Chapter 7 Entropy

Objectives Apply the second law of thermodynamics to processes. Define a new property called entropy to quantify the second-law effects. Establish the increase of entropy principle. Calculate the entropy changes that take place during processes for pure substances, incompressible substances, and ideal gases. Examine a special class of idealized processes, called isentropic processes, and develop the property relations for these processes. Derive the reversible steady-flow work relations. Develop the isentropic efficiencies for various steady-flow devices. Introduce and apply the entropy balance to various systems.

Introduction The second low often leads to expressions that involve inequalities. One more important inequality in thermodynamics: Clausius Inequality (1888)

The Inequality of Clausius For all cycle reversible or irreversible cycle The inequality of Clausius is a consequence of the second law of thermodynamics. Q is the heat transfer to or from the system. T is the absolute temperature at the boundary. The symbol is the cyclic integral

The Inequality of Clausius Energy balance to the combined system: let the system undergo a cycle while the cyclic device undergoes an integral number of cycles. the combined system is exchanging heat with a single thermal energy reservoir while involving (producing or consuming) work WC during a cycle. This violating the Kelvin–Planck statement of the second law, which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy Reservoir Then WC cannot be a work output, and thus it cannot be a positive quantity. Considering that TR is the thermodynamic temperature and thus a positive quantity, we must have the Clausius inequality.

The Inequality of Clausius

The Inequality of Clausius

The cyclic integral The cyclic integral indicates that the integral should be performed over the entire cycle and over all parts of the boundary.

The cyclic integral

Derivation of Clausius Inequality Irreversible Reversible Heat Engine Refrigeration

The cyclic integral of Reversible Heat Engine Since

The cyclic integral of Irreversible Heat Engine We cannot use this It is Irreversible Produced work is larger For same QH Since it was zero for reversible when QL rev.<QL irreversible

The cyclic integral of Reversible Refrigeration Since For reversible Refrigerator

The cyclic integral of Irreversible Refrigeration We cannot use this It is Irreversible Consumed work is larger

Derivation of Clausius Inequality Irreversible Reversible < 0 0 = Heat Engine Refrigeration The equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones.

The Clausius inequality gives the basis for two important ideas Entropy (S) Entropy generation (Sg) These two terms gives quantitative evaluations for systems from second law perspective.

Derivation of Entropy (Reversible Process) For reversible cycle A-B For reversible cycle C-B All paths are arbitrary Subtracting gives Since paths A and C are arbitrary, it follows that the integral of Q/T has the same value for ANY reversible process between the two sates.

Derivation of Entropy (Reversible Process) Clausius chose the name “Entropy” Entropy (the unit) S = entropy (kJ/K); s = specific entropy (kJ/kg K) S2 – S1 depends on the end states only and not on the path,  it is same for any path For Reversible Process

Entropy is a Property The entropy change between two specified states is the same whether the process is reversible or irreversible.

A Special Case: Internally Reversible Isothermal Heat Transfer Processes Recall that isothermal heat transfer processes are internally reversible. Therefore, the entropy change of a system during an internally reversible isothermal heat transfer process can be determined by performing the integration in Eq. 7–5:

Derivation of Entropy (Irreversible Process) Consider 2 cycles AB is reversible and CB is irreversible

Derivation of Entropy (Any Process) This can be written out in a common form as an equality or 2nd law of thermodynamics for a closed system Entropy Balance Equation for a closed system Recall: for cycle In any irreversible process always entropy is generated (Sgen > 0) due to irreversibilities occurring inside the system.

Example (7-1) Entropy change during isothermal process. A friction-less piston-cylinder device contains a liquid-vapor mixture of water at 300 K. During a constant pressure process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of the water during this process. Solution: This is simple problem. No irreversibilities occur within the system boundaries during the heat transfer process. Hence, the process is internally reversible process (Sg = 0).

We computed the entropy change for a system using the RHS of the equation. But we can not get easy form each time. So, we need to know how to evaluate the LHS which is path independent.

Entropy change for different substances (S = S2-S1) We need to find how to compute the left hand side of the entropy balance for the following substances: Pure substance like water, R-134, Ammonia etc.. Solids and liquids Ideal gas

1- S for Pure Substances The entropy of a pure substance is determined from the tables, just as for any other property These values were tabulated after conducting a tedious integration. These values are given relative to an arbitrary reference state. For water its assigned zero at 0.01 C. For R-134 it is assigned a zero at -40 C. Entropy change for a closed system with mass m is given as expected:

EXAMPLE 7–3 Entropy Change of a Substance in a Tank A rigid tank contains 5 kg of refrigerant-134a initially at 20°C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

The volume of the tank is constant so V1=V2

Important Note: The negative sign indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation Sgen that cannot be negative.

7-2: The increase of entropy principle (closed system) There is some entropy generated during an irreversible process such that Entropy change Entropy generation due to irreversibility Entropy transfer with heat This is the entropy balance for a closed system.

The increase of entropy principle (closed system) The entropy change can be evaluated independently of the process details. However, the entropy generation depends on the process, and thus it is not a property of the system. The entropy generation is always a positive quantity or zero and this generation is due to the presence of irreversibilities. The direction of entropy transfer is the same as the direction of the heat transfer: a positive value means entropy is transferred into the system and a negative value means entropy is transferred out of the system.

The increase of entropy principle (closed system) For an isolated (or simply an adiabatic closed system), the heat transfer is zero, then This means that the entropy of an adiabatic system during a process always increases or, In the limiting case of a reversible process, remains constant. In other words, it never decreases. This is called Increase of entropy principle. This principle is a quantitative measure of the second law.

The increase of entropy principle Now suppose the system is not adiabatic. We can make it adiabatic by extending the surrounding until no heat, mass, or work are crossing the boundary of the surrounding. This way, the system and its surroundings can be viewed again as an isolated system. The entropy change of an isolated system is the sum of the entropy changes of its components (the system and its surroundings), and is never less than zero. Now, let us apply the entropy balance for an isolated system:

Summary of the increase of entropy principle Entropy is an extensive property, and thus the total entropy of a system is equal to the sum of the entropies of the parts of the system.

Important Remarks Processes can occur in a certain direction only , not in any direction. A process must proceed in the direction that complies with the increase of entropy principle. A process that violates this principle is impossible. Entropy is a non-conserved property. Entropy is conserved during the idealized reversible process only and increases during all actual processes. The performance of engineering systems is degraded by the presence of irreversibilities, and the entropy generation is a measure of the magnitude of the irreversibilities present during a process. Entropy generation is used to establish criteria for the performance of engineering devices

Example (7-2): entropy generation during heat transfer processes A heat source at 800 K losses 2000 kJ of heat to a sink at (a) 500 K and (b) 750 K. Determine which heat transfer process is more irreversible. Solution: Both cases involve heat transfer via a finite temperature difference and thus are irreversible. Each reservoir undergoes an internally reversible isothermal process (constant temperature).

Take the two reservoirs as your system Take the two reservoirs as your system. Thus they form an adiabatic system and thus Now consider each system alone (isothermal with no irreversibility)

Let us repeat the same with case b. Now consider each system alone Hence the case b involves less irreversibility. The smaller the temperature difference , the smaller the irreversibility ( if DT =0 no irrevocability due heat transfer)

Where does the irreversibility arise from? Tsource > T sink

Where is entropy generated? Finite Temperature difference cause the entropy to generate

What is Entropy? (microscopic point of view) Entropy can be viewed as a measure of molecular disorder, or molecular randomness. As a system becomes more disordered, the positions of the molecules become less predictable and the entropy increases. The entropy of a substance is lowest in the solid phase and highest in the gas phase. In the solid phase, the molecules of a substance continually oscillate about their equilibrium positions, but they cannot move relative to each other, Solid molecules position at any instant can be predicted with good certainty. In the gas phase, however, the molecules move about at random, collide with each other, and change direction, It is extremely difficult to predict accurately the microscopic state of a gas phase system at any instant. Associated with this molecular chaos is a high value of entropy The entropy of a system is related to the total number of possible microscopic states of that system, called thermodynamic probability p, by the Boltzmann relation, expressed as

What is Entropy? From a microscopic point of view, the entropy of a system increases whenever the molecular randomness or uncertainty (i.e., molecular probability) of a system increases. entropy is a measure of molecular disorder, and the molecular disorder of an isolated system increases anytime it undergoes a process.

What is Entropy? The molecules of a substance in solid phase continually oscillate, creating an uncertainty about their position. These oscillations, however, fade as the temperature is decreased, and the molecules supposedly become motionless at absolute zero. This represents a state of ultimate molecular order (and minimum energy). The entropy of a pure crystalline substance at absolute zero temperature is zero since there is no uncertainty about the state of the molecules at that instant, This statement is known as the third law of thermodynamics

What is Entropy? The third law of thermodynamics provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called absolute entropy, and it is extremely useful in the thermodynamic analysis of chemical reactions. Notice that the entropy of a substance that is not pure crystalline (such as a solid solution) is not zero at absolute zero temperature. This is because more than one molecular configuration exists for such substances, which introduces some uncertainty about the microscopic state of the substance. Disorganized energy does not create much useful effect, no matter how large it is.

Isentropic Processes We mentioned earlier that the entropy of a fixed mass can be changed by (1) heat transfer and (2) irreversibilities. Then it follows that the entropy of a fixed mass does not change during a process that is internally reversible and adiabatic A process during which the entropy remains constant is called an isentropic process.

Property diagrams involving entropy In the second-law analysis, it is very helpful to plot the processes on T-s and h-s diagrams for which one of the coordinates is entropy. Recall the definition of entropy Property diagrams serves as great visual aids in the thermodynamic analysis of process. We have used P-v and T-v diagrams extensively in conjunction with the first law of thermodynamics.

Thus This area has no meaning for irreversible processes! It can be done only for a reversible process for which you know the relationship between T and s during a process. Let us see some of them.

Isothermal internally reversible process. Temperature Entropy Isothermal Process 1 2 Q

Adiabatic internally reversible process In this process Q =0, and therefore the area under the process path must be zero. This process on a T-s diagram is easily recognized as a vertical-line. Temperature Entropy 1 2 Q=0 Isentropic Process

T-s Diagram for the Carnot Cycle 2 Isothermal expansion 1 Isentropic compression W=Qin-Qout Qin 3 Isentropic expansion Temperature, T The area inside the box is both the net heat transferred, and the net work Q-W=0 Isothermal compression 4 Qout Entropy, S

Another important diagram is the h-s Diagram This diagram is important in the analysis of steady flow devices such as turbines. In analyzing the steady flow of steam through an adiabatic turbine, for example, The vertical distance between the inlet and the exit states (h) is a measure of the work output of the turbine, The horizontal distance (s) is a measure of the irreversibilities associated with the process.

Example 7-5: Isentropic Expansion of a Steam in Turbine Steam enters an adiabatic turbine at 5 MPa and 450°C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

Example 7-6: T-ds Diagram of Carnot cycle Show the Carnot cycle on a T-S diagram and indicate the areas that represent, the heat supplied QH, heat rejected QL, and the net work output Wnet,out on this diagram. 1) On a T-S diagram, the area under the process curve represents the heat transfer for that process 2) the area A12B represents QH 3) the area A43B represents QL, the difference between these two

Property Relationships

The T-ds relations: Apply the differential form of the first law for a closed stationary system for an internally reversible process

This equation is known as: Divide by the mass, you get This equation is known as: First Gibbs equation or First T-ds relationship Divide by T, .. Although we get this form for internally reversible process, we still can compute Ds for an irreversible process. This is because S is a point function.

Second T-ds (Gibbs) relationship Recall that… Take the differential for both sides Rearrange to find du Substitute in the First Tds relationship Second Tds relationship, or Gibbs equation

Thus We have two equations for ds Divide by T, .. Thus We have two equations for ds To find Ds, we have to integrate these equations. Thus we need a relation between du and T (or dh and T). Now we can find entropy change (the LHS of the entropy balance) for liquids and solids

The T ds relations are valid for both T-ds Eq. 1 &2 are relations between the properties of a unit mass of a simple compressible system as it undergoes a change of state, Even though they were developed for internally reversible process THEN The T ds relations are valid for both reversible and irreversible processes and for both closed and open systems.

7-8: Entropy Change of Liquids and Solids Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use. Thus For solids and liquids Recall also that For solids and liquids,

Only true for solids and liquids!! Integrate to give… What if the process is isentropic? The only way this expression can equal 0 is if, Hence, for solids and liquids (incompressible) , isentropic processes are also isothermal.

Example(7-7): Effect of Density of a Liquid on Entropy Liquid methane is commonly used in various applications. The critical temperature of methane is 191 K (or -82oC), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa using actual data for methane and approximating liquid methane as an incompressible substance. What is the error involved in the later case?

Error % = 12.2

Example (7-19): Entropy generated when a hot block is dropped in a lake A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water. Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine: (a) The entropy change of the iron block, (b) The entropy change of the water lake, (c) the entropy generated during this process.

Tsurr= 285 K (a) The entropy change of the iron block, we need also to find Q coming out of the system. Tsurr= 285 K (b) The entropy change of the water lake, T=500K

(c) the entropy generated during this process. Choose the iron block and the lake as the system and treat it is an isolated system. Tsurr= 285 K T=500K Thus Sg = Stot = Ssys + Slake Sg = Stot = -12.6 + 16.97 = 4.32 System boundary

3- The Entropy Change of Ideal Gases, first relation The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/ into Tds relations: integrating

Second relation integrating A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and /T= R/P into Tds relations: integrating

The integration of the first term on the RHS can be done via two methods: Assume constant Cp and constant Cv (Approximate Analysis) Evaluate these integrals exactly and tabulate the data (Exact Analysis)

Method 1: Constant specific heats (Approximate Analysis) First relation Only true for ideal gases, assuming constant heat capacities Second relation Only true for ideal gases, assuming constant heat capacities

Ru is the universal gas constant Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass kJ/kmol. K kJ/kmol. K Ru is the universal gas constant

Method 2: Variable specific heats (Exact Analysis) We use the second relation We could substitute in the equations for Cv and Cp, and perform the integrations Cp = a + bT + cT2 + dT3 But this is time consuming. Someone already did the integrations and tabulated them for us (table A-17) They assume absolute 0 as the starting point

The integral is expressed as: Where is tabulated in Table A-17 Therefore

Is s = f (T) only? like u for an ideal gas. Let us see Temperature dependence Pressure dependence From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume. The function s° represents only the temperature-dependent part of entropy

How about the other relation We can develop another relation for the entropy changed based on the above relation but this will require the definition of another function and tabulating it which is not practical.

Isentropic Processes The entropy of a fixed mass can be changed by Heat transfer, Irreversibilities It follows that the entropy of a system will not change if we have Adiabatic process, Internally reversible process. Therefore, we define the following:

Isentropic Processes of Ideal Gases Many real processes can be modeled as isentropic Isentropic processes are the standard against which we should measure efficiency We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids

Constant specific heats (1st relation) Recall For the isentropic case, DS=0. Thus Recall also from ch 2, the following relations..… Only applies to ideal gases, with constant specific heats

Constant specific heats (2nd relation) Recall..… or Only applies to ideal gases, with constant specific heats

Since… Third isentropic relationship and HENCE Which can be simplified to… Third isentropic relationship

3- Constant specific heats Full form of Isentropic relations of Ideal Gases Compact form Valid for only for 1- Ideal gas 2- Isentropic process 3- Constant specific heats

That works if the specific heat constants can be approximated as constant, but what if that’s not a good assumption? We need to use the exact treatment This equation is a good way to evaluate property changes, but it can be tedious if you know the volume ratio instead of the pressure ratio

s20 is only a function of temperature!!! Rename the exponential term as Pr , (relative pressure) which is only a function of temperature, and is tabulated on the ideal gas tables ( for air as an ideal gas see table A-17)

You can use either of the following 2 equations This is good if you know the pressure ratio but how about if you know only the volume ratio In this case, we use the ideal gas law where Remember, these relationships only hold for ideal gases and isentropic processes tabulated in A-17

Example (7-10): Isentropic Compression of Air in a Car Engine Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this piston-cylinder device is 8, determine the final temperature of the air. <Answer: 662.7 K> Sol:

Assuming constant specific heat and assuming k we can have alternative solution The k value at this anticipated average temperature is determined from Table A–2b to be 1.391

Reversible Steady Flow Work

For the pump the specific volume is constant so h2-h1 = v (p2-p1)

For compressor the specific volume is function of pressure then we use h2-h1 instead of v (p2-p1) which is only used for a pump

7-10: Reversible Steady Flow Work the energy balance for a steady-flow device If the changes in kinetic and potential energies are negligible, Then

Reversible work relations for steady flow and closed systems.

7–11 MINIMIZING THE COMPRESSOR WORK When the changes in kinetic and potential energies are negligible, the compressor work is given by effect of cooling during the compression process ( PVn = Constant) an isentropic process (involves no cooling), n =k a polytropic process (involves some cooling), 1<n<k an isothermal process (involves maximum cooling). n =1 One common way of cooling the gas during compression is to use cooling jackets around the casing of the compressors.

7–11 MINIMIZING THE COMPRESSOR Work Win decrease

Multistage Compression with Intercooling cooling a gas as it is compressed is desirable since this reduces the required work input to the compressor. often it is not possible to have adequate cooling through the casing of the compressor, effective cooling. One such technique is multistage compression with intercooling, The gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler. Ideally, the cooling process takes place at constant pressure, and the gas is cooled to the initial temperature T1 at each intercooler.

wcomp I,in = wcomp II,in. 4/2=2 4-2=2 8/4=2 8-4=4 The Px value that minimizes the total work is determined by differentiating this expression with respect to Px and setting the resulting expression equal to zero. to minimize compression work during two-stage compression, the pressure ratio across each stage of the compressor must be the same. wcomp I,in = wcomp II,in.

ENTROPY BALANCE The property entropy is a measure of molecular disorder or randomness of a system, The second law of thermodynamics states that entropy can be created but it cannot be destroyed or decreased.

Entropy Change of a System

Mechanisms of Entropy Transfer, Sin and Sout Entropy can be transferred to or from system by two mechanisms: heat transfer and mass flow Entropy transfer by heat transfer: Recall: Entropy Generated by heat & Irreversabilities The quantity Q/T represents the entropy transfer accompanied by heat transfer, and the direction of entropy transfer is the same as the direction of heat transfer since thermodynamic temperature T is always a positive quantity.

Mechanisms of Entropy Transfer, Sin and Sout When the temperature T is not constant, the entropy transfer during a process 1-2 can be determined by integration (or by summation if appropriate) as work is entropy-free, and no entropy is transferred by work. Energy is transferred by both heat and work, whereas entropy is transferred only by heat. That is, where Qk is the heat transfer through the boundary at temperature Tk at location k.

Entropy Transfer by mass flow Mass contains entropy as well as energy, and the entropy and energy contents of a system are proportional to the mass. entropy and energy are carried into or out of a system by streams of matter, and the rates of entropy and energy transport into or out of a system are proportional to the mass flow rate

Entropy Generation , Sgen Irreversibilities such as friction, mixing, chemical reactions, heat transfer through a finite temperature difference, unrestrained expansion, nonquasiequilibrium compression, or expansion always cause the entropy of a system to increase Entropy generation is a measure of the entropy created by such effects during a process. For a reversible process (a process that involves no irreversibilities), the entropy generation is zero and thus the entropy change of a system is equal to the entropy transfer.

Entropy balance for any system undergoing any process or, in the rate form where the rates of entropy transfer by heat transferred at a rate of And mass flowing at a rate of

Mechanisms of entropy transfer for a general system.

Entropy Generation – Closed System The entropy change of a closed system during a process is equal to the sum of the net entropy transferred through the system boundary by heat transfer and the entropy generated within the system boundaries ( No mass flow). For an adiabatic process (Q = 0) then, any closed system and its surroundings can be treated as an adiabatic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts, the entropy balance for a closed system and its surroundings is where and

Entropy Generation – Closed System For an adiabatic process (Q = 0) then, any closed system and its surroundings can be treated as an adiabatic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts, the entropy balance for a closed system and its surroundings is where and

Entropy Generation – Open System or Control Volume In the rate form The rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer, the net rate of entropy transfer into the control volume by mass flow, and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities.

Entropy Generation – Open System or Control Volume turbines, compressors, nozzles, diffusers, heat exchangers, pipes, and ducts operate steadily, reversible and adiabatic, then the entropy remains constant, se = si Reversible adiabatic called Isentropic process

Example (7-17): Entropy Generation in a Wall Steady heat transfer in a wall. The temperatures are shown in the figure. The rate of heat transfer through the wall is 1035 W. Outside Surface temperature = 5 C Inside Surface Temperature = 20 C Outside T=0 C Inside home T= 27 C Q = 1035 W

Example (7-17): Entropy Generation in a Wall-continue The rate form of the entropy balance for the wall

This is a steady state problem. A) Determine the rate of entropy generation in the wall This is a steady state problem. We have to set the CV correctly. For part a of the question, the CV boundary will be as shown in the figure below. CV Tin= 20 C =293 K Tout= 5 C =278 K

B) Determine the rate of total entropy generation for the process We will extend the CV boundary as follows (system+ immediate surrounding CV Boundary Tin= 27 C =300 K Tout= 0 C =273 K = Sgen, total

Special case: Sgen for closed system with constant temperature surroundings Tsurr

Example: water T1= 100 C P1 = 5 Mpa m = 2 kg s= ? A piston/cylinder contains 2 kg of water at 5 MPa, 100°C. Heat is added from a reservoir at 700°C to the water until it reaches 700°C. Find the work, heat transfer, and total entropy production for the system and surroundings. Solution This is a constant pressure process. Hence the work is , W = mP(v2-v1) To get the heat, Q-W=m(u2-u1) =>Q=m(h2-h1)

To get the entropy change for the system, s=m(s2-s1) To get the total entropy production for the system and the surrounding, we apply the entropy balance equation for the extended system (system + the immediate surrounding). So let us begin our solution. State 1 is fixed. Go to the tables and get the following

State 2 is fixed also since the pressure is constant (P2=P1). Go to the tables and get the following

Example 7-18

Example 7-18

Example 7-18

Example 7-20

Example 7-20

Some Remarks about Entropy Processes can occur in certain direction only, a direction that complies with the increase of entropy. Entropy is a non-conserve property. Entropy is conserved during the idealized reversible process only and increasing during all actual processes. The greater the extent of the irreversibilities, the greater the entropy generation. Therefore, it can be used as a quantitative measure of irreversibilities.