API 510 PC 5Mar05 Case Study 2 Thickness Calculations
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Transcript of API 510 PC 5Mar05 Case Study 2 Thickness Calculations
![Page 1: API 510 PC 5Mar05 Case Study 2 Thickness Calculations](https://reader030.fdocuments.in/reader030/viewer/2022021400/577cc5f71a28aba7119d6b6d/html5/thumbnails/1.jpg)
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](https://reader030.fdocuments.in/reader030/viewer/2022021400/577cc5f71a28aba7119d6b6d/html5/thumbnails/2.jpg)
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](https://reader030.fdocuments.in/reader030/viewer/2022021400/577cc5f71a28aba7119d6b6d/html5/thumbnails/3.jpg)
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
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