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

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Presentation on theme: "EGR 334 Thermodynamics Chapter 3: Section 11 Lecture 09: Generalized Compressibility Chart Quiz Today?"— Presentation transcript:

1 EGR 334 Thermodynamics Chapter 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: Homework Assignment: Read Chap 3: Sections From Chap 3: 92, 93, 96, 99

3 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

4 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)

5 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

6 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

7 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).

8 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

9 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.)

10 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

11 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

12 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

13 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

14 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

15 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

16 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.

17 End of Slides for Lecture 09 17


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