Isentropic process

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    Lec 18: Isentropic processes,TdS relations, entropy changes

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    For next time: Read: 7-2 to 7-9 Group project subject selection due on

    November 3, 2003

    Outline: Entropy generation and irreversible processes Entropy as a property Entropy changes for different substances

    Important points: Entropy is a property of a system it is not

    conserved and is generated by irreversibleprocesses Know how to identify an isentropic processes Know how to use the tables to find values for

    entropy

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    Recall we had entropy

    2

    1 revintT

    q

    Rlb

    Btuor

    Kkg

    kJ

    m

    s2 - s1 =

    Units

    are

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    Lets look at a simpleirreversible cycle on a p-v

    diagram with two processes

    P

    1

    2

    .

    .A

    B

    Let A be

    irreversible and Bbe reversible

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    Irreversible cycle

    0)T

    QAB

    By Clausius Inequality

    Evaluate cyclic integral

    0T

    Q

    T

    Q

    T

    Q2

    1 B

    2

    1 Acycle

    (non-rev) (rev)

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    Irreversible cycle

    For the reversible process, B, dS=Q/dT,thus:

    0dST

    Q

    T

    Q2

    1

    2

    1 Acycle

    Rearranging and integrating dS:

    2

    1 AT

    QS

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    Second Law of Thermodynamics

    Entropy is a non-conserved property!

    2

    1 A

    12T

    QSSS

    This can be viewed as a mathematicalstatement of the second law (for aclosed system).

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    We can write entropy change as anequality by adding a new term:

    gen

    2

    1 A

    12 S

    T

    QSS

    entropychange

    entropytransfer

    due toheattransfer

    entropyproduction

    orgeneration

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    Entropy generation

    Sgen 0 is an actual irreversible process.

    Sgen = 0 is a reversible process.

    Sgen 0 is an impossible process.

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    TEAMPLAY

    Consider the equation

    You have probably heard, Entropy alwaysincreases.

    Could it ever decrease? What are theconditions under which this could happen(if it can)?

    gen

    2

    1 A

    12 S

    T

    QSS

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    Entropy transfer and production

    What if heat were transferred from thesystem?

    The entropy can actually decrease if

    gen

    2

    1 A

    ST

    Q

    and heat is being transferred awayfrom the system so that Q is negative.

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    Entropy Production

    Sgen quantifies irreversibilities. Thelarger the irreversibilities, the greaterthe value of the entropy production,Sgen .

    A reversible process will have no entropyproduction.

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    Entropy transfer and production

    S2 S1

    > 0, Q could be + or ; if,

    because Sgen is always positive.

    < 0, if Q is negative and

    = 0 if Q = 0 and Sgen = 0.

    = 0 if Q is negative and

    gen

    2

    1 A

    ST

    Q

    gen

    2

    1 A

    ST

    Q

    gen

    2

    1 A

    ST

    Q

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    Isentropic processes

    Note that a reversible (Sgen = 0),adiabatic (Q = 0) process is alwaysisentropic (S1 = S2)

    But, if the process is merely isentropicwith S1 = S2, it may not be a reversibleadiabatic process.

    For example, if Q 0 and gen

    2

    1 A

    ST

    Q

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    Entropy generation

    Consider

    What if we draw our system boundariesso large that we encompass all heat

    transfer interactions? We wouldthereby isolate the system.

    gen

    2

    1 A

    12 S

    T

    QSS

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    Entropy changes of isolated systems

    And then

    gen

    2

    1 A

    12 ST

    QSS

    0

    gen12 SSS

    But Sgen0. So, the entropy of anisolated system always increases. (Thisis the source of the statement, The world

    is running down.)

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    Entropy

    )ss(xss fgf

    )T(s)p,T(s f

    It is tabulated just like u, v, and h.

    Also,

    And, for compressed or subcooled liquids,

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    The entropy of a pure substance is determined from the tables, just as forany other property

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    Ts Diagram for Water

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    TEAMPLAY

    Use the tables in your book

    Find the entropy of water at 50 kPa and500 C. Specify the units.

    Find the entropy of water at 100 C anda quality of 50%. Specify the units.

    Find the entropy of water at 1 MPa and

    120 C. Specify the units.

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    Ts diagrams

    pdVw

    Work was the area under the curve.

    Recall that the P-v diagram was veryimportant in first law analysis, and that

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    For a Ts diagram

    revintT

    QdS

    TdSQ revint

    2

    1

    revint TdSQ

    Rearrange:

    Integrate:

    If the internally reversible process also isisothermal at some temperature To:

    STdSTQ o

    2

    1

    orevint

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    On a T-S diagram, the area under the process curve represents theheat transfer for internally reversible processes

    d

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    Entropy change of a thermalreservoir

    For a thermal reservoir, heat transfer occursat constant temperaturethe reservoirdoesnt change temperature as heat isremoved or added:

    TQ

    S

    Since T=constant:

    T

    QS

    Applies ONLY tothermalreservoirs!!!!

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    The Tds Equations

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    Derivation ofTds equations:

    dQ dW = dU

    For a simple closedsystem:

    dW = PdV

    The work is given by:

    dQ = dU + PdV

    Substituting gives:

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    More derivation.

    For a reversible process:

    TdS = dQ

    Make the substitution for Q in the energyequation:

    PdV+dU=TdS

    Or on a per unit mass basis:

    Pdv+du=Tds

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    Entropy is a property. The Tds expression

    that we just derived expresses entropy interms of other properties. The propertiesare independent of path.We can use theTds equation we just derived to calculatethe entropy change between any twostates:

    Tds = du +Pdv

    Tds = dh - vdP

    Starting with enthalpy, it is possible todevelop a second Tds equation:

    Tds Equations

    L t l k t th t h

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    Lets look at the entropy changefor an incompressible

    substance:

    dT

    T

    )T(cds

    We start with the first Tds equation:

    Tds = cv(T)dT + Pdv

    For incompressible substances, v const, sodv = 0.

    We also know that cv(T) = c(T), so we canwrite:

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    Entropy change of anincompressible substance

    dTT

    )T(css

    2

    1

    T

    T

    12

    1

    212

    T

    Tlncss

    Integrating

    If the specific heat does not vary with

    temperature:

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    TEAMPLAY

    Work Problem 7-48

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    Entropy change for an ideal gas

    dTcdh p And

    dpp

    RTdTcTds p

    Tds = dh - vdp

    Start with 2nd Tds equation

    Remember dh and v for an ideal gas?

    v=RT/p

    Substituting:

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    Change in entropy for an ideal gas

    p

    dpR

    T

    dTcds p

    Dividing through by T,

    Dont forget, cp=cp(T)..a function oftemperature! Integrating yields

    1

    2

    T

    T

    p12pplnR

    TdT)T(css

    2

    1

    Entropy change of an ideal gas

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    Entropy change of an ideal gasfor constant specific heats:

    approximation

    Now, if the temperature range is solimited that c

    p constant (and c

    v

    constant),

    1

    2pp

    T

    Tlnc

    T

    dTc

    1

    2

    1

    2p12

    p

    plnR

    T

    Tlncss

    Entropy change of an ideal gas

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    Entropy change of an ideal gasfor constant specific heats:

    approximation

    Similarly it can be shown from

    Tds = du + pdv

    that

    1

    2

    1

    2v12

    vvlnR

    TTlncss

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    TEAMPLAY

    Work Problem 7-62

    Entropy change of an ideal gas

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    Entropy change of an ideal gasfor variable specific heats: exact

    analysis

    1

    2

    T

    T

    p12 p

    p

    lnRT

    dT

    )T(css

    2

    1

    2

    1

    T

    T

    p

    T

    dTc

    Integrating..

    To evaluate entropy change, well

    have to evaluate the integral:

    Entropy change of an ideal gas

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    122

    1

    T

    0T

    p

    T

    0T

    p

    T

    T

    pT

    dTcT

    dTcT

    dTc

    )T(s)T(s 1o

    2

    o

    And so(T), the reference entropy, istabulated in the ideal gas tables for areference temperature of T = 0 and p = 1

    atm.

    Entropy change of an ideal gasfor variable specific heats: exact

    analysis

    Evaluation of the integral

    Entropy change of an ideal gas for

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    Entropy change of an ideal gas forvariable specific heats: exact

    analysis

    Only is tabulated. Theis not.

    So,

    dTcp dTcv

    1

    21

    o

    2

    o

    12p

    plnR)T(s)T(sss

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    Entropy change of an ideal gas

    Note that the entropy change of an idealgas, unlike h and u, is a function of twovariables.

    Only the reference entropy, so, is afunction of T alone.

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    Sample Problem

    A rigid tank contains 1 lb of carbon monoxideat 1 atm and 90F. Heat is added until thepressure reaches 1.5 atm. Compute:

    (a) The heat transfer in Btu.

    (b) The change in entropy in Btu/R.

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    Draw diagram:

    State 1:

    P = 1atm

    T = 90oF

    CO:

    m= 1 lbmState 2:

    P = 1.5 atm

    Rigid Tank => volume isconstant

    Heat Transfer

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    Assumptions

    Work is zero - rigid tank

    kinetic energy changes zero

    potential energy changes zero

    CO is ideal gas

    CO in tank is system

    Constant specific heats

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    Apply assumptions to conservationof energy equation

    PE+KE+UWQ

    12v TTmc=Q

    For constant specific heats, weget:

    0 0 0

    Need T2> How do we get it?

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    Apply ideal gas EOS:

    2

    1

    22

    11

    mRT

    mRT

    VP

    VP Cancel common

    terms...

    Solve forT2:

    R825R460901.01.5TPPT 112

    2

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    Solve for heat transfer

    R550825Rlb

    Btu18.0)lb1(Q

    m

    m

    Btu5.49Q

    Now, lets get entropy change...

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    For constant specific heats:

    1

    2

    1

    2v12

    vvRln

    TTlncmSS

    Since v2 = v1

    0

    1

    2v12

    TTlncSS

    R550

    R825ln

    Rlb

    Btu18.0)lb1(SS

    m

    m12

    Btu/R073.0SS 12