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1. EGR 334 ThermodynamicsChapter 3: Section 12-14

   Lecture 10: Ideal Gas Law Quiz Today?
2. Today’s main concepts: Be able to explain the Ideal Gas Law Be able to explain when it is appropriate to use the Ideal Gas Law Be able to use the Ideal Gas Law to determine State Properties Be able to apply the Ideal Gas Law to the solution of 1st Law problems. Reading Assignment: Read Chap 3: Sections 15 Homework Assignment: From Chap 3: 102, 107,115, 125
3. Ideal Gas Law When the general compressibility factor, Z  1,then the following relationship between pressure, temperature, and volume of a gas applies. which can also be written or m = mass n = number of moles R may be found on Table 3.1
4. Sec 3.13 : Ideal gases and u, h, cv, cp If a gas behaves as an ideal gas, then its specific internal energy, u, depends only on temperature. Enthalpy, h, was defined as: For an ideal gas, since then Since and Therefore: If a gas can be treated as an ideal gas, its intensive properties of specific energy and enthalpy are entirely functions of temperature.
5. Sec 3.13 : Ideal gases and u, h, cv, cp If u and h are only functions of temperature, then the specific heats may be used to determine relations between temperature change and energy levels. For an ideal gas, the expressions for u and h can be simplified since u = f(T) and h = f(T) For many cases, the specific heats will be treated as constant values over a limited temperature range and these integrals will be approximated as:
6. Sec 3.13 : Ideal gases and u, h, cv, cp Another important ideal gas equation may be written as When the specific heat ratio is used, this equation may also be written as and For monotonic gases (Ar, Ne, He) with k = 1.4
7. Sec 3.13 : Ideal gases and u, h, cv, cp Temperature Dependence: Specific heats cv and cp are functions of temperature. If possible, you should look up their values from tables which give the specific heat at the temperature indicated. (see Tables A-20 and A-20E) An alternative method is to use a formula to represent cp based on where Table A-21 has values of , , , ,  for different gases
8. Sec 3.13 : Ideal gases and u, h, cv, cp It best to use an actual function of the specific heats to evaluate u and h, by integration But, often this is simplified, evaluating cV and cP at an average T either and or the specific heat at the average temperature may be used. where if cV and cP are treated as constants, then
9. Sec 3.13 : Ideal gases and u, h, cv, cp Summary: 1) Decide if a substance can be treated as an ideal gas…if yes, then or 2) To evaluate changes in internal energy, u, and enthalpy, h: i) Integrate with cv and cp as function of T (see Table A-21) or ii) Use value of cv or cp at an average temperature (see Table A-20) iii) Use k and R to define and then or or iv) Look up temperature dependent values of u and h on property tables. Table A-22 has property values for Air Table A-23 has property values for CO2, C0, H20, O2, and N2.
10. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer.
11. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.111) A piston cylinder assembly contains air at 2 bar, 300K, and a volume of 2 cubic meters. the air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior, determine a) the mass of the air, b) the work, and c) the heat transfer. State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3 State 2: p2 = 1 bar T2 = ? V2 = ? For Ideal Gas: For constant pV: 
12. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.111) continued... State 2: p2 = 1 bar T2 = ? V2 = 4 m3 State 1: p1 = 2 bar T1 = 300 K V1 = 2 m3 find T2:  1st Law of Thermodynamics:
13. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) Two kilograms (2 kg) of air, initially at 5 bar, 350 K and 4 kg of CO initially at 2 bar, 450 K are confined to opposite sides of a rigid, well-insulated container by a partition. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and CO each behave as ideal gases with constant specific heat ratio, k = 1.395. Determine at equilibrium (a) the temperature in K, (b) the pressure, in bar, and (c) the volume occupied by each gas, in m3.
14. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) k = 1.395. CO State 1: Air: State 1: State 2: 1st Law of Thermo: since system is isolated or
15. Sec 3.13 : Ideal gases and u, h, cv, cp CO State 1: Air: State 1: Example: (3.124) continued… For CO: For Air:
16. Sec 3.13 : Ideal gases and u, h, cv, cp CO State 1: Air: State 1: Example: (3.124) continued…
17. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) continued Find Vtotal Then find pfinal
18. Sec 3.13 : Ideal gases and u, h, cv, cp Example: (3.124) continued… The final volumes are then,
19. Solution using IT: Note: The results are slightly different as the cv and cp values that IT pulled out slightly different.
20. End of Slides for Lecture 10