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 & Ideal Gases

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3. Unreadable Self-assessment

4. CHAPTER 2 Properties of Pure Substances- Part 2 Pure substance Send self-assessments to: Thermopre@salam.uitm.edu.my Thermopost@salam.uitm.edu.my Send self-assessments to: Thermopre@salam.uitm.edu.my Thermopost@salam.uitm.edu.my

5. Quotes "Education is an admirable thing, but it is well to remember from time to time that nothing that is worth knowing can be taught taught." - Oscar Wilde we learn by doing "What we have to learn to do, we learn by doing." -Aristotle

6. Introduction 1.Choose the right property table to read and to determine phase and other properties. 2.Derive and use the mathematical relation to determine values of properties in the wet-mix phase 3.Sketch property diagrams with respect to the saturation lines, representing phases, processes and properties of pure substances. Objectives:

7. Introduction 4.Use an interpolation technique to determine unknown values of properties in the superheated vapor region 5.State conditions for ideal gas behaviour 6.Write the equation of state for an ideal gas in many different ways depending on the units. 7.Use all mathematical relations and skills of reading the property table in problem-solving. Objectives:

8. 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

9. Phase Change of Water - Pressure Change Water when pressure is reduced 2 = f@30 °C 4.246 T = 30  C P = 100 kPa T = 30  C P = 100 kPa H 2 O: C. liquid T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa H 2 O: Sat. liquid 100 P, kPa, m 3 /kg 1 30  C 2 = f @ 30 °C 1 = f@ 30 °C T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

10. Phase Change of Water - Pressure Change H 2 O: Sat. Liq. Sat. Vapor 3 T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa H 2 O: Sat. liquid Water when pressure is reduced 2 = f@ 30 °C, m 3 /kg 1 100 P, kPa 4.246 30  C 3 = [ f + x f g ] @ 30 °C 2 = f@30 °C 1 = f@ 30 °C T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

11. Phase Change of Water - Pressure Change 4 = g@ 30 °C H 2 O: Sat. Vapor H 2 O: Sat. Vapor H 2 O: Sat. Liq. Sat. Vapor T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa Water when pressure is reduced 1 3 P, kPa, m 3 /kg 2 = f@ 30 °C 100 4.246 30  C 4 = g@ 30 °C 3 = [ f + x f g ] @ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

12. Phase Change of Water - Pressure Change H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa 2 5 H 2 O: Super Vapor H 2 O: Super Vapor T = 30  C P = 2 kPa T = 30  C P = 2 kPa Water when pressure is reduced 4 = g@ 30 °C 2 = f@100 kPa, m 3 /kg 1 3 2 = f@ 30 °C 1 3 100 P, kPa 4.246 30  C 3 = [ f + x f g ] @ 30 °C 4 = g@ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C 5 = @2kPa, 30 °C T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

13. Phase Change of Water - Pressure Change H 2 O: Sat. Vapor H 2 O: Sat. Vapor T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa H 2 O: Super Vapor H 2 O: Super Vapor T = 30  C P = 2 kPa T = 30  C P = 2 kPa H 2 O: Sat. Liq. Sat. Vapor T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa T = 30  C P = 100 kPa T = 30  C P = 100 kPa H 2 O: C. liquid T = 30  C P = 4.246 kPa T = 30  C P = 4.246 kPa H 2 O: Sat. liquid Water when pressure is reduced T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

14. Phase Change of Water- Pressure Change Compressed liquid: Good estimation for properties by taking y = y f@T where y can be either, u, h or s. 2 = f@ 30 °C 4.246 3 2 5 4 = g@ 30 °C, m 3 /kg 1 100 P, kPa 30  C 3 = [ f + x f g ] @ 30 °C 4 = g@ 30 °C 1 = f@ 30 °C 2 = f@ 30 °C 5 = @2kPa, 30 °C T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa T sat@100 kPa = 99.63  C P sat@30  C = 4.246 kPa

15. Phase Change of Water 1,553.8 200  C 1.2276 10  C g@100  C P,  C, m 3 /kg 101.35 100  C f@100  C

16. Phase Change of Water P,  C, m 3 /kg 101.35 g@100  C 1,553.8 1.2276 200  C 10  C 100  C f@100  C 22,090 P- diagram with respect to the saturation lines

17. T,  C, m 3 /kg T – v diagram - Example 70 = f@70  C = 0.001023 50 kPa P, kPa T,  C 5070 Phase, Y? Compressed Liquid, T P sat, m 3 /kg f@70  C T sat,  C 81.33 P sat, kPa 31.19 81.3 3.240 0.001030

18. P, kPa, m 3 /kg P – v diagram - Example P, kPa T,  C 5070 Phase, Y? Compressed Liquid, P > P sat or T < T sat, m 3 /kg f@70  C 70  C 31.19 5.042 0.001023 T sat,  C 81.33 P sat, kPa 31.19 = f@70  C = 0.001023 50

19. 400  C P – v diagram - Example P, kPa T,  C 200400 P- diagram with respect to the saturation lines Phase, Why? Sup. Vap., T >T sat P sat, kPa T sat,  C NA120.2 P, kPa, m 3 /kg 22,090.0 = 1.5493, m 3 /kg 1.5493 200 f@200 kPa = 0.001061 g@200 kPa = 0.8857 120.2  C

20. 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

21. Property Table Saturated water – Temperature table Temp T sat,  C 10 50 100 200 300 374.14 Specific internal energy, kJ/kg u f, kJ/kgu fg, kJ/kgu g, kJ/kg 42.002347.22389.2 209.322234.22443.5 418.942087.62506.5 850.651744.72593.3 1332.01231.02563.0 2029.60 Specific volume, m 3 /kg f, m 3 /kg g, m 3 /kg 0.001000106.38 0.00101212.03 0.0010441.6729 0.0011570.13736 0.0014040.02167 0.003155 Sat. P. P, kPa 1.2276 12.349 P, MPa 0.10235 1.5538 8.581 22.09

22. Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Given the pressure, P, then T = T sat, y f < y <y g Given the pressure, P, then T = T sat, y f < y <y g Vapor Phase:, V g, m g, g, u g, h g Liquid Phase:, V f, m f, f, u f, h f Mixture:, V, m,, u, h, x Specific volume of mixture?? Since V=m Specific volume of mixture?? Since V=m Mixture’s quality More vapor, higher quality x = 0 for saturated liquid x = 1 for saturated vapor More vapor, higher quality x = 0 for saturated liquid x = 1 for saturated vapor

23. Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Vapor Phase:, V g, m g, g, u g, h g Liquid Phase:, V f, m f, f, u f, h f Mixture:, V, m,, u, h, x Mixture’s quality Divide by total mass, m t wherewhere Given the pressure, P, then T = T sat, y f < y <y g Given the pressure, P, then T = T sat, y f < y <y g

24. Saturated Liquid-Vapor Mixture H 2 O: Sat. Liq. Sat. Vapor Vapor Phase:, V g, m g, g, u g, h g Liquid Phase:, V f, m f, f, u f, h f Mixture:, V, m,, u, h, x Mixture’s quality Given the pressure, P, then T = T sat, y f < y <y g Given the pressure, P, then T = T sat, y f < y <y g wherewhere wherewhere If x is known or has been determined, use above relations to find other properties. If either, u, h are known, use it to find quality, x. u If x is known or has been determined, use above relations to find other properties. If either, u, h are known, use it to find quality, x. u y can be, u, h

25. Interpolation: Example – Refrigerant-134a T,  C, m 3 /kg THTH L TLTL H T = ?? m1m1 m2m2 P, kPa, m 3 /kg 2000.10600 Phase, Why? Sup. Vap., > g P sat, kPa T sat,  C --10.09 T,  C ?? Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

26. Interpolation: Example – Refrigerant-134a P, kPa, m 3 /kg 2000.10600 Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. Phase, Why? Sup. Vap., > g P sat, kPa T sat,  C --10.09 T,  C ?? T,  C, m 3 /kg u, kJ/kg T L = 0 L = 0.10438 u L = 229.23 0 < T L < T H = 0.10600 0 < u L < u H T H = 10 H = 0.10922 u H = 237.05

27. Interpolation: Example – Refrigerant-134a P, kPa, m 3 /kg 2000.10600 P- diagram with respect to the saturation lines Phase, Why? Sup. Vap., > g P sat, kPa T sat,  C --10.09 P, kPa, m 3 /kg 22,090.0 T,  C3.35 f@200 kPa = 0.0007532 g@200 kPa = 0.0993 200 -10.09  C u, kJ/kg231.85 = 0.10600 T = 3.35  C P sat ??

28. Properties of Pure Substances- Ideal Gases Equation of State Ideal Gases

29. Equation of StateEquation of State –An equation relating pressure, temperature and specific volume of a substance. –Predicts P- -T behaviour quite accurately –Any properties relating to other properties –Simplest EQOS of substance in gas phase is ideal-gas (imaginary gas) equation of state

30. Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Boyle’s Law: Pressure of gas is inversely proportional to its specific volume P Equation of State for ideal gasEquation of State for ideal gas –Charles’s Law: At low pressure, volume is proportional to temperature

31. Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Combining Boyles and Charles laws: and where R u is universal gas constant R u = 8.314 kJ/kmol.K and where M is molar mass So,So, EQOS: since the mass m = MN where N is number of moles: So,So, where gas constant R is EQOS: Since the total volume is V = m, so : = V/m

32. Ideal Gases Equation of State for ideal gasEquation of State for ideal gas –Real gases with low densities behaves like an ideal gas P > T cr where,where, Hence real gases satisfying conditions R u = 8.314 kJ/kmol.K, V = m and m = MN Obeys EQOS

33. Ideal Gases Gas Mixtures – Ideal Gases  Low density (mass in 1 m 3 ) gases Molecules are further apart  Real gases satisfying condition P gas > T crit P gas > T crit, have low density and can be treated as ideal gases High density Low density Molecules far apart

34. Ideal Gases Gas Mixtures – Ideal Gases  Equation of State  Equation of State - P- -T behaviour P =RT R T P =RT (energy contained by 1 kg mass) where is the specific volume in m 3 /kg, R is gas constant, kJ/kg  K, T is absolute temp in Kelvin. High density Low density Molecules far apart

35. Ideal Gases Gas Mixtures – Ideal Gases  Equation of State  Equation of State - P- -T behaviour P =RT P =RT, since = V/m then, P(V/m)=RT. So, PV =mRT PV =mRT, in kPa  m 3 =kJ. Total energy of a system. Low density High density

36. Ideal Gases Gas Mixtures – Ideal Gases  Equation of State  Equation of State - P- -T behaviour PV =mRT PV =mRT = NMRT = N(MR)T PV = NR u T Hence, can also write PV = NR u T where N N is no of kilomoles, kmol, M M is molar mass in kg/kmole and R u R u =MR R u is universal gas constant; R u =MR. R u = 8.314 kJ/kmol  K Low density High density

37. 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