ChemE 260 The Clausius Inequality & Entropy Dr. William Baratuci Senior Lecturer Chemical Engineering Department University of Washington TCD 7: A & B CB 6: 1 May 3, 2005

The Clausius Inequality Cyclic Integrals Integrate through all the steps in a cycle and return to the initial state. Inexact Differentials: Q & W Used for path variables, Q and W Evaluating Cyclic Integrals Example 1: Carnot HE Example 2: Carnot HE Hot Reservoir Cold Reservoir HER QH QC,rev Wrev Cyclic integrals are new, but they are not scary or terribly difficult when applied to thermodynamic cycles. All you need to do is integrate through all the steps of the cycle, so that you begin and end in the same state. The funky “” is a common way of indicating that the differential is not an exact differential, “d”, nor is it a partial differential, “”. It is absolutely crucial that you understand these two examples of how to evaluate a cyclic integral. Example 1 In the cyclic integral, our sign convention applies, but in the tie-fighter diagram, both QH and QC are positive quantities. When we evaluate the cyclic integral, we get something like Q12 + Q23 + Q34 + Q41 for a Carnot cycle. In a Carnot Cycle, Steps 2-3 and 4-1 are adiabatic, so Q23 = Q41 = 0. So, the cyclic integral is just Q12 + Q34. Because of the sign convention conflict, QH = Q12 and QC = - Q34. Therefore, the cyclic integral is QH – QC. The 1st Law tells us that QH = QC + WHE Since WHE > 0, QH > QC and finally QH – QC > 0 Example 2 Because the Carnot Cycle is completely reversible, all heat transfer must occur through an infinitessimal temperature difference. Therefore, the temperature of the hot reservoir must be equal to the temperature of the working fluid in the system to which it transfers heat. And the temperature of the cold reservoir must be equal to the temperature of the working fluid in the system from which it receives heat. Since the reservoir temperatures are constant, the temperatures within the system where the heat exchange occurs must also be constant. Baratuci ChemE 260 May 3, 2005

Clausius: Int. Rev. and Irrev. Cycles Reversible Cycle, such as Carnot: Kelvin: or: Therefore: Irreversible Cycles: Hot Reservoir Cold Reservoir HEIrr QH QC,irr Wirr Definition of efficiency : 1st Law : Reversible Cycles The Kelvin Relationship Applies to reversible cycles like Carnot Depends on the fact that the temperature scale we use, the Kelvin Scale, is a thermodynamic temperature scale Irreversible Cycles Consider an irreversible HE… … that operates between the same hot and cold thermal reservoirs … that receives the same amount of heat from the hot reservoir, QH. 1st Carnot Principle tells us that the reversible HE is more efficient than the irreversible HE The thermal efficiency is defined as the ratio of the work output to QH. Since QH is the same for both HE’s and the efficiency of the reversible HE is greater, we conclude that the work output of the irreversible HE must be less than the work output of the reversible HE. The 1st Law allows us to replace the work output with QH – QC for each HE. The QH term on each side of the inequality cancels leaving us with the fact that the irreversible HE must reject more heat to the cold reservoir than the reversible HE does. The Cyclic Integral We can now apply the the cyclic integral of Q/T to the irreversible process. The key here is that we can use the Kelvin Relationship (applied to the reversible HE, not this irreversible one) to replace QH / TH with QC,rev / TC. We just showed that QC,rev < QC,rev , so we conclude that the cyclic integral must be negative for all irreversible cycles ! Conclusion The Clausius Inequality is TRUE ! The equality part of the equation applies for internally reversible cycles and the “less than” part applies for irreversible cycles. Why relax the rule from reversible to just internally reversible ? If the temperatures inside the system are NOT equal to the temperatures of the hot and cold reservoirs, then the exchange is NOT reversible. The system is, at best, internally reversible. But, the Kelvin Relationship applies if the cycle is internally reversible. So, it turns out that as long as the cycle is internally reversible, the equality in the Clausius Inequality applies. It also turns out that the temperature within the system at which heat exchange occurs does not need to be constant for the equality part of Clausius to hold true. It just makes it a WHOLE LOT easier to evaluate the cyclic integral if the temperature of the working fluid is constant while heat is being exchanged (such is in a vaporization or condensation process). All Cycles : Baratuci ChemE 260 May 3, 2005

Entropy Cycle 1-A-B-2: Cycle 1-A-C-2: Subtract Eqns: Cycles 1-A-B-2 and 1-A-C-2 are internally reversible, so the cyclic integral of  Q / T = 0. This means that the integral of  Q / T from state 2 to state 1 must be the same along paths B and C. Since we didn’t specify anything special about paths B and C except that they are internally reversible. We didn’t specify anything special at all about states 1 and 2. We can conclude that the integral of  Q / T must be the same for ALL INTERNALLY REVERSIBLE paths between ANY two states. When the change in a quantity between two states does not depend on the path, we call it a state variable or property. This new property is called entropy and it is defined by the equation in the box. This equation really only tells how to calculate CHANGES in entropy. So, like enthalpy and elevation above sea level, we must choose a reference state for entropy. In the Fluid and Thermal Property Tables of the NIST Webbook, they always list the reference state and it always includes the reference state for entropy. Check it out for yourself. Does not depend on path ! It is a state variable or property ! Definition of Entropy: Baratuci ChemE 260 May 3, 2005

S: Int. Rev. & Irrev. Processes Change in Entropy: Problem: Since Int. Rev. processes do not exist, how do we evaluate S ? Special Case: Reversible, Isothermal Processes Although we have to follow an internally reversible path to evaluate S12, S2 – S1 has the same value whether the process follows an internally reversible path or a IRREVERSIBLE path. This result is enormously important ! In the real world, we won’t analyze any internally reversible processes because they don’t really exist. But, because entropy is a state variable or property, we can use it to analyze real, IRREVERSIBLE processes anyway ! So, how do we evaluate S for processes OTHER THAN reservoirs ? We cannot use reversible processes because they don’t exist ! It turns out that S is related to changes in other properties that are easier to measure. In Lesson D, we will derive the Gibbs Equations and we will see how S can be calculated using the heat capacity and equations of state. Especially useful for evaluating Sreservoir because Treservoir = constant Baratuci ChemE 260 May 3, 2005

Entropy of a Pure Substance S = fxn( T, P, phase ) NIST Webbook Thermophysical Properties of Fluid Systems Specific entropy is listed in the thermodynamic data tables Observations: All substances at all T : IG and many real gases : Specific entropy is just another property that we can look up in tables of thermodynamic data. It is available in the NIST Webbook, in the back of the Schaum’s Outline and in the tables you downloaded from Thermo-CD (most notably the tables for R-134a). A quick look at the thermodynamic tables should convince you that specific entropy increases with increasing T as long as a phase change does not occur. For ideal gases and most real gases at most values of T & P, specific entropy decreases as pressure increases. You can use quality to calculate the specific entropy of a saturated mixture (just as you can for any other specific property). Subcooled Liquids If data is available for subcooled liquids, then you should definitely use it. If data is not available, then you can approximate the specific entropy using the specific entropy of the saturated liquid AT THE SAME TEMPERATURE. This is a reasonable approximation as long as the liquid is incompressible (V^ = constant and CP = CV) This approximation is most accurate if the actual pressure, P, is not a great deal larger than the vapor pressure, P*. Sat’d Mixtures: Subcooled Liquids: Where P* = vapor pressure = Psat Baratuci ChemE 260 May 3, 2005

T-S Diagram The T-S Diagram will be the most useful diagram in the remainder of this course. The good news and the bad news is that it looks very similar to the PV Diagram ! Good news because you will learn how to use the diagram quickly Bad news because it is easy to get the two confused ! TS Diagrams look even more similar to TV Diagrams, but we have not used TV Diagrams much. The TS Diagram has isobars Isobars go from lower left to upper right because specific entropy increases as T increases. Isobars are horizontal in the two-phase envelope because the phase transition entails a large change in the specific entropy, but occurs at a constant temperature. You are probably wondering WHY the TS Diagram is going to be so useful. Let’s see what a Carnot Cycle looks like on a TS Diagram. Baratuci ChemE 260 May 3, 2005

Carnot Cycle Steps Step 1-2: Step 2-3: Step 3-4: Step 4-1: 1-2: Isothermal expansion 2-3: Adiabatic expansion 3-4: Isothermal compression 4-1: Adiabatic compression Step 1-2: Step 2-3: Isentropic ! We use the definition of entropy to evaluate changes in entropy for each step in te Carnot Cycle. Steps 2-3 and 4-1 are both adiabatic and reversible. Therefore there is NO CHANGE in the entropy of the system during these steps. Steps 2-3 and 4-1 are ISENTROPIC ! Steps 1-2 and 3-4 occur at constant temperature. So, the definition of entropy simplifies (just as it did for a reservoir). S is just Q / T. So, what about work ? Step 3-4: Step 4-1: Isentropic ! Baratuci ChemE 260 May 3, 2005

Heat, Work & TS Diagrams 1st Law Cycle Areas under process paths on TS Diagrams represent heat transferred in reversible processes. Application of the 1st Law lets us also associate work with the area enclosed by the cycle. This simple, tangible interpretation of area makes it much easier to understand and compare the performance of complex cycles that we will study later. For example, if we decrease PLO, does the efficiency of this power cycle increase or decrease ? QH = area under path for step 1-2 QC = positive area under path for step 3-4 W = area enclosed by the cycle ! Baratuci ChemE 260 May 3, 2005

Next Class … Principle of Increasing Entropy & Entropy Generation This will let us express the 2nd Law in terms of Entropy This very powerful result will let us perform 2nd Law Analysis on processes as well as cycles. Fundamental Property Relationships These are sometimes called the Gibbs Equations They show us how entropy is related to other properties, such as H, U, P, V and T This will show us how the specific entropy values in the thermodynamic tables were determined. Baratuci ChemE 260 May 3, 2005

Example #1 A piston-and-cylinder device contains saturated R-134a vapor at -5oC. This vapor is compressed in an internally reversible, adiabatic process until the pressure is 1.0 Mpa. Determine the work per kg of R-134a for this process.

Example #2 Steam enters an adiabatic turbine at 5 MPa and 450oC and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per kg of steam flowing through the turbine if the process is reversible and changes in kinetic and potential energies are negligible.

Example #3 Consider a process in which 1.00 kg of saturated water vapor at 100oC is condensed to a saturated liquid in an isobaric process by heat transfer to the surrounding air, which is at 25oC. What is the change in entropy of the water ? What is the change in entropy of the surroundings ? What is the change in the entropy of the universe ?