| 7E-1 : |
Isentropic Compression of Ammonia |
4 pts |
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| Saturated ammonia vapor enters an adiabatic compressor at -10oC and leaves at a pressure of 1.0 MPa. Determine the work requirement per kg of ammonia for the compressor if the process is reversible. |
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| Read: |
| We must apply the 1st Law to the compressor. We can get H1 from the ammonia tables or the NIST Webbook, but we do not know H2. The key to solving this problem is that a process that is both adiabatic and internally reversible is also isentropic. Knowing that S2 = S1 gives us the value of a 2nd intensive variable for state 2. This allows us to use the ammonia tables or NIST Webbook to detemine H2. We can than plug H2 into the 1st Law to determine the work requirement per kg of ammonia. |
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| Given: |
T1 |
-10 |
oC |
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Find: |
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x1 |
1 |
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WS |
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kJ/kg |
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P2 |
1000 |
kPa |
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| Diagram: |
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| Assumptions: |
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1 - |
The compressor is adiabtic and reversible and is therefore isentropic. |
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2 - |
The compressor operates at steady-state. |
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3 - |
Changes in kinetic and potential energies are negligible. |
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4 - |
Shaft work and flow work are the only types of work that cross the system boundary. |
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| Equations / Data / Solve : |
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Apply the 1st Law to the compressor to determine the shaft work requirement. |
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For a steady-state, single-inlet, single outlet system with no heat transfer and negligible kinetic and potential energy changes, the 1st Law is: |
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Eqn 1 |
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We can get H1 from the ammonia tables or the NIST Webbook because we know the temperature and we know it is a saturated vapor: |
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H1 |
1593.9 |
kJ/kg |
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The compressor is isentropic, so S2 = S1 and we can get S1 from the ammonia tables or the NIST Webbook. |
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S1 |
6.2285 |
kJ/kg-K |
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S2 |
6.2285 |
kJ/kg-K |
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| Now, we know the values of two intensive properties at state 2, so we can use the ammonia tables or the NIST Webbook to evaluate any other properties by interpolation. Here, we are interested in H2. |
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At P = 1 MPa: |
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T (oC) |
H (kJ/kg) |
So (kJ/kg- |
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70 |
1751.50 |
6.1829 |
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T2 |
H2 |
6.2285 |
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T2 |
76.26 |
oC |
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80 |
1776.8 |
6.2558 |
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H2 |
1767.3 |
kJ/kg |
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Now, we can plug values back into Eqn 1 : |
WS |
-173.43 |
kJ/kg |
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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: |
WS |
-173 |
kJ/kg |
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