R134A enters an insulated turbine as a saturated vapour at 2

R-134A enters an insulated turbine as a saturated vapour at 2MPa at a rate of 20kg/s and exits at 100kPa

What is the max amount of power the turbine can produce?

No temperaure is given in the question

Solution

Dear Student,

It is given that at input point, R-134A is saturated vapour at 2 MPa, we can find the other data from property table of saturated R134A.

At 2 MPa (2000 KPa), Energy 261.6 kJ/kg, Entropy 0.9020 kJ/kg-K (taken form R134a - TetraFlouroEthane Saturation Properties - Pressure Table (60 kPa - 3 MPa)

Flow rate is 20 kg/s

Also, the system is insulated that means there is no heat transfered, so we have an adiabatic process.

So this is an isentropic (constant Entropy) proecess, so that the final entropy equals the initial entropy.

therefore, at 100 KPa Entropy shall be 0.9020 KJ/Kg.K, which is betwen the value of Entropy of Fluid and Entropy of gas in saturation table at 100 KPa.

so claculate the quality of Vapour to find Enthalpy.

Quality of Vapour = Entropy - Entropy fluid(@ 100 KPa) / Entropy gas and fluid (@ 100 KPa)

At 100 KPa, Entropy fluid 0.0720 kJ/kg-K and Entropy fluid and gas 0.8799 kJ/kg-K (taken form R134a - TetraFlouroEthane Saturation Properties - Pressure Table (60 kPa - 3 MPa)

Quality of Vapour =   (0.9020 KJ/Kg.K - 0.0720  KJ/Kg.K) / 0.8799 kJ/kg-K = 0.94329

Energy at final Vapour = Energy final (@ 100KPa) + Quality of Vapour X Energy fluid and gas (@ 100KPa)

= 17.3 + 0.94329 X 217.2

= 222.18 KJ/ Kg

power = mass/ second X change in Specific internal energy

mass = 20 kg /s

initial Specific internal energy = 261.56 kJ/kg

Final Energy = 222.18  kJ/kg

power = 20 kg /s X (261.56 kJ/kg - 222.18  kJ/kg)

= 787.55 KJ/ s