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Thermodynamics Lecture Series

Thermodynamics Lecture Series. Assoc. Prof. Dr. J.J. Pure substances – Property tables and Property Diagrams & Ideal Gases. Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA.

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Thermodynamics Lecture Series

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  1. Thermodynamics Lecture Series Assoc. Prof. Dr. J.J. Pure substances – Property tables and Property Diagrams & Ideal Gases Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: drjjlanita@hotmail.comhttp://www5.uitm.edu.my/faculties/fsg/drjj1.html

  2. Submitted-Post-assessment

  3. Unreadable Self-assessment

  4. CHAPTER 2 Pure substance Properties of Pure Substances- Part 2 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." - Oscar Wilde "What we have to learn to do, we learn by doing." -Aristotle

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

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

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

  9. P, kPa T = 30 C P = 4.246 kPa T = 30 C P = 100 kPa 1 = f@ 30 °C 2 = f@ 30 °C 30 C 100 H2O: C. liquid H2O: Sat. liquid 4.246 , m3/kg 1 2 = f@30 °C Phase Change of Water - Pressure Change Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

  10. T = 30 C P = 4.246 kPa P, kPa 1 = f@ 30 °C 100 30 C 2 = f@30 °C 4.246 Sat. Vapor H2O: Sat. liquid H2O: Sat. Liq. , m3/kg 1 3 2 = f@ 30 °C Phase Change of Water - Pressure Change 3 = [f + x fg]@ 30 °C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

  11. T = 30 C P = 4.246 kPa 1 = f@ 30 °C P, kPa 2 = f@ 30 °C 3 = [f + x fg]@ 30 °C 100 30 C 4.246 Sat. Vapor H2O: Sat. Vapor H2O: Sat. Liq. , m3/kg 1 3 2 = f@ 30 °C Phase Change of Water - Pressure Change 4 = g@ 30 °C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa 4 = g@ 30 °C Water when pressure is reduced

  12. 1 = f@ 30 °C P, kPa T = 30 C P = 4.246 kPa T = 30 C P = 2 kPa 2 = f@ 30 °C 3 = [f + x fg]@ 30 °C 100 30 C 4 = g@ 30 °C 4.246 H2O: Sat. Vapor H2O: Super Vapor 2 , m3/kg 1 1 5 3 3 2 = f@100 kPa 2 = f@ 30 °C 4 = g@ 30 °C Phase Change of Water - Pressure Change 5= @2kPa, 30 °C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

  13. T = 30 C P = 4.246 kPa T = 30 C P = 4.246 kPa T = 30 C P = 2 kPa T = 30 C P = 100 kPa T = 30 C P = 4.246 kPa Sat. Vapor H2O: C. liquid H2O: Sat. liquid H2O: Sat. Vapor H2O: Sat. Liq. H2O: Super Vapor Phase Change of Water - Pressure Change Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa Water when pressure is reduced

  14. P, kPa 100 30 C 4.246 2 , m3/kg 5 1 3 4 = g@ 30 °C 2 = f@ 30 °C Phase Change of Water- Pressure Change 1 = f@ 30 °C Tsat@100 kPa = 99.63 C Psat@30 C = 4.246 kPa 2 = f@ 30 °C 3 = [f + x fg]@ 30 °C 4 = g@ 30 °C 5= @2kPa, 30 °C Compressed liquid: Good estimation for properties by taking y = yf@T where y can be either , u, h or s.

  15. P, C 200 C 1,553.8 101.35 100 C 1.2276 10 C , m3/kg g@100 C f@100 C Phase Change of Water

  16. P, C 22,090 200 C 1,553.8 101.35 100 C 1.2276 10 C , m3/kg g@100 C f@100 C Phase Change of Water P-  diagram with respect to the saturation lines

  17. T, C 50 kPa 81.3 70 , m3/kg 0.001030 3.240 =f@70 C = 0.001023 T – v diagram - Example

  18. P, kPa 50 31.19 70 C , m3/kg 0.001023 5.042 =f@70 C = 0.001023 P – v diagram - Example

  19. P, kPa 400 C 22,090.0 200 120.2 C , m3/kg f@200 kPa = 0.001061 g@200 kPa = 0.8857  = 1.5493 P – v diagram - Example P-  diagram with respect to the saturation lines

  20. T, C 374.1 1,000 kPa 179.9 , m3/kg f@1,000 kPa = 0.001127  g@1,000 kPa = 0.19444 T – v diagram - Example T-  diagram with respect to the saturation lines  = [f + x fg]@1,000 kPa

  21. Property TableSaturated water – Temperature table

  22. Vapor Phase:, Vg, mg, g, ug,hg Liquid Phase:, Vf, mf, f, uf,hf Sat. Vapor H2O: Sat. Liq. Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Mixture:, V, m, , u,h, x Mixture’s quality More vapor, higher quality x = 0 for saturated liquid x = 1 for saturated vapor Specific volume of mixture?? Since V=m 

  23. Vapor Phase:, Vg, mg, g, ug,hg Liquid Phase:, Vf, mf, f, uf,hf Sat. Vapor H2O: Sat. Liq. Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Mixture:, V, m, , u,h, x Mixture’s quality Divide by total mass, mt where

  24. Vapor Phase:, Vg, mg, g, ug,hg Liquid Phase:, Vf, mf, f, uf,hf Sat. Vapor H2O: Sat. Liq. Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Mixture:, V, m, , u,h, x Mixture’s quality where y can be , u,h where 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.

  25. T, C TH T = ?? TL , m3/kg L  H  Interpolation: Example – Refrigerant-134a Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. m2 m1

  26. Interpolation: Example – Refrigerant-134a Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

  27. P, kPa 22,090.0 Psat ?? T = 3.35 C 200 -10.09 C , m3/kg f@200 kPa = 0.0007532 g@200 kPa = 0.0993  = 0.10600 Interpolation: Example – Refrigerant-134a P-  diagram with respect to the saturation lines

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

  29. Equation of State An equation relating pressure, temperature and specific volume of a substance. Ideal Gases • 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. Equation of State for ideal gas Boyle’s Law: Pressure of gas is inversely proportional to its specific volume P Ideal Gases • Equation of State for ideal gas • Charles’s Law: At low pressure, volume is proportional to temperature

  31. Equation of State for ideal gas Combining Boyles and Charles laws: Ideal Gases and where M is molar mass where gas constant R is • and where Ru is universal gas constant Ru = 8.314 kJ/kmol.K EQOS: Since the total volume is V = m, so :  = V/m So, EQOS: since the massm = MNwhere N isnumber of moles: So,

  32. Equation of State for ideal gas Real gases with low densities behaves like an ideal gas Ideal Gases P << Pcr, T >> Tcr Hence real gases satisfying conditions Obeys EQOS V = m and where, Ru = 8.314 kJ/kmol.K, m = MN

  33. Ideal Gases High density Low density Molecules far apart Gas Mixtures – Ideal Gases • Low density (mass in 1 m3) gases Molecules are further apart • Real gases satisfying condition Pgas << Pcrit; Tgas >> Tcrit , have low density and can be treated as ideal gases

  34. Ideal Gases Gas Mixtures – Ideal Gases • Equation of State - P--T behaviour P=RT (energy contained by 1 kg mass) where  is the specific volume in m3/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 - P--T behaviour P=RT , since = V/m then, P(V/m)=RT. So, PV =mRT, in kPam3=kJ. Total energy of a system. High density Low density

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

  37. T, C 374.1 1,000 kPa 179.9 , m3/kg f@1,000 kPa = 0.001127  g@1,000 kPa = 0.19444 T – v diagram - Example T-  diagram with respect to the saturation lines  = [f + x fg]@1,000 kPa

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