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

2. Two Important Relations (δQ) rev = T dS (δQ) rev = dU + (δW) rev First Law of Thermodynamics T dS = dU + p dVT ds = du + p dv H = U + pV dH = dU + p dV + V dp T dS = dH - V dpT ds = dh - v dp

3. Entropy Change of an Ideal Gas T ds = du + p dv For an ideal gas, du = c v dT, pv = RT

4. Entropy Change of an Ideal Gas T ds = dh - v dp For an ideal gas, dh = c p dT, pv = RT

5. Entropy Change of an Ideal Gas Reference state: 1 atm and 0 K Standard-State Entropy

6. Isentropic Processes of Ideal Gases 1. Constant Specific Heats (a) (b)

7. Isentropic Processes of Ideal Gases 1. Constant Specific Heats R = c p – c v k = c p /c v R/c v = k – 1 (a)

8. Isentropic Processes of Ideal Gases 1. Constant Specific Heats R = c p – c v k = c p /c v R/c p = (k – 1)/k (b)

9. Isentropic Processes of Ideal Gases 1. Constant Specific Heats Polytropic Processes pV n = constant n = 0 constant pressure isobaric processes n = 1 constant temperature isothermal processes n = k constant entropy isentropic processes n = ±∞ constant volume isometric processes p 1 V 1 k = p 2 V 2 k

10. Isentropic Processes of Ideal Gases 2. Variable Specific Heats Relative Pressure p r = exp[sº(T)/R] ► is not truly a pressure ► is a function of temperature

11. Isentropic Processes of Ideal Gases 2. Variable Specific Heats Relative Volume v r = RT/p r (T) ► is not truly a volume ► is a function of temperature

12. Example 1 An insulated tank is filled initially with 5 kg of air at p 1 = 500 kPa and T 1 = 500 K. A leak develops and air slowly escapes until p 2 = 100 kPa. Determine the final mass of air in the tank and its temperature. Insulated no heat transfer, Q = 0 Slow leakage can be approximated as a reversible process The process can be assumed to be an isentropic process.

13. Example 1 (continued) Table A-17 T 1 = 500 K p r1 = 8.411 = (8.411)(100/500) = 1.6822 Table A-17 p r2 = 1.6822 T 2 = 317 K

14. Example 2 Air is compressed in an adiabatic piston-cylinder device from p 1 = 95 kPa and T 1 = 22 ºC in a reversible manner. If V 1 /V 2 = 8, Find the final temperature of air. Adiabatic no heat transfer, Q = 0 Reversible The process can be assumed to be an isentropic process.

15. Example 2 (continued) Table A-17 T 1 = 295 K v r1 = 647.9 = (647.9)(1/8) = 80.99 Table A-17 v r2 = 80.99 T 2 = 662.7 K = (295)(8) 0.391 = 665.2 K Table A-2 T ave = 450 K k = 1.391 Alternative approach