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- 1. Entropy Dr. Rohit Singh Lather
- 2. Introduction • It has already been proved that for a reversible cycle working between source and sink temperatures, the heat supplied and rejected to the engine is related with source and sink temperature as 𝑄1 𝑇1 = 𝑄2 𝑇2 '( )( + '+ )+ = 0 That is, the ratio of energy absorbed to the energy rejected as heat by a reversible engine is equal to the ratio of the temperatures of the source and the sink.
- 3. Replacement of a Reversible process by an equivalent process Pressure Volume Let us consider cyclic changes in a system other than heat engines. If the cycle can be split up into a large number of heat engine cycles then the above observation can be made use of in relating the heat interactions with the absolute temperatures. Any reversible process can be approximated by a series of reversible, isothermal and reversible, adiabatic processes. The same change of a state can be achieved by process Reversible Process 1-2 1-a Reversible adiabatic process a-b-c Isothermal process c-2 Reversible adiabatic process The areas 1-a-b and b-c-2 are equal U2-U1=Q1-a-b-c-2-W1-a-b-c-2
- 4. • Consider a reversible process 1-2 • The same change of a state can be achieved by - Process 1-a (reversible adiabatic process) - Isothermal process a-b-c and a reversible adiabatic process c-2 • The areas 1-a-b and b-c-2 are equal. • From the first law U2 - U1 = Q1-a-b-c-2 - W1-a-b-c-2 • Consider the cycle 1-a-b-c-2-b-1. The net work of the cycle is zero • The heat interaction along the path 1-a-b-c-2 is Q1-a-b-c-2 = Q1-a + Qa-b-c + Qc-2 = Qa-b-c Since 1-a and c-2 are reversible adiabatic paths ∮ 𝑑𝑊 = W1-a-b-c-2 + W2-b-1 = 0 W1-a-b-c-2 = - W2-b-1 = W1-b-2
- 5. .Hence, Application of the first law of the thermodynamics to the process 1-b-2 gives U2 - U1 = Q1-b-2 - W1-b-2 Comparing the two equations Qa-b-c = Q1-b-2 • The heat interaction along the reversible path 1-b-2 is equal to that along the isothermal path a- b-c • Therefore a reversible process can be replaced by a series of reversible adiabatic and reversible isothermal processes. heat transferred in the process 1 – 2 is equal to heat transferred in the isothermal process a – b • So any reversible path can be replaced by a zig-zag path between the same end states • The zig-zag path consists of a reversible adiabatic followed by a reversible isotherm and then by a reversible adiabatic in such a way that will satisfy above equation i.e., the heat transferred during the isothermal process is the same as that transferred during the original process U2 - U1 = Qa-b-c - W1-b-2
- 6. Introduction to Clausius Theorem and Clausius Inequality • When a reversible engine uses more than two reservoirs, the third or higher numbered reservoirs will not be equal in temperature to the original two • Consideration of expression for efficiency of the engine indicates that for maximum efficiency, all the heat transfer should take place at maximum or minimum reservoir temperatures - Any intermediate reservoir used will, therefore, lower the efficiency of the heat engine • Practical engine cycles often involve continuous changes of temperature during heat transfer • A relationship among processes in which these sort of changes occur is necessary • The ideal approach to a cycle in which temperature continually changes is to consider the system to be in communication with a large number of reservoirs in procession • Each reservoir is considered to have a temperature differing by a small amount from the previous one
- 7. • A given cycle may be subdivided by drawing a family of reversible, adiabatic lines • Every two adjacent adiabatic lines may be joined by two reversible isotherms • The heat interaction along the reversible path is equal to the heat interaction along the reversible isothermal path Reversible Cycle Adiabatic Lines Pressure Volume Isothermal LinesThe original reversible cycle thus is a split into a family of Carnot cycles a1-b1-d1-c1 is a Carnot cycle Qa-b = Qa1-b1 and Qc-d = Qc1-d1 Clausius Theorem
- 8. For a1-b1-d1-c1 Carnot cycle - aQ1 is added at contact temperature T1 and aQ2 is rejected at temperature T2 So we can write 𝑄1 𝑇1 = 𝑄2 𝑇2 '( )( + '+ )+ = 0 '( )( + '+ )+ + ….......= 0 '/ )/ + '0 )0 = 0 ∑ ' )2 = 0 ∮ 3' ) = 0 Clausius Theorem • The work interaction along the reversible path is equal to the work interaction along the reversible adiabatic and the reversible isothermal path
- 9. Temperature Volume Let us Consider cycle A-B-C-D - AB is a general process, either reversible or irreversible - Let the cycle be divided into number of elementary cycles as shown 𝜂 = 1 - 𝜹𝑸 𝟐 𝜹𝑸 𝜹𝑸 𝒊𝒔 𝒉𝒆𝒂𝒕 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅 𝒂𝒕 𝑻 𝜹𝑸2 𝒊𝒔 𝒉𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝒂𝒕 𝑻 𝟐 𝜂 ≤ 𝜂rev The efficiency of a general cycle is always less than the efficiency of a reversible cycle Source: P K Nag, “Engineering Thermodynamics", Mc Graw Hill Eductaion,5th Edition, 2016 1 - 𝜹𝑸 𝟐 𝜹𝑸 ≤ (1 - 𝜹𝑸 𝟐 𝜹𝑸 )rev 𝜹𝑸2 𝜹𝑸 Clausius Inequality
- 10. ( 𝜹𝑸 𝜹𝑸 𝟐 )rev = 𝑻 𝑻 𝟐 𝜹𝑸 𝜹𝑸 𝟐 ≤ 𝑻 𝑻 𝟐 𝜹𝑸 𝑻 ≤ 𝜹𝑸 𝟐 𝑻 𝟐 𝒅𝑺 = 𝜹𝑸 𝑻 = 𝜹𝑸 𝟐 𝑻 𝟐 For any process A-B For reversible process A-B 𝜹𝑸 𝑻 ≤ 𝑑𝑆 For any process AB I 𝛿𝑄 𝑇 ≤ I 𝑑𝑆 I 𝜹𝑸 𝑻 ≤ 𝟎 dS = 0 for a cyclic process Clausius Inequality For cyclic process We know that Replacing for reversible - 𝜹𝑸 𝟐 𝜹𝑸 ≤ - ( 𝜹𝑸 𝟐 𝜹𝑸 )rev 𝜹𝑸 𝟐 𝜹𝑸 ≥ ( 𝜹𝑸 𝟐 𝜹𝑸 )rev 𝜹𝑸 𝜹𝑸 𝟐 ≤ ( 𝜹𝑸 𝜹𝑸 𝟐 )rev
- 11. 𝜕Q Reservoir at Variable Temperature T Reversible Heat Engine Reservoir at Temperature Tres Heat rejected 𝜕Q’ 𝜕W’ 𝜕W engine delivers a small amount of work Thermal reservoir at Tres which delivers a small amount of heat δQ′ to a reversible cyclic engine reservoir itself delivers a different small amount of work δW to the surroundings Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
- 12. The first law in differential form for the combined system is δQ is internal and so does not cross the boundary of the combined system and is not present in our first law formulation Rearrange Replace δQ’ Let this configuration undergo a thermodynamic cycle Apply the Kelvin-Planck form of the second law of thermodynamicsW + 𝑊O ≤ 0 We cannot convert all the heat to work, but we can convert all the work to heat Q'O Q' = )RST ) Q'O )RST = Q' ) dE = (𝛿𝑄′) − (𝛿𝑊 + 𝛿𝑊′) (𝛿𝑊 + 𝛿𝑊′) = (𝛿𝑄′) − dE (𝛿𝑊 + 𝛿𝑊′) = (Tres Q' ) ) − dE (∮ 𝛿𝑊 + ∮ 𝛿𝑊′= ∮Tres Q' ) − ∮dE W + 𝑊′= Tres ∮ Q' ) Tres ∮ Q' ) ≤ 0 Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
- 13. This is known as Clausius Inequality And since Tres > 0, we can divide above equation by it, without changing the sense of the inequality to get a mathematical representation of the second law of thermodynamics This is known as Clausius Inequality for a reversible process This is known as Clausius Inequality for a irreversible process ∮ Q' ) ≤ 0 I 𝛿𝑄 𝑇 ≤ 0 I 𝛿𝑄 𝑇 = 0 Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition For every Carnot cycleI 𝛿𝑄 𝑇 = 0
- 14. Whenever a system undergoes a cyclic change, however complex the cycle may be (as long as it involves heat and work interactions), the algebraic sum of all the heat interactions divided by the absolute temperature at which heat interactions are taking place considered over the entire cycle is less than or equal to zero (for a reversible cycle) Source: P K Nag, “Engineering Thermodynamics", Mc Graw Hill Eductaion,5th Edition, 2016 • The integral symbol with a circle in the middle is used to indicate that the integration is to be performed over the entire cycle • Any heat transfer to or from a system can be considered to consist of differential amounts of heat transfer • Then the cyclic integral of dQ/T can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature at the boundary
- 15. Entropy Path-Independent Thermodynamic Property Sketch of P − V diagram for various combinations of processes forming cyclic integrals • Consider starting from 1, proceeding on path A to 2, and returning to 1 via path B. • The cyclic integral δQ/T =0 decomposes to • Perform the same exercise going from 1 to 2 on path A and returning on path C, yielding Volume Pressure 1 2 C B A A, B, & C are arbitrary processes between state 1 and state 2 (∫ Q' ) + ( )A + (∫ Q' ) ) ( + B = 0 (∫ Q' ) + ( )A + (∫ Q' ) ) ( + C = 0 Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
- 16. We can reverse direction and recover the same result • Since paths B and C are different and arbitrary, but ∫ δQ/T is the same on either path, the integral must be path-independent. • It therefore defines a thermodynamic property of the system. • We define that property as entropy, S, an extensive thermodynamic property (∫ Q' ) ) ( + B - (∫ Q' ) ) ( + C = 0 (∫ Q' ) ) ( + B = (∫ Q' ) ) ( + C (∫ Q' ) ) + ( B = (∫ Q' ) ) + ( C S2 – S1 = ∫ Q' ) + ( Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
- 17. scale by the constant mass m to get the corresponding intensive variable s = S/m Integrating Differential Form This is the heat transfer equivalent to s2 – s1= ∫ Q' ) + ( 𝛿𝑠 = 𝛿𝑞 𝑇 𝛿𝑞 = 𝑇. 𝛿𝑠 ] 𝛿𝑞 + ( = ] 𝑇. 𝛿𝑠 + ( 1q2 = ∫ 𝑇. 𝛿𝑠 + ( 1w2= ∫ 𝑇. 𝛿𝑠 + ( Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
- 18. The entropy change between two specific states is the same whether the process is reversible or irreversible Isentropic = Adiabatic + Reversible The heat transfer for a process from 1 to 2 is given by the area under the curve in the T − s plane • If a process lies on a so-called Isentrope: a line on which entropy s is constant, then, 1q2 = 0; thus, the process is adiabatic • Above equation only applies for a reversible process Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 19. Example : For a particular power plant, the heat added and rejected both occur at constant temperature and no other processes experience any heat transfer. The heat is added in the amount of 3150 kJ at 440oC and is 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 • Since the inequality is less than zero, the cycle has at least one irreversible process and the cycle is irreversible Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 20. Example: For a particular power plant, the heat added and rejected both occur at constant temperature; no other processes experience any heat transfer. The heat is added in the amount of 3150 kJ at 440oC and is 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 • Since the cyclic integral is equal to zero, the cycle is made of reversible processes Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 21. Second Law in terms of Entropy – Entropy Change in an Irreversible Process Consider the cycle in the T − S diagram as shown - We start at 1, and proceed to 2 along path I, which represents an irreversible process - We return from 2 to 1 along path R, which represents a reversible process Entropy Temperature 1 2 I R I 𝛿𝑄 𝑇 ≤ 0 S2 – S1 = ∫ Q' ) + ( Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA Statement of second law valid for reversible and irreversible heat transfer Definition of Entropy provided the heat transfers are reversible S2 – S1 = ∫ Q' ) + ( For a reversible process
- 22. S1 – S2 = ∫ Q' ) ( + Since the process is reversible, we reverse and get Now, apply the second law 0 ≥ (∫ Q' ) + ( )I + (∫ Q' ) ) ( + R Now, substitute, to eliminate the integral along R to get 0 ≥ (∫ Q' ) + ( )I + S1 –S2 S2 – S1 ≥ (∫ Q' ) + ( )I More generally, we can write the second law of thermodynamics as, S2 – S1 ≥ ∫ Q' ) + ( If 2 → 1 is reversible, the equality holds; if 1 → 2 is irreversible, the inequality holds
- 23. • For an isolated system, there can be no heat transfer interactions and 𝜕𝑄 = 0, So S2 − S1 ≥ 0, - This implies 2 occurs later in time than 1 - Thus, for isolated systems, the entropy increases as time moves forward Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA Isolated system Example: Two large thermal reservoirs, one at TA and the other at TB, exchange a finite amount of heat Q with no accompanying work exchange. The reservoirs are otherwise isolated and thus form their own universe, when considered as a combined system. Consider the implications for entropy and the second law. A TA Heat transfer from A to B B TB Q
- 24. • Assume for now that positive Q leaves A and enters B • Both A and B are so massive that the respective loss and gain of thermal energy does not alter their respective temperatures. Consider the entropy changes for each system: Now, our universe is the combination of A and B, so the entropy change of the universe is found by adding the entropy changes of the components of the universe SA2 – SA1 = ∫ Q' ) + ( = ( )^ ∮ 𝛿𝑄 = − ' )^ + ( SB2 – SB1 = ∫ Q' ) + ( = ( )_ ∮ 𝛿𝑄 = ' )_ + ( (SA2 + SB2) – (SA1 + SB1) = ' )_ + ' )^ = SU2 = SU1 Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA With the universe entropy SU as SU = SA + SB , we get SU2 – SU1 = Q ( ( )_ - ( )^ )
- 25. The universe is isolated, so the second law holds that SU2 − SU1 ≥ 0; thus, Now, we have assumed Q > 0; therefore, we can divide by Q without changing the sense of the inequality: • We have thus confirmed that our mathematical formulation of the second law in terms of entropy yields a result consistent with the Clausius statement of the second law. • We must have TA ≥ TB in order to transfer positive heat Q from A to B Q ( ( )_ - ( )^ ) ≥ 0 ( )_ - ( )^ ≥ 0 )^` )_ )^)_ ≥ 0 𝑇 𝐴 − 𝑇 𝐵 ≥ 0 𝑇 𝐴 ≥ 𝑇 𝐵 Since TA > 0 and TB > 0, we can multiply both sides by TATB without changing the sense of the inequality to get
- 26. 6-3 The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero Entropy Change of an Isolated System “The entropy of an isolated system either increases or, in the limit, remains constant.” The principle of entropy increase: The entropy of an isolated system can never decrease. It always increases with every irreversible process and remains constant, only when the process is reversible. dSsys = `Q' )TcT dSsur = Q' )TdR dSiso = dSsys + dSsur = 𝛿𝑄 [ ( )TdR - ( )TcT ] > 0 The entropy of an isolated system either increases if it undergoes an irreversible process System Tsys Surroundings Tsur SQ If, Tsys = Tsur, the process would occur reversibly and entropy will remain constant Source: D. S. Kumar, “Engineering Thermodynamics", S.K. Kataria & Sons, 2nd Edition
- 27. • Second law of thermodynamics while applied to a process give rise to a property, called entropy • Absolute entropy (S) or entropy is a measure of energy dispersion in a system • Following are basic features of entropy: - Entropy transfer is associated with heat transfer - Entropy is proportional to mass flow in an open system - Entropy does not transfer with work Entropy Heat to Work Work to Work Work to Heat Entropy Transfer No Entropy Transfer No Entropy Transfer; Entropy Generates
- 28. • If no irreversibilities occur within the system as well as the reversible cyclic device, - Then the cycle undergone by the combined system will be internally reversible - As such, it can be reversed • In the reversed cycle case, all the quantities will have the same magnitude but the opposite sign - Therefore, the work WC, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case - Then it follows that WC,int rev = 0 since it cannot be a positive or negative quantity, and therefore for internally reversible cycles • Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 29. • Entropy change of the system will have the same sign as the heat transfer in a reversible process • Processes can occur in a certain direction only, not in just any direction, such that Sgen ≥ 0 • Entropy is a non-conserved property, and there is no such thing as the conservation of entropy principle • The entropy of the universe is continuously increasing, as all the processes happening in the universe are irreversible • 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 that process • Entropy change is caused by heat transfer and irreversibilities Some Remarks about Entropy Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 30. • Heat transfer to a system increases the entropy; heat transfer from a system decreases it • The effect of irreversibilities is always to increase the entropy - In fact, a process in which the heat transfer is out of the system may be so irreversible that the actual entropy change is positive (Friction is one source of irreversibilities in a system) • The entropy change during a process is obtained by integrating the dS equation over the process • The inequality is to remind us that the entropy change of a system during an irreversible process is always greater than ∮ 𝛿𝑄/𝑇 + ( , called the entropy transfer - That is, some entropy is generated or created during an irreversible process - Generation is due entirely to the presence of irreversibilities. • The entropy generated during a process is called entropy generation and is denoted as Sgen Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 31. • We can remove the inequality by noting the following - Sgen is always a positive quantity or zero - Its value depends upon the process and thus it is not a property - Sgen is zero for an internally reversible process • The integral ∮ 𝛿𝑄/𝑇 + ( is performed by applying the first law to the process to obtain the heat transfer as a function of the temperature. The integration is not easy to perform, in general ∆𝑆 𝑠𝑦𝑠 = 𝑆2 − 𝑆1 = I 𝛿𝑄 𝑛𝑒𝑡 𝑇 + + ( 𝑆 𝑔𝑒𝑛 Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
- 32. 6-9 Entropy Temperature 1 2 dA = TdS = 𝜕Q Internally Reversible Process Heat Transfer as the Area under a T-S Curve For the reversible process, the equation for dS implies that or the incremental heat transfer in a process is the product of the temperature and the differential of the entropy, the differential area under the process curve plotted on the T-S diagram
- 33. 6-9 Heat Transfer for a Internally Reversible Process On a T-S diagram, the area under the process curve represents the heat transfer for internally reversible processes Entropy Temperature 1 2 dA = TdS = 𝜕Q Internally Reversible Process Adiabatic Process Temperature Entropy Isothermal Process Temperature Entropy 𝛿𝑄𝑟𝑒𝑣 = T∫ 𝑑𝑆 = 𝑇 + ( (S2 − S1)
- 34. Volume Pressure Reversible Adiabatic Processes Heat exchange with a single reservoir in the process AB Violates Kelvin Plank Statement • For a reversible adiabatic process, heat interactions are zero so change in entropy is zero • Two adiabatic reversible paths never intersect each other • Entropy is a property and a point function • Through one point, there can only pass one reversible adiabatic line • If, two adiabatic reversible paths intersect each other, then entropy will have two values at that point, this violates Kalvin Planks statement (engine which exchanges heat with a single reservoir)
- 35. 6-11 For adiabatic steady-flow devices, the vertical distance is ∆h on an h-s diagram, and the horizontal distance is ∆s h-s Diagram for Adiabatic Steady Flow Devices Entropy Enthalpy(h) 1 2 ∆h ∆s measure of work measure of irreversibilities
- 36. • Temperature-entropy and enthalpy-entropy diagrams for water • The h-s diagram, called the Mollier diagram, is a useful aid in solving steam power plant problems. Constant Temperature Lines Steam Quality Lines Super-heated Vapor Steam Region Mixed Region of vapor and liquid • The Mollier diagram shows only two regions, the mixed region of vapor and liquid and the super-heated vapor steam region. • The two regions are separated by the downward sloping saturation line, where steam quality is equal to 1
- 37. Constant Volume and Constant Pressure Lines on T-S Diagram 3) 3o = ) pq 3) 3o = ) pr ds= Cv 3) ) ds= Cp 3) ) As Cp > Cv, Constant volume lines on T-s diagram is steeper than constant pressure line, working between the same temperature limits 3) 3o 𝛼 𝑇 For both Constant volume and constant pressure Constant volume Constant pressure Slope of constant volume line on T-S diagram Slope of constant pressure line on T-S diagram
- 38. The level of molecular disorder (entropy) of a substance increases as it melts and evaporates Level of Molecular Disorder (Entropy) During a heat transfer process, the net disorder (entropy) increases Net disorder decreases during heat transfer the increase in the disorder of the cold body more than offsets the decrease in the disorder in the hot body
- 39. • The third law of thermodynamics states that “The entropy of a pure crystalline substance at absolute zero temperature is zero.” • This law provides an absolute reference point for the determination of entropy • The entropy determined relative to this point is called absolute entropy Third Law of Thermodynamics

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