Lecture notes Thermodynamics

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UNIFIED ENGINEERING 2000 Lecture Outlines Ian A. Waitz THERMODYNAMICS: COURSE INTRODUCTION Course Learning Objectives: To be able to use the First Law of Thermodynamics to estimate the potential for thermo- mechanical energy conversion in aerospace power and propulsion systems. Measurable outcomes (assessment method) : 1) To be able to state the First Law and to define heat, work, thermal efficiency and the difference between various forms of energy. (quiz, self-assessment, PRS) 2) To be able to identify and describe energy exchange processes (in terms of various forms of energy, heat and work) in aerospace systems. (quiz, homework, self-assessment, PRS) 3) To be able to explain at a level understandable by a high school senior or non- technical person how various heat engines work (e.g. a refrigerator, an IC engine, a jet engine). (quiz, homework, self-assessment, PRS) 4) To be able to apply the steady-flow energy equation or the First Law of Thermodynamics to a system of thermodynamic components (heaters, coolers, pumps, turbines, pistons, etc.) to estimate required balances of heat, work and energy flow. (homework, quiz, self-assessment, PRS) 5) To be able to explain at a level understandable by a high school senior or non- technical person the concepts of path dependence/independence and reversibility/irreversibility of various thermodynamic processes, to represent these in terms of changes in thermodynamic state, and to cite examples of how these would impact the performance of aerospace power and propulsion systems. (homework, quiz, self-assessment, PRS) 6) To be able to apply ideal cycle analysis to simple heat engine cycles to estimate thermal efficiency and work as a function of pressures and temperatures at various points in the cycle. (homework, self-assessment, PRS) Teaching & Learning Methods 1) Detailed lecture notes are available on the web (for viewing and/or downloading). You should download a copy of these and bring them with you to lecture. 2) Preparation and participation will be important for learning the material. You will be responsible for studying the notes prior to each lecture. Several reading - 1 -assignments will be given to help promote this activity (1/3 of participation grade). 3) Several active learning techniques will be applied on a regular basis (turn-to-your- partner exercises, muddiest part of the lecture, and ungraded concept quizzes). We will make extensive use of the PRS system (2/3 of participation grade). 4) Homework problems will be assigned (approximately one hour of homework per lecture hour). The Unified Engineering collaboration rules apply. - 2 -UNIFIED ENGINEERING 2000 Lecture Outlines Ian A. Waitz THERMODYNAMICS CONCEPTS I. Thermodynamics (VW, S & B: Chapter 1) A. Describes processes that involve changes in temperature, transformation of energy, relationships between heat and work. B. It is a science, and more importantly an engineering tool, that is necessary for describing the performance of propulsion systems, power generation systems, refrigerators, fluid flow, combustion, .... C. Generalization of extensive empirical evidence (however most thermodynamic principles and can be derived from kinetic theory) D. Examples of heat engines Combustion Heat   MechanicalWork     Solar Heat Heat Engine     Electrical Energy   Nuclear Heat    Waste Heat OR MechanicalWork   Heat   Electrical Energy   Waste Heat Fuel Air + fuel Heat Air Fuel Electricity V2, T2 Electricity V1, T1 Air 1. propulsion system 2. power generation 3. Refrigerator - 3 -E. Questions: 1. Describe the energy exchange processes in ___________ (fill in the blank, e.g. a nuclear power plant, a refrigerator, a jet engine). 2. Given that energy is conserved, where does the fuel+oxidizer energy that is used to power an airplane go? 3. Describe the energy exchange processes necessary to use electricity from a nuclear power plant to remove heat from the food in a refrigerator. 4. Describe the energy exchange processes necessary for natural gas to be used to provide electricity for the lights in the room you are in. II. Concept of a thermodynamic system (VW, S & B: 2.1) A. A quantity of matter of fixed identity, boundaries may be fixed or movable, can transfer heat and work across boundary but not mass Force x distance (work) System boundary System boundary Electrical energy Heat (Q) (work) B. Identifiable volume with steady flow in and out, a control volume. Often more useful way to view devices such as engines System boundary m, p ,T 1 1 complex process m, p ,T 2 2 - 4 -III. Thermodynamic state of a system A. The thermodynamic state of a system is defined by specifying a set of measurable properties sufficient so that all remaining properties are determined. Examples of properties: pressure, temperature, density, internal energy, enthalpy, and entropy. B. For engineering purposes we usually want gross, average, macroscopic properties (not what is happening to individual molecules and atoms) thus we consider substances as continua the properties represent averages over small volumes. For 16 3 example, there are 10 molecules of air in 1 mm at standard temperature and pressure. (VW, S & B: 2.2) . Intensive properties do not depend on mass (e.g. p, T, ρ, v=1/ρ, u and h); extensive properties depend on the total mass of the system (e.g. V, M, U and H). Uppercase letters are usually used for extensive properties. (VW, S & B: 2.3) D. Equilibrium: States of a system are most conveniently described when the system is in equilibrium, i. e. it is in steady-state. Often we will consider processes that change “slowly” termed quasi- steady. (VW, S & B: 2.3-2.4) thermally insulated copper Force boundary Gas 1 Gas 1 T T 1 3 Wait Gas 2 Gas 2 Pressure Area T T 3 2 1. mechanical equilibrium 2. thermal equilibrium (force balances pressure times area) (same temperature) - 5 -E. Two properties are needed to define the state of any pure substance undergoing a steady or quasi-steady process. (This is an experimental fact) (VW, S & B: 3.1, 3.3) 1. For example for a thermally perfect gas (this is a good engineering approximation for many situations, but not all (good for pp , and T2T crit crit up to about 4p ). (VW, S & B: 3.4): crit p v = RT v is volume per mol of gas, R is the universal gas constant R = 8.31kJ/Kmol-K. Dividing by molecular weight, p v/M = (R / M ) T where M is the molecular weight of the gas. Most often written as pv = RT or p = ρRT where v is the specific volume and R is the gas constant (which varies depending on the gas. R = 287J/kg - K for air). Thus, if we know p and T we know ρ, if we know T and ρ, we know p, etc. F. For thermodynamic processes we are interested in how the state of a system changes. So typically we plot the behavior as shown below. It is useful to know what a constant temperature line (isotherm) looks like on a p-v diagram, what a constant volume line (isochor) looks like on a T-p diagram, etc. - 6 -300 Increasing Temperature 250 200 150 100 50 0.5 0.75 1.0 1.25 1.5 3 Specific Volume (m /kg) 1.p-v diagram 300 Increasing 250 Specific Volume 200 150 100 50 200 300 400 500 600 Temperature (Kelvin) 2. p-T diagram - 7 - Pressure (kPa) Pressure (kPa)600 550 500 450 400 350 Increasing 300 pressure 250 200 0.5 0.75 1.0 1.25 1.5 3 Specific Volume (m /kg) 3. T-v diagram G. Note that real substances may have phase changes (water to water vapor, or water to ice, for example). Many thermodynamic devices rely on these phase changes (liquid-vapor power cycles are used in many power generation schemes, for example). You will learn more about these in 16.050. In this course we will deal only with single-phase thermodynamic systems. - 8 - Temperature (Kelvin)Pressure-temperature-volume surface for a substance that expands on freezing (fromVW, S & B: 3.7) - 9 -UNIFIED ENGINEERING 2000 Lecture Outlines Ian A. Waitz CHANGING THE STATE OF A SYSTEM WITH HEAT AND WORK - Changes in the state of a system are produced by interactions with the environment through heat and work . - During these interactions, equilibrium (a static or quasi-static process) is necessary for the equations that relate system properties to one-another to be valid. I. Changing the State of a System : Heat (VW, S & B: 4.7-4.9) A. Heat is energy transferred between a system and its surroundings by virtue of a temperature difference only. 1. This transfer of energy can change the state of the system. 2. “Adiabatic” means no heat is transferred. B. Zeroth Law of Thermodynamics (VW, S & B: 2.9-2.10) 1. There exists for every thermodynamic system in equilibrium a property called o o temperature. (Absolute temperature scales: K = 273.15+ C, R = 459.9 + F) 2. Equality of temperature is a necessary and sufficient condition for thermal equilibrium, i.e. no transfer of heat. - 10 -3 1 3 1 2 (thermometer) if T = T and T = T then Q = 0 1 2 2 3 1 3 II. Changing the State of a System: Work (VW, S & B: 4.1-4.6) A. Definition of Work We saw that heat is a way of changing the energy of a system by virtue of a temperature difference only. Any other means for changing the energy of a system is called work. We can have push-pull work (e.g. in a piston-cylinder, lifting a weight), electric and magnetic work (e.g. an electric motor), chemical work, surface tension work, elastic work, etc. In defining work, we focus on the effects that the system (e.g. an engine) has on its surroundings. Thus we define work as being positive when the system does work on the surroundings (energy leaves the system). If work is done on the system (energy added to the system), the work is negative. - 11 -B. Consider a simple compressible substance Work done by system dW = Force ⋅ dl Force   dW = Area ⋅ dl () p p external system   Area dW = Pr essure ⋅ dVolume dW = p dV ext dl Area therefore: V 2 dV W = p ext P ∫ ext V 1 or in terms of the specific volume, v: v 2 W =mp dv ext ∫ v 1 where m is the mass of the system. V V V 1 2 1. If system volume expands against a force, work is done by the system. 2. If system volume contracts under a force, work is done on the system. 3. Why p instead of p ? external system Consider p = 0 (vacuum). No work is done by the system external even though p changes and the system volume changes. system C. Quasi-static processes Use of p instead of p is often inconvenient because it is usually the state of the ext sys system that we are interested in. However, for quasi-static processes p ≈ p . sys ext. = p ± dp Consider p ext sys - 12 -V V V 2 2 2 W = p dV = p ± dp dV = p dV ± dpdV () ext sys sys ∫ ∫ ∫ V V V 1 1 1 therefore V 2 W = p dV sys ∫ V 1 is the work done by the system in a quasi-static process. 1. Can only relate work to system pressure for quasi-static processes. 2. Take a free expansion (p = 0) for example: p is not related to p ( and ext sys ext thus the work) at all the system is not in equilibrium. D. Work is a path dependent process 1. Work depends on path 2. Work is not a function of the state of a system 3. Must specify path if we need to determine work p 1 a 2p 0 p 0 b 2 V 2V V 0 0 Along Path a: W = 2p (2V - V ) = 2p V 0 0 0 0 0 Along Path b: W = p (2V - V ) = p V 0 0 0 0 0 - 13 -isotherm 4. Question: Given a piston filled with air, ice, a bunsen burner, and a stack of small weights, describe how you would use these to move along either path a or path b above. When you move along either path how do you physically know the work is different? 5. Example : Work during quasi-static, isothermal expansion of a thermally perfect gas from p , V to p , V . 1 1 2 2 First, is path specified? Yes isothermal. p V Equation of state for thermally perfect gas pV = nRT n = number of moles R = Universal gas constant V = total system volume V V 2 2   nRT dV V 2 W = dV = nRT = nRT ⋅ ln   ∫ ∫ V V  V  1 V V 1 1 also for T = constant, p V = p V , so the work done by the system is 1 1 2 2     V p 2 1 W = nRT ⋅ ln = nRT ⋅ ln      V   p  1 2 or in terms of the specific volume and the system mass     v p 2 1 W = mRT ⋅ ln = mRT ⋅ ln      v   p  1 2 E. Work vs. heat transfer which is which? 1. Can have one, the other, or both. It depends on what crosses the system boundary. For example consider a resistor that is heating a volume of water. - 14 -2. If the water is the system, then the state of the system will be changed by heat transferred from the resistor. 3. If the system is the water + the resistor, then the state of the system will be changed by (electrical) work. - 15 -UNIFIED ENGINEERING 2000 Lecture Outlines Ian A. Waitz FIRST LAW OF THERMODYNAMICS: CONSERVATION OF ENERGY I. First Law of Thermodynamics (VW, S & B: 2.6) A. There exists for every system a property called energy . E = internal energy (arising from molecular motion - primarily a function of temperature) + kinetic energy + potential energy + chemical energy. 1. Defines a useful property called “energy”. 2. The two new terms in the equation (compared to what you have seen in physics and dynamics, for example) are the internal energy and the chemical energy. For most situtations in this class, we will neglect the chemical energy. 3. Let’s focus on the internal energy, u. It is associated with the random or disorganized motion of the particles. T T 2 1 add heat (molecular motion) (molecular motion) u is a function of the state of the system. Thus u = u (p, T), or u = u (p, v), or u = u(v,T). Recall that for pure substances the entire state of the system is specified if any two properties are specified. We will discuss the equations that relate the internal energy to these other variables later in the class. B. The change in energy of a system is equal to the difference between the heat added to the system and the work done by the system. (This tells what the property energy is useful for.) (VW, S & B: Chapter 5) - 16 -∆ E = Q - W (units are Joules) 1. The signs are important (and sometimes confusing) E is the energy of the system Q is the heat transferred to the system (positive) - if it is transferred from the system Q is negative. (VW, S & B: 4.7-4.8) W is the work done by the system (positive) - if work is done on the system W is negative. (VW, S & B: 4.1-4.4) 2. The equation can also be written on a per unit mass basis ∆ e = q - w (units are J/kg) 3. In many situations the potential energy and the kinetic energy of the system are constant. Then ∆ e = ∆ u, and ∆ u = q - w or ∆ U = Q - W Q and W are path dependent, U is not it depends only on the state of the system, not how the system got to that state. 4. Can also write the first law in differential form: dU = δQ - δW or du = δq - δw Here the symbol “δ” is used to denote that these are not exact differentials but are dependent on path. 5. Or for quasi-static processes dU = δQ - pdV or du = δq - pdv 6. Example: Heat a gas, it expands against a weight. Force (pressure times area) is applied over a distance, work is done. - 17 -weight weight Pressure Area Pressure Heat (Q) 7. We will see later that the First Law can be written for a control volume with steady mass flow in and steady mass flow out (like a jet engine for example). We will call this the Steady-Flow Energy Equation. (VW, S & B: 5.8-5.12) 8. We will spend most of the course dealing with various applications of the first law - in one form or another. II. Corollaries of the First Law A. Work done in any adiabatic (Q=0) process is path independent. a U 1 U 2 0 ∆ U = Q - W b B. 2. For a cyclic process heat and work transfers are numerically equal U = U therefore ∆ U = 0 final initial 1 2 and Q = W or δQ=δW ∫ ∫ - 18 -III. Example applications of the First Law to motivate the use of a property called “enthalpy” (VW, S & B: 5.4-5.5) A. The combination u+pv shows up frequently so we give it a name: “enthalpy” h= u+pv (or H = U+pV). It is a function of the state of the system. The utility and physical significance of enthalpy will be clearer when we discuss the steady flow energy equation in a few lectures. For now, you may wish to think of it as follows (Levenspiel, 1996). When you evaluate the energy of an object of volume V, you have to remember that the object had to push the surroundings out of the way to make room for itself. With pressure p on the object, the work required to make a place for itself is pV. This is so with any object or system, and this work may not be negligible. (Recall, the force of one atmosphere pressure on one square meter is equivalent to the force of a mass of about 10 tons.) Thus the total energy of a body is its internal energy plus the extra energy it is credited with by having a volume V at pressure p. We call this total energy the enthalpy, H. B. Consider a quasi-static process of constant pressure expansion Q = (U - U ) +W 2 1 = (U - U ) + p(V - V ) since p = p = p 2 1 2 1 1 2 + pV ) - (U + pV ) = H - H so Q = (U 2 2 1 1 2 1 C. Consider adiabatic throttling of a gas (gas passes through a flow resistance). What is the relation between conditions before and after the resistance? p ,V p ,V p ,V p ,V 1 1 2 2 1 1 2 2 Q = 0 therefore ∆ U = - ∆ W or U - U = -(p V - p V ) 2 1 2 2 1 1 + p V = U + p V so U 2 2 2 1 1 1 H = H 2 1 - 19 -IV. First Law in terms of enthalpy dU = δQ - δW (for any process, neglecting ∆ KE and ∆ PE) dU = δQ - pdV (for any quasi-static process, no ∆ KE or ∆ PE) H = U + pV therefore dH = dU + pdV + Vdp so dH = δQ - δW + pdV + Vdp (any process) or dH = δQ + Vdp (for any quasi-static process) V. Specific Heats and Heat Capacity (VW, S & B: 5.6) A. Question : Throw an object from the top tier of the lecture hall to the front of the room. Estimate how much the temperature of the room has changed as a result. Start by listing what information you need to solve this problem. B. How much does a given amount of heat transfer change the temperature of a substance? It depends on the substance. In general Q = C∆ T where C is a constant that depends on the substance. 1. For a constant pressure process δQ δq     C = or c =     p p     ∂T ∂T p p and for a constant volume process δQ δq     C = or c =     v v     ∂T ∂T v v we use c and c to relate u and h to the temperature for an ideal gas. p v Expressions for u and h. Remember that if we specify any two properties of 2. the system, then the state of the system is fully specified. In other words we can write u = u(T,v), u=u(p,v) or u=u(p,T) the same holds true for h. (VW, S & B: 5.7) Consider a constant volume process and write u = u(T,v). Then - 20 -

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