1. Thermodynamics Lecture Series email: drjjlanita@hotmail.com http://www5.uitm.edu.my/faculties/fsg/drjj1.htmldrjjlanita@hotmail.com Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA Pure substances – Property tables and Property Diagrams

2. Quotes You do not really understand something unless you can explain it to your grandmother. (Albert Einstein)

3. Introduction Objectives: 1.State the meaning of pure substances 2.Provide examples of pure and non-pure substances. 3.Read the appropriate property table to determine phase and other properties. 4.Sketch property diagrams with respect to the saturation lines, representing phase and properties of pure substances.

4. FIGURE 1–5 Some application areas of thermodynamics. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-1 Application

5. Example: A steam power cycle. Steam Turbine Mechanical Energy to Generator Heat Exchanger Cooling Water Pump Fuel Air Combustion Products System Boundary for Thermodynamic Analysis System Boundary for Thermodynamic Analysis Steam Power Plant

7. FIGURE 1–17 A control volume may involve fixed, moving, real, and imaginary boundaries. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1-5 Open system devices

8. Heat Exchanger Throttle

9. CHAPTER 2 Properties of Pure Substances Title:

10. Pure Substances Pure substancesPure substances –Substance with fixed chemical composition Can be single element: Such as, N 2, H 2, O 2Can be single element: Such as, N 2, H 2, O 2 Compound: Such as Water, H 2 O, C 4 H 10,Compound: Such as Water, H 2 O, C 4 H 10, Mixture such as Air,Mixture such as Air, 2-phase system such as H 2 O.2-phase system such as H 2 O. –Responsible for the receiving and removing dynamic energy (working fluid) Pure substancesPure substances –Substance with fixed chemical composition Can be single element: Such as, N 2, H 2, O 2Can be single element: Such as, N 2, H 2, O 2 Compound: Such as Water, H 2 O, C 4 H 10,Compound: Such as Water, H 2 O, C 4 H 10, Mixture such as Air,Mixture such as Air, 2-phase system such as H 2 O.2-phase system such as H 2 O. –Responsible for the receiving and removing dynamic energy (working fluid)

11. Phase Change of Water H 2 O Sat. liquid Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C Water interacts with thermal energy 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 H 2 O: C. liquid P = 100 kPa T = 30 C P = 100 kPa T = 30 C Q in

12. Phase Change of Water H 2 O Sat. liquid Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C Water interacts with thermal energy H 2 O: Sat. Liq. Sat. Vapor Q in 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 3

13. Phase Change of Water Water interacts with thermal energy 4 = g@100 kPa 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 3 P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor Q in H 2 O: Sat. Liq. Sat. Vapor Q in

14. Phase Change of Water Water interacts with thermal energy 150 5 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 4 = g@100 kPa 3 5 = @100 kPa, 150°C 3 = [ f + x f g ] @100 kPa 1 = f@T1 H 2 O: Super Vapor H 2 O: Super Vapor P = 100 kPa T = 150 C P = 100 kPa T = 150 C Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor Q in

15. Phase Change of Water Water interacts with thermal energy H 2 O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O: Sat. Vapor H 2 O: Sat. Vapor Q in P = 100 kPa T = 150 C P = 100 kPa T = 150 C H 2 O: Super Vapor H 2 O: Super Vapor Q in P = 100 kPa T = 30 C P = 100 kPa T = 30 C H 2 O: C. liquid Q in P = 100 kPa T = 99.6 C P = 100 kPa T = 99.6 C H 2 O Sat. liquid Q in

16. Phase Change of Water 99.6 2 = f@100 kPa T, C 30, m 3 /kg 1 4 = g@100 kPa 3 5 = @100 kPa, 150°C 3 = [ f + x f g ] @100 kPa 1 = f@T1 150 100 kPa 5 Compressed liquid: Good estimation for properties by taking y = y f@T where y can be either, u, h or s.

17. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-1 FIGURE 2-11 T-v diagram for the heating process of water at constant pressure.

18. Phase Change of Water T, C, m 3 /kg 99.6 f@100 kPa g@100 kPa 100 kPa 179.9 45.8 10 kPa1000 kPa

19. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-2 FIGURE 2-16 T-v diagram of constant- pressure phase-change processes of a pure substance at various pressures (numerical values are for water). 99.6 45.8 179.9 T –v diagram: Multiple P

20. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2-3 FIGURE 2-18 T-v diagram of a pure substance. T –v diagram: Multiple P

21. T, C, m 3 /kg T – v diagram - Example 70 = f@70 C = 0.001023 81.3 3.240 50 kPa P, kPa T, C 5070 P sat, kPa T sat, C 81.33 Phase, Y? Compressed Liquid, T < T sat, m 3 /kg f@70 C

22. T – v diagram - Example T, C, m 3 /kg f@200 kPa = 0.001061 200 kPa P, kPa, m 3 /kg 2001.5493 T- diagram with respect to the saturation lines Phase, Why? Sup. V., > g P sat, kPa T sat, C120.2 374.1 400 = 1.5493 120.23 g@200 kPa = 0.8857 T, C400

23. T – v diagram - Example T, C, m 3 /kg 1,000 kPa P, kPau, kJ/kg 1,0002,000 T- diagram with respect to the saturation lines Phase, Why? Wet Mix., u f < u < u g P sat, kPa T sat, C179.9 374.1 f@1,000 kPa = 0.001127 179.9 g@1,000 kPa = 0.19444 T, C179.9 = [ f + x f g ] @1,000 kPa

24. Property Table Saturated water – Pressure table Pressure P, kPa 10 50 P, MPa 0.100 1.00 10 22.09 Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg 191.822246.12437.9 340.442143.42483.9 417.362088.72506.1 761.681822.02583.6 1393.041151.42544.4 2029.60 Specific volume, m 3 /kg f, m 3 /kg g, m 3 /kg 0.00101014.67 0.0010303.240 0.0010431.6940 0.0011270.19444 0.0014520.018026 0.003155 Sat. temp. T sat, C 45.81 81.33 99.63 179.91 311.06 374.14