EGR 334 Thermodynamics Chapter 6: Sections 1-5 Lecture 24: Introduction to Entropy Quiz Today?

Today’s main concepts: Explain key concepts about entropy Learn how to evaluate entropy using property tables Learn how to evaluate changes of entropy over reversible processes Introduction of the temperature-entropy (T-s) diagram Reading Assignment: Read Chapter 6, Sections 6-8 Homework Assignment: Problems from Chap 6: 1,11,21,28

Introducing Entropy Change and the Entropy Balance The entropy change and entropy balance concepts are developed using the Clausius inequality expressed as: where scycle = 0 no irreversibilities present within the system scycle > 0 irreversibilities present within the system scycle < 0 impossible

Defining Entropy Change Consider two cycles, each composed of two internally reversible processes, process A plus process C and process B plus process C, as shown in the figure. Applying Classius Equation to these cycles gives, where scycle is zero because the cycles are composed of internally reversible processes.

Defining Entropy Change Subtracting these equations: Since A and B are arbitrary internally reversible processes linking states 1 and 2, it follows that the value of the integral is independent of the particular internally reversible process and depends on the end states only.

Defining Entropy Change Recalling (from Sec. 1.3.3) that a quantity is a property if, and only if, its change in value between two states is independent of the process linking the two states, we conclude that the integral represents the change in some property of the system. We call this property entropy and represent it by S. The change in entropy is where the subscript “int rev” signals that the integral is carried out for any internally reversible process linking states 1 and 2.

Defining Entropy Change The equation allows the change in entropy between two states to be determined by thinking of an internally reversible process between the two states. But since entropy is a property, that value of entropy change applies to any process between the states – internally reversible or not.

Entropy Facts Entropy is an extensive property. Like any other extensive property, the change in entropy can be positive, negative, or zero: Units of entropy, S, are: in SI  [kJ/K] in US Customary  [Btu/oR] Units for specific entropy, s, are: in SI  [kJ/kg-K] in US Customary  [Btu/lbm-oR]

Finding Entropy: Method 1 For problem solving, specific entropy values are provided in Tables A-2 through A-18. Values for specific entropy are obtained from these tables using the same procedures as for specific volume, internal energy, and enthalpy. For two phase liquid/vapor mixtures, quality, x, may be used as or For slightly compressed liquids, the entropy may be approximated as:

Example 1: find s using property tables a) Given: H20 at T=520 oC and p= 8 MPa find s: From Table A-2…substance is found to be superheated vapor. From Table A-4…..s can be found ad 6.7871 kJ/kg-K b) Given: Ammonia at p =2.5 bar and v = 0.20 m3/kg find s: From Table A14…substance is found to be liquid/vapor mixture. the quality can be found as and then c) Given: R22 at T = 20 oF and p = 80 psi find s: From Table A-7E or A-8E…substance is found to be compressed liquid then s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R. (a) Water p1 = 1000 psi, T1 = 800 °F p2 = 1000 psi, T2 = 1000 °F (b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F saturated vapor, p2 = 40 psi

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R. (a) Water p1 = 1000 psi, T1 = 800 °F p2 = 1000 psi, T2 = 100 °F T 1000 psi Table A-4E @ T1 = 800 °F, p1 = 1000 psi s1 = 1.5665 BTU/lb·°R 1 545°F 2 Table A-5E @ T2 = 100 °F, p2 = 1000 psi s2 = 0.12901 BTU/lb·°R v s2 - s1 = 0.12901 - 1.5665 = -1.43749 BTU/lb·°R

(b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R. (b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F saturated vapor, p2 = 40 psi T Table A-10E, @ T1 = -40 °F hf=0, hg=95.82 BTU/lb 2 -40°F 1 sf=0, sg=0.2283 BTU/lb·°R v Table A-11E, sat. vapor, p2 = 40 psi s2 - s1 = 0.2197 - 0.11415 = 0.1056 BTU/lb·°R

Finding Entropy: Method 2 For problem solving, states often are shown on property diagrams having specific entropy as a coordinate: the Temperature-Entropy (T-s) diagram and the Enthalpy-Entropy (h-s) diagram the h-s diagram is also know as the Mollier diagram These diagrams are in your appendix as Figures 7 and 8 and Figures 7E and 8E.

T-s diagram: Vertical axis: Temperature Horizontal axis: Entropy Shows other constant property lines: Constant Pressure Constant Quality Constant Enthalpy Constant Specific Volume

h-s diagram: Vertical axis: Enthalpy Horizontal axis: Entropy Shows other constant property lines: Constant Temperature Constant Pressure Constant Quality

Example 2: Using T-s diagram: b) Given: H20 at s=1.5 Btu/lbm-R and x=85% find h: a) Given: H20 at T=900 oF and h = 1400 Btu/lbm find s: s ≈ 1.5 Btu/lbm-R h ≈ 1.5 Btu/lbm From Table A-7E or A-8E…substance is found to be compressed liquid then s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR From Table A-7E or A-8E…substance is found to be compressed liquid then s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR

Finding Entropy: Method 3…IT IT can also be used for working with entropy. Once again it’s recommended that you don’t compare individual values from IT with the values you might find on the Appendix tables from you book, but changes in entropy between two states will be consistent and can be compared. For H20, find the difference in entropy between State 1: T = 500 deg C and p = 20 bar State 2: T = 200 deg C and p = 10 bar Δs=0.1992 kJ/kg-K

Sec 6.3 : The TdS Equations How is S related to U? Consider a pure, simple, compressible system undergoing a internally reversible process. Energy Balance: then differentiate where, therefore, 1st TdS equation

How is S related to H? Recall that: and therefore, Sec 6.3 : The TdS Equations How is S related to H? Recall that: and from 1st Tds equation therefore, 2nd T dS equation can also write these on a per mass basis Relationships between p, T, v, u, h, and s Even though this derivation is based on a reversible process, these equations hold for any process… even an irreversible process.

Finding Entropy: Method 4 for incompressible fluid When a substance is compressible (like many liquids), Recall that for incompressible fluids: and where c is the specific heat capacity then if the specific heat is treated as constant Δs for incompressible liquids

Finding Entropy: Method 5 for Ideal Gas Recall that if a substance can be treated as an ideal gas, the following relationships may be applied to the defining the properties: Combining ideal gas relations with the Tds equations give ds = f(T,v) ds = f(T,p)

So for Ideal Gas, change in entropy can be evaluated by integrating Sec 6.4 Idea Gases So for Ideal Gas, change in entropy can be evaluated by integrating these relations and

or constant heat capacities from Tables A20, A22, or A23. Sec 6.4 Idea Gases Options to evaluate: and there are several options to evaluate the heat capacity. Remember that cV & cP are functions of temperature. i) Find values for cp and or cv (Using Table A21) or constant heat capacities from Tables A20, A22, or A23. then use or or ii) Use Tabulated values( Table A22): The s° indicates that it is based on a reference temp

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R. Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm and T2 = 420°F = 880° R, p2 = 1 atm (a) using a constant cv at Tave where: Tave = 230 F = 690 R R = 0.06855 Btu/lb-R cp = 0.242 Btu/lb-R

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R. Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm and T2 = 420°F = 880° R, p2 = 1 atm (c) Air as an ideal gas using so from Table A-22E @ T1 = 500 °R, s° = 0.58233 BTU/lb·°R @ T2 = 880 °R, s° = 0.71886 BTU/lb·°R

end of lecture 24 slides