An adiabatic
turbine lets 10 Mpa steam
down to 2.1 Mpa.
Determine the maximum work output if the inlet temperature is 500oC and changes in kinetic
and potential energies
are negligible. |
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Read : |
The key to solving this problem is to
recognize that any
process that is both adiabatic and reversible is ISENTROPIC. This means that S2 = S1 and this allows you to
fix state 2 and evaluate H2. Use H2 in the 1st Law to evaluate WS. |
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Given: |
T1 |
500 |
oC |
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Find: |
WS |
??? |
kJ/kg |
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P1 |
10000 |
kPa |
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P2 |
2100 |
kPa |
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Diagram: |
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Assumptions: |
1 - |
The turbine is both adiabatic and reversible. |
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2 - |
Changes in kinetic and potential energies are negligible. |
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3 - |
Shaft
work is the only form of work that crosses the system boundary. |
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Equations
/ Data / Solve: |
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Begin by applying the 1st Law to the process, assuming changes in kinetic and potential energies are negligible: |
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Eqn 1 |
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The process is adiabatic so Eqn 1 can be simplified to : |
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Eqn 2 |
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Use the NIST Webbook to obtain properties
for state 1. First we have to determine the phases present. |
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At P1 : |
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Tsat |
311.00 |
oC |
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Since T1 > Tsat, state 1
is a superheated vapor. |
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The superheated Steam Tables and the NIST Webbook yield : |
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H1 |
3375.1 |
kJ/kg |
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S1 |
6.5995 |
kJ/kg-K |
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Because the process is
both reversible and adiabatic, it is isentropic. |
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Therefore, S2 = S1 : |
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S2 |
6.5995 |
kJ/kg-K |
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At this point we know
values of two intensive variables for state 2, so we can use the NIST Webbook to determine the value
of any other property. In this case, we need H2. First, we need to
determine the phases
that exist at state 2. |
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At P2 : |
Tsat |
214.86 |
oC |
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Ssat vap |
6.3210 |
kJ/kg-K |
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Ssat liq |
2.4699 |
kJ/kg-K |
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Because S2 > Ssat vap at P2, we can conclude that state 2 is a superheated vapor. We could have reached
the same conclusion after careful consideration of a TS Diagram. |
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We can get the
following data from the superheated Steam Tables or from the NIST Webbook : |
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At 2.1
MPa : |
T (oC) |
S (kJ/kg-K) |
H (kJ/kg) |
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265 |
6.5903 |
2937.1 |
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T2 |
6.5995 |
H2 |
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270 |
6.6133 |
2949.4 |
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Interpolation yields : |
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T2 |
266.99 |
oC |
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H2 |
2941.99 |
kJ/kg |
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Now, we can plug
values back into Eqn 2 : |
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WS |
433.13 |
kJ/kg |
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Verify: |
None of the
assumptions made in this problem solution can be verified. |
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Answers : |
W |
433 |
kJ/kg |
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