Need help calculating T1, T2 and thickness - API 620 3

[Rafleonard]

Im new working with API620 and need some guidance.
Im designing several storage and mixing vertical cylindrical tanks base on API 620.
One of the tanks has the following data:
Radius = 40 in
Wshell = 3800 lbs.
Wroof = 550 lbs.
Wimpeller = 300 lbs.
Density = 998 kg/m3
Cone bottom.
Opened to atmosphere.

Base on eq. 10 and 11 from 5.10.2.5 section.
T1 = (Rcyl/2)*[P + (W + F)/A]
T2 = P*Rcyl

P is defined as "total pressure" which includes "pressure resulting from the liquid head at the level under consideration in the tank" and "gas pressure." "W" is defined as "total weight of that portion of the tank and its contents.

Base on that P = Pliq + Pair where Pair = O since is opened to atm.
W = Wliq + Wshell + Wroof
F = Wimpeller.

Doing the free body analysis at the point where the cone intersects with the tank shell. Having all loads acting down as negative we can eliminate P and Wliq (am I correct?).
We end up with T1 = -[(Wshell + Wroof + Wimpeller)/2]
Doing the calculations I ended up with T1 = -18.34 Lb/in and T2 = 10.55 lb/in

According to the standard, since T1 is under compression and T2 under tension I have to use Figure 5-1 to find N.
I do not know how to find the computed compression stress to compare it to the allowable compression stress to see if the thickness at that particular point used is ok.

Other doubt is as long as you have a cylindrical tank opened to atmosphere T1 will always be negative?

Thank you for your help.

 

[JStephen]

Assuming you are analyzing near the junction but above any supports, what you have sounds reasonable.
You should find some equations there for allowable compression stress as a function of t/R.
There may be a statement somewhere in there that says if stress in one direction is less than 5 or 10% of allowable, it need not be combined with stress in the other direction- don't have the standard in front of me, though.

 

[HTURKAK]

What is the reason for designing to API 620 ? . The scope of API 620 covers the tanks with pressures not more than 15 lbf/in.2 . If the tank is open to atmosphere API 620 should not be followed.

The radius (80 in), wt of shell and roof implies the tanks shop fabricated atmospheric tanks. If this is case, you should look API 650
APPENDIX J—SHOP-ASSEMBLED STORAGE TANKS.

Pls provide more info. ( dimensions, content, impeller, wind and seismic loading cond...) with a sketch to get more helpful responds.

 

[dig1]

HTURKAK,

The OP listed the tank as having a conical bottom; therefore, API 620 would be correct.

Rafleonard,

JStephen is providing correct advice.

If there is no internal pressure the side wall T1 may very likely always be negative (compressive). I have seen agitators with upward reaction and if this is larger enough it may create tension. If the tank is outside wind may create a small tensile force.

When you cut the FBD below the cone-cylinder junction (CCJ) the conical bottom will be in tension due to the weight of the cone plus the weight of the contents being carried by the cone back to the side wall and to the support. If the support is above the CCJ then a portion of the straight side will be in tension as well.

As a reference I like Structural Design & Analysis of Process Equipment by Jaward and Farr. If you can find it, C.T. Main Engineers developed a in-house design guide for the design of tanks (this was for the pulp and paper industry but fundamentals and fundamentals) has a couple of great examples. C.T. Main is long out of business but there may be a copy on-line or e-bay.

BR,

Patrick

 

[Rafleonard]

JStephen

This is what the standard says "in cases where the unit force acting in compression does not exceed 5% of the coexistent tensile unit force acting perpendicular to it, the designer has the option of permitting a tensile stress of the magnitude specified in 5.5.3.2 instead of complying strictly with the provisions of this paragraph"

I may be wrong but I imagine its talking about T1 and T2 (correct?) if that is the case, T1 is greater that 1.05T2.

you are right on the allowable tensile stress (Sta), the formula is Sta = N*Sts.. Which in this case Sts is 16,000 since I'm working with A36 steel, the value of N is taking from Figure 5-1.
I use t = 3/16" to find the value on x-axis from Figure 5-1.
Using a corrosion allowance of 1/16"
(t - c)/R
That gives me 0.003125
But this is as far as I go. I do not know how to calculate the Scc and Stc (Computed compression and tensile stresses) so I compare it to Sta and Sca, maybe I'm overcomplicating myself.
Can I use the Hoop and axial stress formulas to calculate Scc and Stc?


dig1

Thank you for the reference on the book, I will look for it.

For this particular tank, wind load can be neglected since is inside the building

 

[Geoff13]

The computed stresses (Scc and Stc) are just the computed forces (T1 and T2) divided by the corroded thickness.

You tank is very small, and so your computed forces are also small, so I'd guess that minimum thickness (5.10.4.1) will work.

 

[Rafleonard]

Geoff13

Where did you get that formula for Scc and Stc?
If you do that shouldn't the formula be Scc = T1/(t-c) and Stc = T2/(t-c)? why is the pressure only acting against the corroded thickness and not the thickness of the shell minus the corroded thickness?

 

[Geoff13]

If T1 or T2 is the force in pounds per inch of width, then we need to divide by the thickness over that width to calculate at the stress at that location.

I think we're saying the same thing, but perhaps our terminology is different? t is the new thickness, c is the corrosion allowance, and t-c is the corroded thickness at some future date.

 

[Rafleonard]

Geoff13

You are right, I misunderstood your term "corroded thickness" with corrosion allowance.

Using that eq. I get Scc = 18.34/0.1875... Scc = 98 PSI and Stc = 56 PSI.
Like you were saying, since my tank is very small and pressures are low.
Taking that Scc and doing an analysis with t = 3/16, looking in Figure 5-1, my N is equal to 1, which means that Sta = (1)(16,000)... Sta = 16,000 PSI which is > than Scc, I'm good.

Like you said, I could work with the minimum thickness from 5.10.4.1 but I wanted to do all the calculations for learning purposes.

Thank you Geoff13.