Eng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics.

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1 Eng Thermodynamics I 1 1 Introduction 1.1 Thermodynamics Thermodynamics is the study of the relationships between heat transfer, work interactions, kinetic and potential energies, and the properties of systems. The basic principles involved are: 1. conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics. The traditional emphasis of thermodynamics is on turbines, pumps, engines (i.e. mechanical devices), but it can also be applied to electric devices, or any system that transforms energy from one form to another. Much of the historical difficulty in thermodynamics was due to imprecise definitions. The power and utility of thermodynamics centers around the fact that there are relatively few variables: Properties: pressure volume temperature internal energy entropy Process variables: work heat and these variables can be applied to a wide variety of systems. The following subsections define terms that arise in thermodynamics. 1.2 Systems A system is a region enclosed by an imaginary boundary that may be rigid or flexible (control volume). The system in thermodynamics is similar to the free body diagram in mechanics. Closed System (or Control Mass) No mass crosses the system boundary, however, energy (in the form of heat and/or work) may cross the boundary. e.g. piston-cylinder

2 Eng Thermodynamics I 2 Open System (or Control Volume) permits the transfer of both mass and energy across the system boundary a region in space through which mass and energy pass (control volume) e.g. pump Isolated System neither mass nor energy may cross the system boundary idealization common to all systems surroundings (everything external to the system) system boundary system volume e.g. An oxy-acetylene torch is used to boil water in a cylinder, thus creating steam and raising a weight. Is the system open, closed or isolated? The answer depends on where the system boundary is placed. A frequent error is to assume a system is closed when in fact it is open. Note: a system may consist of a single substance or many substances.

3 Eng Thermodynamics I Property A property is any measurable characteristic of a system, or any quantity that depends on the state of matter in the system. e.g. T, p, V, ρ, µ, k, R, etc. Properties are independent of the past history of matter, they are conditions at an equilibrium state. (Note: a system is in equilibrium if its properties do not change with time.) Intensive Property independent of the amount of mass present e.g. T, p, ρ, v (specific volume) Extensive Property Note: 1.4 State dependent on the amount of mass present e.g. V (volume), E k (kinetic energy), U (internal energy) extensive property divided by mass intensive property convention - lower case intensive, upper case extensive, but there are exceptions (T, m) The properties of a system describe the state of the substance within the system, i.e. a phase may exist at different states (p,t ). A relationship among properties is an equation of state. 1.5 Process The transformation of a system, resulting in a change of properties, from one state to another is a process. Consider a frictionless piston-cylinder assembly in equilibrium. If a weight is removed, there will be a change in the system properties, i.e. p 1, V 1, T 1 p 2, V 2, T 2.

4 Eng Thermodynamics I 4 The path of the succession of states through which the system passes is called a process. In thermodynamics, processes are not considered to occur in a continuous manner, but to move through a series of equilibrium steps which are infinitely small in size. A continuous process gives rise to conceptual difficulties (e.g. kinetic energy of the piston, friction in the cylinder). Equilibrium mechanics - balance maintained by equality of opposing forces thermodynamics (deals with forms of energy) mechanical - balance of forces thermal - no change in temperature phase - balance between phases chemical - no further chemical reactions Remember - properties are conditions at equilibrium What if heat is added to the same piston-cylinder? Note: p 1 = p 2 since the load on the piston has not changed, but V 2 > V 1, and T 2 > T 1.

5 Eng Thermodynamics I 5 The change in properties can be plotted on process diagrams, which are very useful in the visualization of how a change in state occurs during a process. An inherent assumption in the use of process diagrams is that the system passes through a series of equilibrium states (since properties only have meaning at equilibrium). This gives rise to what is called a quasi-equilibrium or internally reversible process. An internally reversible process could be reversed, and the system and surroundings would retrace the same series of equilibrium steps. This requires an infinitesimal process. Real processes are not reversible (finite rate and losses due to friction and heat transfer). Internally reversible (ideal) processes are often used in the analysis of systems (e.g. calculation of maximum efficiencies). Common types of processes: isothermal - constant temperature adiabatic - no heat transfer isobaric - constant pressure isochoric - constant volume 1.6 Point function A property is a point function; the value of any property of any system at any state is independent of the path or process used to reach that state. The integral of the differential change of a property between two states, 1 and 2, is: 2 1 dv = V 2 V 1 (1) d is used to represent the exact differential of a point function.

6 Eng Thermodynamics I Path function The value of a path function is dependent on the path followed during a process, and the path must be specified before a path function can be evaluated (e.g. heat transfer and work). The total integral of the work done between two states is: 2 1 δw = W 12 (2) Note: 2 1 δy Y 2 Y 1 δ is used to represent the inexact differential of a path function. 1.8 Cycle a process or series of processes whose initial and final states are identical The change in a property during a cycle is zero. dx = 0 (3) The cyclic integral of a path function is not necessarily zero, as it depends on the path taken. δy 0 (4) e.g. δw = pdv, and W 12 W 21, therefore, δw 0

7 Eng Thermodynamics I Pressure (p) normal force per unit area acting on the surface of the system intensive property Thermodynamic Pressure results from the cumulative effect of individual molecules striking the walls of a container δf n p = lim (5) δa δa δa where δf n is the normal force component acting on a small area δa, and δa is the smallest area over which the substance can be considered a continuum Dynamic Pressure (ρu 2 /2) pressure due to the motion of a fluid Absolute Pressure (p abs ) absolute force per unit area on a surface (including atmospheric pressure) Gauge Pressure (p g ) pressure read on a gauge (e.g. tire pressure) p g < 0 vacuum units of measure p abs = p g + p atm (6) Pa = 1 N/m 2 1 kpa = 1000 Pa 1 bar = 10 5 Pa 1 atm = 101,325 Pa 1 atm = 760mm Hg 1 atm = 14.7 psi measured using gauges, transducers and manometers 1.10 Continuum Pressure and other properties imply the use of a continuum (or macroscopic view). Classical thermodynamics studies the macroscopic behaviour of matter, and is, therefore, concerned with the time-averaged behaviour of a large collection of atoms (continuum). Statistical thermodynamics takes a microscopic view, and studies the behaviour of individual atoms.

8 Eng Thermodynamics I 8 To use the concept of a continuum, dimensions must be much larger than the mean free path of the molecules under study (i.e. the mean distance between molecules), otherwise the behaviour of individual molecules would be important. In the definition of pressure, δa is chosen large enough to ensure that the influence of individual molecules is negligible. e.g. ρ = m/v or ρ = m/ V : As V is decreased, the curve becomes irregular since the volume sample has been made so small that the number of molecules in the sample would vary with time Temperature (T ) Temperature is a measure of the kinetic energy of the molecules in a substance. With a temperature increase the molecular activity and velocity increase, therefore, E k increases, and if temperature decreases, E k decreases. Temperature is not defined, rather the equality of temperatures is defined. Two bodies are equal in temperature if they do not interchange heat when placed in contact, i.e. they are in thermal equilibrium. Temperature is used as a measure of thermal equilibrium. Zeroth Law of Thermodynamics When two bodies have equality of temperature with a third body, they have equality of temperature with each other. The Zeroth Law allows the use of a thermometer to measure temperature. It does not define temperature. A temperature scale is defined with respect to some references (e.g. freezing, boiling, triple points). units of measure SI: C Celsius K = C Kelvin Imperial: F Fahrenheit R = F Rankine Conversion: C = ( F - 32) 5/9

9 Eng Thermodynamics I Heat Transfer (Q) Heat transfer is the energy transfer across a system boundary due to a temperature difference between the system and its surroundings. There are three modes of heat transfer: 1. conduction - solids and fluids 2. convection - fluids 3. radiation - solids, fluids, and vacuum Heat transfer always occurs from high to low temperatures, and increases for larger temperature differences. The concept of thermal resistance is used when solving heat transfer problems. Thermal resistances are used in methods similar to those applied to electric circuits, therefore, the concepts of good thermal conductors and insulators arise. If no heat transfer occurs during a process (due to large thermal resistances, or infinitesimal temperature difference) the process is adiabatic. Note the difference between adiabatic (no heat transfer) and isothermal (constant temperature) processes. Heat transfer is directional and the standard convention is: Q < 0 for heat transfer from the system to its surroundings; and Q > 0 for heat transfer to the system from its surroundings. units of measure: SI: Q J q J/kg Q J/s or W (heat transfer rate) Imperial: Q Btu (1 Btu = kj) Q Btu/h (1 Btu/h = W)

10 Eng Thermodynamics I 10 A Btu (British thermal unit) is the amount of energy required to raise the temperature of 1 lbm of water at 68 F by 1 F. Heat transfer is a path function, therefore, the amount of heat transfer in a process between states 1 and 2 is: 2 δq = Q 12 (7) 1.13 Work Interactions (W ) 1 A work interaction is an energy transfer across a system boundary that is equivalent to a force acting through a distance. The sign convention used for work is: W < 0 when work is done on the system by its surroundings; and W > 0 when work is done by the system on its surroundings. Note: this convention is opposite to that used for heat transfer. It was chosen so that work produced by a steam engine would be positive. Work is a path function. Equations used to evaluate work only give the magnitude of the work, the sign must be inferred from the direction of the force and the displacement with respect to the system. e.g. force acting through a distance W 12 = 2 1 F d s e.g. electric work (electric force is used to displace electrons) W 12 = 2 1 EI dt

11 Eng Thermodynamics I 11 e.g. rotating shaft work (e.g. pumps, compressors, turbines) W 12 = e.g. pdv work (e.g. piston-cylinder) 2 1 T dθ If the piston is displaced by a distance ds, what is the work done during a quasiequilibrium process? δw = pads = pdv therefore W 12 = 2 1 p dv (8) The pdv work done during a process is equal to the area under the p -V curve drawn for the process.

12 Eng Thermodynamics I 12 If dv > 0 (expansion) work is done by the system on its surroundings (i.e. W > 0). If dv < 0 (compression)work is done on the system by its surroundings (i.e. W < 0). Often, the functional relationship between p and V can be written in the form of a polytropic equation: pv n = c (a constant) (9) such a process is called a polytropic process. If the exponent n is known, one can easily integrate to determine the pdv work: but p = c/v n, therefore: W 12 = W 12 = p dv c dv V n (10) where the integration will depend upon the value of the exponent n (see e.g. 2.1 of Moran and Shapiro). e.g. if n = 1 (an isothermal process of an ideal gas) W 12 = c ln V 2 (11) where c = p 1 V 1 = p 2 V 2. Units of measure: 1.14 Energy Forms of energy: SI: W J or N m Ẇ W or J/s (power) Imperial: W ft lbf or Btu (1 Btu = ft lbf = kj) Ẇ hp or Btu/h (1 hp = 2545 Btu/h = kw) kinetic, potential, chemical, internal, surface tension, magnetic... Microscopic forms of energy are those related to molecules and the interaction of molecules. Macroscopic forms of energy are those related to the gross characteristics of a substance, i.e. they are based on dimensions much larger than the mean free path of the molecules (continuum), e.g. a mass acting at a center of gravity. Total energy is a sum of the macroscopic and microscopic forms of energy: V 1 E = E micro + E macro Thermodynamics studies changes in total energy, however, there are usually significant changes in only a few forms of energy (all other changes are negligible).

13 Eng Thermodynamics I 13 This course will be concerned with chemically non-reacting systems, and, in most cases, magnetic, electrical, and surface tension energy levels will be neglected. The most frequent mode of work encountered will be either pdv or shaft work, therefore, most examples will be concerned with simple compressible substances. Kinetic Energy (E k, e k ) physical property or intensive kinetic energy, e k : Potential Energy (E p, e p ) E k = m V 2 2 e k = V 2 2 (12) (13) physical property E p = mgz (14) where z is the measured elevation of mass, m, above some reference datum in a gravity field g intensive potential energy, e p : Internal Energy (U, u) thermodynamic property microscopic form of energy due to: e p = gz (15) molecular level energies kinetic, rotational, vibrational intermolecular forces All substances possess some level of internal energy. temperature increase increase u change from solid to liquid, or liquid to vapour increase u Internal energy cannot be measured directly, but changes in u are related to other properties (e.g. T, p, v) Ignoring chemical, electrical, surface tension, and magnetic energies, the energy possessed by a substance is: therefore, the intensive energy is: E = E k + E p + U = m V 2 + mgz + mu (16) 2 e = V gz + u (17) The most significant changes in E are often due to changes in U, therefore, the kinetic and potential energies (E k and E p ) are often neglected.

14 Eng Thermodynamics I Enthalpy Enthalpy is a defined thermodynamic property that is obtained from other properties. Often, the sum U + pv arises in thermodynamic analyses. thermodynamic property called enthalpy is defined: To make life easy, a 1.16 Conservation of Energy for a Closed System H U + pv (18) h u + pv (19) The principle of the conservation of energy provides the foundation of thermodynamics, as it supplies the basic framework required to study the relationships among work, heat transfer, and the various forms of energy. The principle of the conservation of energy can be written in the following form: Energy is a conserved property. It can neither be created nor destroyed; only its form can be altered from one form of energy to another. The conservation of energy equation, or energy balance, for a closed system will have to account for the following energies and energy transfers: 1. Energy transfer due to differences in pressure and temperature (heat transfer and work), which are only identifiable at a system boundary (e.g. a system can do work on its surroundings). 2. Energy related to the mass of the substance(s) within the system (i.e. U, E k, E p ).

15 Eng Thermodynamics I 15 The conservation of energy equation may be written in the following form: Total rate at which Total rate at which energy enters a cv across its boundary energy leaves a cv across its boundary = Net rate of increase of energy within the control volume (20) i.e. if an amount of energy enters a control volume, then the same amount of energy must leave the control volume, or it will produce a net change in the energy within the control volume. The first two terms in Eq. (20) represent heat transfer and work interactions. or or Q Ẇ = de sys dt δq δt δw δt = de sys dt δq δw = de sys (21) In a process from state 1 to state 2 during time interval t = t 2 t 1 then: or 2 1 δq 2 1 δw = 2 de sys 1 Q 12 W 12 = (E 2 E 1 ) sys (22) and for a closed system undergoing a cycle: δq δw = de sys = 0 (23)

16 Eng Thermodynamics I 16 2 Properties of Pure Substances 2.1 State Postulate The state postulate defines how many properties are required to specify a state. Note: The number of independent intensive thermodynamic properties required to completely and uniquely specify the thermodynamic state of a homogeneous substance is one more than the number of relevant, reversible modes of work. 1. Not all properties are independent (e.g. v = 1/ρ, and T and p at the boiling point of a substance). 2. The state postulate is for homogeneous substances, i.e. substances that are uniform in physical and chemical structure. Most mechanical systems use simple compressible substances, for which pdv work is the only significant work mode, therefore, only two independent intensive properties are required to specify a state. e.g. If T and v for water are known the state of the water may be determined and other properties evaluated. The state postulate is used as a basis for relationships between difficult to measure properties (e.g. internal energy, u) and those that are easily measured (e.g. T, p, V ). 2.2 Equilibrium Diagrams There is a relationship between the properties which define the state of any substance. Equilibrium diagrams are used to obtain a qualitative feel for the behaviour of substances in thermal equilibrium. Consider a closed system (e.g. frictionless piston cylinder) containing 1 kg of water at 0.1 MPa (i.e. 1 bar) and 20 C. Heat is added to the system in an isobaric process.

17 Eng Thermodynamics I 17 As heat is added to the the liquid water (state 1), the temperature of the water will increase to C, and the volume will increase to m 3 (state 2). If more heat is added, the temperature will remain at C, but the volume will increase as vapour forms (i.e. boiling or vaporization), (e.g. V = 1 m 3, state 3). The temperature will remain at C until all of the water is vaporized, and the volume has increased to m 3 (state 4). Continued heat addition will increase both the temperature and volume (e.g. T = 500 C, V = m 3, state 5). This process can be plotted on a T -v diagram for a constant pressure of p = 0.1 MPa. State 1 Subcooled or Compressed Liquid any addition of heat will increase temperature and volume (slightly) State 2 Saturated Liquid liquid at its boiling point any addition of heat vaporizes the water but does not change its temperature (if pressure is constant) T = C, v = m 3 /kg

18 Eng Thermodynamics I 18 State 3 Liquid-Vapour Saturation Region T = C, m 3 /kg v m 3 /kg State 4 Saturated Vapour vapour (only) at its condensation point any addition of heat will increase temperature and volume T = C, v = m 3 /kg State 5 Superheated Vapour e.g. T = 500 C, v = m 3 /kg The temperature and pressure at which boiling and condensation occur are the saturation temperature, T sat, and pressure, p sat, respectively. State 1 is subcooled, because its temperature is below T sat for this pressure (p sat = 0.1 MPa T sat = C). Similarly, it is compressed, because its pressure is higher than the p sat that corresponds to the given temperature (T sat = 20 C p sat = kpa). The amount of water vapour present varies from 0% at the start of boiling to 100% at the start of superheating. The fraction of water in the vapour form is called the quality, x, (steam quality for water). x m g m f + m g (24) where, m g is the mass of vapour, and m f is the mass of liquid.

19 Eng Thermodynamics I 19 Similar experiments can be performed at different pressures, and the result is: As pressure increases, the length of the saturated liquid-vapour line decreases until it reaches a point (the critical point or state), where the liquid and vapour states coincide. At pressures above the critical pressure, there is no clear distinction between superheated vapour and compressed liquid phases, and a phase change cannot occur at a temperature or pressure higher than the critical state values, therefore, the substance is called a fluid. For water: T crit = K, p crit = MPa (Table A-1, Moran and Shapiro). For air: T crit = 133 K, p crit = 3.77 MPa (Table A-1, Moran and Shapiro), i.e. at standard atmosphere conditions air is a superheated vapour.

20 Eng Thermodynamics I 20 A similar experiment can be performed, but starting with the solid phase. The results will depend on whether the substance contracts on freezing (Fig. 3.2, Moran and Shapiro) or expands on freezing (Fig. 3.1, Moran and Shapiro). For a substance that contracts on freezing: Note: each line separates two phases, except the triple line, which separates three phases. This occurs for all substances. The triple point for water is T t = 0.01 C, p t = kpa. The equilibrium values of p-v-t can be plotted on a 3-D plot (as in Figs. 3.1 and 3.2 of Moran and Shapiro). It is obvious from these plots that a single equation of state cannot be derived tables are used to evaluate properties. It is possible to project the p-v-t diagram to the p-t plane (see Figs. 3.1 and 3.2 of Moran and Shapiro).

21 Eng Thermodynamics I 21 The triple line becomes a point representing the infinite number of states at which the solid, liquid, and vapour states can co-exist. Note : dry ice sublimes at 1 atm, because the pressure is below the triple point value. (CO 2 : T crit = K, p crit = kpa) 2.3 Evaluation of the Properties of Pure Substances The tables to be used in the evaluation of the properties of pure substances are included in Appendix A of Moran and Shapiro. Tables are presented for water (Tables A-2 to A-6), Refrigerant 22, R-22, (Tables A-7 to A-9), Refrigerant 134a, R-134a, (A- 10 to A-12), Ammonia (Tables A-13 to A-15), and Propane (Tables A-16 to A-18). Tables are supplied for the following phases of water: saturated water (liquid and vapour), Tables A-2 and A-3; superheated water vapour, Table A-4; compressed liquid water, Table A-5; and saturated water (solid and vapour), Table A-6. Note: The saturated liquid specific volume, v f, is multiplied by 10 3 in Tables A-2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 16 and 17. So, v f = m 3 /kg for saturated liquid water at 20 C (Table A-2). The examples in the notes are for water, but the methods employed apply to any substance. The first step is to identify the phase(s) present which table is to be used. Assume a pure substance (e.g. water) at a given temperature, pressure, and specific volume, T 1, p 1, and v 1, respectively.

22 Eng Thermodynamics I 22 Superheated Vapour: if T 1 > T sat, assuming p 1 is p sat if p 1 < p sat, assuming T 1 is T sat if v 1 > v g T1 or v 1 > v g p1 (this test also applies to u, h, and s) if T 1 > T crit Any two independent intensive thermodynamic properties are sufficient to fully describe the state of the substance. Compressed (or Subcooled) Liquid if T 1 < T sat, assuming p 1 is p sat (subcooled) if p 1 > p sat, assuming T 1 is T sat (compressed) if v 1 < v f T1 or v 1 < v f p1 (this test also applies to u, h, and s) if T 1 < T crit when p 1 = p crit Any two independent intensive thermodynamic properties are sufficient to fully describe the state of the substance. Liquid-Vapour Saturation Region p sat and T sat are not independent, therefore, another property is required (e.g. v, u, h, s or x) Note: v f v v g in the saturation region. This is also true for u, h and s. Obviously, the properties of mixtures with different qualities will not be equivalent, but the tables only supply properties for the saturated liquid and the saturated vapour states. e.g. Determine the volume of a mixture of saturated liquid and vapour. V = V f + V g (25) = m f v f + m g v g (26) Defining the total mass: m = m f + m g, and the quality x = m g /m: V = m(1 x)v f + mxv g (27) = m(v f + xv fg ) (28) or v = v f + xv fg (29) where v fg = v g v f. Note: this expression also applies for u, h, and s. Any two independent intensive thermodynamic properties are sufficient to fully describe the state of the substance. e.g. Assuming water is the working fluid, determine the phase(s) present, and the desired properties for the following states: 1. p = 100 kpa, T = 500 C, v =?

23 Eng Thermodynamics I T = 400 C, v = m 3 /kg, p =? 3. p = 1.0 MPa, T = 560 C, h =? 4. T = 100 C, p = 105 kpa, u =? 5. T = 120 C, h = 1000 kj/kg, u =? 6. p = 160 kpa, u = 1500 kj/kg, T =?, x =? 2.4 Specific Heats and Latent Heats Using the state postulate (i.e. two independent intensive properties define the state of a simple compressible substance) the internal energy can be written as a function of temperature and specific volume (two easily measured quantities). u = u(t, v) The change in u can then be written as: ( ) u du = dt + T v ( ) u dv v T The first term is defined to be the constant volume specific heat. ( ) u c v T then du = c v dt + and for an isochoric (constant volume) process: which facilitates the evaluation of changes in u. v (30) ( ) u dv (31) v T du = c v dt (32) Enthalpy may be written as a function of pressure and temperature: and the change in enthalpy is: dh = h = h(p, T ) ( ) h dt + T p ( ) h dp p T The first term is defined as the constant pressure specific heat: ( ) h c p T then ( ) h dh = c p dt + dp (34) p T and for an isobaric (constant pressure) process: which facilitates the evaluation of changes in h. p (33) dh = c p dt (35)

24 Eng Thermodynamics I 24 The units of the specific heats are kj/kg K. The values of c p and c v are more strongly dependent upon temperature than pressure. The ratio of the specific heats is often used: k c p c v (36) and it becomes of particular importance for processes involving ideal gases. Consider a closed system receiving heat in a constant volume process: The 1st Law for this closed system is: du = δq δw but δw = pdv = 0, therefore, du = δq or δq = du = c v dt. Consider a closed system receiving heat in a constant pressure process: The work during the constant pressure process can be expressed as: δw = pdv = d(pv) Therefore, the 1st Law can be written as: du = δq d(pv) and rearranged to give: δq = du + d(pv) = d(u + pv) = dh = c p dt

25 Eng Thermodynamics I 25 Note: The amount of heat gained (or lost) by a system undergoing constant volume or constant pressure processes can easily be evaluated using specific heats. If the internal energy and enthalpy of a substance can be defined as functions of temperature only (i.e. u = u(t ), h = h(t )) then the changes in internal energy and enthalpy for that substance can be calculated using specific heats for any process (i.e. du = c v dt, dh = c p dt ). This becomes of particular significance for ideal gases. Latent heat is the amount of heat that must be added to or removed from a substance during a phase change at a constant temperature. h fg latent heat of vaporization h ig latent heat of sublimation h if latent heat of fusion Sensible heat is used to change the temperature of a substance. 2.5 Ideal Gases It has been found from experiment that the p-v-t behaviour of gases can be approximated by the following equation (under certain conditions): where, p is the absolute pressure (kpa), v is the specific volume (m 3 /kg), R is the gas constant (kj/kg K), T is the absolute temperature (K). pv = RT (37) Note: When using the ideal gas law be careful of the units, and use absolute temperature and pressure. The ideal gas law can also be written on a mass basis (m in kg): It can also be written on a molal basis: where, v = V/n is the molal volume (m 3 /kmol), R is the universal gas constant=8.314 kj/kmol K, n = m/m is the number of moles, pv = mrt (38) pv = RT (39) M is the molar mass (kg/kmol), (Table A-1, Moran and Shapiro). Multiplying both sides of Eq. (39) by n gives: so the gas constant is R = R/M. pvn = m M RT pv = mrt

26 Eng Thermodynamics I 26 The ideal gas law is very simple to use (convenient): e.g. Constant mass process p 1 V 1 T 1 = p 2V 2 T 2 (40) e.g. Constant pressure, constant mass process v 1 T 1 = v 2 T 2 (41) The ideal gas law is an idealization of the behaviour of real gases at low pressures. The accuracy of this approximation increases for high temperatures and low molecular weight. The ideal gas law neglects intermolecular forces. Since the mean free path increases for increases in temperature, the intermolecular forces will decrease, and the approximation improves at high temperatures (a similar process happens at low pressures). A gas is more nearly ideal if p < p crit and T > T crit. One check for ideal gas behaviour is to compare the given state to the critical state using the values in Table A-1, Moran and Shapiro. Air at 1 atm and 300 K is highly superheated, and the assumption of ideal gas behaviour is acceptable. Note: air at 300K can be assumed an ideal gas up to pressures of 10 MPa. Strictly, to verify ideal gas behaviour, one should check the compressibility, Z: Z 1 for ideal gas behaviour. Z pv RT (42) Nonideal gas behaviour will result if: 1. The temperature is decreased (or pressure increased) the molecules become more densely packed and attractive forces between the molecules become significant. 2. The molecules become too close, the electron clouds will interact, producing repulsive forces. 3. The pressure is very high, then the space occupied by the molecules becomes a factor, and they can no longer be treated as point masses. Ideal gas behaviour can be assumed in the vapour phase of a pure compressible substance, when the pressure and temperature are not near the critical values. At low pressures ideal gas behaviour can be assumed irrespective of temperature. Ideal gas behaviour can be assumed at temperatures double T crit. e.g. Can the following substances be considered ideal gases? 1. N 2 at 30 C, 3 MPa 2. CO 2 at 30 C, 3 MPa 3. H 2 O at 1300 C, 3 MPa

27 Eng Thermodynamics I H 2 O at 50 C, 10 kpa 5. H 2 O at 30 C, 10 kpa Critical States (Table A-1, Moran and Shapiro): N 2 : T crit = 126 K, p crit = 3.39 MPa CO 2 : T crit = 304 K, p crit = 7.39 MPa H 2 O: T crit = K, p crit = MPa 1. T >> T crit, OK (highly superheated) 2. T T crit, but p << p crit, OK 3. T >> T crit, p << p crit, OK 4. Table A-2: T sat = 50 C, p sat = kpa, superheated vapour, and p << p crit, OK (Note: be careful near the saturated vapour line) 5. Table A-2: T sat = 30 C, p sat = kpa, compressed liquid, not OK For an ideal gas, it can be shown that u = u(t ), therefore, du = c v dt for any process of an ideal gas. Changes in internal energy can be evaluated as follows: u 2 u 1 = 2 1 c v dt (43) Since h = u + pv = u + RT, then h = h(t ) for an ideal gas, and dh = c p dt for any process of an ideal gas: h 2 h 1 = 2 1 c p dt (44) Note: since h = u + RT, then: dh dt du dt = R substituting the expressions for the specific heats, Eqs. (30) and (33): Since k = c p /c v then c p c v = R (45) c p = c v = k k 1 R (46) R k 1 Eqs. (45) through (47) are valid for any ideal gas. An ideal gas that has a constant k is called a perfect gas. If c p and c v are known, problems related to the evaluation of properties can be significantly reduced. In general, c p and c v are functions of temperature: for a small T, c p and c v may be assumed constant (47)

28 Eng Thermodynamics I 28 for a large T, the temperature variation must be taken into account molar zero-pressure specific heats (c p and c v ) or tables c p R = α + βt + γt 2 + δt 3 + ɛt 4 (48) where values of the constants α, β, γ, δ, and ɛ for common gases are given in Table A-21 for the temperature range 300 to 1000 K The constant pressure specific heat is found from: c p = c p R R M (49) Equations (48) and (49) would then be substituted into Eq. (44) to determine the change in enthalpy. To save time, use Tables A-22 and A-23, Moran and Shapiro, to evaluate changes in h, u and s for air, N 2, O 2, H 2 O, CO, CO 2. These tables allow the evaluation of enthalpy through the following equation: h(t ) = T T ref c p (T ) dt + h(t ref ) (50) where T ref is an arbitrary reference temperature, and h(t ref ) is the enthalpy at the reference temperature. Tables A-22 and A-23 use a reference value of h(0 K) = 0 kj/kg. Therefore, the change in enthalpy in air as it changes temperature from 300 K to 600 K is h(600 K) h(300 K) where the enthalpy values are determined from Table A-22 (i.e = kj/kg). To obtain h and u from the h and u values given in Table A-23, divide h and u by M, the molar mass of the gas. The changes in enthapy and internal energy for gases with linearly varying specific heats (or gases undergoing processes with a small temperature difference) can be evaluated as follows: h 2 h 1 = c pav (T 2 T 1 ) (51) u 2 u 1 = c vav (T 2 T 1 ) (52) where c pav and c vav are appropriate mean values of c p and c v determined from Table A-20. Table A-20 may also be used to determine the specific heats at a specified temperature (use linear interpolation if necessary). In summary, changes in enthalpy and internal energy should be evaluated using: Eqs. (73) and (52) when changes in temperature are small (i.e. T < 200 K); Tables A-22 and A-23 for processes with T > 200 K. Note: these tables are valid for all processes, and the enthalpy and internal energy values are referenced to 0 kj/kg at 0 K.

29 Eng Thermodynamics I 29 Note: for monotonic gases (Ar, Ne, He, Xe) the specific heats are constants: 2.6 Polytropic Processes of Ideal Gases c p = 5 2 R (53) c v = 3 2 R (54) Using a p-v diagram, pressure may be written as a function of specific volume: p = p(v) and in some cases a polytropic equation may be written for the process: pv n = c (55) Equation (55) can be used to determine the pdv work of a process (see e.g. 2.1 in Moran and Shapiro): w 12 = 2 1 w 12 = p dv = p 1 v 1 ln v 2 = p 2 v 2 ln v 2 (n = 1) (56) v 1 v p dv = p 2v 2 p 1 v 1 1 n (n > 1) (57) Note: these expressions give specific work since specific volume was used (W 12 = mw 12 ). Polytropic equations are not restricted to ideal gases, however, further relations can be derived using the polytropic equation and the ideal gas law. If pv n = c and pv = RT, then: or p = c T n n 1, therefore p 2 p 1 = also or T v n 1 = c, therefore: ( ) T n p = c p v 2 v 1 = ( T2 T 1 T v vn = c ( T1 T 2 ) n n 1 ) 1 n 1 (58) (59) So, for an ideal gas undergoing a polytropic process: pv n = c (60) p/t n n 1 = c (61) T v n 1 = c (62) the value of the exponent n defines specific processes for an ideal gas:

30 Eng Thermodynamics I Incompressible Substances n = 0 constant pressure 1 constant temperature constant volume k reversible and adiabatic (i.e. isentropic) A substance with constant density (or specific volume) regardless of what happens to the other properties is an incompressible substance. Many substances are assumed incompressibe, e.g. liquid water (Tables A-2 and A-5): T ( C) p (MPa) v (m 3 /kg) h (kj/kg) u (kj/kg) For incompressible substances V = 0, therefore: ( ) u du = c v dt + dv v T = c v dt (63) and Also u 2 u 1 = 2 1 c v dt (64) dh = d(u + pv) = du + vdp + pdv = c v dt + vdp (65) and h 2 h 1 = u 2 u 1 + v f T1 (p 2 p 1 ) (66) To evaluate the properties of a compressed liquid at low pressures, use the the saturated liquid values at the given temperature(s), and use Eq. (66) to evaluate changes in enthalpy for large changes in pressure. Note: for liquid water v 10 3 m 3 /kg, therefore, for a 1 kj/kg change in enthalpy in Eq. (66) a pressure difference of 1 MPa is required. The compressed liquid water tables in Moran and Shapiro begin at 2.5 MPa. For incompressible substances only one specific heat is defined c = c p = c v. Values of c for some liquids and solids are given in Table A-19, Moran and Shapiro.

31 Eng Thermodynamics I 31 3 Conservation of Mass 3.1 Principle of the Conservation of Mass The principle of the conservation of mass can be written as follows: Mass is a conserved property. It can neither be created nor destroyed; only its composition can be altered from one form to another. Or, the mass of a system must always be accounted for or conserved. Note: changes in chemical composition occur during chemical reactions, however, the mass of the constituents and the final products must be the same. 3.2 Conservation of Mass for a Control Volume The conservation of mass statement can be written in the following form for a control volume (cv): Total rate at which mass enters a cv across its boundary Total rate at which mass leaves a cv across its boundary = Net rate of increase of mass within the control volume or, in mathematical form: ṁ i e i ṁ e = dm cv dt (67) This is the general statement of the conservation of mass equation. The total mass within a cv is defined as follows: m cv = ρ dv (68) V

32 Eng Thermodynamics I 32 Since the velocity and density can vary over an inlet (or exit), the mass flow rate at an inlet (or exit) can be expressed as: ṁ = ρv n da (69) A where, A is the area of the inlet (or exit) and V n is the velocity normal to the inlet (or exit), i.e. V n = V n, and n is the normal to the cv boundary. The mass conservation equation can then be rewritten as: ( ) ρv n da ( ) ρv n da = d A i e A e dt i V ρ dv (70) 3.3 Forms of the Mass Conservation Equation Equation (70) is a general form of the mass conservation equation, but it requires information regarding the variation of ρ within the cv and the variation of ρ and V n at all inlets and exits. Often this information is unavailable, and simplfying assumptions are made One-Dimensional (or Uniform) Flow Assuming all intensive properties are uniform with position (i.e. bulk averaged values) over each inlet and exit, and the given velocity is normal to the cv boundary, then: ṁ = ρav (71) and Eq. (70) can be written as: (ρav ) i e i (ρav ) e = dm cv dt (72) The assumption of uniform flow is appropriate in typical engineering applications where flows are turbulent, because the velocity profile is full, and the properties are relatively uniform over a cross-section.

33 Eng Thermodynamics I Steady State When all properties of a control volume are independent of time the control volume or system is said to be at steady state (e.g. a turbine or pump a suitably long time after start up). For steady state conditions, Eq. (67) can be reduced to: ṁ i = e i ṁ e (73) i.e. (mass in) = (mass out). If the flow is incompressible (ρ =const) mass conservation gives (volume in) = (volume out). The steady state assumption is a realistic and useful assumption for mechanical devices. Note: if all properties are independent of time, then dv/dt = 0, and there can be no pdv work in a device operating under steady state conditions Closed Systems By definition, a closed system does not have mass crossing its boundary, therefore, the mass conservation equation, Eq. (67), reduces to: m cv = const (74)

34 Eng Thermodynamics I Transient Analysis When a system operates under transient conditions (e.g. start up, shutdown, varying inlet or exit conditions, varying environment) the mass within the cv can change. Usually, the change in conditions over a time interval t = t 2 t 1 are of interest. Integrating Eq. (67) over this time interval: t2 i t 1 ṁ i dt e t2 t 1 ṁ e dt = t2 t 1 dm cv dt dt gives: m i e i m e = (m 2 m 1 ) cv (75) i.e. the net change in the mass in the cv over a time interval t is equal to the difference between the total mass entering and the total mass leaving the cv during t.

35 Eng Thermodynamics I 35 4 Conservation of Energy (1st Law of Thermodynamics) 4.1 Principle of the Conservation of Energy The principle of the conservation of energy provides the foundation of thermodynamics, as it supplies the basic framework required to study the relationships among work, heat transfer, and the various forms of energy. The principle of the conservation of energy can be written in the following form: Energy is a conserved property. It can neither be created nor destroyed; only its form can be altered from one form of energy to another. This principle is so fundamental to the study of thermodynamics that it is called the 1st Law of Thermodynamics. The general form of the 1st Law will contain the following energies and energy transfers: 1. Energy transfer due to differences in pressure and temperature (i.e. work and heat transfer), which are only identifiable at a system boundary (e.g. a system can do work on its surroundings). 2. Energy related to the mass of a substance (i.e. U, E k, E p ). This would include energy due to the mass within a control volume and that due to the mass being transported across the control volume boundary at inlets and exits. The energy equation (or 1st Law) may be written in the following form: Total rate at which energy enters a cv across its boundary Total rate at which energy leaves a cv across its boundary = Net rate of increase of energy within the control volume (76) i.e. if an amount of energy enters a control volume, then the same amount of energy must leave the control volume, or it will produce a net change in the energy within the control volume. The first two terms in Eq. (76) represent heat transfer, work interactions, and energy transport across the control volume boundary due to mass flow.

36 Eng Thermodynamics I A Mathematical Statement of the 1st Law of Thermodynamics To obtain a mathematical statement of Eq. (76), consider the following control volume: The rate at which mass flows across an elemental area da is ρv n da, therefore, the rate at which energy is transported across the elemental area da by this mass flow is eρv n da. This expression can be integrated over all inlets to give: Total rate of energy transport due to mass entering the cv = i A i e(ρv n ) da (77) Similary, the energy transport by mass flow out of the control volume can be written as: Total rate of energy transport due to = e(ρv n ) da (78) mass leaving the cv e A e The mass within a control volume is V ρ dv, therefore, the total energy contained within a control volume is V eρ dv. We are interested in the time rate of change of the total energy within a control volume: [ ] Net rate of increase of the total = d eρ dv (79) energy within a control volume dt V Using Eqs. (77), (78) and (79), Eq. (76) can be written as follows: Q cv Ẇ + e(ρv n ) da e(ρv n ) da = d i A i e A e dt V eρ dv (80) where Q cv = δq/δt = { Net rate of heat transfer to the control volume from the surroundings

37 Eng Thermodynamics I 37 Net rate of work leaving the control Ẇ = δw/δt = volume due to all work interactions (e.g. pdv, shaft, electric, flow work) Note: both the work and heat transfer are assumed in the positive directions. Flow work is the work that must be performed to cause mass to flow across control volume boundaries. Due to pressure differentials, mass must be pushed into a control volume at inlets, and pushed out of the control volume at exits. This work is usually separated from other forms of work. Consider the element of fluid, shown below, that is being pushed into a control volume. The work done pushing this element of fluid into the control volume is F s, and the rate at which this work is done can be written as: F d s dt = F V = F n V n = pv n da since F n = pda. With this definition of flow work, the net rate at which flow work is done by the control volume can be defined: [ ] Net rate of = pv(ρv flow work n ) da + pv(ρv n ) da i A i e A e The flow work at the inlets has a negative sign, because work is being done on the system at an inlet. Note: each term has been multiplied by ρv (= 1). Defining: Ẇ = Ẇcv + Ẇflow where Ẇcv represents all reversible and irreversible forms of work except flow work, allows the energy equation to be rewritten in the following form: Q cv Ẇcv + (e + pv)(ρv n ) da (e + pv)(ρv n ) da = d eρ dv i A i e A e dt V But e + pv = u + pv + e k + e p = h + e k + e p

38 Eng Thermodynamics I 38 therefore, the energy equation, or 1st Law of thermodynamics, can be written as follows: Q cv Ẇcv + (h+e k +e p )(ρv n ) da (h+e k +e p )(ρv n ) da = d eρ dv i A i e A e dt V (81) for a thermodynamic system, or control volume, consisting of a simple compressible substance. 4.3 The 1st Law of Thermodynamics for a Closed System By definition, no mass crosses the boundary of a closed system, therefore conservation of mass gives: m sys = const (82) and the 1st Law (or conservation of energy) gives: or or Q Ẇ = de sys dt δq δt δw δt = de sys dt δq δw = de sys (83) In a process from state 1 to state 2 during time interval t = t 2 t 1 then: Q 12 W 12 = (E 2 E 1 ) sys (84) and for a closed system undergoing a cycle: δq δw = de sys = 0 (85) 4.4 Forms of the 1st Law for a Control Volume (or an Open System) Since an open system, or control volume, can have mass and energy crossing its boundary the general form of the 1st Law, Eq. (81), must be used. But some simplifying assumptions are usually employed to ease the use of this equation One-dimensional (or Uniform) Flow If one-dimensional flow is assumed, then the fluid properties (ρ, V n, h, e k, and e p ) are uniform at each inlet and exit, therefore: and the 1st Law, Eq. (81) becomes: A i (h + e k + e p )(ρv n ) da = ṁ i (h + e k + e p ) i Q cv Ẇcv + i ṁ i (h + e k + e p ) i e ṁ e (h + e k + e p ) e = de cv dt (86) where (h + e k + e p ) are evaluated at each inlet and exit and multiplied by the appropriate mass flow rate.

39 Eng Thermodynamics I One-Dimensional, Steady State Flow Under steady state conditions all properties are independent of time, therefore, Eq. (86) reduces to: Q cv Ẇcv + i ṁ i (h + e k + e p ) i e ṁ e (h + e k + e p ) e = 0 (87) and there will be a balance between the total rate of transport of energy into and out of the control volume. Equation (87) is the form of the 1st Law most often used in this course. Since dv/dt = 0 there will be no pdv work. Typically, mechanical engineers deal with rotating machinery, and under steady state conditions the only significant mode of work is shaft work. This is why the work in steady state control volume (or open system) analyses is usually termed shaft work Transient Analysis of Open Systems (or Control Volumes) During the transient operation of a device (e.g. start up, shutdown, varying load on a motor, etc.) the time rate of change of properties becomes significant. Usually, the change over a time interval t = t 2 t 1 is of interest, therefore, the energy equation is integrated over this time interval (as for the mass conservation equation). Since Q cv = δq cv /δt, Ẇ cv = δw cv /δt, and ṁ = dm/dt, these three terms can be substituted into the 1st law for uniform flow, Eq. (86), and the resulting equation can be integrated over the time interval t = t 2 t 1 : e t2 t 1 t2 t 1 δq t2 cv δt dt t 1 which can be rewritten as: δw cv dt + δt i dm e dt (h + e k + e p ) e dt = t2 t 1 t2 t 1 de cv dt dm i dt (h + e k + e p ) i dt dt Q 12 W 12 + i Note: t2 t 1 (h+e k +e p ) i dm i e t2 t 1 (h+e k +e p ) e dm e = (E 2 E 1 ) cv (88) 1. The fluid flow entering and/or leaving the control volume is usually assumed uniform (1D), and the flow rate is often assumed constant with time. 2. Perfect mixing (i.e. the state of the substance throughout the control volume is the same (except at inlets and exits)) is often assumed. This is not the same as the assumption used for uniform flow control volume analyses, where the state is uniform at a location, but can vary from location to location.

40 Eng Thermodynamics I Applications of the 1st Law of Thermodynamics Potential and Kinetic Energies Potential energy is one of the forms of energy included in the 1st Law, but it is often neglected. The change in potential energy is defined as e p = g z. To give a change in potential energy of e p = 1 kj/kg will require a change in elevation of z = (1000J/kg)/(9.81m/s 2 ) = 102 m. For a change in enthalpy of h = 1 kj/kg for an ideal gas ( h = c p T ) or an incompressible substance ( h = c av T ): Substance T (K) c p (kj/kg K) Required T ( C) Air Liquid H 2 O Superheated H 2 O Since z is often less than 10 m, but T is significant, then h >> e p, and e p is neglected. Note: this would not be true for hydroelectric plants, or when pumping fluids between reservoirs, where e p is the most significant change in energy. The change in kinetic energy is e k = (V2 2 V 1 2 )/2. The required change in velocity to give e k = 1 kj/kg depends on the magnitudes of the two velocities (i.e. large V requires small V, and vice versa). For example, if V 1 = 10 m/s V 2 = 46 m/s, and if V 1 = 100 m/s V 2 = 110 m/s. So whether kinetic energy should be neglected relative to changes in enthalpy is problem dependent. Often, e k is neglected due to insufficient information, or to obtain a quick solution (i.e. no need to determine flow areas, and velocity profiles) Turbines Turbines are used to convert the energy of a working fluid (e.g. steam, liquid water, combustion gases) to rotating shaft work. In a steam turbine, the decrease in enthalpy of the steam as it flows through the turbine is used to increase the kinetic energy of the steam, and rotate a shaft. In a hydroelectric plant, the decrease in potential energy of the water is used to increase the kinetic energy of the water, and rotate a shaft.