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EGR 334 Thermodynamics Chapter 3: Section Lecture 10: Ideal Gas Law Quiz Today?

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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 1 st Law problems. Reading Assignment: Homework Assignment: Read Chap 3: Sections 15 From Chap 3: 102, 107,115, 125

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When the general compressibility factor, Z 1, then the following relationship between pressure, temperature, and volume of a gas applies. 3 Ideal Gas Law which can also be written or R may be found on Table 3.1 m = mass n = number of moles

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4 Sec 3.13 : Ideal gases and u, h, c v, c p 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.

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5 Sec 3.13 : Ideal gases and u, h, c v, c p 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:

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6 Sec 3.13 : Ideal gases and u, h, c v, c p Another important ideal gas equation may be written as When the specific heat ratio is used, this equation may also be written as For monotonic gases (Ar, Ne, He) with k = 1.4 and

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7 Sec 3.13 : Ideal gases and u, h, c v, c p Temperature Dependence: Specific heats c v and c p are functions of temperature. An alternative method is to use a formula to represent c p based on 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) where Table A-21 has values of , , , , for different gases

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8 Sec 3.13 : Ideal gases and u, h, c v, c p where if c V and c P are treated as constants, then It best to use an actual function of the specific heats to evaluate u and h, by integration But, often this is simplified, evaluating c V and c P at an average T either and or the specific heat at the average temperature may be used.

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1) Decide if a substance can be treated as an ideal gas…if yes, then 9 Sec 3.13 : Ideal gases and u, h, c v, c p Summary: 2) To evaluate changes in internal energy, u, and enthalpy, h: i) Integrate with c v and c p as function of T (see Table A-21) ii) Use value of c v or c p at an average temperature (see Table A-20) 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 CO 2, C0, H 2 0, O 2, and N 2. iii) Use k and R to define and then or

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10 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. Sec 3.13 : Ideal gases and u, h, c v, c p

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11 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. Sec 3.13 : Ideal gases and u, h, c v, c p State 1: p 1 = 2 bar T 1 = 300 K V 1 = 2 m 3 State 2: p 2 = 1 bar T 2 = ? V 2 = ? For Ideal Gas: For constant pV:

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12 Example: (3.111) continued... Sec 3.13 : Ideal gases and u, h, c v, c p State 1: p 1 = 2 bar T 1 = 300 K V 1 = 2 m 3 State 2: p 2 = 1 bar T 2 = ? V 2 = 4 m 3 find T 2 : 1 st Law of Thermodynamics:

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13 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 = Determine at equilibrium (a) the temperature in K, (b) the pressure, in bar, and (c) the volume occupied by each gas, in m 3. Sec 3.13 : Ideal gases and u, h, c v, c p

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14 Example: (3.124) k = Sec 3.13 : Ideal gases and u, h, c v, c p CO State 1: Air: State 1: State 2: 1 st Law of Thermo: since system is isolated or

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15 Example: (3.124) continued… Sec 3.13 : Ideal gases and u, h, c v, c p CO State 1: Air: State 1: For CO: For Air:

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16 Example: (3.124) continued… Sec 3.13 : Ideal gases and u, h, c v, c p CO State 1: Air: State 1:

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17 Example: (3.124) continued Find V total Then find p final Sec 3.13 : Ideal gases and u, h, c v, c p

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18 Example: (3.124) continued… The final volumes are then, Sec 3.13 : Ideal gases and u, h, c v, c p

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Solution using IT: 19 Note: The results are slightly different as the c v and c p values that IT pulled out slightly different.

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End of Slides for Lecture 10 20

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