initial commit / add all books

11 files changed, 389 insertions, 0 deletions

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