Calculation of Icremental Volumes in Storage Tanks 2

[ONEPOINT]

Hi,
I need some help with the following problem:

Given Data

StorageTank(Shell Height=59'-4")and(E=29,000000psi)and the fluid to be stored is Oil.
The Shell Body is divided in 6 Rings.(Butt Welded)..
The first ring has 117.25 inches, and the other 5 Rings have the same width (118.25 inches).
The inside diameter at the bottom of the Tank is 168'-6". There is also a Bottom Crown of 8 and 7/16 inches.

Requested: To calculate the incremental volumes per 1/16 inches in the body of the shell tank.
Regards,

OnePoint

 

[JStephen]

It depends on what you need it for.

We occasionally have customers who want a capacity table (usually not in 1/16" increments) just for informational purposes. In those cases, we make up a table on spreadsheet, based on tank dimensions only. Just cross sectional area times 1/16" times appropriate gallon/feet/inch conversions.

If a certified gauge table is needed, you're looking at something more complicated- check with the people that strap tanks. I think API publishes one or more standards on appropriate procedures. I don't know to what extent they go in preparation of these tables. You could do an elastic shell design and calculate changes in volume due to shell movement, etc.

 

[jte]

Well, if you're looking to evaluate volume to that kind of accuracy, I can just about guarantee that you cannot achieve such accuracy. Based on a 0.0625" change in a 712" high tank, you are looking for an accuracy of roughly 0.01%. Nothing in the real world of tank farms is anywhere close to accurate to 0.01%. As JStephen pointed out in his post, are you expecting to incorporate the change in diameter as the tank strain changes due to changes in the head pressure as the liquid level changes? Don't forget to include shell rotation at the bottom shell to floor seam! How about volumetric expansion/contraction in a 24 hour cycle with a hot day and clear, cool night? How about accounting for the increase in volume as the shell corrodes? How accurate is the as built diameter compared to the drawings?

I'd say the premise of the request for measurements to that kind of accuracy needs to be re-evaluated.

jt

 

[ONEPOINT]

Hi JStephen and jte,
Thanks for your answers.I was asked by my superviser to solve this problem a few days ago. I am quite new in this area of expertise. I have managed so far to do some calculations. I am not sure if I am on the right track.The following is what I have considered. There is also a crown at the bottom of the tank. The height of this crown is 8.43 inches. That crown is within the first ring(117.75 width inches).
How should I extract the volume of the crownn(cone-like shape)?
I would appreciate some suggestions.

Best regards,

OnePoint

________________________________________________________
Given Data

Oil storage tank(Butt Welded Model)
Material(Elasticity modulus: 29X106 psi)
Liquid in tank: Oil, S.G. = 0.86

Shell height: 59’ – 4” = 712”
Inside radius (for ring 1) = 84’ – 3” = 1011”
Bottom crown = 8.4375”
6 rings, heights and thickness as follows:

Ring no Ring width (in) Thickness (in)
1 117.75 1.233
2 118.25 0.977
3 118.25 0.784
4 118.25 0.590
5 118.25 0.396
6 118.25 0.313

1. Inside radii are calculated for all the rings, using the inside radius for Ring 1 and plate thicknesses of each Ring, as follows:

Outside radius, Ring 1 = inside radius1 + plate thichness1 = 1011” + 1.233” = 1012.233”

Assumption: outside diameter for tank is constant.

For rings 2 – 6, inside radius = outside radius – plate thickness

e.g. Ring 2, IR2 = 1012.233”- 0.977” = 1011.256”

2. The inside(underformed/unstressed) circumferences are calculated below,


Ring1 : C1 = 2 x 3.141592483x1011.000”=6352.300” (529.358’)

Ring2: C2 = 2 x 3.141592483x1011.256”= 6353.909” (529.492’)

Calculations for inside radii and inside ring circumferences are summarized in table below:

Ring no Thickness (in) Outside radius (in) Inside radius (in) Circumference (in) C (ft)
1 1.233 1012.233 1011.000 6352.300 529.358
2 0.977 1012.233 1011.256 6353.909 529.492
3 0.784 1012.233 1011.449 6355.121 529.593
4 0.590 1012.233 1011.643 6356.340 529.695
5 0.396 1012.233 1011.837 6357.559 529.797
6 0.313 1012.233 1011.921 6358.084 529.840
3. Incremental volumes per course are calculated for each ring.

The fully stressed circumferences are: C’=C+ ? C

Formula for liquid head stress (correction in feet) using the empty tank circumference(s):

Delta(C)=WxHxC(squered)/12x2x3.141xExT
where:

W = weight of oil, p/ft3
H = height of liquid above ring, ft
C = inside circumference for empty rings, ft
E = modulus of elasticity for steel, psi
T= thickness of the rings

For simplification, a constant K, is taken as:

K=62.3/24x3.141xE
E = 29X106 psi, therefore K = 2.84924 X10-8

Then Delta(C)=Kx(SGxHxsqueredC)/T
Calculations for the circumference corrections (?C) and the fully stressed circumferences (C’) are included in the table below:

Ring no C (ft) ? C (ft) C’ (ft)
1 529.358 0.055 529.413
2 529.492 0.069 529.562
3 529.593 0.086 529.680
4 529.695 0.115 529.810
5 529.797 0.171 529.968
6 529.840 0.217 530.057

The formula for the incremental volume per inch (barrels/inch), using the fully stressed circumference, for each of the rings:
The stressed(Corrected) circumference is C'= 2x3.1415xR'and A'=3.141xR'xR'

Therefore R'=C'/2PI

Inc.Vol. = PI(C'x12/2PI)x(C'x12/2PI)x1/F (inches),
Where correction factor for barrels, F=9702

Results for incremental volumes are attached below:

Ring no Inc. Vol., in barrels/in
1 331.040
2 331.226
3 331.374
4 331.537
5 331.734
6 331.846

Average fully stressed circumference = 529.748’
Average fully stressed diameter = 168.624’

4. Uncorrected incremental volumes (using unstressed circumferences)
Uncorrected Volume=
Ring no C (ft) Barrels/in
1 529.358 330.972
2 529.492 331.139
3 529.593 331.266
4 529.695 331.393
5 529.797 331.520
6 529.840 331.575


The incremental volumes for 1/16” increments:

Ring no Volume, no correction, barrels/in Corrected volume,barrels/in ?V ,barrels per1/16” increments
1 330.972 331.040 0.0043
2 331.139 331.226 0.0054
3 331.266 331.374 0.0068
4 331.393 331.537 0.0090
5 331.520 331.734 0.0134
6 331.575 331.846 0.0170

 

[JStephen]

If you don't have accurate measurements for the tank, there's not a lot of point in going into that much detail in the calculations.

Shell plates can be stacked flush inside, flush outside, or aligned on centerline. Just depends on who did it.

The crown in the floor is intended partly to accommodate settlement. A tank usually settles more in the center than at the shell. So the crown may or may not be there, or may be reduced, once the tank is in service.

If you don't have an actual survey of the tank showing floor profile, exact circumference, etc., just assume a plates are stacked flush inside, ignore shell movement, and you've got a simple spreadsheet to whip up.