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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 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 Z pv RT where p ab so lu te p ressu re T ab so lu te tem p eratu re v m o lar sp ecific vo lu m e and 8 .3 1 4 k J/k m o l K o R 1 .9 8 6 B tu /lb m o l R o 1 5 4 5 ft lb /lb R f m ol 4 Universal Gas Constant R can also be expresses on a per mole basis: R R where M is the molecular weight (see Tables A-1 and A-1E) M Substance Chem. Formula R (kJ/kg-K) R(Btu/lm-R) Air --- 0.2870 0.06855 Ammonia NH3 0.4882 0.11662 Ar 0.2082 0.04972 Carbon Dioxide CO2 0.1889 0.04513 Carbon Monoxide CO 0.2968 0.07090 Helium He 2.0769 0.49613 Hydrogen H2 4.1240 0.98512 Methane CH4 0.5183 0.12382 Nitrogen N2 0.2968 0.07090 Oxygen O2 0.2598 0.06206 H2O 0.4614 0.11021 Argon Water 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 lim P0 Pv R T and the limit as P0 The ideal gas model assumes low P molecules are elastic spheres no forces between molecules 5 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, Pv Z RT Z1 when ideal gas near critical point T >> Tc or (T > 2Tc) Step 1: Thus, analyze Z by first looking at the reduced variables PR TR P PC T TC Pc = Critical Pressure Tc = Critical Pressure 6 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). 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 RT v Z p where v v R M R M The figures also let’s you directly read reduced specific volume where v 'R v R Tc pc Sec 3.11 : Compressibility 9 Summarize: 1) from given information, calculate any two of these: pR p TR pC T TC v v 'R R Tc pc (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 Z pv RT or Z pv RT (Note: M for different gases can be found on Table 3.1 on page 123.) where v v M R R M 8 .3 1 4 k J/k m o l K o R 1 .9 8 6 B tu /lb m o l R o 1 5 4 5 ft lb /lb R f m ol 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 10 Sec 3.11 : Compressibility 11 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 pR TR Z p pC T pv 0 .4 0 3 7 .7 TC RT m 15 180 pc = 37.7 bar View Compressibility Figure Z=0.95 1 .3 5 133 p V m m ZRT RT 1 5 1 0 pV 5 N m 0 .9 5 8 .3 1 4 kmkJo l K 2 2 m3 km o l 2 8 .9 7 kg pV Z R 1kJ 1 8 0 K M T 1J 1000 J 1N m 6 1 .1 kg Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model 12 Equations of State: Relate the state variables T, p, V Ideal Gas pv RT Alternate Expressions pV m RT p v m RT When the gas follows the ideal gas law, Z=1 p << pc and / or T >> Tc u u T and h h T u T pv u T RT Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model 13 Equations of State: Relate the state variables T, P, V Ideal Gas pv RT 2 n a Van der Waals p V n b n R T b volume of particles 2 V a attraction between particles a Redlich–Kwong p R T Vm b T V m V m b Peng-Robinson p virial Z 1 B T RT Vm b p C T a Vm 2bVm b 2 2 p D T p ..... B T C T D T Z 1 ..... 2 3 v v v 2 3 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 Sec 3.12 : Ideal Gas 15 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 pR TR Z p 2 .0 4 pC T TC pv RT V View Compressibility Figure 0 .0 5 5 3 7 .2 530 Z= 1.0 (Figure A-1) 2 .2 2 239 pV mZRT V p mRT (1 0 lb m )(1 .0 ) 1 5 4 5 (3 0 ft lb f lb m o l R lb f 2 ) in mZ R M p 2 8 .9 7 5 3 0 R 1 lb m o l / lb m 1 4 4 in ft T 2 2 6 5 .4 ft 3 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 At State 1, pcr = 33.5 atm, pr,1 = 200/33.5 = 5.97 and Tcr = 126.2 K 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