API 510 PC 5Mar05 Case Study 3 MAWP Analysis
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Transcript of API 510 PC 5Mar05 Case Study 3 MAWP Analysis
CASE STUDY ASME SEC
CASE STUDY 3
MAWP ANALYSIS
DATA:
Following data is presented for a tall process column
Vessel ID (actual) = 8-0. (96)
Vessel height (Tan Tan) = 60
Types of Dished head = 2:1 Ellip.
Corrosion Allowance = Nil
Vessel MAWP (Stamped) = 180 psi
Safe stress for material = 20,000 psi
Vessel thk = 0.55 ( 14.0 mm )
Vessel was fully radiographed.
Due to erosion, three patches were observed at locations A,B,C located at 5, 54 and 57 from top-tan line. The thicknesses measured at locations A,B,C were 0.45 (11.4mm), 0.47(12.0mm), 0.50(12.7mm).
Calculate:
1. Vessel part MAWPs at A,B and C
2. Hydrostatic Pressure at A,B and C
3. Total Pressure at A, B, C
4. Identify Unsafe patch (es), if any
5. Considering limiting patch (es) what is safe vessel MAWP?
SOLUTION:
1 Vessel part MAWP = S E t R+0.6t
MAWP at A = 20000 x 0.45 = 186.4 psi
48 +0.6x0.45
MAWP at B = 20000 x 0.47 = 194.7 psi
48 + 0.6x0.47
MAWP at C = 20000 x 0.5 = 207.0 psi
48 + 0.6x0.5
2 Hydrostatic pressure at any point = h x 0.433 psi
(h = height from top in ft.)
Head depth = D/4 = 8/4 = 2
At A, h = 5 + 2 = 7, Hyd. Pr. = 7x0.433=3.03psi
At B, h = 54 + 2 = 56 Hyd. Pr. = 56 x 0.433 = 24.25 psi
At C, h = 57 + 2 = 59 Hyd. Pr. = 59 x 0.433 = 25.55 psi
3 Total pressure at any point = Vessel MAWP + Hyd. Pressure
At A, 180 + 3.03 = 183.03 psi
At B, 180 + 24.25 = 204.25 psi
At C, 180 + 25.55 = 205.55 psi
4 Since Total pressure at B is > part MAWP at B, it is unsafe.
5 For safety, total pressure at B < part MAWP at B
Take total pressure at B = MAWP = 194.7 psiDeducting hydraulic pressure at B = 24.25 psi
Safe MAWP for vessel = 170.45
Say, 170 psi
D:\Documents and Settings\Administrator\Desktop\IRS-510-Jan- 06\API 510-July Pune\API_510_PC_5Mar05\API_510_PC_5Mar05_Course_Notes_1\API_510_PC_5Mar05_Case_Studies\API_510_PC_5Mar05_Case_Study_3_MAWP_Analysis.doc1/11/2006 / 10:31:53 AM /NCD