Question

Steam enters an adiabatic turbine at 8 MPa and 500°C at a rate of 18 kg/s, and exits at 0.2 MPa and 300°C. Determine the rate of entropy generation (kW/K) in the turbine

Answer 1

the rate of entropy generation = 21.06 kw/k

Step-by-step explanation:

given data in question

Power (P1) = 8 MPa

temperature ( T1 ) = 500°C

mass ( M ) = 18 kg/s

Power (P2) = 0.2 MPa

temperature ( T2 ) = 300°C

To find out

the rate of entropy generation (kW/K) in the turbine

Solution

We know the entropy generation in turbine is

ΔS = m ( S2 - S1 ) ...........1

here S1 and S2 entropy will be calculated by the steam table with the help of P1 = 8 MPa and temperature ( T1 ) = 500°C

S1 will be by table = 6.7245 kJ/kg-k

and for P2 = 0.2 MPa and temperature ( T2 ) = 300°C

S2 will be by the table = 7.8916 kJ/kg-k

now put S1 and S2 and m in the equation 1

ΔS = 18 ( 7.89 - 6.72 )

ΔS = 18 × 1.17 = 21.06 kw/k

so the rate of entropy generation = 21.06 kw/k