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### Ch07a entropy

• 1. 1 Entropy: A measure of Disorder 7CHAPTER
• 2. 7.1. Entropy and the Clausius Inequality 2
• 3. 3 Introduction The second law often leads to expressions that involve inequalities. , , th rev th th rev η η η <  = ,th th revη η≤
• 4. 4 The Inequality of Clausius  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 0≤∫ T Qδ ∫
• 5. 5 The cyclic integral  The cyclic integral indicates that the integral should be performed over the entire cycle and over all parts of the boundary. Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫
• 6. 6 The cyclic integral 0 0H L H L Q Q T T = + − + Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ H L H L Q Q T T = −
• 7. 7 The cyclic integral of Reversible Heat Engine 0 0H L H L Q Q T T = + − + Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ H L H L Q Q T T = − H H L L Q T Q T = 0= Since
• 8. 8 The cyclic integral of Irreversible Heat Engine irr revW W< H L H L Q Q T T = Q T δ ∫ H L H L Q Q T T = − We cannot use this ( ) ( )H L H Lirr rev Q Q Q Q− < − It is Irreversible H L irr H L revQ Q Q Q− < − L irr L revQ Q> H Lirr H L Q Q T T − 0<
• 9. 9 The cyclic integral of Reversible Refrigeration 0 0L H L H Q Q T T = + + − Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ L H L H Q Q T T = − H H L L Q T Q T = 0= Since
• 10. 10 The cyclic integral of Irreversible Refrigeration irr revW W> H L H L Q Q T T = Q T δ ∫ H L H L Q Q T T =− + We cannot use this ( ) ( )H L H Lirr rev Q Q Q Q− > − It is Irreversible H irr L H rev LQ Q Q Q− > − H irr H revQ Q> H irr L H L Q Q T T − + 0<
• 11. 11 IrreversibleReversible < 00=Heat Engine < 00=Refrigeration 0≤∫ T Qδ Q T δ ∫ Derivation of Clausius Inequality
• 12. 12 All paths are arbitrary 0 Q T δ =∫ Derivation of Entropy (Reversible Process) 2 2 1 1A C Q Q T T δ δ    =        ∫ ∫Subtracting gives 2 1 1 2 0 C B Q Q T T δ δ    = + =        ∫ ∫ For reversible cycle A-B 2 1 1 2 0 A B Q Q T T δ δ    = + =        ∫ ∫ For reversible cycle C-B 0 Q T δ =∫ Q the quantity is independent of the path and dependent on the end states only T δ ∴ ∫ Since paths A and C are arbitrary, it follows that the integral of δQ/T has the same value for ANY reversible process between the two sates.
• 13. 13 work & heat are dependent on path Path functions Recall are independent of path properties Point functions and depend on state only      ( ) is a thermodynamic property we call it entropy S δQ T ⇒ ∫ Entropy (the unit) S = entropy (kJ/K); s = specific entropy (kJ/kg K) ∫       =−      ≡ 2 1 12gintegratin revrev T Q SS T Q dS δδ S2 – S1 depends on the end states only and not on the path, ∴ it is same for any path reversible or irreversible Derivation of Entropy (Reversible Process)
• 14. 14 Consider 2 cycles AB is reversible and CB is irreversible 2 1 1 2 for cycle A-B (reversible) 0 A B Q Q Q T T T δ δ δ    = + =        ∫ ∫ ∫ 2 1 1 2 for path C-B (irreversible) 0 C B Q Q Q T T T δ δ δ    = + <        ∫ ∫ ∫ 2 2 1 1 comparing gives A C Q Q T T δ δ         >        ∫ ∫   2 2 2 1 1 1 reversible it is a property but A C A δQ dS dS T   === ===    ∫ ∫ ∫ in general δQ dS T ⇒ ≥ 2 2 1 1C C δQ dS T   ∴ >     ∫ ∫ 2 2 1 1 or δQ S S T − ≥ ∫ equality for reversible inequality for irreversible Derivation of Entropy (Irreversible Process)
• 15. 15 Example (7-1) Entropy change during isothermal process. Solution:  This is simple problem.  No irreversibilities occur within the system boundaries during the heat transfer 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.
• 16. 16 kkJ k kJ T Q S sys /5.2 300 750 ===∆ ( ) 2 2 1 1 1 rev rev Q Q S Q T T T δ δ   ∆= = =    ∫ ∫ Q S T ∆ =  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.
• 17. 7.3. Heat Transfer as the Area Under the T-s Curve 17
• 18. 18 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: 1. Pure substance like water, R-134, Ammonia etc.. 2. Solids and liquids 3. Ideal gas
• 19. 19 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. )( 12 ssmS −=∆ Entropy change for a closed system with mass m is given as expected: For water its assigned zero at 0.01 C. For R-134 it is assigned a zero at -40 C.
• 20. 20 Property diagrams involving 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. T Q dS revδ = TdSQrev =δ 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. W PdVδ =
• 21. 21 ∫= 2 1 TdSQrev This area has no meaning for irreversible processes! Thus 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.
• 22. 22 Isothermal reversible process. Entropy Isothermal Process Q 1 2 0 0T S T m s= ∆ = ∆
• 23. 23 Adiabatic 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. Entropy Isentropic Process 1 2 Q=0
• 24. 24 T-s Diagram for the Carnot Cycle Temperature Entropy 1 Isentropic compression Qin W=Qin-Qout Qout 2 Isothermal expansion 3 Isentropic expansion Isothermal compression 4
• 25. 25 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.
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