Steam is
contained in a piston and cylinder device with a free-floating piston. Initially, the steam occupies a volume of 0.18 m3 at a pressure of 500 kPa. |
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The steam is slowly heated until the temperature is 300oC, while the pressure remains constant. If the cylinder contains 0.65 kg of steam, determine the heat transfer and the work in kJ for this process. |
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Read : |
We know the values of two intensive variables
for state 2: T and P, so we can determine the values of all other properties in this state. |
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Therefore we can
calculate DU directly. |
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We can also use the
definition of work for an isobaric process to evaluate W12. |
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Once we know W12 and DU, we can use the 1st Law to evaluate Q12. |
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Given: |
V1 |
0.18 |
m3 |
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Find: |
Q12 |
??? |
kJ |
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m |
0.65 |
kg |
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W12 |
??? |
kJ |
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P1 |
500 |
kPa |
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P2 |
500 |
kPa |
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T2 |
300 |
oC |
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Diagram: |
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Assumptions: |
1 - |
Changes in kinetic and potential energies are negligible. |
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2 - |
The process is a quasi-equilibrium process. |
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Equations
/ Data / Solve: |
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Choose the water inside the cylinder as the system. |
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Apply the integral form of the 1st Law to the process: |
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Eqn 1 |
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If we assume that changes in kinetic and potential energies are negligible, then Eqn 1 simplifies to : |
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Eqn 2 |
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We can evaluate W12 from the definiton of work applied to an isobaric process. |
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Eqn 3 |
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Let's combine Eqns 2 and 3: |
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Eqn 4 |
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We still need to
lookup the same amount of data in the Steam Tables, V and H, but the calculations are just a
little bit simpler and faster using H than using U. |
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Before we can look up H, we need to determine the state
of the water in the cylinder. |
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Calculate V1 from : |
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V1 |
0.2769 |
m3/kg |
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At 500
kPa : |
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Vsat liq |
0.0010925 |
m3/kg |
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Vsat vap |
0.37481 |
m3/kg |
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Since Vsat liq < V1 < Vsat vap, we conclude that a saturated
mixture exists in the cylinder at state 1. |
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So, we must next
evaluate the quality of the steam. |
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Eqn 5 |
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x |
0.7381 |
kg vap/kg |
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Then, we can use the quality to evaluate the specific enthalpy : |
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Eqn 6 |
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Hsat liq |
640.09 |
kJ/kg |
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Hsat vap |
2748.1 |
kJ/kg |
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H1 |
2196.0 |
kJ/kg |
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Next, we need to
determine the phases present
in State 2. We can do this by comparing T2 to Tsat(P2). |
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In the saturation pressure table of the Steam Tables we find: |
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Tsat(P2) |
151.8 |
oC |
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Because T2 > Tsat(P2), state
2 is a superheated
vapor. |
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From the NIST Webbook or the Superheated Tables of the Steam Tables we obtain the
following data: |
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V2 |
0.52261 |
m3/kg |
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H2 |
3064.6 |
kJ/kg |
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Now, we can plug
values back into Eqns 3 and 4 to evaluate Q12 and W12: |
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W12 |
79.8 |
kJ |
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Q12 |
564.6 |
kJ |
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Verify: |
The assumptions made
in this problem solution cannot be verified. |
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Answers : |
W12 |
80 |
kJ |
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Q12 |
565 |
kJ |
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