ES12 Energy Transfer Fundamentals Unit C: Thermodynamic ROAD MAP... C-1: p-v-t Relations C-2: Thermodynamic Property Tables Unit C-2: List of Subjects Thermodynamic Property Tables Saturated Liquid and Saturated Vapor Saturated 2-Phase (Liquid and Vapor) Mixture Superheated Vapor Compressed Liquid
PAGE 1 of 14 Thermodynamic Property Tables Table A-7 (A-7E) Table A-4 (A-4E) Table A-5 (A-5E) Table A-6 (A-6E) THERMODYNAMICS PROPERTY TABLES Appendix 1: PROPERTY TABLES AND CHARTS (SI UNITS) Appendix 2: PROPERTY TABLES AND CHARTS (ENGLISH UNITS: E ) Water properties (for example) Table A-4 (A-4E): Saturated water Temperature table Table A-5 (A-5E): Saturated water Pressure table Table A-6 (A-6E): Superheated water Table A-7 (A-7E): Compressed liquid water ENTHALPY Enthalpy is a combined property: H U pv Enthalpy per unit mass is: h u pv (kj/kg) Enthalpy is the combination of internal energy and energy added to the system by flow work. Greek word enthalpien means to heat. Enthalpy represents the total heat content of the system.
PAGE 2 of 14 Saturated Liquid and Saturated Vapor Table A-4 (A-4E) SATURATED LIQUID AND SATURATED VAPOR Saturated liquid and saturated vapor properties for water are given in tables A-4 (A-4E) / A-5 (A-5E). Table A-4(A-4E): properties are listed under temperature (temperature table) Table A-5(A-5E): properties are listed under pressure (pressure table) For two-phase mixture, the subscript f is used to denote properties of a saturated liquid, and the subscript g to denote the properties of saturated vapor. Another subscript commonly used is fg, which denotes the difference between the saturated vapor and saturated liquid values of the same property. ENTHALPY OF VAPORIZATION The quantity h fg is called the enthalpy of vaporization This represents the amount of energy needed to vaporize a unit mass of saturated liquid at a given temperature or pressure. This decreases as the temperature or pressure increases, and becomes zero at the critical point.
PAGE of 14 EXERCISE C-2-1 (Do-It-Yourself) A rigid tank contains 50 kg of saturated liquid water at 90 C. Determine the pressure (in kpa ) in the tank and the volume (in m ) of the tank. Solution A rigid tank contains 50 kg of saturated liquid water at 90 C. Determine the pressure in the tank and the volume of the tank. The state of the saturated liquid water is shown on a T-v diagram in the figure. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90 C: p p sat @ 90 o C 70.18 kpa (Table A-4: Saturated water Temperature table) The specific volume of the saturated liquid at 90 C is: v v 0.00106 m /kg (Table A-4: Saturated water Temperature table) f @ 90 o C Therefore, the total volume of the tank becomes: V mv 50 kg0.00106 m /kg 0.0518 m
PAGE 4 of 14 EXERCISE C-2-2 (Do-It-Yourself) A piston-cylinder device contains 2 ft of saturated water vapor at 50 psia. Determine the temperature (in F ) and the mass of the vapor (in lbm ) inside the cylinder. Solution A cylinder contains saturated water vapor. The temperature and the mass of vapor are to be determined. The state of the saturated water vapor is shown on a p-v diagram in the figure. Since the cylinder contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this pressure: o T T sat @ 50 psia 280.99 F (Table A-5E: Saturated water Pressure table) The specific volume of the saturated vapor at 50 psia is: v vg @ 50 psia 8.5175 ft /lbm (Table A-5E: Saturated water Pressure table) Therefore, the mass of water vapor inside the cylinder becomes: m V 2 ft v 8.5175 ft /lbm 0.25 lbm
PAGE 5 of 14 EXERCISE C-2- (Do-It-Yourself) 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kpa. Determine (a) the volume change (in m ) and (b) the amount of energy transferred to the water (in kj ). Solution Saturated liquid water is vaporized at constant pressure. The volume change and the energy transferred are to be determined. (a) The process described is illustrated on a p-v diagram in the figure. The volume change per unit mass during a vaporization process is vfg, which is the difference between vg and vf. Reading these values from Table A-5 (Saturated water Pressure table) at 100 kpa: v fg v g v f 1.6941 m /kg 0.00104 m /kg 1.691 m /kg Therefore, V mv fg 0.2 kg1.691 m /kg 0.86 m (b) The amount of energy needed to vaporize a unit mass of a substance at a given pressure is the enthalpy of vaporization at that pressure, which is: hfg 2257.5 kj/kg from Table A-5 (Saturated water Pressure table) at 100 kpa. Therefore the amount of energy transferred to the water is: mhfg 0.2 kg2257.5 kj/kg 451.5 kj
PAGE 6 of 14 Saturated 2-Phase (Liquid and Vapor) Mixture The relative amounts of liquid and vapor phases in a saturated mixture are specified by the quality x Quality is related to the horizontal distances on P-v and T-v diagrams The value of v of a saturated liquid-vapor mixture lies between the v f and v g values at the specified temperature or pressure SATURATED TWO-PHASE MIXTURE Quality of saturated mixture of liquid-vapor: is the ratio of the mass of vapor to the total mass of the mixture. x m vapor where, mtotal mliquid mvapor mf mg m total x = 0: Saturated liquid x = 1: Saturated vapor AVERAGE PROPERTIES OF MIXTURE A two-phase system can be treated as a homogeneous mixture. The total volume and total mass of two-phase mixture can be determined as: Vt Vf V g and mt mf mg In terms of specific volume: V mv => mtv avg mfv f mgv g (eqn. 1) The average property (specific volume) can be given by: mt mf mg => mf mt mg, this can be substituted into the eqn.1: mtv avg mt mg v f mgv g => v avg 1 xv f xv g (eqn. 2) Note that v g v f v fg, and the eqn. 2 yields: vavg v f xv fg (kj/kg) Similarly: uavg u f xu fg (kj/kg) havg hf xhfg (kj/kg) In general: yavg y f xy fg ( y f yavg y fg ) where, y is v, u, or h
PAGE 7 of 14 EXERCISE C-2-4 (Do-It-Yourself) A rigid tank contains 10 kg of water at 90 C. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine (a) the pressure (in kpa ) in the tank and (b) the volume of the tank (in m ). Solution A rigid tank contains saturated mixture. The pressure and the volume of the tank are to be determined. (a) The state of the saturated liquid-vapor mixture is shown in the figure. Since the two phases coexist in equilibrium, we have a saturated mixture, and the pressure must be the saturation pressure at the given temperature: p p sat @ 90 o C 70.18 kpa (Table A-4: Saturated water Temperature table) (b) At 90 C, we have specific volume of: v f 0.00106 m /kg (Table A-4: Saturated water Temperature table) @ 90 o C v g @ 90 o 2.59 m /kg (Table A-4: Saturated water Temperature table) C One way of finding the volume of the tank is to determine the volume occupied by each phase and then add them: V V f Vg mfv f mgv g 8 kg0.00106 m /kg 2 kg2.59 m /kg 4.7 m Another way is to first determine the quality (x), then the average specific volume, and finally the total volume: mg 2 kg x 0.2 mf 10 kg v v f xv fg 0.00106 m /kg 0.2 2.59 0.00106 m /kg 0.47 m /kg V mv 10 kg0.47 m /kg 4.7 m
PAGE 8 of 14 A 1.8 m rigid tank contains steam at 220 C. 1/ of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure (in kpa ) of the steam, (b) the quality of the saturated mixture, and (c) the density (in kg/m ) of the mixture. Properties From table A-4, at 220 C: vf = 0.001190 m /kg and vg = 0.08609 m /kg (a) Two-phase (liquid-vapor) mixture coexist in equilibrium, therefore this is a saturated liquid-vapor mixture. The pressure of the tank is the saturation pressure at the given temperature: p = Tsat@220 C = 2,19.6 kpa (b) The total mass and the quality can be determined as: 1 1.8 m V f mf 504.2 kg v f 0.001190 m /kg V 2 1.8 m g mg 1.94 kg v g 0.08609 m /kg mg 1.94 kg mt mf mg 504.2 kg 1.94 kg 518.1 kg => x 0.0269 mt 518.1 kg (c) The density is determined from: v v f xv g v f 0.001190 m /kg 0.02690.08609 m /kg 0.001190 m /kg 0.00474 m /kg 1 1 287.8 kg/m v 0.00474 m /kg
PAGE 9 of 14 Superheated Vapor At a specified pressure, superheated vapor exists at a higher h than saturated vapor Table A-6 (A-6E) SUPERHEATED VAPOR In the region to the right of the saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor. Since the superheated region is a single-phase region (vapor only), temperature and pressure are no longer dependent properties and they can conveniently be used as the two independent properties in the tables. In table A-6 (A-6E), the properties are listed against temperature for selected pressures starting with the saturated vapor data. The saturation temperature is given in parentheses following the pressure data. Compared to saturated vapor, superheated vapor is characterized by: Lower pressures (p < psat at a given T) Higher temperatures (T > Tsat at a given p) Higher specific volumes (v > vg at a given p or T) Higher internal energies (u > ug at a given p or T) Higher enthalpies (h > hg at a given p or T)
PAGE 10 of 14 EXERCISE C-2-5 (Do-It-Yourself) Determine the temperature (in C ) of water at a state of pressure 0.5 MPa and enthalpy 2,890 kj/kg. Solution Determine the temperature of water at a state of p = 0.5 MPa and h = 2,890 kj/kg. At 0.5 MPa, the enthalpy of saturated water vapor is: hg = 2,748.1 kj/kg. Since h > hg, as shown in the figure, we have superheated vapor. Under 0.5 MPa in table A-6: By linear interpolation, the temperature can be determined as: o o 2,890 kj/kg 2,855.8 kj/kg o T 250 C 200 C 216. C 2,961 kj/kg 2,855.8 kj/kg
PAGE 11 of 14 A rigid vessel contains 2 kg of refrigerant 14a at 800 kpa and 120 C. Determine the volume (in m ) of the vessel and the total internal energy (in kj ). Properties From table A-1, the properties of R-14a at the given state are: p = 800 kpa & T = 120 C => u = 27.87 kj/kg & v = 0.07625 m /kg The total volume and internal energy can be determined as follows: V mv 2 kg0.07625 m /kg 0.075 m U mu 2 kg27.87 kj/kg 655.7 kj
PAGE 12 of 14 Compressed Liquid A compressed liquid may be approximated as a saturated liquid at the given temperature COMPRESSED LIQUID Compressed liquid tables are not as commonly available: table A-7 (A-7E) is the only exception (compressed liquid of water). The format of table A-7 (A-7E) is very much like the format of the superheated vapor tables. One reason for the lack of compressed liquid data is the relative independence of compressed liquid properties from pressure. In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature: y y at a given T where, y is v, u, or h f Of these three properties, the enthalpy is most sensitive to variations in the pressure. Although the approximation above will provide good results, one can correct the enthalpy by: h h v p p f @ T f @ T sat@ T Compared to saturated liquid, compressed (or subcooled) liquid is characterized by: Higher pressures (p > psat at a given T) Lower temperatures (T < Tsat at a given p) Lower specific volumes (v < vf at a given p or T) Lower internal energies (u < uf at a given p or T) Lower enthalpies (h < hf at a given p or T) REFERENCE STATES AND VALUES The absolute values of u, h, and s cannot be measured directly: hence often a convenient reference state is chosen and zero value is assigned. Water: state of saturated liquid at 0.01 C is taken as a reference state Refrigerant R-14a: state of saturated liquid at 40 C is taken as a reference state
PAGE 1 of 14 EXERCISE C-2-6 (Do-It-Yourself) Determine the internal energy (in kj/kg ) of compressed liquid water at 80 C and 5 MPa, using (a) data from the compressed liquid table and (b) saturated liquid data. What is the error (in % ) involved in the second case? Solution The exact and approximate values of the internal energy of liquid water are to be determined. At 80 C, the saturation pressure of water is 47.416 kpa, and since 5 MPa > psat, we have compressed liquid, as shown in the figure. (a) From the compressed liquid table (table A-7): p = 5 MPa & T = 80 C => u =.82 kj/kg (b) From the saturation table (table A-4): o u u f at T 80 C => u = 4.97 kj/kg The % error can be computed as: 4.97.82 100 0.4%.82
PAGE 14 of 14 A piston-cylinder device initially contains 50 liters of liquid water at 40 C and 200 kpa. Heat is transferred to the water at constant pressure until the entire liquid is vaporized. (a) What is the mass of the water (in kg )? (b) What is the final temperature (in C ) (c) Determine the total enthalpy change (in kj ). (d) Show the process on a T-v diagram with respect to saturation lines. Initially, the cylinder contains compressed liquid (since p > psat@40 C) that can be approximated as a saturated liquid at the specified temperature. o o From table A-4: v1 v f at T 40 C 0.001008 m /kg & h1 hf at T 40 C 167.5 kj/kg (a) The mass is determined from: m V1 0.050 m v1 0.001008 m /kg 49.61 kg (b) At the final state, the cylinder contains saturated vapor; the final temperature must be the saturation temperature at the final pressure: o T = Tsat@200kPa = 120.21 C (c) The final enthalpy is h2 = hg@200kpa = 2,706. kj/kg. Therefore, the total enthalpy change is: H mh2 h1 49.61 kg2,706. kj/kg 167.5 kj/kg 125,94 kj