HW6s - Spring 2006 Thermodynamics Homework #6 Solution 1....

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Spring 2006 1 ME 201 Thermodynamics Homework #6 Solution 1. (10 pts) What is the enthalpy, internal energy, specific volume, and entropy for steam at 1107 ° C and 27 MPa? Solution: Substance Type: Compressible (steam) Problem Type: State We are given steam at 27 MPa and 1107 ° C. First, we must determine the fluid phase. We go to the saturation pressure table, A-5, and find the boiling temperature at 27 MPa. However, we note that the table does not go up that far. This means we are beyond the critical point and can treat the steam as superheated vapor. So now we go to the superheat table, A-6, and find that again we do not have a table entry at either 27 MPa or 1107 ° C. This means we will have to set up a double interpolation. To do so we set up the table shown below and enter the values from the steam tables at 25 MPa, 30 MPa, and 1100 ° C, 1200 ° C. P=25 MPa P=27 MPa P=30 MPa T v u h s v u h s v u h s 1100 0.02517 4206.1 4835.4 7.5825 0.0235 4201.7 0.0210 4195.2 4823.9 7.4906 1107 1200 0.02716 4424.6 5103.5 7.7710 0.0226 4415.3 5094.2 7.6807 We begin our double interpolation by interpolating in pressure, so, for example, at 1100 ° C for specific volume we have MPa 25 - MPa 30 MPa 25 - MPA 27 = MPa) v(25 - MPa) v(30 MPa) v(25 - MPa) v(27 or solving for v(27 MPa) /kg m 0235 . 0 25
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HW6s - Spring 2006 Thermodynamics Homework #6 Solution 1....

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