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Thermodynamics Lecture Series Pure substances – Property tables and Property Diagrams & Ideal Gases Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA email: [email protected] http://www5.uitm.edu.my/faculties/fsg/drjj1.html CHAPTER 2 Pure substance Properties of Pure Substances- Part 2 Send self-assessments to: [email protected] [email protected] 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 Introduction Objectives: 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. Introduction Objectives: 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. Example: A steam power cycle. Steam Power Plant Combustion Products Steam Turbine Fuel Air Pump Mechanical Energy to Generator Heat Exchanger Cooling Water System Boundary for Thermodynamic Analysis Phase Change of Water - Pressure Change T = 30 C T = 30 C P = 100 kPa P = 4.246 kPa P, kPa 1 = [email protected] 30 °C 100 2 = f @ 30 °C H2O: C. liquid H2O: Sat. liquid 4.246 , m3/kg 2 = [email protected] °C 1 [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa Water when pressure is reduced Phase Change of Water - Pressure Change T = 30 C P, kPa P = 4.246 kPa 1 = [email protected] 30 °C 100 2 = [email protected] °C 3 = [f + x f g]@ 30 °C Sat. Vapor H2O: Sat. liquid 4.246 H2O: Sat. Liq. , m3/kg 3 2 = [email protected] 30 °C 1 [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa Water when pressure is reduced Phase Change of Water - Pressure Change T = 30 C 1 = [email protected] 30 °C P, kPa 2 = [email protected] 30 °C P = 4.246 kPa 3 = [f + x f g]@ 30 °C 100 Sat. Vapor H2O: Sat. Liq. 4 = [email protected] 30 °C 4.246 H2O: Sat. Vapor , m3/kg 4 = [email protected] 30 °C 3 2 = [email protected] 30 °C 1 [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa Water when pressure is reduced Phase Change of Water - Pressure Change T = 30 C T = 30 C P = 4.246 kPa P = 2 kPa 1 = [email protected] 30 °C P, kPa 2 = [email protected] 30 °C 3 = [f + x f g]@ 30 °C 100 4 = [email protected] 30 °C 5= @2kPa, 30 °C 4.246 , m3/kg 5 4 = [email protected] 30 °C 3 2 = [email protected] 30 °C [email protected] [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa 2 kPa 1 H2O: Sat. Vapor H2O: Super Vapor Water when pressure is reduced Phase Change of Water - Pressure Change T = 30 C T = 30 C T = 30 C T = 30 C P = 100 kPa P = 4.246 kPa P = 4.246 kPa P = 4.246 kPa T = 30 C P = 2 kPa [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa Sat. Vapor H2O: C. liquid H2O: Sat. liquid H2O: Sat. Liq. H2O: Sat. Vapor Water when pressure is reduced H2O: Super Vapor Phase Change of Water- Pressure Change P, kPa [email protected] kPa = 99.63 C [email protected] C = 4.246 kPa 100 1 = [email protected] 30 °C 2 = [email protected] 30 °C 3 = [f + x f g]@ 30 °C 4 = [email protected] 30 °C 5= @2kPa, 30 °C 4.246 2 , m3/kg 5 4 = [email protected] 30 °C 3 2 = [email protected] 30 °C 1 Compressed liquid: Good estimation for properties by taking y = [email protected] where y can be either , u, h or s. Phase Change of Water P, C 1,553.8 101.35 1.2276 , m3/kg [email protected] C [email protected] C Phase Change of Water P, C P- diagram with respect to the saturation lines 22,090 1,553.8 101.35 1.2276 , m3/kg [email protected] C [email protected] C Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 2-19 P-v diagram of a pure substance. 2-4 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 2-25 P-T diagram of pure substances. 2-7 Property Table Saturated water – Temperature table Temp Tsat, C Sat. P. Specific volume, m3/kg Specific internal energy, kJ/kg P, kPa f, m3/kg g, m3/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 10 1.2276 0.001000 106.38 42.00 2347.2 2389.2 50 12.349 0.001012 12.03 209.32 2234.2 2443.5 P, MPa 100 0.10235 0.001044 1.6729 418.94 2087.6 2506.5 200 1.5538 0.001157 0.13736 850.65 1744.7 2593.3 300 8.581 0.001404 0.02167 1332.0 1231.0 2563.0 22.09 0.003155 0.003155 2029.6 0 2029.6 374.14 Property Table Saturated water – Temperature table Specific internal energy, Specific enthalpy, kJ/kg Temp kJ/kg Tsat, C uf, kJ/kg ufg, kJ/kg ug, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg 10 42.00 2347.2 2389.2 42.01 2377.7 2519.8 50 209.32 2234.2 2443.5 209.33 2382.7 2592.1 100 418.94 2087.6 2506.5 419.04 2257.0 2676.1 300 1332.0 1231.0 2563.0 1332.0 1940.7 2793.2 374.14 2029.6 0 2029.6 2099.3 0 2099.3 Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization. T – v diagram - Example P, kPa T, C Tsat, C Psat, kPa , m3/kg 50 70 81.33 31.19 [email protected] C Phase, Y? T, C 50 kPa Compressed Liquid, T < Tsat or P > Psat 81.3 70 , m3/kg 0.001030 =[email protected] C = 0.001023 3.240 P – v diagram - Example P, kPa T, C Psat, kPa Tsat, C , m3/kg 50 70 31.19 81.33 [email protected] C Phase, Y? P, kPa Compressed Liquid, P > Psat or T < Tsat 50 31.19 , m3/kg 0.001023 =[email protected] C = 0.001023 5.042 P – v diagram - Example P, kPa T, C 200 400 Phase, Why? Psat, kPa Tsat, C , m3/kg Sup. Vap., T >Tsat NA 120.2 1.5493 P, kPa P- diagram with respect to the saturation lines 22,090.0 200 [email protected] kPa [email protected] kPa = 0.001061 = 0.8857 , m3/kg = 1.5493 T – v diagram - Example P, kPa u, kJ/kg 1,000 2,000 Psat, kPa Tsat, C T, C 179.9 179.9 T, C Phase, Why? Wet Mix., uf < u < ug T- diagram with respect to the saturation lines 374.1 179.9 = [f + x f g]@1,000 kPa [email protected],000 kPa = 0.001127 , m3/kg [email protected],000 kPa = 0.19444 Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Vapor Phase:, Vg, mg, g, ug, hg Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x V V f Vg mt m f mg Specific volume of mixture?? Since V=m Sat. Vapor H2O: Sat. Liq. Mixture’s quality x mg mt More vapor, higher quality x = 0 for saturated liquid x = 1 for saturated vapor mt avg m f f mg g Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Vapor Phase:, Vg, mg, g, ug, hg Sat. Vapor H2O: Sat. Liq. Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x Mixture’s quality mt avg m f f mg g mt mg f mg g avg mg 1 mt x mg mt Divide by total mass, mt f x g f x g x f x f fg where fg g f Saturated Liquid-Vapor Mixture Given the pressure, P, then T = Tsat, yf < y <yg Vapor Phase:, Vg, mg, g, ug, hg Sat. Vapor H2O: Sat. Liq. Liquid Phase:, Vf, mf, f, uf, hf Mixture:, V, m, , u, h, x f x fg where y y f xy fg where Mixture’s quality fg g f y fg yg y f x mg mt y can be , u, h 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. Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Psat, kPa Tsat, C T, C Sup. Vap., > g - -10.09 ?? Assume properties are linearly dependent. T, C Perform interpolation in superheated vapor phase. slope m1 slope m2 T TL TH TL L H L TH T = ?? TH TL L T TL H L m2 m1 , m3/kg TL L H Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Psat, kPa Tsat, C T, C Sup. Vap., > g - -10.09 ?? T, C , m3/kg u, kJ/kg TL= 0 L = 0.10438 uL = 229.23 0 < TL < TH = 0.10600 0 < uL < uH TH= 10 H = 0.10922 uH = 237.05 TH TL L T TL H L Assume properties are linearly dependent. Perform interpolation in superheated vapor phase. uH uL L u uL H L Interpolation: Example – Refrigerant-134a P, kPa , m3/kg 200 0.10600 Phase, Why? Psat, kPa Tsat, C T, C Sup. Vap., > g - -10.09 3.35 u, kJ/kg 231.85 P, kPa P- diagram with respect to the saturation lines 22,090.0 Psat ?? 200 [email protected] kPa [email protected] kPa = 0.0007532 = 0.0993 = 0.10600 , m3/kg Property Table Saturated water – Temperature table Temp Tsat, C Sat. P. Specific volume, m3/kg Specific internal energy, kJ/kg P, kPa f, m3/kg g, m3/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 10 1.2276 0.001000 106.38 42.00 2347.2 2389.2 50 12.349 0.001012 12.03 209.32 2234.2 2443.5 P, MPa 100 0.10235 0.001044 1.6729 418.94 2087.6 2506.5 200 1.5538 0.001157 0.13736 850.65 1744.7 2593.3 300 8.581 0.001404 0.02167 1332.0 1231.0 2563.0 22.09 0.003155 0.003155 2029.6 0 2029.6 374.14 Property Table Saturated water – Temperature table Specific internal energy, Specific enthalpy, kJ/kg Temp kJ/kg Tsat, C uf, kJ/kg ufg, kJ/kg ug, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg 10 42.00 2347.2 2389.2 42.01 2377.7 2519.8 50 209.32 2234.2 2443.5 209.33 2382.7 2592.1 100 418.94 2087.6 2506.5 419.04 2257.0 2676.1 300 1332.0 1231.0 2563.0 1332.0 1940.7 2793.2 374.14 2029.6 0 2029.6 2099.3 0 2099.3 Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization. Sample Questions (a)Briefly, give a description that best fit each of the following: i)process ii)cyclic process iii) saturated vapor iv) kinetic energy ANSWER: (2 marks each) (i) A property change or a change of state of a system (ii) A process where the initial and the final state is the same. (iii) Vapor ready to condense. (iv) Energy associated with the movement(speed) of a system. Sample Questions Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively. ANSWER: System P, kPa Water 30 Water 300 Tsat, C f, m3/kg g,m3/kg (12 marks) Sample Questions Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively. ANSWER: System P, kPa Tsat, C f, m3/kg g,m3/kg Water 30 69.1 0.001022 5.229 Water 300 133.55 0.001073 0.6058 (12 marks) Sample Questions Using the property table, determine the missing properties in the table below. [Note: The symbols used are the following: P for pressure, T for temperature, and u for the specific internal energy. When indicating the quality, use NA for the compressed liquid and the superheated vapor phase and use 0 < x < 1 for the wet mix phase. When indicating properties such as , u, h or s in the wet mix phase, just write down an expression on how to get it. NO CALCULATIONS ARE REQUIRED.] Sample Questions Substance P kPa [email protected] kPa Water 200 - T C [email protected] P C Quali ty x u kJ/kg Phase & Reaso n 0.5 (5 marks) Sample Questions ANSWER: Substance Water P kPa 200 [email protected] kPa - T C 120.2 3 [email protected] P C 120. 23 Quali ty x 0.5 u kJ/kg u=uf+x ufg Phase & Reaso n Wet mix 0<x <1 (5 marks) Sample Questions iii) Refrigerant 134a P kPa 200 [email protected] kPa T C [email protected] C Quality x u kJ/kg Phase & Reason -16 (6 marks) Sample Questions ANSWER: P kPa 200 [email protected] kPa 157.48 T C -16 [email protected] C -10.09 Quality x NA u kJ/kg Phase & Reason [email protected]C Comp. Liquid T<Tsat or P>Psat = 29.18 (6 marks) Sample Questions In the question that follows, you need to use the data in the table above along with the property table to sketch property diagrams with respect to the saturation lines to indicate the state of the system. In your diagram, draw and label the temperature or the pressure lines. Place an X mark on the graph to indicate the state. Insert values for T, Tsat, P, Psat, , f and g. Use the space below to draw a T - diagram for the system in part (iii) of the table. Sample Questions Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are: P, kPa [email protected] kPa = 0.0007532 g @200 kPa = 0.0993 [email protected] C = 0.0007428 g @-16 C = 0.1247 200 -10.09 C 157.48 -16 C , m3/kg = f =0.0007428 g = 0.1247 (8 marks) Sample Questions Heat is isothermally transferred into a piston-cylinder device containing water at 100 oC, 400 kPa until the pressure drops to 50 kPa. Use the space below to draw a T - diagram with respect to the saturation lines for this process. Indicate the initial and final states and the direction of the process. Draw and label the temperature lines and insert values for T1, T2, Tsat, P1, P2, 1, 2, f, and g. Sample Questions Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are: [email protected] C = 0.001044 [email protected] kPa = 0.001084 g @100 C = 1.6729 g @400 kPa = 0.4625 [email protected] kPa = 0.001030 g @50 kPa = 3.240 T, C 400 kPa 101.35 kPa 50 kPa 143.63 1 2 100 81.33 1 = [email protected] C = 0.001044 1 = 0.001084 0.4625 2 = 3.418 , m3/kg (12 marks) Ideal Gases Properties of Pure Substances- Ideal Gases Equation of State Ideal Gases • Equation 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 Ideal Gases • Equation of State for ideal gas – Boyle’s Law: Pressure of gas is inversely P 1 / V proportional to its specific volume P • Equation of State for ideal gas – Charles’s Law: At low pressure, volume is V T proportional to temperature Ideal Gases • Equation of State for ideal gas – Combining Boyles and Charles laws: P RT , kJ / kg Ru where gas constant R is R M and where M is molar mass and where Ru is universal gas constant Ru = 8.314 kJ/kmol.K EQOS: Since the total volume is V = m, so : = V/m EQOS: since the mass m = MN where N is number of moles: So, V P RT m So, PV mRT, kJ Ru PV MN T M PV NRuT , kJ Ideal Gases • Equation of State for ideal gas – Real gases with low densities behaves like an ideal gas Hence real gases satisfying conditions P << Pcr, T >> Tcr Obeys EQOS P RT , kJ / kg where, PV mRT, kJ PV NRuT , kJ Ru = 8.314 kJ/kmol.K, m = MN V = m and Ru R M Gas Mixtures – Ideal Gases Real Gases Low density (mass in 1 m3) gases Molecules are further apart High density Real gases satisfying condition Pgas << Pcrit; Tgas >> Tcrit , have low density and can be treated as ideal gases Low density Molecules far Gas Mixtures – Ideal Gases Real Gases Equation of State - P--T behaviour High density P = RT (energy contained by 1 kg mass) where is the specific volume in m3/kg, R is gas constant, kJ/kgK, T is absolute temp in Kelvin. Low density Molecules far apart Gas Mixtures – Ideal Gases Real Gases Equation of State - P--T behaviour P=RT , since = V/m then, P(V/m) =RT. So, PV = mRT, in kPam3=kJ. Total energy of a system. High density Low density Gas Mixtures – Ideal Gases Real 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/kmolK High density Low density Equations of States Other EQOS Van Der Waals Equation of State Considers intermolecular interactions, a/2, which then increases the pressure: P replaced by P+a/2 Considers volume occupied by molecules, b, at high pressures: Replace by -b, then High density a EQOS is: P 2 b RT 2 where 2 27R Tcr RTcr a ; b 64Pcr 8Pcr Low density T – v diagram - Example P, kPa u, kJ/kg 1,000 2,000 Psat, kPa Tsat, C T, C 179.9 179.9 T, C Phase, Why? Wet Mix., uf < u < ug T- diagram with respect to the saturation lines 374.1 179.9 = [f + x f g]@1,000 kPa [email protected],000 kPa = 0.001127 , m3/kg [email protected],000 kPa = 0.19444