Steam expand isentropically through a turbine from 6 Mpa and

Steam expand isentropically through a turbine from 6 Mpa and 600 degree C to 10 MPa. Calculate the power output if the mass flux is 2 ng/s

Solution

Ans

Temperature is T=600 C = 873 K

Mass flux (M_f)= 2 kg/s

Initial pressure (P_i) = 6 MPa = 6 X 10⁶ Pa

Final pressure (P_f) = 10 KPa = 10⁴ Pa

As a steam turbine extracts the thermal energy of the steam and converts into mechanical energy

and the process is specified as an \"Isentropic\" process, one has to calculate the

power output = workdone = M_f X (h₁ - h₂) ----(1)

The values of enthalpies h₁ & h₂ have to be calculated from the steam table for specified temperature and pressure. From the standard steam table , it is found:

h₁= 3706.3 and h₂= 3658.8 KJ/kg

Hence from equ(1):

Power output = ( 2 Kg/s) [(3706.3 - 3658.8) KJ/Kg] = 95.6 KJ/s