- #1

AGiantGolden49er

- 6

- 0

## Homework Statement

Water vapor at 6 MPa, 600C enters a turbine operating at steady state and expands to 10 kPa. The mass flow rate is 2 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the isentropic turbine efficiency and (b) the rate of entropy production within the turbine, in kW/K.

## Homework Equations

dm

_{cv}/ dt = ∑ m

_{i}- ∑ m

_{e}

dE

_{cv}/ dt = Q

_{cv}- W

_{cv}+ ∑ m

_{i}(h

_{i}+ V

_{i}

^{2}/2 + gz

_{i}) - ∑ m

_{e}(h

_{e}+ V

_{e}

^{2}/2 + gz

_{e})

dS

_{cv}/ dt = ∑ Q

_{j}/ T

_{j}+ ∑ m

_{i}s

_{i}- ∑ m

_{e}s

_{e}+ σ

_{cv}

## The Attempt at a Solution

From the problem statement, I know that entry mass flow rate and exit mass flow rate are the same, because it is steady-state. I also know that, because of the steady-state operation, the energy balance equation simplifies to:

W

_{cv}= m(h

_{1}- h

_{2})

Using the steam tables, I found the enthalpy and entropy values at state 1 to be 3658.4 kJ/kg and 7.1677 kJ/kg⋅K, respectively. I can use this enthalpy value, along with the given mass flow rate and work output values to find h

_{2}, but what I need to find is h

_{2s}. Once I find that, I feel like I can crack this problem wide open, but I'm having issues on how to do just that. Since the working fluid isn't an ideal gas, I can't use the relative pressure and volume relations, so I haven't the slightest clue on how to find h

_{2s}. If anyone can steer me in the right direction, that would be really appreciated.