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diff --git a/167/CH13/EX13.1/ex1.sce b/167/CH13/EX13.1/ex1.sce
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+//example 1

+//Mass and Mole Fractions of a Gas Mixture

+clear

+clc

+m=20 //total mass of the mixture in kg

+mfO2=3/m //mass fraction of oxygen

+mfN2=5/m //mass fraction of nitrogen

+mfCH4=12/m //mass fraction of methane

+NO2=3/32 //no.of kilo moles of oxygen

+NN2=5/28 // no.of kilo moles of nitrogen

+NCH4=12/16 //no.of kilo moles of methane

+N=NO2+NN2+NCH4//total no. of moles

+yO2=NO2/N //mole fraction of O2

+yN2=NN2/N //mole fraction of N2

+yCH4=NCH4/N // mole fraction of CH4

+Mm=m/N //average molar mass of gas in kg/kmol

+printf("\n Mass fraction of oxygen is = %.2f. \n",mfO2);

+printf("\n Mass fraction of Nitrogen is = %.2f. \n",mfN2);

+printf("\n Mass fraction of Methane is = %.2f. \n",mfCH4);

+printf("\n Mole fraction of Nitrogen is = %.3f. \n",yN2);

+printf("\n Mole fraction of Oxygen is = %.3f. \n",yO2);

+printf("\n Mole fraction of Methane is = %.3f. \n",yCH4);

+printf("\n Average Molar mass of gas is = %.1f kg/mol. \n",Mm);
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diff --git a/167/CH13/EX13.2/ex2.sce b/167/CH13/EX13.2/ex2.sce
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+//example 2

+//P-v-T Behavior of Nonideal Gas Mixtures

+clear

+clc

+NN2=2 //No.of kmol of N2

+NCO2=6 //No. of kmol of CO2

+Nm=8 // total no. of kmol of mixture

+Ru=8.314 //Universal gas constant in kPa-m^3/kmol-K

+Tm=300//Temp. of mixture in K

+Pm=15000 //Pressure of mixture in kPa

+Vm=Nm*Ru*Tm/Pm //volume of tank on the basis of ideal gas equation in m^3

+printf("\n Hence, the volume of the mixture on the basis of ideal gas equation of state is = %.3f m^3. \n",Vm);

+disp('Now,estimating volume of tank on the basis of Kays rule')

+yN2=NN2/Nm//mole fraction of nitrogen

+yCO2=NCO2/Nm //mole fraction of CO2

+TcrN2=126.2 // critical temop. of N2 in Kelvins

+TcrCO2=304.2 //critical temp. of CO2 in kelvins

+Tcrm=yN2*TcrN2+yCO2*TcrCO2 //pseudo critical temp. of mixture in Kelvins

+PcrN2=3.39 //critical pressure of N2 in MPa

+PcrCO2=7.39 //critical pressure in MPa

+Pcrm=yN2*PcrN2+yCO2*PcrCO2 //pseodo critical pressure of mixture in MPa

+Tm=300 //actual critical temp. of mixture in kelvins

+Pm=15 //actual critical pressure of mixture in MPa

+Tr=Tm/Tcrm //Reduced Temp. of mixture

+Pr=Pm/Pcrm //Reduced pressure of mixture

+Zm1=Tr/Pr //compressibility of the mixture

+Vm1=Zm1*Vm//volume of tank on the basis of Kays rule in m^3

+printf("\n Hence, the volume of the mixture on the basis of Kays rule is = %.3f m^3. \n",Vm1);

+disp('Now, estimating volume of tank on the basis of compressibility factors and Amagats law')

+TrN2=Tm/TcrN2 //Reduced Temp. of N2

+PrN2=Pm/PcrN2 //Reduced Pressure of N2

+ZN2=1.02 //compressibility factor of N2

+TrCO2=Tm/TcrCO2 //Reduced Temperature of CO2

+PrCO2=Pm/PcrCO2 //Reduced pressure of CO2

+ZCO2=0.30 //compressibility factor of CO2

+Zm2=ZN2*yN2+ZCO2*yCO2 //compressibility factor of the mixture

+Vm2=Zm2*Vm //volume of the mixture in m^3

+printf("\n Hence, the volume of the mixture on the basis of compressibility factors and Amagats law is = %.3f m^3. \n",Vm2);

+disp('Now, estimating volume of tank on the basis of compressibility factors and daltons law')

+VrN2=(Vm/NN2)/(Ru*TcrN2/(PcrN2*1000))

+VrCO2=(Vm/NCO2)/(Ru*TcrCO2/(PcrCO2*1000))

+ZN2=0.99 //compressibility factor of N2

+ZCO2=0.56 //compressibility factor of CO2

+Zm3=yN2*ZN2+yCO2*ZCO2 //compressibility factor of the mixture

+Vm3=Zm3*Vm //volume of the mixture in m^3

+disp('This is 33 percent lower than the assumed value. Therefore, we should repeat the calculations, using the new value of Vm. When the calculations are repeated we obtain 0.738 m^3 after the second iteration, 0.678 m^3 after the third iteration, and 0.648 m^3 after the fourth iteration. This value does not change with more iterations. Therefore')

+Vm=0.648 //volume of the mixture in m^3

+printf("\n Hence, the volume of the mixture on the basis of compressibility factors and Daltons law is = %.3f m^3. \n",Vm);
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diff --git a/167/CH13/EX13.3/ex3.sce b/167/CH13/EX13.3/ex3.sce
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+//example 3

+//Mixing Two Ideal Gases in a Tank

+clear

+clc

+disp('We assume both gases to be ideal gases, and their mixtureto be an ideal-gas mixture. This assumption is reasonable since both the oxygen and nitrogen are well above their critical temperatures and well below their critical pressures.')

+CvN2=0.743 //Constant-Volume Specific heat of N2 in kJ/kg-K

+CvO2=0.658 //Constant-Volume Specific heat of O2 in kJ/kg-K 

+disp('This is a closed system since no mass crosses the boundary during the process. We note that the volume of a rigid tank is constant and thus,there is no boundary work done. ')

+T1N2=20 // Temperature of N2 in celsius

+T1O2=40 // Temperature of O2 in celsius

+mN2=4 //mass of N2 in kg

+mO2=7 //mass of O2 in kg

+Tm=(mN2*CvN2*T1N2+mO2*CvO2*T1O2)/(mN2*CvN2+mO2*CvO2) //Temp. of mixture in Celsius

+printf("\n Hence, the temp. of the mixture is = %.1f C. \n",Tm)

+NO2=mO2/32 //No. of kmol of O2

+NN2=mN2/28 //No. of kmol of N2

+Nm=NO2+NN2 //Total No. of kmol of mixture

+Ru=8.314 //Universal Gas Constant in kPa-m^3/kmol-K

+P1O2=100 //Initial Pressure of O2 in kPa

+P1N2=150 //Initial Pressure of N2 in kPa

+VO2=NO2*Ru*(T1O2+273)/P1O2//Initial volume of O2 in m^3

+VN2=NN2*Ru*(273+T1N2)/P1N2 //Initial volume of N2 in m^3

+Vm=VO2+VN2 //total volume of mixture in m^3

+Pm=Nm*Ru*(Tm+273)/Vm //Mixture Pressure after equilbrium in kPa

+printf("\n Hence, the mixture pressure after equilbrium is = %.1f kPa. \n",Pm)
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diff --git a/167/CH13/EX13.4/ex4.sce b/167/CH13/EX13.4/ex4.sce
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+//example 4

+//Exergy Destruction during Mixing of Ideal Gases

+clear

+clc

+disp('We take the entire contents of the tank as the system. This is a closed system since no mass crosses the boundary during the process. We note that the volume of a rigid tank is constant, and there is no energy transfer as heat or work.')

+NO2=3 //No.of kmol of O2

+NCO2=5 //No. of kmol of CO2

+Nm=NO2+NCO2 //total moles of the mixture

+yO2=NO2/Nm //mole fraction of O2

+yCO2=NCO2/Nm //mole fraction of CO2

+Ru=8.314 //Universal Gas Constant in kJ/kmol-K

+dSm=-Ru*(NO2*log(yO2)+NCO2*log(yCO2)) //Entropy change in kJ/K

+To=298 //Temp. of surroundings in kelvins

+X=To*dSm //energy destuction in the process in kJ

+printf("\n Hence, the exergy destruction in the process is = %.1f MJ. \n",X/10^3);
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diff --git a/167/CH13/EX13.5/ex5.sce b/167/CH13/EX13.5/ex5.sce
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+//example 5

+//cooling of non ideal gas mixture

+clear

+clc

+disp('We take the cooling section as the system. This is a control volume since mass crosses the system boundary during the process.The critical properties ')

+TcrN2= 126.2 //Critical Pressure of N2in K 

+PcrN2=3.39 //Critical Pressure of N2 in MPa 

+TcrO2=154.8 //Critical Temp. of O2 in K 

+PcrO2=5.08 //Critical Pressure of O2 in MPa 

+yN2=0.79 //mole fraction of nitrogen

+yO2=0.21 //mole fraction of O2

+T1=220 //Initial Temp. of air in kelvins

+T2=160//Final Temp. of air in kelvins

+Pm=10 //PRessure in MPa

+Ru=8.314 //Universal Gas constant in kJ/kmol-K

+disp('calculating heat transfer per kmol of air using ideal gas approximation')

+h1N2=6391 //Enthalpy of N2 at T1 in kJ/kmol

+h1O2=6404 //Enthalpy of O2 at T1 in kJ/kmol

+h2O2=4657 //Enthalpy of O2 at T2 in kJ/kmol

+h2N2=4648 //Enthalpy of N2 at T2 in kJ/kmol

+qout=yN2*(h1N2-h2N2)+yO2*(h1O2-h2O2) //Heat Transfer in kJ/kmol

+printf("\n Hence, the heat transfer during the process using the ideal gas approximation is = %.0f kJ/kmol. \n",qout);

+disp('calculating heat transfer per kmol of air using Kays law')

+Tcrm2=yN2*TcrN2+yO2*TcrO2 //critical temp. of pseudopure substance

+Pcrm2=yN2*PcrN2+yO2*PcrO2 //critical pressure of pseudopure substance

+Tr1=T1/Tcrm2 //Reduced Temp. at T1

+Tr2=T2/Tcrm2 //Reduced Temp. at T2

+Zh1m=1.0 //Compresibility factor at T1

+Zh2m=2.6 //Compressibility Factor at T2

+Pr=Pm/Pcrm2 //Reduced Pressure 

+h1m=yN2*h1N2+yO2*h1O2 //Enthalpy of the mixture at T1 in kJ/kmol

+h2m=yN2*h2N2+yO2*h2O2 //Enthalpy of the mixture at T2 in kJ/kmol

+qout=(h1m-h2m)-Ru*Tcrm2*(Zh1m-Zh2m)//Heat transfer during the process in kJ/kmol

+printf("\n Hence, the heat transfer during the process using Kays law is = %.0f kJ/kmol. \n",qout);

+disp('calculating heat transfer per kmol of air using Amagats law')

+Zh1N2=0.9 //Compressibility factor of N2 at T1

+Zh2N2=2.4 //Compressibility factor of N2 at T2

+Zh1O2=1.3 //Compressibility factor of O2 at T1

+Zh2O2=4.0 //Compressibility factor of O2 at T2

+dhN2=(h1N2-h2N2)-Ru*TcrN2*(Zh1N2-Zh2N2) //Enthalpy change for N2 in kJ/kmol

+dhO2=(h1O2-h2O2)-Ru*TcrO2*(Zh1O2-Zh2O2) //Enthaloy change for O2 in kJ/kmol

+qout=yN2*dhN2+yO2*dhO2 //kJ/mol //heat transfer during the process in kJ/kmol

+printf("\n Hence, the heat transfer during the process using Amagats law is = %.0f kJ/kmol. \n",qout);
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diff --git a/167/CH13/EX13.6/ex6.sce b/167/CH13/EX13.6/ex6.sce
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+//example 6

+//obtaining fresh water from sea water

+clear

+clc

+Mw=18.0 //molar mass of water kg/kmol

+Ms=58.44 //molar mass of salt kg/kmol

+Rw=0.4615 //gas constant of pure water kJ/kg-K

+mfs=0.0348 //mass fraction of salt

+mfw=1-mfs //mass fraction of water

+Mm=1/((mfs/Ms)+(mfw/Mw)) //molar mass of mixture kg/kmol

+yw=mfw*Mm/Mw //mole fraction of water

+ys=1-yw //mole fraction of salt

+To=288.15 //Temp. of Seawater in kelvins

+Ru=8.314 //Universal Gas constant in kJ/kmol-K

+dm=1028 //density of seawater in kg/m^3

+wminin=-Ru*To*(yw*log(yw)+ys*log(ys)) //minimum work input required to separate 1 kg of seawater completely into pure water and pure salts kJ/kmol

+Wminin=wminin/Mm //minimum work input in kJ/kg seawater

+wminin2=Rw*To*log(1/yw) //minimum work input required to produce 1 kg of fresh water from seawater in kJ/kg fresh water

+Pm=dm*Rw*To*log(1/yw) //the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes in kPa

+printf("\n Hence, the mole fraction of water in the seawater is = %.4f. \n",yw);

+printf("\n Hence, the mole fraction of salt in the seawater is = %.2f percentage. \n",ys*100);

+printf("\n Hence, the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts is = %.2f kJ/kg sea water. \n",Wminin);

+printf("\n Hence, the minimum work input required to produce 1 kg of fresh water from seawater kJ/kg fresh water is = %.2f kJ/kg fresh water. \n",wminin2);

+printf("\n Hence, the the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes is = %.0f kPa. \n",Pm);
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