11 files changed, 320 insertions, 0 deletions
diff --git a/647/CH7/EX7.1/Example7_1.sce b/647/CH7/EX7.1/Example7_1.sce
new file mode 100755
index 000000000..56a70f23a
--- /dev/null
+++ b/647/CH7/EX7.1/Example7_1.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+
+// Example: 7.1
+// Page: 256
+
+printf("Example: 7.1 - Page: 256\n\n");
+
+// Solution
+
+// *****Data******//
+d1 = 0.15;// [inlet dia, m]
+d2 = 0.20;// [outlet dia, m]
+U1 = 7;// [inlet velocity, m/s]
+//****************//
+
+// From Fig. 7.2 (Pg 256)
+// At the inlet:
+A1 = (%pi/4)*d1^2;// [square m]
+// At the outlet:
+A2 = (%pi/4)*d2^2;// [square m]
+Q = A1*U1;// [cubic m/s]
+printf("Flow rate is %.4f m/s\n",Q);
+// Using Continuity Eqn.
+// density1*U1*A1 = Density2*U2*A2
+// For water: Density1 = Density2. Therefore:
+U2 = A1*U1/A2;
+printf("Velocity of water at the outlet is %.3f m/s",U2);
+\ No newline at end of file
diff --git a/647/CH7/EX7.10/Example7_10.sce b/647/CH7/EX7.10/Example7_10.sce
new file mode 100755
index 000000000..386fb15bd
--- /dev/null
+++ b/647/CH7/EX7.10/Example7_10.sce
@@ -0,0 +1,30 @@
+clear;
+clc;
+
+// Example: 7.10
+// Page: 280
+
+printf("Example: 7.10 - Page: 280\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 800;// [kPa]
+T1 = 773;// [K]
+H1 = 3480;// [kJ/kg]
+P2 = 100;// [kPa]
+T2 = 573;// [K]
+H2 = 3074;// [kJ/kg]
+//***************//
+
+// Solution (a)
+// Velocity of the fluid exiting the nozzle:
+// U2 = sqrt(U1^2 + 2*(H1 - H2))
+// Neglecting initial velocity:
+U2 = sqrt(2*(H1 - H2)*1000);// [m/s]
+printf("(a) Final Velocity is %.2f m/s\n",U2);
+
+// Solution (b)
+U1 = 40;// [m/s]
+U2 = sqrt((U1^2 + 2*(H1 - H2))*1000);// [m/s]
+printf("(b) Final Velocity is %.2f m/s\n",U2);
+\ No newline at end of file
diff --git a/647/CH7/EX7.11/Example7_11.sce b/647/CH7/EX7.11/Example7_11.sce
new file mode 100755
index 000000000..564f93cfd
--- /dev/null
+++ b/647/CH7/EX7.11/Example7_11.sce
@@ -0,0 +1,39 @@
+clear;
+clc;
+
+// Example: 7.11
+// Page: 281
+
+printf("Example: 7.11 - Page: 281\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 200;// [OC]
+U1 = 190;// [m/s]
+A1 = 2000/10^4;// [square m]
+U2 = 70;// [m/s]
+P2 = 200;// [kPa]
+Qdot = 100;// [kW]
+V1 = 2.172;// [cubic m/kg]
+H1 = 2875.3;// [kJ/kg]
+//***************//
+
+// Solution (a)
+mdot = U1*A1/V1;// [kg/s]
+printf("Mass flow rate of the steam is %.2f kg/s\n",mdot);
+
+// Solution (b)
+// Amount of heat transferred to the surrounding per unit steam:
+Q = Qdot/mdot;// [kJ/kg]
+// The Enthalpy at the diffuser outlet can be estimated as:
+H2 = Q + H1 + (U1^2 - U2^2)/2;// [kJ/kg]
+// From the steam table:
+T2 = 393.38;// [K]
+V2 = 1.123;// [cubic m/kg]
+printf("The temperature of the steam leaving the outlet is %.2f K\n",T2);
+
+// Solution (c)
+A2 = V2*mdot/U2;// [square m]
+printf("Area of diffuser outlet is %.2f square m\n",A2);
+\ No newline at end of file
diff --git a/647/CH7/EX7.2/Example7_2.sce b/647/CH7/EX7.2/Example7_2.sce
new file mode 100755
index 000000000..4e5ddf68c
--- /dev/null
+++ b/647/CH7/EX7.2/Example7_2.sce
@@ -0,0 +1,32 @@
+clear;
+clc;
+
+// Example: 7.2
+// Page: 257
+
+printf("Example: 7.2 - Page: 257\n\n");
+
+// Solution
+
+//*****Data******//
+d1 = 0.2;// [m]
+d2 = 0.15;// [m]
+d3 = 0.1;// [m]
+U1 = 3;// [m/s]
+U2 = 2.5;// [m/s]
+//**************//
+
+// From Fig. 7.3 (Pg: 257)
+// For pipe I:
+A1 = (%pi/4)*d1^2;// [square m]
+Q1 = A1*U1;// [cubic m/s]
+// For pipe II:
+A2 = (%pi/4)*d2^2;// [square m]
+Q2 = A2*U2;// [cubic m/s]
+// For pipe III:
+A3 = (%pi/4)*d3^2;// [square m]
+// From continuity Eqn.:
+Q3 = Q1 - Q2;// [cubic m/s]
+U3 = Q3/A3;// [m/s]
+printf("Discharge through the 10 cm pipe is %.4f cubic m/sec\n",Q1);
+printf("Average velocity in the 15 cm pipe is %.2f m/s",U3);
+\ No newline at end of file
diff --git a/647/CH7/EX7.3/Example7_3.sce b/647/CH7/EX7.3/Example7_3.sce
new file mode 100755
index 000000000..863ea29be
--- /dev/null
+++ b/647/CH7/EX7.3/Example7_3.sce
@@ -0,0 +1,29 @@
+clear;
+clc;
+
+// Example: 7.3
+// Page: 262
+
+printf("Example: 7.3 - Page: 262\n\n");
+
+// Solution
+
+//*****Data******//
+d1 = 0.3;// [m]
+d2 = 0715;//[m]
+Q = 40/1000;// [cubic m/s]
+Z1 = 8;// [m]
+Z2 = 6;// [m]
+P1 = 5*10^5;// [Pa]
+density = 1000;// [kg/cubic m]
+g = 9.81;// [m/square s]
+//*************//
+
+// From Fig. 7.3 (Pg: 262)
+A1 = (%pi/4)*d1^2;// [square m]
+A2 = (%pi/4)*d2^2;// [square m]
+U1 = Q/A1;// [m/s]
+U2 = Q/A2;// [m/s]
+// Applying Bernoulli's equations at sections 1 & 2:
+P2 =  ((U1^2/(2*g) + Z1 + P1/(density*g)) - (U2^2/(2*g) + Z2))*(density*g);// [Pa]
+printf("Pressure at section 2 is %.2f bar",P2/10^5);
+\ No newline at end of file
diff --git a/647/CH7/EX7.4/Example7_4.sce b/647/CH7/EX7.4/Example7_4.sce
new file mode 100755
index 000000000..ad744d058
--- /dev/null
+++ b/647/CH7/EX7.4/Example7_4.sce
@@ -0,0 +1,32 @@
+clear;
+clc;
+
+// Example: 7.4
+// Page: 268
+
+printf("Example: 7.4 - Page: 268\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 320;// [K]
+P2 = 600;// [kPa]
+T2 = 430;// [K]
+m_dot = 0.03;// [kg/s]
+Qout = 15;// [kJ/kg]
+//*************//
+
+// The energy balance around the compressor:
+// dE_System/dt = Ein - Eout
+// Since it is a steady state process:
+// dE_Sysytem/dt = 0
+// Ein = Eout
+// Win + m_dot*H1 = Qout + m_dot*H2
+// Since, Qout = Qout/m
+// Win = m_dot*(Qout + (H2 - H1))
+// From enthalpy chart of air:
+H1 = 320.20;// [Enthalpy of air at 320 K, kJ/kg]
+H2 = 431.43;// [Enthalpy of air at 430 K, kJ/kg]
+Win = m_dot*(Qout + (H2 - H1));// [kW]
+printf("Power Requirement of the compressor is %.2f kW",Win);
+\ No newline at end of file
diff --git a/647/CH7/EX7.5/Example7_5.sce b/647/CH7/EX7.5/Example7_5.sce
new file mode 100755
index 000000000..74d1a60fb
--- /dev/null
+++ b/647/CH7/EX7.5/Example7_5.sce
@@ -0,0 +1,39 @@
+clear;
+clc;
+
+// Example: 7.5
+// Page: 269
+
+printf("Example: 7.5 - Page: 269\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 250;// [K]
+Q = 0.1;// [cubic m/s]
+P2 = 500;// [kPa]
+M = 44;// [g/mol]
+R = 8.314;// [J/mol K]
+//****************//
+
+// Solution (a)
+// Work done by reversible adiabatic compression, gama = 1.4;
+gama = 1.4;
+T2 = T1*(P2/P1)^((gama - 1)/gama);// [K]
+Wad = (gama*R/(gama - 1))*(T1 - T2);// [J/mol]
+Wad = Wad/M;// [J/g]
+printf("Work done by reversible adiabatic compression when gama = 1.4 is %.2f J/g\n",Wad);
+
+// Solution (b)
+// Work done by isothermal compression:
+Wiso = - (R*T1)*log(P2/P1);// [J/mol]
+Wiso = Wiso/M;// [J/g]
+printf("Work done by isothermal compression is %.2f J/g\n",Wiso);
+
+// Solution (c)
+// Work done in single stage compression, gama = 1.3:
+gama = 1.3;
+V1 = Q;// [cubic m]
+Wsingle_stage = (gama*P1*V1/(gama - 1))*(1-(P2/P1)^((gama - 1)/gama));// [kW]
+printf("Work done in single stage compression is %.2f kW",Wsingle_stage);
+\ No newline at end of file
diff --git a/647/CH7/EX7.6/Example7_6.sce b/647/CH7/EX7.6/Example7_6.sce
new file mode 100755
index 000000000..a63fcd742
--- /dev/null
+++ b/647/CH7/EX7.6/Example7_6.sce
diff --git a/647/CH7/EX7.7/Example7_7.sce b/647/CH7/EX7.7/Example7_7.sce
new file mode 100755
index 000000000..691bb18a0
--- /dev/null
+++ b/647/CH7/EX7.7/Example7_7.sce
@@ -0,0 +1,31 @@
+clear;
+clc;
+
+// Example: 7.7
+// Page: 274
+
+printf("Example: 7.7 - Page: 274\n\n");
+
+// Solution
+
+//*****Data******//
+T_steam1 = 50;// [OC]
+T_steam2 = 30;// [OC]
+msteam_dot = 10;// [kg/min]
+T_water1 = 15;// [OC]
+T_water2 = 25;// [OC]
+//***************//
+
+// Solution (a)
+// From the Stem Table:
+H1 = 2645.9;// [kJ/kg, At 50 OC]
+H2 = 768.2;// [kJ/kg, At 30 OC]
+H3 = 62.982;// [kJ/kg, At 15 OC]
+H4 = 104.83;// [kJ/kg, At 25 OC]
+// The mass & Energy balance of the above flow gives:
+mwater_dot = msteam_dot*(H1 - H2)/(H4 - H3);// [kg/min]
+printf("The mass flow rate of water is %.2f kg/min\n",mwater_dot);
+
+// Solution (b)
+Qdot = mwater_dot*(H4 - H3);// [kJ/min]
+printf("The rate of heat transfer is %.2f kJ/min",Qdot);
+\ No newline at end of file
diff --git a/647/CH7/EX7.8/Example7_8.sce b/647/CH7/EX7.8/Example7_8.sce
new file mode 100755
index 000000000..0dc2298f1
--- /dev/null
+++ b/647/CH7/EX7.8/Example7_8.sce
@@ -0,0 +1,30 @@
+clear;
+clc;
+
+// Example: 7.8
+// Page: 279
+
+printf("Example: 7.8 - Page: 279\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 500;// [kPa]
+T1 = 623;// [K]
+mdot = 12;// [kg/s]
+P2 = 500;// [kPa]
+T2 = 523;// [K]
+Qdot = -120;// [kW]
+H1 = 3168;// [kJ/kg]
+H2 = 2976;// [kJ/kg]
+//************//
+
+Q = Qdot/mdot;// [kJ/kg]
+// By energy balance:
+// (deltaU^2/2) + g*deltaZ + deltaH = Q - Ws
+// Considering negligible change in P.E., deltaZ = 0 & Ws = 0.
+// (U2^2 - U1^2)/2 + deltaH = Q
+deltaH = H2 - H1;// [kJ/kg]
+// Neglecting inlet velocity.
+U2 = sqrt(2*(Q - deltaH)*1000);// [m/s]
+printf("Outlrt velocity is %.1f m/s\n",U2);
+\ No newline at end of file
diff --git a/647/CH7/EX7.9/Example7_9.sce b/647/CH7/EX7.9/Example7_9.sce
new file mode 100755
index 000000000..11277cc48
--- /dev/null
+++ b/647/CH7/EX7.9/Example7_9.sce
@@ -0,0 +1,30 @@
+clear;
+clc;
+
+// Example: 7.9
+// Page: 279
+
+printf("Example: 7.9 - Page: 279\n\n");
+
+// Solution
+
+//*****Data******//
+Pin = 1000;// [kPa]
+Tin = 600;// [K]
+Uin = 50;// [m/s]
+gama = 1.4;
+M = 17;// [g/mol]
+R = 8314;// [kJ/mol K]
+MachNumber = 2;
+//***************//
+
+// Solution (i)
+// Using Eqn. (7.36):
+Critical_Ratio = (2/(gama + 1))^(gama/(gama - 1));
+printf("Critical Ratio is %.2f\n",Critical_Ratio);
+
+// Solution (ii)
+PV_in = R*Tin/M;// [square m]
+Uthroat = sqrt(Uin^2 + (2*gama*PV_in/(gama - 1))*(1-(Critical_Ratio)^((gama - 1)/gama)));// [m/s]
+Uout = MachNumber*Uthroat;// [m/s]
+printf("The discharge velocity is %.2f m/s",Uout);
+\ No newline at end of file