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API 650 Thickness Reinforcement Plate for Nozzles 5
- Thread starter hamidun
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For roof nozzles, see Fig. 5.19 and 5.20, which show 6 mm/ 1/4".
For shell nozzles, the common practice is to use the same thickness and grade as the shell material, and this is assumed in Tables 5.6a and 5.6b. However, this can vary as indicated in 5.7.1.8. I'm not sure if there is actually a minimum thickness stated anywhere, though. Normally, nothing less than 3/16" would be used, though.
For shell nozzles, the common practice is to use the same thickness and grade as the shell material, and this is assumed in Tables 5.6a and 5.6b. However, this can vary as indicated in 5.7.1.8. I'm not sure if there is actually a minimum thickness stated anywhere, though. Normally, nothing less than 3/16" would be used, though.
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JStephen,
Do you mean 5.7a and 5.6b? Because in 5.6a and 5.6b, I can not see, it states about nozzle reinforcement plate thickness.
"If a shell plate thicker than required is used for the product and hydrostatic loading (see 5.6), the excess shell-plate thickness, within a vertical distance both above and below the centerline of the hole in the tank shell plate equal to the vertical dimension of the hole in the tank shell plate, may be considered as reinforcement, and the thickness T of the nozzle reinforcing plate may be decreased accordingly. In such cases, the reinforcement and the attachment welding shall conform to the design limits for reinforcement of shell openings specified in 5.7.2"
I am quoting the statement above from table 5.7a and 5.7b. What does this really mean? If it states that the excess shell-plate thickness may be considered as reinforcement, it means we have to exclude corrosion allowance when determining reinforcement plate thickness?
Actually, according to API 650, is there any calculation needed to determine nozzle reinforcement thickness?
Do you mean 5.7a and 5.6b? Because in 5.6a and 5.6b, I can not see, it states about nozzle reinforcement plate thickness.
"If a shell plate thicker than required is used for the product and hydrostatic loading (see 5.6), the excess shell-plate thickness, within a vertical distance both above and below the centerline of the hole in the tank shell plate equal to the vertical dimension of the hole in the tank shell plate, may be considered as reinforcement, and the thickness T of the nozzle reinforcing plate may be decreased accordingly. In such cases, the reinforcement and the attachment welding shall conform to the design limits for reinforcement of shell openings specified in 5.7.2"
I am quoting the statement above from table 5.7a and 5.7b. What does this really mean? If it states that the excess shell-plate thickness may be considered as reinforcement, it means we have to exclude corrosion allowance when determining reinforcement plate thickness?
Actually, according to API 650, is there any calculation needed to determine nozzle reinforcement thickness?
All holes in the shell for nozzles larger than 2" NPS need reinforcement. Reinforcement calculations are essentially area replacement in the vertical plane of the nozzle. Reinforcement (replacement area) can be found in a repad, excess shell plate thickness, thickened shell plates, the nozzle neck, etc as listed in API 650 5.7.2.3 through 5.7.2.7. In some situations the repad thickness may be reduced to zero and sometimes it is useful to do so. For instance, a 3" nozzle in an uppermost shell course may not need any re-pad at all.
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IFRs,
But I can not find the formula to calculate the thickness of reinforcement pad in API 650, do you have any reference for that?
Well, I have read 5.7.2.3 through 5.7.2.7, but I still do not understand how to calculate the thickness of reinforcement pad. Is that about the calculation of area replacement that you mention?
But I can not find the formula to calculate the thickness of reinforcement pad in API 650, do you have any reference for that?
Well, I have read 5.7.2.3 through 5.7.2.7, but I still do not understand how to calculate the thickness of reinforcement pad. Is that about the calculation of area replacement that you mention?
When you have calculated the area required to reinforce the opening, it will be in square inches ( or square MM ). Divide this by the sum of the reinforcing plate heights above and below the opening ( in inches or MM ) and the result will be a thickness. Section 5.7.2.3 limits the reinforcement height above and below the centerline of the opening to the height of the opening. For a simplified example: if the hole in the shell is 10" round, it's vertical dimension is 10". If the shell is 2" thick, the area removed is 2 x 10 or 20 square inches. You need to replace this area. Assume the nozzle neck is 1/2" thick. The nozzle neck contributes 4" x 1/2" x 1/2" inside, outside, top and bottom = 4 x 1/2 x 1/2 x 4 = 4 square inches. Within the shell the neck contributes 1/2" x 2" twice = 2 square inches. The total contributed by the neck is 6 square inches. In this example the repad has to contribute the rest. Assume the reinforcing plate is at least 20 inches high. Effective reinforcement is limited to 10" above and 10" below the nozzle centerline and this repad extends to those limits. Since the hole in the repad is 10" diameter, there is 10" of available reinforcing plate to contribute to the area (5 inches above and 5 inches below). The area needed is 20 - 4 - 2 = 14 square inches. The vertical distance we are using from the repad is 10". Therefore the required repad thickness is 14 / 10 = 1.4 inches. Your repad in this case would be 1.4 inches thick. If you use the tabled repad sizes and make the repad thickness equal to the tank shell you will never go wrong. If you want to (for some reason) use thinner repads you will need to calculate the required area, take credit for any or all of the places API allows and calculate a new repad thickness. API 650 is for new tanks. It is common to order an extra bottom course shell plate to cut repads from, which makes all the calculations and justifications moot.
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IFRs,
I don't get "the reinforcement height above and below the centerline of the opening to the height of the opening" what is the difference between height above the centerline of the opening and height of the opening? For example I have nozzle 10" with length of
side of reinforcing plate or diameter of reinforcing plate is 585 mm and my shell plate is 1828 mm height.
The reinforcement height above the centerline is 585 mm and the height of the opening is 1828 mm - 585 mm - OD of nozzle (273 mm)? Is that right? Or the height of the opening is the total reinforcement height above and below the centerline of nozzle?
I don't get "the reinforcement height above and below the centerline of the opening to the height of the opening" what is the difference between height above the centerline of the opening and height of the opening? For example I have nozzle 10" with length of
side of reinforcing plate or diameter of reinforcing plate is 585 mm and my shell plate is 1828 mm height.
The reinforcement height above the centerline is 585 mm and the height of the opening is 1828 mm - 585 mm - OD of nozzle (273 mm)? Is that right? Or the height of the opening is the total reinforcement height above and below the centerline of nozzle?
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"what is the difference between height above the centerline of the opening and height of the opening? "
My advice is to make a sketch.
API chose to word it that way. They could have said that reinforcement is only effective for a distance equal to half the opening above and below the opening. For an opening 10" diameter, reinforcement is only counted for 10" above the nozzle centerline or 5" above the top and below the bottom of the hole. Sometimes there is more repad above the hole but it is not all counted. Sometimes there is less repad below the hole and only what is there is counted.
"Or the height of the opening is the total reinforcement height above and below the centerline of nozzle?" This is also true but only for the maximum effective reinforcement (not necessarily the available reinforcement height depending on the nozzle elevation).
My advice is to make a sketch.
API chose to word it that way. They could have said that reinforcement is only effective for a distance equal to half the opening above and below the opening. For an opening 10" diameter, reinforcement is only counted for 10" above the nozzle centerline or 5" above the top and below the bottom of the hole. Sometimes there is more repad above the hole but it is not all counted. Sometimes there is less repad below the hole and only what is there is counted.
"Or the height of the opening is the total reinforcement height above and below the centerline of nozzle?" This is also true but only for the maximum effective reinforcement (not necessarily the available reinforcement height depending on the nozzle elevation).
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The book is entitled Design and Analysis of Openings in Steel Tanks and Pressure Vessels. As the name states, the content is limited to penetrations in steel vessels but it does include large openings formed through conical reductions. Standards 620 and 650, ASME Section VIII Division 1, and to a lesser extent Division 2, are discussed. AWWA D100 and NFPA 22 comprise minor sections of the text. I am hoping for a Fall 2019 publication date.
If the example shown in the link is meant to represent API 650, you may want to note that the hole diameter "dp" in the tank shell should be between 11 1/4" and 11 1/2" ( both per API 650 ) and possibly use 11 1/2" ( you have used 11" ) to show the most conservative calculations.
It is not clear from the example text what allowable stress is to be used - product or hydrotest, API 650 lists both and the ratios may be different ( but probably fairly close ).
It is not clear from the example text what allowable stress is to be used - product or hydrotest, API 650 lists both and the ratios may be different ( but probably fairly close ).
Thank you IFRs for your thoughtful response; you are exactly correct, the required amount of reinforcement should read (11)(1.375) = 15.125 in2. The value of peer-review is priceless.
Can you offer any comments or considerations regarding my 20 Sep 18 query regarding API Standard 650 material groups?
Can you offer any comments or considerations regarding my 20 Sep 18 query regarding API Standard 650 material groups?
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To rww88
Since you say the PDF example is for a textbook I hope you won't mind a few "review" comments. These comments reflect how I've design API nozzles for the past 30+ years. They may or may not reflect what you want your text to demonstrate.
All API references are for 12th Edition. I'll try to use your nomenclature for variables.
In the initial question you state the pipe physicals are 83% and 85% of the shell, but somehow the allowable is only 64%. Since I can't use Table 5.2 for the pipe allowable I'd use the formulas in clause 5.6.2. Thus I'd estimate that a realistic Sp would be closer to 0.84 x 28,000 = 23,500 psi. You'd still be able to demonstrate the effect of the weaker pipe in your A'2 formula.
Having calculated the required thickness (tr = 1.117") I'd use that for calculating the required area (as permitted by clause 5.7.2.1). The intent of the example seems to be to reduce the repad as much as possible, so using t required instead of t nominal will help.
In step 5 we should be doing the calcs in the corroded condition, so the permitted projection would be 4(tn-c). The available area thus becomes A2 = 8(tn-c)2.
In step 9 you suggest there's an Standard rule requiring the repad to be the same thickness as the shell. While it commonly is, since it's easiest to just order some additional ring 1 plate for the repads, I'm not aware it's an API requirement.
This is almost too picky to even mention. You could be very exact and use the repad ID (10.875" per Table 5.6b column 4) instead of the hole cut in the shell (11") in your formula to calculate do.
Ignoring how my comments above will change the required pad size, I wouldn't be comfortable suggesting 13¼" OD. This is barely over 1" wide. The weld spacing between the shell-to-neck weld and the repad outside fillet is too small, particularly for such a high-strength shell material and large weld sizes. I'd be reluctant to go below 3" width due to both fabrication and weld spacing concerns. If this was a more typical low-strength shell I'd try to use a thinner repad to match this additional width (but this tank is probably stuck with first ring material). The outside fillet size is t, so even though the weld circumference increases for a thinner repad, the weld volume also comes down by t2, so the total weld volume would be smaller (less manhours = less dollars)
The big savings isn't the 100 lbs of material (or whatever the changes above come out to) but the reduced welding manhours from the smaller repad outside circumference. I'd guess the manhours savings might be 10 times the material cost savings. In fact when an additional first ring plate is bought there's no material dollar savings, just more scrap at the end of the job.
I hope you find at least some of these comments helpful.
Good luck with your text book.
Geoff
Since you say the PDF example is for a textbook I hope you won't mind a few "review" comments. These comments reflect how I've design API nozzles for the past 30+ years. They may or may not reflect what you want your text to demonstrate.
All API references are for 12th Edition. I'll try to use your nomenclature for variables.
In the initial question you state the pipe physicals are 83% and 85% of the shell, but somehow the allowable is only 64%. Since I can't use Table 5.2 for the pipe allowable I'd use the formulas in clause 5.6.2. Thus I'd estimate that a realistic Sp would be closer to 0.84 x 28,000 = 23,500 psi. You'd still be able to demonstrate the effect of the weaker pipe in your A'2 formula.
Having calculated the required thickness (tr = 1.117") I'd use that for calculating the required area (as permitted by clause 5.7.2.1). The intent of the example seems to be to reduce the repad as much as possible, so using t required instead of t nominal will help.
In step 5 we should be doing the calcs in the corroded condition, so the permitted projection would be 4(tn-c). The available area thus becomes A2 = 8(tn-c)2.
In step 9 you suggest there's an Standard rule requiring the repad to be the same thickness as the shell. While it commonly is, since it's easiest to just order some additional ring 1 plate for the repads, I'm not aware it's an API requirement.
This is almost too picky to even mention. You could be very exact and use the repad ID (10.875" per Table 5.6b column 4) instead of the hole cut in the shell (11") in your formula to calculate do.
Ignoring how my comments above will change the required pad size, I wouldn't be comfortable suggesting 13¼" OD. This is barely over 1" wide. The weld spacing between the shell-to-neck weld and the repad outside fillet is too small, particularly for such a high-strength shell material and large weld sizes. I'd be reluctant to go below 3" width due to both fabrication and weld spacing concerns. If this was a more typical low-strength shell I'd try to use a thinner repad to match this additional width (but this tank is probably stuck with first ring material). The outside fillet size is t, so even though the weld circumference increases for a thinner repad, the weld volume also comes down by t2, so the total weld volume would be smaller (less manhours = less dollars)
The big savings isn't the 100 lbs of material (or whatever the changes above come out to) but the reduced welding manhours from the smaller repad outside circumference. I'd guess the manhours savings might be 10 times the material cost savings. In fact when an additional first ring plate is bought there's no material dollar savings, just more scrap at the end of the job.
I hope you find at least some of these comments helpful.
Good luck with your text book.
Geoff
If you are this close to not needing a repad at all, I'd consider using a thicker neck, possibly of higher strength steel to eliminate it altogether. One concern I always have is how does the next inspector know what you did and will they just flag this nozzle as needing repairs?
Good point.
However once the corrections are made to the calc the repad will be closer to 20"Ø x 1.375". Not much smaller than the standard 23"Ø, but that's what we'd expect with t required being so close to t shell minus corrosion.
I threw in the last couple of paragraphs to provide rww88 with more thoughts for his textbook.
However once the corrections are made to the calc the repad will be closer to 20"Ø x 1.375". Not much smaller than the standard 23"Ø, but that's what we'd expect with t required being so close to t shell minus corrosion.
I threw in the last couple of paragraphs to provide rww88 with more thoughts for his textbook.
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