Consider the rigid tank shown below. It is
divided into two equal volumes by a barrier. The left-hand side (LHS) is a perfect
vacuum and the right-hand side (RHS) contains 5 kg of ammonia at 300 kPa and -10oC. |
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When the barrier is removed,
the ammonia expands and fills the entire
tank. |
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When the ammonia reaches equilibrium, the pressure in the tank is 200 kPa. Calculate ΔS for the ammonia and Q
for this process. |
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Read : |
The key to this
problem is that the mass of ammonia in the system does not change and the volume doubles. We can use the Ammonia Tables to determine the specific volume and specific entropy at state 1 because we know T1 and P1. We can use the specific volume at state 1 and the known mass and volume
relationships to determine the specific volume at state 2. This gives us the
values of two intensive properties at state 2, P2 and specific volume, and allows us to
use the Ammonia Tables
to determine the specific entropy and the total entropy
at state 2. DS = S2 - S1 and we are done. |
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The 2nd Law and the fact that entropy generation must be positive will allow us to determine
the direction of heat transfer or if the process could be adiabatic. |
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Given: |
m |
5 |
kg |
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P2 |
200 |
kPa |
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T1 |
-10 |
oC |
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Eqn 1 |
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P1 |
300 |
kPa |
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Find: |
DS |
??? |
kJ/K |
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Determine whether: |
Q = 0, Q > 0
or Q < 0 |
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Diagram: |
See the problem statement. |
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Assumptions: |
1 - |
The system is the contents of the entire tank. |
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2 - |
No work or
mass crosses the system boundary. |
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3 - |
Changes in kinetic and potential energies are negligible. |
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Equations
/ Data / Solve: |
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The change in entropy can be calculated using: |
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Eqn 2 |
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We know both T1 and P1, so we can look up S1 in the Subcooled Liquid Table of the Ammonia Tables or in the NIST Webbook. |
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S1 |
0.54252 |
kJ/kg-K |
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At state
2, we only know the value of one intensive variable, P2. So, we need to
determine the value of another intensive variable before we can use the Ammonia
Tables to determine S2. |
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We can determine
the
specific volume at state 2 as follows: |
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Eqn 3 |
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We can obtain specific volume at state 1 from the Ammonia Tables or the NIST Webbook : |
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V1 |
0.0015336 |
m3/kg |
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V1 |
0.0076680 |
m3 |
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Then, we can use the
given relationship in Eqn 1
to determine V2: |
V2 |
0.015336 |
m3 |
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Then, determine the specific volume at state 2 using: |
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Eqn 4 |
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V2 |
0.0030672 |
m3/kg |
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Now, we know the
values of two intensive variables, so we can go
back to the Ammonia Tables or NIST Webbook and determine S2 by interpolation. |
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At P2 = 200 kPa : |
Vsat liq |
0.0015068 |
m3/kg |
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Since Vsat liq < V2 < Vsat vap, state 2
is a saturated mixture. |
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Vsat vap |
0.5946 |
m3/kg |
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We can determine x2 from the specific volume, using: |
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Eqn 5 |
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x2 |
0.002631 |
kg vap/kg |
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Then, we can use the quality to determine S2, using: |
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Eqn 6 |
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At P2 = 200 kPa : |
Ssat liq |
0.387505 |
kJ/kg-K |
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Ssat vap |
5.5998 |
kJ/kg-K |
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S2 |
0.40122 |
kJ/kg-K |
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Finally, we can plug
values into Eqn 2 : |
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DS |
-0.70651 |
kJ/K |
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The 2nd Law tells us that: |
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Eqn 7 |
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where Sgen > 0 and T > 0 because it is a thermodynamic temperature scale, such as the Kelvin scale. |
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Therforefore, the only way for DS to be
negative is if dQ < 0. |
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We conclude that heat must have been transferred out of
the system during this
process ! |
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We could apply the 1st Law to evaluate Q, but it is not required. I got Q =
-184.6 kJ. |
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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 : |
Part a.) |
DS |
-0.707 |
kJ/K |
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Part b.) |
Q < 0 |
Heat was transferred out of the system during this process ! |
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