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