initial commit / add all books

10 files changed, 373 insertions, 0 deletions

diff --git a/3012/CH7/EX7.1/Ex7_1.sce b/3012/CH7/EX7.1/Ex7_1.sce
new file mode 100755
index 000000000..77763035a
--- /dev/null
+++ b/3012/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,19 @@
+// Given:-
+v = 2450.00                                    // volume of gaseous products in cm^3
+P = 7.00                                       // pressure of gaseous product in bar
+T = 867.00                                     // temperature of gaseous product in degree celcius
+T0 = 300.00                                    // in kelvin
+P0 = 1.013                                     // in bar
+
+// From table A-22
+u = 880.35                                     // in kj/kg
+u0 = 214.07                                    // in kj/kg
+s0T = 3.11883                                  // in kj/kg.k
+s0T0 = 1.70203                                 // in kj/kg.k
+
+// Calculations
+
+e = (u-u0) + (P0*(8.314/28.97)*(((T+273)/P)-(T0/P0))) - T0*(s0T-s0T0-(8.314/28.97)*log(P/P0))  // kj/kg
+
+// Results
+printf( ' The specific exergy of the gas is %.3f kJ/kg.',e)
diff --git a/3012/CH7/EX7.10/Ex7_10.sce b/3012/CH7/EX7.10/Ex7_10.sce
new file mode 100755
index 000000000..9d703feb2
--- /dev/null
+++ b/3012/CH7/EX7.10/Ex7_10.sce
@@ -0,0 +1,49 @@
+// Given:-
+EfFdot = 100.00                                             // exergy rate of fuel entering the boiler in MW
+cF = 1.44                                                   // unit cost of fuel in cents per kw.h
+Zbdot = 1080.00                                             // the cost of owning and operating boiler in dollars per hour
+Ef1dot = 35.00                                              // exergy rate of exiting steam from the boiler in MW
+p1 = 50.00                                                  // pressure of exiting steam from the boiler in bar
+T1 = 466.00                                                 // temperature of exiting steam from the boiler in degree celcius
+Ztdot = 92.00                                               // the cost of owning and operating turbine in dollars per hour
+p2 = 5.00                                                   // pressure of exiting steam from the turbine in bars
+T2 = 205.00                                                 // temperature of exiting steam from the turbine in degree celcius
+m2dot = 26.15                                               // mass flow rate of exiting steam from the turbine in kg/s
+T0 = 298.00                                                 // in kelvin 
+
+
+// Part(a)
+// From table A-4,
+h1 = 3353.54                                                // in kj/kg
+h2 = 2865.96                                                // in kj/kg
+s1 = 6.8773                                                 // in kj/kg.k
+s2 = 7.0806                                                 // in kj/kg.k
+
+// Calculations
+// From assumption,For each control volume,Qcvdot = 0 and kinetic and potential energy effects are negligible,the mass and energy rate 
+// balances for a control volume enclosing the turbine reduce at steady state to give
+Wedot = m2dot *(h1-h2)/1000                                 // power in MW
+Ef2dot = Ef1dot+m2dot*(h2-h1-T0*(s2-s1))/1000               // the rate exergy exits with the steam in MW
+
+// Results
+printf( ' For the turbine,the power is %.2f MW.',Wedot)
+printf( ' For the turbine,the rate exergy exits with the steam is %.2f MW.',Ef2dot)
+
+// Part(b)
+// Calculations
+c1 = cF*(EfFdot/Ef1dot) + ((Zbdot/Ef1dot)/10**3)*100         // unit cost of exiting steam from boiler in cents/Kw.h
+c2 = c1                                                      // Assigning the same unit cost to the steam entering and exiting the turbine
+ce = c1*((Ef1dot-Ef2dot)/Wedot) + ((Ztdot/Wedot)/10**3)*100  // unit cost of power in cents/kw.h
+
+// Results
+printf('The unit costs of the steam exiting the boiler of exergy is: %.2f cents per kw.h.',c1)
+printf('The unit costs of the steam exiting the turbine of exergy is: %.2f cents per kw.h.',c2)
+printf('Unit cost of power is: %f cents per kw.h.',ce)
+
+// Part(c)
+C2dot = (c2*Ef2dot*10**3)/100                                // cost rate for low-pressure steam in dollars per hour
+Cedot = (ce*Wedot*10**3)/100                                 // cost rate for power in dollars per hour
+
+// Results
+printf( ' The cost rate of the steam exiting the turbine is: %.2f dollars per hour.',C2dot)
+printf( ' The cost rate of the power is: %.2f dollars per hour.',Cedot)
diff --git a/3012/CH7/EX7.2/Ex7_2.sce b/3012/CH7/EX7.2/Ex7_2.sce
new file mode 100755
index 000000000..af9546a57
--- /dev/null
+++ b/3012/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,52 @@
+// Given:-
+mR = 1.11                               // mass of the refrigerant in kg
+T1 = -28.00                             // initial temperature of the saturated vapor in degree celcius
+P2 = 1.4                                // final pressure of the refrigerant in bar
+T0 = 293.00                             // in kelvin
+P0 = 1.00                               // in bar
+
+// Part (a)
+// From table A-10
+u1 = 211.29                             // in kj/kg
+v1 = 0.2052                             // in m^3/kg
+s1 = 0.9411                             // in kj/kg.k
+// From table A-12
+u0 = 246.67                             // in kj/kg
+v0 = 0.23349                            // in m^3/kg
+s0 = 1.0829                             // in kj/kg.k
+
+// From table A-12
+u2 = 300.16                             // in kj/kg
+s2 = 1.2369                             // in kj/kg.k
+v2 = v1
+
+// Calculations 
+E1 = mR*((u1-u0) + P0*(10**5)*(v1-v0)*(10**(-3))-T0*(s1-s0))
+E2 = mR*((u2-u0) + P0*(10**5)*(v2-v0)*(10**(-3))-T0*(s2-s0))
+
+// Results for Part A
+printf( ' Part(a) The initial exergy is %.2f kJ.',E1)
+printf( ' The final exergy is %.2f kJ.',E2)
+printf( ' The change in exergy of the refrigerant is %.2f kj',E2-E1)
+
+
+// Part (b)
+// Calculations
+deltaU = mR*(u2-u1)
+// From energy balance
+deltaPE = -deltaU
+// With the assumption::The only significant changes of state are experienced by the refrigerant and the suspended mass. For the refrigerant, 
+// there is no change in kinetic or potential energy. For the suspended mass, there is no change in kinetic or internal energy. Elevation is 
+// the only intensive property of the suspended mass that changes
+deltaE = deltaPE
+
+// Results for part b
+printf( ' Part(b)The change in exergy of the suspended mass is %.3f kJ',deltaE)
+
+
+// Part(c)
+// Calculations
+deltaEiso = (E2-E1) + deltaE
+
+// Results
+printf( ' Part(c)The change in exergy of an isolated system of the vessel and pulley–mass assembly is  %.2f kJ',deltaEiso)
diff --git a/3012/CH7/EX7.3/Ex7_3.sce b/3012/CH7/EX7.3/Ex7_3.sce
new file mode 100755
index 000000000..1055014ad
--- /dev/null
+++ b/3012/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,47 @@
+// Given :-
+T = 373.15                                              // initial temperature of saturated liquid in kelvin
+T0 = 293.15                                             // in kelvin
+P0 = 1.014                                              // in bar
+
+// Part(a)
+// From table A-2
+ug = 2506.5                                             // in kj/kg
+uf = 418.94                                             // in kj/kg
+vg = 1.673                                              // in m^3/kg
+vf = 1.0435*(10**(-3))                                  // in m^3/kg
+sg = 7.3549                                             // in kj/kg.k
+sf = 1.3069                                             // in kj/kg.k
+
+
+// Calculations 
+// Energy transfer accompanying work
+etaw = 0                                                // since p = p0
+// Exergy transfer accompanying heat
+Q = 2257                                                // in kj/kg,obtained from example 6.1
+etah = (1-(T0/T))*Q
+
+// Exergy destruction
+ed = 0                                                  // since the process is accomplished without any irreversibilities
+deltae = ug-uf + P0*(10**5)*(vg-vf)/(10**3)-T0*(sg-sf)
+
+// Results
+printf( ' Part(a)the change in exergy is %.2f kJ/kg.',deltae)
+printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',etaw)
+printf( ' The exergy transfer accompanying heat is %.2f kJ/kg',etah)
+printf( ' The exergy destruction is %.2f kJ/kg.',ed)
+
+
+// Part(b)
+Deltae = deltae                                          // since the end states are same 
+Etah = 0                                                 // since process is adiabatic
+// Exergy transfer along work
+W = -2087.56                                             // in kj/kg from example 6.2
+Etaw = W- P0*(10**5)*(vg-vf)/(10**3)
+// Exergy destruction
+Ed = -(Deltae+Etaw)
+
+// Results
+printf( ' Part(b)the change in exergy is %.2f kJ/kg.',Deltae)
+printf( ' The exergy transfer accompanying work is %.2f kJ/kg.',Etaw)
+printf( ' The exergy transfer accompanying heat is %.2f kJ/kg.',Etah)
+printf( ' The exergy destruction is %.2f kJ/kg.',Ed)
diff --git a/3012/CH7/EX7.4/Ex7_4.sce b/3012/CH7/EX7.4/Ex7_4.sce
new file mode 100755
index 000000000..20098082a
--- /dev/null
+++ b/3012/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,16 @@
+// Given:-
+T0 = 293.00                                         // in kelvin
+Qdot = -1.2                                         // in KW, from example 6.4a
+Tb = 300.00                                         // temperature at the outer surface of the gearbox in kelvin from example 6.4a
+sigmadot = 0.004                                    // rate of entropy production in KW/k from example 6.4a
+
+// Calculations
+R = -(1-T0/Tb)*Qdot                                  // time rate of exergy transfer accompanying heat
+Eddot = T0*sigmadot                                 // rate of exergy destruction
+
+// Results
+printf( ' Balance sheet');
+printf( '\n Rate of exergy in high speed shaft 60Kw' )
+printf( '\n Disposition of the exergy: Rate of exergy out low-speed shaft %.1f Kw',58.8 )
+printf( '\n Heat transfer is %.3f kw.',R)
+printf( '\n Rate of exergy destruction is %.3f kw',Eddot)
diff --git a/3012/CH7/EX7.5/Ex7_5.sce b/3012/CH7/EX7.5/Ex7_5.sce
new file mode 100755
index 000000000..e6639a76e
--- /dev/null
+++ b/3012/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,27 @@
+// Given:-
+p1 = 3.0                                              // entry pressure in Mpa
+p2 = 0.5                                              // exit pressure in Mpa
+T1 = 320.0                                            // entry temperature in degree celcius
+T0 = 25.0                                             // in degree celcius
+p0 = 1.0                                              // in atm
+
+// From table A-4
+h1 = 3043.4                                           // in kj/kg
+s1 = 6.6245                                           // in kj/kg.k
+h2 = h1                                               // from reduction of the steady-state mass and energy rate balances
+s2 = 7.4223                                           // Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k
+
+// From table A-2
+h0 = 104.89                                           // in kj/kg
+s0 = 0.3674                                           // in kj/kg.k
+
+// Calculations 
+ef1 = h1-h0-(T0+273)*(s1-s0)                          // flow exergy at the inlet
+ef2 = h2-h0-(T0+273)*(s2-s0)                          // flow exergy at the exit
+// From the steady-state form of the exergy rate balance
+Ed = ef1-ef2                                          // the exergy destruction per unit of mass flowing is
+
+// Results 
+printf( ' The specific flow exergy at the inlet is %.2f kJ/kg.',ef1)
+printf( ' The specific flow exergy at the exit is %.2f kJ/kg.',ef2)
+printf( ' The exergy destruction per unit of mass flowing is %.2f kJ/kg.',Ed)
diff --git a/3012/CH7/EX7.6/Ex7_6.sce b/3012/CH7/EX7.6/Ex7_6.sce
new file mode 100755
index 000000000..2820cf276
--- /dev/null
+++ b/3012/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,49 @@
+// Given:-
+T1 = 610.0                                                // temperature of the air entering heat exchanger in kelvin
+p1 = 10.0                                                 // pressure of the air entering heat exchanger in bar
+T2 = 860.0                                                // temperature of the air exiting the heat exchanger in kelvin
+p2 = 9.70                                                 // pressure of the air exiting the heat exchanger in bar
+T3 = 1020.0                                               // temperature of entering hot combustion gas in kelvin
+p3 = 1.10                                                 // pressure of entering hot  combustion gas in bar
+p4 = 1.0                                                  // pressure of exiting hot combustion gas in bar
+mdot = 90.0                                               // mass flow rate in kg/s
+T0 = 300.0                                                // in kelvin
+p0 = 1.0                                                  // in bar
+
+// Part (a)
+// From table A-22
+h1 = 617.53                                               // in kj/kg
+h2 = 888.27                                               // in kj/kg
+h3 = 1068.89                                              // in kj/kg
+
+// Calculations
+h4 = h3+h1-h2
+
+// Using interpolation in table A-22 gives
+T4 = 778                                                  // in kelvin
+
+// Results
+printf( ' The exit temperature of the combustion gas   is %f kelvin.',T4);
+
+// Part(b)
+// From table A-22
+s2 = 2.79783                                              // in kj/kg.k
+s1 = 2.42644                                              // in kj/kg.k
+s4 = 2.68769                                              // in kj/kg.k
+s3 = 2.99034                                              // in kj/kg.k
+
+// Calculations for part b
+
+deltaR = (mdot*((h2-h1)-T0*(s2-s1-(8.314/28.97)*log(p2/p1))))/1000
+deltRc = mdot*((h4-h3)-T0*(s4-s3-(8.314/28.97)*log(p4/p3)))/1000
+
+// Results for part b
+printf( ' The net change in the flow exergy rate from inlet to exit of compressed gas   is %.3f MW.',deltaR)
+printf( ' The net change in the flow exergy rate from inlet to exit of hot combustion gas   is %.3f MW.',deltRc)
+
+// Part(c)
+//From an exergy rate balance
+Eddot = -deltaR-deltRc
+
+// Results 
+printf( ' The rate exergy  destroyed,   is %.3f MW.',Eddot)
diff --git a/3012/CH7/EX7.7/Ex7_7.sce b/3012/CH7/EX7.7/Ex7_7.sce
new file mode 100755
index 000000000..5793d53e0
--- /dev/null
+++ b/3012/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,35 @@
+// Given:-
+p1 = 30.0                                             // pressure of entering steam in bar
+t1 = 400.0                                            // temperature of entering steam in degree celcius
+v1 = 160.0                                            // velocity of entering steam in m/s
+t2 = 100.0                                            // temperature of exiting saturated vapor in degree celcius
+v2 = 100.0                                            // velocity of exiting saturated vapor in m/s
+W = 540.0                                             // rate of work developed in kj per kg of steam
+Tb = 350.0                                            // the temperature on the boundary where heat transfer occurs in kelvin
+T0 = 25.0                                             // in degree celcius
+p0 = 1.0                                              // in atm
+
+// From table A-4
+h1 = 3230.9                                           // in kj/kg
+s1 = 6.9212                                           // in kj/kg.k
+// From table A-2 
+h2 = 2676.1                                           // in kj/kg
+s2 = 7.3549                                           // in kj/kg.k
+// From example 6.6
+Q = -22.6                                             // in kj/kg
+  
+// Calculations
+DELTAef = (h1-h2)-(T0+273)*(s1-s2)+(v1**2-v2**2)/(2*1000)
+// The net exergy carried in per unit mass of steam flowing in kj/kg
+Eq = (1-(T0+273)/Tb)*(Q)                              // exergy transfer accompanying heat in kj/kg
+Ed = ((1-(T0+273)/Tb)*(Q))-W+(DELTAef)                // The exergy destruction determined by rearranging the steady-state form of the exergy 
+                                                      // rate balance
+
+// Results
+printf( ' Balance sheet')
+printf( ' Net rate of exergy %f kJ/kg,',DELTAef)
+printf( ' Disposition of the exergy:')
+printf( '* Rate of exergy out')
+printf( ' Work %f kJ/kg.',W)
+printf( ' Heat transfer %f',-Eq)
+printf( '• Rate of exergy destruction %f kJ/kg.',Ed)
diff --git a/3012/CH7/EX7.8/Ex7_8.sce b/3012/CH7/EX7.8/Ex7_8.sce
new file mode 100755
index 000000000..5cbf01eed
--- /dev/null
+++ b/3012/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,57 @@
+// Given:-
+clc;
+m1dot = 69.78                                             // in kg/s
+p1 = 1.0                                                  // in bar
+T1 = 478.0                                                // in kelvin
+T2 = 400.0                                                // in kelvin
+p2 = 1.0                                                  // in bar
+p3 = 0.275                                                // in Mpa
+T3 = 38.9                                                 // in degree celcius
+m3dot = 2.08                                              // in kg/s
+T4 = 180.0                                                // in degree celcius
+p4 = 0.275                                                // in Mpa
+p5 = 0.07                                                 // in bar
+x5 = 0.93
+Wcvdot = 876.8                                            // in kW
+T0 = 298.0                                                // in kelvin
+
+
+// Part(a)
+// From table A-22
+h1 = 480.35                                               // in kj/kg
+h2 = 400.97                                               // in kj/kg
+s1 = 2.173                                                // in kj/kg
+s2 = 1.992                                                // in kj/kg
+
+// From table A-2E
+h3 = 162.82                                               // in kj/kg
+s3 = 0.5598                                               // in kj/kg.k
+// Using saturation data at 0.07 bars from Table A-3
+h5 = 2403.27                                              // in kj/kg
+s5 = 7.739                                                // in kj/kg.k
+//The net rate exergy carried out by the water stream
+
+// From table A-4
+h4 = 2825.0                                               // in kj/kg
+s4 = 7.2196                                               // in kj/kg.k
+// Calculations
+netRE = m1dot*(h1-h2-T0*(s1-s2-(8.314/28.97)*log(p1/p2))) // the net rate exergy  carried into the control volume
+netREout = m3dot*(h5-h3-T0*(s5-s3))
+// From an exergy rate balance applied to a control volume enclosing the steam generator
+Eddot = netRE + m3dot*(h3-h4-T0*(s3-s4))                  // the rate exergy is destroyed in the heat-recovery steam generator
+
+// From an exergy rate balance applied to a control volume enclosing the turbine
+EdDot = -Wcvdot + m3dot*(h4-h5-T0*(s4-s5))                // the rate exergy is destroyed in the tpurbine
+
+// Results 
+printf( '\n balance sheet')
+printf( '\n- Net rate of exergy in: %f kJ/kg.',netRE)
+printf( '\n Disposition of the exergy:')
+printf( '\n• Rate of exergy out')
+printf( '\n power developed %f kJ/kg.',netRE-netREout-Eddot-EdDot)
+printf( '\n water stream  %f',netREout)
+printf( '\n• Rate of exergy destruction')
+printf( '\n heat-recovery steam generator %f kJ/kg',Eddot)
+printf( '\n turbine %f',EdDot)
+
+// note : answer is slightly different because of rounding off error.
diff --git a/3012/CH7/EX7.9/Ex7_9.sce b/3012/CH7/EX7.9/Ex7_9.sce
new file mode 100755
index 000000000..5dcf9b32e
--- /dev/null
+++ b/3012/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,22 @@
+// Given:-
+T0 = 273.00                                                 // in kelvin
+pricerate = 0.08                                            // exergy value at $0.08 per kw.h
+
+// From example 6.8
+sigmadotComp = 17.5e-4                                      // in kw/k
+sigmadotValve = 9.94e-4                                     // in kw/k
+sigmadotcond = 7.95e-4                                      // in kw/k
+
+// Calculations
+// The rates of exergy destruction
+EddotComp = T0*sigmadotComp                                 // in kw
+EddotValve = T0*sigmadotValve                               // in kw
+Eddotcond = T0*sigmadotcond                                 // in kw
+
+mCP = 3.11                                                  // From the solution to Example 6.14, the magnitude of the compressor power in kW
+
+// Results
+printf( ' Daily cost in dollars of exergy destruction due to compressor irreversibilities = %.3f',EddotComp*pricerate*24)
+printf( ' Daily cost in dollars of exergy destruction due to irreversibilities in the throttling valve = %.3f',EddotValve*pricerate*24)
+printf( ' Daily cost in dollars of exergy destruction due to irreversibilities in the condenser = %.3f ',Eddotcond*pricerate*24)
+printf( ' Daily cost in dollars  of electricity to operate compressor = %.3f',mCP*pricerate*24)