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resolución de ejercicios en Mathcad

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  • RENDIMIENTO DE LA TRANSMISIN

    DATOS OBTENIDOS DEL CATALOGO DE UNA CAMIONETA KIA K2700

    P 83 hp 6.189 104 W=:= n1 4150 rpm 434.5871

    s=:=

    M 17.5 kg m:= n2 2400 rpm 251.3271

    s=:=

    Traccin 4x2

    Relaciones

    i1 4.117:= i2 2.272:= i3 1.425:=

    i4 1.000:= i5 0.871:= ma 3.958:=

    dif 4.444:=

    Neumaticos: 195R14

    Relacion de transmisin en cajasitrans i1 iD:=trans

    iT1 i1 dif:= iT1 18.296=iT2 i2 dif:= iT2 10.097=iT3 i3 dif:= iT3 6.333=iT4 i4 dif:= iT4 4.444=iT5 i5 dif:= iT5 3.871=ima ma dif:= ima 17.589=

    Para el calculo de velocidad el diametro de la rueda:

    Dr 2 0.195 0.8 m 14 in+:= Dr 0.668m=

    Torque maximo

    Tmax M:= Tmax 17.5m kg= para 2400 rpm( ) 251.327 1s

    =

    v n1Dr

    2:= v 145.065 m

    s=

    Vv

    iTc:=

    iTc

    Las velocidades

    V1v

    iT1:= V1 7.929 m

    s=

    V2v

    iT2:= V2 14.367 m

    s=

    V3v

    iT3:= V3 22.907 m

    s=

    V4v

    iT4:= V4 32.643 m

    s=

    V5v

    iT5:= V5 37.478 m

    s=

    Vmav

    ima:= Vma 8.247 m

    s=

  • Calculo del par maximo

    U n2Dr

    2 83.893 m

    s=:= 302.0148 km

    hr 83.893 m

    s=

    U1U

    iT1:= U1 4.585 m

    s=

    U2U

    iT2:= U2 8.309 m

    s=

    U3U

    iT3:= U3 13.248 m

    s=

    U4U

    iT4:= U4 18.878 m

    s=

    U5U

    iT5:= U5 21.674 m

    s=

    UmaU

    ima:= Uma 4.77

    m

    s=

    Coeficiente de adaptacin:

    Kad

    Mmax

    Mn

    :=KadMmax

    Mn

    :=

    Kad 1.08.......1.5( ):=ad M. diesel

    trans c v:=trans

    0.03......0.05( ):= 0.03......0.051 0.98...0.99( ):= 0.98...0.99 1 0.985:=

    c 1( )m 2( )m:=c 2 0.97...0.98( ):= 0.97...0.98 2 0.95:=v 1

    Kad:=Kad

    0.03:=

    Mmax 716.2Ne

    n:=

    eMmax

    716.2 83 0.746 kg m4150 0.736

    := Mmax 14.519m kg=

    KadM

    Mmax:= Kad 1.205=

    v 1

    Kad:= v 0.975=

    c 122

    2:= c 0.876=

    trans c v:= trans 0.854=

    Potencia perdida por resistencia:

  • NT Ne 1 trans( ):=TNt P 1 trans( ):= Nt 9.047 103 W=NT 83 hp 1 trans( ):=T NT 12.19 hp=N

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