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ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma
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Thermodynamic Properties of Water Phase Change of Water T v
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Thermodynamic Properties of Water Phase Change of Water T v
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Thermodynamic Properties of Water Phase Change of Water T v
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Thermodynamic Properties of Water Phase Change of Water T v
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Thermodynamic Properties of Water Phase Change of Water T v ● Critical point Saturated liquid lineSaturated vapor line Subcooled liquid Saturated mixture Superheated vapor
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Thermodynamic Properties of Water Saturation State A state at which a phase change begins or ends Critical Point Intersection of saturated liquid and saturated vapor Single-Phase Regions Subcooled liquid (compressed liquid), superheated vapor Two-Phase Region Liquid-vapor mixture
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States of Water Compressed liquid (Subcooled liquid) ● For a given T, P > P sat ● For a given P, T < T sat ● For a given v, v < v f compressed liquid subcooled liquid Superheated vapor ● For a given P, T > T sat superheated vapor ● For a given T, P < P sat ● For a given v, v > v g
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States of Water Saturated mixture ● For a given T, P = P sat ● For a given P, T = T sat ● For a given v, v f < v < v g Quality 0 ≤ x ≤ 1
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Saturated Mixture of Water V = V f + V g
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Thermodynamic Properties of Water P-v-T Relations of Water watermost substances
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Thermodynamic Properties of Water P-T Diagram of Water ● ● Triple point Critical point Temperature Pressure solid vapor liquid Pressure Temperature ● ● Triple point Critical point liquid solid vapor
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Thermodynamic Property Tables of Water Simple System A system for which there is only one way the system energy can be altered by work as the system undergoes a quasi-equilibrium process. Simple Compressible System The only mode of energy transfer by work is associated with the volume change (expansion or compression).
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Thermodynamic Property Tables of Water State Principle (State Postulate) The number of independent properties required to specify the state of a system is one plus the number of relevant work interactions. Number of independent properties = 1 + Number of work interactions The number of independent properties required to specify the state of a simple compressible system is two.
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Thermodynamic Property Tables of Water Single-Phase Region (Compressed Liquid, Superheated Vapor) ● Temperature and Pressure ● Temperature and Specific Volume ● Pressure and Specific Volume Two-Phase Region (Saturated Mixture) ● Temperature and Specific Volume ● Temperature and Quality ● Pressure and Specific Volume ● Pressure and Quality
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Thermodynamic Property Tables of Water ● Saturated Water – Temperature Table (A-4) ● Saturated Water – Pressure Table (A-5) ● Superheated Vapor (A-6) ● Compressed Liquid (A-7) Linear interpolation is required when the states encountered in problems do not fall exactly on the values provided by the tables.
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Example 1 A rigid tank contains 50 kg of saturated liquid water at 100 ºC. Determine the pressure in the tank and the volume of the tank. Table A-4, P sat = 101.35 kPa, P = P sat = 101.35 kPa Table A-4, v f = 0.001044 m 3 /kg, V = mv f = 50 (0.001044) = 0.052 m 3 A rigid tank contains 50 kg of saturated liquid water at 100 ºC. Determine the pressure in the tank and the volume of the tank.
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Example 2 Table A-5, T sat = 99.63 ºC, T = T sat = 99.63 ºC Table A-5, v g = 1.694 m 3 /kg, m = V/v g = 2/(1.694) = 1.18 kg A piston-cylinder device contains 2 m 3 of saturated water vapor at 100 kPa pressure. Determine the temperature of the vapor and the mass of the vapor Inside the cylinder. A piston-cylinder device contains 2 m 3 of saturated water vapor at 100 kPa pressure. Determine the temperature of the vapor and the mass of the vapor inside the cylinder.
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Example 3 Table A-4, P sat = 70.14 kPa, P = P sat = 70.14 kPa Table A-4, v f = 0.001036 m 3 /kg, v g = 2.361 m 3 /kg V = V f + V g = m f v f + m g v g = 8(0.001036) + 2(2.361) = 4.73 m 3 A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank. A rigid tank contains 10 kg of water at 90 ºC. If 8 kg of the water is in the liquid form and the rest is in the vapor form. Determine the pressure in the tank and the volume of the tank. v = v f + x(v g - v f ) = 0.001036 + 0.2(2.361 – 0.001036) = 0.473 m 3 /kg V = mv = 10(0.473) = 4.73 m 3
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