All,

I have a Nitric acid tank (Pi = 2.5 PSI) with 27 feet dia, 27 feet high with doom roof. I am using Tank software and filling out the spec for the Doom roof. I found the angle - 25 deg. I need help to get the Net area @ Roof/Shell Junction (A).

I am confused with DLR values.

Is there any better way OR formula to calculate A?
Please let me know.

I appreciate your time.

Mayur

According to F.5.1, A is the total required compression area at the roof-to-shell junction, or if you prefer, it is the "compression ring area".
The actual A is a geometrical calculated area- however the boundary of area came from other considerations not explained in API 650, only the rules to calculate A are given in Figure F-2.

In case you are interested on the derivation of the formula, it starts with the calculation of the failure pressure. The failure of the roof-to-shell junction can be expected to occur when the stress in the compression ring area reaches the yield point. This "compression" effect is in a horizontal plane because the roof force (along the tangent at θ angle) can be separated into two forces as the horizontal one, H and the vertical one, V.
The compression force is H (the biggest one) and the vertical force is just V=H*tan(θ).

Pf is given by the roof equilibrium equation on vertical, that's why DLR come into discussion, as well the V force and Pf*PI*D^2/4.
In SI units the failure pressure would be expressed in SI units as:

Pf=8*Fy*A*tanθ/D^2+4/PI*DLR/D^2 - units in [Pa]
(where A is the area subject to stress Fy)

or

Pf=Fy*A*tanθ/(125*D^2)+0.0012732*DLR/D^2 in [kPa]

In order to calculate the correlation between P=the internal design pressure and area A, API considers a safety coefficient of 1.6 applied to the pressure effects term (i.e. the first term), so the maximum design pressure, P is written as:

P=Fy*A*tanθ/(1.6*125*D^2)+0.0012732*DLR/D^2=
=Fy*A*tanθ/(200*D^2)+0.00127*DLR/D^2
which is API 650 F.4.1.

Mathematically this is equivalent with F.5.1.

Best regards.

Yes, You are correct. I appreciate for your reply.

Thanks for the explanation, make it very clear.

Now If I used the formula per F.5.1 (USC unit)

A = D^2(Pi-0.245 Dlr)/D^2/0.962*Fy*(tano).

Units of A = LBF

If I know the weight of the roof plates, say = 4593.8 lb then can I assume the gravity force (f) to get LBF OR How to get LBF?

Please let me know.

Again, I appreciate your time.

Regards,
Mayur

Hello Mayur

API 650 does not really give a formula for computing the surface area of a dome roof. I did it by using the integral calculus to get the formula worked out.

Formulae in F.5.1 are based upon weight minus uplift from internal pressure. Sadly, the code uses USC units and metric units, and they give different results.

Does that answer you question?

A is the total required compression area at the roof-to-shell junction, in inches2.
And the formula you refer should be:
A = D^2(Pi-0.245 DLR/D^2)/(0.962*Fy*(tanθ))
DLR is the nominal weight of roof plate plus any attached structural, in lbf
DLR/D^2 have units of lbf/ft2, is multiplied by 0.19222 to obtain inch H2O units and the formula has a numerical coefficient of 4/3.14159, so finally DLR/D^2 is multiplied by 0.19222*4/3.14159 which is about 0.245, etc.
The coefficient in denominator should be in fact 8/1.6*0.19222=0.961

Is your question as one pound mass in the acceleration due to Earth's gravity exerts one pound force?

Mr. Delaforce,

The surface area of a spherical cap or spherical dome (i.e a portion of a sphere cut off by a plane) can be found in math books. See for example http://mathworld.wolfram.com/SphericalCap.html

What API says is that DLR is the nominal weight of roof plate plus any attached structural, which is correct in my opinion; such force appears in a free-body diagram of the roof.

Mariog,

Thanks for your reply.

I am using the formula you mention earlier for A and I am getting very low number for A.

Please check my input..
1. D = 27 feet
2. Pi = 2.5 Psi
3. Dlr = 4594 lb + 200 lb (for noz attachment) (weight of roof=4594 lb with 0.1875 in thickness)
4. FY= min Yield st = 25,000 Psi for 304L
5. Tan25 = 0.4663

I am getting A=0.06 in^2??

Please check my units for DLR and let me know, what I need to correct.

I appreciate your input and time.

Regards,
Mayur

Ray,

Thanks for your input. I think API 650 does shows the formula for A (under F.5)???

Regards,
Mayur

In API formula, you have to subtract from Pi=2.5 [psig] =69.2 [in H20] the equivalent effect of roof weight which is in your case
0.245 DLR/D^2=0.245*4794/27^2=1.61 [in H20]
In the previous editions of API 650 that term to be subtracted was "8*th", in your case it would be 8*0.1875=1.5 [in H2O].
I would conclude that it is nothing wrong with your "DLR" value.

Of course you shall obtain a bigger value as "A", just follow the correct API formula an pay attention to the units required.
Maybe the mistake is rather the input of Pi as 2.5 inch of water.

What I would add (and probably is of little importance for your case) is that the type "supported dome roof" is not strictly covered by API 650.
By similitude with the definitions given in API 650, 5.10 Roofs, 5.10.1 a supported dome roof without columns may be defined as a roof formed to approximately a spherical surface that is supported principally either by rafters on girders or by rafters on trusses.
I'm not an expert in definitions, in my opinion a better wording would be supported dome plates on self-supporting structure, why not?
Now recalling the last API 650 provisions as valid for this type of roof (quite common in our days, because there is a tendency to rise-up the internal pressure, which means to work with bigger θ than is practical for conical roofs, just trying to keep A as reasonable values), DLR would include all the roof structure not only the plates' weight. This is confirmed by a free-body diagram of roof (a common concept in API 620) and is what I understand from API 650.

Mr. Delaforce,

In USC units API formula of "A" the denominator coefficient with 3 significant decimals should be 0.961 (8/1.6*0.19222156=0.9611078) rather than 0.962. This would be the main source of "errors".

Even so, I checked for Mayur's data and I found that following API formula in USC units the error is -0.095% comparing with a more exact formula working with 8 decimals in all coefficients. Working with the API formula in SI units the error is 0.006% comparing with a more exact formula working with 8 decimals.
When comparing directly USC results vs. SI results, the error is about 0.1%.

For this field of engineering is nothing. I would be happy knowing the "exact formula" has less than 20% inaccuracy versus the physical reality, i.e. Pf as giving actual collapsing of the compression ring- the upper shell. But I guess nobody was interested to experiment destroying a tank just to publish Pf...

Best regards.

Hello mariog

Lamentably, since the advent of the computer, we have become obsessed with numbers. We agonize over the number of decimal places a result should have.

Engineering is an approximation of the real world. No one knows the actual stresses that exist in a component under load. Let me give an example: When computing the hoop stress in a cylinder, we use a simple formula. That is a 'code' stress. The actual stress distribution can only be approximated. The code formulae are based upon many assumptions.

In my early career, I only had a slide rule. There are no decimal points on a slide rule, and the accuracy depended on good eyesight and good judgment. I long for the days when I did all calculations with pen and paper.

This is just my opinion.

I second your opinion.
I still have few slide rules and from time to time I check my eyesight (the good judgment I lost, anyway...)

But my "demonstration" was also a remark on your words "The code uses USC units and metric units, and they give different results."
Is the difference more than-let's say- 0.5%?

Hello mariog

I suppose the difference would vary. I did an exercise with the formula for computing tank wall thickness comparing Metric with USCU. The difference is compounded by the fact that API uses soft conversions, and that the allowable stresses (Sd) have different values in the USCU and metric units in the material tables.

I suppose the simplest solution to all these metric-USCU problems is for America to dump the Imperial system, and align themselves with the rest of the world.

It seems counterproductive to me to have mass and weight in the same units (lb). There is no such confusion in the metric system.

Hi mariog

In API 650 the code takes 1/4 of an inch as 6 mm. What is the error now?

Mariog,

I really appreciate your time and help.

I did use Pi= 69.25 in of H2o (good catch).After finishing up the calculation for A, I am getting A = 4.396 in^2.

Now I am working with Tank software. Under general roof data,

1. Can I use A=4.396 in^2 value (Net area at Roof/Shell (A)in^2?
2. What is your suggestion regarding "Roof live load" for the tank software

regards,
Mayur

Hello Mayur

I shall just stand back and see how this issue pans out. I do have one comment to make - just as an observation:

Pressure in USCU is in inches of water, and in Metric the pressure is in kPa. That to me is strange. Since when is pressure inches of water?

Ray,

I think We already had a conversation regarding this earlier in the post. Also per API CODE 650 - Under F.5 - USC unit - it says that Pi should be in inches of water. just following the unit mention in the code - March 2013

Thanks

Regards,
Mayur

Ok Mayur. I happen not to be an American, and I see thing maybe slightly differently.

Ray,

I am sorry, you may be right. That may be required more detail explanation with the units. Some one can discuss this issue.

Mayur

Historically, devices with water or mercury were used to measure the pressure, so m of water column, mm of water column, inch of water column, mm Hg become "units of pressure". Here people working with either USC and SI abused the units, isn't it?

Of course the conversion 6mm= 1/4" leads to errors- and finally to differences in results, but this is not about error of formulas.

Hi Mariog

In the modern world (as I see it), pressure by definition is Force/Area. Those units are implied by lbf/in^2, kN/mm^2 -> kPa etc. Using inches of liquid disguises that fact, and - again as I see it - has no place in technology.

The metric system has already been abused by the term: kg/cm^2. A kg is a mass, not a force. The cm is used in the kitche*n and in dressmaking.

I'm sure the "height of liquid column " is not so bad. In fact we talk about the force that a column of liquid exerts on the base, divided by the area of base. There are formulas to handle it, in SI the pressure associated with a column of high H is density*g*H.
In hydraulics this unit is still working. A centrifugal pump is usually designed to give the same "H" (discharge head) no matter it pumps water or gasoline. A lot of notions are "heads" in pumps terminology- suction head, NPSH but also in estimating pressure loss when using Darcy equation.
I guess in API 650, Pi is expressed in inch of water by historical reasons. Previously, API counted only the roof plates as helping the roof under pressure and it was quite attractive to count this effect as "8*th" because th inches of steel are equivalent with 8*th inches of water (by densities ratio).

About kg/cm2 as pressure unit, it is in fact kgf/cm2, an old non-SI unit of pressure- technical atmosphere (symbol: at), and kgf and kg as units had the illusory advantage to use the same number for mass and weight on Earth's surface of that mass. Gives the same confusion as lb and lbf (as weight unit) vs the second law of dynamics.

Hi mariog

I am a European. We look at things differently. We have tended to throw off the shackles that have - in our opinion - passed their shelf life.

Now, what do you see as the units of density in your stated equation?

The metric system was built from the ground up during Napoleon's time. It did not depend on the human foot or the length of the first joint of the thumb or the cubit. Mechanical units were constructed from the second, the metre and the kilogramme. Force, pressure, heat transfer etc., were derived from the primary units. The unit of pressure was Newtons(derived unit) = kg(primary unit) multiplied by metres(primary unit) divided by seconds squared(primary unit)

Thus to me (a European) pressure is expressed strictly in force/area units, though it be derived from hydraulic pressure.

In the final analysis, folk will continue with a system in which they are comfortable. I cite as the primary example the resistance of America to adopt what other countries have been using for years. They just don't like it.

I'm an European, too. I am quite familiar with both SI (by school) and USC (by my efforts to understand US engineering works). Mr. Delaforce I think you say that SI is a system that follows the physics and is an "absolute" system of units. It's true, no doubt. However in my earlier school I starting physics with m-kgf-s units. As unit, kilogram force was never SI, but it has been used in technics (probably via German books) and for some reasons it was a unit also in physics books. When they turned to full SI in physics books, I was totally confused when they "revealed" that 1 kilogram of mass weighs 9.81 Newtons rather than 1 kilogram force- for me it was a perfect non-sense. Fortunately, in school you cannot simply declare that you "just don't like it" and I went ahead.

Mayur, your original question was "If I know the weight of the roof plates, say = 4593.8 lb then can I assume the gravity force (f) to get LBF OR How to get LBF? ". You simply know the mass 4593.8 lb hence you know the weight 4593.8 lbf. It is about how your system was constructed- it is a relative (gravitational) system of units. 1 lbm weights (exerts) 1lbf in earth's gravitational field and there is no reason to convert lbm to lbf, or vice-versa; you may simply declare that w=m and skip the "third" letter in lbm and lbf. OK, if you are a scientist working on Earth you may need to consider the local gravity field intensity (g- what we call usually "gravity acceleration") and weight will be slightly different as w = m⋅g/gc≈ m.

Mr. Delaforce, just few historical details. The "metric" system was a child of French revolution and its history is a perfect example of mankind "harmony" that is still driving this Planet. The French Academy approached the British and the Americans in early 1790 with proposals of joint efforts to define a common standard of length based on the length of a pendulum. "The committee" agreed but the proposal brings up the choice of latitude for the pendulum- Thomas Jefferson wanted to be 38°N, Talleyrand for 45°N and Great Britain for London's latitude. French went ahead alone and did it based on other reference. Napoleon, because the French people were quite resistant to the decimal system (and for sure also for other political reasons), introduced in 1812 a new system of measurement "mesures usuelles" (customary measures). Louis Philippe reinstated the original metric system and this took effect in 1840, after a half of century of efforts.

Best regards.

Hello mariog

The final outcome of this correspondence throws light on a very simple problem. By the codes (API, ASME and B31.xx) trying to marry the metric and Imperial systems together in one code has not been completely successful. Other countries have embellished the metric system by introducing non-metric unit such as the kgf and the bar. In my opinion, it has been an unsuccessful marriage.

Industry, again in my opinion, is to blame. It has not taken the lead. The various codes have also avoided taking this bold step.

It is interesting to note, one of the very few countries to use the S.I. Metric system is South Africa. It is legislated. It is completely consistent. It is in their pressure vessel regulations.

What is your suggestion regarding "Roof live load" for the tank software?

Do I need to consider free body dia for the Roof live load or any other formula?

Please let me know. I appreciate your time

By API650, 5.2.1 Loads f): Minimum Roof Live Load (Lr): 1.0 kPa (20 lb/ft2) on the horizontal projected area of the roof. The minimum roof live load may alternatively be determined in accordance with ASCE 7, but shall not be less than 0.72 kPa (15 psf). The minimum roof live load shall be reported to the Purchaser.

Mariog,

Thanks for the information.

I am using Tank software programmed and I am confused with MAWP & MAWV. I was checking App-F. Can some one can help - how to calculate MAWP & MAWV in Tank software

Thanks
Mayur

Hello Mayur

The roof-to-shell junction is limited to the internal pressure it can sustain. The equations are simple, and are basically 'code' equations that can be used quickly. The actual pressure the junction can sustain is anybody's guess. I am puzzled where you found a reference to MAWP and MAWPV. Where did you see that?

Ray,

Thanks for the reply. We have a client and they have couple of storage tanks. Generally every tank have PSV on top. They telling us to find out what is max allowable pressure can tank hold? so they can set PSV less than the MAWP values (PSI).

1, So I did the Tank model and run the analysis. there is a calculation in tank software - Appendix F - Calculated failure pressure - I think this MAWP.

Please let me know the equation so that I can verify (code equation).

2, For MAVP - max allowable working vacuum - That means the external pressure? ((lb/ft^2)


I appreciate your help.

Thanks - Mayur

Hello Mayur

Be careful. API 650 limits the internal pressure to 2-1/2 psi (18 kPa). If this pressure is exceeded, you are in violation of the code. That MAWP calculation is academic only. You cannot use it to set the safety valve.