ENGR 2213 Thermodynamics

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

2. Ideal Gases pV = C1 Robert Boyle (1662) J. Charles and J. Gay-Lussac (1802) At low pressure

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

4. Ideal Gases Equation of State Ru = 8.314 kJ/kmol•K, Universal Gas Constant N = m/M, molar number M molecular weight pV = NRuT Ru=RM, R: Gas constant pV = mRT pv = RT

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. Real Gases PR (= p/pc) reduced pressure TR (= T/Tc) reduced temperature He 5.3 K H2 33.3 K Gas Vapor at a temperature above the critical point Vapor Gas near the state of condensation

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

9. Ideal Gases For ideal gases, internal energy is a function of temperature only. u = u(T) For an ideal gas undergoes a constant volume process, du = q - w = cv dT cv specific heat at constant volume cv = cv (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: p1 = 100 kPa, T1 = 540 K p2 = 200 kPa Find: T2 = ? and v3 = ? p 2 3 1 v

11. Example 1 (continued) 1→2 constant volume heating v1 = v2 2→3 constant temperature expansion T2 = T3 1→2 constant pressure cooling p3 = p1 v2 = v1

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: p1 = 100 kPa, T1 = 540 K p2 = 200 kPa, T2 = 840 K Find: W = ? p 2 1 v

13. Example 2 (continued) (assuming ΔKE = 0, ΔPE = 0) Table A-17 u1 = 389.34 kJ/kg u2 = 624.95 kJ/kg p 2 W = -200 – 2(624.95 – 389.34) = - 671.22 kJ 1 v

14. Example 2 (continued) For air, Tc = 132.5 K, pc = 3.77 MPa pR1 = 0.026, TR1 = 4.08 pR2 = 0.159, TR2 = 6.34 Z ~ 1

15. Example 3 Two tanks are connected by a valve. One tank contains 2kg of CO at T1 = 77 ºC and p1 = 70 kPa. The other tank holds 8 kg of CO at T2 = 27 ºC and p2 = 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 2 Tank 1 m2 = 8 kg p2 = 120 kPa T2 = 300 K m1 = 2 kg p1 = 70 kPa T1 = 350 K Tf = 315 K Q

17. Example 3 (continued)

18. Example 3 (continued) (assuming ΔKE = 0, ΔPE = 0) • Table A-2 • T cv • 0.744 • 350 0.746 Uf = (m1 + m2)u(Tf) cv = 0.745 Ui = m1u(T1) + m2u(T2) Q = m1[u(Tf) – u(T1)] + m2[u(Tf) – u(T2)] = m1cv(Tf – T1) + m2cv(Tf – T2) = (2)(0.745)(315–350) + (8)(0.745)(315–300) = 37.25 kJ

19. Example 3 (continued) (assuming ΔKE = 0, ΔPE = 0) Table A-21 u1 = 7271 kJ/kmol u2 = 6229 kJ/kmol uf = 6541 kJ/kmol Uf = (m1 + m2)uf Ui = m1u1 + m2u2 Q = m1(uf – u1) + m2(uf – u2) = [(2)(6541 - 7271) + (8)(6541 - 6229)]/28 = 37 kJ

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

21. Polytropic Process of an Ideal Gas pVn = constant (pV)Vn-1 = constant (mRT)Vn-1 = constant TVn-1 = constant T1V1n-1 = T2V2n-1 = constant

22. Polytropic Process of an Ideal Gas pVn = constant (pV)np1-n = constant (mRT)np1-n = constant Tnp1-n = constant T1np11-n = T2np21-n = constant