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Transcript
ESS 2410 Engineering Thermodynamics Overview and Review Thermodynamics and Energy Thermodynamics: the science of “energy” Energy: the ability to cause change, the capability to do work Energy can exist in many different forms thermo (heat) ; dynamics (power) Thermodynamics: converts heat to power Thermodynamics: include all aspects of energy, energy transformation, power generation, refrigeration, relationship among properties of matter 1 Conservation of energy principle: Energy can change from one form to another, but the total amount of energy remains constant. Energy is a thermodynamic property fundamental law of nature (1st Law of Thermodynamics) The 2nd Law of Thermodynamics: the energy has quantity as well as quality Energy transfer can occur certain direction but not in the reverse direction. The actual process occurs in the direction of decreasing quality of energy. The ability to do useful work diminishing even though energy is not. Thermodynamics: science of energy and entropy. 2 Forms of Energy Energy can exist in numerous form: thermal, mechanical, kinetic, potential, electric, magnetic, chemical, nuclear….. Thermodynamics provides no information about the absolute value of the total energy. It only deals with the change of total energy. Microscopic forms of energy: those related to the molecular structure of a system and the degree of the molecular activities, and they are independent of outside reference frames. Completely Random; Disorganized form of Energy 3 Energy Microscopic Energy Macroscopic Energy Internal Energy Kinetic Energy Chemical Bonding Energy Nuclear Bonding Energy Potential Energy Thermal Energy Electrical Energy Chemical reaction Nuclear reaction Chemical Energy Kinetic Energy Nuclear Energy friction Kinetic Energy Sensible Energy Heat Generator Potential Energy Mechanical device Internal Thermal Energy Latent Energy Internal Thermal Energy Mechanical device Work Internal Energy: the sum of all the microscopic forms of energy kinetic energy: sensible energy (sensible heat) (proportional to temperature) potential energy: binding force between molecules; strongest in solid, weakest in gas, energy required in phase change latent energy (latent heat) Chemical Energy and Nuclear Energy Macroscopic forms of energy: A system possesses as a whole with respect to some outside reference frame (kinetic and potential energy) Orderly motion: Organized form of Energy 5 Static form of Energy: total energy of a system can be contained or stored in a system. Dynamic form of Energy: energy interactions, energy in transit; defined at system boundary, represent the energy gained or lost by a system during a process Heat : energy interaction driving by temperature difference Organized form of energy is more valuable than the disorganized form of energy. Organized energy can be converted to disorganized energy completely. Only fraction of disorganized energy can be converted into organized energy by specially built devices called heat engines. Thermodynamics: the conversions of disorganized energy into organized energy. 6 Application area of thermodynamics: The science that explains and predicts how much energy we may extract and how efficiently we may do it under a specific situation. Automotive engines; rockets, jet engines, power plant, refrigeration Deals with various properties of substances and the change in these properties as results of energy transformation. 7 Property of System Property: Any characteristic of a system, pressure, temperature, volume, mass, viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficients, color, hardness,….. Not all properties are independent: Two properties are independent if one property can be Varied freely while the other one is constant Some are defined in terms of others: density: mass per unit volume specific properties: property per unit mass specific volume: volume per unit mass (1/density), specific heat, specific power……. 8 Intensive Properties: properties are independent of the size of system pressure, temperature, density…… Specific volume, specific enthalpy, specific internal energy… Extensive Properties: properties depend on the size-or extent- of the system volume, mass, total energy,…… Specific properties are intensive properties Extensive property per unit mass 9 State and Equilibrium State: the condition of system as determined by its properties At a given state, all the properties of a system have fixed values Each of the properties of a substance in a given state has only one definite value, and these properties always have the same value for a given state, regardless of how the substance arrived at that state. How many? A state can be specified by a certain number of properties. Equilibrium: implies a state of balance, no unbalanced potentials (or driving force) experiences no changes when it is isolated from its surroundings. 10 Thermodynamics deals with equilibrium system Thermodynamics Equilibrium thermal equilibrium (no temperature difference mechanical equilibrium (no imbalance force) phase equilibrium (if phase changes involved) chemical equilibrium (no chemical changes) Processes and Cycles Process: any change that a system undergoes from one equilibrium state to another equilibrium state. Path: the series of states through which system passes during a process. To describe a process completely: initial state, path, final state 11 Compressed Process P-V Diagram 12 Process Diagram: employ thermodynamics properties as coordinates to visualize the process, T - P, T - v, P – v, T – s The prefix – iso is often used to designate a process: isothermal : constant temperature; isobaric: constant pressure; isochoric (isometric): constant volume; isentropic: constant entropy Cycles: A system is said to have undergone a cycle if it returns to its initial state at the end of a series of processes. 13 The State Postulate no external force The State Postulate: the state of simple compressible system is completely specified by two independent, intensive properties T, VX T, v P, T (gas, liquid, not during phase change) P, v; T, u; P, h; ….. Two properties are independent if one property can be Varied freely while the other one is constant 14 T-v diagram of the heat addition of water at constant pressure State 5 State 2 State 3 State 4 State 1 15 State 1 water, P = 1 atm, T = 20 ; water exists in liquid phase compressed liquid or subcooled liquid Compressed: Pressure is higher than Psat @ T = 20 C 1.013x105 Pa 2.339x103 Pa Subcooled: Temperature is lower than Tsat @ P = 1 atm 20 C 100 C Compressed or Subcooled: meaning it is not about to vaporized heat addition temperature of liquid increases volume expansion 16 State 2 water reaches 100 C, water is still in liquid form Tsat @ P = 1 atm Saturated liquid: a liquid is about to vaporize Phase change process from liquid to vapor is about to occur Any heat addition will cause some of the liquid to vaporize Under saturated condition, temperature and pressure are not independent. 17 State 3 two- phase mixture, liquid water and water vapor coexist, temperature is 100 C, pressure is 1 atm liquid and vapor are at the same temperature temperature remains the same heat addition amount of vapor increases volume expand significantly boiling vaporization The vaporization will continue until last drop of water is vaporized 18 State 4 entire vessel is filled with saturated water vapor, T = 100 C, P = 1atm Any heat loss will cause some of the vapor to condense Saturated Vapor : a vapor is about to condense State 5 entire vessel is filled with superheated vapor T = 300 C, P = 1 atm Superheated vapor: a vapor that is not about to condense Temperature is greater than Tsat @ given pressure Pressure is lower than Psat @ given temperature heat addition temperature of vapor increases volume expansion 19 T-v diagram of constant pressure phase-change process 20 T-v diagram of a pure substance 21 P-v diagram of a pure substance 22 Saturation temperature and saturation pressure saturation temperature (Tsat): at a given pressure, the temperature at which a pure substance changes phase saturation pressure (Psat): at a given temperature, the pressue at which a pure substance changes phase During a phase change process, pressure and temperature are dependent properties. Tsat f Psat Tsat as Psat Liquid-vapor saturation curve, Clapeyron equation Latent heat: energy absorbed or released during a phase change Table A-2 (760) Saturated Water--temperature table Table A-3 (762) Saturated Water-- pressure table 23 Critical Point: the point at which the saturated liquid and saturated vapor are identical. Critical Point Properties: Tcr, Pcr, vcr Above the critical state, there is no line that separates the compressed liquid region and superheated vapor region. At pressure above the critical pressure, there will be no distinct phase change process. The specific volume of substance will continually increase. (Figure on page 20) Above critical state: Temperature above Tcr: superheated vapor Temperature below Tcr: compressed liquid Saturated liquid line: a line that connects the saturated liquid state Saturated vapor line: a line that connects the saturated vapor state 24 Property Table For most substances, the relationships among thermodynamic properties are presented in the form of tables, Property Table of water Table A-2 Table A-3 Table A-4 Table A-5 Table A-6 Steam Table Saturated Water – Temperature Table (p.760) Saturated Water – Pressure Table (p.762) Superheated Vapor (p.764) Compressed Liquid Water (p.768) Saturated Ice-Water vapor (p. 769) Properties in Property Tables: Pressure, Temperature, Specific Volume, Internal Energy (specific) Enthalpy (specific), Entropy (specific) 25 1a: Saturated liquid and saturated vapor state subscripts f : saturated liquid g : saturated vapor fg : difference between saturated vapor and saturated liquid Need two properties to determine the state the word “saturated” is one of the condition 1b : saturated liquid-vapor mixture (P and T are dependent property) Quality (x) : mass fraction of vapor in the mixture; ratio of mass of vapor to the total mass of the mixture A property 26 Definition of Quality V V f Vg mtot ave m f ave mf mtot f f mg g mg mtot g ave 1 x f x g f ave x ave fg g f f u ave 1 x u f x u g u f x u fg have 1 x h f x hg h f x h fg save 1 x s f x s g s f x s fg 0 x 1 saturated liquid saturated vapor quality has no physical meaning for compressed liquid and superheated vapor In two-phase region (two phase mixture): y f yave y g 27 2 Superheated Vapor Single phase region: P and T are independent property lower pressure (P < Psat @ given T) higher temperature (T > Tsat @ given P) greater specific volume (v > vg @ given P or T) higher internal energy (u > ug @ given P or T) higher enthalpy (h > hg @ given P or T) greater entropy (s > sg @ given P or T) 3. Compressed Liquid (P and T are independent property) higher pressure (P > Psat @ given T) lower temperature (T < Tsat @ given P) smaller specific volume (v < vf @ given P or T) lower internal energy (u < uf @ given P or T) lower enthalpy (h < hf @ given P or T) smaller entropy (s > sf @ given P or T) 28 3. Compressed Liquid (P and T are independent) Not many data exist, relative independence of compressed liquid properties from pressure. Treat compressed liquid at the given temperature. Given P and T f at given T u u f at given T h h f at given T f P Psat Values of u, h, and s can not be measured directly. Calculated using thermodynamics relations (Chapter 11) Relation gives changes in thermodynamic properties related to a reference state. The values of u and s at reference state is set equal to 0.0 the reference state of water is triple point 29 1. water @ P = 150 kPa, T = 111.4 C; find u and v 2. water @ P = 150 kPa, v = 0.0010435 m3/kg; find T and u 3. water @ T = 400 C, v = 0.02641 m3/kg; find P and u 4. water @ T = 120 C, v = 0.44648 m3/kg; find P and u Answers: 1. v = 0.0010528 m3/kg, u = 466.94 kJ/kg (saturated water) 2. T = 100 C, u = 418.94 kJ/kg (compressed liquid) 3. P = 10 MPa, u = 2832.4 kJ/kg (superheated vapor) 4. P = 198.53 kPa, u = 1516.4 kJ/kg (2-Phase mixture) 30 1. water @ P = 150 kPa, T = 111.4 C; find u and v from Table A-3 Tsat @ P = 150 kPa = 111.4 C T= Tsat @ P = 150 kPa; it is saturated liquid from Table A-3 vf @ P=150 kPa = 0.0010528 m3/kg uf @ P=150 kPa = 466.94 kJ/kg 2. water @ P = 150 kPa, v = 0.0010435 m3/kg; find T and u from Table A-3 vf @ P = 150 kPa = 0.0010528 m3/kg v < vf; it is compressed liquid use temperature to determine its properties from Table A-2 (use v find T) vf @ T = 100 C = 0.0010435 m3/kg therefore T = 100 C, and u ~ uf = 418.94 kJ/kg 31 3. water @ T = 400 C, v = 0.02641 m3/kg; find P and u from Table A-2 Tcritical = 374.14 C T > Tcritical ; it superheated vapor use Table A-4 determine its properties First try P = 2.0 MPa , v@ P=2.0 MPa, T = 400 C = 0.1512 m3/kg which is greater than 0.02641 m3/kg ; therefore the Pressure should be higher than 2 MPa. From Table A-4, @ P = 10 MPa, T= 400 C, v = 0.02641 m3/kg Therefore P = 10 MPa, u = 2832.4 kJ/kg 32 4. water @ T = 120 C, v = 0.44648 m3/kg; find P and u From Table A-2, vf (0.0010603)< v(0.44648)<vg(0.8919) It is two-phase mixture Find x (quality) P = 198.53 kPa v vf 0.44648 0.0010603 x 0.5 vg v f 0.8919 0.0010603 u (1 x) u f x u g u f x u fg 503. 50 0.5 2025.8 1516 . 4 (kJ / kg) 33 Internal Energy, Enthalpy, and Specific Heats of Ideal Gas For an ideal gas, the internal energy is function of temperature only. u u T Enthalpy: u u (T ), Pv RT , h u RT h hT u h du dh Cv ;Cp Cv Cp dT dT T v T p du C T dT , u 12 C T dT dh CP T dT , h 12 CP T dT The specific heat of real gases at low pressures are called ideal-gas specific heats or zero-pressure specific heats. C p , C Ideal gas specific heats of monatomic gases (Ar, Ne, He) remain constant over the entire temperature range. 34 35 Three ways to determine the internal energy and enthalpy change of ideal gases. 1. Table A-22 (p. 796) Reference point is 0 K 2. Performing integration. (Table A-21, p. 794) 3. Using average specific heats. u2 u1 Cv ,av T2 T1 Cv ,ave and C p ,ave are evaluated at T2 T1 h h C T T 2 1 P , av 2 1 2 Specific-Heat Relations of Ideal Gases h u RT , dh du rdT C P Cv R, C P Cv Ru specific heat ratio: k C P Cv 36 The Tds Relations Qint rev Wint rev d U Qint rev Td S ; Wint rev P dV or TdS dU P dV Tds du P dv the only possible work mode of a simple compressible system Gibbs equation h u Pv dh du pdv vdp Tds du Pdv Tds dh vdp du Pdv dh vdP ds ; ds T T T T Tds relations related the entropy change of a system to the change of other properties. T, P, u, v, s are properties, therefore, the results obtained are valid 37 for both reversible and irreversible processes. Entropy Change of Ideal Gas du Pdv dh vdP ds ; ds T T T T dT Rdv dT RdP ds Cv ; ds C p T v T P 2 2 dT v2 dT P2 s2 s1 Cv R ln ; s2 s1 C p R ln 1 1 T v1 T P1 1. constant specific heat: approximate analysis T2 v2 T2 P2 s2 s1 Cv,ave ln R ln ; s2 s1 C p,ave ln R ln T1 v1 T1 P1 Good for monoatomic gases, good for most ideal gases if the temperature change is not very large. 38 2 variable specific heats: exact analysis dT P2 s2 s1 C p R ln 1 T P1 2 s T T dT Cp ; T dT C s T s 1 p T 2 2 1 T1 2 Tabulated value, Table A-22 (air) P2 s2 s1 T1 R ln P1 temperature dependency pressure dependency of entropy of entropy s2 T2 s1 39 Entropy Change of an ideal Gas Nitrogen gas is compressed from an initial state of 100 kPa and 17℃ to final state of 600 kPa and 57℃. Determine the entropy change of the nitrogen during this compression process by using (a) property values from the nitrogen table and (b) average specific heats. 40 State 1: P1 = 100 kPa, T1 = 290 K; State 2: P2 = 600 kPa, T2 = 330 K (a) Specific heat is not constant (Table A-23) P2 P1 194.459 190.695 8.314 ln 600 11.133 kJ kmole K 800 s2 s1 (b) s2 s1 s2 T2 s1 T1 Ru ln s2 s1 11.133 0.3974 kJ kg K M 28.013 Specific heat is constant (Table A-20) T2 P2 s2 s1 C p ,ave ln R ln T1 P1 use C p ,ave at 37 C, C p ,ave 1.0394 kJ/kg K 330 600 s2 s1 1.0394 ln 0.297 ln 290 100 0.3978 kJ/kg K 41 Isentropic Process of Ideal Gas Constant Specific Heat: Approximate Treatment T2 v2 s2 s1 Cv ln R ln 0 T1 v1 T2 v2 Cv ln R ln T1 v1 v1 T2 ln ln T1 v2 R Cv R C p Cv ; Define k C p Cv T2 v1 T1 isentropic v2 T2 P2 T1 isentropic P1 R Cv k 1 k v1 v2 k 1 P2 v1 ; P1 isentropic v2 ideal gas, isentropic, constant specific heat k 42 Tv k 1 constant TP 1k k constant ideal gas, isentropic constant specific heat Pv k constant Variable Specific Heat: Exact Treatment s2 s1 s2 s1 P2 R ln 0 P1 s2 s2 s1 exp s2 R P2 exp exp s R P1 R 1 s1 P2 R ln P1 dimensionless function of Define : exp s R as relative pressure Pr temperature only tabulated value P2 P T (Table A-22) r2 2 P1 s constant Pr1 T1 43 P1v1 P2v2 T1 T2 Define : T Pr v2 T2 P1 T2 Pr1 T2 Pr 2 v1 T1 P2 T1 Pr 2 T1 Pr1 as relative specific volume vr v2 vr 2 T2 v1 s constant vr1 T1 dimensionless function of temperature only tabulated value (Table A-22) 44 Isentropic Compression of air in a Car Engine Air is compressed in a car engine from 22℃ and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of piston-cylinder device is 8, determine the final temperature of the air. 45 For closed system: V2 v 1 2 ; V1 v1 8 v2 v1 vr 2 isentropic vr1 1 8 from Table A - 22; vr1 at T1 295 K is 647.9 vr 2 vr1 / 8 80.99 from Table A - 22 T2 662.7 K Alternate Solution: specific heat is constant T2 v1 T1 isentropic v2 v1 T2 T1 v2 665.2 k 1 k 1 29581.3911 K The value of k has to be evaluated at the average of initial and final temperature. Assume the average temperature is 450 K. (665.2+295)/2 = 480.1; need iteration 46 Isentropic Compression of an ideal Gas Helium gas is compressed in an adiabatic compressor from an initial state of 14 psia and 50℉ to a final temperature of 320℉ in a reversible manner. Determine the exit pressure of helium. Specific heat of helium is constant P2 T2 P1 T1 k k 1 T2 P2 P1 T1 k k 1 1.667 0.667 780 P2 14 510 40.5 psia 47 Isentropic Efficiencies of Steady Flow Devices Isentropic process is a model process for most steady-flow device. The more closely the actual process approximates the isentropic process, the better the device will perform. Isentropic efficiency of turbine: actual turbine work isentropic turbine work wa h 1 h 2 a ws h 1 h 2 s T 48 Isentropic Efficiency of a Steam Turbine Steam enters an adiabatic turbine steadily at 3 MPa and 400℃ and leaves at 50 kPa and 100℃. If the power output of the turbine is 2 MW, determine (a) the isentropic efficiency of the turbine and (b) the mass flow rate of the steam flowing through the turbine. 49 next State 1: P1 = 3 MPa T1 = 400 C s1 = 6.9212 kJ/kg K h1 = 3230.9 kJ/kg State 2a: P2a = 50 kPa T2a = 100 C h2a = 2682.5 kJ/kg P2s = 50 kPa s2s = s1 sf = 1.0910 kJ/kg K sg = 7.5939 kJ/kg K State 2s: (a) s f s2 s s g ; x2 s s2 s s f s fg h2 s h f x2 s h fg 2407.4 0.897 kJ / kg h1 h2 a T 66.7% h1 h2 s Figure 6.50 (b) Wa m h1 h2a ; 2 103 m 3230.9 2682.5 m 3.65 kg / s 50 Isentropic efficiency of compressors and pumps: isentropic compressor work actual compressor w ork w h h s 2s 1 wa h2 a h1 c P ws v P2 P1 wa h2 a h1 51 Effect of Efficiency on Compressor Power Input Air is compressed by an adiabatic compressor from 100kPa and 12℃ to a pressure of 800 kPa at a steady rate of 0.2kg/s. If the isentropic efficiency of the compressor is 80 percent, determine (a) the exit temperature of air and (b) the required power input to the compressor. 52 c ws h2 s h1 wa h2a h1 State 1: T1 = 285 K (Table A.22) h1 = 285.14 kJ/kg Pr1= 1.1584 State 2s: P2 = 800 kPa Find state 2 s; isentropic process Pr 2 P2 P2 Pr 2 Pr1 Pr1 P1 P1 from Table A.22; Pr 2 9.2672 State 2a find h2 a ; h2 a 575.03 800 1.1584 9.2672 100 h2 s 517.05 kJ / kg 517.05 285.14 0.80 kJ / kg; h2 a 285.14 from Table A.22 T2 a 569.5 K 53 Isentropic efficiency of nozzle: N actual KE at nozzle exit isentropic KE at nozzle exit V22a V22s h1 h2a h1 h2 s 54 Effect of Efficiency on Nozzle Exit velocity Air at 200 kPa and 950K enters an adiabatic nozzle at low velocity and is discharged at a pressure of 80 kPa. If the isentropic efficiency of the nozzle is 92 percent, determine (a) the maximum possible exit velocity, (b) the exit temperature, and (c) the actual velocity of the air. Assume constant specific heats for air. 55 P1 = 200 kPa, T1 = 950 K; P2 = 80 kPa (a) Find T2s (specific heat is constant) T2 s P2 s T1 P1 k 1 k 80 950 200 748 K 0.354 1.354 V22s V12 h1 h2 s 2 2 V2 s 2h2 s h1 2C p ,ave T2 s T1 666 m / s Guess the average temperature is 800 K; to determine k (748+950)/2 = 849, need iteration C p,ave T1 T2a h h 1 2 a (b) N h1 h2 s C p,ave T1 T2 s 950 T2a 0.92 T2a 764 950 748 (C) N V22a V22s K V2a NV22s 639 m/s 56 Decimal Relationship: 10-12 10-9 Pico(p) nano(n) pm nm 奈米 101 102 deka hecto 10-6 micro( ) m 微米 10-3 10-2 milli(m) centi(c) mm cm 毫米 公分 103 106 109 kilo(k) mega(M) giga (G) km (公里) Mm Gm kW MW GW 千瓦 百萬瓦 十億瓦 10-1 deci(d) dm 公寸 1012 tetra(T) Tm TW 兆瓦 百: hundred; 千: thousand; 萬: ten thousands; 百萬: million 千萬: ten million; 億: hundred million; 十億: billion 兆 (萬億) : trillion 57