API 510 PC 5Mar05 Case Study 2 Thickness Calculations

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CASE STUDY 2 THICKNESS CALCULATIONS DATA: Design pr = 48kg/cm 2 = 683 psig Design temp = 300ºC = 572ºF Inside dia = 1200 mm = 47.25 in .!.C. = 515 gr 70 C"rr. /#rr". $%%"& = 2.5 mm '0.1 in( $%%"&a)%e *tress = 1+400 psi ,engt- ' ( = 315 in '8000mm( adi"grap- = F%% / *p"t Calculate required thickness Desi!n thickness and N"#inal thickness $"r shell and heads c"nsiderin! %"th cases "$ radi"!ra&h'(  /ar/&&&/ap ps/c"ner si"n/tmp/scra tc-5/24176+541.d"c 8/20/2014 / 4303+ /CD

Transcript of API 510 PC 5Mar05 Case Study 2 Thickness Calculations

Page 1: API 510 PC 5Mar05 Case Study 2 Thickness Calculations

 

CASE STUDY 2

THICKNESS CALCULATIONS

DATA:

Design pr = 48kg/cm2 = 683 psig

Design temp = 300ºC = 572ºF

Inside dia = 1200 mm = 47.25 in

.!.C. = 515 gr 70

C"rr./#rr". $%%"& = 2.5 mm '0.1 in(

$%%"&a)%e *tress = 1+400 psi

,engt- '( = 315 in '8000mm(

adi"grap- = F%% / *p"t

Calculate required thickness Desi!n thickness and N"#inal

thickness $"r shell and heads c"nsiderin! %"th cases "$

radi"!ra&h'(

 /ar/&&&/apps/c"nersi"n/tmp/scratc-5/24176+541.d"c

8/20/2014 / 4303+ /CD

Page 2: API 510 PC 5Mar05 Case Study 2 Thickness Calculations

 

683 x 602.519400 x 0.85 – 0.6x683

683 x 1205

2 x 19400 – 0.2 x 683

SOLUTION:

AA) Shell Thickness

t =

= = 21.7……………….for E=1

Required thickne = 21.7!2.5 = 24.2 "". #o"in$% thickne= 26.0 mm.

&' i"i%$r c$%cu%$tion for E=0.85

  t = ! 2.5 = 25.5!2.5=28.0"" (E=0.85)* Say, 28.0 mm.

BB) Dished ead :

!) ems"he#ical t = = = 10.64 ""  (E=1)

  +ddin, -.+* Required thk. = 13.14* #o"on$% thk= !$.0 mm

  ith E = 0.85* t = 12.75"" . +ddin, -.+ * /hk = 15.25. Say, !6.0 mm.

 2) 2:! %lli". ead

ith E =1.0* t = =

= 21.28. +ddin, -.+ of 2.5 ""* thk = 23.78 * Say, 2$ mm.

 /ar/&&&/apps/c"nersi"n/tmp/scratc-5/24176+541.d"c

8/20/2014 / 4303+ /CD

 

RE – 0.6

683 x 602.519400 – 0.6 x 683

x R2E – 0.2

683 x 602.52 x 19400 x 1 –0.2x683

2E – 0.2

Page 3: API 510 PC 5Mar05 Case Study 2 Thickness Calculations

 

0.885 x 683 x 120519400 – 0.1 x 683

0.885 x 683 x 120519400 x 0.85 – 0.1 x 683

&' i"i%$r c$%cu%$tion for E=0.85*

t = == 25.5 "" (E=0.85)

 +ddin, -.+. = 28.0 "". Say, 28 mm "la&e

') T(#is"he#ical head

t = ! - = i$ = 1205 "".

  = = 37.5 !2.5 = 40.0 "" h((se &hk*$0 mm 

t = = 44.35 ! 2.5 = 46.8 "" Say, $+.0 mm. &hk

O-AISON O/ TIN%SS

te"he%%E=1.0

he%%E=0.85

/orih.E=1.0

/orih.E=0.85

21 E%%iE=1.0

21E%%iE=0.85

e"ih.

E=1.0 E=0.85

1)

-$%cu%$ted/hickne

21.7 25.5 37.5 44.35 21.28 25.5 10.64 12.75

2)

/hk ! -.+ 25.0 28.0 40.0 48.0 24.0 28.0 14.0 16.0

 /ar/&&&/apps/c"nersi"n/tmp/scratc-5/24176+541.d"c

8/20/2014 / 4303+ /CD

 

683 x 12052 x 19400 x 0.85 –0.2x683

0.885 x x E – 0.1

Page 4: API 510 PC 5Mar05 Case Study 2 Thickness Calculations