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+//(8.8)  Reconsider the turbine and pump of Example 8.2. Determine for each of these components the rate at which exergy is destroyed, in MW. Express each result as a percentage of the exergy entering the plant with the fuel. Let T0 = 22C, p0 = 1 atm
+
+//solution
+
+T0 = 295                                                                        //in kelvin
+P0 = 1                                                                          //in atm
+
+//analysis
+//from table A-3
+s1 = 5.7432                                                                     //in kj/kg.k
+//Using h2 = 1939.3 kJ/kg from the solution to Example 8.2, the value of s2 can be determined from Table A-3 as
+s2 = 6.2021                                                                     //in kj/kg.k
+
+mdot = 4.449e5                                                                  //in kg/h
+Eddot = mdot*T0*(s2-s1)/(3600*10^3)                                             //the rate of exergy destruction for the turbine in MW
+printf('the rate of exergy destruction for the turbine in MW is:  %f',Eddot)
+//From the solution to Example 8.7, the net rate at which exergy is supplied by the cooling combustion gases is 231.28 MW
+printf('\nThe turbine rate of exergy destruction expressed as a percentage is:  %f ',(Eddot/231.28)*100)
+//However, since only 69% of the entering fuel exergy remains after the stack loss and combustion exergy destruction are accounted for, it can be concluded that
+printf('\npercentage of the exergy entering the plant with the fuel destroyed within the turbine is:  %f',.69*(Eddot/231.28)*100)
+
+//from table A-3
+s3 =.5926                                                                       //in kj/kg.k
+//from solution of example 8.7
+s4 = .5957                                                                      //in kj/kg.k
+EddotP = mdot*T0*(s4-s3)/(3600*10^3)                                            //the exergy destruction rate for the pump
+printf('\n\nthe exergy destruction rate for the pump in MW is:  %f',EddotP)
+printf('  and expressing this as a percentage of the exergy entering the plant as calculated above, we have  %f',(EddotP/231.28)*69 )
+
+printf('\n\nThe net power output of the vapor power plant of Example 8.2 is 100 MW. Expressing this as a percentage of the rate at which exergy is carried into the plant with the fuel,  %f',(100/231.28)*69)
+\ No newline at end of file