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EGR 334 Thermodynamics Chapter 3: Section 11

EGR 334 Thermodynamics Chapter 3: Section 11. Lecture 09: Generalized Compressibility Chart. Quiz Today?. Today’s main concepts:. Universal Gas Constant, R Compressibility Factor, Z. Be able to use the Generalized Compressibility to solve problems

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EGR 334 Thermodynamics Chapter 3: Section 11

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  1. EGR 334 ThermodynamicsChapter 3: Section 11 Lecture 09: Generalized Compressibility Chart Quiz Today?

  2. Today’s main concepts: • Universal Gas Constant, R • Compressibility Factor, Z. • Be able to use the Generalized Compressibility to solve problems • Be able to use Z to determine if a gas can be considered to be an ideal gas. • Be able to explain Equation of State Reading Assignment: • Read Chap 3: Sections 12-14 Homework Assignment: From Chap 3: 92, 93, 96, 99

  3. Limitation: Like cp and cv, today’s topic is about compressible gases….This method does not work for two phase mixtures such as water/steam. It only applies to gases. Compressibility Factor, Z where and

  4. R can also be expresses on a per mole basis: Universal Gas Constant where M is the molecular weight (see Tables A-1 and A-1E)

  5. Sec 3.11 : Compressibility The constant R is called the Universal Gas Constant. Where does this constant come from? For low pressure gases it was noted from experiment that there was a linear behavior between volume and pressure at constant temperature. then and the limit as P0 The ideal gas model assumes low P molecules are elastic spheres no forces between molecules

  6. Sec 3.11 : Compressibility To compensate for non-ideal behavior we can use other equations of state (EOS) or use compressibility Define the compressibility factor Z, Z1 when ideal gas near critical point T >> Tc or (T > 2Tc) Step 1: Thus, analyze Z by first looking at the reduced variables Pc = Critical Pressure Tc = Critical Pressure

  7. Step 2: Using the reduced pressure, pr and reduced temperature, Tr determine Z from the Generalized compressibility charts. (see Figures A-1, A-2, and A-3 in appendix). Fig03_12

  8. Step 3: Use Z to a) state whether the substance behaves as an ideal gas, if Z ≈ 1 b) calculate the specific volume of the gas using where The figures also let’s you directly read reduced specific volume where

  9. Sec 3.11 : Compressibility Summarize: 1) from given information, calculate any two of these: (Note: pc and Tc can be found on Tables A-1 and A-1E) 2) Using Figures A-1, A-2, and A-3, read the value of Z 3) Calculate the missing property using where or (Note: M for different gases can be found on Table 3.1 on page 123.)

  10. Sec 3.11 : Compressibility Example: (3.95) A tank contains 2 m3 of air at -93°C and a gage pressure of 1.4 MPa. Determine the mass of air, in kg. The local atmospheric pressure is 1 atm. • V = 2 m3 • T = -93°C • pgage= 1.4 MPa • patm= 0.101 MPa

  11. Sec 3.11 : Compressibility • Example: (3.95) Determine the mass of air, in kg • V = 2 m3 • T = -93°C = 180 K • p = pgauge + patm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar From Table A-1 (p. 816): For Air: 16) Tc = 133 K pc = 37.7 bar View Compressibility Figure Z=0.95

  12. Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model Ideal Gas Alternate Expressions Equations of State: Relate the state variables T, p, V When the gas follows the ideal gas law, Z = 1 p << pcand / or T >> Tc and

  13. Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model Ideal Gas Van der Waals b  volume of particles Equations of State: Relate the state variables T, P, V a  attraction between particles Redlich–Kwong Peng-Robinson virial B  Two molecule interactions C  Three molecule interactions

  14. Example: (3.105) A tank contains 10 lb of air at 70°F with a pressure of 30 psi. Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions. • m = 10 lb • T = 70°F • p= 30 psi

  15. Sec 3.12 : Ideal Gas Example: (3.105) Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions. For Air, (Table A-1E, p 864) Tc = 239 °R and pc = 37.2 atm • m = 10 lb • T = 70°F = 530°R • p = 30 psi= 2.04 atm View Compressibility Figure Z= 1.0 (Figure A-1)

  16. Example 3: Nitrogen gas is originally at p= 200 atm, T = 252.4 K. It is cooled at constant volume to T = 189.3 K. What is the pressure at the lower temperature? SOLUTION: From Table A-1 for Nitrogen pcr= 33.5 atm, Tcr= 126.2 K At State 1, pr,1= 200/33.5 = 5.97 and Tr,1 = 252.4/126.2 = 2. According to compressibility factor chart , Z = 0.95 vr' = 0.34. Following the constant vr' line until it intersects with the line at Tr,2= 189.3/126.2 = 1.5 gives Pr,2 = 3.55. Thus P2 = 3.55 x 33.5 = 119 atm. Since the chart shows Z drops down to around 0.8 at State 2, so it would not be appropriate to treat it as an ideal gas law for this model.

  17. End of Slides for Lecture 09

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