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.

Answer 2

Following are the calculation of the entropy at the turbine inlet:

Inlet pressure, $P_1 =\overline{20}$

Inlet temperature, $T_1 =400^{\circ} \ C$

Exit pressure, $P_2 =\overline{1.5}$

Using the values of the steam table:

$\to P_1 = \overline{20} \\\\ \to T_1= 400^{\circ}\ C\\\\\to s_1=7.127 \ \frac{KJ}{kg \ K}$

Also at  

$\to P_2= \overline{1.5} \\\\\to S_{f2} =1.4336 \ \frac{kJ}{kg K}\\\\\to S_{fg2} =5.7898 \ \frac{kJ}{kg K}$

Considering the "x" that represents the quality of steam in a turbine for an isentropic operation.

$\to s_1=s_2\\\\\to s_1=s_{f2}+xs_{fg2}\\\\\to x=\frac{s_1-s_{f2}}{s_{fg2}}$

       $= \frac{7.127-1.4336}{5.7898}\\\\=0.9833\\\\= 98.3\%$

Therefore, there is NO, the steam's quality is not $\bold{98\%}$, and the steam has a quality of $\bold{x=98.33\%}$.

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