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Varvara68 [4.7K]
3 years ago
14

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 p

otential 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?
Engineering
1 answer:
kherson [118]3 years ago
4 0

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.

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Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
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Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

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b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

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