Question

Water at 20 bar and 400 C enters a turbine operating at steady state and exits at 1.5 bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read data sheet indicates that the quality, x, at the turbine exit is 98%. What is the value of entropy at the turbine inlet in kJ/kg-K?

Answer 1

Answer:

$s_1=7.1292\frac{kJ}{kg*K}$, nevertheless, the outlet quality is not possible due to negative entropy generation.

Explanation:

Hello,

In this case, with the given conditions, it is possible to extract the entropy at the turbine inlet from the steam overheated tables at 20 bar and 400 °C, realizing that 20 bar equals 2 MPa. In such a way, the entropy at those conditions is:

$s_1=7.1292\frac{kJ}{kg*K}$

Moreover, we should prove that the 98% quality is possible in terms of the entropy generation which MUST be positive based on the entropy balance:

$s_{gen}=s_2-s_1$

Thus, we also compute the entropy at the outlet, looking for liquid-vapor water at 1.5 bar with the given quality:

$s_2=1.3548\frac{kJ}{kg*K}+0.98*5.9187\frac{kJ}{kg*K}\\s_2=7.155\frac{kJ}{kg*K}$

Hence the entropy generation turns out:

$s_{gen}=s_2-s_1=7.155-7.1292=-0.0259\frac{kJ}{kg*K}$

Finally, such value means that the outlet quality is not thermodynamically possible.

Regards.