*ThermodynamicsModule 7: EntropyBernard GalloisGeorge Meade Bond Professor of EngineeringOffice: Burchard 410Telephone: 201-216-5041E-mail: bgallois@stevens.edu

*The Clausius Inequality We saw that an irreversible heat engine was less efficient that a reversible heat engine operating between the same heat reservoirs:

We recall that

So that for an irreversible engine (equality for a reversible engine) :

*The Clausius InequalityThe previous equation can be re-written as

This very important inequality is the Clausius inequality, which has major consequences in thermodynamics. It is more generally expressed as

The cyclic integral of dQ/T is always less than or equal to zero.

*The Clausius Inequality

Here, Q is the net heat added to the system, Qnet...

The equality applies when there are no irreversibilities within the system as it executes the cycle.The inequality applies when internal irreversibilities are present.

*Example 1(a): In a power plant, heat is added in the amount of 3150kJ at 440oC and rejected in the amount of 1950 kJ at 20oC. Is the Clausius inequality satisfied and is the cycle reversible or irreversible?The Clausius inequality is satisfied. The cycle has at least one irreversible process and the cycle is irreversible.

*Example 1(b):Calculate the net work, the cycle efficiency and Carnot efficiency for this cycle.

*Example 2(a): In a power plant, heat is added in the amount of 3150kJ at 440oC and rejected in the amount of 1294.46 kJ at 20oC. Is the Clausius inequality satisfied and is the cycle reversible or irreversible?The Clausius inequality is satisfied. The cycle is made of reversible processes.

*Example 2(b): What cycle could this be? Calculate the net work and cycle efficiency.

*The Definition of EntropyAll processes A, B and C are internally reversible.Consider two cycles executed by a closed system.Cycle 1: A followed by CCycle 2: B followed by CFor the first and second cycles

The integral is the same for processes A and B.

*The Definition of EntropySince A and B are arbitrary, it follows that the integral has the same value for any internally reversible process between the two states. The value of the integral depends only on the initial and final states. The integral represents the change in some property of the system. It is a state function.Clausius called this function the entropy, S and its change is given by:

Entropy is an extensive propertyIts units are J/K or kJ/KThese equations only apply to internally reversible processes.

*Internally Reversible Isothermal Heat Transfer The entropy change during an internally reversible isothermal heat transfer process is given by:

This relation is particularly useful to determine the entropy change of thermal energy reservoirs since they operate at constant temperature. Note that the entropy change can be positive or negative depending on the direction of heat transfer.

*Isentropic ProcessesRecall:

For a reversible process in which dQ = 0,we have dS = 0 and S2 = S1

The reversible adiabatic process is called an isentropic process.

*The Principle of Increase of EntropyThe cycle shown consists of an internally reversible process 2-1 and an arbitrary cycle 1-2. From the Clausius inequality: T absolute temperature> Irreversible process= Reversible process

*The Principle of Increase of EntropyThe entropy change of a closed system is always greater than the entropy transfer.

Entropy is created during an irreversible process, as a result of the irreversibilities.The entropy generation is denoted by Sgen:

The entropy generated is always a positive quantity (or zero for a reversible process).Its value depends on the process: it is not a property.This equation has far reaching implications in thermodynamics and other fields

*Entropy Generation in Isolated Systems:Increase of Entropy Principle Note that the system shown consists of several sub-systems lumped together into an isolated system.The entropy of an isolated system during a process always increases, or in the case of a reversible process, remains constant.A system and its surroundings can be viewed as the two sub-systems of an isolated system. See next slide.

*Example: find the total entropy change, or entropy generation, for the transfer of 1000kJ of energy from a heat reservoir at 1000 K to a heat reservoir at 500 K.

*Entropy Generation in Isolated Systems: Increase of Entropy PrincipleThe entropy change of the isolated system: (system+surroundings) is equal to the entropy generation:

The change in entropy of the surroundings results from the occurrence of the process. Since no actual process is reversible, some entropy is generated during a process. Since the universe can be considered to be an isolated system, the entropy of the universe is continuously increasing. The more irreversible the process, the larger the entropy generated during that process.No entropy is generated during a reversible process.

*Important CaveatThe increase of entropy principle does not imply that the entropy of a process cannot decrease.The entropy change of a system can be negative during a process but entropy generation cannot. The criteria to determine whether a process occurs reversibly, irreversibly or does not occur can be summarized:

*Example: analysis of a closed system exchanging heat with the surroundings.

*Example: analysis of a closed system exchanging heat with the surroundings.

*Further Remarks on EntropyProcesses can occur in certain directions only, not in any direction, such that: DSgen 0Entropy is not a conserved property. There is no conservation of entropy principle and the entropy of the universe is constantly increasing.The performance of engineering systems is degraded by the presence of irreversibilities. Entropy generation is a measure of the magnitudes of the irreversibilities present during a process.

*Entropy Change of Pure Substances: The T-s DiagramValues of the entropy, with respect to a suitable reference state are tabulated for many substances. For the saturated mixture:

For the compressed liquid, the entropy is approximated by the entropy of the saturated liquid:

Values for the superheated vapor are obtained directly from the tables.

*Example: find the entropy and/or temperature of water in the following statesTsat = 263.99 oCNot in tableTsat = 207.15 oC

TsatTsat = 75.87oCsf =1.0259 sg = 7.6700

*

*Example: determine the entropy change of water contained in a closed system as it changes phase from a saturated liquid to a saturated vapor at a constant pressure of 0.1 MPa. Why is the entropy change positive?System: shownProperty relations: steam tablesProcess and diagram: constant pressure in the saturation region so T is constant (Tsat)Conservation Principle

*Example: Steam at 1 MPa and 600oC expands in a turbine to 0.01 MPa. If the process is isentropic, find the final temperature, the final enthalpy of the steam and the turbine work.System: see aboveProperty relations: steam tablesProcess and diagram: isentropic so s1=s2=8.029 kJ/kg.KConservation principle: steady flow, neglect PE and KE, one entrance, one exit

*Example (cont.)Recall mass balance and energy balance from previous chapter for turbineWe get h1 and h2 from the steam tables

At P2 = 0.01 MPa,sf = 0.6491 and sg = 8.151kJ/kg.Ksf < s2

*Example (cont.)Since state 2 is in the saturation region, T2 = Tsat@0.01MPa=45.8oCThe work is now easily determined since:What does the process look like on the T-s diagram?

*Process DiagramWhat does the area under the curve represent?

The area under the process curve under a T-s diagram represents heat transfer during an internally reversible process. (Recall the P-V diagram, where it represented the reversible boundary work done). On a unit mass basis:

*T-S Diagram of the Carnot CycleWhat does this diagram represent?Two isentropic processes: 4-1 and 2-3Two isothermal processes: 1-2 and 3-4What does the area within the rectangle represent?

*The Tds RelationsWrite the first law in differential form on a unit mass basis:

We obtain the first Tds relation:

*The Tds RelationsThe second relation is obtained from the definition of h:

These relations are also used in the following manner:

These relations have many uses in thermodynamics and serve as the starting point to determine changes in entropy (or U or H) for processes.

*Entropy Change of Liquids and SolidsLiquids and solids are almost always approximated as incompressible substances in engineering.

*Entropy Change of Ideal GasesFor an ideal gas

The entropy change is obtained by integration:

Using the enthalpy relation yields:

*Entropy Change of Ideal GasesThere are several ways to calculate the entropy change of ideal gases:Use a constant average value for the specific heatIf a mathematical expression is available, proceed by integration between the two states (we will ignore this method)Use the tabulated values of the entropy

*Constant Average Value of Cv or CPOn a mass basis, we obtain

On a molar basis, after multiplying by the molar mass:

*Variable Specific HeatThe function s0 is defined as:(It is called the standard state entropy.)It is only a function of temperature. Its value is zero at absolute zero temperature. Its values are tabulated as a function of temperature (Tables A-17 to A-25).Now we can express s2-s1 as:

Unlike the internal energy or the enthalpy, the entropy of an ideal gas varies with specific volume or pressure as well as temperature.

*Isentropic Processes of Ideal Gases: s2 = s1Constant Specific HeatsFrom slide 36,

*Isentropic Processes of Ideal Gases: s2=s1Constant Specific HeatsTwo other relations can be derived by using the same approach:In compact form:

*Example: air initially at 0.1 MPa, 27oC, is compressed reversibly to a final state. Find the entropy change of the air when the final state is (a) 0.5 MPa, 227oC and (b) 0.5 MPa, 180oC(c) find the temperature at 0.5 MPa that makes the entropy change zero.(d) show the processes on a T-s diagramAssume air is an ideal gas(a) Second law

*Example(b)

*Example

Why are the pressure line not horizontal as shown on the T-s diagram for water?It is a gas. The constant pressure curve is horizontal only in the saturation region. As T increases at constant pressure, S increasesWhy do the entropy changes have different signs?While the pressure contributions are the same, the temperature contributions are different. (Below 202.4oC and above 202.4oC as shown on the figure).

*Example: nitrogen expands isentropically in a piston-cylinder device from a temperature of 500 K while its volume doubles. What is the final temperature of the nitrogen, and how much work did the nitrogen do against the piston, in kJ/kg?

System: the piston/cylinder deviceProcess: isentropic expansionon T-s and P-v diagrams

*ExampleIsentropic Process

Why did the temperature decrease?The gas expands so that the internal energy decreases and thus T decreases

Now apply the first law to the system:Why is the net work positive?It is an expansion. PdV>0

*Example: a Carnot engine has 1 kg of air as the working fluid. Heat is supplied to the air at 800K and rejected by the air at 300 K. At the beginning of the heat addition process, the pressure is 0.8MPa and during the heat addition the volume triples.(a) Calculate the net cycle work assuming air is an ideal gas with constant specific heats.(b) Calculate the amount of work done in the isentropic expansion process.(c) Calculate the entropy change during the heat rejection process.

*ExampleSystemProcessProperty relations: ideal gas, constant properties(a) The net cycle work is:Wnet = hQH But h = 1-TL/THSo that we need to determine QH. How?

*Example(a) Apply the first law to process 1-2. Wnet,12 is the work done during that process.Note that now we can also determine QL. How?

*Example(b) Apply the first law, closed system to the isentropic expansion 2-3 (meaning Q23 = 0)(c)

We know T3 = T4 = TL We need to find P3 and P4How do we proceed?We know that during the heat addition the volume triples and the pressure is equal to 0.8 MPa. The temperature is 800 K. So.. We have a starting point to determine P2, from which we can determine P3 from which we get P4.

*ExampleWe get P2 from the ideal gas law applied to the isothermal process 12.We get P3 by considering process 23

*ExampleFinally we get P4 by considering process 41

*Entropy and Molecular DisorderAs a substance goes from solid to liquid to vapor, the degree of molecular disorder increases and so does the entropy. At a phase change, in particular, there is a change in entropy: entropy of fusion and entropy of vaporization, which is also an indication that the degree of disorder of the substance has increased.In the same vein, mixing of two substances is also accompanied by an entropy of mixing. The molecular disorder of an isolated system increases each time it undergoes a real process.

*The Third Law of ThermodynamicsAs the temperature of a gas is decreased, it will transform into a liquid (at the melting point). The liquid is less disordered than the gas. Upon further cooling the liquid will crystallize, a more ordered state, as the atoms form a regular three-dimensional array. As the temperature is further reduced, the vibrational activity of the atoms will continue to decrease. At absolute zero, all vibrational activity will have essentially ceased and the entropy of the substance will be zero. The entropy of a pure crystalline substance is zero at absolute zero temperature.The third law provides an absolute reference point for the determination of entropy. Absolute entropy is determined with respect to this reference point and is widely used in the analysis of chemical reactions. The entropy of substances that do not crystallize is not zero at absolute zero.

*Entropy BalanceThe entropy change of a system during a process is equal to the sum of the net entropy transfer through the system boundary and the entropy generated within the system.

*Entropy Change of a SystemThe entropy change of a system is the result of the process occurring within the system:

Entropy change = Entropy in final state Entropy in initial state

*Mechanisms of Entropy TransferEntropy can be transferred in or out of a system by two mechanisms: heat transfer and mass flow. Entropy transfer is recognized at the system boundary as it crosses the boundary, and it represents the entropy gained or lost by a system during a process.The only form of entropy transfer for a closed system is heat transfer.The entropy transfer for an adiabatic system is zero.

*Entropy Transfer by Heat TransferThe ratio of the heat transfer at a location to the absolute temperature T at that location is called the entropy flow or entropy transfer:

Note that Q and S have the same sign.When the temperature T is not constant, the entropy transfer is determined by integration or summation

Where Qk is the heat transfer through the boundary at temperature Tk at location k.

*Entropy Transfer by WorkEntropy transfer by work, Swork = 0Note the distinction between work and heat:An energy interaction that is accompanied by entropy transfer is heat transferAn energy interaction that is not accompanied by entropy transfer is work

*Entropy Transfer by Mass FlowIf the properties of the mass change during the process, the entropy transfer is determined from:Where Ac is the cross-sectional area of the flow and Vn is the local velocity normal to area dAc

*Entropy GenerationEntropy generation is a measure of the entropy created by irreversibilities (friction, etc..).For a reversible process, the entropy generation is zero. The entropy change of the system is equal to the entropy transfer.The entropy balance (kJ/K) for any system undergoing any process is expressed as:

*Rate of Entropy GenerationThe rate of entropy generation (kW/K) is obtained from the previous equation:

On a unit mass basis (kJ/kg.K):

The term sgen represents the entropy generation within the system boundary only and not the entropy generation that could occur outside the system boundary as result of external irreversibilities

*Entropy GenerationThe term Sgen is the entropy generated within the system boundary only and does not include the entropy that may be generated outside the system boundary as a result of external irreversibilities. Note that Sgen = 0 for the internally reversible process.

*Total Entropy Generated by a ProcessThe total entropy generated during a process is determined by applying the entropy balance to an isolated system that includes the system and its surroundings.

*Closed SystemsClosed system

Where Qk is the heat transfer through the boundary at temperature Tk at location k.Adiabatic closed system (Q = 0)

System + Surroundings

Where DSsystem = m(s2 - s1) and DSsurroundings = Qsurr/Tsurr

*Control VolumesThe general entropy balance is given by:

In rate form:

*Steady-flow ProcessSteadyflow control volumes experience no change in entropy. The entropy balance becomes:

For a single stream, it simplifies to:

For a single stream, steady-flow, adiabatic device, it becomes:

If the flow is also reversible, then se = si

*Example: an inventor claims to have developed a water mixing device in which 10kg/s of water at 25oC and 0.1 MPa and 0.5 kg/s of water at 100oC and 0.1 MPa are mixed to produce10.5 kg/s of water as a saturated liquid at 0.1 MPa. If the surroundings to this device are at 20oC, if this process possible? If not, what temperature must the surroundings have for this process to be possible? System: shown hereProcess: steady-flow mixingProperties: steam tables

*ExampleHow do we proceed? We must apply the second law criterion that entropy generation is positive and calculate the terms in the steady-flow entropy balance:

(well simplify this expression later for the problem at hand)How do we get the various terms? We can get the entropy transfers due to mass flow in and out of the system readily since the states of the water is fully specified. We need to determine the entropy transfer due to heat transfer: from the system to the surroundings.How do we do this? By applying the first law to the system: the steady-flow energy balance.

*ExampleWe can neglect changes in PE and KE. There is no work done. We have two inlet streams and one outlet stream.

*ExampleNote that heat is transferred from the surroundings to the system (>0).

*ExampleSince sgen

*Calculation of Ideal WorkFor a completely reversible process

Substitute in the energy balance for a steady-flow device

The shaft work is for a completely reversible process, the ideal work

In most applications in ChE, PE and KE are negligible:

*Calculation of Ideal WorkThe thermodynamic efficiency relates the actual work in the process to the ideal work

The lost work and power are defined by

*Lost WorkLost work is the work that is lost as result of irreversibilities in the system.

And since

We obtain the magnificent result:

*Lost WorkWe can also show that:

Another magnificent result

*Entropy and Molecular DisorderAs a substance goes from solid to liquid to vapor, the degree of molecular disorder increases and so does the entropy. At a phase change, in particular, there is a change in entropy: entropy of fusion and entropy of vaporization, which is also an indication that the degree of disorder of the substance has increased.In the same vein, mixing of two substances is also accompanied by an entropy of mixing. The molecular disorder of an isolated system increases each time it undergoes a real process.

*Entropy from the Microscopic ViewpointWhat is the entropy change that accompanies going from the bags of white and black atoms to the mixture below? Start from Boltzmann relationS = klnWk is Boltzmann constant

*Entropy from the Microscopic ViewpointThe number of configuration, W, of the system is obtained by calculating how many different (distinct) configurations are possible. Assume we have a lattice of N points, n black balls, N-n white balls. For the first ball I pick from the bag, I have N choices to place it on the lattice, the second ball , N-1 positions, the third ball , N-3the last ball, only one choice so that I have N! of placing N balls on the lattice.Assume now that I remove all the black balls. If I put back the first one, I have n choices, the second one, n-1.. The last one only one choice, so that the black balls have n! indistinct configurations. The same is true if I remove all the white balls , numbering N-n and I have (N-n)! Undistinguishable configurations. We conclude that

*Entropy from the Microscopic ViewpointThe math is tedious but easy:

Note that n/N and (N-n)/N are the mole fractions:

The most magnificent equation yet!! Whenever you mix two substances, you create entropy of mixing.