Question

A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW

Answer 1

Solution: Given: Inlet condition of turbine (state 1); superheated state pressure P,=9 MPa Temperature T,=650°C Velocity V=60 m/s Outlet condition of turbine (state 2); liquid vapour mixture quality of steam (i.e. dryness fraction) x=0.94 pressure P =325 KPa velocity V2=15 m/s mass flow rate of steam m=8 kg/s heat transfer from turbine Q =-560 KW Assumption: (1) steady flow process. (2) potential energy change is zero. from superheated steam table: at P, =9 MPa and T,=650°C h=3755.2 KJ/kg 1

from saturated steam table: at P,=325 KPa specific enthalpy of saturated liquid(ht) h=573.19 KJ/kg specific enthalpy of saturated vapour (hr) h =2728.6 KJ/kg specific enthalpy at turbine exit (12) h;=he+x/h - h) h=573.19+0.94 (2728.6–573.19) h = 2599.2754 KJ/kg 1 2 2 152 apply steady flow energy equation (S.F.E.E.) between state 1 and state 2: V2 V2 mh + He =mh. + +W 2000 2000) 602 8 3755.2 + 560=8 2599.2754+ 2000) 2000) 30056 560 20795. 1032 +W W =8700.8968 KW hence, Power produced by turbine is 8700.8969 KW (375 +W

Here heat transfer is taken as negative because heat is transferred from turbine (control volume ) to surrounding and steady flow equation is obtained by assuming that heat is transferred to control volume.