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EGR 334 Thermodynamics Chapter 3: Section 11 Lecture 09: Generalized Compressibility Chart Quiz Today?

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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: Homework Assignment: Read Chap 3: Sections From Chap 3: 92, 93, 96, 99

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Like c p and c v, 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. 3 Limitation: where and Compressibility Factor, Z

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Universal Gas Constant 4 SubstanceChem. FormulaR (kJ/kg-K)R(Btu/lm-R) Air AmmoniaNH ArgonAr Carbon DioxideCO Carbon MonoxideCO HeliumHe HydrogenH2H MethaneCH NitrogenN2N OxygenO2O WaterH2OH2O R can also be expresses on a per mole basis: where M is the molecular weight (see Tables A-1 and A-1E)

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

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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 >> T c or (T > 2T c ) Step 1: Thus, analyze Z by first looking at the reduced variables P c = Critical Pressure T c = Critical Pressure

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Step 2: Using the reduced pressure, p r and reduced temperature, T r determine Z from the Generalized compressibility charts. (see Figures A-1, A-2, and A-3 in appendix).

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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 The figures also let’s you directly read reduced specific volume where where

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

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10 Example: (3.95) A tank contains 2 m 3 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. Sec 3.11 : Compressibility V = 2 m 3 T = -93°C p gage = 1.4 MPa p atm = MPa

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11 Example: (3.95) Determine the mass of air, in kg Sec 3.11 : Compressibility V = 2 m 3 T = -93°C = 180 K p = p gauge + p atm = 1.4 MPa MPa = 1.5 MPa = 15 bar From Table A-1 (p. 816): For Air: 16) T c = 133 K p c = 37.7 bar Z=0.95 View Compressibility Figure

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12 Sec : Equations of State & Sec 3.12 : Ideal Gas Model Ideal Gas Equations of State: Relate the state variables T, p, V Alternate Expressions When the gas follows the ideal gas law, Z = 1 p > T c and

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

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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 ft 3. Verify that ideal gas behavior can be assumed for air under these conditions. m = 10 lb T = 70°F p = 30 psi

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15 Example: (3.105) Determine the volume of the air, in ft 3. Verify that ideal gas behavior can be assumed for air under these conditions. Sec 3.12 : Ideal Gas m = 10 lb T = 70°F = 530°R p = 30 psi= 2.04 atm For Air, (Table A-1E, p 864) T c = 239 °R and p c = 37.2 atm Z= 1.0 (Figure A-1) View Compressibility Figure

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16 Example 3: Nitrogen gas is originally at p = 200 atm, T = K. It is cooled at constant volume to T = K. What is the pressure at the lower temperature? SOLUTION: From Table A-1 for Nitrogen p cr = 33.5 atm, T cr = K At State 1, p r,1 = 200/33.5 = 5.97 and T r,1 = 252.4/126.2 = 2. According to compressibility factor chart, Z = 0.95 v r' = Following the constant v r' line until it intersects with the line at T r,2 = 189.3/126.2 = 1.5 gives P r,2 = Thus P 2 = 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.

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End of Slides for Lecture 09 17

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