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Thermodynamics Lecture Series
Assoc. Prof. Dr. J.J. Pure substances – Property tables and Property Diagrams Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA
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Quotes “You do not really understand something unless you can explain it to your grandmother.” (Albert Einstein)
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State the meaning of pure substances
Introduction Objectives: State the meaning of pure substances Provide examples of pure and non-pure substances. Read the appropriate property table to determine phase and other properties. Sketch property diagrams with respect to the saturation lines, representing phase and properties of pure substances.
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FIGURE 1–5 Some application areas of thermodynamics.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 1–5 Some application areas of thermodynamics. Application 1-1
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Example: A steam power cycle.
Turbine Mechanical Energy to Generator Heat Exchanger Cooling Water Pump Fuel Air Combustion Products System Boundary for Thermodynamic Analysis Steam Power Plant
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Steam Power Plant
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Copyright © The McGraw-Hill Companies, Inc
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 1–17 A control volume may involve fixed, moving, real, and imaginary boundaries. Open system devices 1-5
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Open system devices Throttle Heat Exchanger
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Properties of Pure Substances
CHAPTER 2 Properties of Pure Substances Title:
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Pure substances Substance with fixed chemical composition
Can be single element: Such as, N2, H2, O2 Compound: Such as Water, H2O, C4H10, Mixture such as Air, 2-phase system such as H2O. Responsible for the receiving and removing dynamic energy (working fluid)
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Water interacts with thermal energy
Phase Change of Water T, C 30 , m3/kg 1 H2O: C. liquid P = 100 kPa T = 30 C Qin H2O Sat. liquid Qin P = 100 kPa T = 99.6 C 99.6 2 = kPa Water interacts with thermal energy
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Water interacts with thermal energy
Phase Change of Water 99.6 2 = kPa T, C 30 , m3/kg 1 P = 100 kPa T = 99.6 C H2O: Sat. Liq. Sat. Vapor Qin H2O Sat. liquid Qin 3 Water interacts with thermal energy
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Water interacts with thermal energy
Phase Change of Water P = 100 kPa T = 99.6 C 99.6 2 = kPa T, C 30 , m3/kg 1 3 H2O: Sat. Vapor Qin H2O: Sat. Liq. Sat. Vapor Qin 4 = kPa Water interacts with thermal energy
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Water interacts with thermal energy
Phase Change of Water 5 = kPa, 150°C P = 100 kPa T = 99.6 C H2O: Sat. Vapor Qin H2O: Super Vapor P = 100 kPa T = 150 C Qin 99.6 2 = kPa T, C 30 , m3/kg 1 4 = kPa 3 3 = [f + x f kPa 1 = 150 5 Water interacts with thermal energy
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Water interacts with thermal energy
Phase Change of Water P = 100 kPa T = 30 C H2O: C. liquid Qin P = 100 kPa T = 99.6 C H2O Sat. liquid Qin H2O: Sat. Liq. Sat. Vapor P = 100 kPa T = 99.6 C Qin P = 100 kPa T = 99.6 C H2O: Sat. Vapor Qin P = 100 kPa T = 150 C H2O: Super Vapor Qin Water interacts with thermal energy
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Phase Change of Water T, C 100 kPa 150 99.6 30 , m3/kg
5 T, C 30 , m3/kg 1 5 = kPa, 150°C 3 = [f + x f kPa 1 = 99.6 2 = kPa 3 4 = kPa Compressed liquid: Good estimation for properties by taking y = where y can be either , u, h or s.
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Copyright © The McGraw-Hill Companies, Inc
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 2-11 T-v diagram for the heating process of water at constant pressure. 2-1
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Phase Change of Water T, C 1000 kPa 100 kPa 179.9 10 kPa 99.6 45.8
, m3/kg 10 kPa 179.9 99.6 kPa kPa 45.8
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T –v diagram: Multiple P
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 2-16 T-v diagram of constant-pressure phase-change processes of a pure substance at various pressures (numerical values are for water). T –v diagram: Multiple P 99.6 45.8 179.9 2-2
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T –v diagram: Multiple P
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. FIGURE 2-18 T-v diagram of a pure substance. T –v diagram: Multiple P 2-3
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T – v diagram - Example P, kPa T, C 50 70 Psat, kPa Tsat, C 81.33
, m3/kg C T, C , m3/kg Phase, Y? Compressed Liquid, T < Tsat 50 kPa 81.3 3.240 70 C =
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T- diagram with respect to the saturation lines
T – v diagram - Example P, kPa , m3/kg 200 1.5493 Phase, Why? Sup. V., >g Psat, kPa Tsat, C 120.2 T, C 400 T, C , m3/kg 400 kPa = 200 kPa T- diagram with respect to the saturation lines = 374.1 120.23 kPa =
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T- diagram with respect to the saturation lines
T – v diagram - Example P, kPa u, kJ/kg 1,000 2,000 Psat, kPa Tsat, C 179.9 T, C 179.9 Phase, Why? Wet Mix., uf < u < ug T, C , m3/kg 1,000 kPa T- diagram with respect to the saturation lines 374.1 kPa = 179.9 kPa = = [f + x f kPa
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Property Table Saturated water – Pressure table
P, kPa 10 50 P, MPa 0.100 1.00 22.09 Sat. temp. Tsat, C 45.81 81.33 99.63 179.91 311.06 374.14 Specific volume, m3/kg f, m3/kg g, m3/kg 14.67 3.240 1.6940 Specific internal energy, kJ/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg 191.82 2246.1 2437.9 340.44 2143.4 2483.9 417.36 2088.7 2506.1 761.68 1822.0 2583.6 1151.4 2544.4 2029.6
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