API 650 roof to shell junction

[BlakeThomson]

Hi, I have a question about the new API 650 Nov 2009 eddition.

The new code states in 5.10.5.2

"The participating area at the root to shell joint shall be determined using figure f2 and the nominal material thickness less any corrosion allowance shall equal or exceed the following.

pD^2 / (8 x Fa x tan(theta))"

where p = roof pressure
D = nominal diameter
Fa = permisable stress in junction
theta = roof slope

Firstly am I right to assume that this is the minimum thickness after corrosion?

Secondly this severely affects large diameter tanks. (30+m) for a 40m tank and a loading on the roof of 1.46 kPa and roof slopoe of 9.5 degrees the minimum required thickness is 10.44mm.

This requirement was not in the old code and as such will greatly increase minimum thicknesses. I could accept if there was a minimum thickness if the junction would go into compression, however p is a function of the Live load on the roof (1kPa) self weight of the steel and external pressure. The resulting force would place the junction into tension so thickness limits (especially at this magnitude) do not make sense.

Additionally a dimensional analysis on the equation gives an answer in mm^2 x 1000. So my next question is is this even a thickness?

Looking at the old code it states.

"The participating area at the roof to shell junction shall be determined using figure F2 and shall equal or exceed the following.

D^2 x T / (0.432 x sin (theta) x 2.2)"

Where T = Roof loading
D = Nominal Diameter
theta = roof slope

Now this appears to define the minimum area at the junction.

However a dimensional analysis reveals that the result is in Newtons x 1000.

Can anyone shed some light on this clause as at the moment?

 

[DSB123]

What you are discussing is the required "area" in both cases at the Shell/roof interface.

 

[JStephen]

There should be an "and" right before "shall".

Many of the equations used in API-650 are not dimensionally consistent. The constant term has units that are not shown.

Self-supporting cone roofs would not normally be used on tanks of 30 or 40m diameter, for that matter.

 

[BlakeThomson]

My mistake sorry I was using a previous example I had lying around and grabbed it for numbers.

The equation doesn't hold up for roof area.

If we back the diameter back to say 4m diameter the equation goes to

1.62 x 4^2 / (8 x (2/3 x 280) tan (9.5)

(1 kPa + 8mm roof plate compliant with 5.10.5.1)

Gives a required area of 0.103mm^2

Assuming the diameter should be in mm when calculated the result is 103722mm^2

Is this an imperial units calculation and the SI version hasn't been included yet?

 

[JStephen]

I had to look at that one again. In fact, there are not units on the constant like I supposed. It does require consistent units. So p and Fa should be in the same units of pressure, and the result will be in terms of area using the same units as D^2. This is not the normal API way. Or in metric, you could possibly have some 1000 factors that all cancel out, but if that's the case, they normally give you the US equivalent, also.

Using the old equations, this area would seldom govern the cosntruction on a self-supporting roof, and should be the same here.