api 510 pc 30apr05 ynb case study 2 thickness calculations

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  • 8/17/2019 API 510 PC 30Apr05 YNB Case Study 2 Thickness Calculations

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     _____________________________________________________________________________ 

    Professional Training and ConsultationAPI 510 Preparatory

    CASE STUDY 2THIC!ESS CA"CU"ATI#!S

    DATA:

    Design pressure = 26 kg/cm2 = 370 psig

    Design temperature = 300°C = 572°F

    Inside diameter = 1200 mm = 47.25 in

    .!.C. = 515 gr 70

    C"rr/#rr". a$$"%ance = 3.0 mm

    &$$"%a'$e (tress = 1)400 psi

    *+pe ", supp"rts = (add$e supp"rts

    -engt ** = 315 in 000 mm

    adi"grap+ = Fu$$ / (p"t

    "int e,,icienc+ # = 1 Fu$$# = 0.5 (p"t

    REQUIRED:

    1 Ca$cu$ate reuired ticknesses ,"r se$$ and eads c"nsidering '"tcases ", radi"grap+.

    2 Ca$cu$ate & ,"r se$$ ,"r '"t cases ", radi"grap+.

     /8ar/%%%/apps/c"n8ersi"n/tmp/scratc97/31604)4.d"c

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     _____________________________________________________________________________ 

    Professional Training and ConsultationAPI 510 Preparatory

    SOLUTION TO CASE STUDY 2

    AA) Shell Thickness:

    Design tickness =C 

     pSE 

     PR+

    −   6.0

    ,"r #=1< td =.63.14364.113

    3706.019400

    603370=+=+

    −   x

     x

    sa+< 16.0 mm

    ,"r #=0.5< td = 71.16371.1333706.085.019400603370

    =+=+

    −   x x

     xsa+

    1.0 mm

    F"r & ca$cu$ati"n<a8ai$a'$e ;ett tk t = 16 3 = 13 mm ,"r #=1

    t = 1 3 = 15 mm ,"r # = 0.5

    & =t  R

    SEt 

    6.0+

    F"r #=1< & =136.0603

    13119400

     x x x

    +  = 412.) psig

    F"r # = 0.5< & =156.0603

    1585.019400

     x

     x x

    +  = 404.2 psig

    BB) Dished Head:

    1. >emisperica$ ead

    td = C  P SE  PR

    +

    −   2.02

    F"r #=1< td = 33702.0194002

    603370+

    −   x x

     x= 5.76 ? 3 = .76. sa+ 10.0 mm

    F"r # = 0.5< td = 33702.085.0194002

    603370+

    −   x x x

     x

     = 6.7 ? 3 = ).7 mm

      sa+ 10.0 mm

     /8ar/%%%/apps/c"n8ersi"n/tmp/scratc97/31604)4.d"c

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     _____________________________________________________________________________ 

    Professional Training and ConsultationAPI 510 Preparatory

    2) 2:1 Elli!ical Head

    td = P SE 

     PD

    2.02   − ? C

    F"r #=1< td = 352.1133702.0194002

    1206370+=+

    −   x x

     x= 14.52. sa+ 16.0 mm

    F"r # = 0.5< td = 33702.085.0194002

    1206370+

    −−   x x

     x = 13.56 ? 3 = 16.56 mm

      sa+ 1.0 mm

    ") T#$ishe$ical Head

    t = + C L= Dia = 1203 mm.

     = mm x

     x x39.23339.203

    3701.019400

    1203370885.0=+=+

    −  say, 24.0 mm (for E=1)

      = mm x x x x

    00.27300.2433701.085.019400

    1203370885.0=+=+

    − say, 28 mm (for

    E=0.85)

    COMPARISON OF THICKNESS (mm)

    ItemShe !oris"h. #ea$ 2%1 Ei". #ea$

    #emis"h.#ea$

    E=1.0 E=0.85 E=1.0 E=0.85 E=1.0 E=0.85 E=1.0 E=0.85

    &i'.e*ire$!hi'ess

    11.-4 13.1 20.3/ 24.00 11.52 13.5- 5.- -.8

    Desi'!hi'ess

    14.-3 1-.1 23.3/ 2.00 14.52 1-.5- 8.- /.8

    omi'a!hi'ess

    1-.00 18.00 24.00 28.00 1-.00 18.00 10.0 10.0

     /8ar/%%%/apps/c"n8ersi"n/tmp/scratc97/31604)4.d"c

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    0.885 LSE 0.1