ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

Thermodynamic Properties of Water Phase Change of Water T v

Thermodynamic Properties of Water Phase Change of Water T v

Thermodynamic Properties of Water Phase Change of Water T v

Thermodynamic Properties of Water Phase Change of Water T v

Thermodynamic Properties of Water Phase Change of Water T v ● Critical point Saturated liquid lineSaturated vapor line Subcooled liquid Saturated mixture Superheated vapor

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

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

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

Saturated Mixture of Water V = V f + V g

Thermodynamic Properties of Water P-v-T Relations of Water watermost substances

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

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).

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.

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

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.

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 = kPa, P = P sat = kPa Table A-4, v f = m 3 /kg, V = mv f = 50 ( ) = 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.

Example 2 Table A-5, T sat = ºC, T = T sat = ºC Table A-5, v g = 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.

Example 3 Table A-4, P sat = kPa, P = P sat = kPa Table A-4, v f = m 3 /kg, v g = m 3 /kg V = V f + V g = m f v f + m g v g = 8( ) + 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 ) = (2.361 – ) = m 3 /kg V = mv = 10(0.473) = 4.73 m 3