 # Evaluating entropy changes

## Presentation on theme: "Evaluating entropy changes"— Presentation transcript:

Evaluating entropy changes
Since entropy is a property, the entropy change between two specified states does not depend on the process (reversible or irreversible) that occurs between the two states. However, to evaluate entropy change by the above definition a reversible process has to be imagined connecting the initial and final states and integration has to be carried along its path.

Using entropy change and thermodynamic temperature to graphically interpret the heat supplied in reversible processes P T 1 2 S V 2

Showing a Carnot cycle on a T-s diagram
Isotherms: horizontal lines Isentropes:vertical lines 2 1 T Qnet,in Application: Carnot engine 3 4 s

Isobars and isochores on the T-s diagram for an ideal gas (Sec. 7.9)
P Isochores: Isochores: v Isobars: T s

T-s diagram of a cycle: try yourself
The Brayton cycle used in a gas turbine engine consists of following steps: isentropic compression (1-2). isobaric heat addition (2-3). isentropic expansion (3-4). isobaric heat rejection (4-1). Plot this process on a T-s and P-v diagram.

Use of the Tds relations in calculating entropy changes
On an intensive (per unit mass) basis: First TdS relation Second TdS relation Integrating between initial (1) and the final state (2): Even though the Tds equations are derived using an internally reversible process, the change in entropy during an irreversible process occuring between the same two equilibrium states can be also calculated using the integrals on the left.

Entropy change of liquids/solids (Sec. 7.8)
Liquid/solid can be approximated as incompressible substances (v=constant): i.e. specific volume ( or density) remains constant even when other properties change (dv=0) Internal energy of a solid/liquid is a function of temperature alonecp=cv=c (one specific heat) from First Tds relation if temperature variation of specific heat (c)can be neglected. For solids/liquids, the isentropic process is also isothermal.

Second law for a closed system undergoing a process
Clausius inequality: b: boundary Tb=T for an internally reversible process > for irreversible process 1-2 = internally reversible process 1-2 8 8

Second law for an isolated system
Also, across the boundaries of an isolated system no energy (heat/work) is transferred (by definition) The universe (any system + its surroundings) can be considered an isolated system. “The total entropy of the universe is increasing.” 9 9

Second law for an adiabatic process undergone by a closed system
Second law for a system undergoing adiabatic (irreversible/reversible) process (dQ=0) 10 10

The entropy balance for closed systems
entropy generation within the system entropy change entropy transfer accompanying heat transfer Sgen>0 if irreversibilities present inside the system Sgen=0 for no irreversibilities inside the system The value of Sgen is a measure of the extent of irreversibilities within the system. More irreversibilities Sgen ↑

Entropy generation (Sgen>0) within the system is due to irreversible processes within the system boundary Irreversible processes occurring within the system boundary result in entropy generation. Examples: Friction (solid-solid, solid-fluid) may be present between parts of the system. Hot and cold zones may be present within the system which may be interacting irreversibly through heat transfer. Non-quasi-equilibrium compression/expansion occuring between parts of the system. Other: mixing between substances having different chemical composition, chemical reaction. 12 12

Pf,Tf, (m1+m2) ) to show Sgen>0
Example of entropy generation due to an irreversible process in an isolated system Initial state (1) Pf,Tf, (m1+m2) Final state (2) Remove separator Objective: To show that the process is irreversible. Isolated system has Q=0 To show a process is irreversible ) to show Sgen>0

Example (contd.): The final state of the process
Pf,Tf, (m1+m2) P1,V1, T, m1 P2,V2, T, m2 High pressure Low pressure Final temperature: From the first law Uf=Ui. At the final state: .

Example of entropy generation due to an irreversible process in an isolated system
P1,,,V1, T, m1 P2,V2, T, m2 Pf,Tf, (m1+m2) Low pressure High pressure Entropy balance Entropy change of the system is the entropy change of its parts

Example of entropy generation due to an irreversible process in an isolated system
P1,,,V1, T, m1, P2,V2, T, m2 Pf,Tf, (m1+m2) Second law (entropy balance) from integration of Tds=dh-vdP Similarly

Example of entropy generation due to an irreversible process in an isolated system
P1,,,V1, T, m1, P2,V2, T, m2 Pf,T, (m1+m2) 17

Calculating work done in a reversible steady flow process
Reversible cycles develop the maximum work. Reversible cycles consist of reversible processes through each device. At every stage of the internally reversible process: qrev-wrev=dh +d(ke)+d(pe) Combining: wrev=-vdP-d(ke)-d(pe) Integrating: First law for control volumes

Graphical representation of reversible steady flow work on a p-v diagram
if changes in KE and PE can be neglected (e.g. in turbines, compressor and pumps, but not in nozzles)

Reversible steady flow work for a pump (PUMP HANDLES LIQUIDS)
Since specific volume of a liquid is nearly independent of pressure: (since pump is an adiabatic device) similar to compressor Alternatively

In steam power plants, how big is “turbine work out” compared to “pump work in”?
v vf=1.7 m^3/kg and vf=10^-3 m^3/kg at 100 kPa Specific volume of vapors are orders of magnitude larger than the specific volume of liquid (e.g. vf' 10-3 m3/kg vg' 2 m3/kg at 100 kPa)) Gas turbine pumps consume a significant part of work developed in turbines to run compressor