Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
PSV Sizing: Spherical Tank 3
- Thread starter slntsam
- Start date
-
1
- #2
Typically, the wetted area is calculated up to an elevation of either 30' or 35' (can't remember, it's in API 520) or your plant practices/specifications. So, depending on the height of your support legs and the diameter, normal working level, etc, you work out the wetted area.
Perry's is one reference that has the formulas for the partial surface area of spheres as do many other books.
Perry's is one reference that has the formulas for the partial surface area of spheres as do many other books.
-
1
- #3
Montemayor
Chemical
JStephen:
This is a neat bit of information to have. Where did you obtain it from or can you cite a reference? I'd like to see it's derivation, if possible, for other applications. I've looked through the Web, google, Jeeves, etc. but have not succeeded in finding the relationship between the liquid level in a spherical tank and the wetted area inside.
This is a neat bit of information to have. Where did you obtain it from or can you cite a reference? I'd like to see it's derivation, if possible, for other applications. I've looked through the Web, google, Jeeves, etc. but have not succeeded in finding the relationship between the liquid level in a spherical tank and the wetted area inside.
With a sphere of radius R, measuring depth from the bottom (not the center), the surface area of the wetted part is A = 2*pi*R*h. This comes from integrating 2*pi*Rsin(a)*Rda from zero to angle a, and substituting cos(a) = 1 - h/R. In the integration, Rsin(a) is the radius in a horizontal plane, Rda is the tangential width of a strip around the sphere, and the 2*pi changes radius to circumference. Angle a is measured from the bottom of the sphere up to the point in question.
Of course this same formula gives the area of an umbrella roof, but it is more convenient to use the (1-cos(a)) form shown above for that.
Some of this geometry may be shown in standard math handbooks, such as the CRC handbook of mathematics. Some of these equations are also shown in "Useful Information for Steel Tanks" (not sure of the exact title) available from AISI.
Of course this same formula gives the area of an umbrella roof, but it is more convenient to use the (1-cos(a)) form shown above for that.
Some of this geometry may be shown in standard math handbooks, such as the CRC handbook of mathematics. Some of these equations are also shown in "Useful Information for Steel Tanks" (not sure of the exact title) available from AISI.
Montemayor
Chemical
Thank you JStephen, I appreciate the favor. The information will come in handy.
Regards
Art Montemayor
Montemayor
Chemical
JStephen:
Once again, to thank you & ask if the angle a is the the central angle and measured in radians (as is the usual math custom) or in degreees (as is the engineering custom). Also, my thanks to my uncle Syd and to quark for their contributions on this subject.
Art
Once again, to thank you & ask if the angle a is the the central angle and measured in radians (as is the usual math custom) or in degreees (as is the engineering custom). Also, my thanks to my uncle Syd and to quark for their contributions on this subject.
Art
The angle "a" would be in radians in the derivation, where it is assumed that Rda gives circumferential length. But once you get to the (1-cos(a)) term, it it doesn't matter, just so you set the calculator to match the angle measurement you're using. Angle "a" is measured from the bottom point of the sphere up to the height in question.
The distance "h" is from the bottom of the sphere up to the point in question. From the center of the tank down to that point is R-h. The cosine of the angle is then (R-h)/R or 1-h/R.
The distance "h" is from the bottom of the sphere up to the point in question. From the center of the tank down to that point is R-h. The cosine of the angle is then (R-h)/R or 1-h/R.
-
1
- #11
Hi gangs,
just to help TD2K out with the reference to height above grade, API 521 fourth edition march 1997 page 14 says and i quote;
" To determine the vapor generation, only that portion of the vessel that is wetted by its internal liquid and it is equal to or less than 25 feet above the source of the flame needs to be recognized"
Therefore, the height to be used in calculating the wetted surface area should be the height of the HHL measured from the grade minus the height of the skirt.
Having known the depth, i will recommended that you use GPSA eleventh edition volume I page 6-23. It gives you tables which you can use to determine the surface area if you know the diameter and the depth of the liquid in the sphere.
i hope this helps.
Buchi
just to help TD2K out with the reference to height above grade, API 521 fourth edition march 1997 page 14 says and i quote;
" To determine the vapor generation, only that portion of the vessel that is wetted by its internal liquid and it is equal to or less than 25 feet above the source of the flame needs to be recognized"
Therefore, the height to be used in calculating the wetted surface area should be the height of the HHL measured from the grade minus the height of the skirt.
Having known the depth, i will recommended that you use GPSA eleventh edition volume I page 6-23. It gives you tables which you can use to determine the surface area if you know the diameter and the depth of the liquid in the sphere.
i hope this helps.
Buchi
- Thread starter
- #12
Well this seems to be a very serious message board and I feel flattered that my website was used as a reference in this thread.
slntsam I guess I'll also accept your thanks because, in a way, I did contribute to this thread.
wolf@1728.com
slntsam I guess I'll also accept your thanks because, in a way, I did contribute to this thread.
wolf@1728.com
Mr Montemayor,
Congratulations on being much deserved tip master of the week.
Wolf,
What you did is no small thing and we all are thankful to you(particularly, the lazy ones like me)
The equation given by JStephen is based on Centroid Theorem of Pappus. I just wanted to do the whole derivation for a future reference but I didn't find any better sedative than my engineering mechanics book, now a days. Hope I will do it someday.
Regards,
Congratulations on being much deserved tip master of the week.
Wolf,
What you did is no small thing and we all are thankful to you(particularly, the lazy ones like me)
The equation given by JStephen is based on Centroid Theorem of Pappus. I just wanted to do the whole derivation for a future reference but I didn't find any better sedative than my engineering mechanics book, now a days. Hope I will do it someday.
Regards,
Buchi,
Surely the height of liquid in the sphere to be used for calculating the wetted area for the API RP-521 calc should be the maximum of:
1 25ft minus the distance from the bottom of the sphere to grade (excluding the depth of the burning liquid pool in the bund?)
2 the sphere radius's (based on the wetted area being limited to that below the sphere's equator)
3 HHL if below the sphere's equator.
Or have I forgotten something?
Surely the height of liquid in the sphere to be used for calculating the wetted area for the API RP-521 calc should be the maximum of:
1 25ft minus the distance from the bottom of the sphere to grade (excluding the depth of the burning liquid pool in the bund?)
2 the sphere radius's (based on the wetted area being limited to that below the sphere's equator)
3 HHL if below the sphere's equator.
Or have I forgotten something?
- Status
Similar threads
- Question
- Question
