(1) Determine the power generated by the turbine
Take the turbine as a control volume. The inlet is state 1 and the exit is state 2.
Since the assumptions state that the turbine is adiabatic
and the process is reversible, this means the process is an isentropic process (will
be introduced in the following section). Hence,
S_{1} = S_{2}
The turbine has only one inlet and one exit. That gives
s_{1} = s_{2} where
s = entropy per unit mass
The saturation temperature of the refrigerant R134a at 1 MPa is 39.39 ^{o}C.
Hence, the refrigerant R134a at 1 MPa and 4^{o}C is a compressed
liquid. In the absence of compressed liquid data, the properties of compressed
liquid can be approximated by the saturated liquid data. At 4^{o}C,
the saturated properties of refrigerant R134 are
P_{sat} = 0.25274 MPa
v_{f} = 0.0007644 m^{3}/kg
h_{f} = 44.75 kJ/kg
s = 0.1777 kJ/(kgK)
The properties of state 1 are:
h_{1} = 44.75 + 0.0007644( 1 0.25274)(10^{3})
= 45.32 kJ/kg
s_{1} = 0.1777 kJ/(kgK)
v_{1} = 0.0007644 m^{3}/kg
At state 2, the pressure and the entropy are known, hence, enthalpy
of state 2 can be determined as
h_{2} = 44.75 + 0.0007644( 0.3 0.25274)(10^{3})
= 44.75 kJ/kg
The energy balance of the turbine is:
substituting the mass flow rate and the enthalpies into the above equation yields
= 200(45.3244.75) = 114 kW
