Air enters a turbine with a mass flow rate of 10kg/s at 1500Kand 1000kPa (Stream

Question

Air enters a turbine with a mass flow rate of 10kg/s at 1500Kand 1000kPa (Stream 1) and leaves at 100kPa (Stream 2). The turbinehas an isentropic efficiency of 85%. Find : a) the actual turbine power b)the actual rate of the entropy change for air during thisprocess Thank you. Air enters a turbine with a mass flow rate of 10kg/s at 1500Kand 1000kPa (Stream 1) and leaves at 100kPa (Stream 2). The turbinehas an isentropic efficiency of 85%. Find : a) the actual turbine power b)the actual rate of the entropy change for air during thisprocess Thank you.

Explanation / Answer

mass flow rate m = 10kg/s initial temperature T1 = 1500 K initial pressure p1= 1000 kPa final pressure p2= 100kPa from ideal gas propertis tables at T1 = 1500 Ktake enthalpy h1 = 1635.97 kJ/kg entropy  s10 = 3.44516kJ/kgK pressure pr1= 601.9 pr2s/pr1=p2/p1 pr2s= (601.9)X(100/1000) = 60.19 kPa with corresponding to pr2s value takehr2s value h2s=866.08+(888.27-866.08)X(60.19-57.60)/(63.09-57.60)) =876.55 kJ/kg = (h1 -h2)/(h1-h2s ) actual turbine powerWt=m(h1 -h2)=0.85X(1635.97-876.55)= 645.5X10 = 6455 kW 6455 kW = m(h1 -h2) 6455 kW = 10X(1635.97-h2) h2 = 990.47 kJ/kg with corresponding to h2 value take s2value from ideal das tables s20=2.89748+(2.92128-2.89748)X(990.47-977.92)/(1000.55-977.92)) s20 =2.910678 the actual rate of the entropy changefor air during this process is s1 -s2= s10-s20-Rln(p2/p1 )                                                                                                            =3.44516-2.910678-0.287 ln (100/1000)                                                                                                            = 1.1953   kJ/kgK                            

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