chap4prac

2
Chapter 4 Practice Problems To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 1 of 2 (P4.1) COP = coef. of performance = net S C W Q , Using state numbers of Fig 4.9-4.10. P-H plot will look like Fig 4.10 of pg 151. Use P-H chart pg 653 and table pg 654: state 2 is satV at –40C ! H 2 = 372 kJ/kg (chart) state 3, outlet of the reversible compressor is found by following the isentropic line to 40C, where H 3 ’ = 438 kJ/kg. state 4 is satL at H 4 = 256 kJ/kg (table) state 1, H 1 = H 4 ( ) kg kJ H H Q C / 116 256 372 1 2 = = = W S = (H 3 H 2 ) = 438 – 372 = 66 kJ/kg 76 . 1 66 116 = = = S C W Q COP & & Q C = COP (W S ) = 1.76 (9000 J/day) = 16 kJ/day (P4.2) Methane,. See Fig 4.9-4.10 pg 150-1.Methane chart pg 651. F lb Btu S lbm Btu H MPa P F T O satV satV O evap . / 35 . 2 / 339 032 . 0 , 280 2 2 = = = = Pressure of 3’ isn’t given, however, it will be the same as the condenser outlet (P 4 ) since the condenser will be considered to be isobaric. State 4 will be satL at 40F, so P4 = P3’ = 2 MPa. Follow isentropic line from the compressor inlet state to 2MPa. At P3’=2MPa, and S3’ = 2.35 Btu/lb m F, H3’ = 485 Btu/lbm, and T3’ = 40F. At the outlet of the condenser, the fluid will be saturated liquid at 2 MPa, F lb Btu S lbm Btu H MPa P F T O sat satL O liq . / 54 . 1 / 218 2 , 160 4 4 4 4 = = = 83 . 0 339 485 218 339 2 3 4 2 , 2 3 1 2 = = = = = COP H H H H W Q H H H H COP net S C (P4.3) Rankine Cycle, Fig 4.3 Pg. 143 temperatures C C O O 6 . 99 & 200 , Sat vapor from Turbine outlet can be found in the sat Pressure table p642. State T( O F) P(MPa) H(Btu/lb) S(Btu/lb* O F) 1 -280 0.032 218 2 Sat'd vap -280 0.032 339 2.35 3' 40 2 485 2.35 4 -160 2 218 1.54

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  • Chapter 4 Practice Problems

    To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 1 of 2

    (P4.1) COP = coef. of performance = netS

    C

    WQ

    ,

    Using state numbers of Fig 4.9-4.10. P-H plot will look like Fig 4.10 of pg 151. Use P-H chart pg 653 and table pg 654:

    state 2 is satV at 40C ! H2 = 372 kJ/kg (chart) state 3, outlet of the reversible compressor is found by following the isentropic line to 40C, where H3 = 438 kJ/kg. state 4 is satL at H4 = 256 kJ/kg (table) state 1, H1 = H4

    ( ) kgkJHHQC /11625637212 === WS = (H3 H2) = 438 372 = 66 kJ/kg

    76.166

    116===

    S

    C

    WQ

    COP&

    &

    QC = COP (WS) = 1.76 (9000 J/day) = 16 kJ/day

    (P4.2) Methane,. See Fig 4.9-4.10 pg 150-1.Methane chart pg 651.

    FlbBtuSlbmBtuH

    MPaPFT

    OsatV

    satV

    Oevap

    ./35.2/339

    032.0,280

    2

    2

    =

    =

    ==

    Pressure of 3 isnt given, however, it will be the same as the condenser outlet (P4) since the condenser will be considered to be isobaric. State 4 will be satL at 40F, so P4 = P3 = 2 MPa. Follow isentropic line from the compressor inlet state to 2MPa. At P3=2MPa, and S3 = 2.35 Btu/lbmF, H3 = 485 Btu/lbm, and T3 = 40F. At the outlet of the condenser, the fluid will be saturated liquid at 2 MPa,

    FlbBtuS

    lbmBtuHMPaPFT

    Osat

    satL

    O

    liq./54.1

    /2182,160

    4

    4

    44

    =

    ==

    83.0339485218339

    23

    42

    ,23

    12

    =

    =

    ==

    =

    COPHHHH

    WQ

    HHHHCOP

    netS

    C

    (P4.3) Rankine Cycle, Fig 4.3 Pg. 143 temperatures CC OO 6.99&200 , Sat vapor from Turbine outlet can be found in the sat Pressure table p642.

    State T(OF) P(MPa) H(Btu/lb) S(Btu/lb*OF) 1 -280 0.032 218

    2 Sat'd vap -280 0.032 339 2.35 3' 40 2 485 2.35

    4 -160 2 218 1.54

  • Chapter 4 Practice Problems

    To accompany Introductory Chemical Engineering Thermodynamics J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 2 of 2

    The turbine is assumed to be adiabatic and reversible since no other specifications are given. The inlet state will then have the same entropy as the outlet. Since the boiler/superheater outlet is to operate at T3 = 200 C, and S3 = 7.3592, we must hunt in the superheated tables to find this combination. This will occur between 0.2 and 0.3 MPa. Interpolating, the value of H can be found. Values are tabulated below:

    ( ) kgkJHHW turbineS /1922867267534, === PVVdPWH pumpSpump === ,

    ( )

    ./180.010

    **

    1*

    098.0271.0*104333

    3

    kgkJHJ

    kJMPacm

    Jkg

    cmH

    pump

    pump

    =

    =

    kgkJW netS /82.191180.0192, =+=

    ( )( )

    ( )

    ( )( )

    %83.7

    0783.018.4172867

    82.191/18.41718.0417

    )2867(82.19182.191&

    6

    663

    ,

    =

    =

    =

    =+=

    =

    =

    =

    kgkJHHHHQ

    W

    boiler

    netS

    State TOC P(MPa) H(kJ/kg) S(kJ/kg*K) V(m3/kg) 4 sat vap 99.6 0.098 2675 7.359 5 sat liq. 99.6 0.098 417 1.3028 0.001043

    3 200 0.271 2867 7.3592 6 417.18