initial commit / add all books

13 files changed, 705 insertions, 0 deletions

diff --git a/3012/CH9/EX9.1/Ex9_1.sce b/3012/CH9/EX9.1/Ex9_1.sce
new file mode 100755
index 000000000..64697cf4e
--- /dev/null
+++ b/3012/CH9/EX9.1/Ex9_1.sce
@@ -0,0 +1,58 @@
+// Given:-
+T1 = 300.00                                                           // The temperature at the beginning of the compression process in kelvin
+p1 = 1.00                                                             // the pressure at the beginning of the compression process in bar
+r = 8.00                                                              // compression ratio
+V1 = 560.00                                                           // the volume at the beginning of the compression process in cm^3
+T3 = 2000.00                                                          // maximum temperature during the cycle in kelvin
+
+// Part(a)
+// At T1 = 300k,table A-22 gives
+u1 = 214.07                                                            // in kj/kg
+vr1 = 621.2             
+// Interpolating with vr2 in Table A-22, we get
+T2 = 673.00                                                            // in kelvin
+u2 = 491.2                                                             // in kj/kg
+// At T3 = 2000 K, Table A-22 gives
+u3 = 1678.7                                                            // in kj/kg
+vr3 = 2.776
+// Interpolating in Table A-22 with vr4 gives
+T4 = 1043                                                              // in kelvin
+u4 = 795.8                                                             // in kj/kg
+
+// Calculations
+// For the isentropic compression Process 1–2
+vr2 = vr1/r
+// With the ideal gas equation of state
+p2 = p1*(T2/T1)*(r)                                                     // in bars
+// Since Process 2–3 occurs at constant volume, the ideal gas equation of state gives
+p3 = p2*(T3/T2)                                                         // in bars
+// For the isentropic expansion process 3–4
+vr4 = vr3*(r)
+// The ideal gas equation of state applied at states 1 and 4 gives
+p4 = p1*(T4/T1)                                                         // in bars
+
+// Results
+printf( ' At state1, the pressure is: %f bar.',p1)
+printf( ' At state1, the temperature is %f kelvin.',T1)
+printf( ' At state2, the pressure is : %.3f bar.',p2)
+printf( ' At state2, the temperature is %f kelvin.',T2)
+printf( ' At state3, the pressure is : %.3f bar.',p3)
+printf( ' At state3, the temperature is %f kelvin.',T3)
+printf( ' At state4, the pressure is : %.4f bar.',p4)
+printf( ' At state4, the temperature is %f kelvin.',T4)
+
+// Part(b)
+eta = 1-(u4-u1)/(u3-u2)                                                     // thermal efficiency
+// Result
+printf( ' The thermal efficiency is : %.2f ',eta)
+
+// Part(c)
+R = 8.314                                                                   // universal gas constant, in SI units
+M = 28.97                                                                   // molar mass of air in grams
+// Calculations
+m = ((p1*V1)/((R/M)*T1))*10**-6*10**5*10**-3                                // mass of the air in kg
+Wcycle = m*((u3-u4)-(u2-u1))                                                // the net work per cycle in KJ
+mep = (Wcycle/(V1*(1-1/r)))*10**6*10**3*10**-5                               // in bars
+
+// Result
+printf( ' The mean effective pressure, is : %.4f atm.',mep)
diff --git a/3012/CH9/EX9.11/Ex9_11.sce b/3012/CH9/EX9.11/Ex9_11.sce
new file mode 100755
index 000000000..43b38f205
--- /dev/null
+++ b/3012/CH9/EX9.11/Ex9_11.sce
@@ -0,0 +1,53 @@
+// Given:-
+T1 = 300.00                                                                        // in kelvin
+p1 = 100.00                                                                        // in kpa
+mdot = 5.807                                                                       // in kg/s
+p2 = 300.00                                                                        // in kpa
+p3 = p2
+p4 = 1000.00                                                                       // in kpa
+p5 = p4
+p6 = p4
+T6 = 1400.00                                                                       // in kelvin
+T8 = T6
+p7 = 300.00                                                                        // in kpa
+p8 = p7
+etac = 0.8                                                                         // isentropic efficiency of compressor
+etat = 0.8                                                                         // isentropic efficiency of turbine
+etareg = 0.8                                                                       // regenerator effectiveness
+// Analysis
+// From example 9.9
+h1 = 300.19                                                                        // in kj/kg
+h3 = h1                                                                            // in kj/kg
+h2s = 411.3                                                                        // in kj/kg
+h4s = 423.8                                                                        // in kj/kg
+// From example 9.8
+h6 = 1515.4                                                                        // in kj/kg
+h8 = h6
+h7s = 1095.9                                                                       // in kj/kg
+h9s = 1127.6                                                                       // in kj/kg
+
+// Calculations
+h4 = h3 + (h4s-h3)/etac                                                            // in kj/kg
+h2 = h1 + (h2s-h1)/etac                                                            // in kj/kg
+h9 = h8-etat*(h8-h9s)                                                              // in kj/kg
+h7 = h6-etat*(h6-h7s)                                                              // in kj/kg
+h5 = h4+etareg*(h9-h4)                                                             // in kj/kg
+
+// Part(a)
+// Calculations
+wtdot = (h6-h7)+(h8-h9)                                                            // The total turbine work per unit of mass flow in kj/kg
+wcdot = (h2-h1)+(h4-h3)                                                            // The total compressor work input per unit of mass flow in kj/kg
+qindot = (h6-h5)+(h8-h7)                                                           // The total heat added per unit of mass flow in kj/kg
+eta = (wtdot-wcdot)/qindot                                                         // thermal efficiency
+// Result
+printf( ' The thermal efficiency is:  %.2f',eta)
+
+// Part(b)
+bwr = wcdot/wtdot                                                                  // back work ratio
+// Result
+printf( ' The back work ratio is: %.2f',bwr)
+
+// Part(c)
+Wcycledot = mdot*(wtdot-wcdot)                                                     // net power developed in kw
+// Result
+printf( ' The net power developed, is: %.2f kW.',Wcycledot)
diff --git a/3012/CH9/EX9.12/Ex9_12.sce b/3012/CH9/EX9.12/Ex9_12.sce
new file mode 100755
index 000000000..7c22ca353
--- /dev/null
+++ b/3012/CH9/EX9.12/Ex9_12.sce
@@ -0,0 +1,51 @@
+// Given:-
+Ta = 240.00                                                                      // in kelvin
+pa = 0.8                                                                         // in bar
+Va = 278.00                                                                      // in m/s
+PR = 8.00                                                                        // pressure ratio across the compressor
+T3 = 1200.00                                                                     // in kelvin
+p5 = 0.8                                                                         // in bar
+
+// From table A-22
+ha = 240.02                                                                      // in kj/kg
+h1 = ha + ((Va**2)/2)*10**-3                                                     // in kj/kg
+// Interpolating in Table A-22 gives
+pr1 = 1.070
+pra = .6355
+
+// Interpolating in Table A-22, we get
+h2 = 505.5                                                                        // in kj/kg
+// At state 3 the temperature is given as T3 = 1200 K. From Table A-22
+h3 = 1277.79                                                                      // in kj/kg
+
+
+// Interpolating in Table A-22 with h4, gives
+pr4 = 116.8
+// pr data from table A-22 gives
+pr4 = 116.00
+pr3 = 238.00
+// From table A-22
+h5 = 621.3                                                                        // in kj/kg
+
+// The expansion through the nozzle is isentropic to
+p5 = .8                                                                           // in bars
+
+// Calculations
+p1 = (pr1/pra)*pa                                                                 // in bars
+// With the help of assumption, 'The turbine work output equals the work required to drive the compressor.',
+h4 = h3+h1-h2                                                                     // in kj/kg
+p2 = PR*p1                                                                        // in bars
+// Using assumption 'There is no pressure drop for flow through the combustor', 
+p3 = p2
+p4 = p3*(pr4/pr3)                                                                 // in bars
+pr5 = pr4*(p5/p4)
+V5 = ((2*(h4-h5)*10**3))**(0.5)                                                   // the velocity at the nozzle exit in m/s
+
+// Results
+printf( ' The velocity at the nozzle exit in m/s is: %.2f', V5)
+printf( ' pa in bars =  %.2f',pa)
+printf( ' p1 in bars =  %.2f',p1)
+printf( ' p2 in bars =  %.2f',p2)
+printf( ' p3 in bars =  %.2f',p3)
+printf( ' p4 in bars =  %.2f',p4)
+printf( ' p5 in bars =  %.2f',p5)
diff --git a/3012/CH9/EX9.13/Ex9_13.sce b/3012/CH9/EX9.13/Ex9_13.sce
new file mode 100755
index 000000000..54dba7c11
--- /dev/null
+++ b/3012/CH9/EX9.13/Ex9_13.sce
@@ -0,0 +1,96 @@
+// Given:-
+Wnetdot = 45.00                                                                    // in MW
+T1 = 300.00                                                                        // in kelvin
+p1 = 100.00                                                                        // in kpa
+etac = 0.84                                                                        // The isentropic efficiency of the compressor
+T3 = 1400.00                                                                       // in kelvin
+p2 = 1200.00                                                                       // in kpa
+p3 = p2
+etat = 0.88                                                                        // isentropic efficiency of the turbine
+T5 = 400.00                                                                        // in kelvin
+p4 = 100.00                                                                        // in kpa
+p5 = p4
+T7 = 400.00                                                                        // in degree celcius
+p7 = 8.00                                                                          // in MPa
+etatw =0.9                                                                         // isentropic efficiency of turbine of the vapor cycle
+p8 = 8.00                                                                          // in kpa
+p9 = p8
+etap = 0.8                                                                         // isentropic efficiency of pump of the vapor cycle
+T0 = 300.00                                                                        // in kelvin
+p0 = 100.00                                                                        // -in kpa
+
+// Analysis
+// With procedure similar to that used in the examples of chapters 8 and 9,we can determine following property data
+h1 = 300.19                                                                        // in kj/kg
+h2 = 669.79                                                                        // in kj/kg
+h3 = 1515.42                                                                       // in kj/kg
+h4 = 858.02                                                                        // in kj/kg
+h5 = 400.98                                                                        // in kj/kg
+h6 = 183.96                                                                        // in kj/kg
+h7 = 3138.30                                                                       // in kj/kg
+h8 = 2104.74                                                                       // in kj/kg
+h9 = 173.88                                                                        // in kj/kg
+s1 = 1.7020                                                                        // in kj/kg.k
+s2 = 2.5088                                                                        // in kj/kg.k
+s3 = 3.3620                                                                        // in kj/kg.k   
+s4 = 2.7620                                                                        // in kj/kg.k
+s5 = 1.9919                                                                        // in kj/kg.k
+s6 = 0.5975                                                                        // in kj/kg.k
+s7 = 6.3634                                                                        // in kj/kg.k
+s8 = 6.7282                                                                        // in kj/kg.k
+s9 = 0.5926                                                                        // in kj/kg.k
+
+// Part(a)
+// By applying mass and energy rate balances
+// Calculations
+mvdotbymgdot = (h4-h5)/(h7-h6)                                                     // ratio of mass flow rates of vapor and air
+mgdot = (Wnetdot*10**3)/(((h3-h4)-(h2-h1)) + mvdotbymgdot*((h7-h8)-(h6-h9)))       // mass flow rate of air in kg/s
+mvdot = mvdotbymgdot*mgdot                                                         // mass flow rate of vapor in kg/s
+Wgasdot = mgdot*((h3-h4)-(h2-h1))*10**-3                                           // net power developed by gas turbine in MW
+Wvapdot = mvdot*((h7-h8)-(h6-h9))*10**-3                                           // net power developed by vapor cycle in MW
+
+// Results
+printf( ' Mass flow rate of air is: %.2f kg/s.',mgdot)
+printf( ' Mass flow rate of vapor is: %.2f kg/s.',mvdot)
+printf( ' Net power developed by gas turbine is:  %.2f MW.',Wgasdot)
+printf( ' Net power developed by vapor cycle is:  %.2f MW.',Wvapdot)
+
+
+// Part(b)
+
+// The net rate of exergy increase of the air passing through the combustor is
+Edotf32 = mgdot*(h3-h2-T0*(s3-s2))*10**-3                                          // in MW
+// The net rate exergy is carried out by the exhaust air stream at 5 is
+Edotf51 = mgdot*(h5-h1-T0*(s5-s1))/10**3                                           // in MW
+// The net rate exergy is carried out as the water passes through the condenser is
+Edotf89 = mvdot*(h8-h9-T0*(s8-s9))*10**-3                                          // in MW
+R = 8.314                                                                          // universal gas constant, in SI units
+M = 28.97                                                                          // molar mass of air in grams
+// The rate of exergy destruction for air turbine is
+Eddott = mgdot*T0*(s4-s3-(R/M)*log(p4/p3))/10**3                              // in MW
+// The rate of exergy destruction for compressor is
+Eddotc = mgdot*T0*(s2-s1-(R/M)*log(p2/p1))/10**3                              // in MW
+// The rate of exergy destruction for steam turbine is
+Eddotst = mvdot*T0*(s8-s7)/10**3                                                   // in MW
+// The rate of exergy destruction for pump is
+Eddotp = mvdot*T0*(s6-s9)/10**3                                                    // in MW
+// For heat exchanger
+EddotHE = T0*(mgdot*(s5-s4)+mvdot*(s7-s6))/10**3                                   // in MW
+
+// Results
+printf( ' Balance sheet')
+printf( 'Net exergy increase of the gas passing')
+printf( ' Through the combustor: %.2f MW',Edotf32)
+printf( 'Disposition of the exergy:')
+printf( '• Net power developed')
+printf( 'gas turbine cycle %.2f MW',Wgasdot)
+printf( 'vapor cycle %.2f MW',Wvapdot)
+printf( '• Net exergy lost')
+printf( 'with exhaust gas at state 5 %.2f MW',Edotf51)
+printf( 'from water passing through condenser %.2f MW',Edotf89)
+printf( '• Exergy destruction')
+printf( 'air turbine %.2f MW',Eddott)
+printf( 'compressor %.2f MW',Eddotc)
+printf( 'steam turbine %.2f MW',Eddotst)
+printf( 'pump %.2f MW',Eddotp)
+printf( 'heat exchanger %.2f MW',EddotHE)
diff --git a/3012/CH9/EX9.14/Ex9_14.sce b/3012/CH9/EX9.14/Ex9_14.sce
new file mode 100755
index 000000000..b066ef5e8
--- /dev/null
+++ b/3012/CH9/EX9.14/Ex9_14.sce
@@ -0,0 +1,37 @@
+
+// Given:-
+Tnot = 360.00                                                                      // in kelvin
+pnot = 1.00                                                                        // in MPa
+A2 = 0.001                                                                         // in m^2
+k = 1.4
+
+// Calculations
+pstarbypnot = (1+(k-1)/2)**(k/(1-k))
+pstar =  pstarbypnot*pnot
+
+// Part(a)
+// Since back pressure of 500 kpa is less than critical pressure pstar(528kpa in this case) found above, the nozzle is choked
+// At the exit
+M = 1.00
+p2 = pstar                                                                         // in MPa
+T2 = Tnot/(1+((k-1)/2)*(M**2))                                                     // exit temperature in kelvin
+R = 8.314                                                                          // universal gas constant, in SI units
+Mwt = 28.97                                                                        // molar mass of air in grams
+V2 = ((k*(R/Mwt)*T2*10**3)**0.5)                                                   // exit velocity in m/s
+mdot = (p2/((R/Mwt)*T2))*A2*V2*10**3                                               // mass flow rate in kg/s
+
+// Results
+printf( ' The exit mach number for back pressure of 500kpa is: %.2f',M)
+printf( ' The mass flow rate in kg/s for back pressure of 500kpa is: %.2f',mdot)
+
+// Part(b)
+// Since the back pressure of 784kpa is greater than critical pressure of pstar determined above,the flow throughout the nozzle is subsonic and the exit pressure equals the back pressure,
+p2 = 784.00                                                                        // exit pressure in kpa
+// Calculations
+M2 = (((2.00)/(k-1))*(((pnot*10**3)/p2)**((k-1)/k)-1))**0.5                        // exit mach number
+T2 = Tnot/(1+((k-1)/2)*(M2**2))                                                    // exit temperature in kelvin
+V2 = M2*((k*(R/Mwt)*10**3*T2)**0.5)                                                // exit velocity in m/s
+mdot2 = (p2/((R/Mwt)*T2))*A2*V2                                                    // mass flow rate in kg/s
+// Results
+printf( ' The mass flow rate at the exit for back pressure of 784kpa is: %.2f kg/s.',mdot2)
+printf( ' The exit mach number for back pressure of 784 kpa is: %.2f',M2)
diff --git a/3012/CH9/EX9.15/Ex9_15.sce b/3012/CH9/EX9.15/Ex9_15.sce
new file mode 100755
index 000000000..d497b12f3
--- /dev/null
+++ b/3012/CH9/EX9.15/Ex9_15.sce
@@ -0,0 +1,104 @@
+// Given:-
+// Part(a)
+Mt = 0.7                                                                         // mach mumber at the throat
+At = 6.25                                                                        // throat area in cm^2
+Ae = 15.00                                                                       // exit area in cm^2
+
+// The flow throughout the nozzle, including the exit, is subsonic. Accordingly, with this value for A2byAstar, Table 9.1 gives
+M2 = 0.24
+// For M2 = 0.24,
+T2byTnot = 0.988
+p2bypnot = 0.959
+k = 1.4
+T0 = 280.00                                                                      // in kelvin
+pnot = 6.8                                                                       // in bars
+// Calculations
+// With Mt = 0.7, Table 9.1 gives
+AtbyAstar = 1.09437
+A2byAstar = (Ae/At)*AtbyAstar
+T2 = T2byTnot*T0                                                                 // in kelvin
+p2 = p2bypnot*pnot                                                               // in bars
+V2 = M2*((k*(8.314/28.97)*T2*10**3)**0.5)                                         // velocity at the exit in m/s
+mdot = (p2/((8.314/28.97)*T2))*Ae*V2*10**-2                                      // mass flow rate in kg/s
+// Results
+printf( ' Part(a)  the mass flow rate in kg/s is:  %.2f',mdot)
+printf( ' The exit pressure in bars is:  %.2f',p2)
+printf( ' The exit mach number is:  %.2f',M2)
+
+// Part(b)
+Mt = 1.00                                                                        // mach number at the throat
+// From table 9.1
+M2 = 0.26
+T2byTnot = 0.986                                                                 
+p2bypnot = 0.953
+
+T0 = 280.00                                                                      // in kelvin
+pnot = 6.8                                                                       // in bars
+// Calculations
+T2 = T2byTnot*T0                                                                 // in kelvin
+p2 = p2bypnot*pnot                                                               // in bars
+k = 1.4
+V2 = M2*((k*(8314/28.97)*T2)**0.5)                                                  // exit velocity in m/s
+mdot = (p2/((8.314/28.97)*T2))*Ae*V2*10**-2                                      // mass flow rate in kg/s
+// Results
+printf( ' Part(b)  the mass flow rate  is:  %.f kg/s.',mdot)
+printf( ' The exit pressure is: %f bars. ',p2)
+printf( ' The exit mach number is:  %f',M2)
+
+// Part(c)
+// From part (b), the exit Mach number in the present part of the example is
+M2 = 2.4
+// Using this, Table 9.1 gives
+p2bypnot = 0.0684
+pnot = 6.8                                                                       // in bars
+// Calculation
+p2 = p2bypnot*pnot                                                               // in bars
+// Results
+// Since the nozzle is choked, the mass flow rate is the same as found in part (b).
+printf( ' Part(c)  the mass flow rate is:  %f kg/s.',mdot)
+printf( ' The exit pressure is:  %f bars.',p2)
+printf( ' The exit mach number is:  %f',M2)
+
+// Part(d)
+// Since a normal shock stands at the exit and the flow upstream of the shock is isentropic, the Mach number Mx and the pressure px correspond to the values found in part (c), 
+Mx = 2.4
+px = 0.465                                                                       // in bars
+// Then, from Table 9.2
+My = 0.52
+//py is the exit pressure
+pybypx = 6.5533
+py = px*pybypx
+
+// The pressure downstream of the shock is thus 3.047 bars. This is the exit pressure
+// The mass flow is the same as found in part (b).
+// Results
+printf( ' Part(d)  the mass flow rate is:  %f kg/s.',mdot)
+printf( ' The exit pressure is:  %.3f bars.',py)
+printf( ' The exit mach number is:  %f',My)
+
+// Part(e)
+// A shock stands in the diverging portion where the area is
+Ax = 12.5                                                                       // in cm^2
+// Since a shock occurs, the flow is sonic at the throat, so
+Axstar = 6.25                                                                   // in cm^2
+At = Axstar
+// The Mach number Mx can then be found from Table 9.1, by using AxbyAxstar as
+Mx = 2.2
+
+// Results
+// With Mx = 2.2, the ratio of stagnation pressures is obtained from Table 9.2 as
+pnotybypnotx = 0.62812
+
+// Using this ratio and noting that the flow is subsonic after the shock, Table 9.1 gives
+M2 = 0.43
+// For M2 = 0.43,
+p2bypnoty = 0.88
+// Calculations
+A2byAystar = (Ae/Axstar)*pnotybypnotx
+p2 = p2bypnoty*pnotybypnotx*pnot                                                // in bars
+
+// Results
+// Since the flow is choked, the mass flow rate is the same as that found in part (b).
+printf( ' part(e)  the mass flow rate is: %f kg/s.',mdot)
+printf( ' the exit pressure is:  %f bars',p2)
+printf( ' the exit mach number is:  %f',M2)
diff --git a/3012/CH9/EX9.2/Ex9_2.sce b/3012/CH9/EX9.2/Ex9_2.sce
new file mode 100755
index 000000000..7212c3153
--- /dev/null
+++ b/3012/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,61 @@
+
+// Given :-
+clc;
+r = 18.00                                                                   // compression ratio
+T1 = 300.00                                                                 // temperature at the beginning of the compression process in kelvin
+p1 = 0.1                                                                    // pressure at the beginning of the compression process in MPa
+rc = 2.00                                                                   // cutoff ratio
+
+// Part(a)
+// With T1 = 300 K, Table A-22 gives
+u1 = 214.07                                                                 // in kj/kg
+vr1 = 621.2                             
+// Interpolating in Table A-22, we get
+T2 = 898.3                                                                  // in kelvin            
+h2 = 930.98                                                                 // in kj/kg
+// From Table A-22,
+h3 = 1999.1                                                                 // in kj/kg
+vr3 = 3.97
+
+// Interpolating in Table A-22 with vr4, we get
+u4 = 664.3                                                                  // in kj/kg
+T4 = 887.7                                                                  // in kelvin
+
+// Calculations
+// Since Process 2–3 occurs at constant pressure, the ideal gas equation of state gives
+T3 = rc*T2                                                                  // in kelvin
+// With the ideal gas equation of state
+p2 = p1*(T2/T1)*(r)                                                         // in MPa
+p3 = p2
+// For the isentropic compression process 1–2
+vr2 = vr1/r
+// For the isentropic expansion process 3–4
+vr4 = (r/rc)*vr3
+// The ideal gas equation of state applied at states 1 and 4 gives
+p4 = p1*(T4/T1)                                                             // in MPa
+
+// Results
+printf( '\n At state1, the pressure is : %.2f bar.',p1)
+printf( '\n At state1, the temperature is %.2f kelvin.' ,T1)
+printf( '\n At state2, the pressure in bar is : %.2f bar.',p2)
+printf( '\n At state2, the temperature is  %.2f kelvin.',T2)
+printf( '\n At state3, the pressure in bar is : %.2f bar.',p3)
+printf( '\n At state3, the temperature is %.2f kelvin.',T3)
+printf( '\n At state4, the pressure is: %.2f MPa.',p4)
+printf( '\n At state4, the temperature is %.2f kelvin.',T4)
+
+// Part(b)
+eta = 1- (u4-u1)/(h3-h2)
+printf( '\n The thermal efficiency is : %.2f ',eta)
+
+// Part(c)
+R = 8.314                                                                    // universal gas constant, in SI units
+M = 28.97                                                                    // molar mass of air in grams
+
+// Calculations
+wcycle = (h3-h2)-(u4-u1)                                                     // The net work of the cycle in kj/kg
+v1 = ((R/M)*T1/p1)/10**3                                                      // The specific volume at state 1 in m^3/kg
+mep = (wcycle/(v1*(1-1/r)))*10**3*10**-6                                       // in MPa
+
+// Results
+printf( '\n The mean effective pressure, is : %.2f MPa.',mep)
diff --git a/3012/CH9/EX9.3/Ex9_3.sce b/3012/CH9/EX9.3/Ex9_3.sce
new file mode 100755
index 000000000..bcdaa8d86
--- /dev/null
+++ b/3012/CH9/EX9.3/Ex9_3.sce
@@ -0,0 +1,44 @@
+// Given :-
+T1 = 300.00                                                                     // beginning temperature in kelvin
+p1 = 0.1                                                                        // beginning pressure in MPa
+r = 18.00                                                                       // compression ratio
+pr = 1.5                                                                        // The pressure ratio for the constant volume part of the heating process
+vr = 1.2                                                                        // The volume ratio for the constant pressure part of the heating process
+
+// Analysis
+// States 1 and 2 are the same as in Example 9.2, so 
+u1 = 214.07                                                                     // in kj/kg
+T2 = 898.3                                                                      // in kelvin
+u2 = 673.2                                                                      // in kj/kg
+
+// Interpolating in Table A-22, we get
+h3 = 1452.6                                                                     // in kj/kg
+u3 = 1065.8                                                                     // in kj/kg
+
+// From Table A-22,
+h4 = 1778.3                                                                     // in kj/kg
+vr4 = 5.609
+
+// Interpolating in Table A-22, we get
+u5 = 475.96                                                                     // in kj/kg
+
+// Calculations
+// Since Process 2–3 occurs at constant volume, the ideal gas equation of state reduces to give
+T3 = pr*T2                                                                      // in kelvin
+// Since Process 3–4 occurs at constant pressure, the ideal gas equation of state reduces to give
+T4 = vr*T3                                                                      // in kelvin
+// Process 4–5 is an isentropic expansion, so
+vr5 = vr4*r/vr
+
+// Part(a)
+eta = 1-(u5-u1)/((u3-u2)+(h4-h3))
+// Result
+printf( ' The thermal efficiency is : %.2f',eta)
+
+// Part(b)
+// The specific volume at state 1 is evaluated in Example 9.2 as
+v1 = 0.861                                                                       // in m^3/kg
+mep = (((u3-u2)+(h4-h3)-(u5-u1))/(v1*(1-1/r)))*10**3*10**-6                      // in MPa
+
+// Result
+printf( ' The mean effective pressure, is : %.2f MPa.',mep)
diff --git a/3012/CH9/EX9.4/Ex9_4.sce b/3012/CH9/EX9.4/Ex9_4.sce
new file mode 100755
index 000000000..7c464f298
--- /dev/null
+++ b/3012/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,45 @@
+// Given:-
+T1 = 300.00                                                                     // in kelvin
+AV = 5.00                                                                       // volumetric flow rate in m^3/s
+p1 = 100.00                                                                     // in kpa
+pr = 10.00                                                                      // compressor pressure ratio
+T3 = 1400.00                                                                    // turbine inlet temperature in kelvin
+
+// Analysis
+// At state 1, the temperature is 300 K. From Table A-22,
+h1 = 300.19                                                                     // in kj/kg
+pr1 = 1.386
+
+
+// Interpolating in Table A-22,
+h2 = 579.9                                                                      // in kj/kg
+// From Table A-22
+h3 = 1515.4                                                                     // in kj/kg
+pr3 = 450.5
+
+// Interpolating in Table A-22, we get
+h4 = 808.5                                                                      // in kj/kg
+
+// calculations
+pr2 = pr*pr1
+pr4 = pr3*1/pr
+
+
+// Part(a)
+eta = ((h3-h4)-(h2-h1))/(h3-h2)                                                 // thermal efficiency
+// Result
+printf( ' The thermal efficiency is : %.4f ',eta)
+
+// Part(b)
+bwr = (h2-h1)/(h3-h4)                                                           // back work ratio
+// Result
+printf( ' The back work ratio is : %.4f',bwr)
+
+// Part(c)
+R = 8.314                                                                       // universal gas constant, in SI units
+M = 28.97                                                                       // molar mass of air in grams
+// Calculations
+mdot = AV*p1/((R/M)*T1)                                                         // mass flow rate in kg/s
+Wcycledot = mdot*((h3-h4)-(h2-h1))                                              // The net power developed
+// Result
+printf( ' The net power developed, is : %.2f kW .',Wcycledot)
diff --git a/3012/CH9/EX9.6/Ex9_6.sce b/3012/CH9/EX9.6/Ex9_6.sce
new file mode 100755
index 000000000..97b22bf99
--- /dev/null
+++ b/3012/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,53 @@
+// Given:-
+T1 = 300.00                                                                     // in kelvin
+AV = 5.00                                                                       // volumetric flow rate in m^3/s
+p1 = 100.00                                                                     // in kpa
+pr = 10.00                                                                      // compressor pressure ratio
+T3 = 1400.00                                                                    // turbine inlet temperature in kelvin
+Wt_ms = 706.9                                                                   // kJ/kg
+Wc_m = 279.7                           
+// Analysis
+// At state 1, the temperature is 300 K. From Table A-22,
+h1 = 300.19                                                                     // in kj/kg
+pr1 = 1.386
+
+
+// Interpolating in Table A-22,
+h2 = 579.9                                                                      // in kj/kg
+// From Table A-22
+h3 = 1515.4                                                                     // in kj/kg
+pr3 = 450.5
+
+// Interpolating in Table A-22, we get
+h4 = 808.5                                                                      // in kj/kg
+
+// calculations
+Wtbym = 0.8*Wt_ms
+Wcbym = Wc_m/0.8
+h2 = 300.19 + Wcbym
+
+//pr2 = pr*pr1
+//pr4 = pr3*1/pr
+
+
+// Part(a)
+//eta = ((h3-h4)-(h2-h1))/(h3-h2)                                                 // thermal efficiency
+Qinbym = h3 - h2
+n = (Wtbym-Wcbym)/Qinbym
+// Result
+printf( '\n The thermal efficiency is : %.3f ',n)
+
+// Part(b)
+//bwr = (h2-h1)/(h3-h4)                                                           // back work ratio
+bwr = Wcbym/Wtbym
+// Result
+printf( '\n The back work ratio is : %.3f',bwr)
+
+// Part(c)
+R = 8.314                                                                       // universal gas constant, in SI units
+M = 28.97                                                                       // molar mass of air in grams
+// Calculations
+//mdot = AV*p1/((R/M)*T1)                                                         // mass flow rate in kg/s
+Wcycledot = 5.807*(Wcbym-Wtbym)                                              // The net power developed
+// Result
+printf( '\n The net power developed, is : %.f kW .',-Wcycledot)
diff --git a/3012/CH9/EX9.7/Ex9_7.sce b/3012/CH9/EX9.7/Ex9_7.sce
new file mode 100755
index 000000000..00df35c49
--- /dev/null
+++ b/3012/CH9/EX9.7/Ex9_7.sce
@@ -0,0 +1,26 @@
+// Given:-
+// Part(a)
+etareg = 0.8                                                                     // regenerator effectiveness of 80%.
+// From example 9.4
+h1 = 300.19                                                                      // in kj/kg
+h2 = 579.9                                                                       // in kj/kg
+h3 = 1515.4                                                                      // in kj/kg
+h4 = 808.5                                                                       // in kj/kg
+
+// Calculations
+hx = etareg*(h4-h2)+h2                                                           // in kj/kg
+eta = ((h3-h4)- (h2-h1))/(h3-hx)                                                 // thermal efficiency
+// Result
+printf('The thermal efficiency is:  %.2f',eta)
+
+// Part(b)
+
+etareg = linspace(0,0.8,50)
+for i = 1:50
+    x(i) = (etareg(i)*(h4-h2))+h2   
+    eta(i) = ((h3-h4)- (h2-h1))/(h3-x(i))
+end
+
+plot(etareg,eta)
+xlabel('Regenerator effectiveness')
+ylabel('Thermal efficiency')
diff --git a/3012/CH9/EX9.8/Ex9_8.sce b/3012/CH9/EX9.8/Ex9_8.sce
new file mode 100755
index 000000000..02f1cf4ed
--- /dev/null
+++ b/3012/CH9/EX9.8/Ex9_8.sce
@@ -0,0 +1,31 @@
+// Given:-
+// Analysis
+// States 1, 2, and 3 are the same as in Example 9.4:
+h1 = 300.19                                                                     // in kj/kg
+h2 = 579.9                                                                      // in kj/kg
+h3 =  1515.4                                                                    // in kj/kg
+// The temperature at state b is the same as at state 3, so   
+hb = h3
+
+pa = 300.00                                                                     // in kpa
+p3 = 1000.00                                                                    // in kpa
+// From table A-22
+pr3 = 450.5
+
+// Interpolating in Table A-22, we get
+ha = 1095.9                                                                     // in kj/kg
+p4 = 100.00                                                                     // in kpa
+pb = 300.00                                                                     // in kpa
+// Interpolating in Table A-22, we obtain
+h4 = 1127.6                                                                     // in kj/kg
+
+// Calculions
+pra = pr3*(pa/p3)
+prb = pra
+pr4 = prb*(p4/pb)
+// Since the regenerator effectiveness is 100%,
+hx = h4
+eta = ((h3-ha)+(hb-h4)-(h2-h1))/((h3-hx)+(hb-ha))                               // thermal efficiency
+
+// Result
+printf( ' The thermal efficiency is : %.2f ', eta)
diff --git a/3012/CH9/EX9.9/Ex9_9.sce b/3012/CH9/EX9.9/Ex9_9.sce
new file mode 100755
index 000000000..03e988cb6
--- /dev/null
+++ b/3012/CH9/EX9.9/Ex9_9.sce
@@ -0,0 +1,46 @@
+// Given:-
+T1 = 300.00                                                                     // in kelvin
+p1 = 100.00                                                                     // in kpa
+p2 = 1000.00                                                                    // in kpa
+p3 = p2
+pc = 300.00                                                                     // in kpa
+pd = 300.00                                                                     // in kpa
+Td = 300.00                                                                     // in kelvin
+
+
+// Part(a)
+// From table A-22
+prd = 1.386
+// Interpolating in Table A-22, we get
+T2 = 422                                                                        // in kelvin
+h2 = 423.8                                                                      // in kj/kg
+// Calculations
+pr2 = prd*(p2/pd)
+// Result
+printf( ' The temperature at the exit of the second compressor stage  is : %.2f kelvin.',T2)
+
+// Part(b)
+// From Table A-22 at T1 = 300
+h1 = 300.19                                                                     // in kj/kg
+// Since Td = T1,
+hd = 300.19                                                                     // in kj/kg
+// with pr data from Table A-22 together
+pr1 = 1.386
+// Interpolating in Table A-22, we obtain
+hc = 411.3                                                                      // in kj/kg
+// Calculations
+prc = pr1*(pc/p1)
+wcdot = (hc-h1)+(h2-hd)                                                         // The total compressor work per unit of mass in kj/kg
+// Result
+printf( ' The total compressor work input per unit of mass flow is : %.2f kJ/kg',wcdot)
+
+// Part(c)
+// Interpolating in Table A-22, we get
+T3 = 574                                                                        // in kelvin
+h3 = 579.9                                                                      // in kj/kg
+// Calculations
+pr3 = pr1*(p3/p1)
+wcdot = h3-h1                                                                   // The work input for a single stage of compression in kj/kg
+// Results
+printf( ' For a single stage of compression, the temperature at the exit state is : %.2f kelvin',T3)
+printf( ' For a single stage of compression, the work input is : %.2f kJ.',wcdot)