The principle of corresponding states pcs as stated

Info icon This preview shows pages 19–28. Sign up to view the full content.

View Full Document Right Arrow Icon
The principle of Corresponding States (PCS) as stated by van der Waals : “Substances behave alike at the same reduced states. Substances at same reduced states are at corresponding states.” That is, “Substances at corresponding states behave alike.”
Image of page 19

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
9/22/2010 20 Generalized Compressibility Chart ME200 Therm I Lecture 13, Prof. Mongia
Image of page 20
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 21 p R 1.0
Image of page 21

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 22 p R 10.0
Image of page 22
Generalized Compressibility Chart Sonntag, Borgnakke, and Van Wylen, Fundamentals of Thermodynamics, 5 th Edition, page 763. 23 Z c =[Pv/RT] c Table A-1 or A-1E Z c = 0.22-0.304
Image of page 23

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Generalized Compressibility Chart (cont.) Calculate two reduced properties: p R , T R and v R p R = p/p c T R = T/T c v R = v/ ( RT c /p c ) (pseudoreduced specific volume) Determine Z If p R << 1, then ideal gas at all temperatures If T R >> 2, then ideal gas at all pressures (except when p R >> 1) Near critical point and for saturated vapor, real gas behavior 9/22/2010 24 ME200 Therm I Lecture 13, Prof. Mongia
Image of page 24
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 25 3.96: Butane (C 4 H 10 ) in a piston-cylinder assembly undergoes an isothermal compression at 173 o C from p 1 = 1.9 MPa to p 2 = 2.5 MPa. Determine the work , in kJ/kg. From Table A-1 (p. 816): T c = 425 K, p c = 38 bar M = 58.12, R=0.143kJ/kg.K
Image of page 25

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 26 T R = 446 K/425 K = 1.05 p R1 = 19 bar/38 bar = 0.5 p R2 = 25 bar/38 bar = 0.66 From Fig. A-1, p. 911, Z 1 = 0.835, Z 2 = 0.775 Z av =0.805
Image of page 26
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 27
Image of page 27

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
9/22/2010 ME200 Therm I Lecture 13, Prof. Mongia 28
Image of page 28
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern