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+
+// Theory and Problems of Thermodynamics
+// Chapter 4
+// Energy Analysis of Process
+// Example 5
+
+clear ;clc;
+
+//Given data
+m_f = 1              // mass flow rate in kg/s
+X1 = 0.8             // wet steam quality
+P1 = 3               // Entering steam Pressure in MPa  
+T = 300             // Temperature in C
+vel_1 = 10          // entering Velocity in m/s
+Z1 = 5              // Initial elevation above the ground level
+    //  Data for discharging steam
+X2 = 0.85           // Discharge steam quality
+P2 = 50             // Discharge steam pressure in kPa
+vel_2 = 50          // discharging Velocity in m/s
+Z2 = 10             // Final elevation above the ground level
+Q = -10             // Energy loss as heat from turbine casing in kJ/s
+g = 9.81            // acceleration due to gravity in m/s
+
+// From Super heated steam tables at P = 3.0 MPa and T = 300 c
+h1 = 2993.5     // enthalpy in kJ/kg
+// From Super heated steam tables at P = 50 kPa
+h_f = 340.49     // in kJ/kg
+h_fg = 2305.4    // in kJ/kg 
+    
+h2 = h_f + X2*h_fg
+
+// Power output expression after subjecting
+
+W_s = Q - ((h2 - h1) + (vel_2^2 - vel_1^2)/2*1e-3 + g*(Z2 - Z1)*1e-3 )
+//in above equation term 2 and term 3 are multiplied with 1e-3 to convert to kW
+
+del_KE_PE   = (vel_2^2 - vel_1^2)/2*1e-3 + g*(Z2 - Z1)*1e-3
+
+err_KE_PE   = del_KE_PE*100/W_s
+
+
+
+// Output Results
+mprintf('Turbine output power = %6.2f kW',W_s)
+mprintf('\n Error percentage ignored in KE and PE terms = %4.3f ',err_KE_PE)
+
+
+
+
+