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# Applied thermodynamics

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Applied thermodynamics

Applied thermodynamics

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• 2. SI 2 watt......................................2 wrev steady flow work .......10 ε emissivity ...................... 18universal gas constant......... 4 Wb boundary work ........5, 14 x quality..........................3, 4 ηth thermal efficiency ......... 7V average flow velocity..5, 6 wire ∆E energy ν specific volume ........... 3, 4volume resistive ........................17 gain/loss .....................5, 6 θ methalpy.......................... 6 sphere ...........................19 wnet net work.....................10 ∆s change in entropy per unit ρCp heat capacity ............. 15volume to mass relationship 4 Wnet,in net work input ..........7 mass...................................9W minimum power Wnet,out net work ................10 ∆S entropy change..............9 requirement....................... 7 work ..............................2, 14 α thermal diffusivity ........15 BASIC THERMODYNAMICS 0TH LAW OF THERMODYNAMICS UNITS Two bodies which are each in thermal equilibrium with Energy, work, heat transfer: [J] a third body are in thermal equilibrium with each other. 2 J (joule) = N ·m = V ·C = W ·s = AV ·s = F ·V 2 = C · F 1ST LAW OF THERMODYNAMICS 1 kJ = 0.94782 Btu 1 Btu = 1.055056 kJ The Conservation of Energy Principle Rate of energy, work or heat transfer: [J/s or W] The amount of energy gained by a system is equal to 2 W (watt) = J = N ·m = C ·V = V · A = F ·V = 1 HP the amount of energy lost by the surroundings. ss s s 746 2 2 Pressure: [Pa or N/m or kg/ms ] 2ND LAW OF THERMODYNAMICS Pa (pascal) = N = kg = J = W ·s = 1.45038×10-4 psi Processes occur in a certain direction and energy has m 2 m·s 2 m3 m3 quality as well as quantity. For example, heat flows Density: [kg/m3] from a high temperature place to a low temperature place, not the reverse. Another example, electricity Force: [N or kg·m/s2] flowing through a resistive wire generates heat, but N (newton) = kg ·m = J = C ·V = W ·s heating a resistive wire does not generate electricity. s2 m m m Kelvin-Planck statement: It is impossible for any device Temperature: [°C or K] 0°C = 273.15K that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. Volume: [m3] = 1000 liters Clausius statement: It is impossible to construct a device Note: In this class, we typically use units of KJ, KPa, and that operates in a cycle and produces no effect other than KW. the transfer of heat from a lower temperature body to a higher temperature body. SI UNITS, International System of Units STATE POSTULATE Length meter Temperature: kelvin The state of a simple compressible system is Mass: kilogram Amount: mole completely specified by two independent, intensive Time: second Light intensity: candela. properties. Two properties are independent if one Electric current: ampere property can be varied while the other one is held constant. Properties are intensive if they do not ENERGY [J] depend on size, e.g. the properties of temperature, pressure, entropy, density, specific volume. 1 2 Kinetic energy KE = mv 2 Potential energy PE = mgz Total energy of the system E = U + KE + PE U = internal energy, i.e. sensible energy (translational, rotational, vibrational), latent energy (atomic structure, melting ice), chemical energy (bonding, separating water into hydrogen & oxygen), nuclear. Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 2 of 2
• 3. THERMODYNAMIC PROPERTIES PROPERTIES OF WATERThermodynamic properties are related to the energy of the Compressed liquid: Properties for compressedsystem, i.e. temperature, pressure, mass, volume. liquid are insensitive to pressure. For a givenExtensive properties depend on the size or extent of the temperature use the f subscripted values from tablessystem, e.g. volume, mass, total energy. A-4 and A-5, e.g. v ≈ v f , u ≈ u f , etc. However, in theIntensive properties are independent of size, e.g.temperature, pressure, entropy, density, specific volume. case of enthalpy, h ≈ h f + ( P − Psat ) v . Saturated phases: Properties for the saturated SPECIFIC PROPERTIES phases of water are determined using tables A-4 and A-5 in the back of the book and the formulas below.Extensive properties per unit mass are called specific Note that the fg subscript stands for the differenceproperties. between the g subscripted quantity and the f V subscripted quantity, e.g. ug-uf = ufg, and is providedSpecific volume v = [m3/kg]* for convenience. m Specific volume v = (1 − x ) v f + xvg [m3/kg] ESpecific energy e = [kJ/kg] m Internal energy u = (1 − x ) u f + xu g = u f + xu fg [kJ/kg]Specific internal energy u = U [kJ/kg] Enthalpy h = (1 − x ) h f + xhg = h f + xh fg [kJ/kg] m*We have to be careful with the units for specific volume. By Entropy s = (1 − x ) s f + xsg = s f + xs fg [kJ/(kg·K)] convention, we deal in units of kJ, kW, and kPa for many values. When specific volume or volume is included in an v − vf Quality x= [no units] equation, there is often a factor of 1000 involved. vg − v f R GAS CONSTANT [kJ/(kg·K)] Superheated vapor: Properties for superheated vapor are read directly from table A-6 in the back of the book. R = C p − CvCp = specific heat at constant pressure [kJ/(kg·°C)]Cv = specific heat at constant volume [kJ/(kg·°C)] PHASES OF WATER The different states in which water exists are its R Gas Constant of Selected Materials @300K [kJ/(kg·°C)] phases. We are only concerned with the liquid andAir 0.2870 Carbon monoxide 0.2968 Methane 0.5182 vapor states.Argon 0.2081 Chlorine 0.1173 Neon 0.4119Butane 0.1433 Helium 2.0769 Nitrogen 0.2968 compressed liquid – purely liquid, at less than saturationCarbon dioxide 0.1889 Hydrogen 4.1240 Oxygen 0.2598 temperature (boiling point at pressure), v < vf saturated liquid – purely liquid, but at the saturation temperature (any additional heat will cause some vaporization), v = vf saturated liquid/vapor mixture – a mixture of liquid and vapor at the temperature (and pressure) of saturation, v f < v < vg saturated vapor – purely vapor, but at the saturation temperature (any loss of heat will cause some condensation to occur), v = vg superheated vapor – purely vapor, above the saturation temperature, v > vg ENTHALPY OF VAPORIZATION [hfg] The amount of energy needed to vaporize a unit mass of saturated liquid at a given temperature or pressure. hfg = enthalpy of vaporization [kJ/kg] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 3 of 3
• 4. x QUALITY CONSTANTSThe quality is a value from 0 to 1 describing the ratio Atmospheric pressure: 101.33 kPaof vapor mass to total mass of a pure substance. It isonly applicable at saturation temperature. A quality of Boltzmann constant: 1.380658×10-23 kJ/(kmol·K)0 denotes a saturated liquid and a quality of 1 denotes Critical point, water 22 Mpa, 374°Csaturated vapor. Gas constant R = Ru/M where M is molecular weight (R = 287 J/(kg·K) for air) mg x= Temperature in Kelvin: °C + 273.15 mg + m f Universal gas constant: Ru = 8.314 kJ/(kmol·K)mg = mass of the gas [kg]mf = mass of the fluid (liquid) [kg] ENERGY TRANSFER [kJ] Whether energy transfer is described as heat or work VOLUME TO MASS RELATIONSHIP can be a function of the location of the system boundary. The system boundary may be drawn toFor the saturated state. suite the problem. The area enclosed is also referred to as the control volume. V = V f + Vg = m f v f + mg vg System SystemV = total volume [m3]Vf = volume of the fluid (liquid) [m3]Vg = volume of the gas [m3]mf = mass of the fluid (liquid) [kg] heat transfer workvf = volume density of the fluid (liquid) [m3/kg]mg = mass of the gas [kg]vg = volume density of the gas [m3/kg] I I IDEAL GAS EQUATIONThe ideal gas formula assumes no intermolecularforces are involved. The ideal gas formula may beused as an approximation for the properties of gaseswhich are a high temperatures/low pressures well outof range of their saturation (liquification) values, e.g.air at room temperature and atmospheric pressurecan be considered an ideal gas.. Dont use thisformula for steam, especially near saturation; use thewater property tables. Pν = RT or PV = mRT PV1 PV2In a closed system, m and R are constant, so 1 = 2 T1 T2P = pressure [kPa]ν = V/m specific volume [m3/kg]V = volume [m3]m = mass [kg]R = gas constant (0.287 for air) [kJ/(kg·K)]T = absolute temperature [K] (°C + 273.15) Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 4 of 4
• 5. 1st LAW OF THERMODYNAMICS [kJ] 1st LAW, POWER VERSION [kW]Used in piston-cylinder problems. for open systems System + W Differentiation of the 1st Law of Thermodynamics with respect to time yields the power version. Used for ∆E mixture chamber, heat exchanger, heater in a duct problems. Q Q − W = m ( ∆h + ∆ke + ∆pe ) & & & +Open Systems: Q − W = ∆E = ∆H + ∆KE + ∆PE & & &⎡ V 2 − V1 2 g ( z2 − z1 ) ⎤ Q − W = m ⎢ h2 − h1 + 2 + ⎥ where ∆H = m ( h2 − h1 ) = mC p , avg (T2 − T1 ) ⎣ 2 (1000 ) 1000 ⎦Closed Systems: Q − W = ∆E = ∆U + ∆KE + ∆PE where h2 − h1 = C p , avg (T2 − T1 ) where ∆U = m ( u2 − u1 ) = mCv , avg (T2 − T1 ) & Q = net heat transfer per unit time across system (Closed System means that mass does not enter or leave boundaries, positive when flowing inward [kW or kJ/s] the system.) W& = net work done per unit time in all forms, positive when m ( V2 − V1 2 2 ), flowing outward [kW or kJ/s] ∆KE = ∆PE = mg ( z2 − z1 ) & m = mass flow rate through the control volume [kg/s] 2000 Note that to obtain this value, typically the ideal gasNOTE: Since the piston-cylinder is a closed system, we equation (p4) and the mass flow rate (p6) formulas will normally use the Closed System version of the law. An be used. exception occurs when the piston is allowed to move as ∆h = net change in enthalpy [kJ/kg] the gas expands under constant pressure. In this case, ∆ke = net change in the kinetic energy per unit mass [kJ/kg] there is boundary work Wb, which can be included on the ∆pe = net change in the potential energy per unit mass right-hand side of the equation by using the Open [kJ/kg] Systems version since ∆U + Wb = ∆H. V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s]Q = net heat transfer across system boundaries, positive g = acceleration of gravity 9.807 m/s2 when flowing inward [kJ] z = elevation to some reference point [m]W = net work done in all forms, positive when flowing Cp,avg = specific heat at constant pressure, averaged for the outward [kJ] two temperatures [kJ/(kg·°C)]∆E = net change in the total energy of the system [kJ] T1, T2 = temperature of the fluid at the inlet and outlet∆U = net change in the internal energy of the system [kJ] respectively [°C or K]∆KE = net change in the kinetic energy of the system [kJ]∆PE = net change in the potential energy of the system [kJ]m = mass [kg]u = internal energy [kJ/kg]h = enthalpy [kJ/kg]Cp,avg = specific heat at constant pressure, averaged for the two temperatures [kJ/(kg·°C)]Cv,avg = specific heat at constant volume, averaged for the two temperatures [kJ/(kg·°C)] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 5 of 5
• 6. 1st LAW, UNIT-MASS VERSION [kJ/kg] h SPECIFIC ENTHALPY [kJ/kg] stThe division of the power version of the 1 Law of The per unit mass version of enthalpy (see previous)Thermodynamics equation by the flow rate yields the and often referred to as simply enthalpy, the exactunit-mass version. Used in nozzle, diffuser, turbine, meaning to be determined from context.and compressor problems. ∆h = ∆u + ν∆P incompressible substanceOpen Systems: q − w = ∆h + ∆ke + ∆pe ∆h = ∫ C p ( T ) dT 2 g ( z2 − z1 ) C p , avg ∆T ideal gas V − V1 2 2 q − w = h2 − h1 + 2 + 1 2 (1000 ) 1000 h = u + RT ideal gaswhere h2 − h1 = C p , avg (T2 − T1 ) u = internal energy [kJ/kg] ν = V/m specific volume [m3/kg]Closed Systems: q − w = ∆u + ∆ke + ∆pe P = pressure [kPa] Cp,avg = specific heat at constant pressure, averaged for the V2 2 − V1 2 g ( z2 − z1 ) two temperatures [kJ/(kg·°C)] q − w = u2 − u1 + + 2 (1000 ) 1000 T = absolute temperature [K] (°C + 273.15) R = gas constant (287 for air) [J/(kg·K)]where u2 − u1 = Cv , avg (T2 − T1 )q = heat transfer per unit mass [kJ/kg] θ METHALPY [kJ/kg]w = work done per unit mass [kJ/kg] Methalphy means "beyond enthalpy". The factor ofsee also BERNOULI EQUATION next. 1000 is used to convert m2/s2 to kJ/kg. V2 θ = h + ke + pe = h + + gz BERNOULI EQUATION 2 ×1000For the steady flow of liquid through a device thatinvolves no work interactions (such as a nozzle orpipe section), the work term is zero and we have the & m MASS FLOW RATE [kg/s]expression known as the Bernouli equation. The rate of flow in terms of mass. V2 − V1 2 g ( z2 − z1 ) 2 1 0 = υ ( P2 − P ) + + m= & V1 A1 2 (1000 ) ν 1 1000ν = V/m specific volume [m3/kg] ν = V/m specific volume [m3/kg]P = pressure [kPa] V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s]V = average flow velocity (note 1 kJ/kg = 1000 m2/s2) [m/s] A = cross-sectional area [m2]g = acceleration of gravity 9.807 m/s2z = elevation to some reference point [m] H ENTHALPY [kJ]The sum of the internal energy and the volume-pressure product. If a body is heated withoutchanging its volume or pressure, then the change inenthalpy will equal the heat transfer. We see thismore often in its per unit mass form (see next) calledspecific enthalpy but still referred to as enthalpy. H = U + PVU = internal energy [kJ]P = pressure [kPa]V = volume [m3] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 6 of 6
• 7. Cp, Cv SPECIFIC HEAT [kJ/(kg·°C)] COP COEFFICIENT OF PERFORMANCEDescribes the energy storage capability of a material. A unitless value describing the efficiency of aThe energy required to raise the temperature of a unit refrigerator, of a heat pump.of mass of a substance by one degree under constant & & QL QHpressure (Cp), or under constant volume (Cv). This COPRefrig. = COPH.P. =can be confusing since Cp can be used in problems & Wnet,in & Wnet,ininvolving changing pressure and Cv can be used inproblems involving changing volume. Note that Cp is QL QHused in calculations involving open systems, and Cv COPRefrig. = COPH.P. =is used for closed systems. Cp > Cv because at Q H − QL Q H − QLconstant pressure, the system is allowed to expand Maximum possible COP for a refrigerator, for a heat pump:when heated, requiring additional energy. The valuesfor specific heat increase slightly with increased 1 1temperature. COPRefrig. = COPH.P. = TH / TL − 1 1 − TL / TH C p = Cv + R & QL = heat transfer [kW] u2 − u1 = Cv (T2 − T1 ) h2 − h1 = C p (T2 − T1 ) & & & Wnet,in = QH − QL = net work input [kW]R = gas constant (287 for air) [J/(kg·K)] QH = heat transfer from a high temperature source [kJ]u = internal energy [kJ/kg] QL = heat transfer from a low temperature source [kJ]h = enthalpy [kJ/kg] TH = temperature of high-temperature source [K]Example: The Cp of water at room temperature is 4.18 TL = temperature of low-temperature source [K]kJ/(kg·°C), for iron its 0.45 kJ/(kg·°C). Therefore it takesabout nine times as much energy to heat water as it does toheat iron. EER ENERGY EFFICIENCY RATING specific heat × mass × ∆temp = energy An efficiency rating system used in the United States. The amount of heat removed in Btu’s for 1 Wh of Cp Specific Heat of Selected Materials @300K [kJ/(kg·°C)] electricity consumed. Since 1 Wh = 3.412 Btu, thisAir 1.005 Concrete 0.653 Iron 0.45 works out to:AluminumBrass 0.902 0.400 Copper Glass 0.386 0.800 Steel Wood, hard 0.500 1.26 EER = 3.412 COPRefrig. Wh = watt-hour, a unit of electrical energy Btu = British thermal unit, a unit of thermal energy k SPECIFIC HEAT RATIO [no units] COPRefrig. = coefficient of performance for the refrigerationAn ideal gas property that varies slightly with cycle, an efficiency rating [no units]temperature. For monatomic gases, the value isessentially constant at 1.667; most diatomic gases,including air have a specific heat ratio of about 1.4 at & W MINIMUM POWER REQUIREMENTroom temperature. [kW]Cp = specific heat at constant pressure C The amount of power required to operate a heat [kJ/(kg·°C)] k= p pump/refrigerator.Cv = specific heat at constant volume Cv [kJ/(kg·°C)] & Q & W= L COP ηth THERMAL EFFICIENCY & QL = heat transfer [kW]The efficiency of a heat engine. The fraction of the & & & W = QH − QL = net work input [kW]heat input that is converted to net work output. Wnet,out Qout wnet ηth = = 1− = = 1 − r1− k Qin Qin qHWnet,out = QH - QL = net work output [kW]Qin = heat input [kJ]Qout = heat output [kJ] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 7 of 7
• 8. HEAT TRANSFER LIMIT [kW] Sgen ENTROPY GENERATION [kJ/K]This expression is an equality for a reversible cycle The entropy change of a closed system during anengine (a theoretical device not realizable in practice). irreversible process is always greater than the entropy QH TH transfer. The entropy generation is the entropy ≤ created within the system boundaries due to QL TL irreversibilities. Note that it may be necessary to extend the boundaries of a system in order toQH = magnitude of the heat transferred from a high- consider it a closed system. temperature source [kJ]QL = magnitude of the heat transferred to a low- 2 δQ temperature source [kJ] S2 − S1 = ∫ + Sgen 1 3 { { 2 1 TTH = temperature of high-temperature source [K] Entropy Entropy Entropy change of generationTL = temperature of low-temperature source [K] the system transfer within the with heat system S1 = initial entropy [kJ/K] CLAUSIUS INEQUALITY S2 = final entropy [kJ/K]The cyclic integral of the change in heat transfer δQ = the change in heat transfer [kJ]divided by the absolute temperature is always less T = absolute temperature at the inside surface of thethan or equal to zero. system boundary [K] δQ ∫ T ≤0 ENTROPY BALANCE FOR CONTROL VOLUMES∫ = the integration is to be performed over a full cycle For a control volume, we must consider mass flowδQ = the change in heat transfer [kJ] across the control volume boundaryT = absolute temperature at the inside surface of the dSCV & Qk system boundary [K] dt = ∑ T + ∑ mi si − ∑ me se + { & 1442443 & & Sgen,CV { { k Rate of entropy Rate of entropy Rate of entropy Rate of entropy transport with generation S TOTAL ENTROPY [kJ/K] change of CV transfer with heat mass within CVThe term entropy is used both for the total entropyand the entropy per unit mass s [kJ/(kg·K)]. Entropy is SCV = entropy within the control volume [kJ/K]an intensive property of a system (does not depend & mi = inlet mass flow rate [kg/s]on size). & me = exit mass flow rate [kg/s] ⎛ δQ ⎞ dS = ⎜ ⎟ si = inlet entropy [kJ/(kg·K)] ⎝ T ⎠ se = exit entropy [kJ/(kg·K)]δQ = the change in heat transfer [kJ] & Qk = rate of heat transfer through the boundary at internalT = absolute temperature at the inside surface of the boundary temperature Tk [kJ/s] system boundary [K] Tk = absolute temperature at the inside surface of the system boundary [K] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 8 of 8
• 9. & Sgen RATE OF ENTROPY GENERATION s ENTROPY PER UNIT MASS [kJ/(kg·K)] [kW/K] Entropy change is caused by heat flow, mass flow, and irreversibilities. Irreversibilities always increaseMost steady-state processes such as turbines, entropy.compressors, nozzles, diffusers, heat exchangers,pipes, and ducts, experience no change in their s2 = s1 Isentropic processentropy. This gives the relation: T2 & s2 − s1 = Cavg ln incompressible substances Q & Sgen = ∑ me se − ∑ mi si − ∑ k & & T1 Tk T2 ν ∆s = Cv ,avg ln + R ln 2 ideal gasFor a single-stream (one inlet, one exit) steady-flow device: T1 ν1 Q& T2 P Sgen = m ( se − si ) − ∑ k & & ∆s = C p ,avg ln − R ln 2 ideal gas Tk T1 P1 ν = V/m specific volume [m3/kg]&mi = inlet mass flow rate [kg/s] R = gas constant (287 for air) [J/(kg·K)]&me = exit mass flow rate [kg/s] Cavg = the specific heat at average temperature [kJ/(kg·K)] Cp = specific heat at constant pressure [kJ/(kg·°C)]si = inlet entropy [kJ/(kg·K)] Cv = specific heat at constant volume [kJ/(kg·°C)]se = exit entropy [kJ/(kg·K)] T1, T2 = initial and final temperatures [K]&Qk = rate of heat transfer through the boundary at P1, P2 = initial and final pressure [Pa] temperature Tk [kJ/s]Tk = absolute temperature at the system boundary [K] sgen ENTROPY GENERATION PER UNIT MASS [kJ/(kg·K)] INCREASE OF ENTROPY PRINCIPLE Applies to a single-stream, steady-flow device such asThe entropy of an isolated system during a process a turbine or compressor.always increases or, in the limiting case of a qreversible process, remains constant. sgen = ( se − si ) − ∑ T ∆Sisolated ≥ 0 si = inlet entropy [kJ/(kg·K)] + se = exit entropy [kJ/(kg·K)] System W There is no entropy transfer with work. q = heat transfer per unit mass [kJ/kg] T = absolute temperature at the inside surface of the ∆E system boundary [K] Heat transfer is + Q accompanied by entropy transfer. p THERMODYNAMIC PROBABILITY Molecular randomness or uncertainty. The thermodynamic probability is the number of possible microscopic states for each state of macroscopic equilibrium of a system. It is related to the entropy (disorder) of the system by the Boltzmann relation: S = k ln p ⇒ p = eS / k S = entropy [kJ/K] k = Boltzmann constant 1.3806×10-23 [kJ/(kmol·K)] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 9 of 9
• 10. wrev STEADY-FLOW WORK [kJ/kg] CARNOT CYCLEOne needs to know ν as a function of P in order to Introduced in 1824 by French engineer Sadi Carnot,perform the integration, but when the working fluid is the Carnot cycle is a combination of four reversiblean incompressible fluid, the specific volume ν remains processes that are the basis for the theoretical Carnotconstant during the process and can be taken out of heat engine. The cycle forms a rectangle on the T-sthe integration. For the steady flow of a liquid through plot. Use Cv for specific heat.a device that involves no work (such as nozzle or a The numbered corners T 1-2 • Volume expansion • Reversible isothermal processpipe section), the work term is zero represent the four QH • Heat is added from a states. high temperature area. 2 qin wrev = − ∫ ν dP − ∆ke − ∆pe The area enclosed is TH equal to the work. 1 2 1 wrev = ν ( P − P2 ) − ∆ke − ∆pe wnet 2-3 • Volume expansion • Reversible adiabatic Temperature isentropic process 1 PV k = a constant TLν = V/m specific volume [m3/kg] 4 qout 3 4-1 3-4 • Volume compression • Reversible isothermal process • Volume compression QL • Heat sinks to a low HEAT ENGINES • Reversible adiabatic, isentropic process temperature area. PV k = a constant s1 Entropy s2 s HEAT ENGINESThe conversion of heat to work requires the use ofspecial devices; these are called heat engines and P constant shave the following characteristics: qin constant T 1 • They receive heat from a high-temperature source. 2 • They convert part of this heat to work. Pressure • They reject the remaining waste heat to a low- 4 temperature sink such as the atmosphere or a body of qout 3 water. • They operate on a cycle. Volume v • They usually involve a fluid used in the transfer of heat; TL this is called the working fluid. Thermal efficiency: ηth,Carnot = 1 − TH Wnet,out NET WORK [kJ] Heat transfer and work:The work produced by a heat engine. The net work is qin = TH ( s2 − s1 ) , qout = TL ( s2 − s1 ) [kJ/kg]equal to the area bounded by the cycle as plotted on aT-S diagram. It is also the difference between the heat wnet = qin − qout [kJ/kg]consumed by a heat engine and its waste heat, that Work occurs in all 4 processes of the Carnot cycle (work is 0is, the difference between heat taken from the high- for constant volume processes).temperature source and the heat deposited in the low-temperature sink. wnet = w12 + w23 + w34 + w41 [kJ/kg] Net work: Wnet = QH − QL [kJ] per unit mass: wnet = qin − qout [kJ/kg] per unit time: & & & Wnet = QH − QL [kW]QH = magnitude of the heat transferred from a high- temperature source [kJ]QL = magnitude of the heat transferred to a low- temperature source [kJ]qin = magnitude per kilogram of the heat transferred from a high-temperature source [kJ/kg]qout = magnitude per kilogram of the heat transferred to a low-temperature source [kJ/kg] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 10 of 10
• 11. IDEAL OTTO CYCLE IDEAL DIESEL CYCLEThe Otto cycle is the model for the spark-ignition The Diesel cycle is the model for the compression-reciprocating engine. It consists of four internally ignition reciprocating engine. It consists of fourreversible processes: 1) isentropic compression, 2) internally reversible processes: 1) isentropicheat addition, 3) isentropic expansion, and 4) heat compression, 2) heat addition, 3) isentropicrejection. expansion, and 4) heat rejection. T 2-3 • Heat addition T 2-3 • Heat addition •w =0 3 • Constant pressure TH 3 The area The area enclosed is qin enclosed is equal to the equal to the qin work. work. 2 4 Temperature 4 Temperature 2 3-4 • Isentropic expansion 3-4 • Isentropic 1-2 • Isentropic qout expansion 1-2 • Isentropic qout compression compression 4-1 • Heat 1 rejection TL 4-1 • Heat •w =0 1 rejection •w =0 s1 Entropy s3 s s1 s3 s Entropy P 3 Isentropic P qin P2 2 3 Isentropic qin Constant volume, Pressure so no work 4 Pressure 2 qout 4 Constant volume 1 qout so no work 1 ν2 Volume density ν1 v Volume density ν1 vThe otto cycle is more efficient than the diesel for equalcompression ratios, but the compression ratio is limited due Although less efficient than the otto cycle at a givento spontaneous ignition of the fuel at higher temperatures. compression ratio, higher compression ratios are possible in the diesel engine, enabling greater thermal efficiency than in 1 gasoline engines.Thermal efficiency: ηth,Otto = 1 − r k −1 1 ⎡ rck − 1 ⎤ V1 ν1 Thermal efficiency: ηth,Diesel = 1 − ⎢ ⎥Otto cycle compression ratio is: r= = r k −1 ⎣ k ( rc − 1) ⎦ V2 ν 2 r = compression ratio [no units]r = compression ratio [no units] rc = cutoff ratio [no units]k = Cp/Cv = specific heat ratio k = Cp/Cv = specific heat ratio Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 11 of 11
• 12. IDEAL BRAYTON CYCLE IDEAL JET-PROPULSION CYCLEThe Brayton cycle is the model used for modern gas- The jet-propulsion cycle is the model used for aircraftturbine engines. Although the turbine is an open gas-turbine engines. It consists of six internallysystem, it can be modeled by the Brayton cycle, which reversible processes:1) isentropic compression in ais a closed cycle. It consists of four internally diffuser, 2) isentropic compression in a compressor, 3)reversible processes:1) isentropic compression, 2) heat addition, 4) isentropic expansion in a turbine, 5)heat addition under constant pressure, 3) isentropic isentropic expansion in a nozzle, and 6) heatexpansion, and 4) heat rejection under constant rejection.pressure. 3-4 • Heat addition 4 • Constant pressure T 2-3 • Heat addition T •w =0 4-5 • Isentropic •w =0 3 expansion in turbine qin The area 2-3 • Isentropic 5 V5 ≈ 0 enclosed is equal to the qin compression work. in compressor 5-6 • Isentropic 3 expansion Temperature 4 Temperature 6 in nozzle 2 3-4 • Isentropic expansion V2 ≈ 0 1-2 • Isentropic qout 2 qout compression 4-1 • Heat 6-1 • Heat rejection 1 rejection 1-2 • Isentropic 1 • Constant pressure •w =0 compression •w = 0 by diffuser s1 Entropy s3 s s1 Entropy s6 s P qin 0 − V1 2 2 3 1 − 2 0 = C p (T2 − T1 ) + Isentropic 2000 P3 Pressure wturb,out 2 − 3 rp = wcomp,in P2 qout 4 − 5 wcomp,in = wturb,out → C p (T3 − T2 ) = C p (T4 − T5 ) 1 4 V6 2 − 0 ν2 Volume density ν4 v 5 − 6 0 = C p (T6 − T5 ) + 2000 qin = q23 = C p (T3 − T2 ) , qout = q41 = C p (T4 − T1 ) W & w Propulsive efficiency: ηp = & P = P Compressor work: wcomp,in = C p (T2 − T1 ) Qin qin Turbine work: wturb,out = C p (T3 − T4 ) m ( V6 − V1 ) V1 & ( V6 − V1 ) V1 where & WP = , wP = Net work: wnet,out = wturb,out − wcomp,in = qin − qout 1000 1000 & = mC (T − T ) , q = C (T − T ) Qin & p 4 3 1 P2 in p 4 3 Thermal efficiency: ηth,Brayton = 1 − ( k −1) / k , rp = rp P 1 1 m= & V1 A1Cp = specific heat at constant pressure (1.005 @ 300k) ν [kJ/(kg·°C)] Cp = specific heat at constant pressure (1.005 @ 300k)q = heat transfer per unit mass [kJ/kg] [kJ/(kg·°C)]w = work per unit mass [kJ/kg] rp = pressure ratio [no units]rp = pressure ratio [no units] & WP = propulsive power [kW]k = Cp/Cv = specific heat ratio & Q = heat transfer rate to the working fluid [kW] in & m = mass flow rate [kg/s] T = temperature [K] P = pressure [kPa] V = air velocity [m/s] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 12 of 12
• 13. IDEAL RANKIN CYCLE rc CUTOFF RATIOThe Rankin cycle is the model used for vapor power The ratio of cylinder volumes after and before theplants such as steam-turbine engines. It consists of combustion process. Applies to the diesel cycle.four internally reversible processes: 1) isentropic V3 ν 3compression in a pump (The vertical distance rc = =between 1 and 2 is actually greatly exaggerated on V2 ν 2the diagram below.), 2) heat addition in a boiler atconstant pressure, 3) isentropic expansion in a V = volume [m3]turbine, and 4) heat rejection in a condenser at ν = V/m specific volume [m3/kg]constant pressure. T Saturated liquid/vapor Constant pressure ISENTROPIC RELATIONS mixture lines2-3 • Heat addition 3 3-4 • Isentropic Isentropic means that the entropy does not change. in boiler expansion • Constant in turbine s2 = s1. The following relations apply to ideal gases: pressure k wturb,out ⎛ P2 ⎞ ⎛ ν1 ⎞ Temperature qin ⎜ ⎟=⎜ ⎟ → Pν k = a constant wpump,in 2 ⎝ P ⎠ ⎝ ν2 ⎠ 1 4 k −1 1 ⎛ T2 ⎞ ⎛ ν1 ⎞1-2 • Isentropic qout 4-1 • Heat rejection in condenser ⎜ ⎟=⎜ ⎟ → T ν k −1 = a constant ⎝ T1 ⎠ ⎝ ν 2 ⎠ compression in pump • Constant pressure s1 Entropy s4 s ( k −1) / k ⎛ T2 ⎞ ⎛ P2 ⎞ 1 − 2 wpump,in = ν1 ( P2 − P ) = h2 − h1 TP ( 1− k ) / k 1 ⎜ ⎟=⎜ ⎟ → = a constant where h1 = h f @ P , ν1 = ν f @ P ⎝ T1 ⎠ ⎝ P ⎠ 1 1 1 ν = V/m specific volume [m3/kg] 2 − 3 qin = h3 − h2 T1, T2 = initial and final temperatures [K] 3 − 4 wturb,out = h3 − h4 P1, P2 = initial and final pressure [Pa] Read h3 and s3 from Superheated Water Table k = Cp/Cv = specific heat ratio based on T3 and P3. So s3 = s4 and x4 = (s4-sf)/sfg. Then find h4 = hf+x4hfg. 4 − 1 qout = h4 − h1 POLYTROPIC PROCESS A process in which the compression and expansion of q w real gases have the following pressure/volume Thermal efficiency: η p = 1 − out = net qin qin relationship. where wnet = wturb,out − wpump,in = qin − qout PV n = a constant and Pν n = another constant where n is also a constantw = work per unit mass [kJ/kg] mR (T2 − T1 )q = heat transfer per unit mass [kJ/kg] Work is W= , n ≠1h = enthalpy [kJ/kg] 1− nν = V/m specific volume [m3/kg]P = pressure [kPa] r COMPRESSION RATIO Vmax ν max r= = Vmin ν minV = volume [m3]ν = V/m specific volume [m3/kg] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 13 of 13
• 14. Wb BOUNDARY WORK [kJ] HEAT TRANSFERThe work done in a moving boundary process such as HEAT TRANSFERpiston-cylinder expansion (positive W) and Energy transport due to temperature difference.compression (negative W). The boundary workdepends on the initial and final states as well as the o Conduction – diffusion in a material. In liquids and gases, conduction is due to molecular collisions. Inpath taken between them. In practice, it is often solids, it is due to 1) molecular vibration in the latticeeasier to measure work than to calculate it. and 2) energy transport by free electrons. 2 Wb = ∫ P dV o Convection – by bulk motion of the fluid 1 o Thermal radiation – by electromagnetic waves; doesnt mR (T2 − T1 ) require a mediumFor a polytropic process: W= , n ≠1 The three mechanisms for heat transfer cannot all operate 1− n simultaneously in a medium.In order to include the boundary work in a closed system Gases - are usually transparent to radiation, which canpiston-cylinder operating under constant pressure, it may be occur along with either conduction or convection but notnecessary to use the open system equation since both. In gases, radiation is usually significant compared to conduction, but negligible compared to convection. ∆H = ∆U + Wb Solids - In opaque solids, heat transfer is only byV = volume [m ] 3 conduction. In semitransparent solids, heat transfer is bym = mass [kg] conduction and radiation.n = a constant Fluids - In a still fluid, heat transfer is by conduction; in aT1, T2 = initial and final temperatures [K] flowing fluid, heat transfer is by convection. Radiation may also occur in fluids, usually with a strong absorptionP = pressure [kPa] factor.R = gas constant (0.287 kJ/(kg·K) for air) [kJ/(kg·K)] Vacuum - In a vacuum, heat transfer is by radiation only.∆U = net change in the internal energy of the system [kJ] AIR STANDARD ASSUMPTIONS HEAT TRANSFER AND WORK Heat transfer and work are interactions between aSince air is composed mostly of nitrogen, which system and its surroundings. Both are recognized asundergoes few changes in the combustion chamber, they cross the boundaries of a system. Heat andinternal combustion engines can be modeled as work are transfer phenomena, not properties. Theycontaining air only. are associated with a process, not a state. Both are1) The working fluid is air that continuously circulates in a path functions, meaning that their magnitudes closed loop and behaves as an ideal gas. depend on the path taken as well as the end states.2) All the processes are reversible.3) The combustion process is replaced by a heat addition process from an external source k THERMAL CONDUCTIVITY [W/(m·°C)]4) The exhaust process is replaced by a heat rejection A measure of the ability of a material to conduct heat. process that restores the working fluid to its initial state. k varies with temperature, but we will consider it constant in this class. The conductivity of most solids falls with increasing temperature but for most gases it MEP MEAN EFFECTIVE PRESSURE rises with increasing temperature. For example, theA fictitious pressure which, if it acted on the piston conductivity of copper at 200K is 413, and at 800K isduring the entire power stroke, would produce the 366. The conductivity of air at 200K is 0.0181, and atsame amount of work that is produced during the 800K is 0.0569. The change in conductivity becomesactual cycle. more dramatic below 200K. Thermal Conductivity of Selected Materials @300K [W/(m·°C)] Wnet wnet RT MEP = = , where ν= Air 0.0261 Copper 401 Human skin 0.37 Vmax − Vmin ν max − ν min P Aluminum 237 Diamond 2300 Iron 80.2 Brick 0.72 Fiberglass insul. 0.04 Mercury 8.9 Carbon dioxide 0.0166 Glass 1.4 Plywood 0.12 Concrete 1.4 Gypsum 0.17 Water 0.608 Concrete block 1.1 Helium 0.150 Wood (oak) 0.17 Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 14 of 14
• 15. ρCp HEAT CAPACITY [J/(m3·°C)] Rt THERMAL RESISTANCE OF AThe heat storage capability of a material on a per unit CYLINDRICAL SHAPE [°C/W]volume basis. The term Cp alone expresses heat The thermal resistance per unit length of a cylindricalcapacity per unit mass [J/(kg·°C)]. ρ is density [kg/m3]. shape, e.g. pipe or pipe insulation. r2 ln α THERMAL DIFFUSIVITY [m /s] 2 r1 Through the material: Rt cond. = [°C/W]The ratio of heat conducted to the heat stored per unit 2πlkvolume. 1 heat conducted k Across a boundary: Rt = [°C/W] ∂= heat stored = ρC p ( 2πrl ) h l = length [m]k = thermal conductivity [W/(m·°C)] k = thermal conductivity [W/(m·°C)]ρCp = heat capacity [J/(m3·°C)] h = convection heat transfer coefficient [W/(m2·°C)] r1 = inner radius [m] r2 = outer radius [m] R-value THERMAL RESISTANCEThe R-value is the thermal resistance of a material perunit area. In the United States, the R-value is Rc CONTACT RESISTANCE [°C/W]commonly expressed without units, e.g. R-19 and R- When two surfaces are pushed together, the junction30. These values are obtained by dividing the is typically imperfect due to surface irregularities.thickness of the material in feet by its thermal Numerous tiny air pockets exist at the junctionconductivity in units of Btu/(h·ft·°F). This results in resulting in an abrupt temperature change across theunits of (h·ft2·°F)/Btu. boundary. The value is determined experimentally L and there is considerable scatter of data. Rvalue = [(h·ft2·°F)/Btu] kL = thickness of the material [feet] hconv CONVECTION HEAT TRANSFERk = thermal conductivity [Btu/(h·ft·°F)] COEFFICIENT [W/(m2·°C)] The convection heat transfer coefficient is not a property of the fluid. It is an experimentally Rt THERMAL RESISTANCE [°C/W] determined parameter whose value depends on manyThe resistance of a surface (convection resistance) or variables such as surface geometry, the nature of theof a material (conduction resistance) to heat transfer. fluid motion, the properties of the fluid, and the bulk L fluid velocity. Conduction resistance: Rt cond. = [°C/W] Typical Values of the Convection Heat Transfer Coefficient kA Free convection of gases 2-25 1 Convection resistance: Rt conv. = [°C/W] Free convection of liquids 10-1000 hA Forced convection of gases 25-250L = thickness of the material [m] Forced convection of liquids 50-20,000k = thermal conductivity [W/(m·°C)] Boiling and condensation 2500-100,000h = convection heat transfer coefficient [W/(m2·°C)]A = area [m2] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 15 of 15
• 16. hrad RADIATION HEAT TRANSFER & Q HEAT TRANSFER THROUGH AN 2 COEFFICIENT [W/(m ·°C)] INSULATED PIPE [W]The radiation heat transfer coefficient depends Tin pipestrongly on temperature while the convection heattransfer coefficient does not. A surface exposed to h in insulationthe surrounding air involves parallel convection and r1 kpradiation, so that the two are combined by adding. It r2 kinsis possible for them to be in opposite directions, r3 T∞ ambient temperatureinvolving a subtraction. hout hrad = εσ (Ts2 + Tsurr ) (Ts + Tsurr ) 2 Resistance model: Tin Tsfc T∞ & Q HEAT TRANSFER RATE [W] ln(r2 /r1) ln(r3 /r2) 1 1 hin 2 π r1 kp 2 π kins 2 π hout 2 π r3 & dT T −T Conduction: Qcond = − kA = −kA 2 1 * dx L & Tin − T∞ Q= ln ( r2 / r1 ) ln ( r3 / r2 ) Convection: Qconv = hA (T2 − T1 ) & 1 1 + + + hin 2πr1 k p 2π kins 2π hout 2πr3 & Radiation: Qrad = σT 4 hin = convection heat transfer coefficient, inner wall*Fourier’s law of heat conduction, introduced in 1822 by [W/(m2·°C)] J. Fourier. kp = thermal conductivity of the pipe material [W/(m·°C)]k = thermal conductivity [W/(m·°C)] kins = thermal conductivity of the insulation [W/(m·°C)]h = convection heat transfer coefficient [W/(m2·°C)] hout = convection heat transfer coefficient of the insulation 2A = surface area normal to the direction of heat transfer [m2] surface [W/(m ·°C)]T1, T2 = temperature [°C or K]L = thickness of the material [m] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 16 of 16
• 17. & q HEAT FLUX [W/m2] & g HEAT GENERATION IN A RESISTIVEHeat transfer per unit time per unit area. WIRE [W/m3] T1 − T2 I 2R Through a material: q= & Per unit volume: g = & [W/m3] L/K πr02 L T1 − T2 Heat generation = heat dissipation: Across a boundary: q= & 1/ h Q = I 2 R = gV = g ( πr02 L ) = h (Ts − T∞ )( 2πr0 L ) & & &Example: For a composite wall with convective effects h1 and h2 (wind) at the surfaces: & gr0 Surface temperature: Ts = T∞ + [K] T1 h1 k1 k2 k3 h2 T2 2h & gr02 Temperature at the center: T0 = Ts + L1 L2 L3 [K]The problem can be modeled as a series of resistances: 4k 1 L1 L2 L3 1 I = current [A]T1 h1 Tsfc k1 k2 k3 h2 T2 R = resistance [Ω] r0 = outer radius of wire [m] 1 L L = length of wire [m]Note that for the thermal resistances and we are 2 h k V = volume [m3]using units of [(m ·°C)/W], not [°C/W]. h = convection heat transfer coefficient [W/(m2·°C)] T1 − T2 T∞ = ambient temperature [K]The heat flux is: q= & 1 L1 L2 L3 1 + + + + h1 k1 k2 k3 h2 Lc CHARACTERISTIC LENGTH [m] The equivalent penetration distance of a shape in itsNow the surface temp can be found from the relation: thermodynamic application. For example a large flat T1 − Tsfc plate immersed in a fluid would have a characteristic q= & length equal to half of its thickness, unless the 1/ h1 temperature difference existed across the plate ink = thermal conductivity [W/(m·°C)] which case the characteristic length would be equal toh = convection heat transfer coefficient [W/(m2·°C)] the total thickness. The characteristic length is usedA = surface area [m2] in finding the Biot number.Tsfc = surface temperature [°C or K] V Cylinder: Lc = 1 rT1, T2 = temperature [°C or K] Lc = 2L = thickness of the material [m] A Sphere: Lc = 3 r 1 V = volume [m3] A = surface area [m2] rcr CRITICAL RADIUS [m]Adding insulation to a cylindrical object can reduceheat loss due to the insulating effect, but there is also Bi BIOT NUMBERan increase in surface area. In the case of a small The ratio of the internal resistance of a body to heatdiameter cylinder, the heat loss due to increased conduction to its external resistance to heatsurface area may outweigh the benefit of the added convection. If the Biot number is very small, it meansinsulation. At some point, the two effects are equal that the internal temperature is uniform so that it isand this is called the critical radius of insulation for a possible to use lumped system analysis incylindrical body. This is the outer radius that includes determining thermal behavior.the thickness of the insulation. hLc rcr,cylinder = k Bi = h k V = volume [m3]k = thermal conductivity of the insulation [W/(m·°C)] h = external convection heat transfer coefficient [W/(m2·°C)]h = external convection heat transfer coefficient Lc = characteristic length of the body [m] [W/(m2·°C)] k = thermal conductivity of the insulation [W/(m·°C)] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 17 of 17
• 18. b THERMAL TIME CONSTANT [s-1] STEPHAN-BOLTZMANN LAWThe thermal time constant is used in lumped system The idealized surface that emits radiation at theanalysis (see next) to describe the (exponential) rate maximum rate is called a blackbody. The maximumat which a body approaches thermal equilibrium. It is radiation that can be admitted from a body atthe inverse time it takes for the temperature- temperature Ts is given by the Stephan-Boltzmannequilibrium gap to reduce to 1/e (approximately 1/3) it’s law.previous amount. For example if a 400°C object isplaced in a 100°C medium, it will cool to 200°C in q = σTs4 & [W/m2]approximately 1/b seconds. & Qmax = σATs4 [W ] h σ = Boltzmann constant, 5.6705×10-8 W/(m2·K4) b= A = surface area [m2] ρ C p Lc Ts = surface temperature [K]h = external convection heat transfer coefficient [W/(m2·°C)]ρ = volume density [kg/m3] ε EMISSIVITY [no units]Cp = specific heat at constant pressure (1.005 @ 300k) Radiation from real surfaces is somewhat less than [kJ/(kg·°C)] radiation from a blackbody (idealized heat-radiatingLc = characteristic length of the body [m] surface). The emissivity of a surface is the factor that relates its ability to radiate heat to that of an ideal surface. LUMPED SYSTEM ANALYSISLumped system analysis assumes the interior q = εσTs4 & [W/m2]temperature of a body to be uniform throughout the & Qmax = εσATs4 [W ]body. The method may be used when the Biotnumber is small. ε = emissivity 0 ≤ ε ≤ 1 [no units] Bi ≤ 0.1 σ = Boltzmann constant, 5.6705×10-8 W/(m2·K4)When this condition is met, the temperature variation across A = surface area [m2]the internal area of the body will be slight. The following Ts = surface temperature [K]relation may be used: T ( t ) − T∞ h = e − bt . where b= Ti − T∞ ρ C p LcTo find the time at which the temperature change from Ti toT∞ is 99% complete, set e-bt = 0.01.T(t) = temperature at time t [°C]T∞ = temperature of the surroundings [°C]Ti = initial temperature of the body [°C]b = time constant (see previous) [s-1]t = time [s]h = external convection heat transfer coefficient [W/(m2·°C)]ρ = volume density [kg/m3]Cp = specific heat at constant pressure (1.005 @ 300k) [kJ/(kg·°C)]Lc = characteristic length of the body [m] Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 18 of 18
• 19. GENERAL MATHEMATICAL x + j y COMPLEX NUMBERS EULERS EQUATION Im e jφ = cos φ + j sin φ π j y A and j = e 2 θ 0 x HYPERBOLIC FUNCTIONS Re j sin θ = sinh ( jθ ) x + jy = Ae jθ = A cos θ + jA sin θ j cos θ = cosh ( jθ ) Re x + jy} = x = A cos θ { j tan θ = tanh ( jθ ) I m x + jy} = y = A sin θ { Magnitude { x + jy} = A = x 2 + y 2 PHASOR NOTATION y Phase { x + jy} = θ = tan −1 To express a derivative in phasor notation, replace x ∂ π with jω . For example, the j ∂t j=e 2 ∂V ∂IThe magnitude of a complex number may be written as the Telegraphers equation = −Labsolute value. ∂z ∂t Magnitude { x + jy} = x + jy ∂V becomes = − LjωI .The square of the magnitude of a complex number is the ∂zproduct of the complex number and its complex conjugate. x + jy = ( x + jy )( x + jy ) * = ( x + jy )( x − jy ) 2 SPHERE Area = πd = 4πr2 Volume = 1 πd 3 = 4 πr 3 2 6 3 SERIES 1 GRAPHING TERMINOLOGY 1+ x 1+ x , x 1 With x being the horizontal axis and y the vertical, we have 2 a graph of y versus x or y as a function of x. The x-axis 1 x 3 x 2 5 x 3 35 x 4 represents the independent variable and the y-axis 1− + − + −L, − 1 < x < 1 represents the dependent variable, so that when a graph 1+ x 2 2 2 8 16 128 is used to illustrate data, the data of regular interval (often 1 this is time) is plotted on the x-axis and the corresponding 1 + x2 + x4 + x6 + L , − 1 < x < 1 data is dependent on those values and is plotted on the y- 1 − x2 2 2 axis. 1 1 + 2 x + 3x 2 + 4 x 3 + L , − 1 < x < 1 (1 − x ) 2 2 2 1 1 − x + x2 − x3 + L , − 1 < x < 1 1+ x 2 2 1 1 + x + x2 + x3 + L , − 1 < x < 1 1− x 2 2 Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 19 of 19
• 20. GLOSSARYadiabatic Describes a process in which there is no heat point function A property whose magnitude depends only transfer. This could either be a well-insulated system or a upon the state and not on the path taken. Point functions system that is at the same temperature as the surroundings. have exact differentials designated by the symbol d. Adiabatic is not the same as isothermal because in an pure substance Composed of a single homogeneous adiabatic process the temperature can change by other chemical species, e.g. water (even with ice). Air is often means such as work. considered a pure substance in its gaseous form. However,chemical energy The internal energy associated with the air in the form of a liquid/gas mixture is not a pure substance atomic bonds in a molecule. since the liquid and gas do not have the same chemicalcritical point The pressure and temperature at which a composition. This is because the various gases that substance will change state (between liquid and vapor) comprise air liquefy at different temperatures. instantly without an intermediate state of saturated reversible process An idealized process that can be liquid/vapor mixture. The critical point for water is 22 MPa, reversed without leaving any trace on the surroundings. 374°C. That is, the net heat and work exchanged between theenthalpy (H) The sum of the internal energy and the volume- system and the surroundings is zero for the combined pressure product. If a body is heated without changing its process and reverse process. There is zero entropy volume or pressure, then the change in enthalpy will equal generation in a reversible process. the heat transfer. Units of kJ. Enthalpy also refers to the sensible energy The portion of the internal energy of a more commonly used specific enthalpy or enthalpy per unit system associated with the kinetic energies of the molecule. mass h, which has units of kJ/kg. These kinetic energies are 1) translational kinetic energy, 2)entropy (s) The unavailable energy in a thermodynamic rotational kinetic energy, and 3) vibrational kinetic energy. system, also considered a measure of the disorder of a The degree of activity of the molecules is proportional to the system. Increasing heat energy increases entropy; heat of the gas. Higher temperatures mean greater kinetic increasing pressure yields an increase in energy but little or energy and therefore higher internal energy. no increase in entropy. The entropy of a material will be highest in the gas phase and lowest in the solid phase. Units of kJ/(kg·K).heat engine A heat engine operates on a cycle, receives heat from a high-temperature source, converts part of it to work, and rejects the waste heat to a low-temperature sink.internal energy (u) The sum of all microscopic forms of energy of a system.irreversible process A process that is not reversible—duh. There will be entropy generation in an irreversible process. Factors that can cause a process to be irreversible include friction, unrestrained expansion, mixing of two gases, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions.isentropic Having constant entropy. An isentropic process is an internally reversible adiabatic process in which the entropy does not change.isothermal Describes a process in which temperature remains constant.latent energy The internal energy associated with the phase of a system. When sufficient energy is added to a liquid or solid the intermolecular forces are overcome, the molecules break away, forming a gas. Because of this, the gas is at a higher energy level than the solid or liquid. The internal energy associated with phase change is the latent energy.nuclear energy The rather large amount of internal energy of a system associated with the bonds within the nucleus of the atom itself.pascal A unit of pressure in the MKS system equal to one Newton per square meter.path function A process result whose magnitude depends on the path taken. Heat transfer and work are examples of path functions. Path functions have inexact differentials designated by the symbol δ. Tom Penick tom@tomzap.com www.teicontrols.com/notes AppliedThermodynamics.pdf 10/25/2004 Page 20 of 20