ENGR 2213 Thermodynamics

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

2. Reversible Steady-Flow Work For a steady-flow device undergoing an internally reversible process, (δq)rev - (δw)rev = dh + d(ke) + d(pe) Neglect the changes in kinetic and potential energies, (δq)rev - (δw)rev = dh (δq)rev =T ds = dh - v dp (δw)rev = - v dp

3. Work reversible work in closed systems reversible work associated with an internally reversible process an steady-flow device ► The larger the specific volume, the larger the reversible work produced or consumed by the steady-flow device.

4. Work To minimize the work input during a compression process ►Keep the specific volume of the working fluid as small as possible. To maximize the work output during an expansion process ►Keep the specific volume of the working fluid as large as possible.

5. Work Why does a steam power plant usually have a better efficiency than a gas power plant? Steam Power Plant ►Pump, which handles liquid water that has a small specific volume, requires less work. Gas Power Plant ►Compressor, which handles air that has a large specific volume, requires more work.

6. Work Minimizing the Compressor Work 1. To approach an internally reversible process as much as possible by minimizing the irreversibilities such as friction, turbulence, and non-quasi-equilibrium compression. • To keep the specific volume of the gas as small • as possible by maintaining the gas temperature • as low as possible during the compression • process. This requires that the gas be cooled as it is compressed.

7. Steady-Flow Work Polytropic Processes (pvn = constant)

8. Steady-Flow Work Polytropic Processes (pvn = constant) Isentropic Processes (pvk = constant)

9. p W v Steady-Flow Work Isothermal Processes (pv = constant) 3 2 1 1. n = k 2. 1 < n < k 3. n = 1

10. Example 1 Air entering a compressor at p1 = 100 kPa and T1 = 20 ºC and exiting at p2 = 500 kPa. If the air undergoes a polytropic process with n = 1.3, determine the work and heat transfer per unit mass of flow rate.

11. Example 1 (continued) Polytropic processes = 425 K = - 164.15 kJ/kg

12. Example 1 (continued) Table A-17, T1 = 293 K, h1 = 293.17 kJ/kg T2 = 425 K, h2 = 426.35 kJ/kg = - 164.15 + (426.35 – 293.17) = - 30.97 kJ/kg

13. Isentropic Efficiency for Turbines Isentropic efficiency for a turbine is defined as the ratio of the actual performance of a turbine to the performance that would be achieved by undergoing an isentropic process for the same inlet state and the same exit pressure.

14. 1 h h1 – h2 h1 – h2s 2 2s s Isentropic Efficiency Turbines

15. Example 2 Air enters a turbine at p1 = 300 kPa and T1 = 390 K and exits at p2 = 100 kPa. Given that the actual work output from the turbine is 74 kJ/kg and if the turbine operates adiabatically, determine the isentropic efficiency for the turbine.

16. Example 2 (continued) Table A-17 T1 = 390 K pr1 = 3.481, h1 = 390.88 kJ/kg = 3.481 (100/300) = 1.1603 Table A-17 pr2 = 1.1603, h2s = 285.27 kJ/kg = 390.88 – 285.27 = 105.6 kJ/kg = 0.7

17. Isentropic Efficiency for Compressors Isentropic efficiency for a compressor is defined as the ratio of the performance of a compressor that would be achieved by undergoing an isentropic processto the actual performance for the same inlet state and the same exit pressure.

18. 2 2s h h1 – h2 h1 – h2s 1 s Isentropic Efficiency Compressors

19. Example 3 Air enters an insulated compressor at p1 = 95 kPa and T1 = 22 ºC. Given that p2/p1 = 6 and ηc = 0.82, determine the exit temperature for the air.

20. Example 3 (continued) Table A-17 T1 = 295 K pr1 = 1.3068, h1 = 295.17 kJ/kg = 1.3068 (6) = 7.841 Table A-17 pr2 = 7.841, T2s = 490.29 K h2s = 493.0 kJ/kg Table A-17 h2 = 536.4 kJ/kg, T2 = 532 K

21. Example 4 • 0.5 kilogram of water executes a Carnot power • cycle. During the isothermal expansion, the water • is heated until it is a saturated vapor from an initial • state where the pressure is 1.5 MPa and the quality • is 25%. The vapor then expands adiabatically to • pressure of 100 kPa. Find • the heat addition and rejection from this cycle. • the cycle efficiency.

22. T 1 2 4 3 S Example 4 (continued) Given: p1 = p2 = 1.5 MPa p3 = p4 = 100 kPa x1 = 0.25 W23 = 403.8 kJ/kg Find: Q12 = ? Q34 = ? η = ?

23. Example 4 (continued) Q12 = m(u2 – u1) + mp(v2 – v1) = m(h2 – h1) Table A-5 p1 = p2 = 1.5 MPa, hf = 844.84 kJ/kg, hfg = 1947.3 kJ/kg, hg = 2792.2 kJ/kg sf = 2.315 kJ/kg K, sfg = 4.1298 kJ/kg K, sg = 6.4448 kJ/kg K h1 = hf + x1 hfg = 844.84 + 0.25(1947.3) = 1331.67 kJ/kg s1 = sf + x1 sfg = 2.315 + 0.25(4.1298) = 3.3474 kJ/kg K h2 = hg = 2792.2 kJ/kg s2 = sg = 6.4448 kJ/kg K

24. Example 4 (continued) Q12 = m(h2 – h1) = 0.5(2792.2 – 1331.67) = 730.27 kJ Q34 = m(h4 – h3) s3 = s2 = 6.4448 kJ/kg K s4 = s1 = 3.3474 kJ/kg K Table A-5 p3 = p4 = 100 kPa, hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg, hg = 2675.5 kJ/kg sf = 1.3026 kJ/kg K, sfg = 6.0568 kJ/kg K, sg = 7.3594 kJ/kg K

25. Example 4 (continued) = 417.46 + 0.849(2258) = 2334.5 kJ/kg h3 = hf + x3 hfg h4 = hf + x4 hfg = 417.46 + 0.338(2258) = 1180.66 kJ/kg Q34 = m(h4 – h3) = 0.5(1180.66 – 2334.5) = -576.92 kJ