Question

Water expands in a turbine from state 1 (1 - 4 MPa, T-400°) to state 2 [P2 - 1 bar). The isentropic efficiency is 85. What is the temperature of state 2?

Answer 1

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.