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  • TALLER CICLO COMBINADO MODIFICADOUNIVERSIDAD NACIONAL DE COLOMBIA

    Sede BogotPlantas Temicas

    Ciclo Joue-Brayton

    T1 25 C RPcom 8P1 120 kPa kaire 1.4

    cpaire 1000 Jkg Kmgas 42

    kgs

    P2 =P1 RPcom 9.6 105 Pa P3 P2

    T2 =T1 P2P1

    kaire 1kaire

    540.083 K

    Qcal cpaire mgas (( T3 T2)) Qcal 43 MW

    cpgas 1100 Jkg KT3 =+

    Qcalcpaire mgas

    T2 1.564 103 Kkgas 1.3

    Wturb1 23 MW

    T4 =+Wturb1

    cpgas mgasT3 1.066 103 K

    P4 =

    T3T4

    kgaskgas 1

    P3

    1

    1.824 105 Pa

    Wcomp1 =(( T2 T1)) cpaire mgas 1.016 107 W T5 540 C

    P5 P4

    Ciclo Rankin

    T6 500 CP6 60 bar P7 40 barh6 3422

    Jgm

    P7s P7

    s6 6.88 J

    gm Ks7s s6 h7s 3290

    Jgm

    2

  • sth %90 sth h6 h7h6 h7s

    h7 =h6 sth h6 h7s 3.303 106

    m2

    s2

    T7 438 C

    T8 550 CP8 P7h8 3559

    Jgms8 7.23

    Jgm K

    P9 8 bar stl %75

    s9s s8 h9s 3054 J

    gm

    h9 =h8 h8 h9s stl 3.18 106

    m2

    s2T9 358 C s9 7.43

    Jgm K

    P10 1 bar

    s10s s9 h10s 2688 J

    gm

    h10 =h9 h9 h10s stl 2.811 106

    m2

    s2T10 167 C s10 7.69

    Jgm K

    P11 0.08 bar

    s11s s10 h11s 2407.13 J

    gm

    h11 =h10 h10 h11s stl 2.508 106

    m2

    s2T11 41.5 C s11 8.01

    Jgm K

    Tomando que a lasalida del intercambiadr cerrado (punto 22), liquido saturado

    P22 P10T22 99.6 Ch22 417.5

    Jgm

    h14 h22P14 P11T14 41.5 Csq14 %10.14

    Tomando 16 como liquido saturado

    P16 P9 P17 P6h16 720.864

    Jgm

    T16 170.406 Cv16 0.00111

    m3

    kg

    W h dv p vp

  • W h dv p vp

    h17 =+h16 v16 (( P17 P16)) 7.266 105 m2

    s2T17 171.069 C

    Tomando 19 como liquido saturado

    P19 P7 P20 P6h19 1087.5

    Jgm

    T19 250.35 C Suponiendo que el intercambiador T18 T19

    v19 0.00125 m3

    kgP18 60 barh18 1087.45 J

    gm

    m (( 1 x1)) (( h19 h7)) m x1 (( h18 h17))

    x1 =h7 h19

    +h18 h17 h7 h190.86

    W h dv p vp

    h20 =+h19 v19 (( P20 P19)) 1.09 106 m2

    s2T20 250.88 C

    Suponiendo que el fluido en 12 sale saturado

    P12 P11 P13 P9

    h12 173.84 J

    gm

    v12 0.00101 m3

    kg

    T12 41.5 C

    h13 =+h12 v12 (( P13 P12)) 1.746 105 m2

    s2T13 41.52 C

    Suponiendo que a la salidaa del intercambiador T22=T15

    P15 8 barT15 T22

    h15 418 J

    gm

    m x1 x2 (( 1 x3)) (( h22 h10)) m x1 x2 (( h15 h13))

    x3 =1 (( h15 h13))

    h10 h220.898

  • m ((x1)) h16 +m x1 (( 1 x2)) h9 m x1 x2 h15

    x2 =h16 h9h15 h9

    0.89

    m h21 +m x1 h18 m (( 1 x1)) h20

    h21 =+x1 h18 (( 1 x1)) h20 1.088 106 m2

    s2P21 P6

    T21 250.46 C

    Qrec hrsg cpgas mgas (( T5 T4)) m (( h6 h21)) hrsg %65

    m =hrsg cpgas mgas (( T5 T4))

    h6 h213.254

    kgs

    Q11_14_12 +m (( x1 x2)) h12 (( +m (( x1 x2 (( 1 x3)))) h14 m (( x1 x2 x3)) h11))

    =Q11_14_12 5.286 106 W

    Q78 =m ((x1)) (( h8 h7)) 7.157 105 W

    Wth =m (( h7 h6)) 3.865 105 W

    Wtl =m (( ++x1 (( h9 h8)) x1 x2 (( h10 h9)) x1 x2 x3 (( h11 h10)))) 2.658 106 W

    Wc1 =m x1 x2 (( h13 h12)) 1.993 103 W

    Wc2 =m x1 (( h17 h16)) 1.615 104 W

    Wc3 =m (( 1 x1)) (( h20 h19)) 1.139 103 W

    th =|||

    ++Wturb1 Wth Wtl+++++Qcal Q78 Wcomp1 Wc1 Wc2 Wc3

    |||

    0.483

  • =T1 298.15 K =T11 314.65 K =T21 523.61 K=T2 540.083 K =T12 314.65 K =T22 372.75 K=T3 1.564 103 K =T13 314.67 K=T4 1.066 103 K =T14 314.65 K=T5 813.15 K =T15 372.75 K=T6 773.15 K =T16 443.556 K T17 171.069 C=T7 711.15 K =T17 444.219 K=T8 823.15 K =T18 523.5 K=T9 631.15 K =T19 523.5 K

    =T10 440.15 K =T20 524.03 K T21 250.46 C

    =h6 3.422 106 m2

    s2=P1 1.2 105 Pa=P2 9.6 105 Pa

    =h7 3.303 106 m2

    s2=P3 9.6 105 Pa=P4 1.824 105 Pa

    =h8 3.559 106 m2

    s2=P5 1.824 105 Pa=P6 6 106 Pa

    =h9 3.18 106 m2

    s2=P7 4 106 Pa=P8 4 106 Pa

    =h10 2.811 106 m2

    s2=P9 8 105 Pa

    =P10 1 105 Pa=h11 2.508 106 m

    2

    s2=P11 8 103 Pa=P12 8 103 Pa

    =h12 1.738 105 m2

    s2=P13 8 105 Pa=P14 8 103 Pa

    =h13 1.746 105 m2

    s2=P15 8 105 Pa=P16 8 105 Pa

    =h14 4.175 105 m2

    s2=P17 6 106 Pa=P18 6 106 Pa

    =h15 4.18 105 m2

    s2=P19 4 106 Pa=P20 6 106 Pa

    =h16 7.209 105 m2

    s2=P21 6 106 Pa=P22 1 105 Pa

    =h17 7.266 105 m2

    s2

    =h18 1.087 106 m2

    s2 =h20 1.09 106 m2

    s2=h19 1.088 106 m2

    s2 =h21 1.088 106 m2

    s2

    =h22 4.175 105 m2

    s2