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One kilogram of water contained in a piston–cylinder assembly, initially saturated vapor at 460 kPa, is condensed at constant pressure to saturated liquid. Consider an enlarged system consisting of the water and enough of the nearby surroundings that heat transfer occurs only at the ambient temperature of 25 C. Assume the state of the nearby surroundings does not change during the process, and ignore kinetic and potential energy effects. For the enlarged system, determine the heat transfer, in kJ, and the entropy production, in kJ/K.

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Answer:

[tex]Q = 2118.075\,kJ[/tex], [tex]S_{gen} = 2.0836\,\frac{kJ}{K}[/tex]

Explanation:

The process is modelled after the First and Second Law of Thermodynamic:

[tex]-Q + m\cdot P\cdot (\nu_{1} - \nu_{2}) = m\cdot (u_{2}-u_{1})[/tex]

[tex]-\frac{Q}{T_{surr}} + S_{gen} = m\cdot (s_{2}-s_{1})[/tex]

The properties of the fluid are obtained from steam tables:

Inlet

[tex]\nu = 0.40610\,\frac{m^{3}}{kg}[/tex]

[tex]u = 2557.8\,\frac{kJ}{kg}[/tex]

[tex]s = 6.8490\,\frac{kJ}{kg\cdot K}[/tex]

Outlet

[tex]\nu = 0.001089\,\frac{m^{3}}{kg}[/tex]

[tex]u = 626.03\,\frac{kJ}{kg}[/tex]

[tex]s = 1.8285\,\frac{kJ}{kg\cdot K}[/tex]

The heat transfer is:

[tex]Q = m\cdot [P\cdot (\nu_{1}-\nu_{2})+(u_{1}-u_{2})][/tex]

[tex]Q = (1\,kg)\cdot \left[(460\,kPa)\cdot \left(0.40610\,\frac{m^{3}}{kg} - 0.001089\,\frac{m^{3}}{kg} \right) + \left(2557.8\,\frac{kJ}{kg} - 626.03\,\frac{kJ}{kg} \right)\right][/tex]

[tex]Q = 2118.075\,kJ[/tex]

Lastly, the entropy production is:

[tex]S_{gen} = \frac{Q}{T_{surr}} + m\cdot (s_{2}-s_{1})[/tex]

[tex]S_{gen} = \frac{2118.075\,kJ}{298.15\,K} + (1\,kg)\cdot \left(1.8285\,\frac{kJ}{kg\cdot K}-6.8490\,\frac{kJ}{kg\cdot K} \right)[/tex]

[tex]S_{gen} = 2.0836\,\frac{kJ}{K}[/tex]

For the enlarged system, the heat transfer, in kJ is 2118.075 kJ  and the entropy production, in kJ/K is  2.0836 kJ/kg. K.

What is entropy generation?

The entropy generation is the term used to measure the magnitudes of the irreversibilities present during the whole process.

The outlet properties from the steam table are

specific volume v =0.001089 m³/kg

internal energy u = 626.03kJ/kg

entropy s = 1.8285 kJ/kg.K

The inlet properties from the steam table are

specific volume v =0.40610 m³/kg

internal energy u = 2557.8 kJ/kg

entropy s = 6.8490 kJ/kg.K

Mass of water is 1 kg and the pressure initially at 460kPa

The heat transfer  -Q +m(P (v1 -v2)) = m(u2 -u1)

Q = m[(P (v1 -v2)) -(u2 -u1)]

Substituting the values, we get

Q = 2118.075 kJ

Entropy generation

-Q/Tsurr +Sgen = m (s2 -s1)

Sgen = m (s2 -s1) +Q/Tsurr

Put the values, we have

Sgen = 2.0836 kJ/kg. K

Thus, the heat transfer, in kJ is 2118.075 kJ  and the entropy production, in kJ/K is  2.0836 kJ/kg. K.

Learn more about entropy generation.

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