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diff --git a/3012/CH11/EX11.1/Ex11_1.sce b/3012/CH11/EX11.1/Ex11_1.sce
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index 000000000..b03e76e53
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+// Given:-
+m = 4.00                                                                        // mass of carbon monoxide in kg
+T = 223.00                                                                      // temperature of carbon monoxide in kelvin
+D = 0.2                                                                         // inner diameter of cylinder in meter
+L = 1.00                                                                        // length of the cylinder in meter
+pi=3.14
+// Analysis
+M = 28.00                                                                       // molar mass in kg/kmol
+
+// Calculations
+V = (pi*D**2.00/4.00)*L                                                         // volume occupied by the gas in m^3
+vbar = M*(V/m)                                                                  // The molar specific volume in m^3/kmol
+
+// Part(a)
+// From Table A-1 for CO
+Tc = 133                                                                        // in kelvin
+Pc = 35                                                                         // in bar
+Tr = T/Tc                                                                       // reduced temperature
+Rbar = 8314                                                                     // universal gas constant in N.m/kmol.K
+Z = 0.9
+// Calculations
+vrdash = (vbar*Pc*10**5)/(Rbar*Tc)                                              // pseudoreduced specific volume
+p = (Z*Rbar*T/vbar)*10**-5                                                      // in bar
+// Result
+printf( '\n part(a)the pressure in bar is:  %.2f bar',p)
+
+// Part(b)
+// The ideal gas equation of state gives
+// Calculations
+p = (Rbar*T/vbar)/10**5                                                          // in bar
+// Result
+printf( '\n Part(b)the pressure in bar is: %.2f bar',p)
+
+// Part(c)
+// For carbon monoxide, the van der Waals constants a and b can be read directly from Table A-24
+a = 1.474                                                                        // in (m^3/kmol)^2
+b = 0.0395                                                                       // in m^3/kmol
+// Calculations
+p = (Rbar*T/(vbar-b))/10**5 - a/vbar**2
+// Result
+printf( '\n Part(c)the pressure in bars is:  %.2f bar',p)
+
+// Part(d)
+// For carbon monoxide, the Redlich–Kwong constants can be read directly from Table A-24
+a = 17.22                                                                        // in m^6*K^.5/kmol^2
+b = 0.02737                                                                      // in m^3/kmol
+// Calculations
+p = (Rbar*T/(vbar-b))/10**5 - a/(vbar*(vbar+b)*T**.5)
+// Result
+printf( '\n Part(d)the pressure in bar is:  %.2f bar', p)
diff --git a/3012/CH11/EX11.10/Ex11_10.sce b/3012/CH11/EX11.10/Ex11_10.sce
new file mode 100755
index 000000000..8b1a2eb83
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+++ b/3012/CH11/EX11.10/Ex11_10.sce
@@ -0,0 +1,67 @@
+
+// Given:-
+// Analysis
+V = 0.241                                                                        // volume of the mixture in m^3
+T = 511.00                                                                       // temperature of the mixture in kelvin
+n1 = 0.18                                                                        // number of moles of methane in kmol
+n2 = 0.274                                                                       // number of moles of butane in kmol
+Rbar = 8314                                                                      // universal gas constant in (N.m)/(kmol.K)
+
+// Calculations
+n = n1 + n2                                                                      // The total number of moles of mixture
+y1 = n1/n                                                                        // mole fraction of methane
+y2 = n2/n                                                                        // mole fraction of butane
+vbar = V/(n)                                                                     // The specific volume of the mixture on a molar basis in m^3/kmol
+
+// Part(a)
+p = (Rbar*T/vbar)*10**-5                                                         // in bar
+// Result
+printf( ' The pressure in bar obtained using ideal gas equation is:  %.2f',p)
+
+// Part(b)
+// From table A-1
+Tc1 = 191.00                                                                     // critical temperature for methane in kelvin 
+Pc1 = 46.4                                                                       // critical pressure for methane in bar
+Tc2 = 425.00                                                                     // critical temperature for butane in kelvin
+Pc2 = 38.00                                                                      // critical pressure for butane in bar
+Z = 0.88
+
+
+// Calculations
+Tc = y1*Tc1 + y2*Tc2                                                             // critical temperature in kelvin
+Pc = y1*Pc1 + y2*Pc2                                                             // critical pressure in bar
+TR = T/Tc                                                                        // reduced temperature of the mixture
+vRdash= vbar*Pc/(Rbar*Tc)
+p = ((Z*Rbar*T)/vbar)*10**-5                                                     // mixture pressure in bar
+// Result
+printf( ' Pressure obtained using Kay’s rule together with the generalized compressibility chart, is: %.2f ',p)
+
+// Part(c)
+// Table A-24 gives the following van der Waals constants values for methane
+a1 = 2.293                                                                       // in (m^3/kmol)^2
+b1 = 0.0428                                                                      // in m^3/kmol
+// Table A-24 gives the following van der Waals constants values for butane
+a2 = 13.86                                                                       // in (m^3/kmol)^2
+b2 = 0.1162                                                                      // in m^3/kmol
+
+a = (y1*a1**.5 + y2*a2**.5)**2                                                   // in bar*(m^3/kmol)^2
+b = y1*b1+y2*b2                                                                  // in m^3/kmol
+// From van der Waals equation
+p = ((Rbar*T)/(vbar-b))*10**-5 - a/(vbar**2)
+printf( ' The pressure in bar from van der Waals equation is:   %.2f',p)
+
+// Part(d)
+// For methane
+TR1 = T/Tc1
+vR1dash = (.241/.18)*10**5*Pc1/(Rbar*Tc1)
+Z1 = 1.00
+// For butane
+TR2 = T/Tc2
+vR2dash = (.88*10**5*Pc2)/(Rbar*Tc2)
+Z2 = 0.8
+Z = y1*Z1 + y2*Z2
+// Accordingly, the same value for pressure as determined in part (b) using Kay’s rule results:
+p = 70.4
+
+// Result
+printf( ' The pressure in bar obtained using the rule of additive pressures employing the generalized compressibility chart is:  %.2f',p)
diff --git a/3012/CH11/EX11.3/Ex11_3.sce b/3012/CH11/EX11.3/Ex11_3.sce
new file mode 100755
index 000000000..119bc0631
--- /dev/null
+++ b/3012/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,32 @@
+// Given:-
+// Part(a)
+v = 0.4646                                                                       // specific volume in in m^3/kg
+M = 18.02                                                                        // molar mass of water in kg/kmol
+// At the specified state, the temperature is 513 K and the specific volume on a molar basis is
+vbar = v*M                                                                       // in m^3/kmol
+// From Table A-24
+a = 142.59                                                                       // (m^3/kmol)^2 * K^.5
+b = 0.0211                                                                       // in m^3/kmol
+
+Rbar = 8314.0                                                                    // universal gas constant in N.m/kmol.K
+T = 513.0                                                                        // in kelvin
+delpbydelT = (Rbar/(vbar-b) + a/(2*vbar*(vbar+b)*T**1.5)*10**5)/10**3             // in kj/(m^3*K)
+
+// By The Maxwell relation
+delsbydelv = delpbydelT
+// Result
+printf( ' The value of delpbydelT in kj/(m^3*K) is:   %.2f',delpbydelT);
+
+// Part(b)
+// A value for (dels/delv)T can be estimated using a graphical approach with steam table data, as follows: At 240C, Table A-4 provides the values for specific entropy s and specific volume v tabulated below
+T = 240.0                                                                         // in degree celcius
+// At p =1, 1.5, 3, 5, 7, 10 bar respectively
+y = [7.994, 7.805, 7.477, 7.230, 7.064, 6.882]                  
+x = [2.359, 1.570, 0.781, 0.4646, 0.3292, 0.2275]               
+plot(x,y)
+xlabel("Specific volume")
+ylabel("Specific entropy")
+
+// The pressure at the desired state is 5 bar.The corresponding slope is
+delsbydelv = 1                                                                    // in kj/m^3.K
+printf( ' From the data of the table,delsbydelv = %.2f',delsbydelv);
diff --git a/3012/CH11/EX11.4/Ex11_4.sce b/3012/CH11/EX11.4/Ex11_4.sce
new file mode 100755
index 000000000..0d233d37f
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+++ b/3012/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,35 @@
+// Given:-
+// Analysis
+// For comparison, Table A-2 gives at 100C,
+hgf =2257.00                                                                    // in kj/kg
+ugf = 2087.6                                                                    // in kj/kg
+sgf = 6.048                                                                     // in kj/kg.K
+// Values
+printf( ' From table, hg-hf = %.2f',hgf);
+printf( ' From table, ug-uf = %.2f',ugf);
+printf( ' From table, sg-sf = %.2f',sgf);
+
+// Part(a)
+T = 373.15                                                                      // in kelvin
+// If we plot a graph between temperature and saturation pressure using saturation pressure–temperature data from the steam tables, the desired slope is:
+delpbydelT = 3570.00                                                            // in N/(m^2.K)
+vg = 1.673                                                                      // in m^3/kg
+vf = 1.0435e-3                                                                  // in m^3/kg
+// Calculations
+// From the Clapeyron equation
+hgf = T*(vg-vf)*delpbydelT*10**-3                                               // in kj/kg
+// Result
+printf( '\n Part(a)using Clapeyron equation, hg-hf = %.2f KJ/kg', hgf);
+
+// Part(b)
+psat = 1.014e5                                                                  // in N/m^2
+hgf = 2256.00                                                                   // can be obtained using IT software in kj/kg
+// Calculations
+ugf = hgf - psat*(vg-vf)/10**3                                                  // in kj/kg
+// Result
+printf( '\n Part(b)ug-uf = %.2f KJ/kg',ugf)
+// Part(c)
+// Calculation
+sgf =hgf/T                                                                      // in kj/kg.K         
+// Result
+printf( '\n Part(c)sg-sf = %.2f  KJ/kg. k',sgf)
diff --git a/3012/CH11/EX11.6/Ex11_6.sce b/3012/CH11/EX11.6/Ex11_6.sce
new file mode 100755
index 000000000..acc0b5c1f
--- /dev/null
+++ b/3012/CH11/EX11.6/Ex11_6.sce
@@ -0,0 +1,21 @@
+// Given:-
+// Part(a)
+v = 1.00/998.21                                                             // specific volume of water in m^3/kg
+T = 293.00                                                                  // given temperature in kelvin
+beta = 206.6e-6                                                             // volume expansivity in /K
+k = 45.90e-6                                                                // isothermal compressibility in /bar
+// Interpolating in Table A-19
+cp = 4.188                                                                  // in kj/kg.k
+// Calculations
+cpv = (v*T*beta**2.00/k)*10**2                                              // in kj/kg.k
+cv = cp-cpv                                                                 // in kj/kg.k
+errorPercentage = 100*(cp-cv)/cv
+// Result
+printf( ' The percentage error is:  %.2f',errorPercentage)
+
+// Part(b)
+// Calculations
+K = cp/cv                                                                   // specific heat ratio
+c = ((K*v/k)*10**5)**0.5                                                      // velocity of sound in m/s
+// Result
+printf( ' The velocity of sound is:  %.2f m/s',c)
diff --git a/3012/CH11/EX11.8/Ex11_8.sce b/3012/CH11/EX11.8/Ex11_8.sce
new file mode 100755
index 000000000..91eb2d0d9
--- /dev/null
+++ b/3012/CH11/EX11.8/Ex11_8.sce
@@ -0,0 +1,27 @@
+// Given:-
+p1 = 100.00                                                                     // in bar
+T1 = 300.00                                                                     // in kelvin
+p2 = 40.00                                                                      // in bar
+T2 = 245.00                                                                     // in kelvin
+
+
+// From table A-23
+h1starbar = 8723.00                                                             // in kj/kmol
+h2starbar = 7121.00                                                             // in kj/kmol
+// From Tables A-1
+Tc = 126.00                                                                     // critical temperature in kelvin
+pc = 33.9                                                                       // critical pressure in bar
+M = 28.00                                                                       // molar mass in kg/kmol
+Rbar = 8.314                                                                    // universal gas constant in kj/(kmol.K)
+Term1 = 0.5                                                                      
+Term2 = 0.31
+
+// Calculations
+TR1 = T1/Tc                                                                     // reduced temperature at the inlet
+PR1 = p1/pc                                                                     // reduced pressure at the inlet
+TR2 = T2/Tc                                                                     // reduced temperature at the exit
+PR2 = p2/pc                                                                     // reduced pressure at the exit
+wcvdot = (1.00/M)*(h1starbar-h2starbar-Rbar*Tc*(Term1-Term2))                      // in kj/kg
+
+// Result
+printf( ' The work developed, in kJ per kg of nitrogen flowing is :  %.2f',wcvdot)
diff --git a/3012/CH11/EX11.9/Ex11_9.sce b/3012/CH11/EX11.9/Ex11_9.sce
new file mode 100755
index 000000000..7c8b4f92e
--- /dev/null
+++ b/3012/CH11/EX11.9/Ex11_9.sce
@@ -0,0 +1,35 @@
+
+// Given:-
+// Part(a)
+// With values from Table A-23
+sT2bar = 185.775                                                                // in kj/(kmol.K)
+sT1bar = 191.682                                                                // in kj/(kmol.K)
+Rbar = 8.314                                                                    // universal gas constant
+M = 28.00                                                                       // molar mass in kg/kmol 
+p2 = 40.00                                                                      // in bar
+p1 = 100.00                                                                     // in bar
+Term1 = 0.21
+Term2 = 0.14
+
+// Calculations
+
+S2StarBarMinusS1StarBar = sT2bar-sT1bar-Rbar*log(p2/p1)                    // The change in specific entropy in kj/(kmol.K)
+sigmacvdot = (1.00/M)*(S2StarBarMinusS1StarBar-Rbar*(Term2-Term1))
+// Result
+printf( ' the rate of entropy production in kj/kg.K is:  %.2f', sigmacvdot)
+
+// Part(b)
+// From Table A-23,
+h2starbar = 6654.00                                                             // in kj/kmol
+h1starbar = 8723.00                                                             // in kj/kmol
+Tc = 126.00                                                                     // critical temperature in kelvin
+Term2 = 0.36
+Term1 = 0.5
+wcvdot = 50.1                                                                   // from example 11.8
+
+// Calculations
+wcvdots = (1.00/M)*(h1starbar-h2starbar-Rbar*Tc*(Term1-Term2))                  // isentropic work in kj/kg
+etat = wcvdot/wcvdots                                                           // turbine efficiency
+
+// Result
+printf( ' The isentropic turbine efficiency is: %.2f', etat)