ENGR 2213 Thermodynamics

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

2. Second Law of Thermodynamics First law of Thermodynamics Energy Second law of Thermodynamics Entropy Unlike energy, entropy is a non-conserved property. Clausius Inequality The cyclic integral of δQ/T is always less than or equal to zero. This inequality is valid for all cycles, reversible or irreversible.

3. Reservoir TH QH HE QL Reservoir TL Clausius Inequality For reversible heat engines = 0

4. Clausius Inequality For irreversible heat engines From Carnot principle Wrev > Wirrev QL, rev < QL, irrev QL, irrev = QL, rev+ Qdiff < 0

5. Example 1 • A heat engine receives 600 kJ of heat from a high- • temperature source at 1000 K during a cycle. It • converts 150 kJ of this heat to work and rejects the • remaining 450 kJ to a low-temperature sink at • 300 K. Determine if this heat engine violates the • 2nd law of thermodynamics on the basis of • the Clausius inequality. • (b) the Carnot principle.

6. Example 1 (continued) (a) Clausius inequality = - 0.9 kJ/K < 0 (b) Carnot principle ηth < ηrev

7. 2 A B 1 Second Law of Thermodynamics Internally Reversible Processes A process is called internally reversible if no irreversibilities occur within the boundaries of the system during the process. Clausius Inequality

8. Entropy If a quantity whose integral depends only on the end states and not the process path, then it is a property.

9. Entropy Isothermal Processes Q = T0ΔS

10. 2 A irrev B rev 1 Increase-in-Entropy Principle Clausius Inequality = S1 – S2

11. Increase-in-Entropy Principle The entropy change of a closed system during an irreversible process is greater than the integral of δQ/T evaluated for that process. For an adiabatic process, Q = 0 In the absence of heat transfer, entropy change is due to irreversibilities only, and their effect is always to increase the entropy. (ΔS)adiabatic ≥ 0

12. surroundings system Increase-in-Entropy Principle (ΔS)adiabatic ≥ 0 A system plus its surroundings constitutes an adiabatic system, assuming both can be enclosed by a sufficiently large boundary across which there is no heat or mass transfer. (ΔS)total = (ΔS)system+(ΔS)surroundings ≥ 0

13. Increase-in-Entropy Principle > 0 irreversible processes = 0 reversible processes Sgen = (ΔS)total < 0 impossible processes Causes of Entropy Change ►Heat Transfer ►Irreversibilities Isentropic Process A process involves no heat transfer (adiabatic) and no Irreversibilities within the system (internally reversible).

14. Remarks about Entropy 1. Process can occur in a certain direction only. A process must proceed in the direction that complies with the increase-in-entropy principle. 2. Entropy is a non-conserved property. There is no such thing as the conservation of entropy principle. 3. The quantity of energy is always preserved during an actual process (the first law), but the quality decreases (the second law). The decrease in quality is always accompanied by an increase in entropy.

15. What is Entropy? Entropt can be viewed as a measure of molecular disorder, or molecular randomness From a statistical point of view, entropy is a measure of the uncertainty about the position of molecules at any instant. The entropy of a pure crystalline substance at absolute zero temperature is zero since there is no uncertainty about the state of the molecules at that instant. - the 3rd Law of Thermodynamics

16. Second Law of Thermodynamics Heat is, in essence, a form of disorganized energy and some disorganization (entropy) will flow with heat. Work instead is an organized form of energy, and is free of disorder or randomness and thus free of entropy.

17. p 2 1 v Example 1 • Saturated water at 100 ºC is contained in a piston- • cylinder assembly. The water undergoes an • internally reversible heating process to the • corresponding saturated vapor state. Find • the work per unit mass for the process. • (b) the heat transfer per unit mass for the process.

18. Example 1 (continued) (a) = (101.4)(1.673 – 0.001044) = 170 kJ/kg (b) = 2257 kJ/kg = (373.15)(7.3549 – 1.3069) = 2257 kJ/kg

19. T1 = 450 ºC p1 = 7 MPa p2 = 3 MPa Example 2 Steam at 7 MPa and 450 ºC is throttled through a vavle to 3 MPa. Find the entropy generation through the process.

20. Example 2 (continued) T1 = 450 ºC p1 = 7 MPa h1 = 3287.1 kJ/kg s1 = 6.6327 kJ/kg K Table A-6 Throttling Process h2 = h1 p2 = 3 MPa h2 = 3287.1 kJ/kg s2 = 6.9919 kJ/kg K Table A-6 sgen = Δs = 6.9919 – 6.6327 = 0.3592 kJ/kg K

21. T T 2 4 4 2 1 1 1 1 3 3 3 3 2 4 2 4 W W S S Second Law of Thermodynamics Carnot Cycle in T-S Diagram