2 4. The Second Law 4.1 Reversible vs Irreversible processes Reversible: A process for which a system can be restored to itsinitial state, without leaving a net influence on thesystem or its environment.* idealized, frictionless;* proceeds slowly enough for the system to remain inthermodynamic equilibrium.Irreversible: Not reversible* natural;* proceeds freely, drives the system out of thermodynamicequilibrium;* interacts with environment, can not be exactly reversed
3 Example: Gas-piston system under a constant temperature * Slow expansion and compression* Rapid expansion and compression
4 4.2 Entropy / Carnot’s Theorem * Consider the first law,, orDivide them by T and use pv=RT,orThe specific entropy (s) is defined as,which is a state variable, a property of the system.
5 * The Carnot cycle and Carnot Theorem 1) From state 1 to state 2:Isothermal expansion2) From state 2 to state 3:Adiabatic expansion
6 3) From state 3 to state 4:Isothermal compression4) From state 4 to state 1:Adiabatic compression
7 Why is the Carnot cycle reversible? The net heat transfer and the net work over the Carnot cycle are:Using Poisson’s equation,Then,So the system absorbs heat and performs net work in the Carnot cycle,which behaves as a heat engine.Why is the Carnot cycle reversible?
9 For the Carnot cycle, we can also have: This relationship also hold for the reversed Carnot cycle.This is called the Carnot’s Theorem:(4.1)which shows that the change of entropy is independent of pathunder a reversible process.
10 4.3 The Second Law and its Various Forms To get the second law, we use the Clausius Inequality, i.e.,for a cyclic process,which indicates during the cycle,heat must be rejected to the environment somewhere during a cycle;heat exchange is larger at high temperature than at low temperatureunder reversible conditions;the net heat absorbed is smaller under the irreversible conditionthan under the reversible condition.
11 Now, consider two cycles as shown in the plot. For the cycle which contains onereversible process and one irreversibleprocess,(4.2)For the cycle which have two reversibleprocesses,(4.3)
12 The difference between (4.2) and (4.3) gives Because states 1 and 2 are arbitrary, we have the second law,(4.4)Combine (4.1) and (4.4), we haveIt indicates that the heat absorbed by the system during a processhas an upper limit, which is the heat absorbed during a reversibleprocess.The first law relates the state of a system to work it performs and heatit absorbs.The second law controls how the systems move to the thermodynamicequilibriums, i.e., the direction of processes.
13 Several simplified forms of the second law: 1) For an adiabatic process, (4.4) becomes(4.5)If the adiabatic process is reversible, thenIt is also isentropic (s is constant).2) For an isochoric process, (4.1) becomes(4.6)Because only state variables are involved, it holds for eitherreversible or irreversible processes.(4.5) and (4.6) show that :Irreversible work can only increase entropy;heat transfer can either increase or decrease entropy.
14 4.4 Fundamental Relations / The Maxwell Relations Combine forms of the first law and the second law,or(4.7)Similarly,or(4.8)
15 For reversible processes, the equal signs apply and the equations are called Fundamental Relations,(4.9)(4.10)Because these equations involve only state variables, they do notdepend on path. So they must hold for both reversible andirreversible processes.These identities describe the change in one state variable in termsof changes in two other state variables.
16 Two other state variables can be defined, The Helmholtz function:The Gibbs function:Use these definitions in (4.7)-(4.10),(4.11)(4.12)(4.13)(4.14)
17 * The Maxwell Relations: Recall that the condition for a functionto be exact is,Since the state variables are exact, so for u, fromWe have(4.15)
18 Similarly from the fundamental relations we can show that (4.16)(4.17)(4.18)(4.15)-(4.18) are called the Maxwell relations.
19 * Noncompensated Heat Transfer For an irreversible process,To remove the inequality, a term can be added to the right side of theformulation,Next, use the second law for a reversible process on du,Finally,is called the noncompensated heat transfer, which measuresadditional heat rejection to the environment due to irreversibility.
20 4.5 Thermodynamic Equilibrium * Consider an adiabatic process,the second law becomesFor an irreversible condition,oris the entropy at the initial state.When reaches the maximum,the state is in thermodynamicequilibrium because the entropycan not increase anymore.
21 * Consider an isentropic-isochoric process, From (4.7), we haveFor an irreversible condition,oris the internal energy at the initial state.When reaches the minimum, the state is in thermodynamicequilibrium because the internal energy can not decreaseany further.* The enthalpy, Helmholtz function and Gibbs function must alldecrease as a system reaches thermodynamic equilibrium undercertain processes.
22 4.6 Relationship of Entropy to Potential Temperature * Use the first law and the equation of state for an ideal gasin the second law,* use log derivatives on the potential temperature,
23 So,(4.19)Under the reversible process, we have(4.20)So, the change in entropy can be measured by the change inpotential temperature.Because (4.20) involves only state variables, it is path independentand is valid for both reversible and irreversible processes.
24 4.7 Implications for Vertical Motion 1) Under adiabatic processes:Adiabatic conditions require:a. no heat be transferred between the system and environment;b. no heat exchange between one part of the system and another.So, these exclude the irreversible turbulent mixing and the irreversibleexpansion work-induced mixing in the system.Therefore, the adiabatic process is approximately reversible, i.e.,So, the potential temperature surfaces coincide withIsentropic surfaces An air parcel will remain on a certainIsentropic surface and undergoes no systematic vertical motion.
26 2) Under diabatic processes: An air parcel moves acrossisentropic surfaces followingthe heat transfer with itsenvironment.
27 The displaced motionis sufficiently slow:The air parcel’s temperaturediffers from the environmentonly infinitesimally;Rejection of heat during thepoleward moving is balancedby the absorption of heat duringthe equatorward moving.No net vertical motion, theparcel’s evolution is reversible.The displaced motionis sufficiently fast:Produce the net heattransfer and a verticaldrift of the parcel acrossisentropic surface in acomplete cycle.
28 Meteorology 341 Homework (3) A dry air parcel undergoes a complete Carnot cycle consisting of the steps indicated in (a)-(d). For each individual step,calculate the mechanical work w (per unit mass) done by the air parcel and the heat q added to the parcel.Adiabatic compression from p1=600 hPa and T1=0oC to a temperature T2 of 25oC;Isothermal expansion to a pressure p3 of 700 hPa;Adiabatic expansion to temperature T4 of 0oC;Isothermal compression back to the original pressure p1.Also, compute(e) the total work done and heat added for the complete cycle, and(f) the efficiency of the cycle.2. Two hundred grams of mercury at 100oC is added to 100 g of water at 20oC. If the specific heat capacities of water andmercury are 4.18 and 0.14 JK-1g-1, respectively, determine (a) the limiting temperature of the mixture, (b) the change ofentropy for the mercury, (c) the change of entropy for the water, and (d) the change of entropy for the system as a whole.3. During a cloud-free evening, LW heat transfer with the surface causes an air parcel to descend from 900 to 910 mb andits entropy to decrease by 15 J kg-1 K-1. If its initial temperature is 280 K, determine the parcel’s (a) final temperatureand (b) final potential temperature.