diff options
Diffstat (limited to '680')
157 files changed, 3037 insertions, 0 deletions
diff --git a/680/CH1/EX1.01/1_01.sce b/680/CH1/EX1.01/1_01.sce new file mode 100755 index 000000000..35a5439bd --- /dev/null +++ b/680/CH1/EX1.01/1_01.sce @@ -0,0 +1,16 @@ +//Problem 1.01:
+
+//initializing the variables:
+a1 = 1; // in cm/s2
+h = 3600^2/1; // in s2/h2
+d = 24^2/1; // in h2/day2
+yr = 365^2/1; // in day2/yr2
+in = 1/2.54; // in in/cm
+ft = 1/12; // in ft/in
+mil = 1/5280; // in mile/ft
+
+//calculation:
+a2 = a1*h*d*yr*in*ft*mil
+
+printf("\n\nResult\n\n")
+printf("\n Thus, 1.0 cm/s2 is equal to %.2E miles/yr2\n",a2)
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 diff --git a/680/CH11/EX11.02/11_02.sce b/680/CH11/EX11.02/11_02.sce new file mode 100755 index 000000000..ad4feb726 --- /dev/null +++ b/680/CH11/EX11.02/11_02.sce @@ -0,0 +1,16 @@ +//Problem 11.02:
+
+//initializing the variables:
+Tdb1 = 100; // in deg F
+Twb = 70; // in deg F
+Tdb2 = 80; // in deg F
+p = 1; // in atm
+
+//calculation:
+//from fig. 11.2
+Hi = 0.0090 // lb
+Hf = 0.0133 // lb
+dH = Hf - Hi
+
+printf("\n\nResult\n\n")
+printf("\n moisture added is %.2E lb H2O/lb dry air",dH)
\ No newline at end of file diff --git a/680/CH11/EX11.03/11_03.sce b/680/CH11/EX11.03/11_03.sce new file mode 100755 index 000000000..efe94d662 --- /dev/null +++ b/680/CH11/EX11.03/11_03.sce @@ -0,0 +1,22 @@ +//Problem 11.03:
+
+//initializing the variables:
+T1 = 120; // in deg F
+T2 = 560; // in deg F
+mdte = 9000; // in lb/h
+MW = 30
+MWH2O = 18
+R = 0.73
+
+//calculation:
+//from fig. 11.2
+Hout = 0.0814 // lb H2O/lb bone-dry air
+mdtH2O = Hout*mdte
+mdtt = mdtH2O + mdte
+Yg = (mdte/MW)/(mdtH2O/MWH2O + mdte/MW)
+Ywv = (mdtH2O/MWH2O)/(mdtH2O/MWH2O + mdte/MW)
+MWB = Yg*MW + Ywv*MWH2O
+Pqa = mdtt*R*(460+140)/MWB
+
+printf("\n\nResult\n\n")
+printf("\n the moisture content is %.4f lb H2O/lb dry air, the mass flow rate is %.0f lb/h,the volumetric flow rate of the discharge gas is %.2E ft3/h",Hout, mdtH2O, Pqa)
\ No newline at end of file diff --git a/680/CH11/EX11.04/11_04.sce b/680/CH11/EX11.04/11_04.sce new file mode 100755 index 000000000..ed3f86523 --- /dev/null +++ b/680/CH11/EX11.04/11_04.sce @@ -0,0 +1,12 @@ +//Problem 11.04:
+
+//initializing the variables:
+T1 = 0; // in deg C
+p = 71; // mm of Hg
+P = 1; // in atm
+
+//calculation:
+ymax = p/760
+
+printf("\n\nResult\n\n")
+printf("\n the maximum concentration is %.4f ",ymax)
\ No newline at end of file diff --git a/680/CH11/EX11.05/11_05.sce b/680/CH11/EX11.05/11_05.sce new file mode 100755 index 000000000..dfee3c953 --- /dev/null +++ b/680/CH11/EX11.05/11_05.sce @@ -0,0 +1,16 @@ +//Problem 11.05:
+
+//initializing the variables:
+T1 = 25; // in deg C
+xa = 0.05
+xb = 0.95
+pa = 4150; // in kPa
+pb = 16.1; // in kPa
+
+//calculation:
+P = xa*pa + xb*pb
+ya = xa*pa/P
+yb = 1 - ya
+
+printf("\n\nResult\n\n")
+printf("\n the pressure is %.1f kPa, composition of the first vapor is %.3f ehane and %.3f hexane ",P, ya, yb)
\ No newline at end of file diff --git a/680/CH11/EX11.06/11_06.sce b/680/CH11/EX11.06/11_06.sce new file mode 100755 index 000000000..2ecaa3a8c --- /dev/null +++ b/680/CH11/EX11.06/11_06.sce @@ -0,0 +1,15 @@ +//Problem 11.06:
+
+//initializing the variables:
+T1 = 25; // in deg C
+pa = 111; // in mm of Hg
+pb = 92; // in mm of hg
+P = 100; // in mm of hg
+
+//calculation:
+xa = (P - pb)/(pa - pb)
+ya = xa*pa/P
+
+
+printf("\n\nResult\n\n")
+printf("\nthe composition of the liquid phase is %.3f and of vapour phase %.3f",xa, ya)
\ No newline at end of file diff --git a/680/CH11/EX11.08/11_08.sce b/680/CH11/EX11.08/11_08.sce new file mode 100755 index 000000000..ab2291afb --- /dev/null +++ b/680/CH11/EX11.08/11_08.sce @@ -0,0 +1,13 @@ +//Problem 11.08:
+
+//initializing the variables:
+T1 = 25; // in deg C
+ph2s = 0.01; // in atm
+Pt = 1; // in atm
+Hh2s = 483; // in atm/mole fraction
+
+//calculation:
+xh2s = ph2s/Hh2s
+
+printf("\n\nResult\n\n")
+printf("\nthe maximum mole fraction %.2E",xh2s)
\ No newline at end of file diff --git a/680/CH11/EX11.09/11_09.sce b/680/CH11/EX11.09/11_09.sce new file mode 100755 index 000000000..22e871ca5 --- /dev/null +++ b/680/CH11/EX11.09/11_09.sce @@ -0,0 +1,14 @@ +//Problem 11.09:
+
+//initializing the variables:
+
+//calculation:
+//The partial pressure of ammonia is converted to mole fraction in vapor as shown in Table 11.3.
+// These results of table 11.3 are plotted in Fig. 11.7.
+//Henry’s law constant from the graph is approximately 1.485 at x = 0.095. Since
+Yactual = 0.148
+Ycalculated = 1.485*(0.095)
+Percentagreement = Ycalculated*100/Yactual
+
+printf("\n\nResult\n\n")
+printf("\n from x = 0 to x = %.3f Henry’s law equation, y = 1.485x, predicts the equilibrium vapor content to within 5 percent of the experimental data.",Percentagreement/1000)
diff --git a/680/CH11/EX11.10/11_10.sce b/680/CH11/EX11.10/11_10.sce new file mode 100755 index 000000000..53b60f13e --- /dev/null +++ b/680/CH11/EX11.10/11_10.sce @@ -0,0 +1,34 @@ +//Problem 11.10:
+
+//initializing the variables:
+mdt = 2000; // in acfm
+xCl = 0.13;
+T1 = 80; // in deg F
+xCCl4 = 0.95;
+Tav = -70; // in deg F
+U = 4; // in Btu/ft2.h.deg F
+Cav = 0.125; // Btu/lb.deg F
+Hv = 93.7; // Btu/lb(80 deg F)
+p0 = 7.74; // mm of Hg T 0 deg F
+p18 = 5.64; // mm of Hg T -18 deg F
+p40 = 2.23; // mm of Hg T -40 deg F
+
+//calculation:
+//let
+n = 100; // in lbmol
+nCCl4 = xCl*n
+nair = n - nCCl4
+//for 95%
+nCCl4 = (1 - xCCl4)*nCCl4
+yCCl4 = nCCl4/(nair + nCCl4)
+PCCl4 = yCCl4*760
+mdtair = mdt*(1 - xCl)*60*14.7*29/((460 + T1)*10.73)
+mdtCCl4 = mdt*xCl*60*14.7*154/((460 + T1)*10.73)
+Qair = mdtair*0.24*(-18 - T1)
+QCCl4 = mdtCCl4*Cav*(-18 - T1) + mdtCCl4*xCCl4*(-1*Hv)
+Q = Qair + QCCl4
+LMTD = [(T1 - Tav) - (-18 - Tav)]/(log(150/52))
+A = abs(Q)/(U*LMTD)
+
+printf("\n\nResult\n\n")
+printf("\n the surface area required is %.0f ft^2",A)
\ No newline at end of file diff --git a/680/CH11/EX11.13/11_13.sce b/680/CH11/EX11.13/11_13.sce new file mode 100755 index 000000000..6b25d48b0 --- /dev/null +++ b/680/CH11/EX11.13/11_13.sce @@ -0,0 +1,17 @@ +//Problem 11.13:
+
+//initializing the variables:
+m = 100; // in lb
+p = 40; // in psia
+T1 = 77; // in deg F
+ppm = 10000; // in ppmv
+MCO2 = 44;
+
+//calculation:
+yCO2 = ppm/1E6
+pCO2 = yCO2*p*760*14.7
+//from fig 11.11
+Y = 9.8 // lb CO2/100 lb Seive
+
+printf("\n\nResult\n\n")
+printf("\n the adsorbent capacity is %.1f lb CO2/100 lb Seive",Y)
\ No newline at end of file diff --git a/680/CH11/EX11.14/11_14.sce b/680/CH11/EX11.14/11_14.sce new file mode 100755 index 000000000..d1b99988d --- /dev/null +++ b/680/CH11/EX11.14/11_14.sce @@ -0,0 +1,18 @@ +//Problem 11.14:
+
+//initializing the variables:
+
+//calculation:
+//The equilibrium relationship given in the problem statement can be linearized by taking the log (logarithm) of both sides of the equation. This yields the following equation:
+//log(q) = log(K) + n*log(c)
+//A plot of log(q) vs log(c) yields a straight line if this relationship can be used to represent the experimental data. The slope of this line is equal to n, while the intercept is log(K). The experimental data analyzed using the equation above are plotted in Fig. 11.12, showing that the data fit this linearized isotherm quite well.
+//The equation generated from a regression analysis indicates that
+n = 1.48
+//A = log(K)
+A = -0.456
+K = 10^A
+//provided the units of c and q are mg/L and mg/g, respectively. The describing equation is therefore
+//q = 0.35c^1.48
+
+printf("\n\nResult\n\n")
+printf("\n describing equation is q = %.2f*c^%.2f",K,n)
diff --git a/680/CH12/EX12.01/12_01.sce b/680/CH12/EX12.01/12_01.sce new file mode 100755 index 000000000..20d42ccc2 --- /dev/null +++ b/680/CH12/EX12.01/12_01.sce @@ -0,0 +1,12 @@ +//Problem 12.01:
+
+//initializing the variables:
+xEtOH = 0.3; // mol% ethanol
+
+//calculation:
+// from fig 12.2, for xEtOH = 0.3
+yEtOH = 0.57
+ywater = 1 - yEtOH
+
+printf("\n\nResult\n\n")
+printf("\n the liquid mole fractions is %.2f and vapor mole fraction is %.2f",ywater, yEtOH)
\ No newline at end of file diff --git a/680/CH12/EX12.02/12_02.sce b/680/CH12/EX12.02/12_02.sce new file mode 100755 index 000000000..c762fc19a --- /dev/null +++ b/680/CH12/EX12.02/12_02.sce @@ -0,0 +1,10 @@ +//Problem 12.02:
+
+//initializing the variables:
+xEtOH = 0.3; // mol% ethanol
+
+//calculation:
+xwater = 1 - xEtOH
+
+printf("\n\nResult\n\n")
+printf("\n the liquid mole fractions is %.2f ",xwater)
\ No newline at end of file diff --git a/680/CH12/EX12.06/12_06.sce b/680/CH12/EX12.06/12_06.sce new file mode 100755 index 000000000..d1f1db207 --- /dev/null +++ b/680/CH12/EX12.06/12_06.sce @@ -0,0 +1,24 @@ +//Problem 12.06:
+
+//initializing the variables:
+xb = 0.2; // mol%
+p = 280; // in psia
+a = 0.4;
+
+//calculation:
+BPT = 140; // in deg F
+Kp = 1.15
+Kb = 0.41
+Y = xb*Kb + Kp*(1 - xb)
+DPT = 154; // deg F
+Kp = 1.30
+Kb = 0.50
+Xa = xb/Kb + (1-xb)/Kp
+T = 145; // in deg F
+L = 1 - a
+Kp = 1.20
+Kb = 0.45
+Xz = xb/(L + Kb*a) + (1 - xb)/(L + Kp*a)
+
+printf("\n\nResult\n\n")
+printf("\n bubble point temperature is %.0f deg F, dew point temperature is %.0f deg F and temperature when 40 mole per of the mixture is in the vapor phase is %.0f deg F",BPT, DPT, T)
\ No newline at end of file diff --git a/680/CH12/EX12.08/12_08.sce b/680/CH12/EX12.08/12_08.sce new file mode 100755 index 000000000..195967253 --- /dev/null +++ b/680/CH12/EX12.08/12_08.sce @@ -0,0 +1,31 @@ +//Problem 12.08:
+
+//initializing the variables:
+ZC2 = 0.25;
+ZC4 = 0.15;
+Zy = 0.6;
+p = 120; // in psia
+a = 4;
+T = 40; // in degF
+
+//calculation:
+//Write the componential split equation:
+//Exi = E{zi/[L + Ki(1-L)]} = 1.0
+//Set L and V. The bottoms to boilup ratio is 4/1. Therefore,
+L = 0.80
+V = 1-L
+//Obtain K for ethane (E) and n-butane (B) at 120 psia and 40 degF:
+Ke = 2.60
+Kb = 0.18
+//Calculate xe and xb by employing above Equation:
+//xi = zi/[L + Ki*V]
+//Substituting
+xe = 0.19
+xb = 0.18
+//Set Y as the unknown component (see Table 12.6), and then calculate Xy.
+Xy = 1 - xe-xb
+//Calculate KY by applying Equation (12.4) to component Y
+Ky = (Zy/Xy -L)/V
+
+printf("\n\nResult\n\n")
+printf("\n the chemical name of the unknown componentcorresponding to KY (120 psia and 408F) with a value of %.2f appears to be propane",Ky)
diff --git a/680/CH12/EX12.09/12_09.sce b/680/CH12/EX12.09/12_09.sce new file mode 100755 index 000000000..603eec6f4 --- /dev/null +++ b/680/CH12/EX12.09/12_09.sce @@ -0,0 +1,22 @@ +//Problem 12.09:
+
+//initializing the variables:
+//Antoine Eq Coeff for Methanol
+Am = 16.5938;
+Bm = 3644.3;
+Cm = 239.76;
+//Antoine Eq Coeff for water
+Aw = 16.262;
+Bw = 3799.89;
+Cw = 226.35;
+p = 101.325; // in kpa
+
+//calculation:
+//The saturation temperatures:
+Tsat_m = (Bm/(Am - log(p))) - Cm
+Tsat_w = (Bw/(Aw - log(p))) - Cw
+T = 70
+xm = (p - %e^(Aw - (Bw/(T + Cw))))/((%e^(Am - (Bm/(T + Cm)))) - %e^(Aw - (Bw/(T + Cw))))
+ym = xm*125.07/p
+printf("\n\nResult\n\n")
+printf("\n mole fraction at 70 degC xm = %.3f and ym = %0.3f \n To generate a T-x, y diagram, plot the xm and ym data as the ordinate and temperature as the abscissa. See Fig. 12.6.",xm, ym)
diff --git a/680/CH12/EX12.10/12_10.sce b/680/CH12/EX12.10/12_10.sce new file mode 100755 index 000000000..b27273f92 --- /dev/null +++ b/680/CH12/EX12.10/12_10.sce @@ -0,0 +1,21 @@ +//Problem 12.10:
+
+//initializing the variables:
+//Antoine Eq Coeff for Methanol
+Am = 16.5938;
+Bm = 3644.3;
+Cm = 239.76;
+//Antoine Eq Coeff for water
+Aw = 16.262;
+Bw = 3799.89;
+Cw = 226.35;
+T = 40; // in degC
+
+//calculation:
+xm = 0.3
+pdm = (%e^(Am - (Bm/(T + Cm))))
+pdw = (%e^(Aw - (Bw/(T + Cw))))
+p = xm*pdm + (1 - xm)*pdw
+ym = xm*pdm/p
+printf("\n\nResult\n\n")
+printf("\n mole fraction at 40 degC xm = %.3f and ym = %0.3f \n To generate an x–y diagram, simply plot the xm as the ordinate and ym as the abscissa (Fig. 12.9).\n Again, the convention is to plot only the more volatile compound on phase equilibria diagrams.\n Also, for x–y diagrams, it is standard to plot data on a square coordinate system.",xm, ym)
diff --git a/680/CH12/EX12.10/Ex_12_10_Image.jpg b/680/CH12/EX12.10/Ex_12_10_Image.jpg Binary files differnew file mode 100755 index 000000000..96002c9cf --- /dev/null +++ b/680/CH12/EX12.10/Ex_12_10_Image.jpg diff --git a/680/CH12/EX12.11/12_11.sce b/680/CH12/EX12.11/12_11.sce new file mode 100755 index 000000000..1a7030964 --- /dev/null +++ b/680/CH12/EX12.11/12_11.sce @@ -0,0 +1,45 @@ +//Problem 12.11:
+
+//initializing the variables:
+//Antoine Eq Coeff for ethanol
+Ae = 8.1122;
+Be = 1592.864;
+Ce = 226.184;
+//Antoine Eq Coeff for toulene
+At = 6.95805;
+Bt = 1346.773;
+Ct = 219.693;
+p = 760; // mm of Hg
+R = 1.987;
+
+//calculation:
+//The saturation temperatures:
+Tsat_e = (Be/(Ae - log10(p))) - Ce
+Tsat_t = (Bt/(At - log10(p))) - Ct
+//
+xe = 0.5
+xt = 0.5
+T = xe*Tsat_e + xt*Tsat_t
+//
+pde = 10^(Ae - (Be/(T + Ce)))
+pdt = 10^(At - (Bt/(T + Ct)))
+//
+a = 0.5292
+bet = 713.57
+bte = 1147.86
+//
+tou_et = bet/(R*(T+273))
+tou_te = bte/(R*(T + 273))
+Get = %e^(-1*a*tou_et)
+Gte = %e^(-1*a*tou_te)
+r_e = %e^(0.5^2*(tou_te*(Gte/(xe + xt*Gte))^2 + Get*tou_et/(xt + xe*Get)^2))
+r_t = %e^(0.5^2*(tou_et*(Get/(xt + xe*Get))^2 + Gte*tou_te/(xe + xt*Gte)^2))
+//
+pde = p/(r_e*xe + r_t*xt*pdt/pde)
+//
+Tn = Be/(Ae - log10(pde)) - Ce
+//
+ye = xe*r_e*pde/p
+
+printf("\n\nResult\n\n")
+printf("\n mole fraction at T = %.2f degC, xe = %.3f and ye = %0.3f \n Return to step 2 and use a different value for xe. Continue this until an entire T-x, y diagram is formed. \n A T-x, y diagram for ethanol and toluene, employing the NRTL method can be found in Fig. 12.11\n To generate an x–y diagram, simply plot the xe as the ordinate and ye as the abscissa.",T, xe, ye)
diff --git a/680/CH12/EX12.11/Ex_12_11_image.jpg b/680/CH12/EX12.11/Ex_12_11_image.jpg Binary files differnew file mode 100755 index 000000000..de3c65b6b --- /dev/null +++ b/680/CH12/EX12.11/Ex_12_11_image.jpg diff --git a/680/CH12/EX12.12/12_12.sce b/680/CH12/EX12.12/12_12.sce new file mode 100755 index 000000000..6fd46dd0e --- /dev/null +++ b/680/CH12/EX12.12/12_12.sce @@ -0,0 +1,35 @@ +//Problem 12.12:
+
+//initializing the variables:
+//Antoine Eq Coeff for Methanol
+Am = 16.5938;
+Bm = 3644.3;
+Cm = 239.76;
+//Antoine Eq Coeff for water
+Aw = 16.262;
+Bw = 3799.89;
+Cw = 226.35;
+R = 1.987;
+T = 40; //in degC
+
+//calculation:
+xm = 0.3
+xw = 0.7
+pdm = %e^(Am - (Bm/(T + Cm)))
+pdw = %e^(Aw - (Bw/(T + Cw)))
+//
+a = 0.2994
+bmw = -253.88
+bwm = 845.21
+//
+tou_mw = bmw/(R*(T+273))
+tou_wm = bwm/(R*(T + 273))
+Gmw = %e^(-1*a*tou_mw)
+Gwm = %e^(-1*a*tou_wm)
+r_m = %e^(0.5^2*(tou_wm*(Gwm/(xm + xw*Gwm))^2 + Gmw*tou_mw/(xw + xm*Gmw)^2))
+r_w = %e^(0.5^2*(tou_mw*(Gmw/(xw + xm*Gmw))^2 + Gwm*tou_wm/(xm + xw*Gwm)^2))
+p = xm*pdm + (1 - xm)*pdw
+ym = xm*r_m*pdm/p
+
+printf("\n\nResult\n\n")
+printf("\n mole fraction at T = %.0f degC, xe = %.3f and ye = %0.3f \n To generate a P-x, y diagram, plot xm and ym data as the ordinate and pressure as the abscissa (see Fig. 12.13).",T, xm, ym)
diff --git a/680/CH12/EX12.13/12_13.sce b/680/CH12/EX12.13/12_13.sce new file mode 100755 index 000000000..f754bfb58 --- /dev/null +++ b/680/CH12/EX12.13/12_13.sce @@ -0,0 +1,8 @@ +//Problem 12.13:
+
+//initializing the variables:
+
+//calculation:
+
+printf("\n\nResult\n\n")
+printf("\n Combining the curves generated from both methods into one figure (see Fig. 12.14), \n it can be observed that the plot generated using Raoult’s law gives lower values \n of pressure at the same xm values that the NRTL method gives for higher values. Also the \n bubble point curve from Raoult’s law is (as expected) a straight line compared to the curve generated \n by the NRTL method, which is concave down.")
diff --git a/680/CH12/EX12.14/12_14.sce b/680/CH12/EX12.14/12_14.sce new file mode 100755 index 000000000..684dcf66e --- /dev/null +++ b/680/CH12/EX12.14/12_14.sce @@ -0,0 +1,49 @@ +//Problem 12.14:
+
+//initializing the variables:
+//Antoine Eq Coeff for ethanol
+Ae = 8.1122;
+Be = 1592.864;
+Ce = 226.184;
+//Antoine Eq Coeff for toulene
+At = 6.95805;
+Bt = 1346.773;
+Ct = 219.693;
+p = 760; // mm of Hg
+R = 1.987;
+
+//calculation:
+//The saturation temperatures:
+Tsat_e = (Be/(Ae - log10(p))) - Ce
+Tsat_t = (Bt/(At - log10(p))) - Ct
+//
+xe = 0.5
+xt = 0.5
+T = xe*Tsat_e + xt*Tsat_t
+//
+pde = 10^(Ae - (Be/(T + Ce)))
+pdt = 10^(At - (Bt/(T + Ct)))
+//
+a = 0.5292
+bet = 713.57
+bte = 1147.86
+//
+Ve = 58.68
+Vt = 106.85
+aet = 1556.45
+ate = 210.52
+//
+E_et = (Vt/Ve)*%e^(-aet/(R*(T+273)))
+E_te = (Ve/Vt)*%e^(-ate/(R*(T+273)))
+//
+r_e = %e^(-log(xe + xt*E_et) + xt*(E_et/(xe + xt*E_et) - E_te/(xt + xe*E_te)))
+r_t = %e^(-log(xt + xe*E_te) + xe*(E_te/(xt + xe*E_te) - E_et/(xe + xt*E_et)))
+//
+pde = p/(r_e*xe + r_t*xt*(pdt/pde))
+//
+Tn = Be/(Ae - log10(pde)) - Ce
+//
+ye = xe*r_e*pde/p
+
+printf("\n\nResult\n\n")
+printf("\n mole fraction at T = %.2f degC, xe = %.3f and ye = %0.2f \n Return to step 2 and use a different value for xe. Continue this until an entire T-x, y diagram is formed. \n A T-x, y diagram for ethanol and toluene employingWilson’s method can be found in Fig. 12.15.\n Note that an azeotrope is formed at x = y = 0.8. Generate an x–y diagram from the results obtained above.\n Refer to Table 12.18 for the x–y data. To generate an x–y diagram, simply plot the xe as the ordinate and ye as the abscissa.",T, xe, ye)
diff --git a/680/CH12/EX12.16/12_16.sce b/680/CH12/EX12.16/12_16.sce new file mode 100755 index 000000000..b16a242a3 --- /dev/null +++ b/680/CH12/EX12.16/12_16.sce @@ -0,0 +1,12 @@ +//Problem 12.16:
+
+//initializing the variables:
+pB = 35; // mm of Hg at 0 deg F
+
+//calculation:
+//from example 12.15
+pA = 70.01; // in mm of Hg
+aAB = pA/pB
+
+printf("\n\nResult\n\n")
+printf("\n the relative volatility is %.0f",aAB)
\ No newline at end of file diff --git a/680/CH13/EX13.04/13_04.sce b/680/CH13/EX13.04/13_04.sce new file mode 100755 index 000000000..f9dda3108 --- /dev/null +++ b/680/CH13/EX13.04/13_04.sce @@ -0,0 +1,11 @@ +//Problem 13.04:
+
+//initializing the variables:
+DG0fH2O = -54635; // cal/gmol
+DG0fHCl = -22778; // cal/gmol
+
+//calculation:
+DG0298 = 1*DG0fH2O - 2*DG0fHCl
+
+printf("\n\nResult\n\n")
+printf("\n DG0298 of reaction is %.0f cal/gmol",DG0298)
\ No newline at end of file diff --git a/680/CH13/EX13.05/13_05.sce b/680/CH13/EX13.05/13_05.sce new file mode 100755 index 000000000..96be5a379 --- /dev/null +++ b/680/CH13/EX13.05/13_05.sce @@ -0,0 +1,12 @@ +//Problem 13.05:
+
+//initializing the variables:
+DG0fCH3COOH = -93800; // cal/gmol
+DG0fCH4 = -12140; // cal/gmol
+DG0fCO2 = -94258; // cal/gmol
+
+//calculation:
+DG0298 = DG0fCH3COOH - DG0fCH4 - DG0fCO2
+
+printf("\n\nResult\n\n")
+printf("\n DG0298 of reaction is %.0f cal/gmol",DG0298)
\ No newline at end of file diff --git a/680/CH13/EX13.06/13_06.sce b/680/CH13/EX13.06/13_06.sce new file mode 100755 index 000000000..ef363db49 --- /dev/null +++ b/680/CH13/EX13.06/13_06.sce @@ -0,0 +1,13 @@ +//Problem 13.06:
+
+//initializing the variables:
+DG0 = -20000; // in cal/gmol
+Tf = 70; // in deg F
+R = 1.99; // cal/gmol.K
+
+//calculation:
+Tk = 273 + 5*(Tf - 32)/9
+K = %e^(-1*DG0/(R*Tk))
+
+printf("\n\nResult\n\n")
+printf("\n chemical reaction equilibrium constant K is %.2E",K)
\ No newline at end of file diff --git a/680/CH13/EX13.07/13_07.sce b/680/CH13/EX13.07/13_07.sce new file mode 100755 index 000000000..111969636 --- /dev/null +++ b/680/CH13/EX13.07/13_07.sce @@ -0,0 +1,15 @@ +//Problem 13.07:
+
+//initializing the variables:
+DG0fO2 = 0; // cal/gmol
+DG0fCO = -32781; // cal/gmol
+DG0fCO2 = -94258; // cal/gmol
+Tk = 298; // in K
+R = 1.987; // cal/gmol.K
+
+//calculation:
+DG0 = DG0fCO - 0.5*DG0fO2 - DG0fCO2
+K = %e^(-1*DG0/(R*Tk))
+
+printf("\n\nResult\n\n")
+printf("\n chemical reaction equilibrium constant K is %.2E",K)
\ No newline at end of file diff --git a/680/CH13/EX13.10/13_10.sce b/680/CH13/EX13.10/13_10.sce new file mode 100755 index 000000000..4e3835ed4 --- /dev/null +++ b/680/CH13/EX13.10/13_10.sce @@ -0,0 +1,12 @@ +//Problem 13.10:
+
+//initializing the variables:
+Tk = 1394.3; // in K
+
+//calculation:
+//from problem 13.09
+//lnK = (-33722/T) + 1.560*lnT - 0.00181*T + 2.42E-7*T^2 + 0.4509
+K = %e^((-1*33722)/Tk + 1.560*log(Tk) - 0.00181*Tk + 2.42E-7*Tk^2 + 0.4509)
+
+printf("\n\nResult\n\n")
+printf("\n chemical reaction equilibrium constant K is %.2E",K)
\ No newline at end of file diff --git a/680/CH13/EX13.11/13_11.sce b/680/CH13/EX13.11/13_11.sce new file mode 100755 index 000000000..3202a40ad --- /dev/null +++ b/680/CH13/EX13.11/13_11.sce @@ -0,0 +1,39 @@ +//Problem 13.11:
+
+//initializing the variables:
+A = 0.229E-3;
+B = 7340;
+T = 298; // in K
+R = 1.99; // in cal/gmol.K
+
+//calculation:
+//The following two results are provided from Illustrative Example 13.4:
+DG0298 = -9079 // in cal/gmol
+DH0298 = -13672 // cal/gmol
+//Employ Equation
+//lnK = -DH0/RT + (Da/R)*lnT + (Db/2R)*T + (Dr/6R)*T^-2 + I
+//Next, DH0 and I must be determined.
+//DH0T = DH0298 + int(298,T)(DCpdT)
+//For heat capacities of the form
+//Cp = a + bT + cT^-2
+//Table 7.4 can be employed to generate the following terms:
+Da = (7.30 + 8.85) - [(2*6.27 + 0.5*7.16)]
+Db = 2.46E-3 + 0.16E-3 - [2*1.24E-3 + 0.5*1.0E-3]
+Dc = 0.0 - 0.68E5 - [2*0.30E5 + 0.5*(-0.4E5)]
+//From this, Equation then becomes:
+//DH0T = DH0298 + int(298,T)[Da + (Db)T + (Dc)T^-2] dT
+//or
+//DH0T = DH0298 + Da(T - 298) + (1/2)*Db[T^2 - (298)^2] - Dc(1/T - 1/298)
+//Combining the constant terms into DH0 (as in Chapter 10) yields the following:
+//DH0T = DH0 + (Da)T + (1/2)*(Db)T^2 -(Dc)T6-1
+//where
+DH0 = DH0298 - 298*Da - (1/2)*[(298)^2]*Db + (1/298)*Dc
+//From Equation (13.16)
+lnK = -1*DG0298/(R*T)
+//Therefore,
+I = lnK - DH0/(R*T) - Da/R*log(T) + Db/(2*R)*T + Dc/(2*R)*T^-2
+//The final form of the equation for K is
+//ln(K) = 7048.7/T + (0.0151)*lnT - (9.06E-5)*T - (2.714E4)*T^-2 - 8.09
+
+printf("\n\nResult\n\n")
+printf("\n The final form of the equation for K is \n ln(K) = (%.1f)/T + (%.4f)*lnT - (%.2E)*T - (%.3E)*T^-2 - 8.09",-1*DH0/1.99, Da/1.99, abs(Db)/(2*1.99), abs(Dc)/(2*1.99))
diff --git a/680/CH13/EX13.12/13_12.sce b/680/CH13/EX13.12/13_12.sce new file mode 100755 index 000000000..0027448fe --- /dev/null +++ b/680/CH13/EX13.12/13_12.sce @@ -0,0 +1,12 @@ +//Problem 13.12:
+
+//initializing the variables:
+Tk = 500; // in K
+
+//calculation:
+//from problem 13.11
+//lnK = (7048.7/T) + 0.0151*lnT - 9.06E-5*T - 2.714E4*T^-2 - 8.09
+K = %e^((7048.7/Tk) + 0.0151*log(Tk) - 9.06E-5*Tk - 2.714E4/Tk^2 - 8.09)
+
+printf("\n\nResult\n\n")
+printf("\n chemical reaction equilibrium constant K is %.0f",K)
\ No newline at end of file diff --git a/680/CH13/EX13.13/13_13.sce b/680/CH13/EX13.13/13_13.sce new file mode 100755 index 000000000..3d8931f57 --- /dev/null +++ b/680/CH13/EX13.13/13_13.sce @@ -0,0 +1,47 @@ +//Problem 13.13:
+
+//initializing the variables:
+T = 600; // in K
+P = 1; // atm
+K = 1.5E32;
+DH0 = 103525; // cal/gmol O2
+IR = 3.5
+aC2H4 = 2.830
+aO2 = 6.148
+bC2H4 = 28.601E-3
+bO2 = 3.102E-3
+rC2H4 = -8.726E-6
+rO2 = -0.923E-6
+DH0C2H4O = -39760; // cal/gmol O2
+DH0C2H4 = 12496; // cal/gmol O2
+R = 1.987; // cal/gmol.K
+
+//calculation:
+DH0 = 2*DH0C2H4O - 2*DH0C2H4 // in cal/gmol O2 reacted
+DG0298 = -96484; // in cal/gmol O2 reacted
+DH0 = -103525; // in cal/gmol
+//Write the equation for DH0T at 298K in terms of DH0, Da, Db, and Dr:
+//DH0T = DH0 + Da*T + (Db/2)*T^2 + (Dr/3)*T^3
+//
+//-987 = 298*Da + 44402*Db + 8.82E6*Dr
+//
+//Write the equation for DG0T at 298K in terms of DH0, Da, Db, Dr, and IR. At T = 298K and IR = 3.5,
+//DG0T = DH0 -Da*T*lnT - (Db/2)*T^2 - (Dr/6)*T^3 - IRT
+//
+//-8084 = 1698*Da + 44402*Db + 4.41E6*Dr
+//
+DG0600 = -1*R*T*log(K)
+//at T = 600
+//
+//-17275 = 3839*Da + 1.8E5*Db + 3.6E7*Dr
+//
+//solving these we get
+Da = -5.046
+Db = 1.017E-2
+Dr = 7.406E-6
+aC2H4O = (Da + 2*aC2H4 + aO2)/2
+bC2H4O = (Db + 2*bC2H4 + bO2)/2
+rC2H4O = (Dr + 2*rC2H4 + rO2)/2
+
+printf("\n\nResult\n\n")
+printf("\n a = %.3f, b = %.2E and r = %.2E",aC2H4O, bC2H4O, rC2H4O)
\ No newline at end of file diff --git a/680/CH13/EX13.14/13_14.sce b/680/CH13/EX13.14/13_14.sce new file mode 100755 index 000000000..c12e7d74c --- /dev/null +++ b/680/CH13/EX13.14/13_14.sce @@ -0,0 +1,47 @@ +//Problem 13.14:
+
+//initializing the variables:
+T = 600; // in K
+P = 1; // atm
+K = 1.5E32;
+DH0 = 103525; // cal/gmol O2
+IR = 3.5
+aC2H4 = 2.830
+aO2 = 6.148
+bC2H4 = 28.601E-3
+bO2 = 3.102E-3
+rC2H4 = -8.726E-6
+rO2 = -0.923E-6
+DH0C2H4O = -39760; // cal/gmol O2
+DH0C2H4 = 12496; // cal/gmol O2
+R = 1.987; // cal/gmol.K
+
+//calculation:
+DH0 = 2*DH0C2H4O - 2*DH0C2H4 // in cal/gmol O2 reacted
+DG0298 = -96484; // in cal/gmol O2 reacted
+DH0 = -103525; // in cal/gmol
+//Write the equation for DH0T at 298K in terms of DH0, Da, Db, and Dr:
+//DH0T = DH0 + Da*T + (Db/2)*T^2 + (Dr/3)*T^3
+//
+//-987 = 298*Da + 44402*Db + 8.82E6*Dr
+//
+//Write the equation for DG0T at 298K in terms of DH0, Da, Db, Dr, and IR. At T = 298K and IR = 3.5,
+//DG0T = DH0 -Da*T*lnT - (Db/2)*T^2 - (Dr/6)*T^3 - IRT
+//
+//-8084 = 1698*Da + 44402*Db + 4.41E6*Dr
+//
+DG0600 = -1*R*T*log(K)
+//at T = 600
+//
+//-17275 = 3839*Da + 1.8E5*Db + 3.6E7*Dr
+//
+//solving these we get
+Da = -5.046
+Db = 1.017E-2
+Dr = 7.406E-6
+aC2H4O = (Da + 2*aC2H4 + aO2)/2
+bC2H4O = (Db + 2*bC2H4 + bO2)/2
+rC2H4O = (Dr + 2*rC2H4 + rO2)/2
+
+printf("\n\nResult\n\n")
+printf("\n theoretical values: a = %.3f, b = %.2E and r = %.2E \n Experimental values: \n\ta = 3.364\n\tb = 35.722E-3\n\tr = -12.236E-6 \n\tThe agreement for a and b is excellent; the result is reasonable for r \n in view of its sensitivity to T^3.",aC2H4O, bC2H4O, rC2H4O)
diff --git a/680/CH14/EX14.02/14_02.sce b/680/CH14/EX14.02/14_02.sce new file mode 100755 index 000000000..8263593db --- /dev/null +++ b/680/CH14/EX14.02/14_02.sce @@ -0,0 +1,17 @@ +//Problem 14.02:
+
+//initializing the variables:
+
+//calculation:
+//For this reaction
+//R1 = CO2;
+//R2 = H2;
+//R3 = CH3OH;
+//R4 = H2O;
+n10 = 3
+n20 = 1
+n30 = 0
+n40 = 1
+
+printf("\n\nResult\n\n")
+printf("\n n1 = %.0f - e\n n2 = %.0f - 3e \n n3 = %.0f + e \n n4 = %.0f + e",n10,n20,n30,n40)
diff --git a/680/CH14/EX14.04/14_04.sce b/680/CH14/EX14.04/14_04.sce new file mode 100755 index 000000000..29418e997 --- /dev/null +++ b/680/CH14/EX14.04/14_04.sce @@ -0,0 +1,14 @@ +//Problem 14.04:
+
+//initializing the variables:
+P = 0.5; // in atm
+e = 0.3;
+
+//calculation:
+p1 = ((3-e)/(5 - 2*e))*P
+p2 = ((1-3*e)/(5 - 2*e))*P
+p3 = ((e)/(5 - 2*e))*P
+p4 = ((1+e)/(5 - 2*e))*P
+
+printf("\n\nResult\n\n")
+printf("\n the partial pressures %0.3f, %0.3f, %0.3f and %0.3f",p1,p2,p3,p4)
\ No newline at end of file diff --git a/680/CH14/EX14.05/14_05.sce b/680/CH14/EX14.05/14_05.sce new file mode 100755 index 000000000..4e2afa7fe --- /dev/null +++ b/680/CH14/EX14.05/14_05.sce @@ -0,0 +1,15 @@ +//Problem 14.05:
+
+//initializing the variables:
+P = 0.5; // in atm
+e = 0.3;
+
+//calculation:
+//From Equation (14.27)
+//K = Ky*P^v
+//y1 = (3-e)/(5-2e) \n y2 = (1-3e)/(5-2e) \n y3 = e/(5-2e) \n y4 = (1+e)/(5-2e)
+v = 1+1-3-1
+
+
+printf("\n\nResult\n\n")
+printf("\n K = [(e/(5-2e))^1*(1+e)/(5-2e))^1/((3-e)/(5-2e))^3*((1-3e)/(5-2e))^1]*P^(%.0f)",v)
diff --git a/680/CH14/EX14.06/14_06.sce b/680/CH14/EX14.06/14_06.sce new file mode 100755 index 000000000..60bc332b7 --- /dev/null +++ b/680/CH14/EX14.06/14_06.sce @@ -0,0 +1,15 @@ +//Problem 14.06:
+
+//initializing the variables:
+xCO2 = 0.0314;
+xO2 = 0.0584;
+P = 1; // in atm
+T = 2050; // in deg F
+
+//calculation:
+//from example 13.10, at 2050 deg F
+K = 9.156E-7
+yCO = xCO2*K/xO2^0.5
+
+printf("\n\nResult\n\n")
+printf("\n the mole fraction of CO is %.2E",yCO)
\ No newline at end of file diff --git a/680/CH14/EX14.07/14_07.sce b/680/CH14/EX14.07/14_07.sce new file mode 100755 index 000000000..73ccc9290 --- /dev/null +++ b/680/CH14/EX14.07/14_07.sce @@ -0,0 +1,15 @@ +//Problem 14.07:
+
+//initializing the variables:
+xCO2 = 0.0314;
+xO2 = 0.0584;
+P = 1; // in atm
+T = 250; // in deg F
+K = 1.015E-33;
+
+//calculation:
+yCO = xCO2*K/xO2^0.5
+yCOppm = yCO*1E6
+
+printf("\n\nResult\n\n")
+printf("\n the mole fraction of CO is %.2E and in ppm %.2E ppm",yCO, yCOppm)
\ No newline at end of file diff --git a/680/CH14/EX14.08/14_08.sce b/680/CH14/EX14.08/14_08.sce new file mode 100755 index 000000000..0ec2bfae8 --- /dev/null +++ b/680/CH14/EX14.08/14_08.sce @@ -0,0 +1,34 @@ +//Problem 14.08:
+
+//initializing the variables:
+pCl2 = 0.0;
+pO2 = 0.106;
+pH2O = 0.0292;
+pHCl = 0.146;
+P = 1; // in atm
+T = 1250; // in K
+
+//calculation:
+//By definition,
+//K = Kp = pCl2*pH2O/(pHCl^2*pO2^0.5)
+//At equilibrium
+//pCl2 = pCl2(initial) + x = x
+//The term x represents the increase in the partial pressure of the chlorine due to this equilibrium reaction. In addition,
+//pH2O = pH2O(initial) + x = 0.0292 + x
+//pHCl = pHCl(initial) - 2x = 0.146 - 2x
+//pO2 = pO2(initial) - 0.5x = 0.106 - 0.5x
+//Kp can then be expressed as
+//Kp = (x)*(0.0292 + x)/((0.146 - 2x)^2*(00.106 - 0.5x)^0.5)
+//Now, calculate Kp at 1250K using the result from the equation above.
+lnK = 7048.7/1250 + 0.015*log(1250) - 1250*9.06E-5 - (2.714E4)*1250^-2 - 8.09
+K = %e^lnK
+Kp = K
+//Therefore,
+//0.0842 = (x)*(0.0292 + x)/((0.146 - 2x)^2*(00.106 - 0.5x)^0.5)
+//Solving for x, which is the equilibrium partial pressure of Cl2, by a trial-and-error calculation yields
+x = 0.01050 // in atm
+pCl2 = x
+//Note that approximately 1% of the discharge flue gas is chlorine—a rather sizable amount
+
+printf("\n\nResult\n\n")
+printf("\n the equilibrium partial pressure of Cl2 %.5f atm",pCl2)
diff --git a/680/CH14/EX14.09/14_09.sce b/680/CH14/EX14.09/14_09.sce new file mode 100755 index 000000000..1c3a44ad1 --- /dev/null +++ b/680/CH14/EX14.09/14_09.sce @@ -0,0 +1,17 @@ +//Problem 14.09:
+
+//initializing the variables:
+P = 1; // in atm
+T = 527; // in deg C
+R = 1.987;
+
+//calculation:
+//DG0T = 53424 - 2.6*T*lnT + 0.0005*T^2 - 5.0*T
+Tk = T + 273
+DG0527 = 53424 - 2.6*Tk*log(Tk) + 0.0005*Tk^2 - 5.0*Tk
+K = %e^(-DG0527/(R*Tk))
+Ky = K
+e = ((Ky/4)/(1 + Ky/4))^0.5
+
+printf("\n\nResult\n\n")
+printf("\n the degree of dissociation is %.2E",e)
\ No newline at end of file diff --git a/680/CH14/EX14.10/14_10.sce b/680/CH14/EX14.10/14_10.sce new file mode 100755 index 000000000..9d1465d2e --- /dev/null +++ b/680/CH14/EX14.10/14_10.sce @@ -0,0 +1,18 @@ +//Problem 14.10:
+
+//initializing the variables:
+P = 10; // in atm
+T = 527; // in deg C
+R = 1.987;
+v = 1-2;
+
+//calculation:
+//DG0T = 53424 - 2.6*T*lnT + 0.0005*T^2 - 5.0*T
+Tk = T + 273
+DG0527 = 53424 - 2.6*Tk*log(Tk) + 0.0005*Tk^2 - 5.0*Tk
+K = %e^(-DG0527/(R*Tk))
+Ky = K*P^v
+e = ((Ky/4)/(1 + Ky/4))^0.5
+
+printf("\n\nResult\n\n")
+printf("\n the degree of dissociation is %.2E",e)
\ No newline at end of file diff --git a/680/CH14/EX14.11/14_11.sce b/680/CH14/EX14.11/14_11.sce new file mode 100755 index 000000000..e31e789b2 --- /dev/null +++ b/680/CH14/EX14.11/14_11.sce @@ -0,0 +1,22 @@ +//Problem 14.11:
+
+//initializing the variables:
+P = 2; // in atm
+T1 = 373; // in K
+nA = 1;
+nB = 2.5;
+nC = 2;
+T2 = 273; // in K
+R = 1.987;
+v = 1 - 2 -1;
+
+//calculation:
+n = nA + nB + nC
+yA = nA/n
+yB = nB/n
+yC = nC/n
+K = (yC/(yB*yA^2))*P^v
+DG0273 = R*T2*log(K)
+
+printf("\n\nResult\n\n")
+printf("\n standard free energy change of this reaction is %.2f cal/gmol",DG0273)
\ No newline at end of file diff --git a/680/CH14/EX14.12/14_12.sce b/680/CH14/EX14.12/14_12.sce new file mode 100755 index 000000000..6e992e65c --- /dev/null +++ b/680/CH14/EX14.12/14_12.sce @@ -0,0 +1,14 @@ +//Problem 14.12:
+
+//initializing the variables:
+T = 500; // in K
+R = 1.987;
+
+//calculation:
+//DG0T = -13600 + 4.16*T //cal/gmol of N2O4
+DG0500 = -13600 + 4.16*T
+K = %e^(-1*DG0500/(R*T))
+e = ((K/4)/(1 + K/4))^0.5
+
+printf("\n\nResult\n\n")
+printf("\n conversion is %.2f percent",e*100)
\ No newline at end of file diff --git a/680/CH14/EX14.13/14_13.sce b/680/CH14/EX14.13/14_13.sce new file mode 100755 index 000000000..5c4d2c95c --- /dev/null +++ b/680/CH14/EX14.13/14_13.sce @@ -0,0 +1,28 @@ +//Problem 14.13:
+
+//initializing the variables:
+T = 150; // in deg C
+P = 1.5; // in atm
+nH2O = 0.9
+nC2H4 = 0.1
+DG0150c = 2375 // in cal/gmol
+FH0298k = -10000; // in cal/gmol
+R = 1.987;
+v = 1-1-1
+
+//calculation:
+Tk = T + 273
+K = %e^(-1*DG0150c/(R*Tk))
+//yC2H4 = (1-e)/(10-e)
+//yH2O = (9-e)/(10-e)
+//yC2H5OH = (e)/(10-e)
+//KP = yC2H5OH/(yC2H4*yH2O)
+//K = e*(10-e)/(2*(1 - e)*(9 - e))
+// 1.12*e^2 - 11.2*e + 1.08 = 0
+e = 0.0966
+yC2H4 = (1-e)/(10-e)
+yH2O = (9-e)/(10-e)
+yC2H5OH = (e)/(10-e)
+
+printf("\n\nResult\n\n")
+printf("\n the equilibrium product mole fraction composition of C2H4 is %.4f, of H2O is %.4f and of C2H5OH is %.3f",yC2H4,yH2O,yC2H5OH)
\ No newline at end of file diff --git a/680/CH14/EX14.15/14_15.sce b/680/CH14/EX14.15/14_15.sce new file mode 100755 index 000000000..7ef8fd4a8 --- /dev/null +++ b/680/CH14/EX14.15/14_15.sce @@ -0,0 +1,46 @@ +//Problem 14.15:
+
+//initializing the variables:
+K1 = 152;
+K2 = 665;
+P = 1; // in atm
+nA0 = 1;
+a = 1;
+nB0 = 3;
+b = 4;
+nC0 = 0;
+c = 2;
+g = 1;
+nD0 = 0;
+d =2;
+nE0 = 1;
+e = 2;
+nF0 = 0;
+f = 1;
+
+//calculation:
+//For 1 atm and ideal conditions
+//K1 = YC^c*YD^d/(YA^a*YB^b) = 152
+//K2 = YF^f/(yC^c*YE^e) = 665
+//For this case,
+n0 = nA0 +nB0 + nC0 + nD0 + nE0 + nF0
+//n = n0 + (c+d-b-a)*e1 + (f-e-g)*e2
+//n = 5 - e1 - 2e2
+//with
+//YA = (1-e1)/n
+//YB = (3-4e1)/n
+//YC = (2e1-e2)/n
+//YD = 2e1/n
+//YE = (1-2e2)/n
+//YF = e2/n
+//The reader is left the exercise of solving for the two unknown e1 and e2. Note also that there are \n two equations: one for K1 and one for K2.
+//The following three constraints apply:
+//e1 < 0.512
+//e1 < 0.75
+//e2 < 0.5
+//Answers:
+e1 = 0.622
+e2 = 0.402
+
+printf("\n\nResult\n\n")
+printf("\n e1 = %.3f and e2 = %.3f",e1,e2)
diff --git a/680/CH15/EX15.01/15_01.sce b/680/CH15/EX15.01/15_01.sce new file mode 100755 index 000000000..dc840394d --- /dev/null +++ b/680/CH15/EX15.01/15_01.sce @@ -0,0 +1,14 @@ +//Problem 15.01:
+
+//initializing the variables:
+FECI93 = 391.2
+FECI95 = 425.4
+FECI99 = 434.1
+c93 = 245000; // in $
+
+//calculation:
+c95 = c93*FECI95/FECI93
+c99 = c93*FECI99/FECI93
+
+printf("\n\nResult\n\n")
+printf("\n Cost(1995) is %.0f $ and Cost(1999) is %.0f $",c95,c99)
\ No newline at end of file diff --git a/680/CH15/EX15.02/15_02.sce b/680/CH15/EX15.02/15_02.sce new file mode 100755 index 000000000..ef467de3b --- /dev/null +++ b/680/CH15/EX15.02/15_02.sce @@ -0,0 +1,23 @@ +//Problem 15.02:
+
+//initializing the variables:
+
+//calculation:
+//The first step is to convert the equipment, installation, and operating costs to total\ncosts by multiplying each by the total gas flow, 100,000 acfm. Hence, for the finned exchanger,
+//the total costs are
+Equipmentcost = 100000*3.1 // in $
+Installationcost = 100000*0.80 // in $
+Operatingcost = 100000*0.06 // in $
+//Note that the operating costs are on an annualized basis. The equipment cost and the installation\ncost must then be converted to an annual basis using the CRF. From Equation (15.3)
+CRF = (0.1)*(1+0.1)^20/[(1+0.1)^20 - 1]
+//The annual costs for the equipment and the installation is given by the product of the CRF and\nthe total costs of each:
+Equipmentannualcost = 0.11746*Equipmentcost
+Installationannualcost = 0.11746*Installationcost
+//The calculations for the 4-pass and the 2-pass exchangers are performed in the same manner.\nThe three preheaters can be compared after all the annual costs are added. The tabulated results\nare provided in Table 15.5.
+// total annual costs
+CF = 65000
+C4 = 77385
+C2 = 60111
+
+printf("\n\nResult\n\n")
+printf("\n According to the analysis, Total Annual Costs for Finned exchanger = $%.0f, for 4-Pass Exchanger = $%.0f and for 2-Pass Exchanger = $%.0f.\nTherefore 2-pass exchanger is the most economically attractive device since the annual cost is the lowest.",CF,C4,C2)
diff --git a/680/CH15/EX15.03/15_03.sce b/680/CH15/EX15.03/15_03.sce new file mode 100755 index 000000000..5e231f3e5 --- /dev/null +++ b/680/CH15/EX15.03/15_03.sce @@ -0,0 +1,18 @@ +//Problem 15.03:
+
+//initializing the variables:
+
+//calculation:
+//For both units
+CRF = (0.12)*(1+0.12)^12/[(1+0.12)^12 - 1]
+//Annual capital and installation costs for the liquid injection (LI) unit are
+LIcosts = (2625000 + 1575000)*0.1614
+//Annual capital and installation costs for the rotary kiln (RK) unit are
+RKcosts = (2975000 + 1700000)*0.1614
+//A comparison of costs and credits for both incinerators is given in Table 15.7.
+
+PLI = 272000
+PRK = 420000
+
+printf("\n\nResult\n\n")
+printf("\n Profits for Liquid Injection = $%.0f/yr and for Rotary Kiln = $%.0f/yr.\nTherefore a rotary kiln incinerator is recommended.",PLI,PRK)
diff --git a/680/CH15/EX15.04/15_04.sce b/680/CH15/EX15.04/15_04.sce new file mode 100755 index 000000000..48523113e --- /dev/null +++ b/680/CH15/EX15.04/15_04.sce @@ -0,0 +1,34 @@ +//Problem 15.04:
+
+//initializing the variables:
+A = 10;
+B = 100000;
+
+//calculation:
+// Since there are two contributing factors to the cost model, one may write the \n following equation for the profit, P
+//P = A(t - tc) - B/(TH - t);
+TH = 500
+tc = 100
+//For breakeven operation, set P = 0 so that
+//(t - tc)*(TH - t) = B/A
+//This may be rewritten as
+//t^2 - (TH + tc)*t + [(B/A) + TH*tc] = 0
+//The solution to this quadratic equation for A and B, is
+t1 = (TH+tc)/2 + ([(TH+tc)^2 - 4*1*(B/A + TH*tc)]^0.5)/2
+t2 = (TH+tc)/2 - ([(TH+tc)^2 - 4*1*(B/A + TH*tc)]^0.5)/2
+//To maximize the profit, take the first deriative of P with respect to t and set it equal to zero, i.e.,
+//dP/dt = A - B/(TH - t)^2 = 0
+//Solving,
+//(TH - t)^2 = B/A
+t = TH - (B/A)^0.5
+//Upon analyzing the first derivative with t values greater than and less than 400 degF, one observes\n that the derivative changes sign from + ---> - at about t = 400, indicating a relative maximum.
+//Similarly, for A = 10, B = 4000,
+//tBE = 499 degF, 101 degF
+//tmax = 480 degF
+//For A = 10, B = 400,000,
+//tBE = 300 degF
+//tmax = 300 degF
+//Graphical results for the three scenarios is shown in Fig. 15.2.
+
+printf("\n\nResult\n\n")
+printf("\n Graphical results for the three scenarios is shown in Fig 15.2")
diff --git a/680/CH15/EX15.04/Ex_15_4_Fig_15_2_Heat_Exchanger_Results.jpg b/680/CH15/EX15.04/Ex_15_4_Fig_15_2_Heat_Exchanger_Results.jpg Binary files differnew file mode 100755 index 000000000..0db1c52f4 --- /dev/null +++ b/680/CH15/EX15.04/Ex_15_4_Fig_15_2_Heat_Exchanger_Results.jpg diff --git a/680/CH17/EX17.03/17_03.sce b/680/CH17/EX17.03/17_03.sce new file mode 100755 index 000000000..0f8276a30 --- /dev/null +++ b/680/CH17/EX17.03/17_03.sce @@ -0,0 +1,37 @@ +//Problem 17.03:
+
+//initializing the variables:
+p = 50; // psig
+T = 25; // in deg C
+Tk = 25+273; // in K
+Pi = 5; //in psig
+R = 1.987; // in cal/gmol.K
+
+//calculation:
+//combustion reaction equation for 1 mole of pentane with stoichiometric air:
+//C5H12 + 8O2 + [30.1N2] ---> 5CO2 + 6H2O + [30.1N2]
+// note that
+nO2 = 8
+nC5H12 = 1
+nCO2 = 5
+nH2O = 6
+nN2 = (0.79/0.21)*nO2
+//number of moles initially and finally present, and the change in the number of moles:
+nin = nC5H12 + nO2 + nN2
+nout = nCO2 + nH2O + nN2
+//the constant volume heat capacity as a function of the constant pressure heat capacity
+//(see Table 17.2):
+//Cv = Cp -R
+//the change in internal energy by integrating the internal energy change equation, i.e.,
+//dU = CvdT
+//The integrated form of the left-hand side below is provided on the right-hand side
+//int(Tin,Tout)[E(niCv,i)out]dT = [(4.542)(30.1) + (3.329)(5) + (4.893)(6)](Tout - 298) + [(0.149E-2)(30.1) + (1.42E-2)(5) + (0.345E-2)(6)](Tout^2 - 298^2) + [(-0.0227E-5)(30.1) + (-0.8362E-5)(5) + (-0.0483E5)(6)](Tout^3 - 298^3)/3 + [(1.784E9)(5)](Tout^4 - 298^4)/4
+
+//Solve the equation obtained on the RHS of the above equation for Tout by trial-and-error until the \n equation has the value of the internal energy of reaction at 25 degC given previously:
+//By trial-and-error calculation,
+Tout = 2870
+//the final pressure in the vessel:
+Pf = (14.7+Pi)*(41.1/39.1)*(Tout/Tk) - 14.7
+
+printf("\n\nResult\n\n")
+printf("\n Since the final pressure of %.1f psig is less\nthan the burst pressure of 200 psig, the vessel can withstand the explosion. ",Pf)
diff --git a/680/CH17/EX17.04/17_04.sce b/680/CH17/EX17.04/17_04.sce new file mode 100755 index 000000000..0eaa7b375 --- /dev/null +++ b/680/CH17/EX17.04/17_04.sce @@ -0,0 +1,41 @@ +//Problem 17.04:
+
+//initializing the variables:
+e = 0.5
+m = 2200; // in kg
+sigma = 5.67E-8; // in J/m2.K4.s
+DH0fp = -103.85; // in kJ/gmol
+DH0fo = 0; // in kJ/gmol
+DH0fn = 0; // in kJ/gmol
+DH0fc = -393.51; // in kJ/gmol
+DH0fw = -241.826; // in kJ/gmol
+Cpp = 0; // in J/gmol.K
+Cpo = 33.635; // in J/gmol.K
+Cpn = 31.840; // in J/gmol.K
+Cpc = 50.919; // in J/gmol.K
+Cpw = 39.672; // in J/gmol.K
+p = 50; // psig
+T = 25; // in deg C
+R = 1.987; // in cal/gmol.K
+I = 10; // in kW/m2
+i = 4; // kW/m2
+
+//calculation:
+mO2 = m*2.2*10*32*2/44
+mt = mO2 + m*2.2
+D = 9.56*(mt)^0.325
+t = 0.196*mt^0.349
+DH0 = (3*DH0fc + 4*DH0fw -5*DH0fo - DH0fp)*1000*1000*m/44 // in J
+np = m/44
+no = 5*np
+nc = 3*np
+nw = 4*np
+nn = 500*0.79/0.21
+DT = DH0/((no*Cpo + nw*Cpw + nc*Cpc + nn*Cpn)*1000)
+T2 = abs(DT) + T + 273
+rsq = (D/(3.048*2))^2*(e*sigma*(T2^4)/I)
+r = (rsq*2.7778E-7*3600)^0.5*10/3
+ri = ((D/(3.048*2))^2*(e*sigma*(T2^4)/i)*2.7778E-7*3600)^0.5*10/3
+
+printf("\n\nResult\n\n")
+printf("\n the size of the fireball is %.0f ft and duration of the fireball is %.1f sec, for I = 10 kW/m2 r = %.0f ft and for I = 4 kW/m2 r = %.0f ft ",D,t,r, ri)
diff --git a/680/CH19/EX19.02/19_02.sce b/680/CH19/EX19.02/19_02.sce new file mode 100755 index 000000000..9994a9d0d --- /dev/null +++ b/680/CH19/EX19.02/19_02.sce @@ -0,0 +1,26 @@ +//Problem 19.02:
+
+//initializing the variables:
+T = 400; // in deg F
+P = 3; // in atm
+To = 70; // in deg F
+Po = 1; // in atm
+Cp = 6.986; // Btu/lbmol.degR
+Cv = 5.000; // in Btu/lbmol.degR
+A = 0.5; // in ft2
+m = 100; // in lbf
+R = 0.732; // in ft3.atm/lbmol.degR
+R1 = 1.986; // in Btu/lbmol.degR
+
+//calculation:
+V = 1*R*(T + 460)/P
+Veq = 1*R*(To + 460)/Po
+dV = V - Veq
+dU = 1*Cv*(T - To)
+dS = 1*Cp*log((T + 460)/(To + 460)) + 1*R1*log(Po/P)
+X = dU + Po*2.74*dV - (To+460)*dS
+Peq = Po*14.7 + m/(A*144)
+Veq1 = 1*10.73*(To + 460)/Peq
+
+printf("\n\nResult\n\n")
+printf("\n the exergy of the system is %.0f Btu and Veq = %.1f ft^3 ",X, Veq1)
\ No newline at end of file diff --git a/680/CH19/EX19.03/19_03.sce b/680/CH19/EX19.03/19_03.sce new file mode 100755 index 000000000..6b0cb9d89 --- /dev/null +++ b/680/CH19/EX19.03/19_03.sce @@ -0,0 +1,25 @@ +//Problem 19.03:
+
+//initializing the variables:
+mdt = 40000; // in kg/hr
+Ti = 500; // in deg C
+Pi = 2500; // in kPa
+Qdt = -0.25; // in MW
+To = 175; // in deg C
+Po = 250; // in kPa
+Tr = 25; // in deg C
+Pr = 101.325; // in kPa
+H1 = 3461.7; // in kJ/kg
+H2 = 2816.7; // in kJ/kg
+Heq = 104.8; // in kJ/kg
+S1 = 7.324; // inkJ/kg.K
+Seq = 0.367; // inkJ/kg.K
+
+//calculation:
+mdt = mdt/3600
+Q = Qdt*1000/mdt
+Ws = H2 - H1 - Q
+X1 = H1 - Heq - (Tr+273)*(S1 - Seq)
+
+printf("\n\nResult\n\n")
+printf("\n the shaftwork produced by the turbine is %.1f kJ/kg and the exergy = %.1f kJ/kg ",abs(Ws), X1)
\ No newline at end of file diff --git a/680/CH19/EX19.04/19_04.sce b/680/CH19/EX19.04/19_04.sce new file mode 100755 index 000000000..b820a133a --- /dev/null +++ b/680/CH19/EX19.04/19_04.sce @@ -0,0 +1,24 @@ +//Problem 19.04:
+
+//initializing the variables:
+mdt = 40000; // in kg/hr
+Ti = 500; // in deg C
+Pi = 2500; // in kPa
+Qdt = -0.25; // in MW
+To = 175; // in deg C
+Po = 250; // in kPa
+Tr = 25; // in deg C
+Pr = 101.325; // in kPa
+H1 = 3461.7; // in kJ/kg
+H2 = 2816.7; // in kJ/kg
+Heq = 104.8; // in kJ/kg
+S1 = 7.324; // inkJ/kg.K
+Seq = 0.367; // inkJ/kg.K
+S2 = 7.289; // in kJ/kg.K
+
+//calculation:
+Wsrev = H2 - H1 - (Tr+273)*(S2 - S1)
+
+
+printf("\n\nResult\n\n")
+printf("\n the maximum shaftwork attainable by the steam turbine is %.1f kJ/kg ",abs(Wsrev))
\ No newline at end of file diff --git a/680/CH2/EX2.02/2_02.sce b/680/CH2/EX2.02/2_02.sce new file mode 100755 index 000000000..d8126d82e --- /dev/null +++ b/680/CH2/EX2.02/2_02.sce @@ -0,0 +1,17 @@ +//Problem 2.02:
+
+//initializing the variables:
+f1 = 55; // in °F
+c2 = 55; // in °C
+
+//calculation:
+r1 = f1 + 460
+c1 = (f1 - 32)*5/9
+k1 = (f1 + 460)*5/9
+r2 = 1.8*c2 + 492
+f2 = 1.8*c2 + 32
+k2 = c2 + 273
+
+printf("\n\nResult\n\n")
+printf("\n (a) Rankine = %.0f°R, (b) Celsius = %.1f°C and (c) Kelvin = %.0f K\n",r1, c1, k1)
+printf("\n (a) Fahrenheit = %.1f°F, (b) Rankine = %.0f°R, and (c) Kelvin = %.0f K\n",f2, r2, k2)
diff --git a/680/CH2/EX2.03/2_03.sce b/680/CH2/EX2.03/2_03.sce new file mode 100755 index 000000000..36faea80d --- /dev/null +++ b/680/CH2/EX2.03/2_03.sce @@ -0,0 +1,18 @@ +//Problem 2.03:
+
+//initializing the variables:
+mg = 100; // in lb
+Pg = 35; // in psig
+A = 3; // in in2
+gc = 1; // in lb/lbf
+Pa = 14.7; // in psi
+
+//calculation:
+F = mg/gc
+Pli = F/A // in lbf/in2
+Plf = Pli*144 // in lbf/ft2
+P = Pg + Pa
+
+printf("\n\nResult\n\n")
+printf("\n pressure at the base is %.0f lbf/ft2\n",Plf)
+printf("\n absolute pressure is %.1f psia\n",P)
diff --git a/680/CH2/EX2.04/2_04.sce b/680/CH2/EX2.04/2_04.sce new file mode 100755 index 000000000..5d9969257 --- /dev/null +++ b/680/CH2/EX2.04/2_04.sce @@ -0,0 +1,19 @@ +//Problem 2.04:
+
+//initializing the variables:
+V = 55; // in gal
+W = 20; // in lb
+AWH = 1.008; // in lb/lbmol
+AWO = 15.999; // in lb/lbmol
+Na = 6.023E23; // molecules/gmol
+
+//calculation:
+MW = 2*AWH + AWO
+pmw = W/MW
+gmw = W*454/MW
+nm = gmw*Na
+
+printf("\n\nResult\n\n")
+printf("\n it contain %.2f lbmol of water\n",pmw)
+printf("\n it contain %.1f gmol of water\n",gmw)
+printf("\n it contain %.3E molecules of water\n",nm)
diff --git a/680/CH2/EX2.05/2_05.sce b/680/CH2/EX2.05/2_05.sce new file mode 100755 index 000000000..6e8cc6362 --- /dev/null +++ b/680/CH2/EX2.05/2_05.sce @@ -0,0 +1,11 @@ +//Problem 2.05:
+
+//initializing the variables:
+sgm = 0.92;
+dw = 62.4; // in lb/ft3
+
+//calculation:
+dm = sgm*dw
+
+printf("\n\nResult\n\n")
+printf("\n the density of methanol is %.1f lb/ft3\n",dm)
diff --git a/680/CH2/EX2.06/2_06.sce b/680/CH2/EX2.06/2_06.sce new file mode 100755 index 000000000..0d89cd2c5 --- /dev/null +++ b/680/CH2/EX2.06/2_06.sce @@ -0,0 +1,13 @@ +//Problem 2.06:
+//initializing the variables:
+sg = 0.8
+abvis = 0.02;// in cP
+pref = 62.43; // in lb/ft3
+
+//calculation:
+p = sg*pref
+u = abvis*6.720E-4; // in lb/ft.sec
+v = u/p
+
+printf("\n\nResult\n\n")
+printf("\n kinematic viscosity of a gas is %.3E ft2/sec\n",v)
diff --git a/680/CH2/EX2.07/2_07.sce b/680/CH2/EX2.07/2_07.sce new file mode 100755 index 000000000..4b06f059b --- /dev/null +++ b/680/CH2/EX2.07/2_07.sce @@ -0,0 +1,9 @@ +//Problem 2.07:
+//initializing the variables:
+hc = 0.61 // in cal/g-°C
+
+//calculation:
+hce = hc*452/(252*1.8) // in Btu/lb-°F
+
+printf("\n\nResult\n\n")
+printf("\n Heat capacity in english units is %.2f Btu/lb.°F\n",hce)
diff --git a/680/CH2/EX2.08/2_08.sce b/680/CH2/EX2.08/2_08.sce new file mode 100755 index 000000000..01173f014 --- /dev/null +++ b/680/CH2/EX2.08/2_08.sce @@ -0,0 +1,10 @@ +//Problem 2.08:
+
+//initializing the variables:
+tc = 0.0512 // in cal/m.s.°C
+
+//calculation:
+k = tc*0.3048*3600/(252*1.8) // in Btu/ft.h.°F
+
+printf("\n\nResult\n\n")
+printf("\n thermal conductivity in english units is %.3f Btu/ft.h.°F\n",k)
diff --git a/680/CH2/EX2.09/2_09.sce b/680/CH2/EX2.09/2_09.sce new file mode 100755 index 000000000..7fac1f612 --- /dev/null +++ b/680/CH2/EX2.09/2_09.sce @@ -0,0 +1,17 @@ +//Problem 2.09:
+
+//initializing the variables:
+D = 5/12; // in ft
+p = 50; // in lb/ft3
+v = 10; // in fps
+u = 0.65*6.720E-4 ; //in lb/ft.sec
+
+//calculation:
+Re = D*p*v/u
+if (Re>2100) then
+ s = 'turbulent';
+else
+ s = 'laminar';
+end
+printf("\n\nResult\n\n")
+printf("\n Reynolds number for a fluid flowing is %.2E and the flow is %s\n",Re, s)
diff --git a/680/CH2/EX2.10/2_10.sce b/680/CH2/EX2.10/2_10.sce new file mode 100755 index 000000000..6335adbb1 --- /dev/null +++ b/680/CH2/EX2.10/2_10.sce @@ -0,0 +1,11 @@ +//Problem 2.10:
+
+//initializing the variables:
+pH = 1
+
+//calculation:
+CH = 10^(-1*pH)
+COH = 10^(-14)/CH
+
+printf("\n\nResult\n\n")
+printf("\n hydroxyl ion concentration of an aqueous solution is %.2E g.ion/L and of hydrogen ion is %.1f g.ion/L\n",COH, CH)
diff --git a/680/CH2/EX2.11/2_11.sce b/680/CH2/EX2.11/2_11.sce new file mode 100755 index 000000000..9c28db56c --- /dev/null +++ b/680/CH2/EX2.11/2_11.sce @@ -0,0 +1,18 @@ +//Problem 2.11:
+
+//initializing the variables:
+w= 5000; // in gal
+C = 50000; // in gal
+Cs = 45000; // in gal
+pHmin = 6;
+pHn = 7;
+
+//calculation:
+CHn = 10^(-1*pHn)
+CH = 10^(-1*pHmin)
+X = (C/w)*[CH - Cs*CHn/C]
+pH = -1*log10(X)
+
+printf("\n\nResult\n\n")
+printf("\n the pH of the most acidic waste shipment is %.2f \n",pH)
+printf("\n This is the final correct answer, final answer in book is wrong\n")
\ No newline at end of file diff --git a/680/CH3/EX3.01/3_01.sce b/680/CH3/EX3.01/3_01.sce new file mode 100755 index 000000000..833e7f76e --- /dev/null +++ b/680/CH3/EX3.01/3_01.sce @@ -0,0 +1,12 @@ +//Problem 3.01:
+
+//initializing the variables:
+Tc = 100; // in °F
+Tf = 300; // in °F
+qi = 3500; // in acfm
+
+//calculation:
+qf = qi*(Tf + 460)/(Tc + 460)
+
+printf("\n\nResult\n\n")
+printf("\n the final (f) volumetric flow rate of a gas is %.0f acfm\n",qf)
\ No newline at end of file diff --git a/680/CH3/EX3.02/3_02.sce b/680/CH3/EX3.02/3_02.sce new file mode 100755 index 000000000..6bd7bf4fd --- /dev/null +++ b/680/CH3/EX3.02/3_02.sce @@ -0,0 +1,12 @@ +//Problem 3.02:
+
+//initializing the variables:
+Pi = 1.0; // in atm
+Pf = 3.0; // in atm
+qi = 3500; // in acfm
+
+//calculation:
+qf = qi*Pi/Pf
+
+printf("\n\nResult\n\n")
+printf("\n the final (f) volumetric flow rate of a gas is %.0f acfm\n",qf)
\ No newline at end of file diff --git a/680/CH3/EX3.03/3_03.sce b/680/CH3/EX3.03/3_03.sce new file mode 100755 index 000000000..edd2e989a --- /dev/null +++ b/680/CH3/EX3.03/3_03.sce @@ -0,0 +1,14 @@ +//Problem 3.03:
+
+//initializing the variables:
+Pi = 1.0; // in atm
+Pf = 3.0; // in atm
+Tc = 100; // in °F
+Tf = 300; // in °F
+qi = 3500; // in acfm
+
+//calculation:
+qf = qi*(Pi/Pf)*((Tf + 460)/(Tc + 460))
+
+printf("\n\nResult\n\n")
+printf("\n the final (f) volumetric flow rate of a gas is %.0f acfm\n",qf)
\ No newline at end of file diff --git a/680/CH3/EX3.04/3_04.sce b/680/CH3/EX3.04/3_04.sce new file mode 100755 index 000000000..50a81a689 --- /dev/null +++ b/680/CH3/EX3.04/3_04.sce @@ -0,0 +1,23 @@ +//Problem 3.04:
+
+//initializing the variables:
+T1 = 75; // in °F
+Pa = 14.7; // in psia
+MWair = 29;
+T2 = 60; // in °F
+T3 = 20;// in °C
+P = 1.2; // in atm
+MWgas = 29;
+Rf = 10.73; // in ft^3.psi/lbmol.°R
+Rc = 82.06; // in cm^3.atm/lbmol.K
+
+n = 1; // in lbmol
+//calculation:
+p1 = Pa*MWair/((T1 + 460)*Rf)
+V = n*Rf*(T2 + 460)/Pa
+p2 = P*MWgas/(Rc*(T3 + 273))
+
+printf("\n\nResult\n\n")
+printf("\n density of air is %.4f lb/ft^3",p1)
+printf("\n the volume is %.0f ft^3",V)
+printf("\n density of gas is %.5f g/cm^3",p2)
\ No newline at end of file diff --git a/680/CH3/EX3.05/3_05.sce b/680/CH3/EX3.05/3_05.sce new file mode 100755 index 000000000..3d3bde117 --- /dev/null +++ b/680/CH3/EX3.05/3_05.sce @@ -0,0 +1,13 @@ +//Problem 3.05:
+
+//initializing the variables:
+Vsp = 10.58; // in ft3/lb
+Pa = 14.7; // in psia
+T = 70; // in °F
+R = 10.73; // in ft^3.psi/lbmol.°R
+
+//calculation:
+MW = R*(460 + T)/(Vsp*Pa)
+
+printf("\n\nResult\n\n")
+printf("\n the molecular weight of the gas is %.2f lb/lbmol",MW)
\ No newline at end of file diff --git a/680/CH3/EX3.06/3_06.sce b/680/CH3/EX3.06/3_06.sce new file mode 100755 index 000000000..2785f97c8 --- /dev/null +++ b/680/CH3/EX3.06/3_06.sce @@ -0,0 +1,13 @@ +//Problem 3.06:
+
+//initializing the variables:
+qs = 30000; // in scfm at 60 °F
+P = 1; // in atm
+Ts = 60; // in °F
+Ta = 1100; // in °F
+
+//calculation:
+qa = qs*(Ta + 460)/(Ts + 460)
+
+printf("\n\nResult\n\n")
+printf("\n flow rate in actual cubic feet per minute is %.0f acfm",qa)
\ No newline at end of file diff --git a/680/CH3/EX3.07/3_07.sce b/680/CH3/EX3.07/3_07.sce new file mode 100755 index 000000000..af863f35e --- /dev/null +++ b/680/CH3/EX3.07/3_07.sce @@ -0,0 +1,15 @@ +//Problem 3.07:
+
+//initializing the variables:
+qs = 1000; // in acfm
+A = 1; // in ft2
+P = 1; // in atm
+Ts = 70; // in °F
+Ta = 300; // in °F
+
+//calculation:
+qa = qs*(Ta + 460)/(Ts + 460)
+v = qa/P
+
+printf("\n\nResult\n\n")
+printf("\n velocity of the gas is %.0f ft/min",v)
\ No newline at end of file diff --git a/680/CH3/EX3.08/3_08.sce b/680/CH3/EX3.08/3_08.sce new file mode 100755 index 000000000..3f17ef383 --- /dev/null +++ b/680/CH3/EX3.08/3_08.sce @@ -0,0 +1,12 @@ +//Problem 3.08:
+
+//initializing the variables:
+pco = 0.15; // in mm of Hg
+P = 760; // in mm of Hg
+
+//calculation:
+yco = pco/P
+ppm = yco*1E6
+
+printf("\n\nResult\n\n")
+printf("\n the parts per million of CO in the exhaust is %.0f ppm",ppm)
\ No newline at end of file diff --git a/680/CH3/EX3.09/3_09.sce b/680/CH3/EX3.09/3_09.sce new file mode 100755 index 000000000..57cbf7600 --- /dev/null +++ b/680/CH3/EX3.09/3_09.sce @@ -0,0 +1,17 @@ +//Problem 3.09:
+
+//initializing the variables:
+T = 230; // in deg celcius
+P = 2500; // in psia
+Pa = 14.7; // in psia
+
+//calculation:
+//critical values
+Tc = 417 // in K
+Pc = 76 // in atm
+w = 0.074 // acentric factor
+Tr = (T + 273)/Tc
+Pr = P/(Pa*Pc)
+
+printf("\n\nResult\n\n")
+printf("\n the reduced temperature is %.2f and reduced pressure is %.2f",Tr,Pr)
\ No newline at end of file diff --git a/680/CH3/EX3.10/3_10.sce b/680/CH3/EX3.10/3_10.sce new file mode 100755 index 000000000..4a776cbf9 --- /dev/null +++ b/680/CH3/EX3.10/3_10.sce @@ -0,0 +1,20 @@ +//Problem 3.10:
+
+//initializing the variables:
+B = -0.159; // in m3/kgmol
+C = 0.009; // in (m3/kgmol)2
+T = 400; // in K
+P = 40; // in atm
+
+//calculation:
+//Virial equation.
+//Z = PV/RT = 1 + B/V + C/V^2
+//Insert the appropriate values of the terms and coefficients. Use R = 0.082 Latm/gmolK = 82.06 cm3.atm/gmol K
+//40*V=(0.082)(400) = 1 + (-0.159)/V +(0.009)/V^2
+//(1.22)(V) = 1 + (-0.159)/V + (0.009)/V^2
+//Note that the equation cannot simply be explicitly solved for V. A trial-and-error solution is \n required and any suitable numerical (or analytical) technique may be employed.
+//V is approximately 0.635 L/gmol
+V = 0.635
+
+printf("\n\nResult\n\n")
+printf("\n the specific volume is %.3f L/gmol",V)
diff --git a/680/CH3/EX3.12/3_12.sce b/680/CH3/EX3.12/3_12.sce new file mode 100755 index 000000000..40245d656 --- /dev/null +++ b/680/CH3/EX3.12/3_12.sce @@ -0,0 +1,21 @@ +//Problem 3.12:
+
+//initializing the variables:
+Pc = 45.4; // in atm
+Tc = 343; // in deg R
+T = 373; // in K
+P = 10; // in atm
+w = 0.007
+
+//calculation:
+//Redlich–Kwong equation in terms of a, b, and V.
+//P = [RT/(V - b)] - a/[T^0.5 * V(V + b)]
+T = T*1.8
+//10 = [(0.73)(671)/(V - b)] - a/[671^0.5*V(V + b)]
+a = 10933
+b = 0.478
+//from these we get and By trial-and-error
+V = 48.8; // in ft^3
+
+printf("\n\nResult\n\n")
+printf("\n the molar volume is %.1f ft^3",V)
diff --git a/680/CH3/EX3.13/3_13.sce b/680/CH3/EX3.13/3_13.sce new file mode 100755 index 000000000..736a0392d --- /dev/null +++ b/680/CH3/EX3.13/3_13.sce @@ -0,0 +1,40 @@ +//Problem 3.13:
+
+//initializing the variables:
+Y1 = 0.4;
+Y2 = 0.1;
+Y3 = 0.3;
+Y4 = 0.07;
+Y5 = 0.07;
+Y6 = 0.06;
+T = 60; // in deg F
+P = 1; // in atm
+w = 0.020
+
+//calculation:
+//The reduced properties are therefore
+Tr = 660/(268.8*1.8)
+Pr = 350/(46.54*14.7)
+//For standard conditions
+Trs = 1.074
+Prs = 0.021
+//Employing the “B” approach:
+B0 = 0.083-[0.422/(1.36^1.6)]
+B1 = 0.139 - [0.172/(1.36^4.2)]
+Za = 1 + (B0+w*B1)*0.511/1.36
+//therefore
+qs1 = 3000*520*350/(14.7*660*Za)
+//The problem can also be solved using the “Z” approach. First note that Tr>1, Pr>1. The following equations from Table 3.4 that are given below are to be employed to solve this problem.
+//Z0 = 1.156 - 0.351e^(-Tr) - 0.0885e^Pr
+//and
+//Z1 = -0.200 + 0.018*Pr*Tr + 0.2*(1-0.2*Pr/Tr + (Pr/Tr)^2 -(Pr/Tr)^3 + (Pr/Tr)^4 - (Pr/Tr)^5)
+//For this approach
+//Z = Z0 + w*Z1
+Z0 = 1.156 - 0.351*%e^(-Tr) - 0.0885*%e^Pr
+Z1 = -0.200 + 0.018*Pr*Tr + 0.2*(1-0.2*Pr/Tr + (Pr/Tr)^2 -(Pr/Tr)^3 + (Pr/Tr)^4 - (Pr/Tr)^5)
+Z = Z0 + w*Z1
+//therefore
+qs2 = 3000*520*350/(14.7*660*Z)
+
+printf("\n\nResult\n\n")
+printf("\n the volume flow by B approach is %.0f ft^3/day and \n by the equations provided by Van Vliet and Domato is %.0f ft^3/day ",qs1,qs2)
diff --git a/680/CH4/EX4.01/4_01.sce b/680/CH4/EX4.01/4_01.sce new file mode 100755 index 000000000..668a51aee --- /dev/null +++ b/680/CH4/EX4.01/4_01.sce @@ -0,0 +1,16 @@ +//Problem 4.01:
+
+//initializing the variables:
+mdt = 0.15; // in kg/sec
+v = 420; // in m/sec
+
+//calculation:
+vxin = v
+vxout = 0
+vyin = 0
+vyout = v
+Fxgc = mdt*(vxout - vxin)
+Fygc = mdt*(vyout - vyin)
+
+printf("\n\nResult\n\n")
+printf("\n The x-direction supporting force is %.1f N and The y-direction supporting force is %.1f N",Fxgc,Fygc)
\ No newline at end of file diff --git a/680/CH4/EX4.02/4_02.sce b/680/CH4/EX4.02/4_02.sce new file mode 100755 index 000000000..7d8c664ea --- /dev/null +++ b/680/CH4/EX4.02/4_02.sce @@ -0,0 +1,18 @@ +//Problem 4.02:
+
+//initializing the variables:
+mdt = 0.15; // in kg/sec
+v = 420; // in m/sec
+
+//calculation:
+vxin = v
+vxout = 0
+vyin = 0
+vyout = v
+Fxgc = mdt*(vxout - vxin)
+Fygc = mdt*(vyout - vyin)
+Fres = (Fxgc^2 + Fygc^2)^0.5
+theta = (atan(Fygc/Fxgc))*180/%pi + 180
+
+printf("\n\nResult\n\n")
+printf("\n resultant supporting force is %.1f N and direction is %.0f degree",Fres,theta)
\ No newline at end of file diff --git a/680/CH4/EX4.03/4_03.sce b/680/CH4/EX4.03/4_03.sce new file mode 100755 index 000000000..4ad40d089 --- /dev/null +++ b/680/CH4/EX4.03/4_03.sce @@ -0,0 +1,13 @@ +//Problem 4.03:
+
+//initializing the variables:
+rb = 10000; // in lb/h
+rair = 20000; // in lb/h
+rm = 2000; // in lb/h
+
+//calculation:
+mdtin = rb + rair + rm
+mdtout = mdtin
+
+printf("\n\nResult\n\n")
+printf("\n the product gases exit the incinerator at %.0f lb/h",mdtout)
\ No newline at end of file diff --git a/680/CH4/EX4.04/4_04.sce b/680/CH4/EX4.04/4_04.sce new file mode 100755 index 000000000..b892a5a56 --- /dev/null +++ b/680/CH4/EX4.04/4_04.sce @@ -0,0 +1,25 @@ +//Problem 4.04:
+
+//initializing the variables:
+r1 = 5000; // in scfm
+r2 = 3000; // in scfm
+T1 = 60; // in deg F
+T2 = 70; // in deg F
+Ti = 2000; // in F
+To = 180; // in F
+MWchcl = 112.5;
+MWair = 29;
+
+//calculation:
+//convert scfm to acfm using Charle's law
+R1 = r1*(460 + T2)/(460 + T1)
+R2 = r2*(460 + T2)/(460 + T1)
+ndt1 = R1/387
+ndt2 = R2/387
+mdt1 = ndt1*MWchcl*60
+mdt2 = ndt2*MWair*60
+mdtin = mdt1 + mdt2
+mdtout = mdtin
+
+printf("\n\nResult\n\n")
+printf("\n products exit the cooler at %.0f lb/h",mdtout)
\ No newline at end of file diff --git a/680/CH4/EX4.05/4_05.sce b/680/CH4/EX4.05/4_05.sce new file mode 100755 index 000000000..2af33c734 --- /dev/null +++ b/680/CH4/EX4.05/4_05.sce @@ -0,0 +1,14 @@ +//Problem 4.05:
+
+//initializing the variables:
+e1 = 0.65;
+e2 = 0.98;
+mdtin = 76; // in lb
+
+//calculation:
+mdtout1 = (1 - e1)*mdtin
+mdtout2 = (1 - e2)*mdtout1
+E = 1 - mdtout2/mdtin
+
+printf("\n\nResult\n\n")
+printf("\n overall fractional efficiency is %.3f",E)
\ No newline at end of file diff --git a/680/CH4/EX4.06/4_06.sce b/680/CH4/EX4.06/4_06.sce new file mode 100755 index 000000000..cf9c65813 --- /dev/null +++ b/680/CH4/EX4.06/4_06.sce @@ -0,0 +1,15 @@ +//Problem 4.06:
+
+//initializing the variables:
+e1 = 0.65;
+e2 = 0.98;
+mdtin = 76; // in lb
+
+//calculation:
+mdtout1 = (1 - e1)*mdtin
+mdtout2 = (1 - e2)*mdtout1
+E = 1 - mdtout2/mdtin
+perE = E*100
+
+printf("\n\nResult\n\n")
+printf("\n overall fractional efficiency at percent basis is %.1f percent",perE)
\ No newline at end of file diff --git a/680/CH4/EX4.07/4_07.sce b/680/CH4/EX4.07/4_07.sce new file mode 100755 index 000000000..89453e345 --- /dev/null +++ b/680/CH4/EX4.07/4_07.sce @@ -0,0 +1,14 @@ +//Problem 4.07:
+
+//initializing the variables:
+mdt1 = 1000; // in lb/min
+mdt2 = 1000; // in lb/min
+mdt3 = 200; // in lb/min
+
+//calculation:
+mdt5 = mdt1 + mdt2 - mdt3
+mdt6 = mdt2
+mdt = mdt5 - mdt6
+
+printf("\n\nResult\n\n")
+printf("\n amount of water lost by evaporation is %.0f lb/min",mdt)
\ No newline at end of file diff --git a/680/CH4/EX4.09/4_09.sce b/680/CH4/EX4.09/4_09.sce new file mode 100755 index 000000000..5676fc343 --- /dev/null +++ b/680/CH4/EX4.09/4_09.sce @@ -0,0 +1,14 @@ +//Problem 4.09:
+
+//initializing the variables:
+m = 1800; // in kg
+v = 40; // in km/h
+F = 5000; // in N
+
+//calculation:
+KE1 = (1/2)*m*(v*5/18)^2
+KE2 = 0;
+s = KE1/F
+
+printf("\n\nResult\n\n")
+printf("\n distance the car will travel before it comes to a stop is %.1f m",s)
\ No newline at end of file diff --git a/680/CH4/EX4.10/4_10.sce b/680/CH4/EX4.10/4_10.sce new file mode 100755 index 000000000..ed4c54504 --- /dev/null +++ b/680/CH4/EX4.10/4_10.sce @@ -0,0 +1,12 @@ +//Problem 4.10:
+
+//initializing the variables:
+m = 2000; // in lb
+d = 1200; // in ft
+
+//calculation:
+PE = m*d/2
+PEbtu = PE/778.17
+
+printf("\n\nResult\n\n")
+printf("\n the change in potential energy is %.0f Btu",PEbtu)
\ No newline at end of file diff --git a/680/CH4/EX4.11/4_11.sce b/680/CH4/EX4.11/4_11.sce new file mode 100755 index 000000000..cf9d068ee --- /dev/null +++ b/680/CH4/EX4.11/4_11.sce @@ -0,0 +1,15 @@ +//Problem 4.11:
+
+//initializing the variables:
+m = 2000; // in lb
+v1 = 8; // in ft/s
+v2 = 30; // in ft/s
+
+//calculation:
+KE1 = m*v1^2/(2*32.2)
+KE2 = m*v2^2/(2*32.2)
+delKE = KE1 - KE2
+delKEbtu = delKE/778.17
+
+printf("\n\nResult\n\n")
+printf("\n the change in Kinetic energy is %.3f Btu",delKEbtu)
\ No newline at end of file diff --git a/680/CH4/EX4.12/4_12.sce b/680/CH4/EX4.12/4_12.sce new file mode 100755 index 000000000..5799ad350 --- /dev/null +++ b/680/CH4/EX4.12/4_12.sce @@ -0,0 +1,14 @@ +//Problem 4.12:
+
+//initializing the variables:
+r = 500000; // in gpm
+e = 0.30;
+d = 3000; // in ft
+
+//calculation:
+mdt = r*0.00378*1000/60 // in kg/sec
+delPE = mdt*9.8*d*0.3048
+P = e*delPE
+
+printf("\n\nResult\n\n")
+printf("\n actual power output is %.2E W",P)
\ No newline at end of file diff --git a/680/CH5/EX5.01/5_01.sce b/680/CH5/EX5.01/5_01.sce new file mode 100755 index 000000000..f21bb7b47 --- /dev/null +++ b/680/CH5/EX5.01/5_01.sce @@ -0,0 +1,16 @@ +//Problem 5.01:
+
+//initializing the variables:
+
+//calculation:
+//chemical equation provides a variety of qualitative and quantitative information \n essential for the calculation of the quantity of reactants reacted and products formed \n in a chemical process. A balanced chemical equation, as noted above, must have the same \n number of atoms of each type in the reactants and products. Thus, the balanced equation for \n butane is
+//C4H10 + (13/2)O2 ---> 4CO2 + 5H2O
+//number of carbons in reactants = number of carbons in products = 4
+//number of oxygens in reactants = number of oxygens in products = 13
+//number of hydrogens in reactants = number of hydrogens in products = 10
+//number of moles of reactants is 1 mol C4H10 + 6.5 mol O2 = 7.5 mol total
+//number of moles of products is 4 mol CO2 + 5 mol H2O = 9 mol total
+ratio = 9/7.5
+
+printf("\n\nResult\n\n")
+printf("\n the mole ratio of reactants to products is %.2f ",ratio)
diff --git a/680/CH5/EX5.03/5_03.sce b/680/CH5/EX5.03/5_03.sce new file mode 100755 index 000000000..635771167 --- /dev/null +++ b/680/CH5/EX5.03/5_03.sce @@ -0,0 +1,25 @@ +//Problem 5.03:
+
+//initializing the variables:
+MWCS2 = 76.14
+MWSO2 = 64.07
+MWCO2 = 44
+WCS2 = 500; // iin lb
+WO2 = 225; // in lb
+
+//calculation:
+MWO2 = 2*16
+//The initial molar amounts of each reactant is
+MACS2 = WCS2/MWCS2
+MAO2 = WO2/MWO2
+//The amount of O2 needed to consume all the CS2, i.e., the stoichiometric amount, is then
+O2 = MACS2*3
+
+if (O2 > MAO2) then
+ a = 'O2 is Limiting Reactant'
+else
+ a = 'CS2 is Limiting Reactant'
+end
+
+printf("\n\nResult\n\n")
+printf("\n %s", a)
diff --git a/680/CH5/EX5.04/5_04.sce b/680/CH5/EX5.04/5_04.sce new file mode 100755 index 000000000..77ec2f18a --- /dev/null +++ b/680/CH5/EX5.04/5_04.sce @@ -0,0 +1,22 @@ +//Problem 5.04:
+
+//initializing the variables:
+MWCS2 = 76.14;
+MWSO2 = 64.07;
+MWCO2 = 44;
+wCS2 = 500; // in lb
+wO2 = 225; // in lb
+MWO2 = 32;
+
+//calculation:
+mr1 = wCS2/MWCS2
+mr2 = wO2/MWO2
+mp1 = mr2/3
+m1r = mp1*MWCS2
+mp2 = mr2*1/3
+m2p = mp2*MWCO2
+mp3 = mr2*2/3
+m3p = mp3*MWSO2
+
+printf("\n\nResult\n\n")
+printf("\n %.0f lb CO2 produced and %.0f lb SO2 produced",m2p,m3p)
\ No newline at end of file diff --git a/680/CH5/EX5.06/5_06.sce b/680/CH5/EX5.06/5_06.sce new file mode 100755 index 000000000..3798bdba5 --- /dev/null +++ b/680/CH5/EX5.06/5_06.sce @@ -0,0 +1,16 @@ +//Problem 5.06:
+
+//initializing the variables:
+P = 1; // in atm
+tm = 68.6
+
+//calculation:
+perO2bymol = 7*100/tm
+perHClbymol = 1*100/tm
+perH2Obymol = 2*100/tm
+ppO2 = perO2bymol/100
+ppHCl = perHClbymol/100
+ppH2O = perH2Obymol/100
+
+printf("\n\nResult\n\n")
+printf("\n patial pressures (for O2 = %.3f atm for HCl = %.4f atm and for H2O = %.4f atm)",ppO2, ppHCl, ppH2O)
\ No newline at end of file diff --git a/680/CH5/EX5.07/5_07.sce b/680/CH5/EX5.07/5_07.sce new file mode 100755 index 000000000..09b6d3c89 --- /dev/null +++ b/680/CH5/EX5.07/5_07.sce @@ -0,0 +1,17 @@ +//Problem 5.07:
+
+//initializing the variables:
+P = 1; // in atm
+tm = 68.6;
+pS = 0.005;
+W = 112.5;
+MWS = 32;
+
+//calculation:
+wS = pS*W
+nS = wS/MWS
+perSO2bymol = nS*100/tm
+ppSO2 = perSO2bymol/100
+
+printf("\n\nResult\n\n")
+printf("\n patial pressures for SO2 = %.2E atm ",ppSO2)
\ No newline at end of file diff --git a/680/CH5/EX5.08/5_08.sce b/680/CH5/EX5.08/5_08.sce new file mode 100755 index 000000000..3323f64be --- /dev/null +++ b/680/CH5/EX5.08/5_08.sce @@ -0,0 +1,18 @@ +//Problem 5.08:
+
+//initializing the variables:
+P = 1; // in atm
+tm = 68.6;
+pS = 0.005;
+W = 112.5;
+MWS = 32;
+
+//calculation:
+wS = pS*W
+nS = wS/MWS
+perSO2bymol = nS*100/tm
+ppSO2 = perSO2bymol/100
+ppm = ppSO2*1E6
+
+printf("\n\nResult\n\n")
+printf("\n ppm of SO2 = %.0f",ppm)
\ No newline at end of file diff --git a/680/CH5/EX5.09/5_09.sce b/680/CH5/EX5.09/5_09.sce new file mode 100755 index 000000000..49e7ee86e --- /dev/null +++ b/680/CH5/EX5.09/5_09.sce @@ -0,0 +1,27 @@ +//Problem 5.09:
+
+//initializing the variables:
+nCO2 = 7.5
+nCO = 1.3
+nO2 = 8.1
+nN2 = 83.1
+
+//calculation:
+//Determine the amount of oxygen fed for combustion. Since nitrogen does not react (key component), using the ratio of oxygen to nitrogen in air will provide the amount of oxygen fed:
+O2f = (21/79)*83.1
+//A balanced equation for the combustion of the hydrocarbon in terms of N moles of the hydrocarbon and n hydrogen atoms in the hydrocarbon yields
+//NC3Hn + 22.1O2 ---> 7.5CO2 + 1.3CO + 8.1O2 + N(n/2)H2O
+//The moles of hydrocarbon, N, is obtained by performing an elemental carbon balance:
+//3N = 7.5 + 1.3
+N = 8.8/3
+//Similarly, the moles of water formed is obtained by performing an elemental oxygen balance:
+//2(22.1) = 2(7.5) + 1.3 + 2(8.1) + N(n/2)
+//A = N(n/2)
+A = 44.2 - 15 - 1.3 - 16.2
+//The number of hydrogen atoms, n, in the hydrocarbon is then
+n = 2*A/N
+//Since n = 8, the hydrocarbon is C3H8, propane.
+
+printf("\n\nResult\n\n")
+printf("\n n= %.0f\n",n)
+printf("\n the hydrocarbon is C3H8, propane")
diff --git a/680/CH5/EX5.11/5_11.sce b/680/CH5/EX5.11/5_11.sce new file mode 100755 index 000000000..ebfd57fdb --- /dev/null +++ b/680/CH5/EX5.11/5_11.sce @@ -0,0 +1,17 @@ +//Problem 5.11:
+
+//initializing the variables:
+wr = 5; // in ton/hr
+pcl = 0.02
+x = 2000
+MWHCl = 36.5
+MWCl = 35.5
+y = 0.99
+
+//calculation:
+Clfeed = wr*pcl*x
+HCl = Clfeed*MWHCl/MWCl
+maxrate = HCl*(1-y)
+
+printf("\n\nResult\n\n")
+printf("\n maximum permissible mass emission rate of HCl = %.2f lb HCl/h",maxrate)
\ No newline at end of file diff --git a/680/CH5/EX5.13/5_13.sce b/680/CH5/EX5.13/5_13.sce new file mode 100755 index 000000000..20bc347b8 --- /dev/null +++ b/680/CH5/EX5.13/5_13.sce @@ -0,0 +1,16 @@ +//Problem 5.13:
+
+//initializing the variables:
+tm = 50.85; // total lbmol from problem 5.12
+T = 500+460;
+P = 1; // in atm
+R = 0.7302
+
+//calculation:
+//Noting that 100 lb of fuel was used as a basis, the total lbmol of flue gas produced per pound of oil burned is
+n = tm/100
+//the total volume of flue gas
+V = n*R*T/P
+
+printf("\n\nResult\n\n")
+printf("\n the total volume of flue gas = %.2f ft3/lboil",V)
\ No newline at end of file diff --git a/680/CH5/EX5.14/5_14.sce b/680/CH5/EX5.14/5_14.sce new file mode 100755 index 000000000..c876cdcd8 --- /dev/null +++ b/680/CH5/EX5.14/5_14.sce @@ -0,0 +1,15 @@ +//Problem 5.14:
+
+//initializing the variables:
+lbmolCO2 = 7.38
+lbmolSO2 = 0.00125
+lbmolN2 = 38.03
+
+//calculation:
+//total lbmol dry flue gas
+tn = lbmolCO2 + lbmolSO2 + lbmolN2
+//volume percentage of CO2 in the dry flue gas
+perCO2 = lbmolCO2*100/tn
+
+printf("\n\nResult\n\n")
+printf("\n volume percentage of CO2 in the dry flue gas = %.2f percent",perCO2)
\ No newline at end of file diff --git a/680/CH5/EX5.15/5_15.sce b/680/CH5/EX5.15/5_15.sce new file mode 100755 index 000000000..32a444169 --- /dev/null +++ b/680/CH5/EX5.15/5_15.sce @@ -0,0 +1,15 @@ +//Problem 5.15:
+
+//initializing the variables:
+MWDCB = 147;
+MWTCB = 290
+
+//calculation:
+//for 1 lb of dichlorobenzene (DCB), the following mass of HCl is produced:
+HCLpd1 = 2/MWDCB
+//for 1lb of tetrachlorobiphenyl (TCB), the following mass of HCl is produced
+HCLpd2 = 4/MWTCB
+x = (HCLpd2 - HCLpd1)*100/HCLpd1
+
+printf("\n\nResult\n\n")
+printf("\n the consumption of soda ash be increased by %.2f percent",x)
\ No newline at end of file diff --git a/680/CH6/EX6.01/6_01.sce b/680/CH6/EX6.01/6_01.sce new file mode 100755 index 000000000..954ce79b6 --- /dev/null +++ b/680/CH6/EX6.01/6_01.sce @@ -0,0 +1,17 @@ +//Problem 6.01:
+
+//initializing the variables:
+mc = 20; // in lb
+T1 = 100; // in degrees C
+T2 = 25; // in Deg C
+mw = 6; // in gallons
+Cpc = 0.092; // Btu/lb.degF
+Cpw = 1.0; // Btu/lb.degF
+
+//calculation:
+T = (mc*Cpc*T1 + mw*8.33*Cpw*T2)/(mc*Cpc + mw*8.33*Cpw)
+Tk = T + 273
+dS = mc*Cpc*log(Tk/373)
+
+printf("\n\nResult\n\n")
+printf("\n the entropy change of the copper is %.3f Btu/deg F",dS)
\ No newline at end of file diff --git a/680/CH6/EX6.02/6_02.sce b/680/CH6/EX6.02/6_02.sce new file mode 100755 index 000000000..f14a32d48 --- /dev/null +++ b/680/CH6/EX6.02/6_02.sce @@ -0,0 +1,17 @@ +//Problem 6.02:
+
+//initializing the variables:
+mc = 20; // in lb
+T1 = 100; // in degrees C
+T2 = 25; // in Deg C
+mw = 6; // in gallons
+Cpc = 0.092; // Btu/lb.degF
+Cpw = 1.0; // Btu/lb.degF
+
+//calculation:
+T = (mc*Cpc*T1 + mw*8.33*Cpw*T2)/(mc*Cpc + mw*8.33*Cpw)
+Tk = T + 273
+dS = mw*8.33*Cpw*log(Tk/298)
+
+printf("\n\nResult\n\n")
+printf("\n the entropy change of the water is %.3f Btu/deg F",dS)
\ No newline at end of file diff --git a/680/CH6/EX6.03/6_03.sce b/680/CH6/EX6.03/6_03.sce new file mode 100755 index 000000000..423b20f75 --- /dev/null +++ b/680/CH6/EX6.03/6_03.sce @@ -0,0 +1,19 @@ +//Problem 6.03:
+
+//initializing the variables:
+mc = 20; // in lb
+T1 = 100; // in degrees C
+T2 = 25; // in Deg C
+mw = 6; // in gallons
+Cpc = 0.092; // Btu/lb.degF
+Cpw = 1.0; // Btu/lb.degF
+
+//calculation:
+T = (mc*Cpc*T1 + mw*8.33*Cpw*T2)/(mc*Cpc + mw*8.33*Cpw)
+Tk = T + 273
+dSw = mw*8.33*Cpw*log(Tk/298)
+dSc = mc*Cpc*log(Tk/373)
+dSt = dSw + dSc
+
+printf("\n\nResult\n\n")
+printf("\n the total entropy change is %.3f Btu/deg F",dSt)
\ No newline at end of file diff --git a/680/CH6/EX6.05/6_05.sce b/680/CH6/EX6.05/6_05.sce new file mode 100755 index 000000000..a83ba2622 --- /dev/null +++ b/680/CH6/EX6.05/6_05.sce @@ -0,0 +1,18 @@ +//Problem 6.05:
+
+//initializing the variables:
+n = 5; // in lbmol
+T1 = 100; // in degrees F
+P1 = 1; // in atm
+T2 = 400; // in degrees F
+P2 = 10; // in atm
+Cpg = 5; // Btu/lb.degF
+R = 1.987;
+
+//calculation:
+T1 = T1 + 460
+T2 = T2 + 460
+dS = n*R*log(P1/P2) + n*Cpg*log(T2/T1)
+
+printf("\n\nResult\n\n")
+printf("\n the entropy for the irreversible process is %.2f Btu/deg R",dS)
\ No newline at end of file diff --git a/680/CH6/EX6.06/6_06.sce b/680/CH6/EX6.06/6_06.sce new file mode 100755 index 000000000..aa9047c3d --- /dev/null +++ b/680/CH6/EX6.06/6_06.sce @@ -0,0 +1,18 @@ +//Problem 6.06:
+
+//initializing the variables:
+n = 5; // in lbmol
+T1 = 100; // in degrees F
+P1 = 1; // in atm
+T2 = 400; // in degrees F
+P2 = 10; // in atm
+Cpg = 5; // Btu/lb.degF
+R = 1.987;
+
+//calculation:
+T1 = T1 + 460
+T2 = T2 + 460
+dS = n*R*log(P1/P2) + n*Cpg*log(T2/T1)
+
+printf("\n\nResult\n\n")
+printf("\n the entropy for the irreversible process is %.2f Btu/deg R",dS)
\ No newline at end of file diff --git a/680/CH6/EX6.07/6_07.sce b/680/CH6/EX6.07/6_07.sce new file mode 100755 index 000000000..bc7bee773 --- /dev/null +++ b/680/CH6/EX6.07/6_07.sce @@ -0,0 +1,20 @@ +//Problem 6.07:
+
+//initializing the variables:
+n = 5; // in lbmol
+T1 = 100; // in degrees F
+P1 = 1; // in atm
+T2 = 400; // in degrees F
+P2 = 10; // in atm
+Cpg = 5; // Btu/lb.degF
+R = 1.987;
+
+//calculation:
+T1 = T1 + 460
+T2 = T2 + 460
+dS = n*R*log(P1/P2) + n*Cpg*log(T2/T1)
+dSt = 0
+dSsur = dSt - dS
+
+printf("\n\nResult\n\n")
+printf("\n the entropy for the surrounding is %.2f Btu/deg R",dSsur)
\ No newline at end of file diff --git a/680/CH6/EX6.09/6_09.sce b/680/CH6/EX6.09/6_09.sce new file mode 100755 index 000000000..15e0c6020 --- /dev/null +++ b/680/CH6/EX6.09/6_09.sce @@ -0,0 +1,25 @@ +//Problem 6.09:
+
+//initializing the variables:
+H0 = 28; // in Btu/lb
+H1 = 1151; // in Btu/lb
+Qh = 700; // in Btu/lb
+S0 = 0.056; // in Btu/lb deg R
+S1 = 1.757; // in Btu/lb deg R
+Th = 300; // in deg F
+Tc = 60; // in deg F
+P1 = 1; // in atm
+T1 = 212; // in deg F
+T0 = 60; // in deg F
+
+
+//calculation:
+Qc = H1 - H0 - Qh
+//the entropy change of the steam
+dSs = S0 - S1
+dSh = Qh/(Th + 460)
+dSc = Qc/(Tc + 460)
+dSt = dSs + dSh + dSc
+
+printf("\n\nResult\n\n")
+printf("\n total entropy change is %.3f Btu/lb deg R",dSt)
\ No newline at end of file diff --git a/680/CH6/EX6.10/6_10.sce b/680/CH6/EX6.10/6_10.sce new file mode 100755 index 000000000..ac926e23e --- /dev/null +++ b/680/CH6/EX6.10/6_10.sce @@ -0,0 +1,21 @@ +//Problem 6.10:
+
+//initializing the variables:
+T1 = 540; // in deg F
+T0 = 300; // in deg F
+T2 = 300; // in deg F
+T3 = 60; // in deg F
+m = 1;
+Cp = 1;
+
+//calculation:
+dSh = m*Cp*log((T0 + 460)/(T1 + 460))
+dSc = m*Cp*log((T2 + 460)/(T3 + 460))
+//for one exchanger
+dSa = dSh + dSc
+//there are two similar exchangers
+dSb = dSa
+dStot = dSa + dSb
+
+printf("\n\nResult\n\n")
+printf("\n total entropy change is %.4f Btu/deg R",dStot)
\ No newline at end of file diff --git a/680/CH6/EX6.11/6_11.sce b/680/CH6/EX6.11/6_11.sce new file mode 100755 index 000000000..429a4fbee --- /dev/null +++ b/680/CH6/EX6.11/6_11.sce @@ -0,0 +1,22 @@ +//Problem 6.11:
+
+//initializing the variables:
+T1 = 540; // in deg F
+T0 = 300; // in deg F
+T2 = 300; // in deg F
+T3 = 60; // in deg F
+TDDF = 0;
+m = 1;
+Cp = 1;
+
+//calculation:
+dShc = m*Cp*log((T0 + 460)/(T1 + 460))
+dScc = m*Cp*log((T2 + 460)/(T3 + 460))
+//for one exchanger
+dSc = dShc + dScc
+//exchanger D
+dSd = 0
+dStot = dSc + dSd
+
+printf("\n\nResult\n\n")
+printf("\n total entropy change is %.4f Btu/deg R",dStot)
\ No newline at end of file diff --git a/680/CH6/EX6.12/6_12.sce b/680/CH6/EX6.12/6_12.sce new file mode 100755 index 000000000..1a9a16ca5 --- /dev/null +++ b/680/CH6/EX6.12/6_12.sce @@ -0,0 +1,22 @@ +//Problem 6.12:
+
+//initializing the variables:
+T1 = 540; // in deg F
+T0 = 300; // in deg F
+T2 = 180; // in deg F
+T3 = 60; // in deg F
+m1 = 1;
+m2 = 2;
+Cp = 1;
+
+//calculation:
+dSh = m1*Cp*log((T0 + 460)/(T1 + 460))
+dSc = m2*Cp*log((T2 + 460)/(T3 + 460))
+//for one exchanger
+dSe = dSh + dSc
+//exchanger F
+dSf = dSe
+dStot = dSe + dSf
+
+printf("\n\nResult\n\n")
+printf("\n total entropy change is %.4f Btu/deg R",dStot)
\ No newline at end of file diff --git a/680/CH6/EX6.14/6_14.sce b/680/CH6/EX6.14/6_14.sce new file mode 100755 index 000000000..1dc2c8f2c --- /dev/null +++ b/680/CH6/EX6.14/6_14.sce @@ -0,0 +1,16 @@ +//Problem 6.14:
+
+//initializing the variables:
+T = 298; // in deg F
+na = 1;
+nb = 3;
+nc = 2;
+Sa = 26.3; // in Btu/lbmol deg R
+Sb = 21.0; // in Btu/lbmol deg R
+Sc = 29.9; // in Btu/lbmol deg R
+
+//calculation:
+dS = nc*Sc - nb*Sb - na*Sa
+
+printf("\n\nResult\n\n")
+printf("\n entropy change is %.1f Btu/deg R",dS)
\ No newline at end of file diff --git a/680/CH7/EX7.01/7_01.sce b/680/CH7/EX7.01/7_01.sce new file mode 100755 index 000000000..ed8c2c106 --- /dev/null +++ b/680/CH7/EX7.01/7_01.sce @@ -0,0 +1,11 @@ +//Problem 7.01:
+
+//initializing the variables:
+C = 1;
+P = 1;
+
+//calculation:
+F = C - P + 2
+
+printf("\n\nResult\n\n")
+printf("\n the number of degrees of freedom is %.0f",F)
\ No newline at end of file diff --git a/680/CH7/EX7.02/7_02.sce b/680/CH7/EX7.02/7_02.sce new file mode 100755 index 000000000..0f33183ed --- /dev/null +++ b/680/CH7/EX7.02/7_02.sce @@ -0,0 +1,13 @@ +//Problem 7.02:
+
+//initializing the variables:
+H200 = 1170; // in Btu/lbmol
+H2000 = 14970; // in Btu/lbmol
+n = 20000; // in scfm
+
+//calculation:
+ndt = n*1/379
+Q = ndt*(H2000 - H200)
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.2E Btu/min",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.03/7_03.sce b/680/CH7/EX7.03/7_03.sce new file mode 100755 index 000000000..f5e3bb653 --- /dev/null +++ b/680/CH7/EX7.03/7_03.sce @@ -0,0 +1,17 @@ +//Problem 7.03:
+
+//initializing the variables:
+H200 = 1170; // in Btu/lbmol
+H2000 = 14970; // in Btu/lbmol
+n = 20000; // in scfm
+Cpav = 7.53; // in Btu/lbmol
+T1 = 200; // in deg F
+T2 = 2000; // in deg F
+
+//calculation:
+dT = T2 - T1
+ndt = n*1/379
+Q = ndt*Cpav*dT
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.2E Btu/min",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.04/7_04.sce b/680/CH7/EX7.04/7_04.sce new file mode 100755 index 000000000..f3af16475 --- /dev/null +++ b/680/CH7/EX7.04/7_04.sce @@ -0,0 +1,14 @@ +//Problem 7.04:
+
+//initializing the variables:
+mdt = 1200; // in lb/min
+Cpav = 0.26; // in Btu/lbmol
+T1 = 200; // in deg F
+T2 = 1200; // in deg F
+
+//calculation:
+dT = T2 - T1
+Q = mdt*Cpav*dT
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.2E Btu/min",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.05/7_05.sce b/680/CH7/EX7.05/7_05.sce new file mode 100755 index 000000000..d53275a4a --- /dev/null +++ b/680/CH7/EX7.05/7_05.sce @@ -0,0 +1,13 @@ +//Problem 7.05:
+
+//initializing the variables:
+Qt = 18.7E6; // in Btu/h
+mdt = 72000; // in lb/h
+Cpav = 0.26; // in Btu/lb.degF
+T1 = 1200; // in deg F
+
+//calculation:
+T2 = [-1*Qt/(mdt*Cpav)]+T1
+
+printf("\n\nResult\n\n")
+printf("\n the outlet temperature of the gas stream is %.0f degF",T2)
diff --git a/680/CH7/EX7.06/7_06.sce b/680/CH7/EX7.06/7_06.sce new file mode 100755 index 000000000..2f8111b23 --- /dev/null +++ b/680/CH7/EX7.06/7_06.sce @@ -0,0 +1,20 @@ +//Problem 7.06:
+
+//initializing the variables:
+n = 1000; // in lb/h
+MWCO2 = 44;
+T1 = 200; // in deg F
+T2 = 3200; // in deg F
+a = 10.57;
+b = 2.10E-3;
+c = -1*2.06E5;
+
+//calculation:
+T1 = (T1 + 460)/1.8
+T2 = (T2 + 460)/1.8
+dT = T2 - T1
+ndt = n/MWCO2
+Q = ndt*1.8*(a*dT +(b/2)*(T2^2 - T1^2) + c*dT/(T2*T1))
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.2E Btu/h",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.07/7_07.sce b/680/CH7/EX7.07/7_07.sce new file mode 100755 index 000000000..a50bd6ffb --- /dev/null +++ b/680/CH7/EX7.07/7_07.sce @@ -0,0 +1,20 @@ +//Problem 7.07:
+
+//initializing the variables:
+n = 1000; // in lb/h
+MWCO2 = 44;
+T1 = 200; // in deg F
+T2 = 3200; // in deg F
+a = 6.214;
+b = 10.396E-3;
+c = -3.545E-6;
+
+//calculation:
+T1 = (T1 + 460)/1.8
+T2 = (T2 + 460)/1.8
+dT = T2 - T1
+ndt = n/MWCO2
+Q = ndt*1.8*(a*dT +(b/2)*(T2^2 - T1^2) + (c/3)*(T2^3 - T1^3))
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.2E Btu/h",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.08/7_08.sce b/680/CH7/EX7.08/7_08.sce new file mode 100755 index 000000000..05437125e --- /dev/null +++ b/680/CH7/EX7.08/7_08.sce @@ -0,0 +1,20 @@ +//Problem 7.08:
+
+//initializing the variables:
+x = 0.5;
+T1 = 10; // in deg F
+T2 = 60; // in deg F
+
+//calculation:
+//qout = 0.5*qin
+//Tout = Tup - 60
+//Tmix = Tup - 10
+//qbyp = qup - qin
+//qmix = qup - 0.5*qin
+//(qup - qin)*Cp*p*(Tup - Tref) + 0.5*qin*Cp*p*(Tup + 60 - Tref) = (qup - 0.5*qin)*Cp*p*(Tup - Tref + 60)
+//Tref = 0
+//soving above we get r = qin/qup
+r = 10/35
+
+printf("\n\nResult\n\n")
+printf("\n %.1f percent of the river flow, qup, is available for cooling",r*100)
\ No newline at end of file diff --git a/680/CH7/EX7.09/7_09.sce b/680/CH7/EX7.09/7_09.sce new file mode 100755 index 000000000..5c63cb071 --- /dev/null +++ b/680/CH7/EX7.09/7_09.sce @@ -0,0 +1,41 @@ +//Problem 7.09:
+
+//initializing the variables:
+F1 = 50000; // in lb/h
+F2 = 60000; // in lb/h
+F3 = 80000; // in lb/h
+F4 = 60000; // in lb/h
+F5 = 40000; // in lb/h
+F6 = 35000; // in lb/h
+Cp1 = 0.65; // in Btu/lb.degF
+Cp2 = 0.58; // in Btu/lb.degF
+Cp3 = 0.78; // in Btu/lb.degF
+Cp4 = 0.70; // in Btu/lb.degF
+Cp5 = 0.52; // in Btu/lb.degF
+Cp6 = 0.60; // in Btu/lb.degF
+Tin1 = 70; // in deg F
+Tin2 = 120; // in deg F
+Tin3 = 90; // in deg F
+Tin4 = 420; // in deg F
+Tin5 = 300; // in deg F
+Tin6 = 240; // in deg F
+Tout1 = 300; // in deg F
+Tout2 = 310; // in deg F
+Tout3 = 250; // in deg F
+Tout4 = 120; // in deg F
+Tout5 = 100; // in deg F
+Tout6 = 90; // in deg F
+
+//calculation:
+Duty1 = F1*Cp1*(Tout1 - Tin1)
+Duty2 = F2*Cp2*(Tout2 - Tin2)
+Duty3 = F3*Cp3*(Tout3 - Tin3)
+Duty4 = F4*Cp4*abs(Tout4 - Tin4)
+Duty5 = F5*Cp5*abs(Tout5 - Tin5)
+Duty6 = F6*Cp6*abs(Tout6 - Tin6)
+heat = Duty1 + Duty2 + Duty3
+cool = Duty4 + Duty5 + Duty6
+steam = heat - cool
+
+printf("\n\nResult\n\n")
+printf("\n As a minimum %.0f Btu/h will have to be supplied by steam or another hot medium",steam)
\ No newline at end of file diff --git a/680/CH7/EX7.11/7_11.sce b/680/CH7/EX7.11/7_11.sce new file mode 100755 index 000000000..d9dc751a7 --- /dev/null +++ b/680/CH7/EX7.11/7_11.sce @@ -0,0 +1,20 @@ +//Problem 7.11:
+
+//initializing the variables:
+n = 1200; // in lb/h
+MWCaO = 56;
+T1 = 42; // in deg F
+T2 = 68; // in deg F
+a = 11.67;
+b = 1.08E-3;
+c = -1*1.565E5;
+
+//calculation:
+T1 = (T1 + 460)/1.8
+T2 = (T2 + 460)/1.8
+dT = T2 - T1
+ndt = n/MWCaO
+Q = ndt*(a*dT +(b)*(T2^2 - T1^2) + (c)*dT/(T1*T2))*1.8
+
+printf("\n\nResult\n\n")
+printf("\n the heat transfer rate is %.0f Btu/h",Q)
\ No newline at end of file diff --git a/680/CH7/EX7.12/7_12.sce b/680/CH7/EX7.12/7_12.sce new file mode 100755 index 000000000..793ae970a --- /dev/null +++ b/680/CH7/EX7.12/7_12.sce @@ -0,0 +1,14 @@ +//Problem 7.12:
+
+//initializing the variables:
+T = 1000; // in K
+a = 0.4687;
+b = 9.4870E-2;
+c = -1*0.5553E-4;
+d = 0.02284E-6;
+
+//calculation:
+Cp0 = a + b*T + c*T^2 + d*T^3
+
+printf("\n\nResult\n\n")
+printf("\n the heat capacity is %.2f Btu/lbmol.degR",Cp0)
\ No newline at end of file diff --git a/680/CH8/EX8.01/8_01.sce b/680/CH8/EX8.01/8_01.sce new file mode 100755 index 000000000..4aaf7c348 --- /dev/null +++ b/680/CH8/EX8.01/8_01.sce @@ -0,0 +1,13 @@ +//Problem 8.01:
+
+//initializing the variables:
+T = 0; // in deg C
+A = 15.03;
+B = 2817;
+
+//calculation:
+T = T + 273
+P = %e^(A - B/T)
+
+printf("\n\nResult\n\n")
+printf("\n the vapor pressure is %.2f mm of Hg",P)
\ No newline at end of file diff --git a/680/CH8/EX8.02/8_02.sce b/680/CH8/EX8.02/8_02.sce new file mode 100755 index 000000000..728344cee --- /dev/null +++ b/680/CH8/EX8.02/8_02.sce @@ -0,0 +1,14 @@ +//Problem 8.02:
+
+//initializing the variables:
+T = 0; // in deg C
+A = 16.65;
+B = 2940;
+C = -35.93;
+
+//calculation:
+T = T + 273
+P = %e^(A - B/(T + C))
+
+printf("\n\nResult\n\n")
+printf("\n the vapor pressure is %.2f mm of Hg",P)
\ No newline at end of file diff --git a/680/CH8/EX8.04/8_04.sce b/680/CH8/EX8.04/8_04.sce new file mode 100755 index 000000000..eaa2ebbac --- /dev/null +++ b/680/CH8/EX8.04/8_04.sce @@ -0,0 +1,24 @@ +//Problem 8.04:
+
+//initializing the variables:
+T1 = 250; // in deg C
+T2 = 260; // in deg C
+T3 = 270; // in deg C
+T4 = 280; // in deg C
+T5 = 290; // in deg C
+P1 = 22.01; // in atm
+P2 = 24.66; // in atm
+P3 = 27.13; // in atm
+P4 = 29.79; // in atm
+P5 = 32.42; // in atm
+vl3 = 0.0408; // in ft3/lb
+vg3 = 0.192; // in ft3/lb
+
+//calculation:
+dpdT = (P4 - P2)/(T4 - T2)
+dv = vg3 - vl3
+T3 = T3 + 460
+dH = T3*dv*dpdT*14.7*144/778
+
+printf("\n\nResult\n\n")
+printf("\n the enthalpy of vaporization is %.2f Btu/lb",dH)
\ No newline at end of file diff --git a/680/CH8/EX8.05/8_05.sce b/680/CH8/EX8.05/8_05.sce new file mode 100755 index 000000000..9a7b2eef9 --- /dev/null +++ b/680/CH8/EX8.05/8_05.sce @@ -0,0 +1,24 @@ +//Problem 8.05:
+
+//initializing the variables:
+T1 = 250; // in deg C
+T2 = 260; // in deg C
+T3 = 270; // in deg C
+T4 = 280; // in deg C
+T5 = 290; // in deg C
+P1 = 22.01; // in atm
+P2 = 24.66; // in atm
+P3 = 27.13; // in atm
+P4 = 29.79; // in atm
+P5 = 32.42; // in atm
+vl3 = 0.0408; // in ft3/lb
+vg3 = 0.192; // in ft3/lb
+
+//calculation:
+dpdT = (P5 - P1)/(T5 - T1)
+dpdT13 = (P3 - P1)/(T3 - T1)
+dpdT35 = (P5 - P3)/(T5 - T3)
+dpdTav = (dpdT13+dpdT35)/2
+
+printf("\n\nResult\n\n")
+printf("\n the p` vs T derivative is %.3f",dpdTav)
diff --git a/680/CH8/EX8.06/8_06.sce b/680/CH8/EX8.06/8_06.sce new file mode 100755 index 000000000..07843c911 --- /dev/null +++ b/680/CH8/EX8.06/8_06.sce @@ -0,0 +1,16 @@ +//Problem 8.06:
+
+//initializing the variables:
+Tc = 647; // in K
+Tn = 100 + 273; // in K
+Pc = 217.6; // in atm
+dHe = 970; // in Btu/lb
+
+//calculation:
+Tm = Tn/Tc
+dH = 2.17*Tn*((log(Pc)) - 1.0)/(0.930 - Tm)
+dHn = dH*454/(252*18)
+perdiff = (dHn - dHe)*100/dHe
+
+printf("\n\nResult\n\n")
+printf("\n the enthalpy of vaporization is %.1f percent",perdiff)
\ No newline at end of file diff --git a/680/CH8/EX8.07/8_07.sce b/680/CH8/EX8.07/8_07.sce new file mode 100755 index 000000000..9d7ab0d8e --- /dev/null +++ b/680/CH8/EX8.07/8_07.sce @@ -0,0 +1,12 @@ +//Problem 8.07:
+
+//initializing the variables:
+Tn = 212 + 460; // in deg R
+R = 1.987
+Tr = 12; // say
+
+//calculation:
+dH = Tr*Tn*R
+
+printf("\n\nResult\n\n")
+printf("\n the enthalpy of vaporization is %.0f Btu/lbmol",dH)
\ No newline at end of file diff --git a/680/CH8/EX8.08/8_08.sce b/680/CH8/EX8.08/8_08.sce new file mode 100755 index 000000000..8a2a5f27d --- /dev/null +++ b/680/CH8/EX8.08/8_08.sce @@ -0,0 +1,15 @@ +//Problem 8.08:
+
+//initializing the variables:
+T1 = 100; // in deg C
+P = 101370; // in Pa
+dH100 = 2256.9; // in J/g
+T2 = 200; // in deg C
+
+//calculation:
+Tr100 = (T1 + 273)/647
+Tr200 = (T2 + 273)/647
+dH200 = dH100*((1 - Tr200)/(1 - Tr100))^0.38
+
+printf("\n\nResult\n\n")
+printf("\n the enthalpy of vaporization is %.0f J/g",dH200)
\ No newline at end of file diff --git a/680/CH8/EX8.09/8_09.sce b/680/CH8/EX8.09/8_09.sce new file mode 100755 index 000000000..4e6067a10 --- /dev/null +++ b/680/CH8/EX8.09/8_09.sce @@ -0,0 +1,14 @@ +//Problem 8.09:
+
+//initializing the variables:
+Tn = 100; // in deg C
+P = 101370; // in Pa
+dHn = 2200; // in kJ/Kg
+Tc = 370; // in deg C
+T = 250; // in deg C
+
+//calculation:
+dH250 = dHn*(1 - ((T - Tn)/(Tc - Tn))^2)
+
+printf("\n\nResult\n\n")
+printf("\n the enthalpy of vaporization is %.0f kJ/kg",dH250)
\ No newline at end of file diff --git a/680/CH8/EX8.10/8_10.sce b/680/CH8/EX8.10/8_10.sce new file mode 100755 index 000000000..51e946940 --- /dev/null +++ b/680/CH8/EX8.10/8_10.sce @@ -0,0 +1,21 @@ +//Problem 8.10:
+
+//initializing the variables:
+Tw = 60; // in deg F
+mdt1 = 1000; // in lb/h
+Pw = 1; // in atm
+Ps = 40; // in atm
+Ts = 1000; // in deg F
+Pd = 20; // in atm
+Td = 600; // in deg F
+H40 = 1572; // in Btu/lb
+H20 = 1316; // in Btu/lb
+H1 = 1151; // in Btu/lb
+Ht60 = 28.1; // in Btu/lb
+
+//calculation:
+dHv = mdt1*(H1 - Ht60)
+mdt2 = dHv/(H40 - H20)
+
+printf("\n\nResult\n\n")
+printf("\n mass flowrate of the utility steam required is %.0f lb/h",mdt2)
\ No newline at end of file diff --git a/680/CH8/EX8.11/8_11.sce b/680/CH8/EX8.11/8_11.sce new file mode 100755 index 000000000..4b1a69833 --- /dev/null +++ b/680/CH8/EX8.11/8_11.sce @@ -0,0 +1,21 @@ +//Problem 8.11:
+
+//initializing the variables:
+Ts = 90; // in deg F
+Cp = 1; // in Btu/lb.deg F
+Cpv = 1030; // in Btu/lb
+Tr = 115; // in deg F
+D1 = 12000000; // in Btu/h
+D2 = 6000000; // in Btu/h
+D3 = 23000000; // in Btu/h
+D4 = 17000000; // in Btu/h
+D5 = 31500000; // in Btu/h
+d = 8.32; // in lb/gal
+
+//calculation:
+Qdt = D1 + D2 + D3 + D4 + D5
+F = Qdt/((Tr - Ts)*Cp)
+Fgpm = F/(60*d)
+
+printf("\n\nResult\n\n")
+printf("\n total flowrate of cooling water is %.0f gal/min",Fgpm)
\ No newline at end of file diff --git a/680/CH8/EX8.12/8_12.sce b/680/CH8/EX8.12/8_12.sce new file mode 100755 index 000000000..a1506d806 --- /dev/null +++ b/680/CH8/EX8.12/8_12.sce @@ -0,0 +1,25 @@ +//Problem 8.12:
+
+//initializing the variables:
+Ts = 90; // in deg F
+Cp = 1; // in Btu/lb.deg F
+Cpv = 1030; // in Btu/lb
+Tr = 115; // in deg F
+D1 = 12000000; // in Btu/h
+D2 = 6000000; // in Btu/h
+D3 = 23000000; // in Btu/h
+D4 = 17000000; // in Btu/h
+D5 = 31500000; // in Btu/h
+d = 8.32; // in lb/gal
+a = 0.05;
+
+//calculation:
+Qdt = D1 + D2 + D3 + D4 + D5
+F = Qdt/((Tr - Ts)*Cp)
+B = a*F
+V = Qdt/Cpv
+M = B + V
+Mgpm = 0.002*M
+
+printf("\n\nResult\n\n")
+printf("\n the sum of blowdown and the amount evaporated is %.0f gal/min",Mgpm)
\ No newline at end of file diff --git a/680/CH8/EX8.13/8_13.sce b/680/CH8/EX8.13/8_13.sce new file mode 100755 index 000000000..6d71a1ba9 --- /dev/null +++ b/680/CH8/EX8.13/8_13.sce @@ -0,0 +1,28 @@ +//Problem 8.13:
+
+//initializing the variables:
+P =500; // in psig
+UHD1 = 10000000; // in Btu/h
+UHD2 = 8000000; // in Btu/h
+UHD3 = 12000000; // in Btu/h
+UHD4 = 20000000; // in Btu/h
+T1 = 250; // in deg F
+T2 = 450; // in deg F
+T3 = 400; // in deg F
+T4 = 300; // in deg F
+Ps1 = 75; // in psig
+Ps2 = 500; // in psig
+Ts1 = 320; // in deg F
+Ts2 = 470; // in deg F
+dHv1 = 894; // in Btu/lb
+dHv2 = 751; // in Btu/lb
+
+//calculation:
+mdtb1 = UHD1/dHv2
+mdtb2 = UHD2/dHv2
+mdtb3 = UHD3/dHv2
+mdtb4 = UHD4/dHv2
+mdtb = mdtb1 + mdtb2 + mdtb3 + mdtb4
+
+printf("\n\nResult\n\n")
+printf("\n steam required is %.0f lb/h",mdtb)
\ No newline at end of file diff --git a/680/CH9/EX9.01/9_01.sce b/680/CH9/EX9.01/9_01.sce new file mode 100755 index 000000000..5079dd5f4 --- /dev/null +++ b/680/CH9/EX9.01/9_01.sce @@ -0,0 +1,17 @@ +//Problem 9.01:
+
+//initializing the variables:
+w = 50; // in lb
+Ws = 200; // in lb
+a = 0.5;
+Ts = 25;// in deg C
+
+//calculation:
+WH2SO4 = w + Ws*a
+WH2O = Ws*a
+perH2SO4 = (WH2SO4/(WH2SO4 + WH2O))*100
+//Referring to Fig. 9.3, construct a straight line between the 50% solution and pure H2SO4 at 25 deg C (77 deg F). Estimate the final temperature in deg F:
+T = 140;// in deg F
+
+printf("\n\nResult\n\n")
+printf("\n the adiabatic temperature change is %.0f deg F",T)
\ No newline at end of file diff --git a/680/CH9/EX9.02/9_02.sce b/680/CH9/EX9.02/9_02.sce new file mode 100755 index 000000000..85382e54b --- /dev/null +++ b/680/CH9/EX9.02/9_02.sce @@ -0,0 +1,21 @@ +//Problem 9.02:
+
+//initializing the variables:
+w = 50; // in lb
+Ws = 200; // in lb
+a = 0.5;
+Ts = 25;// in deg C
+
+//calculation:
+WH2SO4 = w + Ws*a
+WH2O = Ws*a
+perH2SO4 = (WH2SO4/(WH2SO4 + WH2O))*100
+//Referring to Fig. 9.3, construct a straight line between the 50% solution and pure H2SO4 at 25 deg C (77 deg F). Estimate the final temperature in deg F:
+T = 140;// in deg F
+H140 = -86; // in Btu/lb
+H77 = -121.5; // in Btu/lb
+Wt = WH2SO4 + WH2O
+Q = Wt*(H77 - H140)
+
+printf("\n\nResult\n\n")
+printf("\n the heat is %.0f Btu",Q)
\ No newline at end of file diff --git a/680/CH9/EX9.03/9_03.sce b/680/CH9/EX9.03/9_03.sce new file mode 100755 index 000000000..c018bfdaa --- /dev/null +++ b/680/CH9/EX9.03/9_03.sce @@ -0,0 +1,17 @@ +//Problem 9.03:
+
+//initializing the variables:
+w = 65; // in lb
+Ws = 125; // in lb
+a = 0.45;
+Ts = 75;// in deg C
+
+//calculation:
+T = 9*Ts/5 - 32
+wf = (w*0 + a*Ws)/(Ws + w)
+//From Fig. 9.4
+Hfinal = 156 // in Btu/lb
+Tf = 208 // in deg F
+
+printf("\n\nResult\n\n")
+printf("\n the adiabatic temperature change is %.0f deg F",Tf)
\ No newline at end of file diff --git a/680/CH9/EX9.04/9_04.sce b/680/CH9/EX9.04/9_04.sce new file mode 100755 index 000000000..24c444594 --- /dev/null +++ b/680/CH9/EX9.04/9_04.sce @@ -0,0 +1,19 @@ +//Problem 9.04:
+
+//initializing the variables:
+w = 65; // in lb
+Ws = 125; // in lb
+a = 0.45;
+Ts = 75;// in deg C
+
+//calculation:
+T = (9*Ts/5) + 32
+wf = (w*0 + a*Ws)/(Ws + w)
+//From Fig. 9.4
+Hfinal = 156 // in Btu/lb
+Tf = 208 // in deg F
+//From Fig. 9.4, the absolute temperature change, DT, is
+DT = Tf - T
+
+printf("\n\nResult\n\n")
+printf("\n the adiabatic temperature change is %.0f deg F",DT)
\ No newline at end of file diff --git a/680/CH9/EX9.05/9_05.sce b/680/CH9/EX9.05/9_05.sce new file mode 100755 index 000000000..b2e8b08f1 --- /dev/null +++ b/680/CH9/EX9.05/9_05.sce @@ -0,0 +1,19 @@ +//Problem 9.05:
+
+//initializing the variables:
+w = 65; // in lb
+Ws = 125; // in lb
+a = 0.45;
+Ts = 75;// in deg C
+
+//calculation:
+T = (9*Ts/5) + 32
+wf = (w*0 + a*Ws)/(Ws + w)
+//From Fig. 9.4
+Hf208 = 156 // in Btu/lb
+Hf167 = 118 // in Btu/lb
+Q = Hf167 - Hf208
+Qr = Q*(Ws + w)
+
+printf("\n\nResult\n\n")
+printf("\n the the heat effect is %.0f Btu",Qr)
\ No newline at end of file diff --git a/680/CH9/EX9.07/9_07.sce b/680/CH9/EX9.07/9_07.sce new file mode 100755 index 000000000..63a349d5a --- /dev/null +++ b/680/CH9/EX9.07/9_07.sce @@ -0,0 +1,13 @@ +//Problem 9.07:
+
+//initializing the variables:
+
+//calculation:
+//Since enthalpy is a point function, it is reasonable to assume that the temperature effects are additive. Therefore, the temperature increases are:
+//Scenario 1: DT = DT1.5 - DT0.0
+DT1 = 10 - 0
+//Scenario 2: DT = DT3.0-DT1.5
+DT2 = 15 - 10
+
+printf("\n\nResult\n\n")
+printf("\n the discharge temperature increase for scenario 1 is %.0f degC and for scenario 2 is %.0f degC",DT1,DT2)
diff --git a/680/CH9/EX9.08/9_08.sce b/680/CH9/EX9.08/9_08.sce new file mode 100755 index 000000000..f450653de --- /dev/null +++ b/680/CH9/EX9.08/9_08.sce @@ -0,0 +1,19 @@ +//Problem 9.08:
+
+//initializing the variables:
+Z = 10000; // in lb/h
+x = 0.1;
+y = 0.75;
+Ts = 75;// in deg C
+
+//calculation:
+X = Z*x/y
+Y = Z - X
+//from Fig. 9.4:
+Hz = 81 // in Btu/lb
+Hx = 395 // in Btu/lb
+Hy = 1150 // in Btu/lb
+Q = Hy*Y + Hx*X - Hz*Z
+
+printf("\n\nResult\n\n")
+printf("\n the evaporator heat required, Q is %.0f Btu/h",Q)
\ No newline at end of file diff --git a/680/CH9/EX9.09/9_09.sce b/680/CH9/EX9.09/9_09.sce new file mode 100755 index 000000000..086f9d548 --- /dev/null +++ b/680/CH9/EX9.09/9_09.sce @@ -0,0 +1,22 @@ +//Problem 9.09:
+
+//initializing the variables:
+Z = 10000; // in lb/h
+x = 0.1;
+y = 0.75;
+Ts = 340;// in deg F
+T = 212; // in deg F
+U = 500; // in Btu/h.ft2.deg F
+
+//calculation:
+X = Z*x/y
+Y = Z - X
+//from Fig. 9.4:
+Hz = 81 // in Btu/lb
+Hx = 395 // in Btu/lb
+Hy = 1150 // in Btu/lb
+Q = Hy*Y + Hx*X - Hz*Z
+A = Q/(U*(Ts - T))
+
+printf("\n\nResult\n\n")
+printf("\n the area requirement in the evaporator is %.1f ft^2",A)
\ No newline at end of file |
