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Thermodynamics Review Craig Bradshaw, Lambert Fellow and PhD Candidate Slides provided by Prof. S. F. Son, and M. Mathison Adapted from Prof. G. A. Risha and other sources such as Kaplan AEC Education Thermodynamics is the transformation of energy THE FIRST LAW OF THERMODYNAMICS • Energy cannot be created or destroyed, but transformed into different forms • Applies to systems classified as either closed or open Thermodynamic Systems Energy Balance: Finite Time E stored E in E out E gen Energy Balance: Rate E E E E stored in out gen Energy Transfer by Work and Heat: Qout< 0 Win < 0 Qin > 0 Wout > 0 Engines burn fuel to put heat into the system and produce work (out of the system) What is your system? Closed (fixed mass) or open (fixed volume) Fixed Mass: Closed System dEcv Q W dt ΔE stored Q W ΔU ΔKE ΔPE Q W Note: Closed systems are a subset of open systems Systems and Sign Convention Consider a system that contains a lightbulb powered by electricity. Which of the following is true? a) Positive work, positive heat transfer b) Positive work, negative heat transfer c) Negative work, positive heat transfer d) Negative work, negative heat transfer dEcv Q W 0 Q W + dt Which of the following would be considered a closed system? a) A pump b) A tire c) A turbine d) A jet nozzle Closed System: Work W Fdx P Boundary Work: Wboundary work pdV m pdv Note that if volume is constant, W=0 Constant Pressure: W pV pV2 V1 mpv pv 2 v1 Polytropic: pv n constant p 2 V2 p1V1 V2 W if n 1, W p1V1ln if n 1 1 n V1 Work: Special Cases (Ideal Gas) Ideal Gas Law: pV mRT pv RT Constant Temperature: pv constant v2 p1 W mRTln mRTln v1 p2 Isentropic: s2 = s1 pv k constant p 2 V2 p1V1 mRT1 p 2 W 1 1 k k 1 p1 specific heat ratio, k cp cv k 1 k Closed System Work When using the previous equations for boundary work, it must be assumed that a quasi-equilibrium process exists. If a quasi-equilibruim process exists, we have assumed a) The pressure at any instant to be everywhere constant. b) An isothermal process. c) The heat transfer to be small. d) The boundary motion to be infinitesimally small. e) That no friction exists. Control Volume: Open System h in Energy entering the control volume in the forms of kinetic energy, potential energy, and enthalpy h in Vin2 gz in 2 Energy exiting the control volume in the forms of kinetic energy, potential energy, and enthalpy (h = u + pv) Vin2 gz in 2 2 dE CV Vout Vin2 in h in out h out Q W m gz in m gz out dt 2 2 dE CV 0 At steady-state, dt Nozzles and Diffusers Assumptions: (1) adiabatic, Q = 0 (2) no volume changes, W = 0 (3) steady-state, d/dt = 0 (4) change in potential energy negligible dE CV QW dt Vin2 in h in m gz in 2 2 Vout Vin2 h in h out 2 2 or 2 Vin2 Vout h h out - h in 2 2 2 Vout out h out m gz out 2 Nozzles and Diffusers h-s Diagram for Nozzle Nozzle Efficiency: Compares the performance of a real nozzle or diffuser to the performance of an ideal, isentropic nozzle or diffuser operating between the same pressures nozzle h actual V h ideal V 2 outlet 2 outlet,s V V 2 2 2 inlet 2 inlet Actual process Isentropic process with same initial state and final pressure Turbines, Pumps, and Compressors Assumptions: (1) adiabatic, Q = 0 (2) change in potential energy negligible (3) steady-state, d/dt = 0 (4) change in kinetic energy negligible dE CV QW dt Vin2 in h in m gz in 2 W h in h out m 2 Vout out h out m gz out 2 Turbines, Pumps, and Compressors Turbine and compressor/pump efficiencies both compare the performance of an actual device to the performance of an ideal, isentropic device operating between the same pressures • Turbine Efficiency: In turbines, the actual power generation will be less than the ideal power generation h actual h inlet h outlet actual t h ideal hinlet houtlet,s s isentropic • Pump/Compressor Efficiency: In pumps/compressors, the actual power consumption will be greater than the ideal power consumption isentropic h outlet h inlet,s s h ideal c h actual h outlet h inlet actual Throttling Valves Assumptions: (1) adiabatic, Q = 0 (2) no volume changes, W = 0 (3) steady-state, d/dt = 0 (4) change in potential energy negligible (5) change in kinetic energy negligible dE CV QW dt Vin2 in h in m gz in 2 2 Vout out h out m gz out 2 h in h out Boilers, Condensers, and Evaporators Assumptions: (1) constant volume, W = 0 (2) change in potential energy negligible (3) steady-state, d/dt = 0 (4) change in kinetic energy negligible dE CV QW dt Vin2 m in h in gz in 2 2 Vout m out h out gz out 2 Q h out h in m Heat Exchangers Assumptions: (1) constant volume, W = 0 (2) change in potential energy negligible (3) steady-state, d/dt = 0 (4) change in kinetic energy negligible (5) adiabatic, Q = 0 dE CV QW dt Vin2 in h in m gz in 2 2 Vout out h out m gz out 2 m in h in m out h out Example Problem 1 Given: Gas initially at 1 MPa and 150 oC receives 7.2 MJ of work while 1.5 kW of heat are removed from the system. Calculate the internal energy change for the system over a period of one hour. Analysis: negligible E U KE PE Q W U Q W Work done ON system: W = -7.2 MJ -1.5 kW OUT of the system: Q Heat -1.5kW(3600s) -5.4MJ Q t Q U 5.4 MJ (7.2 MJ) U 1.8 MJ Example Problem 2 Given: Calculate the power required to compress 10 kg/s of air from 1 atm and 37 oC to 2 atm and 707 oC. For low pressure air: T = 310 K; h = 290.4 kJ/kg T = 980 K; h = 1023 kJ/kg Analysis: dE CV QW dt Vin2 in h in m gz in 2 2 Vout out h out m gz out 2 m (hin h out ) W 10 W kg kJ (290.4 1023) kg s 7326 W kJ s 7326 kW 7326 kW W c Done ON system Example Problem 3 P Given: A gas goes through the following thermodynamic B cycle. A to B: constant-temperature compression B to C: constant-volume cooling C to A: constant-pressure expansion The pressure and volume at state C are 1.4 bar and 0.028 m3, respectively. The net work from C to A is 10.5 kJ. What is the net work from one complete cycle A-B-C? Analysis: Wcyc WAB WBC WCA V 0 B C A Wcyc pdV pdV pdV pdV A B C A C V Example Problem 3 P B Wcyc pdV pdV 10.5 kJ B A Constant Temperature: VB W RTln VA A C V Need volume @ A, constant pressure A to C W pA VA VC Compression, W<0 WCA VA VC pA 2 WCA 10500 J 1 bar 1 Pa m 1Nm 3 VA VC 0.028 m 5 pA 1.4 bar 1x10 Pa 1 N 1J VA 0.103 m3 Example Problem 3 P Constant Temperature: B WAB VB RTln VA C Ideal gas and constant mass WAB A V VB 1x10 5 Pa 1 N 1J 0.028 3 1.4 bar pVA ln 0.103 m ln 2 V 1 bar 1 N m 0.103 1 Pa m A WA B -18.8 kJ Wcyc 18.8 kJ 10.5 kJ Wcyc 8.3 kJ PROPERTIES OF PURE SUBSTANCES • Use of Steam Tables • Use of R-134a Tables, NH3 Tables, and P-h Diagrams Pure Substances • What is a pure substance? • Materials with unchanging chemical composition • Three common phases of pure substances: • Liquid, vapor, and solid • Also consider liquid-vapor mixtures, which are a combination of liquid and vapor where pressure and temperature are not independent • State Postulate: Two independent, intensive properties are required to fix a state – Intensive: Independent of system size (usually per mass) – Extensive: Dependent on size of system T-v Phase Diagram Diagram courtesy of Jerry M. Seitzman, 2001. T-v Phase Diagram @ Constant Pressure T > Tsat “superheated vapor” T = Tsat “two-phase liquid-vapor” T < Tsat “compressed liquid” P-v Phase Diagram @ Constant Temperature P < Psat “superheated vapor” P = Psat “two-phase liquid-vapor” P > Psat “compressed liquid” Common Properties A set of properties that completely describes a system, specifies the state of the system. • Mass, m – Extensive – Units [kg] • Volume, V – Extensive – Units [m3] • Pressure, P – Intensive – Absolute units [Pa, psia] – Gage, Pgage = Pabs - Patm • Temperature, T – Absolute units [K or oR] – T[K] = T[oC] + 273.15 – T[oR] = T[oF] + 459.67 • Specific Volume, v – Intensive – Units [m3/kg] – v = 1/r = V/m • Internal energy, u – Intensive – Units [kJ/kg] – U = m*u [kJ] • Enthalpy, h – Intensive – Units [kJ/kg] – h = u + Pv; H = U + P V – H = m*h Properties [cont’d] • Entropy, s – S, extensive [kJ/K] – s, intensive [kJ/kg-K] • Enthalpy – – – – H, extensive [kJ] H, intensive [kJ/kg] H = U + PV h = u + Pv] • Gibbs’ Function – – – – G, extensive [kJ] g, intensive [kJ/kg] g = h - Ts = u + Pv - Ts G = H - TS = U + PV – TS s Q rev T Saturated Liquid Vapor Mixture (SLVM) Quality: mass of vapor x m vap m liq m vap mg Quality is a function of the horizontal distances on P-v and T-v diagrams mf mg mass of liquid h mix h f xh fg h f x h g h f s mix s f xs fg s f x s g s f u mix u f x u g u f v mix v f x v g v f x u mix u f h mix h f s mix s f v mix v f ug uf hg hf sg sf vg vf Example • The Purdue Physics department did a demonstration of collapsing a barrel by boiling a small amount of water for several minutes, then sealing the barrel and letting it cool (see images below). Room temperature is 25oC. What is the maximum pressure difference (between the outside and inside of the barrel) possible that could be realized if the barrel were strong enough NOT to collapse? • • Assume saturated vapor at atmospheric conditions. Table A-4, for p=101.42 kPa, T=100oC, yields v1 = vg =1.6720 m3/kg At state 2: v2 = v1 =1.6720 m3/kg – The MAXIMUM pressure difference possible would be for the water vapor to be cooled to room temperature. At room temperature (25oC), vg = 45.3 m3/kg and vf = 0.001003 m3/kg. – Therefore, the water is a SLVM at 1.6720 m3/kg and 25oC. The SLVM is at the saturation pressure at 25oC, which is 3.1698 kPa. – For this process, ΔP = 101.42 – 3.1698 kPa. • Thus, the maximum pressure difference is 98.25 kPa. IDEAL GASES • Equations of State • Enthalpy and Internal Energy Changes Ideal Gas Equation of State P pressure [Pa] V volume [m3] m mass [kg] R gas constant [J/kg-K] T temperature [K] r density [kg/m3] v specific volume [m3/kg] n number of moles [mol] Ru universal gas constant [J/kmol-K] MWgas molecular weight [kg/kmol] Ru = 8.314 kJ/kmol-K = 8314 J/kmol-K = 1.986 cal/mol-K PV mRT Pv RT P ρRT PV nR u T Ru R MWgas Ideal Gas Relationships Specific Heats: constant volume specific heat [kJ/kg-K] cp – cv = R gas constant [kJ/kg-K] constant pressure specific heat [kJ/kg-K] Internal Energy: u = cvT [kJ/kg] Enthalpy: (Assuming constant specific heat) h = cpT [kJ/kg] Entropy: T2 p 2 s c p R ln T1 p1 (Assuming constant specific heat) T2 v 2 s c v R ln T1 v1 Ideal Gas Relationships Constant Temperature: p1v1 p2 v2 Constant Pressure: T1 v1 T2 v 2 Constant Volume: T1 p1 T2 p 2 Isentropic Ideal Gas Relationships Isentropic with constant specific heats: p1v1k p 2 v k2 k = specific heat ratio k T1 v 2 T2 v1 k 1 T1 p 2 T2 p1 1k k cp cv Non-ideal Gases • Gases behave ideally at low pressures and high temperatures • What is meant by low pressure and high temperature? – Compressibility factor quantifies the deviation of a pure substance from ideal gas behavior at a given temperature and pressure – Z = 1 (ideal gas) – For a real gas, Z > 1 or Z < 1 – For Z very close to unity, we can assume ideal gas behavior, most of the time Compressibility Factor • Reduced pressure and temperature: – PR = reduced pressure; TR = reduced temperature – Pcr = critical pressure; Tcr = critical temperature • Principle of Corresponding States: The compressibility factor is approximately the same at the same PR and TR for all fluids – Therefore, a generalized compressibility chart can be used to find the compressibility factor of any fluid Generalized Compressibility Chart Sonntag, Borgnakke, and Van Wylen, Fundamentals of Thermodynamics, 5th Edition, page 763. Property Evaluation • STEP 1: ALWAYS start in the saturated or “two-phase liquidvapor” tables • STEP 2: Identify two, independent, intensive properties – Pick one property and find its location; compare the second using the following rules: – If the properties are pressure and temperature, and • Tsat < T or Psat > P, then “superheated vapor” • Tsat > T or Psat < P, then “compressed liquid” or “sub-cooled” • Tsat = T or Psat = P, then “liquid-vapor” and need another property – If one of the properties is not pressure or temperature • Let b be your property • b > bg, then “superheated vapor” • b < bf , then “compressed liquid” or “sub-cooled” • bf< b < bg, then “liquid-vapor” Property Evaluation • STEP 3: Evaluate the remaining properties – If “superheated vapor,” then go to superheated tables – If “compressed liquid” or “sub-cooled,” then go to compressed liquid tables • If data does not exist, assume the following: v = vf h = hf u = uf s = sf where the saturated liquid property is evaluated at the given temperature since pressure does not impact liquids that much – If “liquid-vapor,” then continue using the “two-phase liquid-vapor” tables and find quality Saturated Mixture Table Superheated Steam Superheated Steam [cont’d] Mollier Diagram (P vs h) Example 1 Given: Steam at 2.0 kPa is saturated at 17.5 oC. In what state will the steam be at 40 oC if the pressure is 2.0 kPa? Analysis: @ P = 2.0 kPa = 0.02 bar, and T = 40oC Psat > P “superheated vapor” Example 2 Given: What is the change in internal energy of air (assumed to be an ideal gas) cooled from 550 oC to 100 oC (assume constant cv= 0.718 kJ/kg-K) ? Analysis: “ideal gas” u = cvT = cv (T2-T1) given cv= 0.718 kJ/kg-K u = 0.718 kJ/kg-K (550-100)K u = 323.1 kJ/kg Example 3 Given: A boy on the beach holds a spherical balloon filled with air. At 10:00 a.m., the temperature on the beach is 20 oC and the balloon has a diameter of 30 cm. Two hours later, the balloon diameter is 30.5 cm. Assuming the air is an ideal gas and that no air was lost or added, what is the temperature on the beach at noon? Analysis: “ideal gas” PV mRT constant mass, constant pressure, and same gas, V1 V2 T1V2 T1D32 T2 3 T1 T2 V1 D1 T2 20 273 K(30.5 cm)3 (30 cm)3 308 K 35 o C Example 4 Given: Steam initially at 1 MPa and 200 oC expands in a turbine to 40 oC and 83% quality. What is the change in entropy? Analysis: @ P1=1 MPa = 10 bar, and T1=200 oC >> Psat > P “superheated vapor” @ T=40 oC and x2 = 0.83 “two-phase liquid-vapor” Example 4 - Cont’d Given: Steam initially at 1 MPa and 200 oC expands in a turbine to 40 oC and 83% quality. What is the change in entropy? Analysis: s1 = 6.6940 kJ/kg-K (from superheated table for 1 MPa, 200 oC) @ 40 oC, s2,f = 0.5725 kJ/kg-K; s2,g = 8.257 kJ/kg-K s2 = s2,f +x2(s2,g-s2,f) = 0.5725+0.83*(8.257- 0.5725) s2 = 6.9506 kJ/kg-K s2 - s1 = (6.9506 – 6.694) kJ/kg-K s2 - s1 = 0.2566 kJ/kg-K Example 5 Given: A 3 kg mixture of water and water vapor at 70 oC is held at constant pressure while heat is added. The enthalpy of the water increases by 50 kJ/kg. What is the change in entropy? Heat flow = enthalpy change since Analysis: s Q rev Q 50 kJ/kg s T To (70 273)K s = 0.1458 kJ/kg-K integrate Q H Example 6 Given: Through an isentropic process, a piston compresses 2 kg of an ideal gas at 150 kPa and 35 oC in a cylinder to a pressure of 300 kPa. The specific heat of the gas for constant pressure processes is 5 kJ/kg-K; for constant volume processes, the specific heat is 3 kJ/kg-K. What is the final temperature of the gas? Analysis: “ideal gas and constant entropy” cp = 5 kJ/kg-K; cv = 3 kJ/kg-K T2 T1 p 2 T2 p1 k cp cv 1k k T2 5 1.667 3 35 273 K 406.44 K 133.44 o C 300 150 1-1 .6 6 7 1 .6 6 7 T1 p2 p1 1-k k Example 7 Given: 3 kg of steam with a quality of 30% has a pressure of 12.056 bar. At that pressure, the specific volume of a saturated fluid is vf = 1.5289 cm3/g. The specific volume of the vapor is vg = 14.1889 cm3/g. What is the specific volume of the steam? Analysis: vf = 1.5289 cm3/g vg = 14.1889 cm3/ vmix = vf +x(vg-vf) = 1.5289 +0.30*(14.1889 - 1.5289) cm3/g vmix = 5.3269 cm3/g Second Law • THE SECOND LAW OF THERMODYNAMICS • ENTROPY Thermo in a Nutshell 1. There is a game. (0th law, thermal equilibrium) 2. The best you can do is tie. (1st Law, energy balance) 3. You can never tie. (2nd Law, entropy increase principal) 4. You can’t quit. (3rd Law, absolute zero) 2nd Law of Thermodynamics Generally, there are two ways to state the second law of thermodynamics: (1) Kelvin-Planck: It is impossible to build a cyclic engine that will have a thermal efficiency of 100% * All work can be converted to heat, but all heat CANNOT be converted into work * Maximum possible thermal efficiency for a heat engines is the Carnot cycle efficiency (2) Clausius Statement: It is impossible to devise a cycle such that its only effect is the transfer of heat from a low-temperature body to a high-temperature body * An input of work is always required for refrigeration cycles Entropy Changes Net entropy must ALWAYS increase: • The equality applies for reversible processes T2 dQ S T T1 • The inequality applies for irreversible processes, in which entropy production contributes to entropy change For a constant temperature, reversible process, entropy change is given by: Q S S2 S1 To Isentropic process, entropy change is given by: S S2 S1 0 Adiabatic process, entropy change is given by: S S2 S1 0 discharging energy to reservoir temperature of the reservoir Entropy Changes Evaluating the property of entropy: • For incompressible solids and liquids, constant specific heats, T2 s 2 s1 cln T1 • For ideal gases with constant specific heats, T2 v2 s 2 s1 c v ln Rln v1 T1 T2 p2 s 2 s1 c p ln Rln p1 T1 Problem 4 Given: 1 kg of steam is initially at 400 oC and 800 kPa. The steam expands adiabatically to 200 oC and 400 kPa in a closed process, performing 450 kJ of work, given the following information. Which law does this process violate: (zeroth law, first law, second law, first and second law)? at 400 oC and 800 kPa u = 2959.7 kJ/kg h = 3267.1 kJ/kg s = 7.5716 kJ/kg-K at 200 oC and 400 kPa u = 2646.8 kJ/kg h = 2860.5 kJ/kg s = 7.1706 kJ/kg Analysis: (1) Zeroth law not applicable, does not deal with thermal equilibrium (2) Check first law negligible E U KE PE Q W Problem 4 Cont’d at 400 oC and 800 kPa u = 2959.7 kJ/kg h = 3267.1 kJ/kg s = 7.5716 kJ/kg-K at 200 oC and 400 kPa u = 2646.8 kJ/kg h = 2860.5 kJ/kg s = 7.1706 kJ/kg U Q W Q W Q W u u m m m m W > 0 done BY system Q 2646.8 kJ kg 2959.7 kJ kg 450 kJ kg m Q 137.1 kJ kg m 0 NOT adiabatic Q = 0, as stated Therefore, violates 1st Law Problem 4 Cont’d at 400 oC and 800 kPa s = 7.5716 kJ/kg-K at 200 oC and 400 kPa s = 7.1706 kJ/kg (3) Check 2nd Law s 0 for adiabatic process s 7.1706 kJ kg-K 7.5716 kJ kg-K s -0.401 kJ kg-K < 0 violates 2nd Law Violates both 1st and 2nd Laws CYCLES • Rankine Cycle (Steam) • Vapor Compression Cycle (Refrigeration) • Otto Cycle (Gasoline Engine) Basic Cycles Power Generation (Heat Engine) Refrigeration, AirConditioning, Heat Pump Basic Cycles: Definitions Cycle: Series of processes that eventually brings the system back to its original state Power Cycle: • A cycle that uses heat energy to do work on the surroundings • Performance is measured by thermal efficiency, which is the ratio of useful work output to the energy input required to run the cycle: work done BY system Wnet Wout Win th Q in Q in Q net Q in Q out Q in Q in work done ON system heat OUT OF system heat INTO system Carnot Cycle Carnot Cycle: (1) Theoretical implementation of a cycle without any irreversibilities (2) Work output is the maximum possible for any heat engine due to reversible processes A B D C Processes A to B: isothermal expansion of saturated liquid to saturated vapor B to C: isentropic expansion of vapor (Q = 0; s = 0) C to D: isothermal compression of vapor D to A: isentropic compression of vapor (Q = 0; s = 0) Carnot: P-v and T-s Diagrams A B B A D C D Carnot Efficiency: th,Carnot TH TL TL 1 TH TH C cold reservoir temperature hot reservoir temperature Problem 1 Given: A Carnot engine receives 100 kJ of heat from a hot reservoir at 370 oC and rejects 37 kJ of heat. Determine the temperature of the cold reservoir. Analysis: th Q net Q H Q L 100 kJ 37 kJ Q in QH 100 kJ th 0.63 or 63% TH QH W Use absolute temperatures! TH TL T 1 L TH TH TL TH TH th,Carnot th,Carnot TL (370 273) 370 273 0.63 TL = 237.91 K = -35.09 oC QL TL Problem 2 Given: What is the maximum thermal efficiency possible for a power cycle operating between 600 oC and 110 oC? Analysis: Carnot yields maximum possible efficiency TL TH 110 273K 1 600 273K th,Carnot 1 th,Carnot th,Carnot 0.561 or 56.1% Problem 3 Given: A heat pump takes heat from groundwater at 7 oC and maintains a room at 21oC. What is the maximum coefficient of performance possible for this heat pump? upper limit of heat pump is set by a Carnot heat pump cycle Analysis: COPheat pump TH QH W QL TL QH QH TH W Q H Q L TH TL 21 273 K COPheat pump 21 273K 7 273K COPheat pump 21 How does a refrigerator COP differ? Rankine Cycle Rankine Cycle: • Vapor-power cycle commonly used in power plants with water as the working fluid • Efficiency is ratio of useful output to required input: Wnet h1 h 2 h 4 h 3 th h1 h 4 Q in Processes 1 to 2: Isentropic expansion of the working fluid through the turbine from saturated vapor at state 1 to the condenser pressure (Q = 0; s = 0) 2 to 3: Heat transfer from the working fluid as it flows at constant pressure through the condenser with saturated liquid at state 3 3 to 4: Isentropic compression in the pump to state 4 in the compressed liquid region. (Q = 0; s = 0) 4 to 1: Heat transfer to the working fluid as it flows at constant pressure through the boiler to complete the cycle Rankine-Superheat Superheat: Used to increase the temperature of the working fluid entering the low-pressure turbine to reduce the wear on turbine blades (point 3) Combustion Power Cycles • Differ from vapor cycles since they cannot return to their initial conditions • Due to significant complexities with computing mixtures of fuel and air, combustion power cycles are often analyzed as air-standard cycles – Air-Standard Otto Cycle: Hypothetical closed system using air as the working fluid to simplify the chemistry due to combustion Air-Standard Otto Cycle Air-Standard Otto Cycle: Hypothetical closed system using air as the working fluid to simplify the chemistry due to combustion Processes 1 to 2: 2 to 3: 3 to 4: 4 to 1: Isentropic compression of the working fluid (Q = 0; s = 0) Constant volume heat addition Isentropic expansion of the working fluid (Q = 0; s = 0) Constant volume heat rejection Air-Standard Otto Cycle: Performance Compression Ratio: V1 V4 Vb rv V2 V3 Va compression from A to B expansion from C to D Ideal Thermal Efficiency: specific heat ratio th 1 rv1k compression ratio Refrigeration Cycles • In refrigeration cycles, heat is transferred from a low-temperature area (i.e. inside the refrigerator) to a high-temperature area (e.g., in the kitchen) • Since heat spontaneously flows only from high to low temperature areas, work is required to force heat transfer • Opposite of heat engines • Systems: • Refrigerator: Heat is removed from air inside • Air conditioner: Heat is removed from air in an occupied space • Heat Pump: Heat is supplied to air in an occupied space • Chiller: Heat is removed from water Refrigeration Cycles: Performance • In refrigeration cycles, the coefficient of performance (COP) is used in place of thermal efficiency to measure performance • The COP is always the ratio of useful energy transfer to the work input COPrefrigerat or,AC COPheat pump QL W QH W heat transfer from lowtemperature reservoir work input heat transfer from hightemperature reservoir MISCELLANEOUS • Mixture of Gases • Heat Transfer – Conduction – Convection – Radiation Modes of Heat Transfer • All heat transfer driven by temperature difference • Heat transfer occurs via 3 modes – conduction (solid/fluid; molecular collision) • depends on temperature gradient – convection (fluid; motion) • depends on temperature difference – radiation (no medium needed – electromagnetic waves) • depends on T4 Conductive Heat Transfer • Conduction heat transfer – Transfer of energy by putting a hotter object in contact with a cooler system) Convective Heat Transfer • Convection heat transfer – transfer of energy between a solid surface and a moving fluid Newton’s law of cooling, h = heat transfer coefficient. Convective Heat Transfer • The “wind chill” factor is an example • The air flow makes a given temperature “feel” (higher heat loss) like a lower temperature Radiant Heat Transfer • Radiation heat transfer – transfer of energy due to emission and absorption of electromagnetic radiation Comfort in a Room • At the same air temperature in a room in a house it “feels” cooler in the winter than the summer – Why? Thermal Circuit Model • A model used often to calculate the heat transfer through a 1-D system is called the thermal circuit model • In this model, each layer is replaced by an equivalent resistor called the thermal resistance – For conduction, – For convection, Mass and Mole Fraction Mass Fraction: ratio of component’s mass to the total mass of the mixture mi wi m tot mass fraction of ith species m mass of ith species m i total mass of mixture w i 1 Mole Fraction: ratio of component’s moles to the total moles of the mixture liquid-phase gas-phase moles of ith species Ni yi x i N tot mole fraction of ith species x i 1 y total moles of mixture i 1 Mass and Mole Fraction [cont’d] total mass of mixture Molecular Weight of Mixture: MWmix mass fraction of ith species m tot N tot x MW i total moles of mixture Mass to Mole Fraction Conversion: xi wi MWi wi MWi Mole to Mass Fraction Conversion: wi x i MWi x i MWi i molecular weight of ith species mole fraction of ith species Non-Reacting Ideal Gases mass of ith species Partial Pressure: mi R i T pi xip V partial pressure of ith species total pressure of mixture mole fraction of ith species Partial Volume: partial volume of ith species Vi mi R i T p Amagat’s Law: total volume is the sum of partial volumes V Mole Fraction: xi p i Vi p V V i mass fraction u w u internal energy i i h enthalpy w h i i assume adiabatic and reversible s entropy w s i i The End Appendices A: SI Prefixes B: Constants C: Unit Conversions Appendix A: SI Prefixes Appendix B: Constants Appendix B: Constants [cont’d] Appendix C: Unit Conversions Appendix C: Unit Conversions [cont’d] Appendix C: Unit Conversions [cont’d] Extra Examples Liquid/Vapor Tables & Charts Solution Solution (continued) Liquid/Vapor Tables & Charts Solution Solution (continued) Enthalpy Solution Solution (continued) Solution (continued) Closed Systems Solution Steady Flow/Steady State Solution Reversible Processes Solution Solution (continued) Reversible Processes Solution Three-Process Cycles Solution Solution (continued) Otto Cycle Solution Solution (continued) Solution (continued) Reverse Cycles Solution Solution (continued) Non-Condensable Gas Mixtures Solution