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1 The Second Law of Thermodynamics Consider the following process: A rock spontaneously rises by lowering its temperature such that mC p T = mgh so that U = 0. mgh Since energy is conserved, this type of process is not forbidden by the first law, but we know this never happens! There is a natural flow of things or direction for spontaneous processes to occur. We also know that various forms of work can be completely converted into heat, e.g., rubbing of two rocks together in a heat reservoir such that they undergo no temperature change so that W = Q and U = 0. In general work of any kind can be done on a system in contact with a reservoir giving rise to a flow of heat without altering the state of a system, W = Q. Work can be converted entirely in to heat by a suitable dissipative process.

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2 Heat Engines - The conversion of heat into work In order to convert heat into work we require a machine that will consume heat and produce work. The machine itself must not suffer any permanent change; it must play a passive role in that following the process it must return to its initial state. The machine must pass through a cycle. def. Thermal efficiency, Applying the first law to the operation of the machine or engine, W = Q 1 - Q 2 where Q 2 corresponds to any heat rejected from the engine, system Q1Q1 Q2Q2 W Heat engine

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3 Carnot Cycle The Carnot cycle is a reversible cycle operating between two temperatures. A B: Isothermal expansion adsorbing heat Q 1. B C: Adiabatic expansion decreasing T from T 1 to T 2. C D: Isothermal compression rejecting heat Q 2. D A: Adiabatic compression increasing T from T 2 to T 1. *Note that if the cycle is operated in reverse refrigerator. Carnot cycle for a gas p V A B C D T1T1 T2T2 Q1Q1 Q2Q2 W Q1Q1 Q2Q2 T1T1 T2T2 W Since all the steps are reversible U = 0, W = Q 1 - Q 2 and

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4 The 2 nd Law Kelvin Statement - No process is possible whose sole result is the complete conversion of heat into work. This addresses the efficiency of conversion. Clausius Statement - No process is possible whose sole result is the transfer of heat from a colder to a hotter body. Spontaneity of processes and the irreversibility of nature. Kelvin Statement Clausius Statement Carnot’s Theorem: No engine operating between two given reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs.

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5 Proof of Carnot’s Theorem CH W C = Q C1 - Q C2 W H = Q H1 - Q H2 T1T1 T2T2 Q H1 Q H2 Q C2 Q C1 Assume the existence of a Hypothetical engine such that, Since the Carnot engine is reversible we can drive it backwards using the mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so that in one cycle it uses exactly as much work as H produces.

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6 CH W C = Q C1 - Q C2 W H = Q H1 - Q H2 T1T1 T2T2 Q H1 Q H2 Q C2 Q C1 Proof of Carnot’s Theorem Composite Engine Now consider C and H as a Composite Engine. This composite engine produces no net work but simply extracts heat from a cold reservoir and delivers an amount of heat, to a hotter reservoir. This is a violation of the Clausius statement of the second law! Carnot’s Theorem:

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7 Corollary: All reversible engines operating between the same temperature reservoirs are equally efficient. Thus the efficiency of any reversible engine operating between the same reservoirs are equally efficient. For any reversible engine, Q n+2 T n+2 W n+2 T n+1 TnTn T n-1 C n+2 C n+1 CnCn Q n+1 Q n-1 QnQn W n+1 WnWn Consider a series of Carnot Engines

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8 For the composite engine, Q n+2 T n+2 W n+2 T n+1 TnTn T n-1 C n+2 C n+1 CnCn Q n+1 Q n-1 QnQn W n+1 WnWn Then, This can only be true if the f’s factorize such that Therefore:The ratio of the temperatures of the reservoirs is equal to the ratios of heat exchanged by a reversible engine operating between the same reservoirs.

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9 According to Carnot’s theorem and its corollary we can make the following statements: Therefore. Taking the heat entering the system as positive, we can say For any closed cycle,, where the equality necessarily holds for a reversible cycle.

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10 Entropy def: We can now define a new variable, the entropy S, by the relation dS = for an infinitesimal reversible change. This definition holds for reversible changes only. For a finite reversible change of state, the change in entropy is given by, Q rev T

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11 Entropy in Irreversible Processes Since entropy is a state function, the change in entropy accompanying a state change must always be the same regardless of how the state change occurs. Only when the state change occurs reversibly is the entropy change related to the heat transfer by the equation A B R X x Consider an irreversible change A B. Construct any reversible path R thus forming an irreversible cycle ABRA. For the irreversible cycle the Clausius theorem says, Determination of the change in entropy for an irreversible change

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12 Taking the integral in two parts, i.e., But by definition of entropy. Thus or Thus we have this general result for a differential irreversible change. For a thermally isolated system Q = 0 and we have the general result known as the law of increase of entropy.

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13 Some Interesting Examples Isothermal expansion of an Ideal Gas Combined 1 st & 2 nd Law Adiabatic Free expansion of an Ideal Gas (Joule expansion - no Q or W exchanged with surroundings) This is an irreversible process, but we can always use the combined Law and integrate from the initial to final state by a convenient reversible path.

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14 P 1, V 1 Diathermal Walls Reservoir (T) Isothermal dissipation of Work Electrical work is dissipated isothermally by heat flow into a reservoir. There is no entropy change of the system because it’s thermodynamic coordinates do not change. The reservoir adsorbs Q = W units of heat at temperature T so its entropy change is Adiabatic dissipation of Work P, T 1 Adiabatic Walls Electrical work is dissipated in a thermally isolated system maintained at fixed pressure. The T of the system increases irreversibly. The coordinates of the system change from P,T 1 to P,T 2. The entropy change can be calculated by Composite system

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15 Examples An inventor claims to have developed a power cycle capable of delivering a net work output of 410 kJ for an energy input by heat transfer of 1000 kJ. The system undergoing the cycle receives heat transfer from hot gases at T = 500 K and discharges energy by heat transfer to the atmosphere at T = 300 K. Evaluate this claim. The thermal efficiency is The maximum efficiency of any power cycle is No good!

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16 Different Forms of the Combined 1 st and 2 nd Law Using the definition of entropy Since enthalpy is defined as, H = U + pV Rearranging these equations and writing them on a unit mass basis, We can use these forms to determine the entropy change of an ideal gas subjected to changes in p, v, T. We already know that for an ideal gas,

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17 Entropy Production A B R X x Determination of the change in entropy for an irreversible change Recall the determination of entropy change for an irreversible process. It is convenient to define a quantity such that is necessarily a positive quantity called entropy production. Def:

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18 Entropy Production Rewriting the expression we obtain, If the end states are fixed the entropy change on the left hand side of this equation can be determined. The 2 terms on the RHS of the equation are path dependant. The 1 st term on the RHS of the equation is the entropy transfer associated with heat transfer. The direction or sign of the entropy transfer is the same as heat transfer. The 2 nd term is the entropy production term.

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19 can not be less than zero Entropy Production By contrast the change in the entropy of the system can be positive, negative or zero: The entropy balance can be expressed in various forms. If heat transfer takes place along several locations on the boundary of the system where the temperatures do not vary with position or time, Here Q j /T j is the amount of entropy transferred to the portion of the boundary at temperature T j. On a time rate basis for a closed system

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20 Examples Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly. The water undergoes a process to the corresponding saturated vapor during which the piston moves freely in the cylinder. If the change of state is brought about by heating the water as it undergoes an internally reversible process at constant pressure and temperature determine the work and the heat transfer per unit of mass in kJ/kg At constant pressure the work is simply,Table A-2

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21 Examples Since the process is reversible and occurs at constant temperature This could also have been calculated our old way

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22 Examples The figure shows a system receiving heat Q from a reservoir. By definition the reservoir is free of irreversibilities, but the system is not, fluid friction, etc. Let’s determine the entropy change of the system and that of the reservoir. For the system, For the reservoir

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23 Examples Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly. The water undergoes a process to the corresponding saturated vapor during which the piston moves freely in the cylinder. There is no heat transfer with the surroundings. If the change in state is brought about by the action of a paddle wheel, determine the net work per unit mass in kJ/kg and the amount of entropy produced per unit mass in kJ/kg-K. As the volume of the system increases during this process, there is an energy transfer by work from the system during the expansion as well as an energy transfer by work to the system done by the paddle wheel. The net work is evaluate from the change in internal energy.

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24 Examples The minus sign indicates that the work input by stirring is greater in magnitude than the work done by the water as it expands. From the 1 st Law, U = -W. On a unit mass basis we have, The entropy produced is evaluated by applying the entropy balance, On a unit mass basis,

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25 Entropy Diagrams S T isotherm isentrop isobar isochor isenthalp

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26 S T Carnot cycle on a T – S diagram. CW – power cycle CCW – refrigeration cycle Entropy Diagrams Area representation of heat transfer for an internally reversible process of a closed system.

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27 Thermodynamic Potentials Combined 1 st and 2 nd Law dU = TdS - pdV Potential Function in terms of S and p, Enthalpy Lengendre Transform subtract a -d(pV) term from dU dU + d(pV) = TdS - pdV + d(pV) d(U + pV) = TdS + Vdp where H = (U + pV) is the Enthalpy and H = H (S, p) Enthalpy is a function of S and p Potential Function in terms of T and p, Gibbs Free Energy Lengendre Transform add a -d(TS) term to dH d ( H-TS ) = Vdp - SdT where G = (H-TS) is the Gibbs Free Energy and G = G (p, T)

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28 Potential Function in terms of T and V, Helmholtz Free Energy Lengendre Transform subtract a -d(TS) term from dU d(U-TS) = -pdV - SdT = dA where A = A(V, T) is the Helmholtz Free Energy Four Fundamental Thermodynamic Potentials dU = TdS - pdV dH = TdS + Vdp dG = Vdp - SdT dA = -pdV - SdT The appropriate thermodynamic potential to use is determined by the constraints imposed on the system. For example, since entropy is hard to control (adiabatic conditions are difficult to impose) G and A are more useful. Also in the case of solids p is a lot easier to control than V so G is the most useful of all potentials for solids. The Maxwell relations are useful in that the relate quantities that are difficult or impossible to measure to quantities that can be measured.

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29 Some important bits of information For a mechanically isolated system kept at constant temperature and volume the A = A(V, T) never increases. Equilibrium is determined by the state of minimum A and defined by the condition, dA = 0. For a mechanically isolated system kept at constant temperature and pressure the G = G(p, T) never increases. Equilibrium is determined by the state of minimum G and defined by the condition, dG = 0. Consider a system maintained at constant p. Then

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