Simulating Finite-Time Isothermal Processes with Superconducting Quantum Circuits
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Isobaric Expansion Engines: New Opportunities in Energy Conversion for Heat Engines, Pumps and Compressors
energies Concept Paper Isobaric Expansion Engines: New Opportunities in Energy Conversion for Heat Engines, Pumps and Compressors Maxim Glushenkov 1, Alexander Kronberg 1,*, Torben Knoke 2 and Eugeny Y. Kenig 2,3 1 Encontech B.V. ET/TE, P.O. Box 217, 7500 AE Enschede, The Netherlands; [email protected] 2 Chair of Fluid Process Engineering, Paderborn University, Pohlweg 55, 33098 Paderborn, Germany; [email protected] (T.K.); [email protected] (E.Y.K.) 3 Chair of Thermodynamics and Heat Engines, Gubkin Russian State University of Oil and Gas, Leninsky Prospekt 65, Moscow 119991, Russia * Correspondence: [email protected] or [email protected]; Tel.: +31-53-489-1088 Received: 12 December 2017; Accepted: 4 January 2018; Published: 8 January 2018 Abstract: Isobaric expansion (IE) engines are a very uncommon type of heat-to-mechanical-power converters, radically different from all well-known heat engines. Useful work is extracted during an isobaric expansion process, i.e., without a polytropic gas/vapour expansion accompanied by a pressure decrease typical of state-of-the-art piston engines, turbines, etc. This distinctive feature permits isobaric expansion machines to serve as very simple and inexpensive heat-driven pumps and compressors as well as heat-to-shaft-power converters with desired speed/torque. Commercial application of such machines, however, is scarce, mainly due to a low efficiency. This article aims to revive the long-known concept by proposing important modifications to make IE machines competitive and cost-effective alternatives to state-of-the-art heat conversion technologies. Experimental and theoretical results supporting the isobaric expansion technology are presented and promising potential applications, including emerging power generation methods, are discussed. -
Muddiest Point – Entropy and Reversible I Am Confused About Entropy and How It Is Different in a Reversible Versus Irreversible Case
1 Muddiest Point { Entropy and Reversible I am confused about entropy and how it is different in a reversible versus irreversible case. Note: Some of the discussion below follows from the previous muddiest points comment on the general idea of a reversible and an irreversible process. You may wish to have a look at that comment before reading this one. Let's talk about entropy first, and then we will consider how \reversible" gets involved. Generally we divide the universe into two parts, a system (what we are studying) and the surrounding (everything else). In the end the total change in the entropy will be the sum of the change in both, dStotal = dSsystem + dSsurrounding: This total change of entropy has only two possibilities: Either there is no spontaneous change (equilibrium) and dStotal = 0, or there is a spontaneous change because we are not at equilibrium, and dStotal > 0. Of course the entropy change of each piece, system or surroundings, can be positive or negative. However, the second law says the sum must be zero or positive. Let's start by thinking about the entropy change in the system and then we will add the entropy change in the surroundings. Entropy change in the system: When you consider the change in entropy for a process you should first consider whether or not you are looking at an isolated system. Start with an isolated system. An isolated system is not able to exchange energy with anything else (the surroundings) via heat or work. Think of surrounding the system with a perfect, rigid insulating blanket. -
Section 15-6: Thermodynamic Cycles
Answer to Essential Question 15.5: The ideal gas law tells us that temperature is proportional to PV. for state 2 in both processes we are considering, so the temperature in state 2 is the same in both cases. , and all three factors on the right-hand side are the same for the two processes, so the change in internal energy is the same (+360 J, in fact). Because the gas does no work in the isochoric process, and a positive amount of work in the isobaric process, the First Law tells us that more heat is required for the isobaric process (+600 J versus +360 J). 15-6 Thermodynamic Cycles Many devices, such as car engines and refrigerators, involve taking a thermodynamic system through a series of processes before returning the system to its initial state. Such a cycle allows the system to do work (e.g., to move a car) or to have work done on it so the system can do something useful (e.g., removing heat from a fridge). Let’s investigate this idea. EXPLORATION 15.6 – Investigate a thermodynamic cycle One cycle of a monatomic ideal gas system is represented by the series of four processes in Figure 15.15. The process taking the system from state 4 to state 1 is an isothermal compression at a temperature of 400 K. Complete Table 15.1 to find Q, W, and for each process, and for the entire cycle. Process Special process? Q (J) W (J) (J) 1 ! 2 No +1360 2 ! 3 Isobaric 3 ! 4 Isochoric 0 4 ! 1 Isothermal 0 Entire Cycle No 0 Table 15.1: Table to be filled in to analyze the cycle. -
Thermodynamics I - Enthalpy
CHEM 2880 - Kinetics Thermodynamics I - Enthalpy Tinoco Chapter 2 Secondary Reference: J.B. Fenn, Engines, Energy and Entropy, Global View Publishing, Pittsburgh, 2003. 1 CHEM 2880 - Kinetics Thermodynamics • An essential foundation for understanding physical and biological sciences. • Relationships and interconversions between various forms of energy (mechanical, electrical, chemical, heat, etc) • An understanding of the maximum efficiency with which we can transform one form of energy into another • Preferred direction by which a system evolves, i.e. will a conversion (reaction) go or not • Understanding of equilibrium • It is not based on the ideas of molecules or atoms. A linkage between these and thermo can be achieved using statistical methods. • It does not tell us about the rate of a process (how fast). Domain of kinetics. 2 CHEM 2880 - Kinetics Surroundings, boundaries, system When considering energy relationships it is important to define your point of reference. 3 CHEM 2880 - Kinetics Types of Systems Open: both mass and Closed: energy can be energy may leave and enter exchanged no matter can enter or leave Isolated: neither mass nor energy can enter or leave. 4 CHEM 2880 - Kinetics Energy Transfer Energy can be transferred between the system and the surroundings as heat (q) or work (w). This leads to a change in the internal energy (E or U) of the system. Heat • the energy transfer that occurs when two bodies at different temperatures come in contact with each other - the hotter body tends to cool while the cooler one warms -
Isothermal Process It Is the Process in Which Other Physical Quantities Might Change but the Temperature of the System Remains Or Is Forced to Remain Constant
Sajit Chandra Shakya Department of Physics Kathmandu Don Bosco College New Baneshwor, Kathmandu Isothermal process It is the process in which other physical quantities might change but the temperature of the system remains or is forced to remain constant. For example, the constant temperature of human body. Under constant temperature, the volume of a gas system is inversely proportional to the pressure applied, the phenomena being called Boyle's Law, written in symbols as 1 V ∝ P 1 Or, V = KB× where KB is a constant quantity. P Or, PV = KB …………….. (i) This means whatever be the values of volume and pressure, their product will be constant. So, P1V1 = KB, P2V2 = KB, P3V3 = KB, etc, Or, P1V1 = P2V2 = P3V3, etc. The requirements for an isothermal process are as follows: 1. The process should be carried very slowly so that there is an ample time for compensation of heat in case of any loss or addition. 2. The boundaries of the system should be highly conducting so that there is a path for heat to flow into or flow away from a closed space in case of any energy loss or oversupply. 3. The boundaries should be made very thin because the resistance of the substance for heat conduction will be less for thin boundaries. Since in an isothermal change, the temperature remains constant, the internal energy also does not change, i.e. dU = 0. So if dQ amount of heat is given to a system which undergoes isothermal change, the relation for the first law of thermodynamics would be dQ = dU + dW Or, dQ = 0 + PdV Or, dQ = PdV This means all the heat supplied will be utilized for performing external work and consequently its value will be very high compared to other processes. -
The First Law of Thermodynamics Continued Pre-Reading: §19.5 Where We Are
Lecture 7 The first law of thermodynamics continued Pre-reading: §19.5 Where we are The pressure p, volume V, and temperature T are related by an equation of state. For an ideal gas, pV = nRT = NkT For an ideal gas, the temperature T is is a direct measure of the average kinetic energy of its 3 3 molecules: KE = nRT = NkT tr 2 2 2 3kT 3RT and vrms = (v )av = = r m r M p Where we are We define the internal energy of a system: UKEPE=+∑∑ interaction Random chaotic between atoms motion & molecules For an ideal gas, f UNkT= 2 i.e. the internal energy depends only on its temperature Where we are By considering adding heat to a fixed volume of an ideal gas, we showed f f Q = Nk∆T = nR∆T 2 2 and so, from the definition of heat capacity Q = nC∆T f we have that C = R for any ideal gas. V 2 Change in internal energy: ∆U = nCV ∆T Heat capacity of an ideal gas Now consider adding heat to an ideal gas at constant pressure. By definition, Q = nCp∆T and W = p∆V = nR∆T So from ∆U = Q W − we get nCV ∆T = nCp∆T nR∆T − or Cp = CV + R It takes greater heat input to raise the temperature of a gas a given amount at constant pressure than constant volume YF §19.4 Ratio of heat capacities Look at the ratio of these heat capacities: we have f C = R V 2 and f + 2 C = C + R = R p V 2 so C p γ = > 1 CV 3 For a monatomic gas, CV = R 3 5 2 so Cp = R + R = R 2 2 C 5 R 5 and γ = p = 2 = =1.67 C 3 R 3 YF §19.4 V 2 Problem An ideal gas is enclosed in a cylinder which has a movable piston. -
Thermodynamic Potentials
THERMODYNAMIC POTENTIALS, PHASE TRANSITION AND LOW TEMPERATURE PHYSICS Syllabus: Unit II - Thermodynamic potentials : Internal Energy; Enthalpy; Helmholtz free energy; Gibbs free energy and their significance; Maxwell's thermodynamic relations (using thermodynamic potentials) and their significance; TdS relations; Energy equations and Heat Capacity equations; Third law of thermodynamics (Nernst Heat theorem) Thermodynamics deals with the conversion of heat energy to other forms of energy or vice versa in general. A thermodynamic system is the quantity of matter under study which is in general macroscopic in nature. Examples: Gas, vapour, vapour in contact with liquid etc.. Thermodynamic stateor condition of a system is one which is described by the macroscopic physical quantities like pressure (P), volume (V), temperature (T) and entropy (S). The physical quantities like P, V, T and S are called thermodynamic variables. Any two of these variables are independent variables and the rest are dependent variables. A general relation between the thermodynamic variables is called equation of state. The relation between the variables that describe a thermodynamic system are given by first and second law of thermodynamics. According to first law of thermodynamics, when a substance absorbs an amount of heat dQ at constant pressure P, its internal energy increases by dU and the substance does work dW by increase in its volume by dV. Mathematically it is represented by 풅푸 = 풅푼 + 풅푾 = 풅푼 + 푷 풅푽….(1) If a substance absorbs an amount of heat dQ at a temperature T and if all changes that take place are perfectly reversible, then the change in entropy from the second 풅푸 law of thermodynamics is 풅푺 = or 풅푸 = 푻 풅푺….(2) 푻 From equations (1) and (2), 푻 풅푺 = 풅푼 + 푷 풅푽 This is the basic equation that connects the first and second laws of thermodynamics. -
THERMODYNAMICS Entropy: Entropy Is Defined As a Quantitative Measure of Disorder Or Randomness in a System. the Heat Change, Dq
Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com THERMODYNAMICS Entropy: Entropy is defined as a quantitative measure of disorder or randomness in a system. The heat change, dq and the temperature T are thermodynamic quantities. A thermodynamic function whose change (dq/T) is independent of the path of the system is called entropy 푑푆 = 푑푞 푇 1. Entropy is a state function whose magnitude depends only on the parameters of the system and can be expressed in terms of (P,V,T) 2. dS is a perfect differential. Its value depends only on the initial and final states of the system 3. Absorption of heat increases entropy of the system. In a reversible adiabatic change dq=0, the entropy change is zero 4. For carnot cycle ∮ 푑푆 = 0 5. The net entropy change in a reversible process is zero, ∆푆푢푛푖푣푒푟푠푒 = ∆푆푠푦푠푡푒푚 + ∆푆푠푢푟푟표푢푛푑푖푛푔 = 0 In irreversible expansion ∆푆푢푛푖푣 = +푣푒 cyclic processs, ∆푆푢푛푖푣 > 0 All natural process will take place in a direction in which the entropy would increase. A thermodynamically irreversible process is always accompanied by an increase in the entropy of the system and its surroundings taken together while in a thermodynamically reversible process, the entropy of the system and its surroundings taken together remains constant. Entropy change in isothermal and reversible expansion of ideal gas: In isothermal expansion of an ideal gas carried out reversibly, there will be no change in internal energy, ∆U=0, hence from the first law q = -W In such case, the work done in the expansion of n moles of a gas from volume V1 to V2 at constant temperature T is given by – 푊 = 푛푅푇푙푛(푉2) 푉1 푉2 푞푟푒푣 = −푊 = 푛푅푇푙푛( ) 푉1 Hence 푑푆 = 푑푞 = 1 푛푅푇푙푛 (푉2) = 푛푅푙푛 (푉2) 푇 푇 푉1 푉1 (Problem: 5 moles of ideal gas expand reversibly from volume of 8 dm3 to 80 dm3 at a temperature of 27oC. -
Ch 19. the First Law of Thermodynamics
Ch 19. The First Law of Thermodynamics Liu UCD Phy9B 07 1 19-1. Thermodynamic Systems Thermodynamic system: A system that can interact (and exchange energy) with its surroundings Thermodynamic process: A process in which there are changes in the state of a thermodynamic system Heat Q added to the system Q>0 taken away from the system Q<0 (through conduction, convection, radiation) Work done by the system onto its surroundings W>0 done by the surrounding onto the system W<0 Energy change of the system is Q + (-W) or Q-W Gaining energy: +; Losing energy: - Liu UCD Phy9B 07 2 19-2. Work Done During Volume Changes Area: A Pressure: p Force exerted on the piston: F=pA Infinitesimal work done by system dW=Fdx=pAdx=pdV V Work done in a finite volume change W = final pdV ∫V initial Liu UCD Phy9B 07 3 Graphical View of Work Gas expands Gas compresses Constant p dV>0, W>0 dV<0, W<0 W=p(V2-V1) Liu UCD Phy9B 07 4 19-3. Paths Between Thermodynamic States Path: a series of intermediate states between initial state (p1, V1) and a final state (p2, V2) The path between two states is NOT unique. V2 W= p1(V2-V1) +0 W=0+ p2(V2-V1) W = pdV ∫V 1 Work done by the system is path-dependent. Liu UCD Phy9B 07 5 Path Dependence of Heat Transfer Isotherml: Keep temperature const. Insulation + Free expansion (uncontrolled expansion of a gas into vacuum) Heat transfer depends on the initial & final states, also on the path. -
Thermodynamic Propertiesproperties
ThermodynamicThermodynamic PropertiesProperties PropertyProperty TableTable -- from direct measurement EquationEquation ofof StateState -- any equation that relates P,v, and T of a substance jump ExerciseExercise 33--1212 A bucket containing 2 liters of R-12 is left outside in the atmosphere (0.1 MPa) a) What is the R-12 temperature assuming it is in the saturated state. b) the surrounding transfer heat at the rate of 1KW to the liquid. How long will take for all R-12 vaporize? See R-12 (diclorindifluormethane) on Table A-2 SolutionSolution -- pagepage 11 Part a) From table A-2, at the saturation pressure of 0.1 MPa one finds: • Tsaturation = - 30oC 3 • vliq = 0.000672 m /kg 3 • vvap = 0.159375 m /kg • hlv = 165KJ/kg (vaporization heat) SolutionSolution -- pagepage 22 Part b) The mass of R-12 is m = Volume/vL, m = 0.002/0.000672 = 2.98 kg The vaporization energy: Evap = vap energy * mass = 165*2.98 = 492 KJ Time = Heat/Power = 492 sec or 8.2 min GASGAS PROPERTIESPROPERTIES Ideal -Gas Equation of State M PV = nR T; n = u mol Universal gas constant is given on Ru = 8.31434 kJ/kmol-K = 8.31434 kPa-m3/kmol-k = 0.0831434 bar-m3/kmol-K = 82.05 L-atm/kmol-K = 1.9858 Btu/lbmol-R = 1545.35 ft-lbf/lbmol-R = 10.73 psia-ft3/lbmol-R EExamplexample Determine the particular gas constant for air (28.97 kg/kmol) and hydrogen (2.016 kg/kmol). kJ 8.1417 kJ = Ru = kmol − K = 0.287 Rair kg M 28.97 kg − K kmol kJ 8.1417 kJ = kmol − K = 4.124 Rhydrogen kg 2.016 kg − K kmol IdealIdeal GasGas “Law”“Law” isis aa simplesimple EquationEquation ofof StateState PV = MRT Pv = RT PV = NRuT P V P V 1 1 = 2 2 T1 T2 QuestionQuestion …...….. -
Heat Capacity Ratio of Gases
Heat Capacity Ratio of Gases Carson Hasselbrink [email protected] Office Hours: Mon 10-11am, Beaupre 360 1 Purpose • To determine the heat capacity ratio for a monatomic and a diatomic gas. • To understand and mathematically model reversible & irreversible adiabatic processes for ideal gases. • To practice error propagation for complex functions. 2 Key Physical Concepts • Heat capacity is the amount of heat required to raise the temperature of an object or substance one degree 풅풒 푪 = 풅푻 • Heat Capacity Ratio is the ratio of specific heats at constant pressure and constant volume 퐶푝 훾 = 퐶푣 휕퐻 휕퐸 Where 퐶 = ( ) and 퐶 = ( ) 푝 휕푇 푝 푣 휕푇 푣 • An adiabatic process occurs when no heat is exchanged between the system and the surroundings 3 Theory: Heat Capacity 휕퐸 • Heat Capacity (Const. Volume): 퐶 = ( ) /푛 푣,푚 휕푇 푣 3푅푇 3푅 – Monatomic: 퐸 = , so 퐶 = 푚표푛푎푡표푚푖푐 2 푣,푚 2 5푅푇 5푅 – Diatomic: 퐸 = , so 퐶 = 푑푖푎푡표푚푖푐 2 푣,푚 2 • Heat Capacity Ratio: 퐶푝,푚 = 퐶푣,푚 + 푅 퐶푝,푚 푅 훾 = = 1 + 퐶푣,푚 퐶푣,푚 푃 [ln 1 ] 푃2 – Reversible: 훾 = 푃 [ln 1 ] 푃3 푃 [ 1 −1] 푃2 – Irreversible: 훾 = 푃 [ 1 −1] 푃3 • Diatomic heat capacity > Monatomic Heat Capacity 4 Theory: Determination of Heat Capacity Ratio • We will subject a gas to an adiabatic expansion and then allow the gas to return to its original temperature via an isochoric process, during which time it will cool. • This expansion and warming can be modeled in two different ways. 5 Reversible Expansion (Textbook) • Assume that pressure in carboy (P1) and exterior pressure (P2) are always close enough that entire process is always in equilibrium • Since system is in equilibrium, each step must be reversible 6 Irreversible Expansion (Lab Syllabus) • Assume that pressure in carboy (P1) and exterior pressure (P2) are not close enough; there is sudden deviation in pressure; the system is not in equilibrium • Since system is not in equilibrium, the process becomes irreversible. -
12/8 and 12/10/2010
PY105 C1 1. Help for Final exam has been posted on WebAssign. 2. The Final exam will be on Wednesday December 15th from 6-8 pm. First Law of Thermodynamics 3. You will take the exam in multiple rooms, divided as follows: SCI 107: Abbasi to Fasullo, as well as Khajah PHO 203: Flynn to Okuda, except for Khajah SCI B58: Ordonez to Zhang 1 2 Heat and Work done by a Gas Thermodynamics Initial: Consider a cylinder of ideal Thermodynamics is the study of systems involving gas at room temperature. Suppose the piston on top of energy in the form of heat and work. the cylinder is free to move vertically without friction. When the cylinder is placed in a container of hot water, heat Equilibrium: is transferred into the cylinder. Where does the heat energy go? Why does the volume increase? 3 4 The First Law of Thermodynamics The First Law of Thermodynamics The First Law is often written as: Some of the heat energy goes into raising the temperature of the gas (which is equivalent to raising the internal energy of the gas). The rest of it does work by raising the piston. ΔEQWint =− Conservation of energy leads to: QEW=Δ + int (the first law of thermodynamics) This form of the First Law says that the change in internal energy of a system is Q is the heat added to a system (or removed if it is negative) equal to the heat supplied to the system Eint is the internal energy of the system (the energy minus the work done by the system (usually associated with the motion of the atoms and/or molecules).