From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 479/CH2/EX2.1/Example_2_1.sce | 18 +++++++++ 479/CH2/EX2.10/Example_2_10.sce | 57 +++++++++++++++++++++++++++++ 479/CH2/EX2.2/Example_2_2.sce | 11 ++++++ 479/CH2/EX2.3/Example_2_3.sce | 11 ++++++ 479/CH2/EX2.4/Example_2_4.sce | 11 ++++++ 479/CH2/EX2.5/Example_2_5.sce | 53 +++++++++++++++++++++++++++ 479/CH2/EX2.6/Example_2_6.sce | 49 +++++++++++++++++++++++++ 479/CH2/EX2.7/Example_2_7.sce | 31 ++++++++++++++++ 479/CH2/EX2.8/Example_2_8.sce | 63 ++++++++++++++++++++++++++++++++ 479/CH2/EX2.9/Example_2_9.sce | 81 +++++++++++++++++++++++++++++++++++++++++ 479/CH2/EX2.9/Example_2_9.txt | 4 ++ 11 files changed, 389 insertions(+) create mode 100755 479/CH2/EX2.1/Example_2_1.sce create mode 100755 479/CH2/EX2.10/Example_2_10.sce create mode 100755 479/CH2/EX2.2/Example_2_2.sce create mode 100755 479/CH2/EX2.3/Example_2_3.sce create mode 100755 479/CH2/EX2.4/Example_2_4.sce create mode 100755 479/CH2/EX2.5/Example_2_5.sce create mode 100755 479/CH2/EX2.6/Example_2_6.sce create mode 100755 479/CH2/EX2.7/Example_2_7.sce create mode 100755 479/CH2/EX2.8/Example_2_8.sce create mode 100755 479/CH2/EX2.9/Example_2_9.sce create mode 100755 479/CH2/EX2.9/Example_2_9.txt (limited to '479/CH2') diff --git a/479/CH2/EX2.1/Example_2_1.sce b/479/CH2/EX2.1/Example_2_1.sce new file mode 100755 index 000000000..2f010bdf9 --- /dev/null +++ b/479/CH2/EX2.1/Example_2_1.sce @@ -0,0 +1,18 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.1 +clear; +clc; + +//Given +m = 140;//m is the mass of N2 in Kg +P = 4.052*(10^5);//P is the pressure of the system in Pa +V = 30;//V is the volume of the system in m^3 +R = 8314.4;// R is the gas constant + +//To determine temperature required +T = P*V/((m/28)*R);//T is the temperature of the system in K +mprintf('Temperature of the system is %f K',T); +//end \ No newline at end of file diff --git a/479/CH2/EX2.10/Example_2_10.sce b/479/CH2/EX2.10/Example_2_10.sce new file mode 100755 index 000000000..a135dcfee --- /dev/null +++ b/479/CH2/EX2.10/Example_2_10.sce @@ -0,0 +1,57 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.10 +clear; +clc; + +//Given +yN2 = 1/4;//mole faction of N2 in the mixture +yH2 = 3/4;//mole fraction of H2 in the mixture +V = 5.7;//V is the rate at which mixture enters in m^3 in 1 hour +P = 600;//P is in atm +T = 298;//T is in K +TcN2 = 126;//critical temp of N2 in K +TcH2 = 33.3;//critical temp of H2 in K +TcNh3 = 406.0;//critical temp of NH3 in K +PcN2 = 33.5;//critical pressure of N2 in atm +PcH2 = 12.8;//critical pressure of H2 in atm +PcNH3 = 111.0;//critical pressure of NH3 in atm +R = 0.082;//gas constant + +//To calculate the amount of ammonia leaving the reactor and the velocity of gaseous product leaving the reactor +//(i)Calculation of amount of NH3 leaving the reactor +Tcm = (TcN2*yN2)+(TcH2*yH2);//critical temperature of the mixture +Pcm = (PcN2*yN2)+(PcH2*yH2);//critical pressure of the mixture +Trm = T/Tcm; +Prm = P/Pcm; +//From figure A.2.3 +Zm = 1.57;//compressibility factor of the mixture +N = (P*V)/(Zm*R*T);//Kg mole of the mixture +N1 = 0.25*N;//Kg mole of N2 in feed +//N2+3H2 - 2NH3 +W = 2*0.15*N1*17; +mprintf('(i)Ammonia formed per hour is %f Kg',W); + +//(ii)Calculation of velocity +N1 = 0.25*N-(0.25*N*0.15);//Kg mole of N2 after reactor +N2 = 0.75*N-(0.75*N*0.15);//Kg mole of H2 after reactor +N3 = 0.25*N*2*0.15;//Kg mole of NH3 after reactor +Nt = N1+N2+N3;//total Kg moles after reactor +y1NH3 = N3/Nt;//mole fraction of NH3 after reactor +y1N2 = N1/Nt;//mole fraction of N2 after reactor +y1H2 = N2/Nt;//mole fraction of H2 after reactor +T1cm = (TcN2*y1N2)+(TcH2*y1H2); +P1cm = (PcN2*y1N2)+(PcH2*y1H2); +T1 = 448;//in K +P1 = 550;//in atm +T1rm = T1/T1cm; +P1rm = P1/P1cm; +//From Figure A.2.2 +Zm1 = 1.38; +V1 = (Zm1*Nt*R*T1)/P1; +d = 5*(10^-2);//diameter of pipe +v = V1/((%pi/4)*(d^2)*3600); +mprintf('\n (ii)Velocity in pipe is %f m/sec',v); +//end \ No newline at end of file diff --git a/479/CH2/EX2.2/Example_2_2.sce b/479/CH2/EX2.2/Example_2_2.sce new file mode 100755 index 000000000..e01570f74 --- /dev/null +++ b/479/CH2/EX2.2/Example_2_2.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.2 +clear; +clc; + +//Given +//This example is a theoretical problem and does not involve any numerical computation +//end \ No newline at end of file diff --git a/479/CH2/EX2.3/Example_2_3.sce b/479/CH2/EX2.3/Example_2_3.sce new file mode 100755 index 000000000..10f075406 --- /dev/null +++ b/479/CH2/EX2.3/Example_2_3.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.3 +clear; +clc; + +//Given +//This example is a theoretical problem and does not involve any numerical computation +//end \ No newline at end of file diff --git a/479/CH2/EX2.4/Example_2_4.sce b/479/CH2/EX2.4/Example_2_4.sce new file mode 100755 index 000000000..b94ffe17f --- /dev/null +++ b/479/CH2/EX2.4/Example_2_4.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.4 +clear; +clc; + +//Given +//This example is a theoretical problem and does not involve any numerical computation +//end \ No newline at end of file diff --git a/479/CH2/EX2.5/Example_2_5.sce b/479/CH2/EX2.5/Example_2_5.sce new file mode 100755 index 000000000..728348458 --- /dev/null +++ b/479/CH2/EX2.5/Example_2_5.sce @@ -0,0 +1,53 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.5 +clear; +clc; + +//Given +n = 1;//n is Kg moles of methane +T = 423;//T is the temperatue of the system in kelvin +P = 100;//P is the pressure of the system in atm +Tc = 191;//Tc is the critical temperature of the system in K +Pc = 45.8;//Pc is the critical pressure of the system in atm +R = 0.08206;//R is the gas constant in (m^3 atm/Kg mole K) + +//To calculate the volume of methane +//(i)Using ideal gas equation +V1 = (n*R*T)/P;//V1 is the volume of the gas in m^3 +mprintf('(i)Volume of the gas using ideal gas equation is %f cubic meter',V1); + +//(ii)Using Vander Waals' equation +a = (27*(R^2)*(Tc^2))/(64*Pc);//Vander Waais constant +b = (R*Tc)/(8*Pc);//Vander Waais constant +v = poly(0, 'v'); +q = -((a*b)/P)+(a/P)*v-(((R*T)+(b*P))/P)*v^2+(v^3);//According to Vander Waals equation +r = roots(q); +mprintf('\n (ii)Volume of the gas using Vander Waals equation is %f cubic meter',r(1)); + +//(iii)Using generalized Z chart +Tr = T/Tc;//Tr is the reduced temperatue +Pr = P/Pc;//Pr is the reduced pressure +//From the figure A.2.2, +Z = 0.97;//Z is the compressibility factor +V = (Z*R*T)/P; +mprintf('\n (iii)Volume of the gas using Z chart is %f cubic meter',V); + +//(iv)Using molar polarisation method +//From Table 2.2 +Pmc = 6.82;//Pmc is the molar polarisation for methane +//From figure A.2.4 +Z0 = .965; +Z1 = 14.8*(10^-4); +Z = Z0+(Z1*Pmc); +V = (Z*R*T)/P; +mprintf('\n (iv)Volume of the gas using molar polarisation method is %f cubic meter',V); + +//(v)From experiment +//Given +Z = 0.9848; +V = (0.9848*n*R*T)/P; +mprintf('\n (v)Volume of the gas calculated by experimental Z value is %f cubic meter',V); +//end \ No newline at end of file diff --git a/479/CH2/EX2.6/Example_2_6.sce b/479/CH2/EX2.6/Example_2_6.sce new file mode 100755 index 000000000..6de8f8e07 --- /dev/null +++ b/479/CH2/EX2.6/Example_2_6.sce @@ -0,0 +1,49 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.6 +clear; +clc; + +//Given +P1 = 266; +T1 = 473.16;//Initial temperature in Kelvin +T2 = 273.16;//Final temperature in Kelvin +V1 = 80; V2 = 80;//Initial & final volume in litres +N1 = (14.28/28); N2 = (14.28/28);//Initial and final Kg moles are equal +Tc = 126;//Critical temperature of N2 in K +Pc = 33.5;//Critical pressure of N2 in atm + +//To calculate the final pressure achieved +//(i)Using ideal gas law +p2 = (P1*V1*N2*T2)/(V2*N1*T1); +mprintf('(i)Final pressure of N2 using ideal gas law is %f atm',p2); + +//(ii)Using generalized Z chart +Tr1 = T1/Tc;//reduced initial temp in k +Pr1 = P1/Pc;//reduced initial press in K +//From the Z-chart compressibility factor coressponding to the above Tr1 &Pr1 is +Z1 = 1.07; +P2 = [125,135,150]; +Z2 = [0.95, 0.96, 0.98]; +F = [0,0,0]; +for i = 1:3 + F(i) = (P2(i)/(Z2(i)*T2))-(P1/(Z1*T1)); +end +clf; +plot(P2,F); +xtitle("P2 vs F","P2","F"); +P3 = interpln([F;P2],0); +mprintf('\n (ii)Final pressure of N2 from Z chart is %f atm',P3); + +//(iii)Using Pseudo reduced density chart +R = 0.082;//gas constant +v = V1/N1;//Volume per moles of nitrogen in m^3/Kg mole +Dr = (R*Tc)/(Pc*v); +Tr2 = T2/Tc;//final reduced temp in K +//From figure A.2.1, reduced pressure coressponding to this Dr and Tr2 is +Pr2 = 4.1//final reduced pressure in atm +p2_ = Pr2*Pc; +mprintf('\n (iii)Final pressure achieved using Dr chart is %f atm',p2_); +//end \ No newline at end of file diff --git a/479/CH2/EX2.7/Example_2_7.sce b/479/CH2/EX2.7/Example_2_7.sce new file mode 100755 index 000000000..fd7d89932 --- /dev/null +++ b/479/CH2/EX2.7/Example_2_7.sce @@ -0,0 +1,31 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.7 +clear; +clc; + +//Given +n = 1;//n is the Kg mole of methane gas +T = 298;//T is the constant temperature in K +P1 = 1;//P1 is the initial pressure of the system +P2 = 100;//P2 is the final pressure of the system +R = 8314.4;//R is the gas constant in Nm/Kgmole deg K + +//To compute the work required +//(i)Using ideal gas law +W = R*T*log(P1/P2); +mprintf('(i)Work done by the system if the gas obeys ideal gas law is %4.2e Nm',W); + +//(ii)Using Vander Waals' equation +//Given +//For methane +a = 2.32*(10^5);//Vander Wals' constant a in N/m^2 +b = 0.0428;//Vanderwaals' constant b in m^3 +//V1 and V2 are evaluated by trial and error using Vanderwaals' equation as P1 and P2 are known +V1 = 11.1;//initial volume of the gas in m^3 +V2 = 0.089;//final volume of the gas in m^3 +W = (R*T*log((V2-b)/(V1-b)))+(a*((1/V2)-(1/V1))) +mprintf('\n (ii)Work done by the system if the gas obeys Vander Waals equation is %4.2e Nm',W); +//end \ No newline at end of file diff --git a/479/CH2/EX2.8/Example_2_8.sce b/479/CH2/EX2.8/Example_2_8.sce new file mode 100755 index 000000000..22019a4ec --- /dev/null +++ b/479/CH2/EX2.8/Example_2_8.sce @@ -0,0 +1,63 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.8 +clear; +clc; + +//Given +V = 27*(10^-3);//Volume of the container in m^3 +n = (15/70.91);//n is the Kg moles of chlorine +T = 293;//T is the temperature in K +R = 0.08206; +P = 10^(4.39-(1045/293));//P is the vapour pressure of chlorine +Pc = 76.1;//Critical pressure of Chlorine +Tc = 417;//Critical temperature of Chlorine +Pr = P/Pc;//Reduced pressure of Chlorine +Tr = T/Tc;//Critical temperature of Chlorine +M = 70.91;//Molecular weight of the Chlorine + +//To determine the vapour pressure of chlorine, amount of liquid Cl2 and temperature required +//(i)Specific volume of liquid Chlorine +//From figure A.2.2 +Zg = 0.93; +//From figure A.2.6 +Zl = 0.013; +vl = ((Zl*R*T)/P); +mprintf('(i)Specific volume of liquid Chlorine from compressibility chart is %f cubic meter /Kgmole',vl); + +//From Francis relation, taking the constants from Table 2.3 +D = (1.606-(216*(10^-5)*20)-(28/(200-20)))*10^3;//Density of liq Cl2 in Kg/m^3 +Vl = M/D; +mprintf('\n Specific volume of liquid Chlorine from Francis relation is %f cubic meter /Kgmole',Vl); + +//(ii)Amount of liquid Cl2 present in the cylinder +vg = ((Zg*R*T)/P); +V1 = V-vg;//V1 is the volume of liquid Chlorine +Vct = 0.027;//volume of the container +Vg = (0.212-(Vct/vl))/((1/vg)-(1/vl));//By material balance +W = ((V-Vg)*70.9)/vl; +mprintf('\n\n (ii)Weight of Chlorine at 20deg cel is %f Kg',W); + +//(iii)Calculation of temperature required to evaporate all the liquid chlorine +//log P' = 4.39 - 1045/T (given) +//Assume the various temperature +Ng = 0.212;//total Kg moles of gas +Ta = [413 415 417]; +N = [0,0,0]; +for i = 1:3 + Tr(i) = Ta(i)/Tc;//reduced temperature in K + P(i) = 10^(4.39-(1045/Ta(i))); + Pr(i) = P(i)/Pc;//reduced pressure in K +//From the compressibility factor chart,Z values coressponding to the above Tr &Pr are given as +Z = [0.4 0.328 0.208]; +N(i) = (P(i)*Vct)/(Z(i)*R*Ta(i)); +end + +clf; +plot(N,Ta); +xtitle("Ta vs N","N","Ta"); +T1 = interpln([N;Ta],0.212);//in K +mprintf('\n (iii)The temperature required to evaporate all the liquid chlorine is %f deg celsius',T1-273); +//end \ No newline at end of file diff --git a/479/CH2/EX2.9/Example_2_9.sce b/479/CH2/EX2.9/Example_2_9.sce new file mode 100755 index 000000000..45519a7c2 --- /dev/null +++ b/479/CH2/EX2.9/Example_2_9.sce @@ -0,0 +1,81 @@ +//Chemical Engineering Thermodynamics +//Chapter 2 +//P-V-T Relations + +//Example 2.9 +clear; +clc; + +//Given +N1 = 0.7;//Kg mole of CH4 +N2 = 0.3;//Kg mole of N2 +R = 0.08206;//Gas constant +T = 323;//Temperature in Kelvin +V = 0.04;//Volume in m^3 +a1 = 2.280; b1 = 0.0428;//Vanderwaals constants for CH4 +a2 = 1.345;b2 = 0.0386;//Vanderwaals constants for N2 +Tc1 = 191; Pc1 = 45.8;//Critical temperature in K and pressure of CH4 in atm +Tc2 = 126;Pc2 = 33.5;//Critical temperature in K and pressure of N2 in atm + +//To find Approx Value +function[A]=approx(V,n) + A=round(V*10^n)/10^n;//V-Value n-To what place + funcprot(0) +endfunction + +//To calculate the pressure exerted by the gas mixture +//(i)Using ideal gas law +P = (N1+N2)*((R*T)/V); +mprintf('(i) Pressure exerted by the gas mixture using ideal gas law is %d atm',P); + +//(ii)Using Vander waal equation +P1 = ((N1*R*T)/(V-(N1*b1)))-((a1*(N1^2))/(V^2));//Partial pressure of CH4 +P2 = ((N2*R*T)/(V-(N2*b2)))-((a2*(N2^2))/(V^2));//Partial pressure of N2 +Pt = P1+P2; +mprintf('\n(ii) Pressure exerted by the gas mixture using Vander waal equation is %f atm', Pt); + +//(iii)Using Zchart and Dalton's law +Tra = T/Tc1;//reduced temperature of CH4 +Trb = T/Tc2;//reduced temperature of N2 +//Asssume the pressure +P = [660 732 793 815 831]; +for i = 1:5 + Pa(i) = N1*P(i);// partial pressure of CH4 for the ith total pressure + Pb(i) = N2*P(i);// partial pressure of N2 for the ith total pressure + Pra(i) = Pa(i)/Pc1;//reduced pressure of CH4 for the ith total pressure + Prb(i) = Pb(i)/Pc2;//reduced pressure of N2 for the ith total pressure +end + +//For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as +Za = [1.154 1.280 1.331 1.370 1.390];//Z values of CH4 +Zb = [1 1 1 1 1];//Z values of N2 +V3 = 0.0421; +for i = 1:5 + Pa(i) = Za(i)*N1*((R*T)/V);//partial pressure of CH4 coressponding to the ith total presure + Pb(i) = Zb(i)*N2*((R*T)/V);//partial pressure of N2 coressponding to the ith total pressure + Pt(i) = Pa(i)+Pb(i);//total pressure of the gas mixture + if Pt(i)-P(i) < 15 + mprintf('\n(iii) pressure exerted by the gas mixture using Z chart and Dalton Law is %d atm',Pt(i)); + else + end +end + +//(iv)Using Amagat's law and Z chart +P = [1000 1200 1500 1700]; +for i=1:4 + Pra(i) = P(i)/Pc1; + Prb(i) = P(i)/Pc2; +end +//For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as +Za = [1.87 2.14 2.52 2.77]; +Zb = [1.80 2.10 2.37 2.54]; +for i = 1:4 + Va(i) = approx((N1*Za(i)*((R*T)/P(i))),4); + Vb(i) = approx((N2*Zb(i)*((R*T)/P(i))),4); + V1(i) = approx((Va(i)+Vb(i)),4); + if V1(i)-V <= 0.003 + mprintf('\n(iv) Pressure exerted by the gas mixture using Amagat law and Zchart is %d atm ',P(i)); + else +end +end +//end diff --git a/479/CH2/EX2.9/Example_2_9.txt b/479/CH2/EX2.9/Example_2_9.txt new file mode 100755 index 000000000..791f69feb --- /dev/null +++ b/479/CH2/EX2.9/Example_2_9.txt @@ -0,0 +1,4 @@ +For example 2.9, +In the (iii) part, i am getting answer as P = 843 atm. +But the answer given in the book is 833 atm. +There might be some calculation mistke in the book. \ No newline at end of file -- cgit