1 CHAPTER 2 Properties of Pure Substances What is Pure Substances? A substance that has a fixed chemical composition throughout is called a pure substance. A pu pure re subs substan tance ce does not have to be of a single chemical element or compound, however. A mi mixtu xture re of va vari riou ouss ch chem emica icall elem emeents or com compound ndss also qualifies as a pure substance as long as the mixture is homogeneous. homogeneous. 2 A mixture of liquid and water vapor is a pure substance, but a mixture of liquid and gaseous air is not. Examples: Water (solid, liquid, and vapor phases) Mixture of liquid water and water vapor Carbon dioxide, CO2 Nitrogen, N2 Mixtures of gases, such as air air,, as long as there is no change of phase. 3 Phases of A Pure Substance The substances exist in dif different ferent phases, e.g. at room temperature and pressure, copper is solid and mercury is a liquid. It can exist in different phases under variations of condition. There are 3 Principal phases solid Liquid gas Each with different molecular structures. • • • 4 Phase-change Processes Phase-change Processes of Pur Pure e Substances There are m many any practical situations where two phases of a pure substances coexist in equilibrium. E.g. water exists as a m mixture ixture of liquid and vapor in the boiler and etc. Solid: strong intermolecular bond Liquid: intermediate intermolecular bonds Gas: weak intermolecular bond Solid Liquid Gas 5 Phase-change Processes Compressed liquidWater exist in the liquid phase Saturated Satura ted vaporvapor- T remains constant until the last drop of Saturated liquidwater exist as liquid and ready to vaporize Saturated liquid-vapor mixture part of saturated liquid vaporize – Superheated vapor T of vapor start to rise. liquid vaporized 6 Compressed Liquid / Subcooled Liquid • Water that is not vaporize Saturated Liquid • A liquid that is about to vaporize Saturated vapor Saturated liquidvapor mixture Superheated vapor • • • A vapor that is about to condense Liquid and vapor phases coexist coexist in equilibrium at these states. A vapor that is not about to condense This constant pressure heating process can be illustrated as: s 8 Property Diagram 9 Saturation Saturation is defined as a condition in which a mixture of va vapo porr an and d li liqu quid id ca can n ex exis istt to toge geth ther er at a giv iven en temperature and pressure. Saturation pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature For a pure substance there is a definite relationship between saturation andthe higher saturation temperature. The higher pressure the pressure, the saturation temperature 10 The graphical representation of this relationship between temperature and pressure at saturated conditions is called the vapor pressure curve 11 Satur Saturated ated and Sub-co Sub-cooled oled Liquids If a substance exists as a liq liquid uid at the saturation temperature and pressure, it is called a satur sat urat ated ed li liqu quid id If the temperature of the liquid is lower than temperature the for the saturation existing pressure, it is called either a subcooled liquid or a compressed liquid 12 Satura Saturated ted and Superheated Superheated Vapors If a su subs bsta tanc ncee ex exis ists ts en enti tire rely ly as vapor at saturation temperature, it is called saturated vapor. When the vapor is at a temperature greater than the saturation temperature, it is said to exist as superheated vapor. The The pr pres essu sure re an and d te temp mpera eratu ture re of superheated vapor are independent properties, since the temperature may ma y in incr crea ease se wh while ile the pr pres essu sure re remains constant 13 Latent Heat Latent Late nt he heat at: The amount of energy absorbed or released during a phase-change process. Latent heat of fusion: The amount of energy absorbed during melting. melting. It is equivalen equivalentt to the amou amount nt of ener energy gy released during freezing. Latent heat of vaporization: vaporization: The amount of energy absorbed during vaporization and it is equivalent to the energy released during condensation. At 1 atm pr pressur essure, e, the latent heat of fusion of water is 333.7 kJ/kg and the latent heat of vapor vaporizatio ization n is 2256.5 kJ/kg. 14 Quality When a substance exists as part liquid and part vapor at saturation conditions, its quality (x) is defined as the ratio of the mass of the vapor to the total mass of both vapor and liquid. The quality is zero for the saturated liquid and one for the saturated vapor (0 ≤ x ≤ 1) For example, if the mass of vapor is 0.2 g and the mass of the liquid is 0.8 g, then the quality is 0.2 or 20%. x masssaturat saturated ed vapor masstotal mg mf mg 15 Quality A two phase system can be treated as a homogenous mixture for convinience The relative amounts of liquid and vapor phase in saturated mixture are specified by the quality x 16 Moisture Content Moisture The m moi oist stur uree co cont nten entt of a subs su bsta tanc ncee is th thee op oppo possit itee of it itss quality. Moisture is defined as the ratio of the mass of the liquid to the total mass of both liquid and vapor Recall the definition of quality x x mg m Then mf m mg mf m mg m 1 mg x 17 Moisture Content Moisture Take spe specifi cificc volum volumee as an exam example ple.. The specif specific ic volum volumee of the saturated mixture becomes v (1 x)v f xv g vf ) The form that is most often used v vf x (v g Let Y be any extensive property and let y be the corresponding intensive property property,, Y/m Y/m,, then y where y fg Y m yf yg yf x y fg yf x( yg yf ) 18 Property Table Fo Forr ex exam ampl plee if th thee pr pres esssur uree and specific volume are specified, three questions are asked: For the given pressure, 19 Property Table If answ firs t question questi theethest th staaanswer te ier s ito n the the the first compr om preesse sson ed isliqu iqyes, uid region, and the compressed liquid table is used us ed to find find the the prop proper erti tiees. (or (or usin using g temperaturee table) saturation temperatur table) If the the answ answer er to the the seco second nd questi question on is yes, the state is in the saturation region, and either the saturation temperature table or the saturation pressure table is used. If the answer answer to the third third ques question tion is yes, yes, the state is in the superheated region and the superheated table is used. v vf vf vg v vg v 20 e l b a T y t r e p o r P 21 Example 2.1 Determine the saturated pressure, specific volume, internal energy and enthalpy for saturated water vapor at 45 oC and 50oC. 22 Example 2.2 Determine the saturated pressure, specific volume, vol ume, internal energy and enthalpy for saturated water vapor at 47⁰ C . 23 Solu So lutti on: Extract data from steam table T satt P sa v u h 45 9.5953 15.251 2436.1 2582.4 47 Psat v u h 50 12. 35 2 1 2. 026 24 42. 7 2591 . 3 Interpolation for Psat P sat 9.5953 12.352 9.5953 P sat @ 47 47 45 45 50 45 10.698 kPa Interpolation Scheme for Psat Do the ssame ame principal to others!!!! 24 EXAMPLE • A RIGID TANK CONTAINS 50KG OF SATURATED LIQUID WATER AT 900C. DETERMINE THE PRESSURE IN THE TANK AND THE VOLUME OF THE TANK. • A PISTON CYLINDER DEVICE CONTAINS 0.057 M3 SATURATED WATER VAPOR AT 350KP 350KPA A PRESSURE PRESSURE.. DETERMINED THE TEMPERA TEMPER ATURE AND THE MASS OF THE VAPOR CYLINDER. Exercises 1. Fill in the bla blank nk usi using ng R-1 R-134a 34a 2. Determine the saturated temperature, saturated pressure and enthalpy for water at specific volume of saturated vapor at 10.02 m3/kg . 26 Example 2.3 v vf Is Is Dete De term rmin inee the en entha thalp lpy y of 1. 1.5 5 kg of water contained in a volume of 1.2 m3 at 200 kPa. f v vg Is ? No ? Yes g v v v ? No Solu So lutti on: Find the quality v v f x (v g v f ) x Specific volume for water v vf vg v f v 3 Volume mass 1.2 m 1.5 kg vf 0.001061 vg 0.8858 m3 kg 0.8 0.8 0.001061 kg kg 0.8858 0.001061 From table A-5: m3 m3 0.903 (What does this mean?) The enthalpy h hf x h fg 504.7 504 .7 (0.903 (0.903)(220 )(2201.6 1.6)) 2492.7 kJ kg 27 Example 2.4 Determine the internal energy of refrigerant-134a at a temperature t emperature of 0 C and a quality of 60%. Solu So lutti on: From table A-5: uf ug 51.63 The The in inte tern rnal al ene enerrgy o off R 1 134 34aa at given condition: kJ kg 230.16 kJ kg u uf x (u g uf ) 51.6 51 .63 3 (0.6 (0.6)(2 )(230 30.1 .16 6 51.6 51.63 3) kJ 158.75 kg 28 Example 2.5 Consider the closed, rigid container of water as shown. The pressure is 700 kPa, the mass of the saturated liquid is 1.78 kg, and the mass of the saturated vaporr is 0.2 vapo 0.22 2 kg. He Heat at is ad adde ded d to the water until the pressure increases to 8 MPa. Find the final temperature, enth en thal alpy py,, and and in inte tern rnal al en ener ergy gy of th thee water mg, Vg Sat. Vapor mf, Vf Sat. Liquid 29 Solu So lutti on: v 2 State 2: Sta Theoretically: v 1 The quality before pressure increased (ssttate 1). Information : P2 8 MP MPa a v2 m3 0.031 kg From table A-5: 3 x1 mg1 m f 1 mg1 0.22 kg (1.78 1.78 0.22 0.22)) kg v f,2 v g,2 0.11 Specific volume at sta state 1 v1 v f 1 x1 (vg1 v f 1 ) 0.001 0.0 0110 108 8 (0.11 (0.11)(0 )(0.2 .272 728 8 0.001 0.00110 108) 8) 0.031 kg 0.02352 m kg vg 2 v2 m3 kg Since that it is in superheated region, use table A-6: T2 361.8o C h2 3024 kJ kg kJ m3 0.001384 u2 2776 kg 30 Exercises 3 1. Fou Fourr kg of wat water er is plac placed ed in an en enclos closed ed vo volum lumee of 1m . Heat is added until the temperature is 150° 150°C. Find ( a ) the pressure, ( b )the mass of vapor, and ( c ) the volume of the vapor. 2. A piston-cylin piston-cylinder der device contain containss 0.1 m3 of liquid water and 0.9 m3 of water vapor in equilibrium at 800 kPa. Heat is transferred at constant pressure until the temperature reaches 350°°(a) 350 C. what is the initial temperature of the water, (b) determine the total t otal mass of the water, (c) calculate the final volume, and 31 Exercises 3 3. For a spe specif cific ic volum volumee o off 0.2 m /kg, find the quality of steam if the absolute pressure is (a) 40 kPa and ( b ) 630 kPa. What is the temperature of each case? 4. Water is cconta ontaine ined d in a rig rigid id ve vesse ssell of 5 m3 at a quality of 0.8 and a pressure of 2 MPa. If the a pressure is reduced to 400 kPa by cooling the vessel, find the final mass of vapor mg and mass of liquid mf. 32 Important Definition o Critical point - the temperature and pressure above which there is no distinction between the liquid and vapor phases. o Triple point - the temperature and pressure at which all three phases can exist exist in equilibrium. equilibrium. o Sublimation - change of phase from solid to vapor. vapor. o Vaporization - change of phase from liquid to vapor. vapor. o Condensation - change of phase from vapor to liquid. o Fusion or melting - change of phase from solid to liquid. 33 34 Ideal Gas Law Robert Boyle formulates a well-known law that states the pressure of a gas expanding at constant temperature varies inversely to the volume, or P1V1 P2V2 constant As the result of experimentation, Charles concluded that the pressure of a ga gass va vari ries es di dire rect ctly ly wi with th te tem mpe pera ratu ture re wh when en th thee vo volu lum me is hel eld d constant, and the volume varies directly with temperature when the pressure is held constant, or V1 V 2 T1 T 2 or P1 P 2 T1 T 2 35 By co com mbini binin ng th thee resu result ltss of Charles' and experim imeent ntss, the Boyle's follo low wing Pv T constant relationship can be obtained The constant in the above equation is called the ideal gas constant and is designated by Pv RT R T or PV mRT mR T R;; thus the ideal gas equation R becomes In orde orderr to ma make ke the equa equatio tion n applicable to all ideal gas, a univer uni versal sal gas con consta stant nt RU is R RU M introduced 36 For example example the ideal gas cons constant tant for air, Rair Rair ( RU ) air ( M ) air 8.3144 28.96 0.2871k J / k g.K The amount of energy needed to raise the tem temperature perature of a unit of mass of a substance by one degree is called the specific heat at constant volume Cv for a constant-volume process and the specific heat at constant pressure Cp for a constant pressure process. They are defined as Cv u T v and an d CP h T P 37 Using the definition of enthalpy (h ( h = u + Pv) and writing the differential of enthalpy, the relationship between the specific heats for ideal gases is h dh u Pv du RT CP dt CP CV CV dt RdT R The specific heat ratio, k is defined as k CP Cv 38 For ideal gases u u,, h h,, Cv, and Cp are functions of temperature alone. The Δu and Δh of ideal gases can be expressed as u u 2 u1 Cv (T2 T1 ) h h2 h1 CP (T2 T1 ) 39 Example 2.6 Solu So lutti on: gi ven giv An ideal gas is contained in a closed assembly with an initial pressure and temperature of 220 kPa and 700C res respec pective tively ly.. If thee vol th olu ume of the the system is inccrea in eassed 1.5 tim imees and th thee temperature drops to 150C, determine the final pressure of the gas. state 1 P1 220 kPa T1 70 2 27 73K state 2 T2 15 273 V2 343 K 288 K 1.5V1 From ideal-gas law: PV 1 1 T1 P2 P2V2 T2 288 3 220 10 1.5V1 343 V1 123.15 kPa 40 Example 2.7 Solu So lutti on: giv ven gi state 1 A assembly kg of closed air at an initial contains pressure 2and temperature of 140 kPa and 2100C respectively. If the volume of the system is doubled and temperature drops to 37 0C, determine the final pressure of the air. Air can be modeled as an ideal gas. P1 140 kPa T1 210 2 27 73 K 483 K state 2 T2 37 273 V2 2 V1 310 K From ideal-gas law: PV 1 1 T1 P2 P2V2 T2 310 3 1 4 0 1 0 2V1 483 V1 44.93 kPa 41 Example 2.8 An automobile tire with a volume of 0.6 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if the temperature is 20° 20°C. Solu So lutti on: giv gi ven m state 1 P T From ideal-gas law: 200 100 kPa 20 2 27 73 K 293 K PV RT 300 103 287 Nm kg . K N m3 0.6m 2 293K 2.14 kg 42 Supplementary Problems 1. The pressu pressure re in an an automobi automobile le tire tire depends depends on the temperat temperature ure of of the air air in the tire. When the air temperature is 25° 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 50° 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. temperature. Assume the atmospheric atmospheric pressure is 100 kPa. [ 26 kPa, 0.007 kg ] 2. A 1-m3 tank containing air at 25° 25°C and 500 kPa is connected through a valve to another tank containing 5 kg of air at 35° 35 °C and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20° 20°C. Determine the volume of the second tank and the final equilibrium pressure of air. air. [ 2.21 m3, 284.1 kPa kPa] 43 3. A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5 m3 with propane at 250 kPa, 400 K. The valve is opened and the two tanks come to a uniform state at 325 K. What is the final pressure? [ 139.9 kPa] kPa] 4. A cyli cylind ndri rica call gas gas tank ank 1 m long long,, insi nside diam diamet eter er of 20 cm, cm, is evacuated and then filled with carbon dioxide gas at 25° 25°C. To what pressure should it be charged charged if there should be 1.2 kg of carbon dioxide? [ 2152 kPa] kPa] 44 The ideal gas equation of state is used us ed when (1) the pressure is small compared to the critical pressure or (2) when the temperature is twice t wice the critical temperature and the pressure is less than 10 10 times the critical pressure. The critical point is that state where there is an instantaneous change from the liquid phase to the t he vapor phase for a substance. Critical point data are given in T Table able A-1. Compressibility Factor To understand the above criteria and to determine how much the ideal gas equation of state deviates from the actual gas behavior, behavior, we introduce the compressibility factor Z as follows. Pv Z Ru T or Z Pv Ru T For an ideal gas Z = 1, 1, and the deviation of Z from unity measures the deviation of the actual P-V-T relation from the ideal gas gas equation of state. The compressibility factor is expressed as a function f unction of the reduced pressure and the reduced temperature. The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure, which are defined as TR where P cr T Tcr and PR P Pcr respectively.. The and Tcr are the critical pressure and temperature, respectively critical constant data for various substances are given in Table A-1 A-1. This is known as the principle of corresponding corresponding states. Figure 3-51 gives a comparison of Z factors for various gases and supports the principle of corresponding states. When either T is unknown, Zed can be determined from theas compressibility chart with the help P ofor the pseudo-reduced pseudo-reduc specific volume, defined vR vactual R Tcr Pcr Figure A-15 presents the generalized compressibility chart based on data for a large number of gases. These charts show the conditions for which Z = 1 and the gas behaves as an ideal gas: 1. 1.P PR < 10 and TR > 2 or P < 10P 10Pcr and T > 2T 2Tcr 2. 2.P PR << 1 or P << Pcr Note: When PR is small, we must make sure that the state is not in the compressed liquid region region for the given temperature. temperature. A compressed liquid state is certainly not an ideal gas state. For instance the critical pressure and temperature for oxygen are 5.08 MPa and 154.8 K, respectively. respectively. For temperatures greater greater than 300 K and pressures less than 50 MPa (1 atmosphere pressure is 0.10135 MPa) oxygen is considered c onsidered to be an ideal gas. Example 2-6 Calculate the specific volume of nitrogen at 300 K and 8.0 MPa and compare the result with the value given in a nitrogen table as v = 0.011133 m 3/kg. From Table A.1 for nitrogen 48 Tcr = 126.2 K, Pcr = 3.39 MPa R = 0.2968 k kJ/kg-K J/kg-K TR PR 300K T 126.2 K Tcr 8.0 MPa P 2.38 Pcr 2.36 3.39 MPa Since T > 2T 2Tcr and P < 10P 10Pcr, we use the ideal gas equation of state Pv v RT 0.2968 RT P kJ (300 K ) m3 MPa kg K 8.0 MPa 103 kJ 0.01 0111 113 3 m3 kg Nitrogen is clearly an ideal gas at this state. If the system pressure is low enough and the temperature high enough (P (P and T are compared to the critical values), gases will behave as ideal gases. Consider the T-v diagram for water. water. The figure below shows the percentage of error for the volume 49 ([|v ([| vtable videal|/ |/v vtable]x100) for assuming water (superheated steam) to be an ideal gas. – We see that the region for which water behaves as an ideal gas is in the superheated region and depends on both T and P. We must be cautioned that in this course, when water is the working fluid, the ideal gas assumption may not be used tto o solve problems. We must use the real gas relations, i.e., the property tables. 50 Other Equations of State Many attempts have been made to keep the simplicity of the ideal gas equation of state but yet account for the intermolecular forces and volume occupied by the particles. Three of these are van der Waals: (P a v 2 )(v b) R T where a 27 R 2 Tcr2 64 Pcr and b RTcr 8 Pcr 51 Beattie-Bridgeman: where The constants a, b, c, Ao, Bo for various substances are found in Table 3-4. Benedict-Webb-Rubin:: Benedict-Webb-Rubin The constants for various substances appearing in the t he Benedict-Webb-Rub Benedict-Webb-Rubin in equation are given in Table 3-4. 52
