eduardoeh
Mechanical
- Jan 30, 2015
- 11
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, E[sub]h[/sub] = 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?
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, E[sub]h[/sub] = 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?
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.