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+//(9.11)  A regenerative gas turbine with intercooling and reheat operates at steady state. Air enters the compressor at 100 kPa, 300 K with a mass flow rate of 5.807 kg/s. The pressure ratio across the two-stage compressor is 10. The pressure ratio across the two-stage turbine is also 10. The intercooler and reheater each operate at 300 kPa. At the inlets to the turbine stages, the temperature is 1400 K. The temperature at the inlet to the second compressor stage is 300 K. The isentropic efficiency of each compressor and turbine stage is 80%. The regenerator effectiveness is 80%. Determine (a) the thermal efficiency, (b) the back work ratio, (c) the net power developed, in kW.
+
+
+//solution
+
+//variable initialization
+T1 = 300                                                                        //in kelvin
+p1 = 100                                                                        //in kpa
+mdot = 5.807                                                                    //in kg/s
+p2 = 300                                                                        //in kpa
+p3 = p2
+p4 = 1000                                                                       //in kpa
+p5 = p4
+p6 = p4
+T6 = 1400                                                                       //in kelvin
+T8 = T6
+p7 = 300                                                                        //in kpa
+p8 = p7
+etac = .8                                                                       //isentropic efficiency of compressor
+etat = .8                                                                       //isentropic efficiency of turbine
+etareg = .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
+
+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)
+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
+printf('the thermal efficiency is:  %f',eta)
+
+//part(b)
+bwr = wcdot/wtdot                                                               //back work ratio
+printf('\nthe back work ratio is:  %f',bwr)
+
+//part(c)
+Wcycledot = mdot*(wtdot-wcdot)                                                  //net power developed in kw
+printf('\nthe net power developed, in kW is:  %f',Wcycledot)
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