API 650 - 5.10.5.2 units & angle ring

[andyecho]

Hi

I have a query regarding unit of required area given by equation in 5.10.5.2

I have a very small tank design to API 650, diameter = 3m height = 4m
If I use units for equation, D (m), p (kPa) (value 1.87), Fa (N/mm2/ Mpa) (value 160) I get a required area of 0.08mm2
Can this be correct? seems very small
there are no units given in the definitions given below the equation
If I use mm for all length units ie D =3000, p = 0.00187, Fa = 160 this becomes 88mm2 - more reasonable, but not the units of D & p used generally through the code

A secondary question - if this area (whichever it is) can be satisified by say detail I of F2 do I still need the angle specified in 5.1.5.9 e)?

many thanks

Andyecho

 

[JStephen]

I don't have the code here in front of me. You might try converting the units to the US equivalents, and see how that compares.
Usually, that calculation will not control the area, even for a larger tank, so the very small value may be correct.
In the section dealing with minimum top angle sizes, there is a statement that where an Annex F detail has equivalent area to a standard top angle detail, it can be used instead. But it's the standard top angle construction that is the limiting factor, not the small area from the supported cone roof calculation.

 

[Geoff13]

The units for 5.10.5.2 are the same as 5.10.5.1. Thus D is in metres.
The units for p would be the same as the definitions in 5.2.1, and thus p is in kPa.

Your tank is very, very small, and thus I wouldn't be surprised at a very, very small required area.

 

[andyecho]

Thanks to both of you
Thought it was worth checking, I noramlly design pressure vessels so didn't have a feel for "normal" values

 

[HTURKAK]

- No !!!... this can not be correct...


- If the units not defined explicitly , you MUST USE CONSISTENT UNITS !!! So, if Fa (Mpa or with other way N/mm2), You must use D mm, p ( N/mm2) and the final result will be in mm2. If you use Fa (psi), D must be in Inches, and p must be psi.



I understand you want to use ( Detail i )..Figure F-2 shows all Permissible Details of Compression Rings.. Yes you can use ( Detail i ) .. You do not need the angle specified in 5.1.5.9 e) anymore..
5.1.5.9 ( Roof and Top-Angle Joints ) the minimum angels define at item ( 5.1.5.9 e) is not applicable for ( for open-top tanks in 5.9, for tanks with frangible joints per 5.10.2.6, for self-supporting roofs in 5.10.5 and 5.10.6, and for tanks with the flanged roof-to-shell detail described in Item f )



If you want to get more info, pls .post the roof angle, other loadings, is the roof frangible ? etc..

 

[Geoff13]

API 650 formulas use the units specified within the Code, and these are not consistent units.

API deals with this by having suitable correction factors within the equation. That's why there are typically two different equations for the SI and Imperial units with different constants.

 

[HTURKAK]

I assume this is to answer to my post...

This is copy and paste of API 5.10.5.2 ...

Can you show us the applicable units?

If dimensional analysis is performed,

L^2 = (F/L^2)*(L^2)/(F/L^2) so , the figure 8 is dimensionless.... Tan θ is also dimensionless.. in order to obtain the participating area in any unit , at least , WE SHALL CHOOSE THE UNIT of Diameter D and the units of p and Fa SHALL be the same to cancel each other...


I know the derivation of this formula and explicitly saying again, the use of CONSISTENT UNITS IS NECESSARY...

 

[andyecho]

Got side tracked with other issues
thanks for replies. My gut was that it had to be consistent units and that is what I applied

 

[JStephen]

The total vertical load on the entire roof is pi/4*D^2 * p.
The vertical load per unit circumference is that divided by (pi*D) or D/4*p
The load per unit circumference in the plate is then D/4*p/sin(theta)
The radial load per unit circumference in the plate is the D/4*p/sin(theta)*cos(theta) = D/4*p/tan(theta)
The hoop force in the tension ring is then D/4*p/tan(theta) * D/2.
The required area is then D/4*p/tan(theta) * D/2 /Fa = pD^2/8/tan(theta)/Fa.
So yes, that checks provided consistent units are used for p and for Fa.
However, if, say, the roof pressure is pascals and the yield pressure is megapascals, that will throw the answer off by a factor of 1,000,000.
However, that will also be the conversion factor to go from square meters to square millimeters, which I would expect was the way it was to be interpreted. IE, I think unit conversion factors just happened to cancel out.