initial commit / add all books

16 files changed, 349 insertions, 0 deletions

diff --git a/680/CH10/EX10.01/10_01.sce b/680/CH10/EX10.01/10_01.sce
new file mode 100755
index 000000000..13d4a983d
--- /dev/null
+++ b/680/CH10/EX10.01/10_01.sce
@@ -0,0 +1,14 @@
+//Problem 10.01:

+

+//initializing the variables:

+DH0co2 = -94052; // in cal/gmol

+DH0h2o = -57798; // in cal/gmol

+DH0ch4 = -17889; // in cal/gmol

+DH0o2 = 0; // in cal/gmol

+T = 298; // in K

+

+//calculation:

+DH0298 = DH0co2 + 2*DH0h2o - 2*DH0o2 - DH0ch4

+

+printf("\n\nResult\n\n")

+printf("\n the standard enthalpy of reaction is %.0fcal/gmol",DH0298)
\ No newline at end of file
diff --git a/680/CH10/EX10.02/10_02.sce b/680/CH10/EX10.02/10_02.sce
new file mode 100755
index 000000000..c4bb25a60
--- /dev/null
+++ b/680/CH10/EX10.02/10_02.sce
@@ -0,0 +1,19 @@
+//Problem 10.02:

+

+//initializing the variables:

+DH0no = 21570; // in cal/gmol

+DH0h2o = -68317; // in cal/gmol

+DH0c3h8 = -24820; // in cal/gmol

+DH0ch4 = -17889; // in cal/gmol

+DH0c2h4= 12496; // in cal/gmol

+DH0no2 = 7930; // in cal/gmol

+DH0hno3 = -41404; // in cal/gmol

+T = 298; // in K

+

+//calculation:

+DH02981 = 2*DH0no

+DH02982 = DH0ch4 + DH0c2h4 - DH0c3h8

+DH02983 = DH0no + 2*DH0hno3 - 3*DH0no2 - DH0h2o

+

+printf("\n\nResult\n\n")

+printf("\n Standard heat of reaction 1 is %.0fcal/gmol N2 of reaction 2 is %.0f cal/gmol C3H8 and of rection 3 is %.0f cal/gmol H2O ",DH02981, DH02982, DH02983)
\ No newline at end of file
diff --git a/680/CH10/EX10.03/10_03.sce b/680/CH10/EX10.03/10_03.sce
new file mode 100755
index 000000000..130385342
--- /dev/null
+++ b/680/CH10/EX10.03/10_03.sce
@@ -0,0 +1,20 @@
+//Problem 10.03:

+

+//initializing the variables:

+

+//calculation:

+//From Table 10.1,

+//DH0chex = -1,005,570 cal/gmol

+//First, write the combustion reaction:

+//C6H14 + 9.5O2 ---> 6CO2 + 7H2O(l)

+//From Table 10.1, one obtains

+DH0fC6H14 = -36960 //cal/gmol

+DH0fCO2 = -94052 //cal/gmol

+DH0fH2O = -68317 //cal/gmol

+//Thus,

+//DH0c = E(DH0f,p) - E(DH0f,r) 

+DH0c = 6*DH0fCO2 + 7*DH0fH2O - 1*DH0fC6H14

+//The calculation process is verified.

+

+printf("\n\nResult\n\n")

+printf("\n From Table 10.1 DH0c(n-hexane) = -1,005,570 cal/gmol, we obtains by calculations DH0c = %.0f cal/gmol \n The calculation process is verified.",DH0c)

diff --git a/680/CH10/EX10.04/10_04.sce b/680/CH10/EX10.04/10_04.sce
new file mode 100755
index 000000000..f4296f5cf
--- /dev/null
+++ b/680/CH10/EX10.04/10_04.sce
@@ -0,0 +1,20 @@
+//Problem 10.04:

+

+//initializing the variables:

+

+//calculation:

+//The standard heat of combustion for this organic is obtained directly from \nTable 10.1, noting that the H2O and HCl formed are in the liquid and gaseous states,\n respectively:

+DH0c = -1600 //kcal/gmol

+//First, write a balanced stoichiometric equation for this combustion reaction:

+//C14H9Cl5 + 15O2 ---> 14CO2 + 2H2O(l) + 5HCl(g)

+//For this reaction,

+//DH0c = 14DH0f,CO2 + 2DH0f ,H2O(l) + 5DH0f ,HCl(g) - DH0f ,C14H9Cl5

+//From Table 10.1,

+DH0fCO2 = -94.052 //kcal/gmol

+DH0fH2O = -68.317 //kcal/gmol

+DH0fHCl = -22.063 //kcal/gmol

+//Solving this equation for DH0f ,C14H9C15 yields

+DH0fC14H9Cl5 = -1*DH0c + 14*DH0fCO2 + 2*DH0fH2O + 5*DH0fHCl //kcal=gmol

+

+printf("\n\nResult\n\n")

+printf("\n DH0fC14H9Cl5 = %.3f kcal/gmol ",DH0fC14H9Cl5)

diff --git a/680/CH10/EX10.05/10_05.sce b/680/CH10/EX10.05/10_05.sce
new file mode 100755
index 000000000..d726f79ae
--- /dev/null
+++ b/680/CH10/EX10.05/10_05.sce
@@ -0,0 +1,31 @@
+//Problem 10.05:

+

+//initializing the variables:

+

+//calculation:

+//TThe standard heat of combustion for chlorobenzene is obtained from the heats of formation data in Table 10.1. Since

+//C6H5Cl + 7O2 ---> 6CO2 + 2H2O + HCl(g)

+DH0c = 6*(-94052) + 2*(-57789) - 22063 - 12390

+//This stoichiometric reaction is now written for combustion in air. First note that there are 7.0(79/21) or 26.33 lbmol of nitrogen present in the theoretical combustion air

+//C6H5Cl + 7O2 + [26.33N2] ---> 6CO2 + 2H2O(g) + HCl(g) + [26.33N2]

+//The heat capacity for the flue gas products in the form

+//CP = a + b*T + c*T^-2

+Da = 264.12

+Db = 0.0425

+Dc = -1.522E6

+//DCp = Da + Db*T + Dc*T^-2 cal/gmol.K or Btu/lbmol.degR

+//Equation (10.22) applies in calculating the adiabatic flame temperature. The energy liberated \non combustion appears as sensible energy in heating the flue (product) gas. The sum of \nthese two effects is zero if the operation is conducted adiabatically, i.e.,

+//DH0c + DHp = DH = 0

+//Since 25 degC = 298K, the enthalpy change associated with heating the flue products is given by

+//DHp = int(298,T2)[DCp]dT

+//T2 = theoretical adiabatic temperature (K)

+//Substituting DCp obtained previously and integrating yields

+//DHp = Da*(T2 - 298) + (Db/2)*(T2^2 - 298^2) - Dc*(1/T2 - 1/298)

+DH0c = -714361 // cal/gmol

+DH0p = -1*DH0c

+//by solving the equation trial and error method

+T2 = 2519 // K

+T2f = 4074 // deg F

+

+printf("\n\nResult\n\n")

+printf("\n theoretical adiabatic flame temperature is %.0f degF",T2f)

diff --git a/680/CH10/EX10.06/10_06.sce b/680/CH10/EX10.06/10_06.sce
new file mode 100755
index 000000000..bd9295e80
--- /dev/null
+++ b/680/CH10/EX10.06/10_06.sce
@@ -0,0 +1,31 @@
+//Problem 10.06:

+

+//initializing the variables:

+

+//calculation:

+//TThe standard heat of combustion for chlorobenzene is obtained from the heats of formation data in Table 10.1. Since

+//C6H5Cl + 7O2 ---> 6CO2 + 2H2O + HCl(g)

+DH0c = 6*(-94052) + 2*(-57789) - 22063 - 12390

+//This stoichiometric reaction is now written for combustion in air. First note that there are 7.0(79/21) or 26.33 lbmol of nitrogen present in the theoretical combustion air

+//C6H5Cl + 7O2 + [26.33N2] ---> 6CO2 + 2H2O(g) + HCl(g) + [26.33N2]

+//The heat capacity for the flue gas products in the form

+//CP = a + b*T + r*T^2

+Da = 230.305

+Db = 0.1003175

+Dr = -20.36033E-6

+//DCp = Da + Db*T + Dr*T^2 cal/gmol.K or Btu/lbmol.degR

+//Equation (10.22) applies in calculating the adiabatic flame temperature. The energy liberated \non combustion appears as sensible energy in heating the flue (product) gas. The sum of \nthese two effects is zero if the operation is conducted adiabatically, i.e.,

+//DH0c + DHp = DH = 0

+//Since 25 degC = 298K, the enthalpy change associated with heating the flue products is given by

+//DHp = int(298,T)[DCp]dT

+//T = theoretical adiabatic temperature (K)

+//Substituting DCp obtained previously and integrating yields

+//DHp = Da*(T - 298) + (Db/2)*(T^2 - 298^2) - Dr*(T^3 - 298^3)

+DH0c = -714361 // cal/gmol

+DH0p = -1*DH0c

+//by solving the equation trial and error method

+T = 2511.5 // K

+Tf = 4061 // deg F

+

+printf("\n\nResult\n\n")

+printf("\n theoretical adiabatic flame temperature is %.0f degF",Tf)

diff --git a/680/CH10/EX10.07/10_07.sce b/680/CH10/EX10.07/10_07.sce
new file mode 100755
index 000000000..06c3e8ebb
--- /dev/null
+++ b/680/CH10/EX10.07/10_07.sce
@@ -0,0 +1,31 @@
+//Problem 10.07:

+

+//initializing the variables:

+

+//calculation:

+//TThe standard heat of combustion for chlorobenzene is obtained from the heats of formation data in Table 10.1. Since

+//C6H5Cl + 14O2 ---> 6CO2 + 2H2O + HCl(g) + 7O2

+DH0c = 6*(-94052) + 2*(-57789) - 22063 - 12390

+//This stoichiometric reaction is now written for combustion in air. First note that there are 7.0(79/21) or 26.33 lbmol of nitrogen present in the theoretical combustion air

+//C6H5Cl + 14O2 + [52.6N2] ---> 6CO2 + 2H2O(g) + HCl(g) + 7O2 + [52.6N2]

+//The heat capacity for the flue gas products in the form

+//CP = a + b*T + c*T^-2

+Da = 493.67

+Db = 0.0731

+Dc = -2.12E6

+//DCp = Da + Db*T + Dc*T^-2 cal/gmol.K or Btu/lbmol.degR

+//Equation (10.22) applies in calculating the adiabatic flame temperature. The energy liberated \non combustion appears as sensible energy in heating the flue (product) gas. The sum of \nthese two effects is zero if the operation is conducted adiabatically, i.e.,

+//DH0c + DHp = DH = 0

+//Since 25 degC = 298K, the enthalpy change associated with heating the flue products is given by

+//DHp = int(298,T2)[DCp]dT

+//T2 = theoretical adiabatic temperature (K)

+//Substituting DCp obtained previously and integrating yields

+//DHp = Da*(T2 - 298) + (Db/2)*(T2^2 - 298^2) - Dc*(1/T2 - 1/298)

+DH0c = -714361 // cal/gmol

+DH0p = -1*DH0c

+//by solving the equation trial and error method

+T2 = 1579 // K

+T2f = 2382 // deg F

+

+printf("\n\nResult\n\n")

+printf("\n theoretical adiabatic flame temperature is %.0f degF \n therefore the operating temperature does exceed the permit requirement of 2100 degF",T2f)

diff --git a/680/CH10/EX10.08/10_08.sce b/680/CH10/EX10.08/10_08.sce
new file mode 100755
index 000000000..12e7c4545
--- /dev/null
+++ b/680/CH10/EX10.08/10_08.sce
@@ -0,0 +1,17 @@
+//Problem 10.08:

+

+//initializing the variables:

+EA = 150

+T = 298; // in K

+

+//calculation:

+DHc = -1580.56 // in cal/gmol

+//DCp = 14*Cpco2 + 2*Cph2o + 5*Cphcl + 22.5*Cpo2 + 141.05*Cpn2

+//DCp = 1318.60 + 0.1899*T - 5.327E6*T^-2

+//DHc = 1318.60(T2 - T) + 0.5*0.1899*(T2^2 - T^2) + 5.327E6(T2^-1 - 298^-1)

+//solving this, we get

+T2 = 1377 // in k

+Tf = 9*(T2 - 273)/5 + 32 // in deg F

+

+printf("\n\nResult\n\n")

+printf("\n the operating temperature is %.0fdeg F",Tf)
\ No newline at end of file
diff --git a/680/CH10/EX10.09/10_09.sce b/680/CH10/EX10.09/10_09.sce
new file mode 100755
index 000000000..f3f8141f1
--- /dev/null
+++ b/680/CH10/EX10.09/10_09.sce
@@ -0,0 +1,23 @@
+//Problem 10.09:

+

+//initializing the variables:

+EA = 10

+x = 0.10

+Da = 117

+Db = 0.04521

+Dr = -6.53E-6

+DH0co2 = -94054; // in cal/gmol

+DH0h2o = -57798; // in cal/gmol

+DH0c2h6o = -56240; // in cal/gmol

+

+//calculation:

+DH0298 = 2*DH0co2 + 3*DH0h2o - DH0c2h6o

+DHP = -1*DH0298

+DHPl = (1 - x)*DHP

+//DHPl = Da*(T - 298) + (Db/2)*(T^2 - 298^2) + (Dr/3)*(T^3 - 298^3)

+//solving this, we get

+T = 2025 // in k

+Tf = 9*(T - 273)/5 + 32 // in deg F

+

+printf("\n\nResult\n\n")

+printf("\n the flame temperature is %.0fdeg F",Tf)
\ No newline at end of file
diff --git a/680/CH10/EX10.10/10_10.sce b/680/CH10/EX10.10/10_10.sce
new file mode 100755
index 000000000..b6edbe033
--- /dev/null
+++ b/680/CH10/EX10.10/10_10.sce
@@ -0,0 +1,46 @@
+//Problem 10.10:

+

+//initializing the variables:

+

+//calculation:

+//C2H3Cl + 2.5O2 + [9.4N2] ---> 2CO2 + H2O(g) + HCl(g) + [9.4N2]

+//C3H8 + 5O2 + [18.8N2] ---> 3CO2 + 4H2O + [18.8N2]

+DH0cC2H3Cl = 2*(-94052) + 1*(-57789) - 22063 + 8400

+DH0cC3H8 = 3*(-94052) + 4*(-57789) + 24820

+//The heat capacity for the flue gas products in the form

+//CP = a + b*T + c*T^-2

+Da1 = 87.7416

+Db1 = 35.273E-3

+Dr1 = -6.446E-6

+Da2 = 170.317

+Db2 = 63.820E-3

+Dr2 = -9.5218E-6

+//Calculate the mole fraction of 1-chloroethylene and propane:

+MWoflChloroethylene = 62.5; //lb/lbmol

+MWofpropane = 44; //lb/lbmol

+//Converting from lb to lbmols on a total 100 lb basis,

+mols1chloroethylene =  1.2

+molspropane =0.57

+totalmols = 1.77

+//Converting mols to mole fraction,

+n1 = 0.679

+n2 = 0.321

+//DCp = Da + Db*T + Dr*T^2 cal/gmol.K or Btu/lbmol.degR

+//Equation (10.22) applies in calculating the adiabatic flame temperature. The energy liberated \non combustion appears as sensible energy in heating the flue (product) gas. The sum of \nthese two effects is zero if the operation is conducted adiabatically, i.e.,

+//DH0c + DHp = DH = 0

+//Since 25 degC = 298K, the enthalpy change associated with heating the flue products is given by

+//DHp = int(298,T2)[DCp]dT

+//T = theoretical adiabatic temperature (K)

+//Substituting DCp obtained previously and integrating yields

+//DH0p = Da*(T - 298) + (Db/2)*(T^2 - 298^2) - Dr*(T^3 - 298^3)

+Da = n1*Da1 + n2*Da2

+Db = n1*Db1 + n2*Db2

+Dr = n1*Dr1 + n2*Dr2

+DH0c = n1*DH0cC2H3Cl + n2*DH0cC3H8 // cal/gmol

+DH0p = -1*DH0c

+//by solving the equation trial and error method

+T = 2406 // K

+Tf = 3871 // deg F

+

+printf("\n\nResult\n\n")

+printf("\n theoretical adiabatic flame temperature is %.0f degF ",Tf)

diff --git a/680/CH10/EX10.11/10_11.sce b/680/CH10/EX10.11/10_11.sce
new file mode 100755
index 000000000..984072e41
--- /dev/null
+++ b/680/CH10/EX10.11/10_11.sce
@@ -0,0 +1,13 @@
+//Problem 10.11:

+

+//initializing the variables:

+DH0T = -12236; // in cal/gmol

+

+//calculation:

+//DH0T = -9140 - 7.596*T + 4.243E-3*T^2 - 0.742E-6*T^3

+//solving this, we get

+T = 570 // in k

+Tc = T - 273 // in deg C

+

+printf("\n\nResult\n\n")

+printf("\n the temperature is %.0fdeg C",Tc)
\ No newline at end of file
diff --git a/680/CH10/EX10.12/10_12.sce b/680/CH10/EX10.12/10_12.sce
new file mode 100755
index 000000000..eb6e1162c
--- /dev/null
+++ b/680/CH10/EX10.12/10_12.sce
@@ -0,0 +1,14 @@
+//Problem 10.12:

+

+//initializing the variables:

+T = 250; // in Deg C

+

+//calculation:

+Tk = T + 273

+//DH0T = -9140 - 7.596*T + 4.243E-3*T^2 - 0.742E-6*T^3

+//solving this, we get

+DH0523 = -9140 - 7.596*Tk + 4.243E-3*Tk^2 - 0.742E-6*Tk^3

+Q = DH0523

+

+printf("\n\nResult\n\n")

+printf("\n heat must be added to or removed from a flow reactor per gmole of product formed is %.0f cal/gmol",Q)
\ No newline at end of file
diff --git a/680/CH10/EX10.13/10_13.sce b/680/CH10/EX10.13/10_13.sce
new file mode 100755
index 000000000..e4d8812ca
--- /dev/null
+++ b/680/CH10/EX10.13/10_13.sce
@@ -0,0 +1,15 @@
+//Problem 10.13:

+

+//initializing the variables:

+ndt = 8; // gmol/h

+T = 250; // in Deg C

+

+//calculation:

+Tk = T + 273

+//DH0T = -9140 - 7.596*T + 4.243E-3*T^2 - 0.742E-6*T^3

+//solving this, we get

+DH0523 = -9140 - 7.596*Tk + 4.243E-3*Tk^2 - 0.742E-6*Tk^3

+Qdt = ndt*DH0523

+

+printf("\n\nResult\n\n")

+printf("\n heat rate must be added to or removed from a flow reactor is %.0f cal/h",Qdt)
\ No newline at end of file
diff --git a/680/CH10/EX10.15/10_15.sce b/680/CH10/EX10.15/10_15.sce
new file mode 100755
index 000000000..ad2fd1a20
--- /dev/null
+++ b/680/CH10/EX10.15/10_15.sce
@@ -0,0 +1,19 @@
+//Problem 10.15:

+

+//initializing the variables:

+xin2 = 0.0515

+xich4 = 0.8111

+xic2h6 = 0.0967

+xic3h8 = 0.0351

+xic4h10 = 0.0056

+HVgn2 = 0; // in Btu/scf

+HVgch4 = 1013; // in Btu/scf

+HVgc2h6 = 1792; // in Btu/scf

+HVgc3h8 = 2590; // in Btu/scf

+HVgc4h10 = 3370; // in Btu/scf

+

+//calculation:

+HVg = xin2*HVgn2 + xich4*HVgch4 + xic2h6*HVgc2h6 + xic3h8*HVgc3h8 + xic4h10*HVgc4h10

+

+printf("\n\nResult\n\n")

+printf("\n the gross heating value of the gas mixture is %.0f Btu/scf",HVg)
\ No newline at end of file
diff --git a/680/CH10/EX10.16/10_16.sce b/680/CH10/EX10.16/10_16.sce
new file mode 100755
index 000000000..6fc858dca
--- /dev/null
+++ b/680/CH10/EX10.16/10_16.sce
@@ -0,0 +1,23 @@
+//Problem 10.16:

+

+//initializing the variables:

+c = 0.25

+mo = 0.35

+w = 0.15

+in = 0.25

+ql = 0.05

+co2 = 0.118

+co = 13; // in ppm

+o2 = 0.104

+NHVc = 14000; // in Btu/lb

+NHVmo = 25000; // in Btu/lb

+NHVw = 0; // in Btu/lb

+NHVin = -1000; // in Btu/lb

+

+//calculation:

+NHV = c*NHVc + w*NHVw + mo*NHVmo + in*NHVin

+EA = (1 - ql)*o2*100/(21-o2*100)

+T = 60 + NHV/(0.325*[1 + (1+EA)*7.5E-4*NHV])

+

+printf("\n\nResult\n\n")

+printf("\n the theoretical flame temperature is %.0f deg F",T)
\ No newline at end of file
diff --git a/680/CH10/EX10.17/10_17.sce b/680/CH10/EX10.17/10_17.sce
new file mode 100755
index 000000000..c6e04dd74
--- /dev/null
+++ b/680/CH10/EX10.17/10_17.sce
@@ -0,0 +1,13 @@
+//Problem 10.17:

+

+//initializing the variables:

+T = 1900;// in deg F

+ea0 = 0

+ea100 = 1

+

+//calculation:

+NHV0 = 0.3*(T-60)/(1 - (1+ ea0)*7.5E-4*0.3*(T-60))

+NHV100 = 0.3*(T-60)/(1 - (1+ea100)*7.5E-4*0.3*(T-60))

+

+printf("\n\nResult\n\n")

+printf("\n NHV for 0 percent Excess air is %.0f Btu/lb and for 100 percent is %.0f Btu/lb",NHV0, NHV100)
\ No newline at end of file