MAWP Calculation

July 12, 2017 | Author: Muhammad Zeeshan Wasi | Category: Pipe (Fluid Conveyance), Pressure, Corrosion, Gases, Applied And Interdisciplinary Physics
Share Embed Donate


Short Description

Maximum Allowable Working Pressure Calculation For Process Piping...

Description

Example 1 - Inspection Interval Determination An NPS 16 piping system has been in operation for 10 years and has been taken out of service for its first thorough inspection. The following information is given: • • • •

Pipe service - Gas with 3.5% H2S Minimum required thickness - 0.28 in. Originally installed thickness - 0.375 in. Thicknesses measured at five locations: 0.36, 0.32, 0.33, 0.34, 0.32

Based on the information provided, what maximum thickness measurement interval should be used for this system? Solution:

The pipe service places this system into Class I. Therefore, the maximum interval cannot be more than 5 years based only on the service. Now check the remaining life criterion. CR/Maximum =

0.375 − 0.32 = 5.5 x 10-3 in./yr. 10

Available corrosion allowance = (0.32 - 0.28) = 0.04 in. Maximum Interval =

0.04 = 3.6 years < 5 years 2 x 5.5 x 10 −3

∴ Maximum thickness measurement interval is 3.6 years. MAWP Determination

The MAWP of a piping system must be determined based on the requirements of the applicable piping code (i.e., ASME B31.3 in the case of process plant piping systems). The MAWP of the system is that of the weakest component within the system. Thus, in addition to the pipe itself, all other system components must be considered (e.g., flanges, valves, etc.). If the pipe material is unknown, the MAWP calculations must be based on the lowest grade (i.e., weakest) material and lowest weld joint efficiency that would be permitted by the code. The MAWP calculation is based on: • • •

The actual thicknesses determined by inspection. Double the estimated corrosion loss until the next inspection is done. Additional allowances that might be necessary in specific cases to account for applied loadings other than pressure.

The following examples illustrate calculation of the MAWP. Note that in both cases, only the pipe thickness is considered. 50

Example 2 – MAWP Determination Design Pressure

500 psig

Design Temperature

400°F

Pipe Material

A 106 Gr. B

Pipe Size

NPS 16

Allowable Stress

20,000 psi (from B31.3)

Longitudinal Weld Efficiency

1.0 (A 106 Gr. B is seamless pipe)

Thickness Measured During Inspection

0.32 in.

Observed Corrosion Rate

0.01 in./year

Next Planned Inspection

5 years

Estimated Thinning Until Next Inspection

5 x 0.01 = 0.05 in.

MAWP =

2 S Et D

MAWP =

2 x 20,000 x 1 x (0.32 − 2 x 0.05 ) 16

(from B31.3)

MAWP = 550 psig > 500 psig Since the MAWP exceeds the system design pressure, the system may remain in service at the design pressure without repairs, replacements, or rerating.

51

View more...

Comments

Copyright ©2017 KUPDF Inc.
SUPPORT KUPDF