A piston-and-cylinder device with a free-floating piston contains 2.6 kg of saturated water vapor at 150oC. The water loses heat
to the surroundings
until the cylinder
contains saturated liquid
water. |
The surroundings are at 20oC. Calculate… a.) ΔSwater, b.) ΔSsurroundings, c.) ΔSuniverse |
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Read : |
Calculating DS for
the water in the cylinder is straightforward. The
key to calculating DSsurr is the fact that
the surroundings
behave as a thermal reservoir. The temperature of the surroundings does not
change and there are no irreversibilities in the surroundings that are associated with the process. The key to calculating DSuniv is the fact that the universe is made up of the combination of the system and the surroundings. Consequently, DSuniv = DSsys + DSsurr. |
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Given: |
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2.60 |
kg |
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Find: |
DSsys |
??? |
kJ/K |
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T1 |
150 |
oC |
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DSsurr |
??? |
kJ/K |
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x1 |
1 |
kg vap/kg |
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DSuniv |
??? |
kJ/K |
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Tsurr |
20 |
oC |
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P1 =
P2 |
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x2 |
0 |
kg vap/kg |
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Diagram: |
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Assumptions: |
1 - |
Changes in kinetic and potential energies are negligible. |
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2 - |
Boundary
work is the only form of work that crosses the system boundary. |
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3 - |
The surroundings behave as a thermal reservoir. |
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Equations
/ Data / Solve: |
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Part a.) |
Because both states are saturated we can obtain the specific
entropies directly from the Steam Tables or the NIST Webbook. |
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At T1 = 150oC : |
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S1 |
6.8371 |
kJ/kg-K |
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S2 |
1.8418 |
kJ/kg-K |
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Therefore : |
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Eqn 1 |
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DSsys |
-4.9953 |
kJ/kg-K |
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-12.988 |
kJ/K |
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Part b.) |
The surroundings
behave as a thermal reservoir. |
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We can calculate DSsurr from: |
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Eqn 2 |
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We can determine Qsys by applying the 1st Law using the water within the cylinder as the system. |
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Eqn 3 |
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Eqn 4 |
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Because the process is
isobaric, the boundary work is : |
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Eqn 5 |
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Now, substitute Eqn 5 into Eqn 4 to get : |
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Eqn 6 |
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We can look up enthalpy values for states 1 & 2 in the Saturated
Steam Table or in the NIST
Webbook. |
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At T1 = 150oC : |
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H1 |
2745.9 |
kJ/kg |
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H2 |
632.18 |
kJ/kg |
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Next plug H1 and H2 into Eqn 6. Qsurr is equal
in magnitude, but opposite in sign to Qsys because the heat leaving the system enters
the surroundings. |
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Qsys |
-2113.7 |
kJ/kg |
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Qsurr |
2113.7 |
kJ/kg |
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Now, we can plug
numbers into Eqn 2 to
calculate DSsurr. |
DSsurr |
7.2105 |
kJ/kg-K |
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18.747 |
kJ/K |
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Part c.) |
The universe is made up of the combination of the system and the surroundings. Therefore : |
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Eqn 7 |
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So, all we need to do
is plug values into Eqn 7
that we determined in parts (a) and (b). |
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DSuniv |
5.7595 |
kJ/K |
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DSuniv > 0 because the heat transfer to the surroundings was not reversible. |
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Verify: |
None of the
assumptions made in this problem solution can be verified. |
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Answers : |
DSsys |
-12.988 |
kJ/K |
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DSsurr |
18.747 |
kJ/K |
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DSuniv |
5.7595 |
kJ/K |
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