Calculate DSuniverse for the power cycle shown below. Is this cycle reversible, irreversible or impossible? |
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Read : |
The key to this
problem is that the sign of ΔSuniv determines whether a process is impossible, reversible or irreversible. Use the definition of entropy to evaluate ΔS for each
reservoir and for the cycle and add them up to get ΔSuniv. |
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Diagram: |
See the problem
statement. |
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QH |
700 |
kJ |
Given: |
TH |
450 |
K |
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QC |
-350 |
kJ |
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TC |
280 |
K |
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Find: |
Is this cycle reversible, irreversible or impossible? |
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Assumptions: |
1 - |
The cycle only exchanges heat with the hot and cold reservoirs shown. |
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Equations
/ Data / Solve: |
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In this problem, the universe consists of the cycle, the hot reservoir and the cold reservoir. We can calculate ΔSuniv from: |
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Eqn 1 |
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Because the cycle begins and ends
in the same state, Sinit = Sfinal and ΔScycle = 0. |
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By definition, the temperatures of the thermal reservoirs remain constant and there are no irreversibilities within
the reservoirs because
no process takes place in either reservoir. As a result, it is relatively simple to calculate ΔShot and ΔScold using the following
simplifications of the
definition of entropy. |
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Eqn 2 |
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Eqn 3 |
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ΔShot |
-1.556 |
kJ/K |
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ΔScold |
1.250 |
kJ/K |
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Now, we can plug
values back into Eqn 1 to
complete this problem. |
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ΔSuniv |
-0.306 |
kJ/K |
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If the ΔSuniv is … |
... negative, the cycle is impossible |
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... zero, the cycle is reversible |
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... positive, the cycle is irreversible |
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This cycle is impossible because ΔSuniv < 0. |
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Verify: |
The assumptions made
in the solution of this problem cannot be verified with the given
information. |
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Answers : |
This cycle is impossible because ΔSuniv < 0. |
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