Applying Corrosion Allowance for Thickness/MAWP
Applying Corrosion Allowance for Thickness/MAWP
(OP)
Hello everyone,
I've started reading on pressure vessel design, specifically "Pressure Vessel Handbook" 9th Ed. by Megyesy, and have come up with doubts on when to apply the Corrosion allowance while calculating the required/minimum thickness and MAWP.
EXAMPLE from the book
GIVEN DATA
Design press, P = 100 psi
Stress value of material, S = 17500 psi of SA-575-70 plate @ 650 F
Long seam joint efficiency, E = 0.85
Head joint efficiency, Eh = 1.00
Outside radius, R = 48 in
Outside diameter, D = 96 in
Corrosion allowance = 0.125 in
t = PR / (SE + 0.6P) = 0.325 in
0.325 + (corrosion allowance) = 0.450 in
Use 0.500 in thick plate
P = SEt / (R - 0.6t) = 155 psi, where t = 0.500 in
MAWP = 154 psi
QUESTIONS
1) I saw another example online (link below) where they subtracted the INNER radius by the corrosion allowance, calculated the required thickness and added back the C.A.. Does it matter when to factor in the C.A. prior calculation?
http://blog.mechguru.com/machine-design/simplified...
2) According to the Engineering Handbook, calculations for the MAWP must be done in WORST CONDITIONS, this includes in CORRODED CONDITION. So, why did it calculate the MAWP with 0.5 in thickness (in new condition) and not in corroded condition which would be t = 0.5 - 0.125 = 0.475 in?
Thank you for anyone that responds.
I've started reading on pressure vessel design, specifically "Pressure Vessel Handbook" 9th Ed. by Megyesy, and have come up with doubts on when to apply the Corrosion allowance while calculating the required/minimum thickness and MAWP.
EXAMPLE from the book
GIVEN DATA
Design press, P = 100 psi
Stress value of material, S = 17500 psi of SA-575-70 plate @ 650 F
Long seam joint efficiency, E = 0.85
Head joint efficiency, Eh = 1.00
Outside radius, R = 48 in
Outside diameter, D = 96 in
Corrosion allowance = 0.125 in
t = PR / (SE + 0.6P) = 0.325 in
0.325 + (corrosion allowance) = 0.450 in
Use 0.500 in thick plate
P = SEt / (R - 0.6t) = 155 psi, where t = 0.500 in
MAWP = 154 psi
QUESTIONS
1) I saw another example online (link below) where they subtracted the INNER radius by the corrosion allowance, calculated the required thickness and added back the C.A.. Does it matter when to factor in the C.A. prior calculation?
http://blog.mechguru.com/machine-design/simplified...
2) According to the Engineering Handbook, calculations for the MAWP must be done in WORST CONDITIONS, this includes in CORRODED CONDITION. So, why did it calculate the MAWP with 0.5 in thickness (in new condition) and not in corroded condition which would be t = 0.5 - 0.125 = 0.475 in?
Thank you for anyone that responds.





RE: Applying Corrosion Allowance for Thickness/MAWP
MAWP Hot and Corroded (H&C) would be performed with corrosion allowance deducted and allowable stress at operating temperature. This is normally what is meant by MAWP.
Note for a formed component such as a head, forming allowance must be deducted in both cases. Pipe tolerance as well.
MAWP often is calculated both ways. Which is which must be specified.
General rule is to completely perform the calculation in question, add the CA, then add the allowances. If evaluating a given thickness, deduct the allowances, then the CA before performing the caclulations.
Regards,
Mike
RE: Applying Corrosion Allowance for Thickness/MAWP
Made complete sense! I did think it was N&C in the back of my head, but the example made no mention of this and it specifically says MAWP while using a non corroded thickness.
I'm assuming a vendor would supply the forming tolerance, right?
RE: Applying Corrosion Allowance for Thickness/MAWP
Regards,
Mike