Answer:
The temperature of state 2 is 104.86 °C.
Step-by-step explanation:
To solve this problem, we can use the isentropic efficiency equation for turbines:
η = (h1 - h2s) / (h1 - h2)
where
η = isentropic efficiency
h1 = enthalpy at state 1
h2s = enthalpy at state 2 assuming an isentropic process
h2 = actual enthalpy at state 2
We can assume that the process is reversible and adiabatic, so the entropy remains constant:
s1 = s2s
From steam tables, we can find the enthalpies at states 1 and 2 using the given pressures and temperatures:
h1 = 3303.3 kJ/kg
s1 = 6.9122 kJ/kg-K
s2s = s1
P2 = 1 bar
Using the steam tables, we can look up the enthalpy at state 2s for P2 and s2s:
h2s = 191.81 kJ/kg
Now we can solve for h2 using the isentropic efficiency equation:
0.85 = (3303.3 - 191.81) / (3303.3 - h2)
h2 = 3256.1 kJ/kg
Finally, we can look up the temperature at state 2 using the steam tables and the enthalpy at state 2:
T2 = 104.86 °C
Therefore, the temperature of state 2 is 104.86 °C.