1. ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

2. Ideal Gases At low pressure Robert Boyle (1662) J. Charles and J. Gay-Lussac (1802) pV = C 1

3. Ideal Gases Gas A Gas B Gas C Gas D p RuRu

4. Ideal Gases Equation of State pv = RT pV = NR u T R u = 8.314 kJ/kmolK, Universal Gas Constant N = m/M, molar number Mmolecular weight pV = mRT R u =RM, R: Gas constant

5. Ideal Gases ● The gas consists of molecules that are in random motion and obey the laws of mechanics. ● The total number of molecules is large, but the volume of the molecules is a negligibly small fraction of the volume occupied by the gas. ● No appreciable forces act on the molecules except during the collisions.

6. Real Gases

7. P R (= p/p c ) reduced pressure T R (= T/T c ) reduced temperature Gas Vapor Vapor at a temperature above the critical point Gas near the state of condensation He 5.3 K H 2 33.3 K

8. Real Gases Van der Waals Equation of State Intermolecular attraction forces Volume occupied by molecules Virial Equation of State

9. Ideal Gases For ideal gases, internal energy is a function of temperature only. For an ideal gas undergoes a constant volume process, u = u(T) du = q - w = c v dTc v specific heat at constant volume c v = c v (T)

10. Example 1 1 kg of air contained in a piston-cylinder assembly undergoes a series of processes. 1→2 constant volume heating 2→3 constant temperature expansion 1→2 constant pressure cooling Given: p 1 = 100 kPa, T 1 = 540 K p 2 = 200 kPa Find: T 2 = ? and v 3 = ? p v 1 2 3

11. Example 1 (continued) 1→2 constant volume heating v 1 = v 2 2→3 constant temperature expansion T 2 = T 3 1→2 constant pressure cooling p 3 = p 1 v 2 = v 1

12. Example 2 2 kg of air contained in a piston-cylinder assembly undergoes a process. During this process, there is heat transfer Q = -20 kJ. Given: p 1 = 100 kPa, T 1 = 540 K p 2 = 200 kPa, T 2 = 840 K Find: W = ? p v 1 2

13. Example 2 (continued) p v 1 2 (assuming ΔKE = 0, ΔPE = 0) Table A-17 u 1 = 389.34 kJ/kg u 2 = 624.95 kJ/kg W = -200 – 2(624.95 – 389.34) = - 671.22 kJ

14. Example 2 (continued) For air, T c = 132.5 K, p c = 3.77 MPa p R1 = 0.026,T R1 = 4.08 p R2 = 0.159,T R2 = 6.34 Z ~ 1

15. Example 3 Two tanks are connected by a valve. One tank contains 2kg of CO at T 1 = 77 ºC and p 1 = 70 kPa. The other tank holds 8 kg of CO at T 2 = 27 ºC and p 2 = 120 kPa. The valve is opened now. There is heat transferred from the surrounding. If the final temperature of CO in the tanks is 42 ºC, find the final pressure of CO and the amount of heat transferred.

16. Example 3 (continued) Tank 1 Tank 2 m 1 = 2 kg p 1 = 70 kPa T 1 = 350 K m 2 = 8 kg p 2 = 120 kPa T 2 = 300 K Q T f = 315 K

17. Example 3 (continued)

18. (assuming ΔKE = 0, ΔPE = 0) U f = (m 1 + m 2 )u(T f ) Q = m 1 [u(T f ) – u(T 1 )] + m 2 [u(T f ) – u(T 2 )] = m 1 c v (T f – T 1 ) + m 2 c v (T f – T 2 ) U i = m 1 u(T 1 ) + m 2 u(T 2 ) Table A-2 T c v 300 0.744 350 0.746 c v = 0.745 = (2)(0.745)(315–350) + (8)(0.745)(315–300) = 37.25 kJ

19. Example 3 (continued) (assuming ΔKE = 0, ΔPE = 0) U f = (m 1 + m 2 )u f Q = m 1 (u f – u 1 ) + m 2 (u f – u 2 ) U i = m 1 u 1 + m 2 u 2 Table A-21 u 1 = 7271 kJ/kmol u 2 = 6229 kJ/kmol u f = 6541 kJ/kmol = [(2)(6541 - 7271) + (8)(6541 - 6229)]/28 = 37 kJ

20. Polytropic Process of an Ideal Gas pV n = constant n ≠ 1 n = 1

21. Polytropic Process of an Ideal Gas pV n = constant (pV)V n-1 = constant TV n-1 = constant (mRT)V n-1 = constant T 1 V 1 n-1 = T 2 V 2 n-1 = constant

22. Polytropic Process of an Ideal Gas pV n = constant (pV) n p 1-n = constant T n p 1-n = constant (mRT) n p 1-n = constant T 1 n p 1 1-n = T 2 n p 2 1-n = constant