produced by the following reaction:
+//CO(g)+2H2(g)--->CH3OH(g)
+T0=298.15;//standard temperature in K
+T=500;//temperature in K
+P=5;//pressure in bar
+del_Hv=37.988;//enthalpy of vapourization of CH3OH at 298.15K in kJ/mol
+R=8.314;//universal gas constant in J/molK
+del_Gf=[-161.781;-137.327;0]//Standard Gibbs free energies of formation of CH3OH(g) from Example(14.2),CO(g) and H2(g) respectively in kJ
+del_Hf=[-238.648;-110.532;0]//Standard enthalpies of formation of CH3OH(l), CO(g) and H2(g) respectively in kJ
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+a=[18.382;28.068;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+b=[101.564*10^-3;4.631*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+c=[-28.683*10^-6;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+e=[0;-0.258*10^5;0.690*10^5];
+n=[1;-1;-2];//stoichiometric coefficients of CH3OH(g),CO(g) and H2(g) respectively (no unit)
+//The degree of conversion has been determined for 3 different feed conditions:
+//a) an equimolar mixture of CO(g) and H2(g) is fed to the reactor
+//b) stoichiometric mixture of CO(g) and H2(g) is fed to the reactor
+//c) CO(g) and H2(g) in the ratio 1:4 enter the reactor
+m_a=[0;1;1];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively for condition (a))
+m_b=[0;1;2];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively for condition (b))
+m_c=[0;1;4];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively for condition (c))
+
+//CALCULATION
+del_Hf_CH3OH_g=del_Hf(1,:)+del_Hv;//calculation of the standard enthalpy of formation of CH3OH(g) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=del_Hf_CH3OH_g+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant (no unit)
+I=(1/(R*T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(I*R*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+Ky=Ka/((P)^del_n);//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (K_phi=1.0,assuming ideal gas behaviour)
+mtot_a=m_a(1,:)+m_a(2,:)+m_a(3,:);//calculation of the total mole number of feed entering (no unit) (for condition (a))
+mtot_b=m_b(1,:)+m_b(2,:)+m_b(3,:);//calculation of the total mole number of feed entering (no unit) (for condition (b))
+mtot_c=m_c(1,:)+m_c(2,:)+m_c(3,:);//calculation of the total mole number of feed entering (no unit) (for condition (c))
+//To determine the degree of conversion, the inbuilt function fsolve is used to solve the equation given by Ky=(y_CH3OH)/(y_CO*y_H2^2), where y_CH3OH,y_CO,y_H2 are the mole fractions of CH3OH,CO,H2 respectively. Let the equilibrium conversion be denoted as E. This is done for all the three conditions (a,b and c)
+E_guess=0.1;//taking a guess value for the degree of conversion,to be used in the inbuilt function fsolve (no unit)
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+//For condition (a)
+function[fn]=solver_func1(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m_a(1,:)+(n(1,:)*Ei))/(mtot_a+(del_n*Ei)))^n(1,:))*(((m_a(2,:)+(n(2,:)*Ei))/(mtot_a+(del_n*Ei)))^n(2,:))*(((m_a(3,:)+(n(3,:)*Ei))/(mtot_a+(del_n*Ei)))^n(3,:)));
+endfunction
+[E_a]=fsolve(E_guess,solver_func1,tol)//using inbuilt function fsolve for solving the system of equations
+//For condition (b)
+function[fn]=solver_func2(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m_b(1,:)+(n(1,:)*Ei))/(mtot_b+(del_n*Ei)))^n(1,:))*(((m_b(2,:)+(n(2,:)*Ei))/(mtot_b+(del_n*Ei)))^n(2,:))*(((m_b(3,:)+(n(3,:)*Ei))/(mtot_b+(del_n*Ei)))^n(3,:)));
+endfunction
+[E_b]=fsolve(E_guess,solver_func2,tol)//using inbuilt function fsolve for solving the system of equations
+//For condition (c)
+function[fn]=solver_func3(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m_c(1,:)+(n(1,:)*Ei))/(mtot_c+(del_n*Ei)))^n(1,:))*(((m_c(2,:)+(n(2,:)*Ei))/(mtot_c+(del_n*Ei)))^n(2,:))*(((m_c(3,:)+(n(3,:)*Ei))/(mtot_c+(del_n*Ei)))^n(3,:)));
+endfunction
+[E_c]=fsolve(E_guess,solver_func3,tol)//using inbuilt function fsolve for solving the system of equations
+
+//OUTPUT
+mprintf('The degree of conversion at 500K and 5bar pressure, for an equimolar mixture of CO(g) and H2(g) as feed=%f\n',E_a);
+mprintf('The degree of conversion at 500K and 5bar pressure, for a stoichiometric mixture of CO(g) and H2(g) as feed=%0.4f\n',E_b);
+mprintf('The degree of conversion at 500K and 5bar pressure, for a feed of CO(g) and H2(g) in the ratio of 1:4=%f\n',E_c);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH14/EX14.9/Chap14_Ex9.sce b/611/CH14/EX14.9/Chap14_Ex9.sce new file mode 100755 index 000000000..bdbe5d2fe --- /dev/null +++ b/611/CH14/EX14.9/Chap14_Ex9.sce @@ -0,0 +1,66 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-14,Example 9,Page 499
+//Title: Degree of conversion
+//================================================================================================================
+clear
+clc
+
+//INPUT
+//Industrial methanol is produced by the following reaction:
+//CO(g)+2H2(g)--->CH3OH(g)
+T0=298.15;//standard temperature in K
+T=500;//temperature in K
+P=5;//pressure in bar
+del_Hv=37.988;//enthalpy of vapourization of CH3OH at 298.15K in kJ/mol
+R=8.314;//universal gas constant in J/molK
+del_Gf=[-161.781;-137.327;0]//Standard Gibbs free energies of formation of CH3OH(g) from Example(14.2),CO(g) and H2(g) respectively in kJ
+del_Hf=[-238.648;-110.532;0]//Standard enthalpies of formation of CH3OH(l), CO(g) and H2(g) respectively in kJ
+//The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+a=[18.382;28.068;27.012];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+b=[101.564*10^-3;4.631*10^-3;3.509*10^-3];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+c=[-28.683*10^-6;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+d=[0;0;0];
+//coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for CH3OH(g),CO(g),H2(g) respectively)
+e=[0;-0.258*10^5;0.690*10^5];
+n=[1;-1;-2];//stoichiometric coefficients of CH3OH(g),CO(g) and H2(g) respectively (no unit)
+m=[0.02;1;2];//mole number in feed (for CH3OH(g),CO(g),H2(g) respectively)
+
+//CALCULATION
+del_Hf_CH3OH_g=del_Hf(1,:)+del_Hv;//calculation of the standard enthalpy of formation of CH3OH(g) in kJ
+del_G=(n(1,:)*del_Gf(1,:))+(n(2,:)*del_Gf(2,:))+(n(3,:)*del_Gf(3,:));//calculation of the Gibbs free energy of reaction in kJ
+del_H=del_Hf_CH3OH_g+(n(2,:)*del_Hf(2,:))+(n(3,:)*del_Hf(3,:));//calculation of the enthalpy of the reaction in kJ
+//Framing the isobaric molar heat capacity expression
+del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:));
+del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:));
+del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:));
+del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:));
+del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:));
+//Using Eq.14.21 to compute the value of del_H0 in kJ
+del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3;
+//Using Eq.14.23 to compute the integration constant (no unit)
+I=(1/(R*T0))*((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3));
+//Using Eq.14.23 to compute the Gibbs free energy of the reaction at T in kJ
+del_G_T=((del_H0*10^3)-(del_a*T*log(T))-((del_b/2)*T^2)-((del_c/6)*T^3)-((del_d/12)*T^4)-((del_e/2)*(1/T))-(I*R*T))*10^-3;
+Ka=exp((-del_G_T*10^3)/(R*T));//calculation of the equilibrium constant (no unit)
+del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit)
+Ky=Ka/((P)^del_n);//calculation of the equilibrium constant in terms of the mole fractions using Eq.(14.30) (no unit) (K_phi=1.0,assuming ideal gas behaviour)
+mtot=m(1,:)+m(2,:)+m(3,:);//calculation of the total mole number of feed entering (no unit)
+//To determine the degree of conversion, the inbuilt function fsolve is used to solve the equation given by Ky=(y_CH3OH)/(y_CO*y_H2^2), where y_CH3OH,y_CO,y_H2 are the mole fractions of CH3OH,CO,H2 respectively. Let the equilibrium conversion be denoted as E
+E_guess=0.1;//taking a guess value for the degree of conversion,to be used in the inbuilt function fsolve (no unit)
+tol=1e-6;//tolerance limit for convergence of the system when using fsolve
+function[fn]=solver_func(Ei)
+ //Function defined for solving the system
+ fn=Ky-((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)));
+endfunction
+[E]=fsolve(E_guess,solver_func,tol)//using inbuilt function fsolve for solving the system of equations
+
+
+//OUTPUT
+mprintf('The degree of conversion at 500K and 5bar pressure=%f\n',E);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH2/EX2.1/Chap2_Ex1_R1.sce b/611/CH2/EX2.1/Chap2_Ex1_R1.sce new file mode 100755 index 000000000..3b2735af5 --- /dev/null +++ b/611/CH2/EX2.1/Chap2_Ex1_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-2,Example 1,Page 29
+//Title:Work done by gas
+//================================================================================================================
+clear
+clc
+
+//INPUT
+R=8.314;//universal gas constant in J/molK
+t1=300;//initial temperature of gas in K
+p1=0.1;//initial pressure of the gas in MPa
+p2=0.5;//pressure of gas after heating in MPa
+p3=0.1;//pressure of gas after expansion in MPa
+n=1;//number of moles of gas in mole
+
+//CALCULATION
+w1=0*n;//calculation of work done by the gas during heating in kJ (since dv=0)
+t2=t1*(p2/p1);//calculation of temperature of gas after heating in K
+t3=t2;//calculation of temperature of gas after expansion in K (constant temperature expansion)
+volume2=p2/p3;//calculation of ratio of volume of gas after expansion to the volume of gas after heating (no unit)
+w2=(R*t2*log(volume2)*n)/1000;//calculation of work done by the gas during expansion in kJ
+volume3= t3/t1;//calculation of ratio of volume of gas after expansion to the final volume of gas (no unit)
+w3=(R*t1*(1-volume3)*n)/1000;//calculation of work done by the gas during constant pressure compression in kJ
+work_net=w1+w2+w3;//calculation of net work done by the gas for the process in kJ
+
+//OUTPUT
+mprintf('\n Work done during heating process:Work from 1-2= %d kJ \n',w1);
+mprintf('\n Work done during constant temperature expansion: Work from 2-3= %f kJ \n',w2);
+mprintf('\n Work done during constant pressure compression: Work from 3-1= %f kJ \n',w3);
+mprintf('\n Net work done by the gas during the process= %f kJ \n',work_net);
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH2/EX2.2/Chap2_Ex2_R1.sce b/611/CH2/EX2.2/Chap2_Ex2_R1.sce new file mode 100755 index 000000000..c283eda2d --- /dev/null +++ b/611/CH2/EX2.2/Chap2_Ex2_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-2,Example 2,Page 30
+//Title:Work done by gas in piston cylinder assembly
+//================================================================================================================
+clear
+clc
+
+//INPUT
+v1=0.1;//volume of gas initially present in the cylinder in m^3
+p1=0.1;//initial pressure of gas in MPa
+p_atm=0.1;//atmospheric pressure acting on the piston in MPa
+v2=0.3;//volume of gas after heating in m^3
+p2=0.6;//pressure of gas after heating in MPa
+
+//CALCULATION
+work=((p1+p2)*(v2-v1)*10^6)/(2*1000);//calculation of work done by the gas in kJ
+//calculation is done by using reversible work done as integral of Pdv followed by a force balance taken on the piston
+
+//OUTPUT
+mprintf('\n The work done by the gas = %d kJ',work);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.1/Chap3_Ex1.sce b/611/CH3/EX3.1/Chap3_Ex1.sce new file mode 100755 index 000000000..85fbb09a3 --- /dev/null +++ b/611/CH3/EX3.1/Chap3_Ex1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 1,Page 48
+//Title:Specific volume and Specific internal energy
+//================================================================================================================
+clear
+clc
+
+//INPUT
+X=0.8;//Quality of wet steam (no unit)
+T=150;//Temperature of the wet steam in degree celsius
+vf=0.0010908;//molar volume of saturated liquid in m^3/kg
+vg=0.3924;//molar volume of saturated vapour in m^3/kg
+uf=631.63;//molar internal energy of the saturated liquid in kJ/kg
+ug=2558.6;//molar internal energy of the saturated vapour in kJ/kg
+
+//CALCULATION
+V=(X*vg)+((1-X)*vf);//calculation of specific volume in m^3/kg using Eq.(3.3)
+U=(X*ug)+((1-X)*uf);//calculation of specific internal energy in kJ/kg using Eq.(3.6)
+
+//OUTPUT
+mprintf('\n The specific volume of wet steam= %0.4f m^3/kg \n',V);
+mprintf('\n The specific internal energy of wet steam= %0.1f kJ/kg \n',U);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH3/EX3.10/Chap3_Ex10.sce b/611/CH3/EX3.10/Chap3_Ex10.sce new file mode 100755 index 000000000..1d38a33e2 --- /dev/null +++ b/611/CH3/EX3.10/Chap3_Ex10.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 10,Page 67
+//Title:Acentric factor
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=180;//temperature of water in degree celsius
+P=1.0027;//saturation pressure of water in MPa
+Tc=647.3;//critical temperature of water in K
+Pc=221.2;//critical pressure of water in bar
+Tr=0.7;//reduced temperature at which acentric factor was defined by Pitzer
+
+//CALCULATION
+T1=Tr*Tc;//calculating temperature in K using reduced temperature value
+T1=T1-273.15;//conversion to degree celsius
+Pr=(P*10)/Pc;//calculation of reduced pressure (no unit) using saturation pressure at t1. In this case t1 equals t, therefore the given saturation pressure is taken
+w=-log10(Pr)-1.0;//calculation of acentric factor using Eq.(3.62)
+
+////OUTPUT
+mprintf('\n The acentric factor of water= %f \n',w);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.11/Chap3_Ex11_R1.sce b/611/CH3/EX3.11/Chap3_Ex11_R1.sce new file mode 100755 index 000000000..e4835b4d4 --- /dev/null +++ b/611/CH3/EX3.11/Chap3_Ex11_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 11,Page 72
+//Title:Volume using two paramter and three parameter compressibility factor correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=409.41;//temperature of n-octane in degree celsius
+P=4.98;//pressure in bar
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+Tr=(T+273.15)/Tc;//calculation of reduced temperature (no unit)
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+z0=0.96;//value of compressibilty factor at tr and pr obtained from Fig.3.11
+V1=(z0*R*(T+273.15))/(P*10^5);//calculation of volume in m^3/mol using the two parameter compressibilty factor correlation
+z1=0.01;//value of compressibilty factor at tr and pr obtained from Fig.3.12
+z=z0+(w*z1);//calculation of compressibility factor using Eq.3.64
+V2=(z*R*(T+273.15))/(P*10^5);//calculation of volume in m^3/mol using the three parameter compressibility factor correlation
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane obtained by the two parameter compressibilty factor correlation= %f m^3/mol\n',V1);
+mprintf('\n The volume occupied by n-octane obtained by the three parameter compressibility factor correlation= %f m^3/mol\n',V2);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.12/Chap3_Ex12_R1.sce b/611/CH3/EX3.12/Chap3_Ex12_R1.sce new file mode 100755 index 000000000..644f4c2a7 --- /dev/null +++ b/611/CH3/EX3.12/Chap3_Ex12_R1.sce @@ -0,0 +1,34 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 12,Page 72
+//Title:Pressure developed using two paramter compressibility factor correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1;//volume of the tank in m^3
+m=180;//mass of carbon dioxide in kg
+T=91.8;//temperature of the tank in degree celsius after it is placed in the vicinity of a furnace
+Tc=304.2;//critical temperature of carbon dioxide in K
+Pc=73.87;//critical pressure of carbon dioxide in bar
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+mwt=44*10^-3;//molecular weight of carbon dioxide in kg/mol
+n=m/mwt;//calculation of number of moles of carbon dioxide in the tank in moles
+MV=V/n;//calculation of molar volume in m^3/mol
+slope=(MV*Pc*10^5)/(R*(T+273.15));//slope of the straight line formed when z0 is plotted against Pr formed by using the relation z0=(V*Pc*Pr)/(R*T)
+Tr=(T+273.15)/Tc;//calculation of reduced temperature (no unit)
+
+//At the given conditions, the state of CO2 must lie on the curve corresponding to the obtained value of Tr. To determine the state of CO2, a straight line passing through the origin, with the obtained value of slope is drawn on the z0 vs Pr plot of Fig.3.12 and the point of intersection of this straight line with the Tr curve is obtained to get the value of z0
+
+z0=0.725;//value of compressibilty factor obtained by doing the above mentioned procedure
+P=(z0*R*10^-6*(T+273.15))/(MV)//calculation of pressure in MPa using Eq.(3.59)
+
+//OUTPUT
+mprintf('\n The pressure developed by carbon dioxide= %.0f MPa\n',P);
+
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.13/Chap3_Ex13_R1.sce b/611/CH3/EX3.13/Chap3_Ex13_R1.sce new file mode 100755 index 000000000..6d1153b07 --- /dev/null +++ b/611/CH3/EX3.13/Chap3_Ex13_R1.sce @@ -0,0 +1,44 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 13,Page 73
+//Title:Pressure developed using three paramter compressibility factor correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1;//volume of the tank in m^3
+m=180;//mass of carbon dioxide in kg
+T=91.8;//temperature of the tank in degree celsius after it is placed in the vicinity of a furnace
+Tc=304.2;//critical temperature of carbon dioxide in K
+Pc=73.87;//critical pressure of carbon dioxide in bar
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+w=0.239;//acentric factor (no unit)
+
+//CALCULATION
+mwt=44*10^-3;//molecular weight of carbon dioxide in kg/mol
+n=m/mwt;//calculation of number of moles of carbon dioxide in the tank in mol
+MV=V/n;//calculation of molar volume in m^3/mol
+Tr=(T+273.15)/Tc;//calculation of reduced temperature (no unit)
+z0=0.725;//the value of z0 as computed in Example 3.12 (no unit)
+z_init=z0;//taking the initial guess value of z as z0
+slope=(MV*Pc*10^5)/(R*(T+273.15));//slope of the straight line formed when z is plotted against Pr formed by using the relation z0=(V*Pc*Pr)/(R*T)
+Prguess=z_init/slope;//Calculation of an initial guess value of reduced pressure (no unit) to compute the value of z
+z1guess=0.1;//z1 read from Fig.3.14 for the value of Tr and Prguess (no unit)
+tolerance=1e-6;//framing the tolerance limit for the convergence of the equation
+function[fn]=solver_function(zi)
+ fn=zi-(z0+(w*z1guess));//Function defined for solving the system using Eq.(3.64)
+endfunction
+[z]=fsolve(z1guess,solver_function,tolerance)//using inbuilt function fsolve for solving the system of equations
+Pr=z/slope;//calculation of the proper reduced pressure (no unit)
+P=((Pc*10^5)*Pr)*10^-6;//calculation of pressure exerted by carbon dioxide in MPa
+
+//OUTPUT
+mprintf('\n The pressure developed by carbon dioxide= %f MPa\n',P);
+
+
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH3/EX3.14/Chap3_Ex14_R1.sce b/611/CH3/EX3.14/Chap3_Ex14_R1.sce new file mode 100755 index 000000000..4024868e3 --- /dev/null +++ b/611/CH3/EX3.14/Chap3_Ex14_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 14,Page 75
+//Title:Volume using generalized form of the Redlich-Kwong equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-ocatne vapour in MPa
+Tc=569.4;//critical temperature of n-octane in K
+Pc=2.497;//critical pressure of n-octane in MPa
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+z_init=1;//taking a guess value of z (compressibilty factor) to get a value of h for solving the system
+h=(0.08664*Pr)/(z_init*Tr);//calculation of h using Eq.(3.68)
+tolerance=1e-6;//Framing the tolerance limit for the convergence of the equation
+function[fn]=solver_func(zi)
+ fn=zi-((1/(1-h))-((h/(1+h))*(4.93398/(Tr^(3/2)))));//Function defined for solving the system using Eq.(3.67)
+endfunction
+[z]=fsolve(h,solver_func,tolerance)//using inbuilt function fsolve for solving the system of equations
+V=(z*R*T)/(P*10^6);//calculation of volume in m^3/mol using Eq.(3.59)
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane vapour obtained by the generalized form of Redlich-Kwong equation of state= %f m^3/mol\n',V);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.15/Chap3_Ex15_R1.sce b/611/CH3/EX3.15/Chap3_Ex15_R1.sce new file mode 100755 index 000000000..cde67f246 --- /dev/null +++ b/611/CH3/EX3.15/Chap3_Ex15_R1.sce @@ -0,0 +1,66 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 15,Page 77
+//Title:Volume using Soave-Redlich-Kwong equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+//The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily
+//The general form of the cubic equation is (Z^3)+(alpha*Z^2)+(beeta*Z)+gaamma=0, where alpha,beeta and gaamma are determined using established relations
+
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=(P*10^6)/(Pc*10^5);//calculation of reduced pressure (no unit)
+S=0.48+(1.574*w)-(0.176*w^2);//calculation of S using Eq.(3.73)
+alpha1=(1+(S*(1-sqrt(Tr))))^2;//calculation of alpha1 using Eq.(3.72)
+a=(0.42748*R^2*Tc^2*alpha1)/(Pc*10^5);//calculation of the Soave-Redlich-Kwong constant in (m^6 Pa mol^-2) using Eq.(3.70)
+b=(0.08664*R*Tc)/(Pc*10^5);//calculation of the Soave-Redlich-Kwong constant in m^3/mol using Eq.(3.71)
+A=(a*P*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1;//calculation of alpha for Soave-Redlich-Kwong equation of state using Table (3.2)
+beeta=A-B-B^2;//calculation of beeta for Soave-Redlich-Kwong equation of state using Table (3.2)
+gaamma=-(A*B);//calculation of gaamma for Soave-Redlich-Kwong equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ vf=((Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=((Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+
+ end
+end
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane (saturated vapour) obtained by Soave-Redlich-Kwong equation of state= %f m^3/mol\n',vg);
+mprintf('\n The volume occupied by n-octane (saturated liquid) obtained by Soave-Redlich-Kwong equation of state= %f m^3/mol\n',vf);
+
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.16/Chap3_Ex16_R1.sce b/611/CH3/EX3.16/Chap3_Ex16_R1.sce new file mode 100755 index 000000000..5b51c354c --- /dev/null +++ b/611/CH3/EX3.16/Chap3_Ex16_R1.sce @@ -0,0 +1,66 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 16,Page 78
+//Title:Volume using Peng-Robinson equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+//The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily
+//The general form of the cubic equation is (Z^3)+(alpha*Z^2)+(beeta*Z)+gaamma=0, where alpha,beeta and gaamma are determined using established relations
+
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=(P*10^6)/(Pc*10^5);//calculation of reduced pressure (no unit)
+S=0.37464+(1.54226*w)-(0.26992*w^2);//calculation of S using Eq.(3.79)
+alpha1=(1+(S*(1-sqrt(Tr))))^2;//calculation of alpha1 using Eq.(3.78)
+a=(0.45724*R^2*Tc^2*alpha1)/(Pc*10^5);//calculation of the Peng-Robinson constant in (m^6 Pa mol^-2) using Eq.(3.76)
+b=(0.07780*R*Tc)/(Pc*10^5);//calculation of the Peng-Robinson constant in m^3/mol using Eq.(3.77)
+A=(a*P*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1+B;//calculation of alpha for Peng-Robinson equation of state using Table (3.2)
+beeta=A-(2*B)-(3*B^2);//calculation of beeta for Peng-Robinson equation of state using Table (3.2)
+gaamma=-(A*B)+(B^2)+(B^3);//calculation of gaamma for Peng-Robinson equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ vf=((Z)*R*t)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=((Z)*R*t)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+
+ end
+end
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane (saturated vapour) obtained by Peng-Robinson equation of state= %f m^3/mol\n',vg);
+mprintf('\n The volume occupied by n-octane (saturated liquid) obtained by Peng-Robinson equation of state= %f m^3/mol\n',vf);
+
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.2/Chap3_Ex2_R1.sce b/611/CH3/EX3.2/Chap3_Ex2_R1.sce new file mode 100755 index 000000000..e72ddfe92 --- /dev/null +++ b/611/CH3/EX3.2/Chap3_Ex2_R1.sce @@ -0,0 +1,21 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 2,Page 49
+//Title:Quality of wet steam
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1.42;//specific volume of wet steam in m^3/kg
+T=100;//temperature of wet steam in degree celsius
+vf=0.0010437;//molar volume of saturated liquid in m^3/kg
+vg=1.673;//molar volume of saturated vapour in m^3/kg
+
+//CALCULATION
+X=(V-vf)/(vg-vf);//calculation of the quality of wet steam using Eq.(3.3) (no unit)
+
+//OUTPUT
+mprintf('\n The quality of wet steam= %0.4f \n',X);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.3/Chap3_Ex3_R1.sce b/611/CH3/EX3.3/Chap3_Ex3_R1.sce new file mode 100755 index 000000000..7221c5ca8 --- /dev/null +++ b/611/CH3/EX3.3/Chap3_Ex3_R1.sce @@ -0,0 +1,22 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 3,Page 49
+//Title:Volume ratio
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=100;//temperature inside the vessel in degree celsius
+V=0.00317;//specific volume of water at the critical point in m^3/kg
+vf=0.0010437;//molar volume of saturated liquid in m^3/kg
+vg=1.673;//molar volume of saturated vapour in m^3/kg
+
+//CALCULATION
+X=(V-vf)/(vg-vf);//calculation of the quality of wet steam using Eq.(3.3) (no unit)
+ratio=(X*vg)/((1-X)*vf);//calculation of volume ratio of saturated vapour to the saturated liquid (no unit)
+
+//OUTPUT
+mprintf('\n The volume ratio of saturated vapour to the saturated liquid= %0.2f \n',ratio);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.4/Chap3_Ex4_R1.sce b/611/CH3/EX3.4/Chap3_Ex4_R1.sce new file mode 100755 index 000000000..a71a1f73a --- /dev/null +++ b/611/CH3/EX3.4/Chap3_Ex4_R1.sce @@ -0,0 +1,20 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 4,Page 49
+//Title:Mass ratio
+//================================================================================================================
+clear
+clc
+
+//INPUT
+U=2000;//specific internal energy of liquid-vapour mixture in kJ/kg
+uf=850.6;//specific internal energy of saturated liquid in kJ/kg
+ug=2593.2;//specific internal energy of saturated vapour in kJ/kg
+
+//CALCULATION
+mass_ratio=(U-uf)/(ug-U);//calculation of the mass ratio of vapour to liquid using the lever rule (no unit)
+
+//OUTPUT
+mprintf('\n The mass ratio of vapour to liquid= %0.4f \n',mass_ratio);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.5/Chap3_Ex5_R1.sce b/611/CH3/EX3.5/Chap3_Ex5_R1.sce new file mode 100755 index 000000000..04088a619 --- /dev/null +++ b/611/CH3/EX3.5/Chap3_Ex5_R1.sce @@ -0,0 +1,21 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 5,Page 52
+//Title:Volume using ideal gas law
+//================================================================================================================
+clear
+clc
+
+//INPUT
+n=1;//number of moles of n-octane vapour in mol
+T=427.85;//tempearture of n-octane vapour in K
+P=0.215;//pressure n-octane vapour in MPa
+R=8.314;//universal gas constant in (kPa m^3)/(kmol K)
+
+//CALCULATION
+V=((n*10^-3)*R*T)/(P*10^3);//calculation of volume using ideal gas law-Eq.(3.9) in m^3
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane vapour= %f m^3 \n',V);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.6/Chap3_Ex6_R1.sce b/611/CH3/EX3.6/Chap3_Ex6_R1.sce new file mode 100755 index 000000000..9518fcc7a --- /dev/null +++ b/611/CH3/EX3.6/Chap3_Ex6_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 6,Page 54
+//Title:Volume using van der Waals equation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+n=1;//number of moles occupied by n-octane vapour in mol
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+a=3.789;//van der Waals constant in Pa(m^3/mol)^2
+b=2.37*10^-4;//van der Waals constant in m^3/mol
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+Vguess=(n*R*T)/(P*10^6);//taking the vguess as volume from ideal gas (in m^3/mol) for the iteration process in van der Waals equation
+Vnew=(R*T)/((P*10^6)+(a/Vguess^2))+b;//getting the initial value of volume (in m^3/mol) using van der Waals equation to start the iteration process
+tolerance=1e-6;//defining the tolerance limit for the convergence of the iteration process
+while abs(Vguess-Vnew)>tolerance
+ Vguess=Vnew;
+ Vnew=(R*T)/((P*10^6)+(a/Vguess^2))+b;//the iteration process to solve the system of equation
+end
+V=Vnew;//The volume calculated using the van der Waals equation in m^3/mol
+//OUTPUT
+mprintf('\n The volume occupied by n-octane vapour obtained by van der Waals equation= %f m^3/mol\n',V);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.7/Chap3_Ex7_R1.sce b/611/CH3/EX3.7/Chap3_Ex7_R1.sce new file mode 100755 index 000000000..30033add2 --- /dev/null +++ b/611/CH3/EX3.7/Chap3_Ex7_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 7,Page 55
+//Title:Volume of liquid using van der Waals equation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+n=1;//number of moles occupied by n-octane liquid in mol
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+a=3.789;//van der Waals constant in Pa(m^3/mol)^2
+b=2.37*10^-4;//van der Waals constant in m^3/mol
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+Vguess=b;//taking vguess (in m^3/mol) for the iteration process in van der Waals equation
+Vnew=(R*T)/((P*10^6)+(a/Vguess^2))+b;//getting the initial value of volume (in m^3/mol) using van der Waals equation to start the iteration process
+tolerance=1e-6;//defining the tolerance limit for the convergence of the iteration process
+while abs(Vguess-Vnew)>tolerance
+ Vguess=Vnew;
+ Vnew=(R*T)/((P*10^6)+(a/Vguess^2))+b;//the iteration process to solve the system of equation
+end
+V=Vnew;//The volume calculated using the van der Waals equation in m^3/mol
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane liquid obtained by van der Waals equation= %e m^3/mol\n',V);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH3/EX3.8/Chap3_Ex8_R1.sce b/611/CH3/EX3.8/Chap3_Ex8_R1.sce new file mode 100755 index 000000000..c7ec5ce33 --- /dev/null +++ b/611/CH3/EX3.8/Chap3_Ex8_R1.sce @@ -0,0 +1,59 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 8,Page 57
+//Title:Volume using Cardan's method
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+a=3.789;//van der Waals constant in Pa(m^3/mol)^2
+b=2.37*10^-4;//van der Waals constant in m^3/mol
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+
+//CALCULATION
+//The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily
+//The general form of the cubic equation is (Z^3)+(alpha*Z^2)+(beeta*Z)+gaamma=0, where alpha,beeta and gaamma are determined using established relations
+
+A=(a*P*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1-B;//calculation of alpha for van der Waals equation of state using Table (3.2)
+beeta=A;//calculation of beeta for van der Waals equation of state using Table (3.2)
+gaamma=-(A*B);//calculation of gaamma for van der Waals equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ vf=((Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=((Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+
+ end
+end
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane (saturated liquid) obtained by Cardans method= %e m^3/mol\n',vf);
+mprintf('\n The volume occupied by n-octane (saturated vapour) obtained by Cardans method= %f m^3/mol\n',vg);
+
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE COMPUTED VALUE OF Z2 IS 0.0213 AND NOT 0.0187 AS PRINTED IN THE TEXTBOOK. THIS HAS BEEN CORRECTED IN THE ABOVE PROGRAM.
diff --git a/611/CH3/EX3.9/Chap3_Ex9_R1.sce b/611/CH3/EX3.9/Chap3_Ex9_R1.sce new file mode 100755 index 000000000..1b409b12c --- /dev/null +++ b/611/CH3/EX3.9/Chap3_Ex9_R1.sce @@ -0,0 +1,61 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-3,Example 9,Page 60
+//Title:Volume using Redlich-Kwong equation of state by implementing Cardan's method
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature in K
+P=0.215;//saturation pressure in MPa
+R=8.314;//universal gas constant in (Pa m^3)/(mol K)
+Pc=24.97;//critical pressure of n-octane in bar
+Tc=569.4;//critical temperature of n-octane in K
+
+//CALCULATION
+a=(0.42748*R^2*Tc^2.5)/(Pc*100*10^3*sqrt(T));//calculation of Redlich-Kwong constant in (m^6 Pa mol^-2)
+b=(0.08664*R*Tc)/(Pc*100*10^3);//calculation of Redlich-Kwong constant in m^3/mol
+//The Cardan's method simplifies the equation of state into a cubic equation which can be solved easily
+//The general form of the cubic equation is (Z^3)+(alpha*Z^2)+(beeta*Z)+gaamma=0, where alpha,beeta and gaamma are determined using established relations
+
+A=(a*P*10^6)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P*10^6)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1;//calculation of alpha for Redlich-Kwong equation of state using Table (3.2)
+beeta=A-B-B^2;//calculation of beeta for Redlich-Kwong equation of state using Table (3.2)
+gaamma=-(A*B);//calculation of gaamma for Redlich-Kwong equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+ vf=((Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=((Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Z=[Z1 Z2 Z3];
+ vf=(min(Z)*R*T)/(P*10^6);//Volume of saturated liquid calculated as vf=(Z*R*T)/P in m^3/mol
+ vg=(max(Z)*R*T)/(P*10^6);//Volume of saturated vapour calculated as vg=(Z*R*T)/P in m^3/mol
+
+ end
+end
+
+//OUTPUT
+mprintf('\n The volume occupied by n-octane (saturated vapour) using Redlich-Kwong equation of state= %f m^3/mol\n',vg);
+mprintf('\n The volume occupied by n-octane (saturated liquid) using Redlich-Kwong equation of state= %f m^3/mol\n',vf);
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER:THE COMPUTED VALUE OF Z2 IS 0.0147 AND NOT 0.0163 AS PRINTED IN THE TEXTBOOK. THIS HAS BEEN CORRECTED IN THE ABOVE PROGRAM.
diff --git a/611/CH4/EX4.1/Chap4_Ex1_R1.sce b/611/CH4/EX4.1/Chap4_Ex1_R1.sce new file mode 100755 index 000000000..25c3ac6ff --- /dev/null +++ b/611/CH4/EX4.1/Chap4_Ex1_R1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 1,Page 94
+//Title:Net work done by the system
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Q1=50;//Energy added as heat in kJ when the system undergoes a process 1-2
+W1=30;//Work done by the system in kJ during the process 1-2
+Q2=-40;//Energy rejected as heat in kJ during the process 2-3
+W2=-50;//Work done on the system in kJ during the process 2-3
+Q3=0;//System undergoes an adiabatic process to return to initial state
+
+//CALCULATION
+U2_1=Q1-W1;//calculation of net change in energy in kJ during process 1-2 using Eq.(4.5)
+U3_2=Q2-W2;//calculation of net change in energy in kJ during process 2-3 using Eq.(4.5)
+U1_3=(-U2_1)-(U3_2);//calculation of net change in energy in kJ during process 3-1 using Eq.(4.5)
+W3=Q3-U1_3;//calculation of work by the system in kJ using Eq.(4.5)
+net_work=W1+W2+W3;//calculation of net work done in kJ
+
+//OUTPUT
+mprintf('\n The net work done by the system= %d kJ\n',net_work);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.10/Chap4_Ex10_R1.sce b/611/CH4/EX4.10/Chap4_Ex10_R1.sce new file mode 100755 index 000000000..465c5ce82 --- /dev/null +++ b/611/CH4/EX4.10/Chap4_Ex10_R1.sce @@ -0,0 +1,37 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 10,Page 105
+//Title:Final temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+N=100;//number of moles of carbon dioxide in mol
+T1=298;//initial temperature in K
+Q=1;//energy added as heat in MJ
+a=45.369;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+b=8.688*10^-3;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+e=-9.619*10^5;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+//Where Cp is in J/molK
+
+//CALCULATION
+delh=Q*10^6/N;//calculation of enthalpy in J
+Tguess=520;//The final temperature guess value in K used for solving the system of equations
+//The system of equations are defined by :
+//T2=T1+(delh/Cpm)--->A
+//Cpm=a+(b*Tm)+(e/T1T2)--->B
+Cpm_guess=a+(b*((T1+Tguess)/2))+(e/(T1*Tguess));//calculation of Cpm guess (in J/molK) to be used to determine T2 from Equation A
+T2_guess=T1+(delh/Cpm_guess);//calculation of T2 using Equation A (in K) based on the value of Cpm guess computed using Equation B
+tolerance=1e-6;//defining the tolerance limit to obtain convergence
+while abs(T2_guess-Tguess)>tolerance
+ Tguess=T2_guess;
+ Cpm_guess=a+(b*((T1+Tguess)/2))+(e/(T1*Tguess));
+ T2_guess=T1+(delh/Cpm_guess);//the iteration process to solve the system of equations
+end
+T2=T2_guess;//value of the final temperature of CO2 obtained upon solving the system of equations(A and B) in K
+
+//OUTPUT
+mprintf('\n The final temperature= %0.1f K\n',T2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.11/Chap4_Ex11_R1.sce b/611/CH4/EX4.11/Chap4_Ex11_R1.sce new file mode 100755 index 000000000..4502883b9 --- /dev/null +++ b/611/CH4/EX4.11/Chap4_Ex11_R1.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 11,Page 107
+//Title:Final temperature,Pressure and work done in adiabatic process
+//================================================================================================================
+clear
+clc
+
+//INPUT
+volume_ratio=1/15;//final volume to the initial volume of air at the end of compression stroke (no unit)
+gaamma=1.4;//ratio of the molar heat capacities at constant pressure and constant volume for air (no unit)
+T1=300;//initial temperature of air in K
+P1=0.1;//initial pressure of air in MPa
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+T2=T1*((1/volume_ratio)^(gaamma-1));//calculation of final temperature in K using Eq.(4.32)
+P2=P1*((1/volume_ratio)^(gaamma));//calculation of final pressure in MPa using Eq.(4.34)
+W=(R*(T1-T2)*10^-3)/(gaamma-1);//calculation of work to be done on the system in kJ/mol using Eq.(4.31)
+
+//OUTPUT
+mprintf('\n The final temperature= %0.2f K\n',T2);
+mprintf('\n The final pressure= %0.4f MPa\n',P2);
+mprintf('\n Work done per mole of air= %0.3f kJ/mol\n',W);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH4/EX4.12/Chap4_Ex12_R1.sce b/611/CH4/EX4.12/Chap4_Ex12_R1.sce new file mode 100755 index 000000000..b4d226f8e --- /dev/null +++ b/611/CH4/EX4.12/Chap4_Ex12_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 12,Page 110
+//Title:Final temperature Pressure work done and heat interaction in polytropic process
+//================================================================================================================
+clear
+clc
+
+//INPUT
+volume_ratio=1/15;//final volume to the initial volume of ideal gas at the end of compression (no unit)
+T1=300;//initial temperature of ideal gas in K
+P1=0.1;//initial pressure of ideal gas in MPa
+R=8.314;//universal gas constant in J/molK
+n=1.2;//index of expansion (no unit)
+gaamma=1.4;//ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)
+
+//CALCULATION
+P2=P1*((1/volume_ratio)^n);//calculation of final pressure in MPa using Eq.(4.37)
+T2=T1*(P2/P1)*(volume_ratio);//calculation of final temperature in K (since the gas is taken to be ideal, (P1*V1)/T1=(P2*V2)/T2))
+W=(R*(T1-T2)*10^-3)/(n-1);//calculation of work to be done on the system in kJ/mol using Eq.(4.38)
+del_u=(R*(T2-T1)*10^-3)/(gaamma-1);//calculation of the change in the internal energy in kJ/mol using Eq.(4.28 and 4.29) (del_u=Cv*(T2-T1) and Cv=R/(gaamma-1))
+q=del_u+W;//calculation of the heat interaction during the process in kJ/mol using the first law of thermodynamics
+
+//OUTPUT
+mprintf('\n The final pressure= %0.3f MPa\n',P2);
+mprintf('\n The final temperature= %0.1f K\n',T2);
+mprintf('\n Work done on the gas= %f kJ/mol\n',W);
+mprintf('\n Heat interaction during the process= %f kJ/mol\n',q);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH4/EX4.13/Chap4_Ex13_R1.sce b/611/CH4/EX4.13/Chap4_Ex13_R1.sce new file mode 100755 index 000000000..b2a219109 --- /dev/null +++ b/611/CH4/EX4.13/Chap4_Ex13_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 13,Page 112
+//Title:Final temperature and amount of gas entering the tank
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1;//volume of tank in m^3
+T0=300;//initial temperature of ideal gas in K
+P0=0.1;//initial pressure of ideal gas in MPa
+T=500;//temperature of ideal gas in the pipeline in K
+P=3;//pressure of ideal gas in the pipeline in MPa
+R=8.314;//universal gas constant in J/molK
+gaamma=1.4;//ratio of the molar heat capacities at constant pressure and constant volume for ideal gas (no unit)
+
+//CALCULATION
+Pf=3;//final pressure reached in the tank in MPa
+//calculation of final temperature of the gas in the tank in K using Eq.(4.44) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+Tf=(Pf*10^6)/((((Pf*10^6)-(P0*10^6))/(gaamma*T))+((P0*10^6)/T0));
+//calculation of the moles of ideal gas entering into the tank using Eq.(4.44) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+N=(V/R)*(((Pf*10^6)/Tf)-((P0*10^6)/T0));
+
+//OUTPUT
+mprintf('\n The final temperature= %0.1f K\n',Tf);
+mprintf('\n The amount of gas that has entered the tank= %0.2f mol\n',N);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH4/EX4.14/Chap4_Ex14_R1.sce b/611/CH4/EX4.14/Chap4_Ex14_R1.sce new file mode 100755 index 000000000..6193e7134 --- /dev/null +++ b/611/CH4/EX4.14/Chap4_Ex14_R1.sce @@ -0,0 +1,47 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 14,Page 113
+//Title:Final state and mass of steam that entered the tank
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=3;//volume of tank in m^3
+T0=100;//initial temperature of steam in degree celsius
+T=300;//temperature of superheated steam in the pipeline in degree celsius
+P=3;//pressure of superheated steam in the pipeline in MPa
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Ps=101.33;//pressure of saturated steam in kPa from steam tables corresponding to T0
+vg=1.673;//specific volume of saturated vapour in m^3/kg obtained from steam tables corresponding to T0
+hg=2676.0;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T0
+h=2995.1;//specific enthalpy of superheated steam in kJ/kg obtained from superheated steam tables corresponding to T and P
+u0=((hg*10^3)-(Ps*10^3*vg))*10^-3;//calculation of initial internal energy of steam in kJ/mol using the first law of thermodynamics for the adiabatic charging of a tank
+m0=V/vg;//calculation of mass of steam initially in the tank in kg
+//The first law of thermodynamics for the adiabatic charging of a tank is given by:
+//mfuf-m0u0=(mf-m0)h. This equation is to be solved to determine mf
+
+Tf=418;// assuming final temperature of superheated steam in degree celsius
+//For superheated steam at P and Tf
+vf=0.102329;//specific volume of superheated steam in m^3/kg
+uf=2965.78;//internal energy of the superheated steam in kJ/kg
+
+mf_guess=V/vf;//taking a guess value for the mass of steam inside the tank at the end of the charging operation,in kg
+
+function[fn]=solver_func(ui)
+//Function defined for solving the system to determine the internal energy of steam inside the tank at the end of the charging operation in kJ/kg using Eq.(4.44, where Q=0 as the process is adiabatic)
+ fn=(mf_guess*ui)-(m0*u0)-((mf_guess-m0)*h);
+endfunction
+[uf_solved]=fsolve(mf_guess,solver_func,1e-6)//using inbuilt function fsolve for solving the system of equations
+mf=mf_guess//mass of the steam inside the tank at the end of the charging operation, in kg
+mass=mf-m0;//calculation of mass of steam that entered the tank in kg
+
+//OUTPUT
+mprintf("\n The final state of steam(superheated),Pressure=%d MPa\n",P);
+mprintf("\n The final state of steam(superheated),Temperature=%d degree celsius\n",Tf);
+mprintf("\n The mass of steam that entered the tank=%0.3f kg\n",mass);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH4/EX4.15/Chap4_Ex15_R1.sce b/611/CH4/EX4.15/Chap4_Ex15_R1.sce new file mode 100755 index 000000000..d90e03327 --- /dev/null +++ b/611/CH4/EX4.15/Chap4_Ex15_R1.sce @@ -0,0 +1,37 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 15,Page 115
+//Title:Final temperature and amount of gas escaping the cylinder
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=0.1;//volume of cylinder in m^3
+T0=300;//initial temperature of nitrogen in K
+P0=14;//initial pressure of nitrogen in MPa
+P=0.1;//ambient pressure in MPa
+Pf=2;//final pressure of nitrogen in MPa
+R=8.314;//universal gas constant in J/molK
+gaamma=1.4;//ratio of the molar heat capacities at constant pressure and constant volume for nitrogen (no unit)
+
+//CALCULATION
+//calculation of final temperature of the gas in the tank in K using Eq.(4.51) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+function[fn]=solver_func(Ti)
+ //Function defined for solving the system to determine the final temperature
+ fn=((P0*10^6)-(Pf*10^6))-((gaamma/2)*(T0+Ti)*(((P0*10^6)/T0)-((Pf*10^6)/Ti)));
+endfunction
+Tguess=300;//The final temperature guess value in K used for solving the system of equations
+[Tf]=fsolve(Tguess,solver_func,1e-6)//using inbuilt function fsolve for solving the system of equations
+
+//calculation of the moles of nitrogen escaping from the tank using Eq.(4.51) (and applying u=Cv*T, h=Cp*T and N=P*V/R*T as the gas is taken to be ideal)
+N=(V/R)*(((P0*10^6)/T0)-((Pf*10^6)/Tf));
+
+//OUTPUT
+mprintf('\n The final temperature= %0.1f K\n',Tf);
+mprintf('\n The amount of gas that has escaped from the cylinder= %0.2f mol\n',N);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH4/EX4.16/Chap4_Ex16_R1.sce b/611/CH4/EX4.16/Chap4_Ex16_R1.sce new file mode 100755 index 000000000..e4844eabb --- /dev/null +++ b/611/CH4/EX4.16/Chap4_Ex16_R1.sce @@ -0,0 +1,40 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 16,Page 118
+//Title:Percentage error
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T0=300;//initial temperature of superheated steam in degree celsius
+P0=3;//initial pressure of superheated steam in MPa
+Xe=0.85;//quality of steam leaving the turbine (no unit)
+Tf=45;//final temperature of steam leaving the turbine in degree celsius
+Vi=10;//velocity of steam at the entrance in m/s
+Ve=40;//exit velocity of steam in m/s
+Zi=10;//elevation at the entrance in m
+Ze=4;//elevation at the exit in m
+m=1;//mass flow rate of steam through turbine in kg/s
+g=9.81;//accleration due to gravity in m/s^2
+
+//CALCULATION
+hi=2995.1;//specific enthalpy of superheated steam in kJ/kg obtained from superheated steam tables corresponding to T0 and P0
+hf=188.35;//specific enthalpy of saturated liquid in kJ/kg obtained from steam tables corresponding to Tf
+hg=2583.3;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to Tf
+he=((1-Xe)*hf)+(Xe*hg);//calculation of specific enthalpy of steam at the exit in kJ/kg using Eq.(3.6)
+Q=0;//adiabatic process
+enthalpy_change=(he*10^3)-(hi*10^3);//calculation of the enthalpy change between the entrance and exit in J/kg
+KE_change=((Ve^2)-(Vi^2))/2;//calculation of the kinetic energy change between the entrance and exit in J/kg
+PE_change=g*(Ze-Zi);//calculation of the potential energy change between the entrance and exit in J/kg
+Ws=Q-(m*(enthalpy_change+KE_change+PE_change)*10^-3);//calculation of power output in kW using Eq.(4.61)
+err_KE=((KE_change)/(Ws*10^3))*100;//calculation of percentage error when kinetic energy change is ignored
+err_PE=((abs (PE_change)/(Ws*10^3)))*100;//calculation of percentage error when potential energy change is ignored
+err=err_KE+err_PE;//calculation of percentage error when both potential kinetic energy changes are ignored
+
+//OUTPUT
+mprintf('\n The percentage error when Kinetic energy change is ignored= %0.3f \n',err_KE);
+mprintf('\n The percentage error when Potential energy change is ignored= %0.4f \n',err_PE);
+mprintf('\n The percentage error when both Kinetic and Potential energy changes are ignored= %f \n',err);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.17/Chap4_Ex17_R1.sce b/611/CH4/EX4.17/Chap4_Ex17_R1.sce new file mode 100755 index 000000000..1a07029ae --- /dev/null +++ b/611/CH4/EX4.17/Chap4_Ex17_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 17,Page 119
+//Title:Exit velocity
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Pi=5;//pressure of dry saturated steam at the entrance in bar
+Pe=2;//pressure of dry saturated steam at the exit in bar
+Vi=3;//velocity of dry saturated steam at the entrance in m/s
+m=1;//flow rate of steam through the nozzle in kg/s
+g=9.81;//acceleration due to gravity in m/s^2
+
+//CALCULATION
+hi=2747.5;//specific enthalpy in kJ/kg of the dry saturated steam at the entrance taken from steam tables corresponding to Pi
+he=2706.3;//specific enthalpy in kJ/kg of the dry saturated steam at the exit taken from steam tables corresponding to Pe
+ve=0.8854;//specific volume in m^3/kg of the dry saturated steam at the exit taken from steam tables corresponding to Pe
+Zi=0;//assuming that the nozzle is horizontal
+Ze=0;//assuming that the nozzle is horizontal
+Q=0;//adiabatic process
+Ws=0;//since no shaft work is done
+Ve=sqrt (2*(((Q-Ws)/m)-(g*(Zi-Ze))-((he*10^3)-(hi*10^3)))+(Vi^2));//calculation of velocity at the exit in m/s using Eq.(4.61)
+A=(m*ve)/Ve;//calculation of cross sectional area of the nozzle at the exit in m^2
+
+//OUTPUT
+mprintf('\n The velocity of dry saturated steam at the exit= %0.2f m/s\n',Ve);
+mprintf('\n The cross sectional area of the nozzle at the exit= %0.3e m^2\n',A);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH4/EX4.18/Chap4_Ex18_R1.sce b/611/CH4/EX4.18/Chap4_Ex18_R1.sce new file mode 100755 index 000000000..26ceb0867 --- /dev/null +++ b/611/CH4/EX4.18/Chap4_Ex18_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 18,Page 123
+//Title:Quality of wet steam
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T1=270;//temperature of wet steam in degree celsius
+T2=120;//final temperature of superheated steam in degree celsius
+P=0.1;//pressure of superheated steam in MPa
+
+//CALCULATION
+hf=1185.2;//specific enthaply of saturated liquid in kJ/kg obtained from steam tables corresponding to T1
+hg=2789.9;//specific enthalpy of saturated vapour in kJ/kg obtained from stean tables corresponding to T1
+he=2716.04;//specific enthalpy of superheated steam in kJ/kg obtained from superheated steam tables corresponding to T2 obtained by interpolation
+Xi=(he-hf)/(hg-hf);//calculation of quality of steam using Eq.(3.6) (no unit)
+
+//OUTPUT
+mprintf('\n The quality of wet steam= %0.3f \n',Xi);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH4/EX4.20/Chap4_Ex20_R1.sce b/611/CH4/EX4.20/Chap4_Ex20_R1.sce new file mode 100755 index 000000000..d0438ac0a --- /dev/null +++ b/611/CH4/EX4.20/Chap4_Ex20_R1.sce @@ -0,0 +1,20 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 20,Page 128
+//Title:Standard enthalpy change
+//================================================================================================================
+clear
+clc
+
+//INPUT
+del_H=-90.135;//standard enthalpy change for the reaction CO(g)+2H2(g)--->CH3OH(g) at 298.15K in kJ
+
+//CALCULATION
+del_H1=2*del_H;//calculation of standard enthalpy change for the reaction 2CO(g)+4H2(g)--->2CH3OH(g) at 298.15K in kJ
+del_H2=(1/2)*del_H;//calculation of standard enthalpy change for the reaction (1/2)CO(g)+H2(g)--->(1/2)CH3OH(g) at 298.15K in kJ
+
+//OUTPUT
+mprintf('\n The standard enthalpy change for the reaction 2CO(g)+4H2(g)---->2CH3OH(g) at 298.15K= %0.2f kJ\n',del_H1);
+mprintf('\n The standard enthalpy change for the reaction (1/2)CO(g)+H2(g)---->(1/2)CH3OH(g) at 298.15K= %0.4f kJ\n',del_H2 );
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.22/Chap4_Ex22_R1.sce b/611/CH4/EX4.22/Chap4_Ex22_R1.sce new file mode 100755 index 000000000..4658c0ebd --- /dev/null +++ b/611/CH4/EX4.22/Chap4_Ex22_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 22,Page 130
+//Title:Standard enthalpy change for the reaction from standard enthalpies of formation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+del_Hf_C4H10=-74.943;//standard enthalpy of formation of C4H10(g) at 298.15K in kJ
+del_Hf_CO2=-393.978;//standard enthalpy of formation of CO2(g) at 298.15K in kJ
+del_Hf_H2O=-241.997;//standard enthalpy of formation of H2O(g) at 298.15K in kJ
+
+//CALCULATION
+
+//calculation of the standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)--->4CO2(g)+5H2O(g) at 298.15K in kJ
+// by using the standard enthalpy of formation data where the formation reactions are:
+// 4C(s)+5H2(g)--->C4H10(g)--->A
+//C(s)+O2(g)--->CO2(g)---->B
+//H2(g)+(1/2)O2(g)--->H2O(g)---->C
+//del_Hr=5(C)+4(B)-(A)
+
+del_Hr=(5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);
+
+//OUTPUT
+mprintf('\n The standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)---->4CO2(g)+5H2O(g) at 298.15K= %0.3f kJ\n',del_Hr);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.23/Chap4_Ex23_R1.sce b/611/CH4/EX4.23/Chap4_Ex23_R1.sce new file mode 100755 index 000000000..2d5c7753d --- /dev/null +++ b/611/CH4/EX4.23/Chap4_Ex23_R1.sce @@ -0,0 +1,42 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 23,Page 131
+//Title:Standard enthalpy change for the reaction from standard enthalpies of formation (2)
+//================================================================================================================
+clear
+clc
+
+//INPUT
+del_Hf_C4H10=-74.943;//standard enthalpy of formation of C4H10(g) at 298.15K in kJ
+del_Hf_CO2=-393.978;//standard enthalpy of formation of CO2(g) at 298.15K in kJ
+del_Hf_H2O=-241.997;//standard enthalpy of formation of H2O(g) at 298.15K in kJ
+del_H_vap=43.966;//enthalpy of vaporization of H2O at 298.15K in kJ/mol
+
+//CALCULATION
+
+//calculation of the standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)--->4CO2(g)+5H2O(l)--->A at 298.15K in kJ
+//The above reaction A can be expressed as a sum of the following two reactions:
+//C4H10(g)+(13/2)O2(g)--->4CO2(g)+5H2O(g) --->B
+//5H2O(g)--->5H2O(l)--->C
+//Reaction C represents the physical change H2O((g);25 degree celsius,1 bar)--->H2O((l);25 degree celsius,1 bar), which can be expressed as:
+//a-->H2O((g);25 degree celsius,1 bar)--->H2O((g);25 degree celsius,Ps)---->del_H1
+//b-->H2O((g);25 degree celsius,Ps)--->H2O((l);25 degree celsius,Ps)--->del_H2
+//c-->H2O((l);25 degree celsius,Ps)--->H2O((l);25 degree celsius,1 bar)--->del_H3, where Ps is the saturation pressure at 25 degree celsius
+//The overall enthalpy change therefore is given as del_H0=del_H1+del_H2+del_H3
+
+del_H1=0;//vapour phase at low pressures behaves like an ideal gas therefore the enthalpy change is zero
+del_H2=5*(-del_H_vap);//calculation of the enthalpy of reaction b in kJ
+del_H3=0;//effect of pressure on the enthalpy of liquids is negligible
+
+//calculation of the standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)--->4CO2(g)+5H2O(g) at 298.15K in kJ, from Example (4.22)
+//by using the standard enthalpy of formation data where the formation reactions are:
+//4C(s)+5H2(g)--->C4H10(g)
+//C(s)+O2(g)--->CO2(g)
+//H2(g)+(1/2)O2(g)--->H2O(g)
+del_H=(5*del_Hf_H2O)+(4*del_Hf_CO2)-(del_Hf_C4H10);
+del_net_H=(del_H)+(del_H1)+(del_H2)+(del_H3);//calculation of the standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)-->4CO2(g)+5H2O(l) at298.15K in kJ
+
+//OUTPUT
+mprintf('\n The standard enthalpy change for the reaction C4H10(g)+(13/2)O2(g)---->4CO2(g)+5H2O(l) at 298.15K= %0.3f kJ\n',del_net_H);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.24/Chap4_Ex24_R1.sce b/611/CH4/EX4.24/Chap4_Ex24_R1.sce new file mode 100755 index 000000000..6390a3efc --- /dev/null +++ b/611/CH4/EX4.24/Chap4_Ex24_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 24,Page 132
+//Title:Standard enthalpy change of formation of n-butane gas
+//================================================================================================================
+clear
+clc
+
+//INPUT
+del_H_comb=2880.44;//gross heating value of n-buatne gas at 298.15K in kJ/mol
+del_Hf_CO2=-393.978;//standard enthalpy of formation of CO2(g) at 298.15K in kJ
+del_Hf_H2O=-285.958;//standard enthalpy of formation of H2O(l) at 298.15K in kJ
+del_Hf_O2=0;//standard enthalpy of formation of O2(g) at 298.15K in kJ
+//CALCULATION
+
+//The combustion reaction is given by:
+//C4H10(g)+(13/2)O2(g)--->4CO2(g)+5H2O(l)
+//del_H_comb=(4*del_Hf_CO2)+(5*del_Hf_H2O)-(del_Hf_C4H10), from which del_Hf_C4H10 is computed
+n_CO2=4;//stoichiometric coefficient (no unit)
+n_H2O=5;//stoichiometric coefficient (no unit)
+n_O2=-13/2;//stoichiometric coefficient (no unit)
+n_C4H10=-1;//stoichiometric coefficient (no unit)
+//computation of the standard enthaply of formation of n-butane gas at 298.15K in kJ
+del_Hf_C4H10=(n_CO2*del_Hf_CO2)+(n_H2O*del_Hf_H2O)+(n_O2*del_Hf_O2)-(-del_H_comb);
+
+
+//OUTPUT
+mprintf('\n The standard enthalpy of formation of n-butane gas at 298.15K= %0.3f kJ\n',del_Hf_C4H10);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.25/Chap4_Ex25_R1.sce b/611/CH4/EX4.25/Chap4_Ex25_R1.sce new file mode 100755 index 000000000..353bc40d0 --- /dev/null +++ b/611/CH4/EX4.25/Chap4_Ex25_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 25,Page 133
+//Title:Standard enthalpy change
+//================================================================================================================
+clear
+clc
+
+//INPUT
+
+//The reaction is: CH4(g)+H2O(g)--->CO(g)+3H2(g)
+//The standard enthalpy change for the above reaction is determined by using the individual combustion reactions
+//The combustion reactions are:
+//A--->CH4(g)+2O2(g)--->CO2(g)+2H2O(l)--->del_Hc_A
+//B--->CO(g)+(1/2)O2(g)--->CO2(g)--->del_Hc_B
+//C--->H2(g)+(1/2)O2(g)--->H2O(l)--->del_Hc_C
+//D--->H2O(g)--->H2O(l)--->del_H_vap
+
+del_Hc_A=-890.94;//enthalpy change accompanying reaction A in kJ
+del_Hc_B=-283.18;//enthalpy change accompanying reaction B in kJ
+del_Hc_C=-286.03;//enthalpy change accompanying reaction C in kJ
+del_H_vap=-43.966;//enthalpy change of vaporization of H2O at 298.15K in kJ/mol
+
+//CALCULATION
+del_H0=(del_Hc_A)-(del_Hc_B)-(3*del_Hc_C)+(del_H_vap);//calculation of the standard enthalpy change of the reaction in kJ
+
+//OUTPUT
+mprintf('\n The standard enthalpy change at 298.15K for the reaction CH4(g)+H2O(g)--->CO(g)+3H2(g)= %0.3f kJ\n',del_H0);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH4/EX4.26/Chap4_Ex26.sce b/611/CH4/EX4.26/Chap4_Ex26.sce new file mode 100755 index 000000000..80ba859ea --- /dev/null +++ b/611/CH4/EX4.26/Chap4_Ex26.sce @@ -0,0 +1,54 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 26,Page 135
+//Title:Standard enthalpy change at 400K
+//================================================================================================================
+clear
+clc
+
+//INPUT
+
+//The reaction is : C2H4(g)+H2O(g)--->C2H5OH(g)
+
+del_H_vap=43.82;//enthalpy of vaporization of ethanol at 298.15K in kJ/mol
+
+//Data taken from Appendix tables A.3 and A.4 have been given below:
+
+del_Hf=[52.335;-241.997;0;-277.819];//standard enthalpies of formation of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) at 298.15K in kJ
+a=[4.196;28.850;20.691;0]//coefficients to compute isobaric molar heat capacity of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) in J/molK
+b=[154.565*10^-3;12.055*10^-3;205.346*10^-3;0]//coefficients to compute isobaric molar heat capacity of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) in J/molK
+c=[-81.076*10^-6;0;-99.793*10^-6;0]//coefficients to compute isobaric molar heat capacity of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) in J/molK
+d=[16.813*10^-9;0;18.825*10^-9;0]//coefficients to compute isobaric molar heat capacity of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) in J/molK
+e=[0;1.006*10^5;0;0];//coefficients to compute isobaric molar heat capacity of C2H4(g),H2O(g),C2H5OH(g),C2H5OH(l) in J/molK
+//where Cp0=a+bT+cT^2+dT^3+eT^-2
+
+T1=298.15;//Ambient temeprature in K
+T2=400;//temperature at which the standard enthalpy change has to be determined in K
+n_C2H4=-1;//stoichiometric coefficient (no unit)
+n_H2O=-1;//stoichiometric coefficient (no unit)
+n_C2H5OH=1;//stoichiometric coefficient (no unit)
+
+//CALCULATION
+
+//The standard enthalpy of formation of C2H5OH(g) can be obtained from the following reactions:
+//2C(s)+3H2(g)+(1/2)O2(g)--->C2H5OH(l)
+//C2H5OH(l)--->C2H5OH(g)
+
+del_Hf_C2H5OH_g=del_Hf(4,:)+del_H_vap;//calculation of standard enthalpy of formation of C2H5OH(g) at 298.15K in kJ
+del_Hr=(n_C2H5OH*del_Hf_C2H5OH_g)+(n_C2H4*del_Hf(1,:))+(n_H2O*del_Hf(2,:));//calculation of standard enthalpy change of reaction in kJ
+del_a=(n_C2H4*a(1,:))+(n_H2O*a(2,:))+(n_C2H5OH*a(3,:));//calculation of del_a using Eq.(4.83)
+del_b=(n_C2H4*b(1,:))+(n_H2O*b(2,:))+(n_C2H5OH*b(3,:));//calculation of del_b using Eq.(4.83)
+del_c=(n_C2H4*c(1,:))+(n_H2O*c(2,:))+(n_C2H5OH*c(3,:));//calculation of del_c using Eq.(4.83)
+del_d=(n_C2H4*d(1,:))+(n_H2O*d(2,:))+(n_C2H5OH*d(3,:));//calculation of del_d using Eq.(4.83)
+del_e=(n_C2H4*e(1,:))+(n_H2O*e(2,:))+(n_C2H5OH*e(3,:));//calculation of del_e using Eq.(4.83)
+del_H0=(del_Hr*10^3)-((del_a*T1)+((del_b/2)*T1^2)+((del_c/3)*T1^3)+((del_d/4)*T1^4)-(del_e/T1));//calculation del_H0 in kJ using Eq.(4.82)
+//calculation of the standard enthalpy of reaction at 400K in kJ
+del_Hr_T2=(del_H0+((del_a*T2)+((del_b/2)*T2^2)+((del_c/3)*T2^3)+((del_d/4)*T2^4)-(del_e/T2)))*10^-3;
+
+//OUTPUT
+mprintf('\n The standard enthalpy change at 400K for the reaction C2H4(g)+H2O(g)--->C2H5OH(g)= %f kJ\n',del_Hr_T2);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
diff --git a/611/CH4/EX4.28/Chap4_Ex28.sce b/611/CH4/EX4.28/Chap4_Ex28.sce new file mode 100755 index 000000000..655e3bd76 --- /dev/null +++ b/611/CH4/EX4.28/Chap4_Ex28.sce @@ -0,0 +1,61 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 28,Page 137
+//Title:Flame temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+
+//The combustion reaction of methane is given by:
+// CH4(g)+2O2(g)--->CO2(g)+2H2O(g)
+
+n_O2=2;// stoichiometric amount of oxygen required for combustion
+n_CH4=1;//number of moles of CH4(g) in moles
+n_CO2=1;//number of moles of CO2(g) formed in the combustion reaction in moles
+n_H2O=2;//number of moles of H2O(g) formed in the combustion reaction in moles
+del_Hf=[-74.943;0;-393.978;-241.997];// standard enthalpies of formation of CH4(g),O2(g),CO2(g),H2O(g) at 298.15K in kJ
+a=[45.369;28.850;30.255;27.270];//coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK
+b=[8.688*10^-3;12.055*10^-3;4.207*10^-3;4.930*10^-3];//coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK
+c=[0;0;0;0];//coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK
+d=[0;0;0;0];//coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK
+e=[-9.619*10^5;1.006*10^5;-1.887*10^5;0.333*10^5];//coefficients to compute isobaric molar heat capacity of CO2(g),H2O(g),O2(g),N2(g) in J/molK
+per_excess_air=50; //percentage excess of air supplied to the adiabatic burner
+T_amb=298.15;// temperature at which air and methane enter the burner in K
+per_N2=79;//percentage of N2 in the air supplied
+per_O2=21;//percentage of O2 in the air supplied
+
+//CALCULATION
+n_O2_actual=(1+(per_excess_air/100))*n_O2;//calculation of the number of moles of oxygen actually present in the system in moles
+n_N2=n_O2_actual*(per_N2/per_O2);//calculation of the number of moles of nitrogen actually present in the system in moles
+n_O2_residual=n_O2_actual-n_O2;//calculation of excess oxygen leaving as product in moles
+// The actual combustion reaction can be written as:
+// CH4(g)+3O2(g)+11.286N2(g)--->CO2(g)+2H2O(g)+O2(g)+11.286N2(g)
+
+
+del_Hr=(n_CO2*del_Hf(3,:))+(n_H2O*del_Hf(4,:))-(n_O2*del_Hf(2,:))-(n_CH4*del_Hf(1,:));//standard enthalpy of reaction at 298.15K in kJ
+
+//-delH_r=del_Hp
+//deriving an expression for del_Hp:
+
+del_a=(n_CO2*a(1,:))+(n_H2O*a(2,:))+(n_O2_residual*a(3,:))+(n_N2*a(4,:));//calculation of del_a using Eq.(4.83)
+del_b=(n_CO2*b(1,:))+(n_H2O*b(2,:))+(n_O2_residual*b(3,:))+(n_N2*b(4,:));//calculation of del_b using Eq.(4.83)
+del_c=(n_CO2*c(1,:))+(n_H2O*c(2,:))+(n_O2_residual*c(3,:))+(n_N2*c(4,:));//calculation of del_c using Eq.(4.83)
+del_d=(n_CO2*d(1,:))+(n_H2O*d(2,:))+(n_O2_residual*d(3,:))+(n_N2*d(4,:));//calculation of del_d using Eq.(4.83)
+del_e=(n_CO2*e(1,:))+(n_H2O*e(2,:))+(n_O2_residual*e(3,:))+(n_N2*e(4,:));//calculation of del_a using Eq.(4.83)
+tguess=500;//giving a guess value of temperature in K for using the inbuilt solver, fsolve to solve the system of equations below
+function[fn]=solver_func(ti)
+ //Function defined for solving the system
+fn=(-(del_Hr*10^3))-((del_a*(ti-T_amb))+((del_b/2)*((ti^2)-(T_amb^2)))+((del_c/3)*((ti^3)-(T_amb^3)))+((del_d/4)*((ti^4)-(T_amb^4)))+(del_e*((1/T_amb)-(1/ti))));
+endfunction
+[T]=fsolve(tguess,solver_func,1e-6)//using inbuilt function fsolve for solving the system of equations to get the flame temperaure in K
+
+ //OUTPUT
+mprintf('\n The flame temperature when methane is burned with 50 percent excess air in an adiabatic burner= %f K\n',T);
+
+//===============================================END OF PROGRAM===================================================
+
+
+
+
diff --git a/611/CH4/EX4.29/Chap4_Ex29_R1.sce b/611/CH4/EX4.29/Chap4_Ex29_R1.sce new file mode 100755 index 000000000..9cf0a1957 --- /dev/null +++ b/611/CH4/EX4.29/Chap4_Ex29_R1.sce @@ -0,0 +1,72 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 29,Page 139
+//Title: Amount of energy transferred as heat in the boiler
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_exit=550;//temperature in K at which the combustion products leave the boiler
+percent_molar_comp_prdct=[6.28;3.14;7.85;82.73];// percentage molar composition of the combustion products CO2(g),CO(g),O2(g),N2(g) on dry basis
+T_ent=298.15;//temperature in K at which Propane and air enter the combustion chamber
+del_Hf=[-393.978;-110.532;-241.997;0;0;-103.833];//standard enthalpies of formation of CO2(g),CO(g),H2O(g),O2(g),N2(g),C3H8(g) at 298.15K in kJ
+a=[45.369;28.068;30.255;27.270;28.850];//coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK
+//coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK
+b=[8.688*10^-3;4.631*10^-3;4.207*10^-3;4.930*10^-3;12.055*10^-3];
+c=[0;0;0;0;0];//coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK
+d=[0;0;0;0;0];//coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK
+e=[-9.619*10^5;-0.258*10^5;-1.887*10^5;0.333*10^5;1.006*10^5];//coefficients to compute isobaric molar heat capacity of CO2(g),CO(g),O2(g),N2(g),H2O(g) in J/molK
+per_N2=79;//percentage of nitrogen in air
+per_O2=21;//percentage of oxygen in air
+molar_mass_propane=44*10^-3;//molar mass of propane in kg/mole
+
+//CALCULATION
+// TAKE BASIS AS 100 mol OF DRY COMBUSTION PRODUCTS
+n_CO2=percent_molar_comp_prdct(1,:);//number of moles of CO2(g) in the product stream
+n_CO=percent_molar_comp_prdct(2,:);//number of moles of CO(g) in the product stream
+n_O2=percent_molar_comp_prdct(3,:);//number of moles of O2(g) in the product stream
+n_N2=percent_molar_comp_prdct(4,:);//number of moles of N2(g) in the product stream
+
+//The combustion reaction can be given as:
+// x C3H8+ y O2+ (79/21)y N2--->6.28CO2+3.14CO+7.85O2+82.73N2+ zH2O
+
+//Determination of x, y, z
+
+//carbon atom balance:
+x=(n_CO2+n_CO)/3;
+
+//Nitrogen atom balance:
+y=(2*n_N2)/(2*(per_N2/per_O2));
+
+//oxygen atom balance
+z=(2*y)-(2*n_CO2)-(n_CO)-(2*n_O2);
+
+//The actual combustion reaction becomes:
+//3.14C3H8+ 21.992O2+ 82.73N2--->6.28CO2+3.14CO+7.85O2+82.73N2+ 12.584H2O
+
+n_H2O=z;//number of moles of H2O determined after the balance done on the carbon,oxygen,and nitrogen atoms
+n_C3H8=x;//number of moles of C3H8 determined after the balance done on the carbon,oxygen,and nitrogen atoms
+
+//calculation of the standard enthalpy of the reaction at 298.15K in kJ
+//del_hf=0, for oxygen and nitrogen,therefore they are omitted in the expression
+del_Hr=(n_CO2*del_Hf(1,:))+(n_CO*del_Hf(2,:))+(n_H2O*del_Hf(3,:))-(n_C3H8*del_Hf(6,:));
+
+del_a=(n_CO2*a(1,:))+(n_CO*a(2,:))+(n_O2*a(3,:))+(n_N2*a(4,:))+(n_H2O*a(5,:));//calculation of del_a using Eq.(4.83)
+del_b=(n_CO2*b(1,:))+(n_CO*b(2,:))+(n_O2*b(3,:))+(n_N2*b(4,:))+(n_H2O*b(5,:));//calculation of del_b using Eq.(4.83)
+del_c=(n_CO2*c(1,:))+(n_CO*c(2,:))+(n_O2*c(3,:))+(n_N2*c(4,:))+(n_H2O*c(5,:));//calculation of del_c using Eq.(4.83)
+del_d=(n_CO2*d(1,:))+(n_CO*d(2,:))+(n_O2*d(3,:))+(n_N2*d(4,:))+(n_H2O*d(5,:));//calculation of del_d using Eq.(4.83)
+del_e=(n_CO2*e(1,:))+(n_CO*e(2,:))+(n_O2*e(3,:))+(n_N2*e(4,:))+(n_H2O*e(5,:));//calculation of del_e using Eq.(4.83)
+
+//calulation of del_Hp J
+ del_Hp=(del_a*(T_exit-T_ent))+((del_b/2)*((T_exit^2)-(T_ent^2)))+((del_c/3)*((T_exit^3)-(T_ent^3)))+((del_d/4)*((T_exit^4)-(T_ent^4)))-(del_e*((1/T_exit)-(1/T_ent)));
+ del_H=((del_Hr*10^3)+(del_Hp))*10^-3;//calculation of total energy transferred as heat in kJ
+ mass_propane=n_C3H8*molar_mass_propane;//calculation of amount of propane in the system in kg
+ energy=(-(del_H*10^3)/mass_propane)*10^-6;//energy transferred as heat per kg propane in MJ
+
+//OUTPUT
+mprintf('\n The energy transferred as heat per kg propane= %f MJ\n',energy);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH4/EX4.3/Chap4_Ex3_R1.sce b/611/CH4/EX4.3/Chap4_Ex3_R1.sce new file mode 100755 index 000000000..dd983af6d --- /dev/null +++ b/611/CH4/EX4.3/Chap4_Ex3_R1.sce @@ -0,0 +1,35 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 3,Page 96
+//Title:Final temperature and final pressure
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=1;//volume of tank in m^3;
+N=200;//number of moles of carbon dioxide in tank in mol
+T1=25;//ambient temperature in degree celsius
+I=5;//current in amperes passed through the resistor place in the tank
+Voltage=440;//voltage in volts existing across the resistor
+t=10;//time in minutes for which the current is passed
+a=363.077*10^-3;//van der waals constant in Pa (m^3/mol)^2
+b=0.043*10^-3;//van der waals constant in m^3/mol
+Cv=32.34;//molar heat capacity at constant volume in J/molK
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+MV=V/N;//calculation of molar volume in m^3/mol
+Q=0;//energy transfer as heat during the process
+W_Pdv=0;//mechanical work done by the system
+W_elec=-(Voltage*I*t*60)*(10^-6);//calculation of electrical work done on the system in MJ
+U2_1=Q-(W_Pdv+W_elec);//calculation of internal energy in MJ using Eq.(4.5)
+T2=((U2_1*10^6)/(N*Cv))+(T1+273.15);//calculation of final temperature in K using the relation du=CvdT+(a/v^2)dv, where dv is zero
+P=(((R*T2)/(MV-b))-(a/(MV^2)))*10^-3;//calculation of final pressure in kPa using Eq.(3.18)
+
+//OUTPUT
+mprintf('\n The final pressure= %0.3f kPa \n',P);
+mprintf('\n The final temperature= %0.2f K\n',T2);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.4/Chap4_Ex4_R1.sce b/611/CH4/EX4.4/Chap4_Ex4_R1.sce new file mode 100755 index 000000000..da2129791 --- /dev/null +++ b/611/CH4/EX4.4/Chap4_Ex4_R1.sce @@ -0,0 +1,40 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 4,Page 97
+//Title:Energy transferred and final state masses of liquid and vapour
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=0.1;//volume of tank in m^3
+T1=200;//initial temperature of saturated steam inside the tank in degree celsius
+T2=150;//temperature in degree celsius that the tank attains after some time due to poor insulation
+P1=15.549;//pressure in bar obtained from steam tables corresponding to T1
+vg1=0.1272;//specific volume of saturated vapour in m^3/kg obtained from steam tables corresponding to T1
+hg1=2790.9;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1
+P2=4.76;//pressure in bar obtained from steam tables corresponding to T2
+vf=0.0010908;//specific volume of saturated liquid in m^3/kg obtained from steam tables corresponding to T2
+vg2=0.3924;//specific volume of saturated vapour in m^3/kg obtained from steam tables corresponding to T2
+hf=632.15;//specific enthalpy of saturated liquid in kJ/kg obtained from steam tables corresponding to T1
+hg2=2745.4;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables corresponding to T1
+
+//CALCULATION
+ug1=((hg1*10^3)-(P1*10^5*vg1))*10^-3;//calculation of internal energy of vapour corresponding to T1 in kJ/kg
+uf=((hf*10^3)-(P2*10^5*vf))*10^-3;//calculation of internal energy of liquid corresponding to T2 in kJ/kg
+ug2=((hg2*10^3)-(P2*10^5*vg2))*10^-3;//calculation of internal energy of vapour corresponding to T2 in kJ/kg
+v2=vg1;//since constant volume process
+X2=(v2-vf)/(vg2-vf);//calculation of the final quality of steam (no unit)
+u2=(X2*ug2)+((1-X2)*uf);//calculation of the internal energy at the final state in kJ/kg
+m=V/vg1;//calculation of the mass of steam in kg
+Q=m*(u2-ug1);//calculation of energy transferred as heat in kJ, using the first law of thermodynamics
+mf=m*(1-X2);//calculation of mass of liquid in the tank in the final state in kg
+mg=m*X2;//calculation of mass of vapour in the tank in the final state in kg
+
+ //OUTPUT
+mprintf('\n The energy transferred as heat= %f kJ\n',Q);
+mprintf('\n The mass of liquid in the tank in the final state= %0.3f kg\n',mf);
+mprintf('\n The mass of vapour in the tank in the final state= %0.3f kg\n',mg);
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH4/EX4.5/Chap4_Ex5_R1.sce b/611/CH4/EX4.5/Chap4_Ex5_R1.sce new file mode 100755 index 000000000..416c31743 --- /dev/null +++ b/611/CH4/EX4.5/Chap4_Ex5_R1.sce @@ -0,0 +1,41 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 5,Page 102
+//Title:Work done and energy transferred
+//================================================================================================================
+clear
+clc
+
+//INPUT
+W=1;//weight of steam in kg in the piston cylinder assembly
+X=0.8;//quality of steam (no unit)
+T1=150;//initial temperature of steam in degree celsius
+T2=200;//final temperature of steam in degree celsius
+P1=476;//pressure in kPa obatined from steam tables (corresponding to T1)
+vf=0.0010908;//specific volume of saturated liquid in m^3/kg obatined from steam tables (corresponding to T1)
+vg=0.3924;//specific volume of satuarted vapour in m^3/kg obatined from steam tables (corresponding to T1)
+hf=632.15;//specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)
+hg=2745.4;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)
+
+//CALCULATION
+V1=(X*vg)+((1-X)*vf);//calculation of specific volume of steam in m^3/kg
+h1=(X*hg)+((1-X)*hf);//calculation of specific enthalpy of steam in kJ/kg
+P2=0.476;//Pressure in MPa is held constant during the process
+P_int1=0.4;//Pressure in MPa from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+P_int2=0.5;//Pressure in MPa from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+V_int1=0.5343;//specific volume in m^3/kg at P_int1 obtained from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+V_int2=0.4250;//specific volume in m^3/kg at P_int2 obtained from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+h_int1=2860.4;//specific enthalpy in kJ/kg at P_int1 obtained from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+h_int2=2855.1;//specific enthalpy in kJ/kg at P_int2 obtained from steam tables at T2 taken for interpolation to find V2 and h2 corresponding to P2
+V2=(((P2-P_int1)/(P_int2-P_int1))*(V_int2-V_int1))+V_int1;//specific volume of superheated steam in m^3/kg obtained by interpolation (corresponding to T2 and P2)
+h2=(((P2-P_int1)/(P_int2-P_int1))*(h_int2-h_int1))+h_int1;//specific enthalpy of superheated steam in kJ/kg obtained by interpolation (corresponding to T2 and P2)
+Q=(h2-h1)*W;//calculation of net heat transferred in kJ using Eq.(4.15)
+W=P1*(V2-V1)*W;//calculation of work done by steam in kJ using Eq.(4.14)
+
+//OUTPUT
+mprintf('\n The work done by steam= %0.2f kJ \n',W);
+mprintf('\n The net energy transferred as heat = %0.2f kJ\n',Q);
+mprintf('\n The final state of superheated steam,Pressure=%0.3f MPa \n',P2);
+mprintf('\n The final state of superheated steam,Temperature=%d degree celsius \n',T2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.6/Chap4_Ex6_R1.sce b/611/CH4/EX4.6/Chap4_Ex6_R1.sce new file mode 100755 index 000000000..dac1b2f30 --- /dev/null +++ b/611/CH4/EX4.6/Chap4_Ex6_R1.sce @@ -0,0 +1,43 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 6,Page 103
+//Title:Work done and final temperature
+//================================================================================================================
+clear
+clc
+
+//INPUT
+W=1;//weight of steam in kg in the piston cylinder assembly
+X=0.8;//quality of steam (no unit)
+T1=150;//initial temperature of steam in degree celsius
+I=5;//current passed in Amperes
+V=220;//voltage in volts across the resistor
+t=10;//time for which the current is passed in minutes
+P1=476;//pressure in kPa obatined from steam tables (corresponding to T1)
+vf=0.0010908;//specific volume of saturated liquid in m^3/kg obatined from steam tables (corresponding to T1)
+vg=0.3924;//specific volume of satuarted vapour in m^3/kg obatined from steam tables (corresponding to T1)
+hf=632.15;//specific enthalpy of saturated liquid in kJ/kg obtained from steam tables (corresponding to T1)
+hg=2745.4;//specific enthalpy of saturated vapour in kJ/kg obtained from steam tables (corresponding to T1)
+
+//CALCULATION
+V1=(X*vg)+((1-X)*vf);//calculation of specific volume of steam in m^3/kg
+h1=(X*hg)+((1-X)*hf);//calculation of specific enthalpy of steam in m^3/kg
+Ws=-V*I*t*60*10^-3;//calculation of electrical work done on the system in kJ
+h2=h1-Ws;//calculation of the specific enthalpy of steam in the final state in kJ/kg
+P2=0.476;//Pressure in MPa is held constant during the process
+T_int1=200;//Temperature in degree celsius obtained from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+T_int2=300;//Temperature in degree celsius obtained from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+V_int1=0.4512;//specific volume in m^3/kg at T_int1 from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+V_int2=0.5544;//specific volume in m^3/kg at T_int2 from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+h_int1=2856.37;//specific enthalpy in kJ/kg at T_int1 obtained from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+h_int2=3065.38;//specific enthalpy in kJ/kg at T_int2 obtained from steam tables at P2 taken for interpolation to find V2 and T2 corresponding to P2
+V2=(((h2-h_int1)/(h_int2-h_int1))*(V_int2-V_int1))+V_int1;//specific volume of superheated steam in m^3/kg obtained by interpolation (corresponding to T2 and P2)
+//Temperature of superheated steam in degree celsius obtained by interpolation (corresponding to T2 and P2)
+T2=(((h2-h_int1)/(h_int2-h_int1))*(T_int2-T_int1))+T_int1;
+W=(P1*10^3*(V2-V1)*W)*10^-3;//calculation of work done by steam in kJ using Eq.(4.14)
+
+//OUTPUT
+mprintf('\n The work done by steam= %0.2f kJ \n',W);
+mprintf('\n The final temperature= %0.2f degree celsius\n',T2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.7/Chap4_Ex7_R1.sce b/611/CH4/EX4.7/Chap4_Ex7_R1.sce new file mode 100755 index 000000000..1ca3312c2 --- /dev/null +++ b/611/CH4/EX4.7/Chap4_Ex7_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 7,Page 104
+//Title:Amount of energy
+//================================================================================================================
+clear
+clc
+
+//INPUT
+N=1;//number of moles of carbon dioxide in kmol
+T1=298;//initial temperature in K
+T2=600;//final raised temperature in K
+a=45.369;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+b=8.688*10^-3;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+e=-9.619*10^5;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+//Where Cp is in J/molK
+
+//CALCULATION
+Q=N*10^3*((a*(T2-T1)+((b/2)*(T2^2-T1^2))-(e*((1/T2)-(1/T1)))))*10^-6;//calculation of the amount of energy to be transferred in MJ using Eq.(4.25)
+
+//OUTPUT
+mprintf('\n The amount of energy to be transferred as heat= %0.3f MJ\n',Q);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH4/EX4.8/Chap4_Ex8_R1.sce b/611/CH4/EX4.8/Chap4_Ex8_R1.sce new file mode 100755 index 000000000..7b2d1be3b --- /dev/null +++ b/611/CH4/EX4.8/Chap4_Ex8_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 8,Page 104
+//Title:Isobaric molar heat capacity
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T1=298;//initial temperature in K
+T2=600;//final raised temperature in K
+a=45.369;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+b=8.688*10^-3;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+e=-9.619*10^5;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+//Where Cp is in J/molK
+
+//CALCULATION
+Cpm=((a*(T2-T1))+((b/2)*(T2^2-T1^2))-(e*((1/T2)-(1/T1))))/(T2-T1);//calculation of isobaric molar heat capacity in J/molK using Eq.(4.26)
+
+//OUTPUT
+mprintf('\n The isobaric molar heat capacity= %0.2f J/molK\n',Cpm);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH4/EX4.9/Chap4_Ex9_R1.sce b/611/CH4/EX4.9/Chap4_Ex9_R1.sce new file mode 100755 index 000000000..377eac40e --- /dev/null +++ b/611/CH4/EX4.9/Chap4_Ex9_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-4,Example 9,Page 105
+//Title:Amount of energy transferred using isobaric molar heat capacity
+//================================================================================================================
+clear
+clc
+
+//INPUT
+N=1;//number of moles of carbon dioxide in kmol
+T1=298;//initial temperature in K
+T2=600;//final raised temperature in K
+a=45.369;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+b=8.688*10^-3;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+e=-9.619*10^5;//coefficient in the specific heat capacity expression where Cp=a+bT+eT^-2
+//Where Cp is in J/molK
+
+//CALCULATION
+Cpm=((a*(T2-T1))+((b/2)*(T2^2-T1^2))-(e*((1/T2)-(1/T1))))/(T2-T1);//calculation of isobaric molar heat capacity in J/molK using Eq.(4.26)
+Q=N*10^3*Cpm*(T2-T1)*10^-6;//calculation of the amount of energy to be transferred in MJ using Eq.(4.25)
+
+//OUTPUT
+mprintf('\n The amount of energy to be transferred as heat= %0.3f MJ \n',Q);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.10/Chap5_Ex10_R1.sce b/611/CH5/EX5.10/Chap5_Ex10_R1.sce new file mode 100755 index 000000000..a2c74ab03 --- /dev/null +++ b/611/CH5/EX5.10/Chap5_Ex10_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 10,Page 173
+//Title: Change in entropy of steel and water
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m_steel=10;//mass of steel casting in kg
+T_steel=800;//temperature of steel casting in degree celsius
+m_water=100;//mass of water used for quenching in kg
+T_water=30;//temperature of water used for quenching in degree celsius
+Cp_steel=0.461;//heat capacity of steel in kJ/kgK
+Cp_water=4.23;//heat capacity of water in kJ/kgK
+
+//CALCULATION
+Ti_steel=T_steel+273.15;//conversion of temperature in K
+Ti_water=T_water+273.15;//conversion of temperature in K
+//calculation of final temperature of steel and water usung the first law of thermodynamics in K
+T_final=((m_steel*Cp_steel*Ti_steel)+(m_water*Cp_water*Ti_water))/((m_steel*Cp_steel)+(m_water*Cp_water));
+del_S_steel=m_steel*Cp_steel*log(T_final/Ti_steel);//calculation of the entropy change of steel using Eq.(5.32) in kJ/K
+del_S_water=m_water*Cp_water*log(T_final/Ti_water);//calculation of the entropy change of water using Eq.(5.32) in kJ/K
+
+//OUTPUT
+mprintf("\n The change in entropy of steel = %0.4f kJ/K\n",del_S_steel);
+mprintf("\n The change in entropy of water = %f kJ/K\n",del_S_water);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.11/Chap5_Ex11_R1.sce b/611/CH5/EX5.11/Chap5_Ex11_R1.sce new file mode 100755 index 000000000..6fcdebaf4 --- /dev/null +++ b/611/CH5/EX5.11/Chap5_Ex11_R1.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 11,Page 175
+//Title: Entropy change of the gas
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=2;//volume of insulated tank in m^3
+Ta=400;//temperature of gas in compartment (a) in K
+Pa=3;//pressure of gas in compartment (a) in MPa
+Tb=600;//temperature of gas in compartment (b) in K
+Pb=1;//pressure of gas in compartment (b) in MPa
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Va=V/2;//calculation of volume of compartment (a) in m^3
+Vb=V/2;//calculation of volume of compartment (b) in m^3
+Na=(Pa*10^6*Va)/(R*Ta);//calculation of number of moles of gas in compartment (a) in mol
+Nb=(Pb*10^6*Vb)/(R*Tb);//calculation of number of moles of gas in compartment (b) in mol
+//From the first law of thermodynamics, del_U=Q-W=0; since Q=0 and W=0. This implies that Na*Cv*(T-Ta)+Nb*Cv*(T-Tb)=0, therefore, Na*(T-Ta)+Nb*(T-Tb)=0
+T=((Na*Ta)+(Nb*Tb))/(Na+Nb);//calculation of final temperature using the above equation in K
+N=Na+Nb;//calculation of total number of moles of gas in mol
+P=((N*R*T)/V)*10^-6;//calculation of final pressure of gas in MPa
+Cp=(5/2)*R;//calculation of isobaric molar heat capacity as given in the problem statement in J/molK
+del_S=((Na*((Cp*log(T/Ta))-(R*log(P/Pa))))+(Nb*((Cp*log(T/Tb))-(R*log(P/Pb)))))*10^-3;//calculation of the change in entropy using Eq.(5.43) in kJ/K
+
+//OUTPUT
+mprintf("\n Entropy change of the gas=%0.2f kJ/K\n",del_S);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.12/Chap5_Ex12_R1.sce b/611/CH5/EX5.12/Chap5_Ex12_R1.sce new file mode 100755 index 000000000..7e556bce4 --- /dev/null +++ b/611/CH5/EX5.12/Chap5_Ex12_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 12,Page 177
+//Title: Minimum work to be done for separation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+N=1;//amount of air to be separated into its components in kmol
+P=0.1;//pressure of air in MPa
+T=300;//temperature of air in K
+per_oxygen=21;//percentage of oxygen in air
+per_nitrogen=79;//percentage of nitrogen in air
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+
+//From the first law of thermodynamics, del_U=Q-W. As air is considered as an ideal gas, U is a function of temperature alone.Here, both pure nitrogen and
+//pure oxygen are at the same temperature in the initial and the final states. Therefore, del_U=0. This implies that W=Q
+
+//From the second law of thermodynamics ,Eq.(5.32), ds=(dQ/T), therefore, Q=T*del_S, hence, W=T*del_s
+x1=per_nitrogen/100;//calculation of mole fraction of nitrogen (no unit)
+x2=per_oxygen/100;//calculation of mole fraction of oxygen (no unit)
+W=(T*N*10^3*R*((x1*log (x1))+(x2*log (x2))))*10^-3;//calculation of the work to be done based on the above formula in kJ, del_S is computed using Eq.(5.47)
+
+//OUTPUT
+mprintf("\n Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen and nitrogen at the same temperature and pressure=%0.2f kJ\n",abs(W));
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.13/Chap5_Ex13_R1.sce b/611/CH5/EX5.13/Chap5_Ex13_R1.sce new file mode 100755 index 000000000..69dcdb41f --- /dev/null +++ b/611/CH5/EX5.13/Chap5_Ex13_R1.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 13,Page 179
+//Title: Change in the entropy of the mixture
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m_ice=10;//mass of the block of ice in kg
+T_ice=0;//temperature of the ice in degree celsius
+m_water=100;//mass of watre in the tank in kg
+T_water=30;//temperature of the water in the tank in degree celsius
+Cp=4.23;//heat capacity of water in kJ/kgK
+lambda_melting=333.44;//latent heat of melting of ice in kJ/kg
+
+//CALCULATION
+Ti_ice=T_ice+273.15;//conversion of temperature in K
+Ti_water=T_water+273.15;//conversion of temperature in K
+
+//applying the first law of thermodynamics, an energy balance on the system is established from which the final temperature of water is determined
+
+T_final=((m_water*Cp*Ti_water)+(m_ice*Cp*Ti_ice)-(m_ice*lambda_melting))/((m_ice*Cp)+(m_water*Cp));//calculation of final temperature of water in K
+del_S_ice=((m_ice*lambda_melting)/(Ti_ice))+(m_ice*Cp*log (T_final/Ti_ice));//calculation of entropy change of ice in kJ/K
+del_S_water=m_water*Cp*log (T_final/Ti_water);//calculation of entropy change of water in kJ/K
+del_S_G=del_S_ice+del_S_water;//calculation of entropy generated using Eq.(5.54) in kJ/K
+
+//OUTPUT
+mprintf("\n The change in entropy of ice = %f kJ/K\n",del_S_ice);
+mprintf("\n The change in entropy of water = %f kJ/K\n",del_S_water);
+mprintf("\n The entropy generated= %f kJ/K\n",del_S_G);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.14/Chap5_Ex14_R1.sce b/611/CH5/EX5.14/Chap5_Ex14_R1.sce new file mode 100755 index 000000000..1453c379f --- /dev/null +++ b/611/CH5/EX5.14/Chap5_Ex14_R1.sce @@ -0,0 +1,35 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 14,Page 182
+//Title: Power output of turbine
+//================================================================================================================
+clear
+clc
+
+//INPUT
+P=3;//pressure of superheated steam in MPa
+T_enter=300;//entrance temperature of superheated steam in degree celsius
+T_exit=45;//final temperature at which the steam leaves in degree celsisus
+m=1;//mass flow rate of steam in kg/s
+
+//CALCULATION
+
+//From steam tables corresponding to P and T_enter
+si=6.5422;//entropy of steam at the entrance in kJ/kgK
+hi=2995.1;//entahlpy of steam at the entrance in kJ/kg
+
+//From steam tables corresponding to T_exit
+sf=0.6383;//entropy of saturated liquid in kJ/kgK
+hf=188.35;//enthalpy of saturated liquid in kJ/kg
+sg=8.1661;//entropy of saturated vapour in kJ/kgK
+hg=2583.3;//entahlpy of saturayed vapour in kJ/kg
+
+Xe=(si-sf)/(sg-sf);//calculation of quality of steam at the exit (no unit)
+he=((1-Xe)*hf)+(Xe*hg);//calculation of enthalpy of steam at the exit in kJ/kg
+Ws=-m*(he-hi);//calculation of power output from turbine using the first law of thermodynamics on the control-volume in kW
+
+//OUTPUT
+mprintf("\n The power output from the turbine=%0.1f kW\n",Ws);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.15/Chap5_Ex15.sce b/611/CH5/EX5.15/Chap5_Ex15.sce new file mode 100755 index 000000000..44fd21bf6 --- /dev/null +++ b/611/CH5/EX5.15/Chap5_Ex15.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 15,Page 183
+//Title: Exit velocity of steam
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Pi=3;//pressure of dry saturated steam when it enters the nozzle in bar
+Pe=2;//pressure of dry saturated steam at the exit in bar
+
+//CALCULATION
+//From steam tables corresponding to Pi
+si=6.9909;//entropy of steam at the entrance in kJ/kgK
+hi=2724.7;//entahlpy of steam at the entrance in kJ/kg
+
+//From steam tables corresponding to Pe
+sf=1.5301;//entropy of saturated liquid in kJ/kgK
+hf=504.70;//enthalpy of saturated liquid in kJ/kg
+sg=7.1268;//entropy of saturated vapour in kJ/kgK
+hg=2706.3;//entahlpy of saturayed vapour in kJ/kg
+
+se=6.9909;//From Eq.(5.67), se=si (i.e. entropy of the fluid remains constant), where se is in kJ/kgK
+Xe=(se-sf)/(sg-sf);//calculation of the quality of steam at the exit (no unit)
+he=((1-Xe)*hf)+(Xe*hg);//calculation of enthalpy of steam at the exit in kJ/kg
+Ve=sqrt (2*(hi-he)*10^3);//calculation of exit velocity of steam in m/s by applying the first law of thermodynamics
+
+//OUTPUT
+mprintf("\n The exit velocity of steam=%f m/s\n",Ve);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.16/Chap5_Ex16_R1.sce b/611/CH5/EX5.16/Chap5_Ex16_R1.sce new file mode 100755 index 000000000..e1bcc11a3 --- /dev/null +++ b/611/CH5/EX5.16/Chap5_Ex16_R1.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 16,Page 183
+//Title: Rate at which entropy is generated
+//================================================================================================================
+clear
+clc
+
+//INPUT
+N_glycerol=100;//molar flow rate of glycerol in mol/s
+Ti_gly=227;//inlet temperature of glycerol in degree celsius
+Te_gly=40;//outlet temperature of glycerol in degree celsius
+Ti_water=25;//inlet temperature of cooling water in degree celsius
+Te_water=50;//outlet temperature of cooling water in degree celsius
+Cp_gly=280;//heat capacity of glycerol in J/molK
+Cp_water=77;//heat capacity of water in J/molK
+
+//CALCULATION
+Ti_gly=Ti_gly+273.15;//conversion of temperature in K
+Te_gly=Te_gly+273.15;//conversion of temperature in K
+Ti_water=Ti_water+273.15;//conversion of temperature in K
+Te_water=Te_water+273.15;//conversion of temperature in K
+//calculation of the molar flow rate of water in mol/s by applying the first law of thermodynamics on the control-volume
+N_water=-(N_glycerol*Cp_gly*(Te_gly-Ti_gly))/(Cp_water*(Te_water-Ti_water));
+del_S_gly=N_glycerol*Cp_gly*log (Te_gly/Ti_gly)*10^-3;//calculation of change in entropy of glycerol in kJ/K s
+del_S_water=N_water*Cp_water*log (Te_water/Ti_water)*10^-3;//calculation of change in entropy of water in kJ/K s
+S_G=del_S_gly+del_S_water;//calculation of the rate at which entropy is generated in the heat exchanger in kJ/K s
+
+//OUTPUT
+mprintf("\n The rate at which entropy is generated in the heat exchanger=%0.3f kJ/K s\n",S_G);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.17/Chap5_Ex17_R1.sce b/611/CH5/EX5.17/Chap5_Ex17_R1.sce new file mode 100755 index 000000000..f4e9e50d4 --- /dev/null +++ b/611/CH5/EX5.17/Chap5_Ex17_R1.sce @@ -0,0 +1,51 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 17,Page 183
+//Title: Device and its feasibility
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_i=150;//temperature of saturated steam taken up by the device in degree celsius
+T_e=200;//temperature of superheated steam delivered by the device in degree celsius
+P_e=0.2;//pressure of superheated steam delivered by the device in MPa
+me2=0.949;//mass of superheated steam leaving the device in kg
+me1=0.051;//mass of saturated liquid leaving the device in kg
+T_liq=100;//temperature of saturated liquid leaving the device in degree celsius
+mi=1;//mass of saturated steam fed to the device in kg
+
+//CALCULATION
+//From steam tables corresponding to T_i
+hi=2745.4//enthalpy of saturated vapour in kJ/kg
+si=6.8358;//entropy of saturated vapour in kJ/kgK
+
+//For saturated liquid at T_liq
+he1=419.06;//enthalpy of saturated liquid in kJ/kg
+se1=1.3069;//entropy of saturated vapour in kJ/kgK
+
+//For superheated steam at P_e and T_e
+he2=2870.5;//enthalpy of superheated steam in kJ/kg
+se2=7.5072;//entropy of superheated steam in kJ/kgK
+
+//Test to see if the device obeys the first law of thermodynamics
+//Application of the first law of thermodynamics to the flow device gives: mi*hi=(me1*he1)+(me2*he2)
+LHS=mi*hi;
+RHS=(me1*he1)+(me2*he2);
+
+//Test to see if the device obeys the second law of thermodynamics
+//Application of the second law of thermodynamics to the flow device gives: (Ne1*se1)+(Ne2*se2)-(Ni*si)>|0
+S_G=(me1*se1)+(me2*se2)-(mi*si);
+
+//OUTPUT
+mprintf("\n The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied= %0.1f kJ\n",LHS);
+mprintf("\n The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied=%0.1f kJ\n",RHS);
+mprintf("\n The entropy generated by applying the second law of thermodynamics to the flow device=%0.4f kJ/kgK\n",S_G);
+if int(LHS)== int(RHS)& S_G>0 | S_G==0 then
+ mprintf("\n As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \n");
+else
+mprintf("\n As both the first and second law or either the first or second law of thermodynamics are not satisfied, the device is not feasible \n");
+end
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.18/Chap5_Ex18.sce b/611/CH5/EX5.18/Chap5_Ex18.sce new file mode 100755 index 000000000..325cb1c2a --- /dev/null +++ b/611/CH5/EX5.18/Chap5_Ex18.sce @@ -0,0 +1,36 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 18,Page 185
+//Title: Isentropic efficiency
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Pi=30;//pressure of superheated steam entering the turbine in bar
+Ti=300;//temperature of superheated steam entering the turbine in degree celsius
+Pe=0.1;//pressure at which steam exits the turbine in bar
+Xe=0.9;//quality of steam at the exit (no unit)(for the actual turbine)
+
+//CALCULATION
+//For superheated steam at Pi and Ti
+hi=2995.1;//enthalpy of superheated steam at the entrance in kJ/kg
+si=6.5422;//entropy of superheated steam at the entrance in kJ/kgK
+
+//For steam at Pe
+hf=191.83;//enthalpy of saturated liquid in kJ/kg
+hg=2584.8;//enthalpy of saturated vapour in kJ/kg
+sf=0.6493;//entropy of saturated liquid in kJ/kgK
+sg=8.1511;//entropy of saturated vapour in kJ/kgK
+
+//For isentropic turbine s1=s2 i.e. si=se, where si is the entropy at the entrance and se is the entropy at the exit
+X2=(si-sf)/(sg-sf);//calculation of the quality of steam at the exit for the isentropic process (no unit)
+h2=(hf*(1-X2))+(X2*hg);//calculation of the enthalpy of steam at the exit for the isentropic process in kJ/kg
+he=(hf*(1-Xe))+(Xe*hg);//calculation of the enthalpy of steam at the exit for the actual turbine in kJ/kg
+n_T=(hi-he)/(hi-h2);//calculation of isentropic efficiency of the turbine using Eq.(5.68) (no unit)
+
+//OUTPUT
+mprintf("\n The isentropic efficiency of the turbine=%f \n",n_T);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.19/Chap5_Ex19_R1.sce b/611/CH5/EX5.19/Chap5_Ex19_R1.sce new file mode 100755 index 000000000..2c72ed36f --- /dev/null +++ b/611/CH5/EX5.19/Chap5_Ex19_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 19,Page 186
+//Title: Power consumed by the compressor
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Ti=25;//temperature of air taken in by the adiabatic air compressor in degree celsius
+Pi=0.1;//pressure of air taken in by the adiabatic air compressor in MPa
+Pe=1;//discharge pressure of air in MPa
+n_c=0.8;//isentropic efficiency of the compressor (no unit)
+gaamma=1.4;//ratio of molar specific heat capacities (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Ti=Ti+273.15;//conversion of temperature in K
+Te=Ti*(((Pe*10^6)/(Pi*10^6))^((gaamma-1)/gaamma));//calculation of the discharge temperature of air using Eq.(4.35) in K (for reversible and adiabatic compression)
+W_s=(((R*gaamma)/(gaamma-1))*(Te-Ti))*10^-3;//calculation of the power consumed by the isentropic compressor using Eq.(5.69) in kW
+Ws=W_s/n_c;//calculation of the power consumed by an actual compressor per mole of air using Eq.(5.68)in kW
+Te_actual=((Ws*10^3*(gaamma-1))/(R*gaamma))+Ti;//calculation of the exit temperature of air in K
+
+//OUTPUT
+mprintf("\n The exit temperature of air=%0.2f K\n",Te_actual);
+mprintf("\n The power consumed by the compressor =%f kW/mol\n",Ws);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.2/Chap5_Ex2_R1.sce b/611/CH5/EX5.2/Chap5_Ex2_R1.sce new file mode 100755 index 000000000..b6ea1b7d2 --- /dev/null +++ b/611/CH5/EX5.2/Chap5_Ex2_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 2,Page 161
+//Title: Inventor and the heat engine
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Q=1000;//amount of energy absorbed by the heat engine in kJ/s
+W=650;//work delivered by the heat engine in kW
+T_source=500;//temperature of the source in degree celsius
+T_sink=25;//temperature of the sink in degree celsius
+
+//CALCULATION
+n_claimed=W/Q;//calculation of the efficiency of the heat engine invented by the inventor (no unit)
+T1=T_source+273.15;//conversion of source temperature in K
+T2=T_sink+273.15;//conversion of sink temperature in K
+n_carnot=1-(T2/T1);//calculation of the efficiency of a carnot engine from Eg.(5.1) (no unit)
+
+//OUTPUT
+mprintf("\n The efficiency of the Carnot engine=%0.3f \n",n_carnot);
+mprintf("\n The efficiency of the engine claimed by the inventor=%0.2f \n",n_claimed);
+if n_claimed<n_carnot then
+ mprintf("\n The claimed heat engine is theoretically feasible as the efficiency of the engine is lesser than that of a Carnot engine\n");
+else
+ mprintf("\n The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\n");
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.20/Chap5_Ex20_R1.sce b/611/CH5/EX5.20/Chap5_Ex20_R1.sce new file mode 100755 index 000000000..436190fba --- /dev/null +++ b/611/CH5/EX5.20/Chap5_Ex20_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 20,Page 187
+//Title: Power consumed by the pump
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Ti=30;//temperature of saturated liquid water in degree celsius
+m=500;//mass flow rate of water being pumped in kg/s
+P2=3;//preesure maintained in the boiler in MPa
+n_p=0.75;//isentropic efficiency of the pump (no unit)
+
+//CALCULATION
+//For saturated liquid water at Ti
+vf=0.0010043;//specific volume in m^3/kg
+P1=4.241;//pressure in kPa
+
+//Assuming that the liquid is incompressible, the power input required for an isentropic pump is calculated as:
+Ws_m=(vf*((P2*10^6)-(P1*10^3)))*10^-3;//power input required by the isentropic pump in kJ/kg
+Ws_act_m=Ws_m/n_p;//power input required by an actual pump in kJ/kg
+P=((Ws_act_m*10^3)*m)*10^-6;//calculation of the total power consumed by the pump in MW
+
+//OUTPUT
+mprintf("\n The power consumed by the pump = %d MW\n",P);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.21/Chap5_Ex21_R1.sce b/611/CH5/EX5.21/Chap5_Ex21_R1.sce new file mode 100755 index 000000000..8fd49fc1f --- /dev/null +++ b/611/CH5/EX5.21/Chap5_Ex21_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 21,Page 188
+//Title: Isentropic efficiency of nozzle
+//================================================================================================================
+clear
+clc
+
+//INPUT
+Pi=3;//pressure of dry saturated steam entering the nozzle in bar
+Xe=0.98;//quality of steam exiting the nozzle (no unit)
+Pe=2;//pressure of steam exiting the nozzle in bar
+
+//CALCULATION
+//For steam at Pi
+hi=2724.7;//enthalpy of dry saturated steam in kJ/kg
+he=2652.8;//enthalpy of steam at the exit for an isentropic nozzle,from Example 5.15,in kJ/kg
+V2_2_s=hi-he;//calculation of the specific kinetic energy of steam at the nozzle exit (for isentropic nozzle) in kJ/kg
+
+//For steam at Pe
+hf=504.70;//enthalpy of saturated liquid in kJ/kg
+hg=2706.3;//enthalpy of saturated vapour in kJ/kg
+he_act=((1-Xe)*hf)+(Xe*hg);//calculation of enthalpy of steam at the exit in kJ/kg
+V2_2=hi-he_act;//calculation of the actual specific kinetic energy of steam leaving the nozzle in kJ/kg
+n_N=(V2_2)/(V2_2_s);//calculation of isentropic efficiency of the nozzle using Eq.(5.72) (no unit)
+
+//OUTPUT
+mprintf("\n The isentropic efficiency of the nozzle=%0.3f \n",n_N);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.3/Chap5_Ex3.sce b/611/CH5/EX5.3/Chap5_Ex3.sce new file mode 100755 index 000000000..8efe9a198 --- /dev/null +++ b/611/CH5/EX5.3/Chap5_Ex3.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 3,Page 165
+//Title: Minimum power required
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_source_summer=42;//temperature in the summer months in degree celsius
+T_sink_winter=0;//temperature in the winter months in degree celius
+T_amb=25;//temperature at which the house is to be maintained during both the months in degree celsius
+//rate of energy loss from the windows,walls and roof, in kW per degree celsius difference between the ambient temperature and the conditions inside the room
+energy_loss=0.5;
+
+//CALCULATION
+T_H_summer=T_source_summer+273.15;//conversion of temperature in K
+T_L_summer=T_amb+273.15;//conversion of temperature in K
+T_H_winter=T_amb+273.15;//conversion of temperature in K
+T_L_winter=T_sink_winter+273.15;//conversion of temperature in K
+W_summer=(energy_loss*((T_H_summer-T_L_summer)^2))/(T_L_summer);//calculation of the minimum power required to operate the device in summer using Eq.(5.20) in kW
+W_winter=(energy_loss*((T_H_winter-T_L_winter)^2))/(T_H_winter);//calculation of the minimum power required to operate the device in winter using Eq.(5.21) in kW
+
+//OUTPUT
+mprintf("\n The minimum power required to operate the device in summer=%f kW \n",W_summer);
+mprintf("\n The minimum power required to operate the device in winter=%f kW \n",W_winter);
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE ANSWER GIVEN FOR THE MINIMUM POWER REQUIRED TO OPERATE THE DEVICE IN WINTER, IS NUMERICALLY INCORRECT IN THE TEXTBOOK.
diff --git a/611/CH5/EX5.4/Chap5_Ex4_R1.sce b/611/CH5/EX5.4/Chap5_Ex4_R1.sce new file mode 100755 index 000000000..789da876f --- /dev/null +++ b/611/CH5/EX5.4/Chap5_Ex4_R1.sce @@ -0,0 +1,24 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 4,Page 166
+//Title: Minimum work and maximum possible COP
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_L=4.25;//normal boiling point of helium in K
+Q_L=0.083;//latent heat of vaporization of helium in kJ/mol
+n=1;//amount of liquid helium to be produced in kmol
+T_amb=42;//ambient temperature in summer in degree celsius
+
+//CALCULATION
+T_H=T_amb+273.15;//conversion of temperature in K
+COP=(T_L)/(T_H-T_L);//calculation of COP of the refrigerator using Eq.(5.20)(no unit)
+W=(Q_L)/COP;//calculation of work to be done on the refrigerator unit using Eq.(5.20) in kJ
+
+//OUTPUT
+mprintf("\n The maximum possible COP of the unit=%0.4f \n",COP);
+mprintf("\n The minimum amount of work to be done on the refrigerating unit=%f kJ \n",W);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.5/Chap5_Ex5_R1.sce b/611/CH5/EX5.5/Chap5_Ex5_R1.sce new file mode 100755 index 000000000..775678902 --- /dev/null +++ b/611/CH5/EX5.5/Chap5_Ex5_R1.sce @@ -0,0 +1,36 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 5,Page 166
+//Title: Minimum power and maximum efficiency
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_ice=0;//temperature of the ice to be produced in degree celsius
+m=5000;//rate at which ice is to be produced in kg/hour
+T_water=0;//temperature of water used to produce ice in degree celsius
+T_amb=40;//ambient temperature in degree celsius
+T_source=100;//temperature of the source for operating heat engine in degree celsius
+lambda_fusion=6.002;//latent heat of fusion of water in kJ/mol at 0 degree celsius
+molar_mass=18*10^-3;//molar mass of water in kg/mol
+
+//CALCULATION
+T_L=T_water+273.15;//sink temperature of the refrigerating unit in K
+T_H=T_amb+273.15;//source temperature of the refrigerating unit in K
+COP=(T_L)/(T_H-T_L);//calculation of COP of the refrigerating unit using Eq.(5.20) (no unit)
+Q_L=((m/3600)/molar_mass)*(lambda_fusion);//calculation of the energy from the sink taken up by the refrigerator in kW
+W=(Q_L)/(COP);//calculation of the minimum power required to operate the refrigerator using Eq.(5.20)in kW
+T1=T_source+273.15;//temperature of the source of the heat engine in K
+T2=T_amb+273.15;//temperature of the sink of the heat engine in K
+n_heatengine=(T1-T2)/T1;//calculation of the efficiency of heat engine using Eq.(5.18) (no unit)
+Q1=W/n_heatengine;//calculation of the energy absorbed by the heat engine using Eq.(5.1) in kW
+//calculation of the ratio of energy rejected by both the devices to ambient atmosphere to the energy absorbed by the refrigerator (no unit)
+energy_ratio=(Q1+Q_L)/Q_L;
+
+//OUTPUT
+mprintf("\n The minimum power required to operate the refrigerator=%0.2f kW\n",W);
+mprintf("\n The maximum possible efficiency of the heat engine=%0.4f \n",n_heatengine);
+mprintf("\n Ratio of the energy rejected to the ambient atmosphere to the energy absorbed from the water=%0.4f \n",energy_ratio);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.6/Chap5_Ex6_R1.sce b/611/CH5/EX5.6/Chap5_Ex6_R1.sce new file mode 100755 index 000000000..de77a7240 --- /dev/null +++ b/611/CH5/EX5.6/Chap5_Ex6_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 6,Page 169
+//Title: Inventor and the claim
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T1=800;//temperature of reservoir 1 in K
+T2=400;//temperature of reservoir 2 in K
+Q1=1000;//energy absorbed from reservoir maintained at T1 in kJ
+Q2=400;//energy absorbed from reservoir maintained at T2 in kJ
+W=1000;//work delivered by the heat engine in kJ
+T3=300;//temperature of the sink in K
+
+//CALCULATION
+Q3=(Q1+Q2)-W;//calculation of the energy rejected to the sink using the first law of thermodynamics in kJ
+clausius_inequality=(Q1/T1)+(Q2/T2)-(Q3/T3);//application of the second law of thermodynamics in the form of the Clausius inequality using Eq.(5.28)
+
+//OUTPUT
+mprintf("\n The LHS of the Clausius inequality=%0.4f \n",clausius_inequality);
+if clausius_inequality<0 | clausius_inequality==0 then
+ mprintf("\n The given process does not violate the second law of thermodynamics, therefore the claim is correct\n");
+else
+ mprintf("\n This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\n");
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.7/Chap5_Ex7_R1.sce b/611/CH5/EX5.7/Chap5_Ex7_R1.sce new file mode 100755 index 000000000..4f6fd1cf0 --- /dev/null +++ b/611/CH5/EX5.7/Chap5_Ex7_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 7,Page 172
+//Title: Change in the entropy of the reactor contents
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T_system=200;//temperature of the contents of reactor in degree celsius
+t=15;//operation time of agitator in minutes
+P=750;//power of the operating motor in W
+
+//CALCULATION
+dQ=P*t*60*10^-3;//calculation of energy added as heat in kJ
+T=T_system+273.15;//conversion of temperature in K
+del_S=dQ/T;//calculation of entropy change using Eq.(5.32) in kJ/K
+
+//OUTPUT
+mprintf("\n The change in the entropy of the reactor contents=%0.4f kJ/K \n",del_S);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH5/EX5.8/Chap5_Ex8.sce b/611/CH5/EX5.8/Chap5_Ex8.sce new file mode 100755 index 000000000..b0747cff4 --- /dev/null +++ b/611/CH5/EX5.8/Chap5_Ex8.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 8,Page 172
+//Title: Entropy change
+//================================================================================================================
+clear
+clc
+
+//INPUT
+P=0.101325;//pressure in the piston cylinder assembly in MPa
+T1=300;//temperature of the piston cylinder assembly in K
+T2=400;//final temperature of the piston cylinder assembly in K
+a=45.369;//coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK
+b=8.688*10^-3;//coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK
+e=-9.619*10^5;//coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK
+//Cpo for CO2(g) is given as a+bT+eT^-2
+
+//CALCULATION
+
+del_S=(a*log(T2/T1))+(b*(T2-T1))-((e/2)*((1/T2^2)-(1/T1^2)));//calculation of entropy change for the constant pressure expansion in J/molK
+
+//OUTPUT
+mprintf("\n The change in entropy of CO2=%f J/molK\n",del_S);
+
+
+//===============================================END OF PROGRAM===================================================
+//DISCLAIMER: THE AUTHOR HAS NOT DIVIDED 'e' IN THE INTEGRATED EXPRESSION USED TO COMPUTE del_S BY 2, WHICH IS AN ERROR.THE INTEGRATION OF (eT^-3)dT IS -(e/2)*T^-2 THIS ERROR HAS BEEN RECTIFIED IN THIS PROGRAM.
diff --git a/611/CH5/EX5.9/Chap5_Ex9_R1.sce b/611/CH5/EX5.9/Chap5_Ex9_R1.sce new file mode 100755 index 000000000..19f8cce73 --- /dev/null +++ b/611/CH5/EX5.9/Chap5_Ex9_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-5,Example 9,Page 173
+//Title: Change in entropy of water
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m=1;//amount of saturated liquid water in kg
+T_initial=100;//initial temperature of water in degree celsius
+T_body=500;//temperature of body which is brought into contact with the cylinder in degree celsius
+hfg=2256.94;//enthalpy of vaporization taken from steam tables corresponding to T1 in kJ/kg
+
+//CALCULATION
+T=T_initial+273.15;//conversion of temperature in K
+del_S=hfg/T;//calculation of the entropy change during the process using Eq.(5.34) in kJ/kgK
+
+//OUTPUT
+mprintf("\n The change in entropy of water=%0.4f kJ/kgK\n",del_S);
+
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH6/EX6.10/Chap6_Ex10_R1.sce b/611/CH6/EX6.10/Chap6_Ex10_R1.sce new file mode 100755 index 000000000..3ff21a30f --- /dev/null +++ b/611/CH6/EX6.10/Chap6_Ex10_R1.sce @@ -0,0 +1,36 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-6,Example 10,Page 223
+//Title: Maximum work obtained from steam
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m=0.1;//mass of superheated steam in the piston cylinder assembly in kg
+P1=3;//initial pressure of superheated steam in MPa
+T1=300;//initial temperature of superheated steam in degree celsius
+T0=300;//temperature of the reservoir which is placed in thermal contact with the piston-cylinder assembly in degree celsius
+P2=0.1;//pressure of steam after expansion in MPa
+
+// ADDITIONAL DATA PROVIDED
+//For steam at P1 and T1:
+h1=2995.1;//specific enthalpy of steam in kJ/kg
+v1=0.08116;//specific volume of steam in m^3/kg
+s1=6.5422;//entropy of steam in kJ/kgK
+
+//For steam at P2 and T2:
+h2=3074.5;//specific enthalpy of steam in kJ/kg
+v2=2.6390;//specific volume of steam in m^3/kg
+s2=8.2166;//entropy of steam in kJ/kgK
+
+//CALCULATION
+T0=T0+273.15;//conversion of temperature in K
+
+//The maximum work can be carried out if the process is carried out reversibly, which is given by:
+W=m*(h1-h2-(((P1*v1)-(P2*v2))*10^3)-(T0*(s1-s2)));//calculation of maximum work obtained from the steam using Eq.(6.90) in kJ
+
+//OUTPUT
+mprintf("\n The maximum work obtained from steam=%0.2f kJ\n",W);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH6/EX6.12/Chap6_Ex12_R1.sce b/611/CH6/EX6.12/Chap6_Ex12_R1.sce new file mode 100755 index 000000000..e586d548f --- /dev/null +++ b/611/CH6/EX6.12/Chap6_Ex12_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-6,Example 12,Page 226
+//Title: Minimum power for compression
+//================================================================================================================
+clear
+clc
+
+//INPUT
+P1=0.1;//pressure at which air enters the compressor in MPa
+T1=300;//temperature at which air enters the compressor in K
+P2=1;//pressure at which air leaves the compressor in MPa
+T2=300;//temperature at which air leaves the compressor in K
+T0=300;//ambient temperature in K
+N=1;//molar flow rate of air in mol/s
+gaamma=1.4;//ratio of specific heat capacities (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+
+//T0=T1=T2 and h2-h1=Cp*(T2-T1)=0 as T2=T1
+
+Ws=(-N*T0*(-R*log (P2/P1)))*10^-3;//calculation of minimum power required for compression using Eq.(6.99) in kW
+
+//OUTPUT
+mprintf("\n The minimum power required to compress one mole per second of air=%0.3f kW\n",Ws);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH6/EX6.6/Chap6_Ex6_R1.sce b/611/CH6/EX6.6/Chap6_Ex6_R1.sce new file mode 100755 index 000000000..57b0df692 --- /dev/null +++ b/611/CH6/EX6.6/Chap6_Ex6_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-6,Example 6,Page 218
+//Title: Work done by steam
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m=0.1;//mass of superheated steam in the piston cylinder assembly in kg
+P1=1;//initial pressure of superheated steam in MPa
+T1=300;//initial temperature of superheated steam in degree celsius
+P2=0.1;//pressure of steam after expansion in MPa
+T2=200;//temperature of steam after expansion in degree celsius
+
+//CALCULATION
+//For steam at P1 and T1:
+h1=3052.1;//specific enthalpy of steam in kJ/kg
+v1=0.2580;//specific volume of steam in m^3/kg
+
+//For steam at P2 and T2:
+h2=2875.4;//specific enthalpy of steam in kJ/kg
+v2=2.1720;//specific volume of steam in m^3/kg
+
+del_u=(((h1*10^3)-(P1*10^6*v1))-((h2*10^3)-(P2*10^6*v2)))*10^-3;//calculation of the change in internal energy of the steam in kJ/kg
+W=m*(del_u);//calculation of the work done by using Eq.(6.77)steam in kJ
+
+//OUTPUT
+mprintf("\n The work done by steam=%0.2f kJ\n",W);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH6/EX6.8/Chap6_Ex8_R1.sce b/611/CH6/EX6.8/Chap6_Ex8_R1.sce new file mode 100755 index 000000000..d736a261e --- /dev/null +++ b/611/CH6/EX6.8/Chap6_Ex8_R1.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-6,Example 8,Page 220
+//Title: Power output of the turbine
+//================================================================================================================
+clear
+clc
+
+//INPUT
+P=3;//pressure of superheated steam in MPa
+Ti=300;//temperature at which the steam enters the turbine in degree celsius
+m=1;//mass flow rate of steam in kg/s
+Te=60;//temperature of dry saturated steam when it leaves the turbine in degree celsius
+
+//CALCULATION
+//For steam at P and Ti:
+h1=2995.1;//specific entahlpy of steam in kJ/kg
+
+//For saturated steam at Te:
+h2=2609.7;//specific enthalpy of saturated vapour in kJ/kg
+
+Ws=m*(h1-h2);//calculation of the power output of the turbine using Eq.(6.83) in kW
+
+//OUTPUT
+mprintf("\n The power output of the turbine=%0.1f kW\n",Ws);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH7/EX7.10/Chap7_Ex10_R1.sce b/611/CH7/EX7.10/Chap7_Ex10_R1.sce new file mode 100755 index 000000000..321e8aeee --- /dev/null +++ b/611/CH7/EX7.10/Chap7_Ex10_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-7,Example 10,Page 259
+//Title: Pressure at which boiler is to be operated
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T2=150;//temperature at which water it is desired to boil water in degree celsius
+P1=0.10133;//ambient pressure in MPa
+T1=100;//temperature at which water boils corresponding at pressure P1 in degree celsius
+del_hv=2256.94;//enthalpy of vaporization in kJ/kg
+R=8.314;//universal gas constant in J/molK
+M=18*10^-3;//molar mass of water in kg/mol
+
+//CALCULATION
+T1=T1+273.15;//conversion of temperature in K
+T2=T2+273.15;//conversion of temperature in K
+P2=P1*(exp (((del_hv*10^3*M)*((1/T1)-(1/T2)))/(R)));//calculation of the approximate pressure at which the boiler is to be operated using Eq.(7.92) in MPa
+
+//OUTPUT
+mprintf("\n The approximate pressure at which the boiler is to be operated=%0.3f MPa\n",P2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH7/EX7.11/Chap7_Ex11_R1.sce b/611/CH7/EX7.11/Chap7_Ex11_R1.sce new file mode 100755 index 000000000..a2fc816ee --- /dev/null +++ b/611/CH7/EX7.11/Chap7_Ex11_R1.sce @@ -0,0 +1,34 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-7,Example 11,Page 259
+//Title: The skating problem
+//================================================================================================================
+clear
+clc
+
+//INPUT
+m=60;//mass of the person who wants to skate in kg
+T=-2;//temperature of the ice in degree celsius
+A=15;//area of contact between the skate edges and ice in mm^2
+vs=1.091*10^-3;//specific volume of ice in m^3/kg (at Tref)
+vf=1.0*10^-3;//specific volume of water in m^3/kg (at Tref)
+del_hf=6.002;//enthalpy of melting of ice in kJ/mol
+g=9.81;//accleration due to gravity in m/s^2
+Tref=0;//reference temperature at which the specific enthalpy of ice and water are taken in degree celsius
+
+//CALCULATION
+Tref=Tref+273.15;//conversion of temperature in K
+del_P=((m*g)/(A*10^-6))*10^-6;//calculation of the pressure exerted on the ice by the skater in MPa
+del_v=(vf-vs)*(18*10^-3);//calculation of the change in volume in m^3/mol
+del_T=(del_P*10^6)/((del_hf*10^3)/(Tref*del_v));//calculation of the reduction in melting point of ice using Eq.(7.86)(Clapeyron equation) in degree celsius
+
+//OUTPUT
+mprintf("\n The temperature of ice originally = %d degree celsius \n",T);
+mprintf("\n The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation = %0.2f degree celsius \n",del_T);
+if del_T<T then
+ mprintf ("\n The ice can melt due to the additional pressure and therefore it will be possible to skate \n");
+else
+ mprintf ("\n The ice will not melt and therefore it will be difficult to skate \n");
+end
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH7/EX7.12/Chap7_Ex12_R1.sce b/611/CH7/EX7.12/Chap7_Ex12_R1.sce new file mode 100755 index 000000000..2f817122d --- /dev/null +++ b/611/CH7/EX7.12/Chap7_Ex12_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-7,Example 12,Page 261
+//Title: Enthalpy of vaporization
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T1=100;//temperature of water in degree celsius
+del_hv1=2256.94;//enthalpy of vaporization at T1 in kJ/kg
+T2=150;//temperature at which the enthalpy of vaporization is to be determined in degree celsius
+Cp_f=4.26;//isobaric heat capacity of liquid in kJ/kgK
+Cp_g=1.388;//isobaric heat capacity of vapour in kJ/kgK
+
+//CALCULATION
+del_hv2=((Cp_g-Cp_f)*(T2-T1))+del_hv1;//calculation of the enthalpy of vaporization at T2 using Eq.(7.98) (Kirchhoff equation) in kJ/kg
+
+//OUTPUT
+mprintf("\n The enthalpy of vaporization at 150 degree celsius=%0.2f kJ/kg\n",del_hv2);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH7/EX7.13/Chap7_Ex13_R1.sce b/611/CH7/EX7.13/Chap7_Ex13_R1.sce new file mode 100755 index 000000000..9b1bca508 --- /dev/null +++ b/611/CH7/EX7.13/Chap7_Ex13_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-7,Example 13,Page 261
+//Title: Enthalpy of vaporization using Watson's correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T1=100;//temperature of water in degree celsius
+del_hv1=2256.94;//enthalpy of vaporization at T1 in kJ/kg
+T2=150;//temperature at which the enthalpy of vaporization is to be determined in degree celsius
+del_hv_kirchoff=2113.34;//enthalpy of vaporization predicted by the Kirchhoff relation taken from Example 7.12 for comparison, in kJ/kg
+del_hv_steam_tables=2113.25;//enthalpy of vaporization taken from the steam tables corresponding to T2,for comparison, in kJ/kg
+Tc=647.3;//critical temperature of water in K
+
+//CALCULATION
+T1=T1+273.15;//conversion of temperature in K
+T2=T2+273.15;//conversion of temperature in K
+Tr1=T1/Tc;//calculation of reduced temperature corresponding to state 1 (no unit)
+Tr2=T2/Tc;//calculation of reduced temperature corresponding to state 2 (no unit)
+del_hv2=del_hv1*(((1-Tr2)/(1-Tr1))^0.38);//calculation of enthalpy of vaporization at T2 using Eq.(7.101) in kJ/kg
+
+//OUTPUT
+mprintf("\n The enthalpy of vaporization at 150 degree celsius using \n");
+mprintf("\n Watson correlation \t = %f kJ/kg\n",del_hv2);
+mprintf("\n Kirchhoffs relation \t = %f kJ/kg\n",del_hv_kirchoff);
+mprintf("\n From steam tables \t = %f kJ/kg\n",del_hv_steam_tables);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH7/EX7.14/Chap7_Ex14_R1.sce b/611/CH7/EX7.14/Chap7_Ex14_R1.sce new file mode 100755 index 000000000..c3a0210a3 --- /dev/null +++ b/611/CH7/EX7.14/Chap7_Ex14_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-7,Example 14,Page 262
+//Title: Enthalpy of vaporization using Riedel's correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=373.15;//normal boiling point of water in K (temperature at which the enthalpy of vaporization is to be determined)
+Pc=221.2//critical pressure of water in bar
+Tc=647.3;//critical temperature of water in K
+R=8.314;//universal gas constant in J/molK
+del_hvn_steam_tables=2256.94;//enthalpy of vaporization at the normal boiling point taken from the steam tables, for comparison, in kJ/kg
+
+//CALCULATION
+Tbr=T/Tc;//calculation of the reduced normal boiling point (no unit)
+//calculation of the enthalpy of vaporization at the normal boiling point using Eq.(7.102) in kJ/kg
+del_hvn=((1.093*R*Tc*(Tbr*((log (Pc)-1.013)/(0.930-Tbr))))*10^-3)/(18*10^-3);
+err=abs ((del_hvn-del_hvn_steam_tables)/del_hvn_steam_tables)*100;//calculation of percentage error
+
+//OUTPUT
+mprintf("\n The enthalpy of vaporization at the normal boiling point \n");
+mprintf("\n Using Riedels correlation \t = %f kJ/kg\n",del_hvn);
+mprintf("\n From the steam tables \t \t = %f kJ/kg\n",del_hvn_steam_tables);
+mprintf("\n Error \t \t \t \t = %f %% \n",err);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH8/EX8.10/Chap8_Ex10_R1.sce b/611/CH8/EX8.10/Chap8_Ex10_R1.sce new file mode 100755 index 000000000..e79c004d8 --- /dev/null +++ b/611/CH8/EX8.10/Chap8_Ex10_R1.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 10,Page 299
+//Title: Enthalpy and entropy departure using the generalized virial coefficient correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=339.7;//temperature of ethylene in K
+P=1;//pressure of ethylene in bar
+Tc=283.1;//critical temperature of ethylene in K
+Pc=51.17;//critical pressure of ethylene in bar
+w=0.089;//acentric factor (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+dep_h=R*Tc*Pr*((0.083-(1.097/(Tr^1.6)))+(w*(0.139-(0.894/(Tr^4.2)))));//calculation of the enthalpy departure using Eq.(8.75) in J/mol
+dep_s=-Pr*R*((0.675/(Tr^2.6))+(w*(0.722/(Tr^5.2))));//calculation of the entropy departure using Eq.(8.76) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for ethylene using the generalized virial coefficient correlation = %f J/mol\n",dep_h);
+mprintf("\n The entropy departure for ethylene using the generalized virial coefficient correlation = %e J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH8/EX8.11/Chap8_Ex11_R1.sce b/611/CH8/EX8.11/Chap8_Ex11_R1.sce new file mode 100755 index 000000000..9aede604b --- /dev/null +++ b/611/CH8/EX8.11/Chap8_Ex11_R1.sce @@ -0,0 +1,104 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 11,Page 299
+//Title: Volume,Enthalpy and entropy departure using the Peng-Robinson equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+T_ref=0;//reference state saturated liquid temperature in degree celsius
+h0=0;//enthalpy of saturated liquid in J/mol (reference state)
+s0=0;//entropy of saturated liquid in J/molK (reference state)
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+NBP=398.8;//normal boiling point of n-octane (saturated liquid)
+[Cp]=[6.907;741.770*10^-3;-397.204*10^-6;82.629*10^-9;0];//coefficients to compute the isobaric molar heat capacity, where Cp=a+bT+cT^2+dT^3+eT^-2,in J/molK
+S=0.9457;//value of S taken from Example (3.16)
+b=1.4750*10^-4;//value of the Peng-Robinson constant in m^3/mol from Example (3.16)
+v=15.14*10^-3;//volume of saturated vapour in m^3/mol from Example (3.16)
+R=8.314;//universal gas constant in J/molK
+P_amb=101.325;//pressure at which the normal boiling point is computed in kPa
+
+//CALCULATION
+
+//Step a: Vaporization of n-octane at T_ref
+T_ref=T_ref+273.15;//conversion of temperature in K
+//calculation of the enthalpy of vaporization using Eq.(7.92)(Clausius-Clayperon equation) in kJ/mol
+del_hv=((R*log ((Pc*10^5)/(P_amb*10^3)))/((1/NBP)-(1/Tc)))*10^-3;
+P2=P_amb*exp (((del_hv*10^3)/(R))*((1/NBP)-(1/T_ref)));//calculation of the vapour pressure at T_ref using Eq.(7.92)(Clausius-Clayperon equation) in kPa
+Tbr=NBP/Tc;//calculation of the reduced normal boiling point (no unit)
+//calculation of the enthalpy of vaporization at the normal boiling point using Eq.(7.102)(Riedel's correlation) in kJ/mol
+del_hvn=(1.093*R*Tc*(Tbr*(((log (Pc))-1.013)/(0.930-Tbr))))*10^-3;
+Tr2=T_ref/Tc;//calculation of the reduced temperature with reference to T_ref (no unit)
+//calculation of the enthalpy of vaporization (Step a)at T_ref using Eq.(7.101)(Watson's correlation) in kJ/mol
+del_ha=((del_hvn*10^3)*(((1-Tr2)/(1-Tbr))^(0.38)))*10^-3;
+del_sa=(del_ha*10^3)/T_ref;//calculation of the entropy change in the conversion from saturated liquid to saturated vapour (Step a) at T_ref in J/molK
+
+//Step b:Change from real state of n-octane at T_ref and P2 to ideal gas state at the same temperature and pressure
+alpha=(1+(S*(1-sqrt (Tr2))))^2;//calculation of alpha so as to compute the Peng-Robinson constant (a)
+a=(0.45724*(R^2)*(Tc^2)*alpha)/(Pc*10^5);//calculation of the Peng-Robinson constant using Eq.(3.76) in Pa(m^3/mol)^2
+//Using the Cardan's method to evaluate Z
+A=(a*P2*10^3)/(R*T_ref)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P2*10^3)/(R*T_ref);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1+B;//calculation of alpha for Peng-Robinson equation of state using Table (3.2)
+beeta=A-(2*B)-(3*B^2);//calculation of beeta for Peng-Robinson equation of state using Table (3.2)
+gaamma=-(A*B)+(B^2)+(B^3);//calculation of gaamma for Peng-Robinson equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ else
+ theta=acos((-(q)/2)*(sqrt((-27)/(((p)^3)))));//calculation of theta in radians using Eq.(3.37)
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ end
+end
+da_dT=(-a*S)/(sqrt (alpha*T_ref*Tc));//calculation of da/dT using Eq.(8.47)
+//calculation of the enthalpy departure using Eq.(8.45) in J/mol
+dep_h=(R*T_ref*(Z-1))+(((((T_ref*da_dT)-a)/(2*sqrt(2)*b)))*(log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2)))))));
+//calculation of the entropy departure using Eq.(8.46)in J/molK
+dep_s=(R*log (Z-B))+((1/(2*sqrt (2)*b))*(da_dT)*(log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2)))))));
+del_hb=-dep_h;//calculation of the enthalpy change corresponding to Step b in J/mol
+del_sb=-dep_s;//calculation of the entropy change corresponding to Step b in J/molK
+
+//Step c:Change n-octane in the ideal gas state at T_ref and P2 to n-octane in the ideal gas state at T and P
+//Calculation of the enthalpy change corresponding to Step c using Eq.(4.25) in kJ/mol
+ del_hc=((Cp(1,:)*(T-T_ref))+(((Cp(2,:))/2)*((T^2)-(T_ref^2)))+(((Cp(3,:))/3)*((T^3)-(T_ref^3)))+(((Cp(4,:))/4)*((T^4)-(T_ref^4)))-((Cp(5,:))*((1/T)-(1/T_ref))))*10^-3;
+del_sc=((Cp(1,:))*log (T/T_ref))+((Cp(2,:))*(T-T_ref))+(((Cp(3,:))/2)*((T^2)-(T_ref^2)))+(((Cp(4,:))/3)*((T^3)-(T_ref^3)))-(((Cp(5,:))/2)*((1/(T^2))-(1/(T_ref^2))))-(R*log ((P*10^6)/(P2*10^3)));//calculation of the entropy change corresponding to Step c using Eq.(5.43) in J/molK
+
+//Step d: Change n-octane from the ideal gas state to the real state at T and P
+Z=0.9151;//compressibility factor taken from Example (3.16) (no unit)
+da_dT=(-a*S)/(sqrt (alpha*T*Tc));//calculation of da/dT using Eq.(8.47)
+//calculation of the enthalpy change corresponding to Step d using Eq.(8.45) in J/mol
+del_hd=(R*T*(Z-1))+((((T*da_dT)-a)/(2*sqrt(2)*b))*log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2))))));
+//calculation of the entropy change corresponding to Step d using Eq.(8.46)in J/molK
+del_sd=(R*log (Z-B))+((1/(2*sqrt (2)*b))*(da_dT)*(log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2)))))));
+
+h=h0+del_ha+(del_hb*10^-3)+del_hc+(del_hd*10^-3);//calculation of the enthalpy of n-octane vapour at T and P in kJ/mol
+s=s0+del_sa+del_sb+del_sc+del_sd;//calculation of the entropy of n-octane vapour at T and P in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state = %f kJ/mol\n",h);
+mprintf("\n The entropy of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state = %f J/mol K\n",s);
+mprintf("\n The volume of n-octane vapour at 427.85K and 0.215MPa using the Peng-Robinson equation of state = %f m^3/mol\n",v)
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE VOLUME OF n-OCTANE VAPOUR AS COMPUTED IN EXAMPLE 3.16 IS 15.14*10^-3 m^3/mol AND NOT 15.41*10^-3 m^3/mol AS PRINTED IN THIS EXAMPLE IN THE TEXTBOOK.
+
diff --git a/611/CH8/EX8.2/Chap8_Ex2_R1.sce b/611/CH8/EX8.2/Chap8_Ex2_R1.sce new file mode 100755 index 000000000..72678dbe6 --- /dev/null +++ b/611/CH8/EX8.2/Chap8_Ex2_R1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 2,Page 275
+//Title: Enthalpy and entropy departure
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+a=3.789;//van der Waals constant in Pa (m^3/mol)^2
+b=2.37*10^-4;//van der Waals constant in m^3/mol
+v=15.675*10^-3;//volume occupied by n-octane vapour taken from Example (3.8) in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//n-octane obeys the van der Waals equation of state
+dep_h=(P*10^6*v)-(R*T)-(a/v);//calculation of the enthalpy departure using Example(8.1) in J/mol
+dep_s=R*log ((P*10^6*(v-b))/(R*T));//calculation of the entropy departure using Example(8.1) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for n-octane vapour = %0.2f J/mol\n",dep_h);
+mprintf("\n The entropy departure for n-octane vapour = %0.4f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH8/EX8.3/Chap8_Ex3_R1.sce b/611/CH8/EX8.3/Chap8_Ex3_R1.sce new file mode 100755 index 000000000..d6106cc2f --- /dev/null +++ b/611/CH8/EX8.3/Chap8_Ex3_R1.sce @@ -0,0 +1,48 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 3,Page 276
+//Title: Enthalpy departure using Beattie-Bridgman equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=100;//temperature of carbon dioxide in degree celsius
+P=10;//pressure of carbon dioxide in MPa
+A0=0.5073;//Beattie-Bridgman constant for carbon dioxide in (Pa m^3)/mol^2
+B0=104.76*10^-6;//Beattie-Bridgman constant for carbon dioxide in m^3/mol
+a=71.32*10^-6;//Beattie-Bridgman constant for carbon dioxide in m^3/mol
+b=72.35*10^-6;//Beattie-Bridgman constant for carbon dioxide in m^3/mol
+C=660.0;//Beattie-Bridgman constant for carbon dioxide in (m^3 K^3)/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//The virial form of the Beattie-Bridgman equation of state from Eq.(8.25) is given as:
+//P=(A1/v)+(A2/v^2)+(A3/v^3)+(A4/v^4)
+
+T=T+273.15;//conversion of temperature in K
+A1=(R*T);//calculation of A1 using Eq.(8.27)
+A2=(B0*R*T)-A0-((C*R)/T^2);//calculation of A2 using Eq.(8.28)
+A3=(a*A0)-(b*B0*R*T)-((B0*C*R)/T^2);//calculation of A3 using Eq.(8.29)
+A4=((b*C*B0*R)/T^2);//calculation of A4 using Eq.(8.30)
+vguess=0.01;//taking a guess value of the volume,to be used for solving Eq.(8.25) using the function defined below, in m^3/mol
+tolerance=1e-6;//Framing the tolerance limit for the convergence of the equation
+function[fn]=solver_func(vi)
+ fn=(P*10^6)-((A1/vi)+(A2/vi^2)+(A3/vi^3)+(A4/vi^4));//Function defined for solving the system given by Eq.(8.25)
+endfunction
+[v]=fsolve(vguess,solver_func,tolerance)//using inbuilt function fsolve for solving the system of equations, to determine the volume in m^3/mol
+Z=(P*10^6*v)/(R*T);//calculation of compressibility factor (no unit)
+//calculation of the enthalpy departure using Eq.(8.37) in J/mol
+dep_h=(((B0*R*T)-(2*A0)-((4*C*R)/(T^2)))*(1/v))+((((3/2)*a*A0)-(b*B0*R*T)-((5*B0*C*R)/(2*(T^2))))*(1/(v^2)))+((2*b*C*B0*R)/((T^2)*(v^3)));
+
+//OUTPUT
+mprintf("\n Molar volume of CO2 at %0.f MPa and %0.2f K = %f m^3/mol \n",P,T,v);
+mprintf("\n The compressibility factor=%f \n",Z);
+mprintf("\n The enthalpy departure for carbon dioxide using the Beattie-Bridgman equation of state = %f J/mol\n",dep_h);
+
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE PROBLEM STATEMENT GIVES THE TEMPERATURE AS 100 DEGREE CELSIUS, WHICH CORRESPONDS TO A TEMPERATURE OF 373.15K. HOWEVER, IN THE COMPUTATION OF THE SECOND TERM IN THE ENTHALPY DEPARTURE EXPRESSION, THE AUTHOR HAS TAKEN THE TEMPERATURE TO BE 313.15K, WHICH CLEARLY IS A PRINTING ERROR. THE CODE ABOVE HAS BEEN WRITTEN FOR THE TEMPERATURE OF 373.15K, AS GIVEN IN THE PROBLEM STATEMENT.
+
+
diff --git a/611/CH8/EX8.4/Chap8_Ex4_R1.sce b/611/CH8/EX8.4/Chap8_Ex4_R1.sce new file mode 100755 index 000000000..560e8a63b --- /dev/null +++ b/611/CH8/EX8.4/Chap8_Ex4_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 4,Page 278
+//Title: Entropy departure using Beattie-Bridgman equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=100;//temperature of carbon dioxide in degree celsius
+P=10;//pressure of carbon dioxide in MPa
+B0=104.76*10^-6;//Beattie-Bridgman constant for carbon dioxide in m^3/mol
+b=72.35*10^-6;//Beattie-Bridgman constant for carbon dioxide in m^3/mol
+C=660.0;//Beattie-Bridgman constant for carbon dioxide in (m^3 K^3)/mol
+R=8.314;//universal gas constant in J/molK
+v=0.233*10^-3;//volume calculated in Example (8.3) in m^3/mol
+Z=0.751;//compressibility factor as calculated in Example (8.3) (no unit)
+
+//CALCULATION
+T=T+273.15;//conversion of temperature in K
+//calculation of entropy departure using Eq.(8.38) in J/molK
+dep_s=(R*log (Z))-(((B0*R)+((2*C*R)/(T^3)))*(1/v))+(((b*B0*R)-((2*C*B0*R)/(T^3)))*(1/(2*(v^2))))+((2*b*C*B0*R)/(3*(T^3)*(v^3)));
+
+//OUTPUT
+mprintf("\n The entropy departure for carbon dioxide using the Beattie-Bridgman equation of state = %f J/mol K\n",dep_s);
+
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE PROBLEM STATEMENT MENTIONS THE TEMPERATURE TO BE 100 DEGREE CELSIUS, WHICH CORRESPONDS TO A TEMPERATURE OF 373.15K. HOWEVER, THE AUTHOR HAS EVALUATED THE ENTROPY DEPARTURE FOR A TEMPERATURE OF 313.15K, WHICH IS CLEARLY A PRINTING ERROR. THE CODE ABOVE HAS BEEN WRITTEN FOR THE TEMPERATURE OF 373.15K AS GIVEN IN THE PROBLEM STATEMENT.
diff --git a/611/CH8/EX8.5/Chap8_Ex5_R1.sce b/611/CH8/EX8.5/Chap8_Ex5_R1.sce new file mode 100755 index 000000000..eb10b789c --- /dev/null +++ b/611/CH8/EX8.5/Chap8_Ex5_R1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 5,Page 281
+//Title: Enthalpy and entropy departure using the generalized Redlich-Kwong equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+a=4.426;//Redlich-Kwong constant taken from Example(3.9) in (m^6 Pa mol^-2)
+b=164.3*10^-6;//Redlich-Kwong constant taken from Example(3.9) in m^3/mol
+Z=0.9308;//compressibility factor taken from Example(3.9) (no unit)
+B=9.9306*10^-3;//value of B, used in the Cardan's method in Example (3.9)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+dep_h=(R*T*(Z-1))-(((3*a)/(2*b))*log ((Z+B)/Z));//calculation of the enthalpy departure using Eq.(8.39) in J/mol
+dep_s=(R*log(Z-B))-((a/(2*b*T))*log((Z+B)/Z));//calculation of the entropy departure using Eq.(8.40) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for n-octane vapour using the generalized Redlich-Kwong equation of state = %0.2f J/mol\n",dep_h);
+mprintf("\n The entropy departure for n-octane vapour using the generalized Redlich-Kwong equation of state = %0.4f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH8/EX8.6/Chap8_Ex6_R1.sce b/611/CH8/EX8.6/Chap8_Ex6_R1.sce new file mode 100755 index 000000000..ace688753 --- /dev/null +++ b/611/CH8/EX8.6/Chap8_Ex6_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 6,Page 281
+//Title: Enthalpy and entropy departure using the SRK equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+S=1.0786;//constant used in the SRK equation of state,from Example(3.15)
+alpha=1.3079;//constant used in the SRK equation of state,from Example(3.15)
+a=5.0180;//constant used in the SRK equation of state,from Example(3.15) in (m^6 Pa mol^-2)
+b=1.6426*10^-4;//constant used in the SRK equation of state,from Example(3.15) in m^3/mol
+B=9.9282*10^-3;//factor used in the Cardan's method for solving the SRK equation of state,from Example(3.15) (no unit)
+Z=0.9191;//compressibility factor taken from Example (3.15) (no unit)
+R=8.314;//universal gas constant in J/molK
+Tc=569.4;//critical temperature of n-octane in K
+
+//CALCULATION
+da_dT=(-a*S)/(sqrt (alpha*T*Tc));//calculation of da/dT using Eq.(8.44)
+dep_h=(R*T*(Z-1))+((((T*da_dT)-a)/b)*log ((Z+B)/Z));//calculation of the enthalpy departure using Eq.(8.42) in J/mol
+dep_s=(R*log (Z-B))+((1/b)*(da_dT)*log ((Z+B)/Z));//calculation of the entropy departure using Eq.(8.43) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for n-octane vapour using the SRK equation of state = %f J/mol\n",dep_h);
+mprintf("\n The entropy departure for n-octane vapour using the SRK equation of state = %0.4f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
+
+// DISCLAIMER: NUMERICAL ERROR OBSERVED IN ENTHALPY DEPARTURE FUNCTION CALCULATION. FOR THE GIVEN INPUT DATA, THE ANSWER GIVEN IN THE BOOK "-890.22 J/mol" WAS FOUND TO BE NUMERICALLY INCORRECT.
diff --git a/611/CH8/EX8.7/Chap8_Ex7_R1.sce b/611/CH8/EX8.7/Chap8_Ex7_R1.sce new file mode 100755 index 000000000..e5d5a8699 --- /dev/null +++ b/611/CH8/EX8.7/Chap8_Ex7_R1.sce @@ -0,0 +1,31 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 7,Page 282
+//Title: Enthalpy and entropy departure using the Peng-Robinson equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+S=0.9457;//constant used in the Peng-Robinson equation of state,from Example(3.16)
+alpha=1.2677;//constant used in the Peng-Robinson equation of state,from Example(3.16)
+a=5.2024;//constant used in the Peng-Robinson equation of state,from Example(3.16) in (m^6 Pa mol^-2)
+b=1.4750*10^-4;//constant used in the Peng-Robinson equation of state,from Example(3.16) in m^3/mol
+B=8.9151*10^-3;//factor used in the Cardan's method for solving the Peng-Robinson equation of state,from Example(3.16) (no unit)
+Z=0.9151;//compressibility factor taken from Example (3.16) (no unit)
+R=8.314;//universal gas constant in J/molK
+Tc=569.4;//critical temperature of n-octane in K
+
+//CALCULATION
+da_dT=(-a*S)/(sqrt (alpha*T*Tc));//calculation of da/dT using Eq.(8.47)
+//calculation of the enthalpy departure using Eq.(8.45) in J/mol
+dep_h=(R*T*(Z-1))+(((((T*da_dT)-a)/(2*sqrt(2)*b)))*(log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2)))))));
+dep_s=(R*log (Z-B))+((1/(2*sqrt (2)*b))*(da_dT)*(log ((Z+(B*(1+sqrt (2))))/(Z+(B*(1-sqrt (2)))))));//calculation of the entropy departure using Eq.(8.46)in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for n-octane vapour using the Peng-Robinson equation of state = %0.1f J/mol\n",dep_h);
+mprintf("\n The entropy departure for n-octane vapour using the Peng-Robinson equation of state = %0.3f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH8/EX8.8/Chap8_Ex8_R1.sce b/611/CH8/EX8.8/Chap8_Ex8_R1.sce new file mode 100755 index 000000000..00f1ac0ff --- /dev/null +++ b/611/CH8/EX8.8/Chap8_Ex8_R1.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 8,Page 284
+//Title: Enthalpy and entropy departure using the Edmister charts
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=339.7;//temperature of ethylene in K
+P=30.7;//pressure of ethylene in bar
+Tc=283.1;//critical temperature of ethylene in K
+Pc=51.17;//critical pressure of ethylene in bar
+w=0.089;//acentric factor (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+del_h0=0.45;//value of ((h0-h)/RTc)_0 read from Fig.(8.2) corresponding to Pr and Tr (no unit)
+del_h1=0.18;//value of ((h0-h)/RTc)_1 read from Fig.(8.3) corresponding to Pr and Tr (no unit)
+del_s0=0.26;//value of ((s0-s)/R)_0 read from Fig.(8.4) corresponding to Pr and Tr (no unit)
+del_s1=0.20;//value of ((s0-s)/R)_1 read from Fig.(8.5) corresponding to Pr and Tr (no unit)
+dep_h=((del_h0)+(w*del_h1))*R*Tc;//calculation of the enthalpy departure using Eq.(8.52) in J/mol
+dep_s=((del_s0)+(w*del_s1))*R;//calculation of the entropy departure using Eq.(8.56) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for ethylene using the Edmister charts = %0.3f J/mol\n",dep_h);
+mprintf("\n The entropy departure for ethylene using the Edmister charts = %0.4f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH8/EX8.9/Chap8_Ex9_R1.sce b/611/CH8/EX8.9/Chap8_Ex9_R1.sce new file mode 100755 index 000000000..2cd472c74 --- /dev/null +++ b/611/CH8/EX8.9/Chap8_Ex9_R1.sce @@ -0,0 +1,33 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-8,Example 9,Page 297
+//Title: Enthalpy and entropy departure using the Lee-Kesler data
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=339.7;//temperature of ethylene in K
+P=30.7;//pressure of ethylene in bar
+Tc=283.1;//critical temperature of ethylene in K
+Pc=51.17;//critical pressure of ethylene in bar
+w=0.089;//acentric factor (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+del_h0=0.474;//value of ((h0-h)/RTc)_0 read from Fig.(8.6) corresponding to Pr and Tr (no unit)
+del_h1=0.232;//value of ((h0-h)/RTc)_1 read from Fig.(8.8) corresponding to Pr and Tr (no unit)
+del_s0=0.277;//value of ((s0-s)/R)_0 read from Fig.(8.10) corresponding to Pr and Tr (no unit)
+del_s1=0.220;//value of ((s0-s)/R)_1 read from Fig.(8.12) corresponding to Pr and Tr (no unit)
+dep_h=((del_h0)+(w*del_h1))*R*Tc;//calculation of the enthalpy departure using Eq.(8.62) in J/mol
+dep_s=((del_s0)+(w*del_s1))*R;//calculation of the entropy departure using Eq.(8.65) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure for ethylene using the Lee-Kesler data = %f J/mol\n",dep_h);
+mprintf("\n The entropy departure for ethylene using the Lee-Kesler data = %f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH9/EX9.1/Chap9_Ex1_R1.sce b/611/CH9/EX9.1/Chap9_Ex1_R1.sce new file mode 100755 index 000000000..2522476e1 --- /dev/null +++ b/611/CH9/EX9.1/Chap9_Ex1_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 1,Page 313
+//Title: Partial molar volume
+//================================================================================================================
+clear
+clc
+
+//INPUT
+per_ethanol=60;//mole percent of ethanol in a ethanol-water system
+per_water=40;//mole percent of water in a ethanol-water system
+v1=57.5*10^-6;//partial molar volume of ethanol in the ethanol-water system in m^3
+rho=849.4;//density of the mixture in kg/m^3
+M_ethanol=46*10^-3;//molar mass of ethanol in kg/mol
+M_water=18*10^-3;//molar mass of ethanol in kg/mol
+
+//CALCULATION
+X1=per_ethanol/100;//calculation of the mole fraction of ethanol (no unit)
+X2=per_water/100;//calculation of the mole fraction of water (no unit)
+M=(X1*M_ethanol)+(X2*M_water);//calculation of the molar mass of the ethanol-water mixture in kg/mol
+v=M/rho;//calculation of the molar volume of the mixture in m^3/mol
+v2=(v-(X1*v1))/(X2);//calculation of the partial molar volume of water using Eq.(9.10) in m^3/mol
+
+//OUTPUT
+mprintf("\n The partial molar volume of water = %f m^3/mol\n",v2);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH9/EX9.10/Chap9_Ex10_R1.sce b/611/CH9/EX9.10/Chap9_Ex10_R1.sce new file mode 100755 index 000000000..1e622f701 --- /dev/null +++ b/611/CH9/EX9.10/Chap9_Ex10_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 10,Page 334
+//Title: Molar volume of an equimolar mixture using pseudocritical properties
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+Tcm=(y1*Tc(1,:))+(y2*Tc(2,:));//calculation of pseudocritical temperature of mixture using Eq.(9.100) in K
+Pcm=(y1*Pc(1,:))+(y2*Pc(2,:));//calculation of pseudocritical pressure of mixture using Eq.(9.101) in bar
+Trm=T/Tcm;//calculation of pseudoreduced temperature of the mixture using Eq.(9.102) (no unit)
+Prm=P/Pcm;//calculation of pseudoreduced pressure of the mixture using Eq.(9.103) (no unit)
+Zm0=0.9;//value of Zm0 is taken from the generalized compressibility chart, Figure(3.11) corresponding to Trm and Prm (no unit)
+vm=(Zm0*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+
+//OUTPUT
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane using the pseudocritical properties estimated through Kays rule = %0.2e m^3/mol\n",vm);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH9/EX9.11/Chap9_Ex11_R1.sce b/611/CH9/EX9.11/Chap9_Ex11_R1.sce new file mode 100755 index 000000000..d982fc412 --- /dev/null +++ b/611/CH9/EX9.11/Chap9_Ex11_R1.sce @@ -0,0 +1,34 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 11,Page 335
+//Title: Molar volume of mixture using Prausnitz-Gunn rule
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+vc=[255.0*10^-6;486.0*10^-6];//critical molar volume of n-butane and n-octane in m^3/mol
+Zc=[0.274;0.256];//compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+Tcm=(y1*Tc(1,:))+(y2*Tc(2,:));//calculation of pseudocritical temperature of mixture using Eq.(9.104) in K
+Pcm=((R*((y1*Zc(1,:))+(y2*Zc(2,:)))*Tcm)/((y1*vc(1,:))+(y2*vc(2,:))))*10^-5;//calculation of the pseudocritical pressure of mixture using Eq.(9.105) in bar
+Trm=T/Tcm;//calculation of pseudoreduced temperature using Eq.(9.102) (no unit)
+Prm=P/Pcm;//calculation of pseudoreduced pressure using Eq.(9.103) (no unit)
+Zm0=0.89;//value of Zm0 is taken from the generalized compressibility chart, Figure(3.11) corresponding to Trm and Prm (no unit)
+vm=(Zm0*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+
+//OUTPUT
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar estimated using the Prausnitz-Gunn rule = %0.2e m^3/mol\n",vm);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH9/EX9.12/Chap9_Ex12_R1.sce b/611/CH9/EX9.12/Chap9_Ex12_R1.sce new file mode 100755 index 000000000..fd383d6c1 --- /dev/null +++ b/611/CH9/EX9.12/Chap9_Ex12_R1.sce @@ -0,0 +1,51 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 12,Page 335
+//Title: Molar volume of mixture using van der Waals equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+a_m=2.4405;//van der Waals constant for the mixture as determined in Example 9.8 in Pa(m^3/mol)^2
+b_m=0.1767*10^-3;//van der Waals constant for the mixture as determined in Example 9.8 in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//The problem is solved by using the Cardan's method
+A=(a_m*P*10^5)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b_m*P*10^5)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1-B;//calculation of alpha for van der Waals equation of state using Table (3.2)
+beeta=A;//calculation of beeta for van der Waals equation of state using Table (3.2)
+gaamma=-(A*B);//calculation of gaamma for van der Waals equation of state using Table (3.2)
+p=beeta-((alpha^2)/3);//calculation of p to determine the roots of the cubic equaton using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equaton using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=(((-(q)/2)+(sqrt(D)))^(1/3))+(((-(q)/2)-(sqrt(D)))^(1/3))-(alpha/3);//One real root given by Eq.(3.32)
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ end
+end
+vm=(Z*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+
+//OUTPUT
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane at 600K and 16bar found using the van der Waals equation of state = %e m^3/mol\n",vm);
+
+//===============================================END OF PROGRAM===================================================
+// DISCLAIMER: VALUE OF Z COMPUTED IN PROGRAM IS NOT AS THAT REPORTED IN THE TEXTBOOK. HOWEVER, VALUES OF ALL OTHER PERTINENT VARIABLES A, B, alpha, beeta, p, q etc. AGREE WELL WITH THE TEXTBOOK ANSWER. COMPUTATION WAS ALSO VERIFIED MANUALLY AND GAVE THE ANSWER AS COMPUTED IN PROGRAM. ONE POSSIBLE REASON FOR DEVIATION COULD BE ROUND OFF ERROR.
diff --git a/611/CH9/EX9.13/Chap9_Ex13_R1.sce b/611/CH9/EX9.13/Chap9_Ex13_R1.sce new file mode 100755 index 000000000..184e76718 --- /dev/null +++ b/611/CH9/EX9.13/Chap9_Ex13_R1.sce @@ -0,0 +1,22 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 13,Page 336
+//Title: Molar volume of mixture using the generalized virial coefficient correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Bm=-309*10^-6;//second virial coefficient for the mixture taken from Example(9.7) in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Zm=1+((Bm*P*10^5)/(R*T));//calculation of the compressibility factor for the mixture (no unit)
+vm=(Zm*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+
+//OUTPUT
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane found using the generalized virial coefficient correlation = %0.4e m^3/mol\n",vm);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.14/Chap9_Ex14_R1.sce b/611/CH9/EX9.14/Chap9_Ex14_R1.sce new file mode 100755 index 000000000..165a74522 --- /dev/null +++ b/611/CH9/EX9.14/Chap9_Ex14_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 14,Page 337
+//Title: Enthalpy and entropy departure
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+am=2.4405;//van der Waals constant for the mixture taken from Example 9.8 in Pa(m^3/mol)^2
+bm=0.1767*10^-3;//van der Waals constant for the mixture taken from Example 9.8 in m^3/mol
+vm=2.8933*10^-3;//molar volume of the mixture taken from Example 9.12 in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+dep_h=((P*10^5*vm)-(R*T)-(am/vm))*10^-3;//calculation of the enthalpy departure using Example(8.1) in kJ/mol
+dep_s=R*(log ((P*10^5*(vm-bm))/(R*T)));//calculation of the entropy departure using Example(8.1) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure of an equimolar mixture of n-butane and n-octane = %0.3f kJ/mol\n",dep_h);
+mprintf("\n The entropy departure of an equimolar mixture of n-butane and n-octane = %0.3f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.15/Chap9_Ex15_R1.sce b/611/CH9/EX9.15/Chap9_Ex15_R1.sce new file mode 100755 index 000000000..691494f6d --- /dev/null +++ b/611/CH9/EX9.15/Chap9_Ex15_R1.sce @@ -0,0 +1,36 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 15,Page 338
+//Title: Enthalpy and entropy departure using the generalized compressibility factor correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tcm=497.3;//pseudocritical temperature of mixture taken from Example(9.10) in K
+Pcm=31.47;//pseudocritical pressure of mixture taken from Example(9.10) in bar
+Trm=1.21;//pseudoreduced temperature of the mixture taken from Example(9.10) (no unit)
+Prm=0.51;//pseudoreduced pressure of the mixture taken from Example(9.10) (no unit)
+w_butane=0.199;//acentric factor for n-butane (no unit)
+w_octane=0.398;//acentric factor for n-octane (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+wm=(y1*w_butane)+(y2*w_octane);//calculation of the acentric factor for the mixture (no unit)
+del_h0=0.380;//value of ((h0-h)/RTcm)_0 read from Fig.(8.6) corresponding to Prm and Trm (no unit)
+del_h1=0.188;//value of ((h0-h)/RTcm)_1 read from Fig.(8.8) corresponding to Prm and Trm (no unit)
+del_s0=0.22;//value of ((s0-s)/R)_0 read from Fig.(8.10) corresponding to Prm and Trm (no unit)
+del_s1=0.18;//value of ((s0-s)/R)_1 read from Fig.(8.12) corresponding to Prm and Trm (no unit)
+dep_h=((R*Tcm)*(del_h0+(wm*del_h1)))*10^-3;//calculation of the enthalpy departure using Eq.(8.62) in kJ/mol
+dep_s=(R)*(del_s0+(wm*del_s1));//calculation of the entropy departure using Eq.(8.65) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation = %0.3f kJ/mol\n",dep_h);
+mprintf("\n The entropy departure of an equimolar mixture of n-butane and n-octane using the generalized compressibility factor correlation = %f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.16/Chap9_Ex16_R1.sce b/611/CH9/EX9.16/Chap9_Ex16_R1.sce new file mode 100755 index 000000000..7187ff4ec --- /dev/null +++ b/611/CH9/EX9.16/Chap9_Ex16_R1.sce @@ -0,0 +1,47 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 16,Page 339
+//Title: Enthalpy and entropy departure using the virial coefficient correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+w=[0.199;0.398];//acentric factor of n-butane and n-octane (no unit)
+Tr1=1.411;//reduced temperature of n-butane (no unit) taken from Example (9.7)
+Tr2=1.054;//reduced temperature of n-octane (no unit) taken from Example (9.7)
+Tr_12=1.24;//reduced temperature for computing the mixture interaction virial coefficient (no unit) taken from Example(9.7)
+Pc_12=2.978;//Pc_ij in MPa taken from Example(9.7)
+Tc_12=483.9;//Tc_ij in K taken from Example(9.7)
+w_12=0.2985;// w_ij (no unit) taken from Example(9.7)
+Bm=-309*10^-6;//second virial coefficient in m^3/mol taken from Example (9.7)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+dB0_dTr1=0.675/(Tr1^2.6);//calculation of dBij0/dTrij using Eq.(8.73) (no unit)
+dB0_dTr2=0.675/(Tr2^2.6);//calculation of dBij0/dTrij using Eq.(8.73) (no unit)
+dB1_dTr1=0.722/(Tr1^5.2);//calculation of dBij1/dTrij using Eq.(8.74) (no unit)
+dB1_dTr2=0.722/(Tr2^5.2);//calculation of dBij1/dTrij using Eq.(8.74) (no unit)
+dB0_dTr12=0.675/(Tr_12^2.6);//calculation of dBij0/dTrij using Eq.(9.114) (no unit)
+dB1_dTr12=0.722/(Tr_12^5.2);//calculation of dBij1/dTrij using Eq.(9.115) (no unit)
+dB1_dT=(R/(Pc(1,:)*10^5))*((dB0_dTr1)+(w(1,:)*(dB1_dTr1)));//calculation of dBij/dT using Eq.(9.112) (m^3/molK)
+dB2_dT=(R/(Pc(2,:)*10^5))*((dB0_dTr2)+(w(2,:)*(dB1_dTr2)));//calculation of dBij/dT using Eq.(9.112) (m^3/molK)
+dB12_dT=(R/(Pc_12*10^6))*((dB0_dTr12)+(w_12*(dB1_dTr12)));//calculation of dBij/dT using Eq.(9.112) (m^3/molK)
+dBm_dT=((y1^2)*(dB1_dT))+((2*y1*y2)*(dB12_dT))+((y2^2)*(dB2_dT));//calculation of dBm/dT using Eq.(9.110) (m^3/molK)
+dep_h=((Bm-(T*dBm_dT))*P*10^5)*10^-3;//calculation of enthalpy departure using Eq.(8.69) in kJ/mol
+dep_s=-P*10^5*(dBm_dT);//calculation of entropy departure using Eq.(8.70) in J/molK
+
+//OUTPUT
+mprintf("\n The enthalpy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation = %f kJ/mol\n",dep_h);
+mprintf("\n The entropy departure of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation = %0.3f J/mol K\n",dep_s);
+
+//===============================================END OF PROGRAM===================================================
+
+
diff --git a/611/CH9/EX9.17/Chap9_Ex17_R1.sce b/611/CH9/EX9.17/Chap9_Ex17_R1.sce new file mode 100755 index 000000000..553771c20 --- /dev/null +++ b/611/CH9/EX9.17/Chap9_Ex17_R1.sce @@ -0,0 +1,27 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 17,Page 340
+//Title: Fugacity and fugacity coefficient using van der Waals equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+a_m=2.4405;//van der Waals constant (a_m) in Pa(m^3/mol)^2 taken from Example(9.8)
+b_m=0.1767*10^-3;//van der Waals constant (b_m) in m^3/mol taken from Example(9.8)
+Z=0.928;//compressibility factor taken from Example(9.12)
+vm=2.8933*10^-3;//molar volume of the equimolar mixture in m^3/mol taken from Example(9.12)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+phi=exp (Z-1-log ((P*10^5*(vm-b_m))/(R*T))-(a_m/(R*T*vm)));//calculation of the fugacity coefficient (f/P) using the expression derived in Example 9.3 (no unit)
+f=phi*P;//calculation of fugacity using Eq.(9.37) in bar
+
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state = %0.4f \n",phi);
+mprintf("\n The fugacity of an equimolar mixture of n-butane and n-octane using the van der Waals equation of state = %0.2f bar\n",f);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.18/Chap9_Ex18_R1.sce b/611/CH9/EX9.18/Chap9_Ex18_R1.sce new file mode 100755 index 000000000..f91810486 --- /dev/null +++ b/611/CH9/EX9.18/Chap9_Ex18_R1.sce @@ -0,0 +1,29 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 18,Page 341
+//Title: Fugacity and fugacity coefficient using the pseudocritical constants method
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tcm=497.3;//pseudocritical temperature of mixture in K taken from Example(9.10)
+Pcm=31.47;//pseudocritical pressure of mixture in bar taken from Example(9.10)
+Trm=1.21;//pseudoreduced temperature of the mixture (no unit) taken from Example(9.10)
+Prm=0.51;//pseudoreduced pressure of the mixture (no unit) taken from Example(9.10)
+w=[0.199;0.398];//acentric factor of n-butane and n-octane (no unit)
+
+//CALCULATION
+wm=(w(1,:)+w(2,:))/2;//calculation of the acentric factor for the mixture (no unit)
+log_phi0=-0.042;//value of log_phi0 taken from Figure(9.2) (no unit)
+log_phi1=0.01;//value of log_phi1 taken from Figure(9.4) (no unit)
+phi=10^(log_phi0+(wm*log_phi1));//calculation of the fugacity coefficient using Eq.(9.54) (no unit)
+f=P*phi;//calculation of the fugacity using Eq.(9.37) in bar
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method = %0.3f \n",phi);
+mprintf("\n The fugacity of an equimolar mixture of n-butane and n-octane using the pseudocritical constants method = %f bar\n",f);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.19/Chap9_Ex19_R1.sce b/611/CH9/EX9.19/Chap9_Ex19_R1.sce new file mode 100755 index 000000000..12bb442f6 --- /dev/null +++ b/611/CH9/EX9.19/Chap9_Ex19_R1.sce @@ -0,0 +1,26 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 19,Page 341
+//Title: Fugacity and fugacity coefficient using the virial coefficient correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Bm=-309*10^-6;//second virial coefficient in m^3/mol taken from Example (9.7)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//Using Eq.(3.91) and Eq.(9.58) ln(phi)=BP/RT, which is used to compute phi
+phi=(exp((Bm*P*10^5)/(R*T)));//calculation of the fugacity coefficient using the above expression (no unit)
+f=phi*P;//calculation of the fugacity using Eq.(9.37) in bar
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation = %f \n",phi);
+mprintf("\n The fugacity of an equimolar mixture of n-butane and n-octane using the virial coefficient correlation = %f bar\n",f);
+
+//===============================================END OF PROGRAM===================================================
+
+//DISCLAIMER: THE VALUE OF FUGACITY COEFFICIENT AS CALCULATED IN THE TEXTBOOK IS WRONG.THIS HAS BEEN CORRECTED IN THIS PROGRAM.
diff --git a/611/CH9/EX9.2/Chap9_Ex2_R1.sce b/611/CH9/EX9.2/Chap9_Ex2_R1.sce new file mode 100755 index 000000000..1f4dcf7e6 --- /dev/null +++ b/611/CH9/EX9.2/Chap9_Ex2_R1.sce @@ -0,0 +1,32 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 2,Page 313
+//Title: Volumes to be mixed
+//================================================================================================================
+clear
+clc
+
+//INPUT
+V=3;//volume of mixture to be prepared in m^3
+per_ethanol=60;//mole percent of ethanol in a ethanol-water system
+per_water=40;//mole percent of water in a ethanol-water system
+v1=57.5*10^-6;//partial molar volume of ethanol in the ethanol-water system in m^3/mol
+v2=16*10^-6;//partial molar volume of water in the ethanol-water system in m^3/mol
+v1_pure=57.9*10^-6;//molar volume of pure ethanol in m^3/mol
+v2_pure=18*10^-6;//molar volume of pure water in m^3/mol
+
+//CALCULATION
+X1=per_ethanol/100;//calculation of the mole fraction of ethanol (no unit)
+X2=per_water/100;//calculation of the mole fraction of water (no unit)
+v=(X1*v1)+(X2*v2);//calculation of the molar volume of the solution using Eq.(9.10) in m^3/mol
+N=V/v;//calculation of the mole number of solution required in mol
+N1=N*X1;//calculation of the mole number of ethanol in solution in mol
+N2=N*X2;//calculation of the mole number of water in solution in mol
+V1=N1*v1_pure;//calculation of the volume of pure ethanol required in m^3
+V2=N2*v2_pure;//calculation of the volume of pure water required in m^3
+
+//OUTPUT
+mprintf("\n The volume of pure ethanol required = %0.3f m^3\n",V1);
+mprintf("\n The volume of pure water required = %0.3f m^3\n",V2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.20/Chap9_Ex20_R1.sce b/611/CH9/EX9.20/Chap9_Ex20_R1.sce new file mode 100755 index 000000000..88726b6db --- /dev/null +++ b/611/CH9/EX9.20/Chap9_Ex20_R1.sce @@ -0,0 +1,65 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 20,Page 344
+//Title: Fugacity coefficients of the components in a mixture using Redlich-Kwong Equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+a1=(0.42748*R^2*Tc(1,:)^2.5)/(Pc(1,:)*10^5*sqrt(T));//calculation of Redlich-Kwong constant for n-butane in (m^6 Pa mol^-2)
+a2=(0.42748*R^2*Tc(2,:)^2.5)/(Pc(2,:)*10^5*sqrt(T));//calculation of Redlich-Kwong constant for n-octane in (m^6 Pa mol^-2)
+b1=(0.08664*R*Tc(1,:))/(Pc(1,:)*10^5);//calculation of Redlich-Kwong constant for n-butane in m^3/mol
+b2=(0.08664*R*Tc(2,:))/(Pc(2,:)*10^5);//calculation of Redlich-Kwong constant for n-octane in m^3/mol
+//Set Kij=0 and evaluate a using Eq.(9.64)
+a=((y1^2)*a1)+(2*y1*y2*sqrt(a1*a2))+((y2^2)*a2);//calculation of Redlich-Kwong constant for the mixture using Eq.(9.64) in (m^6 Pa mol^-2)
+b=(y1*b1)+(y2*b2);//calculation of Redlich-Kwong constant for the mixture using Eq.(9.65) in m^3/mol
+
+//The Cardans method can be used to determine Z
+A=(a*P*10^5)/(R*T)^2;//calculation of A to determine alpha,beeta and gaamma by using Eq.(3.25)
+B=(b*P*10^5)/(R*T);//calculation of B to determine alpha,beeta and gaamma by using Eq.(3.26)
+alpha=-1;//calculation of alpha for Redlich-Kwong equation of state using Table (3.2)
+beeta=A-B-B^2;//calculation of beeta for Redlich-Kwong equation of state using Table (3.2)
+gaamma=-(A*B);//calculation of gaamma for Redlich-Kwong equation of state using Table (3.2)
+p=beeta-(alpha^2)/3;//calculation of p to determine the roots of the cubic equation using Eq.(3.29)
+q=((2*alpha^3)/27)-((alpha*beeta)/3)+gaamma;//calculation of q to determine the roots of the cubic equation using Eq.(3.30)
+D=(((q)^2)/4)+(((p)^3)/27);//calculation of D to determine the nature of roots using Eq.(3.31)
+
+if D>0 then
+ Z=((-q/2)+(sqrt(D)))^(1/3)+((-q/2)-(sqrt(D)))^(1/3)-(alpha/3);//One real root given by Eq.(3.32)
+else if D==0 then
+ Z1=((-2*(q/2))^(1/3))-(alpha/3);//Three real roots and two equal given by Eq.(3.33)
+ Z2=((q/2)^(1/3))-(alpha/3);
+ Z3=((q/2)^(1/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ else
+ r=sqrt((-(p^3)/27));//calculation of r using Eq.(3.38)
+ theta=acos((-(q)/2)*(1/r));//calculation of theta in radians using Eq.(3.37)
+ Z1=(2*(r^(1/3))*cos(theta/3))-(alpha/3);
+ Z2=(2*(r^(1/3))*cos(((2*%pi)+theta)/3))-(alpha/3);//Three unequal real roots given by Eqs.(3.34,3.35 and 3.36)
+ Z3=(2*(r^(1/3))*cos(((4*%pi)+theta)/3))-(alpha/3);
+ Za=[Z1 Z2 Z3];
+ Z=max(Za);
+ end
+end
+//calculation of the fugacity coefficient of n-butane in the mixture using Eq.(9.126) (no unit)
+phi1=exp (((b1/b)*(Z-1))-log(Z-B)+((a/(b*R*T))*((b1/b)-(2*sqrt(a1/a)))*log((Z+B)/Z)));
+//calculation of the fugacity coefficient of n-octane in the mixture using Eq.(9.126) (no unit)
+phi2=exp (((b2/b)*(Z-1))-log(Z-B)+((a/(b*R*T))*((b2/b)-(2*sqrt(a2/a)))*log((Z+B)/Z)));
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of n-butane in the equimolar mixture using the Redlich-Kwong Equation of state = %0.4f \n",phi1);
+mprintf("\n The fugacity coefficient of n-octane in the equimolar mixture using the Redlich-Kwong Equation of state = %0.4f \n",phi2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.21/Chap9_Ex21_R1.sce b/611/CH9/EX9.21/Chap9_Ex21_R1.sce new file mode 100755 index 000000000..eb7930fa2 --- /dev/null +++ b/611/CH9/EX9.21/Chap9_Ex21_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 21,Page 346
+//Title: Fugacity coefficients of the components in a mixture using the Virial Equation of state
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+B_11=-131*10^-6;//pure component (n-butane) second virial coefficient in m^3/mol taken from Example(9.7)
+B_22=-577*10^-6;//pure component (n-octane) second virial coefficient in m^3/mol taken from Example(9.7)
+B_12=-264*10^-6;//mixture interaction virial coefficient in m^3/mol taken from Example(9.7)
+Bm=-309*10^-6;//second virial coefficient in m^3/mol taken from Example(9.7)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+Zm=(1/2)*(1+sqrt(1+((4*Bm*P*10^5)/(R*T))));//calculation of compressibility for the mixture(Zm) using Eq.(9.136) (no unit)
+phi1=exp((((2*P*10^5)/(Zm*R*T))*((y1*B_11)+(y2*B_12)))-log(Zm));//calculation of the fugacity coefficient of n-butane in the mixture using Eq.(9.135) (no unit)
+phi2=exp((((2*P*10^5)/(Zm*R*T))*((y1*B_12)+(y2*B_22)))-log(Zm));//calculation of the fugacity coefficient of n-octane in the mixture using Eq.(9.135) (no unit)
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of n-butane in the equimolar mixture using the Virial Equation of state = %0.3f \n",phi1);
+mprintf("\n The fugacity coefficient of n-octane in the equimolar mixture using the Virial Equation of state = %f \n",phi2);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.22/Chap9_Ex22_R1.sce b/611/CH9/EX9.22/Chap9_Ex22_R1.sce new file mode 100755 index 000000000..5868899fe --- /dev/null +++ b/611/CH9/EX9.22/Chap9_Ex22_R1.sce @@ -0,0 +1,23 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 22,Page 349
+//Title: Fugacity of liquid n-octane
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+Psat=0.215;//saturation pressure of n-octane vapour at T in MPa
+P=1;//pressure at which the fugacity of liquid n-octane is to be determined in MPa
+f_sat=0.2368;//fugacity of n-octane vapour at T and Psat taken from Example(9.5) in MPa
+vl=0.2003*10^-3;//molar volume of n-octane liquid at T and Psat taken from Example(3.16) in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+f_l=(0.2368*exp((vl*(P-Psat)*10^6)/(R*T)));//calculation of fugacity of n-octane liquid using Eq.(9.150) in MPa
+
+//OUTPUT
+mprintf("\n The fugacity of liquid n-octane at 427.85K and 1MPa = %0.4f MPa\n",f_l);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.3/Chap9_Ex3_R1.sce b/611/CH9/EX9.3/Chap9_Ex3_R1.sce new file mode 100755 index 000000000..1d5ec83fe --- /dev/null +++ b/611/CH9/EX9.3/Chap9_Ex3_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 3,Page 318
+//Title: Fugacity and fugacity coefficient
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+a=3.789;//van der Waals constant in Pa(m^3/mol)^2
+b=2.37*10^-4;//van der Waals constant in m^3/mol
+v=15.675*10^-3;//molar volume of n-octane saturated vapour taken from Example 3.8 in m^3/mol
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+Z=(P*10^6*v)/(R*T);//calculation of the compressibility factor (no unit)
+//calculation of the fugacity coefficient (f/P) using the expression derived in Example 9.3 (no unit)
+phi=exp (Z-1-log (((P*10^6)*(v-b))/(R*T))-a/(R*T*v));
+f=(P*10^6*phi)*10^-6;//calculation of fugacity using Eq.(9.37) in MPa
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of n-octane vapour = %0.2f \n",phi);
+mprintf("\n The fugacity of n-octane vapour = %0.4f MPa\n",f);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH9/EX9.5/Chap9_Ex5_R1.sce b/611/CH9/EX9.5/Chap9_Ex5_R1.sce new file mode 100755 index 000000000..b6b5eb841 --- /dev/null +++ b/611/CH9/EX9.5/Chap9_Ex5_R1.sce @@ -0,0 +1,30 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 5,Page 322
+//Title: Fugacity and fugacity coefficient from the Lee-Kesler data
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=427.85;//temperature of n-octane vapour in K
+P=0.215;//pressure of n-octane vapour in MPa
+Tc=569.4;//critical temperature of n-octane in K
+Pc=24.97;//critical pressure of n-octane in bar
+w=0.398;//acentric factor (no unit)
+
+//CALCULATION
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=(P*10^6)/(Pc*10^5);//calculation of reduced pressure (no unit)
+log_phi0=-0.032;//value of log_phi0 taken from Figure(9.2) (no unit)
+log_phi1=-0.025;//value of log_phi1 taken from Figure(9.4) (no unit)
+phi=10^(log_phi0+(w*log_phi1));//calculation of the fugacity coefficient using Eq.(9.54) (no unit)
+f=P*phi;//calculation of the fugacity using Eq.(9.37) in MPa
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of n-octane vapour = %f \n",phi);
+mprintf("\n The fugacity of n-octane vapour = %f MPa\n",f);
+
+//===============================================END OF PROGRAM===================================================
+
+// DISCLAIMER: THE VALUE OF FUGACITY COEFFICIENT AS CALCULATED IN THE TEXTBOOK UPON TAKING THE ANTILOG IS WRONG. THE ANTILOG OF -0.042 IS A VALUE LESS THAN 1. THIS HAS BEEN CORRECTED ACCORDINGLY IN THIS PROGRAM AND THE VALUE OF FUGACITY COEFFICIENT AND FUGACITY HAVE BEEN COMPUTED.
diff --git a/611/CH9/EX9.6/Chap9_Ex6_R1.sce b/611/CH9/EX9.6/Chap9_Ex6_R1.sce new file mode 100755 index 000000000..27365294b --- /dev/null +++ b/611/CH9/EX9.6/Chap9_Ex6_R1.sce @@ -0,0 +1,28 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 6,Page 327
+//Title: Fugacity and fugacity coefficient using the virial coefficient correlation
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=339.7;//temperature of ethylene in K
+P=1;//pressure of ethylene in bar
+Tc=283.1;//critical temperature of ethylene in K
+Pc=51.17;//critical pressure of ethylene in bar
+w=0.089;//acentric factor (no unit)
+
+//CALCULATION
+Tr=T/Tc;//calculation of reduced temperature (no unit)
+Pr=P/Pc;//calculation of reduced pressure (no unit)
+B0=0.083-(0.422/(Tr^1.6));//calculation of B0 using Eq.(3.95) so as to compute fugacity coefficient using Eq.(9.58)
+B1=0.139-(0.172/(Tr^4.2));//calculation of B1 using Eq.(3.96) so as to compute fugacity coefficient using Eq.(9.58)
+phi=exp ((B0+(w*B1))*(Pr/Tr));//calculation of the fugacity coefficient using Eq.(9.58) (no unit)
+f=P*phi;//calculation of the fugacity using Eq.(9.37) in bar
+
+//OUTPUT
+mprintf("\n The fugacity coefficient of ethylene = %0.4f \n",phi);
+mprintf("\n The fugacity of ethylene = %0.4f bar\n",f);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.7/Chap9_Ex7_R1.sce b/611/CH9/EX9.7/Chap9_Ex7_R1.sce new file mode 100755 index 000000000..4ef9de16e --- /dev/null +++ b/611/CH9/EX9.7/Chap9_Ex7_R1.sce @@ -0,0 +1,46 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 7,Page 330
+//Title: Second virial coefficient
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+vc=[255.0*10^-6;486.0*10^-6];//critical molar volume of n-butane and n-octane in m^3/mol
+Zc=[0.274;0.256];//compressibility factor of n-butane and n-octane corresponding to Tc,Pc (no unit)
+w=[0.199;0.398];//acentric factor of n-butane and n-octane (no unit)
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+K_12=1-((8*((vc(1,:)*vc(2,:))^(1/2)))/((((vc(1,:))^(1/3))+((vc(2,:))^(1/3)))^3));//calculation of the binary interaction parameter using Eq.(9.94)
+Tc_12=(((Tc(1,:))*(Tc(2,:)))^(1/2))*(1-K_12);//calculation of Tc_ij using Eq.(9.89) in K
+w_12=(w(1,:)+w(2,:))/2;//calculation of w_ij using Eq.(9.92) (no unit)
+Zc_12=(Zc(1,:)+Zc(2,:))/2;//calculation of Zc_ij using Eq.(9.91) (no unit)
+vc_12=((((vc(1,:))^(1/3))+((vc(2,:))^(1/3)))/2)^3;//calculation of vc_ij using Eq.(9.90) in m^3/mol
+Pc_12=((Zc_12*R*Tc_12)/vc_12)*10^-6;//calculation of Pc_ij using Eq.(9.93) in MPa
+Tr_12=T/Tc_12;//calculation of reduced temperature for computing the mixture interaction virial coefficient (no unit)
+B_12_0=0.083-(0.422/(Tr_12^(1.6)));//calculation of B_ij0 using Eq.(9.87)
+B_12_1=0.139-(0.172/(Tr_12^(4.2)));//calculation of B_ij1 using Eq.(9.88)
+B_12=((R*Tc_12)/(Pc_12*10^6))*(B_12_0+(w_12*B_12_1));//calculation of the mixture interaction virial coefficient using Eq.(9.86) in m^3/mol
+Tr1=T/Tc(1,:);//calculation of reduced temperature of n-butane (no unit)
+B_11_0=0.083-(0.422/(Tr1^(1.6)));//calculation of B_i0 for the pure component (n-butane) using Eq.(9.87)
+B_11_1=0.139-(0.172/(Tr1^(4.2)));//calculation of B_i1 for the pure component (n-butane) using Eq.(9.88)
+B_11=((R*Tc(1,:))/(Pc(1,:)*10^5))*(B_11_0+(w(1,:)*B_11_1));//calculation of the pure component (n-butane) second virial coefficient using Eq.(9.86) in m^3/mol
+Tr2=T/Tc(2,:);//calculation of reduced temperature of n-octane (no unit)
+B_22_0=0.083-(0.422/(Tr2^(1.6)));//calculation of B_i0 for the pure component (n-octane) using Eq.(9.87)
+B_22_1=0.139-(0.172/(Tr2^(4.2)));//calculation of B_i1 for the pure component (n-octane) using Eq.(9.88)
+B_22=((R*Tc(2,:))/(Pc(2,:)*10^5))*(B_22_0+(w(2,:)*B_22_1));//calculation of the pure component (n-octane) second virial coefficient using Eq.(9.86) in m^3/mol
+Bm=((y1^2)*B_11)+((2*y1*y2)*B_12)+((y2^2)*B_22);//calculation of the second virial coefficient using Eq.(9.85) in m^3/mol
+
+//OUTPUT
+mprintf("\n The second virial coefficient for an equimolar mixture of n-butane and n-octane at 600K = %f m^3/mol\n",Bm);
+
+//===============================================END OF PROGRAM===================================================
+
diff --git a/611/CH9/EX9.8/Chap9_Ex8_R1.sce b/611/CH9/EX9.8/Chap9_Ex8_R1.sce new file mode 100755 index 000000000..0f971bfee --- /dev/null +++ b/611/CH9/EX9.8/Chap9_Ex8_R1.sce @@ -0,0 +1,25 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 8,Page 331
+//Title: van der Waals constants
+//================================================================================================================
+clear
+clc
+
+//INPUT
+a=[1.3874;3.7890];//van der Waals constant of n-butane and n-octane in Pa(m^3/mol)^2
+b=[0.1163*10^-3;0.237*10^-3];//van der Waals constant of n-butane and n-octane in m^3/mol
+
+//CALCULATION
+//For convenience, n-butane is taken as 1 and n-octane as 2
+//Set K_ij=0 in Eq.(9.64) to compute the van der Waals constant (a_m) for an equimolar mixture of n-butane and n-octane
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+a_m=((y1^2)*a(1,:))+((2*y1*y2)*sqrt (a(1,:)*a(2,:)))+((y2^2)*a(2,:));//calculation of the van der Waals constant (a_m) using Eq.(9.64) in Pa(m^3/mol)^2
+b_m=(y1*b(1,:))+(y2*b(2,:));//calculation of the van der Waals constant (b_m) using Eq.(9.65) in m^3/mol
+
+//OUTPUT
+mprintf("\n The van der Waals constant for an equimolar mixture of n-butane and n-octane, a_m = %0.4f Pa(m^3/mol)^2\n",a_m);
+mprintf("\n The van der Waals constant for an equimolar mixture of n-butane and n-octane, b_m = %f m^3/mol\n",b_m);
+
+//===============================================END OF PROGRAM===================================================
diff --git a/611/CH9/EX9.9/Chap9_Ex9_R1.sce b/611/CH9/EX9.9/Chap9_Ex9_R1.sce new file mode 100755 index 000000000..ddd86b713 --- /dev/null +++ b/611/CH9/EX9.9/Chap9_Ex9_R1.sce @@ -0,0 +1,61 @@ +// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India.
+
+//Chapter-9,Example 9,Page 333
+//Title: Molar volume of an equimolar mixture
+//================================================================================================================
+clear
+clc
+
+//INPUT
+T=600;//temperature of the equimolar n-butane and n-octane mixture in K
+P=16;//pressure of the equimolar n-butane and n-octane mixture in bar
+Tc=[425.2;569.4];//critical temperature of n-butane and n-octane in K
+Pc=[37.97;24.97];//critical pressure of n-butane and n-octane in bar
+R=8.314;//universal gas constant in J/molK
+
+//CALCULATION
+//Calculation by Amagat's law of additive volumes
+//For convenience, n-butane is taken as 1 and n-octane as 2
+y1=0.5;//mole fraction of n-butane in the equimolar mixture
+y2=0.5;//mole fraction of n-octane in the equimolar mixture
+Tr1=T/Tc(1,:);//calculation of reduced temperature of n-butane (no unit)
+Pr1=P/Pc(1,:);//calculation of reduced pressure of n-butane (no unit)
+Z1_0=0.95;//value of Z1_0 taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr1 and Pr1 (no unit)
+Tr2=T/Tc(2,:);//calculation of reduced temperature of n-octane (no unit)
+Pr2=P/Pc(2,:);//calculation of reduced pressure of n-octane (no unit)
+Z2_0=0.785;//value of Z2_0 taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr2 and Pr2 (no unit)
+//Part a: Using the Amagat's law of additive volumes
+Zm=(y1*Z1_0)+(y2*Z2_0);//calculation of the compressibility of the mixture using Eq.(9.96) (no unit)
+vm=(Zm*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+
+//Calculation by Dalton's law of additive pressures
+//The compressibility factors for the pure components are to be evaluated at the mixture temperature and pure component pressure in Dalton's law of additive pressures. As the pure component pressures are not known, they are assumed to be equal to the partial pressure
+P1=y1*P;//pure component pressure in bar
+P2=y2*P;//pure component pressure in bar
+Pr1=P1/Pc(1,:);//calculation of reduced pressure of n-butane (no unit)
+Pr2=P2/Pc(2,:);//calculation of reduced pressure of n-octane (no unit)
+Z1_0=0.97;//value of Z1_0 taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr1 and Pr1 (no unit)
+Z2_0=0.91;//value of Z2_0 taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr2 and Pr2 (no unit)
+Zm=(y1*Z1_0)+(y2*Z2_0);//calculation of the compressibility of the mixture using Eq.(9.96) (no unit)
+vm_dalton=(Zm*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+//Recalculation of P1 and P2 to verify the initial assumption
+P1=((Z1_0*y1*R*T)/(vm_dalton))*10^-2;//P1 recalculated in bar
+P2=((Z2_0*y2*R*T)/(vm_dalton))*10^-2;//P2 recalculated in bar
+Pr1=P1/Pc(1,:);//recalculation of reduced pressure of n-butane (no unit)
+Pr2=P2/Pc(2,:);//recalculation of reduced pressure of n-octane (no unit)
+Z1_0_new=0.97;//value of Z1_0_new taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr1 and Pr1 (no unit)
+Z2_0_new=0.91;//value of Z2_0_new taken from the generalized compressibility chart, Figure(3.11) corresponding to Tr2 and Pr2 (no unit)
+if Z1_0_new==Z1_0 & Z2_0_new==Z2_0 then
+ vm_new=vm_dalton;//molar volume of the equimolar mixture in m^3/mol
+else
+ Zm=(y1*Z1_0_new)+(y2*Z2_0_new);//calculation of the compressibility of the mixture using Eq.(9.96) (no unit)
+ vm_new=(Zm*R*T)/(P*10^5);//calculation of the molar volume of the equimolar mixture in m^3/mol
+end
+
+//OUTPUT
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane found using the Amagats law of additive volumes = %0.4e m^3/mol\n",vm);
+mprintf("\n The molar volume of an equimolar mixture of n-butane and n-octane found using the Daltons law of additive pressures = %0.2e m^3/mol\n",vm_new);
+
+//===============================================END OF PROGRAM===================================================
+
+
|
