initial commit / add all books

7 files changed, 188 insertions, 0 deletions

diff --git a/3012/CH14/EX14.1/Ex14_1.sce b/3012/CH14/EX14.1/Ex14_1.sce
new file mode 100755
index 000000000..724e5a6d3
--- /dev/null
+++ b/3012/CH14/EX14.1/Ex14_1.sce
@@ -0,0 +1,48 @@
+
+// Given:-
+// The reaction is CO + .5O2  ---   CO2
+// Part(a)
+T = 298.0                                                                         // in kelvin
+Rbar = 8.314                                                                      // universal gas constant in SI units
+// From table A-25
+
+hfbarCO2 = -393520.0                                                              // in kj/kmol
+hfbarCO = -110530.0                                                               // in kj/kmol
+hfbarO2 = 0                                                                       // in kj/kmol
+deltahbarCO2 = 0                                                                  // in kj/kmol
+deltahbarCO = 0                                                                   // in kj/kmol
+deltahbarO2 = 0                                                                   // in kj/kmol
+sbarCO2 = 213.69                                                                  // in kj/kmol.K
+sbarCO = 197.54                                                                   // in kj/kmol.K
+sbarO2 = 205.03                                                                   // in kj/kmol.K
+// From table A-27
+logKtable = 45.066
+// Calculations
+deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)
+lnK = -deltaG/(Rbar*T)
+logK = (1/log(10))*lnK
+// Results
+printf( ' Part(a) the value of equilibrium constant expressed as log10K is:  %f',logK);
+printf( ' The value of equilibrium constant expressed as log10K from table A-27 is: %f ',logKtable);
+
+// Part(b)
+T = 2000.0                                                                        // in kelvin
+// From table A-23
+hfbarCO2 = -393520.0                                                              // in kj/kmol
+hfbarCO = -110530.0                                                               // in kj/kmol
+hfbarO2 = 0                                                                       // in kj/kmol
+deltahbarCO2 = 100804-9364                                                        // in kj/kmol
+deltahbarCO = 65408 - 8669                                                        // in kj/kmol
+deltahbarO2 = 67881 - 8682                                                        // in kj/kmol
+sbarCO2 = 309.210                                                                 // in kj/kmol.K
+sbarCO = 258.6                                                                    // in kj/kmol.K
+sbarO2 = 268.655                                                                  // in kj/kmol.K
+// Calculations
+deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)
+lnK = -deltaG/(Rbar*T)
+logK = (1/log(10))*lnK
+// From table A-27
+logKtable = 2.884
+// Results
+printf( ' Part(b) the value of equilibrium constant expressed as log10K is: %f ',logK);
+printf( ' The value of equilibrium constant expressed as log10K from table A-27 is: %f ',logKtable);
diff --git a/3012/CH14/EX14.10/Ex14_10.sce b/3012/CH14/EX14.10/Ex14_10.sce
new file mode 100755
index 000000000..c18a8d761
--- /dev/null
+++ b/3012/CH14/EX14.10/Ex14_10.sce
@@ -0,0 +1,17 @@
+
+// Given:-
+// With data from Table A-2 at 20C,
+vf = 1.0018e-3                                                                  // in m^3/kg
+psat = 0.0239                                                                   // in bar
+p = 1.0                                                                         // in bar
+T = 293.15                                                                      // in kelvin
+Rbar = 8.314                                                                    // universal gas constant in SI units
+M = 18.02                                                                       // molat mass of water in kg/kmol
+e=2.715
+
+// Calculations
+pvbypsat = e**(vf*(p-psat)*10**5/((1000*Rbar/M)*T))
+percent = (pvbypsat-1)*100
+
+// Result
+printf( ' The departure, in percent, of the partial pressure of the water vapor from the saturation pressure of water at 20 is: %.3f',percent)
diff --git a/3012/CH14/EX14.2/Ex14_2.sce b/3012/CH14/EX14.2/Ex14_2.sce
new file mode 100755
index 000000000..23b2160e2
--- /dev/null
+++ b/3012/CH14/EX14.2/Ex14_2.sce
@@ -0,0 +1,34 @@
+
+// Given:-
+// Applying conservation of mass, the overall balanced chemical reaction equation is
+// CO + .5O2    ------    zCO + (z/2)O2 + (1-z)CO2
+
+// At 2500 K, Table A-27 gives
+log10K = -1.44
+// Part(a)
+p = 1.0                                                                         // in atm
+// Calculations
+K = (10.0)**(log10K)                                                            // equilibrium constant
+// Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
+z = 0.129
+yCO = 2.0*z/(2.0 + z)
+yO2 = z/(2.0 + z)
+yCO2 = 2.0*(1.0 - z)/(2.0 + z)
+
+// Results
+printf( ' Part(a) mole fraction of CO is: %.3f ',yCO)
+printf( ' Mole fraction of O2 is:  %.3f',yO2)
+printf( ' Mole fraction of CO2 is:  %.3f',yCO2)
+
+// Part(b)
+p = 10.0                                                                          // in atm
+// Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
+z = 0.062
+yCO = 2.0*z/(2.0 + z)
+yO2 = z/(2.0 + z)
+yCO2 = 2.0*(1.0 - z)/(2.0 + z)
+
+// Results
+printf( ' Part(b) mole fraction of CO is:  %.3f',yCO)
+printf( ' Mole fraction of O2 is:  %.3f',yO2)
+printf( ' Mole fraction of CO2 is: %.3f ',yCO2)
diff --git a/3012/CH14/EX14.3/Ex14_3.sce b/3012/CH14/EX14.3/Ex14_3.sce
new file mode 100755
index 000000000..4eec016b6
--- /dev/null
+++ b/3012/CH14/EX14.3/Ex14_3.sce
@@ -0,0 +1,15 @@
+
+// Given:-
+yCO = 0.298
+p = 1                                                                           // in atm
+pref = 1                                                                        // in atm
+// With this value of K, table A-27 gives
+T = 2881
+
+// Calculations
+// Solving yCO = 2z/(2 + z)
+z = 2*yCO/(2 - yCO)
+K = (z/(1-z))*(z/(2 + z))**.5*(p/pref)**.5
+
+// Result
+printf( ' The temperature T of the mixture in kelvin is:  %f',T);
diff --git a/3012/CH14/EX14.4/Ex14_4.sce b/3012/CH14/EX14.4/Ex14_4.sce
new file mode 100755
index 000000000..71190fea4
--- /dev/null
+++ b/3012/CH14/EX14.4/Ex14_4.sce
@@ -0,0 +1,24 @@
+
+// Given:-
+// For a complete reaction of CO with the theoretical amount of air
+// CO + .5 O2 + 1.88N2  ----  CO2 + 1.88N2    
+// Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is
+// CO + .5O2 + 1.88N2 --  zCO + z/2 O2 + (1-z)CO2 + 1.88N2
+
+K = 0.0363                                                                       // equilibrium constant the solution to Example 14.2
+p =1.0                                                                           // in atm
+pref = 1.0                                                                       // in atm
+
+// Calculations
+// Solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives
+z = 0.175
+yCO = 2.0*z/(5.76 + z)
+yO2 = z/(5.76 + z)
+yCO2 = 2.0*(1.0-z)/(5.76 + z)
+yN2 = 3.76/(5.76 + z)
+
+// Results
+printf( ' The mole fraction of CO is: %.3f ',yCO)
+printf( ' The mole fraction of O2 is: %.3f ',yO2)
+printf( ' The mole fraction of CO2 is:  %.3f',yCO2)
+printf( ' The mole fraction of N2 is:  %.3f',yN2)
diff --git a/3012/CH14/EX14.5/Ex14_5.sce b/3012/CH14/EX14.5/Ex14_5.sce
new file mode 100755
index 000000000..0d2047373
--- /dev/null
+++ b/3012/CH14/EX14.5/Ex14_5.sce
@@ -0,0 +1,28 @@
+
+// Given:-
+// Applying the conservation of mass principle, the overall dissociation reaction is described by
+// CO2  ---  zCO2 + (1-z)CO + ((1-z)/2)O2
+
+p = 1.0                                                                           // in atm
+pref = 1.0                                                                        // in atm
+// At 3200 K, Table A-27 gives
+log10k = -.189
+// Solving k = ((1-z)/2)*((1-z)/(3-z))^.5 gives
+z = 0.422
+
+// Calculations
+k = 10**log10k
+// From tables A-25  and A-23
+hfbarCO2 = -393520.0                                                              // in kj/kmol
+deltahbarCO2 = 174695-9364                                                        // in kj/kmol
+hfbarCO = -110530.0                                                               // in kj/kmol
+deltahbarCO = 109667-8669                                                         // in kj/kmol
+hfbarO2 = 0                                                                       // in kj/kmol
+deltahbarO2 = 114809-8682                                                         // in kj/kmol
+hfbarCO2r =  -393520.0                                                            // in kj/kmol
+deltahbarCO2r = 0                                                                 // in kj/kmol
+
+Qcvdot = 0.422*(hfbarCO2 + deltahbarCO2) + 0.578*(hfbarCO + deltahbarCO) + 0.289*(hfbarO2 + deltahbarO2)- (hfbarCO2r + deltahbarCO2r)  
+
+// Result
+printf( ' The heat transfer to the reactor, in kJ per kmol of CO2 entering is:  %f', Qcvdot);
diff --git a/3012/CH14/EX14.8/Ex14_8.sce b/3012/CH14/EX14.8/Ex14_8.sce
new file mode 100755
index 000000000..3298fa8fb
--- /dev/null
+++ b/3012/CH14/EX14.8/Ex14_8.sce
@@ -0,0 +1,22 @@
+
+// Given:-
+// The ionization of cesium to form a mixture of Cs, Cs+, and e- is described by
+// Cs  ---  (1-z)Cs + zCs+ + Ze-
+
+K = 15.63
+z = 0.95
+pref =1                                                                         // in atm
+// Calculation
+p = pref*K*((1-z**2)/z**2)
+
+// Results
+printf( ' The pressure if the ionization of CS is 95 percent complete is: %f atm',p);
+
+x = linspace(0,10,100)
+for i = 1:100
+    y(i)= 100*((1/(1+x(i)/K))**0.5)
+end
+
+plot(x,y)
+xlabel("Pressure (atm)")
+ylabel("Ionization")