Hello,
The tank I am designing has an external pressure of 1 milli bar g (0.145 psi). On running the calculation, the results suggests to add 2 stiffener rings at various elevations.

Is there any provision in "TANK" to add the stiffener ring?

Thanks.

No, TANK determines the ring size and location.

1 mbar=0.40146308 inches water= 0.1 kPa= 0.014503774 psi

If indeed the external pressure is 1 mbarg, you would consider API 650, 5.2.5 External Pressure:

"See Appendix V for the provisions for the design of tanks subject to partial internal vacuum exceeding 0.25 kPa (1 in. of water).
Tanks that meet the requirements of this Standard may be subjected to a partial vacuum of 0.25 kPa (1 in. of water), without the need to provide any additional supporting calculations."

[1 inch water=2.4908891 mbar]

Best regards.

Just few remarks about this subject.

We don't need to provide any additional calculation for external pressure less than 0.25 kPa (1 in. of water) because API650 already considered 5 psf (aprox 1 inch water) as external pressure on shell, combined with velocity pressure.

Indeed, following the explanations given by API 650, Note 2 a:
- The velocity pressure is considered 1.48 kPa (31 lbf/ft2) with V = 3-second gust design wind speed = 190 km/h (120 mph) at 10 m (33 ft) above ground/
- A 0.24 kPa (5 lbf/ft2) internal vacuum is added for inward drag on open-top tanks or for external pressure on closed top tanks for a total of 1.72 kPa (36 lbf/ft2).
- see also API 650/ 5.9.7.1 Note c; "the modified U.S. Model Basin formula for the critical uniform external pressure on thin-wall tubes free from end loadings, subject to the total pressure specified in Item a.", so the total of 1.72 kPa (36 lbf/ft2) should be considered as p_cr in "the modified U.S. Model Basin formula" which is of form
p_cr=[2.42E/[(1-m^2)^0.75)*(L/D)]]*(t/D)^2.5
E is elastic modulus, m is Poisson coef, etc

Well, identifying now L with H=The maximum height of the unstiffened shell, and considering P_cr=1.72 kPa (36 lbf/ft2) we obtain something closed to formula written in the beginning of 5.9.7.1. The numerical coefficient is 9.53 instead of 9.47, but we may maintain as 9.47- giving just a little more conservative result.

In a later stage API 650 tried to amend this formula considering the case when velocity is not 120 mhr/ 190 kmph and case when the velocity pressure is different due to different factors Kz, Kzt, I- etc. They considered the velocity correction as (190/V)^2 in 5.9.7.1 formula and introduced a new Note d. "When other factors are specified by the Purchaser that are greater than the factors in Items a – c, the total load on the shell shall be modified accordingly, and H shall be decreased by the ratio of 1.72 kPa (36 lbf/ft2) to the modified total pressure."

Strange enough, API failed to provide proper corrections.
The "problem" is that the total pressure has two components: velocity pressure and external pressure (vacuum equivalent)- the last one is not V^2 proportional.

IMO, the correction should be as following:
- for a total of 1.72 kPa (36 lbf/ft2) consisting of 1.48 kPa (31 lbf/ft2) velocity pressure and 0.24 kPa (5 lbf/ft2) external pressure,
H=9.47*t*sqrt[(t/D)^3] in SI units as explained in 5.9.7.1
- for a velocity V [in specific units], factors as considered in 5.9.7.1 Note a and .24 kPa (5 lbf/ft2) external pressure, the result should be amended by factor:
1.72/[1.48*(V/190)^2+0.24]
or 36/[[31*(V/120)^2+5]
- for a velocity V [in specific units], factors as considered in 5.9.7.1 Note a and EP [in specific units] as external pressure, the result should be amended by factor:
1.72/[1.48*(V/190)^2+EP]
or 36/[[31*(V/120)^2+EP]
- for a velocity V [in specific units], other factors than considered in 5.9.7.1 Note a- giving a Velocity_pressure [in specific units] plus EP [in specific units] as external pressure, the result should be amended by factor:
1.72/[Velocity_pressure +EP]
or 36/[[Velocity_pressure +EP]

Maybe it appears as a complicated procedure; we would simplify it keeping in mind that the formula
H=9.47*t*sqrt[(t/D)^3] in SI units
considers in denominator 1.72 kPa as total pressure.
When we have the total pressure Velocity_pressure +EP we need to consider Velocity_pressure +EP in denominator; this is equivalent to consider in the first phase 1.72 kPa in denominator and to multiply the results with 1.72/[Velocity_pressure +EP] later.

Best regards.

Thank you Mariog. But, the external pressure is 10 millibar g (1 Kilo pascal or 4" water column) I had mistyped it as 1 millibar. Please let me know how I should add the stiffeners.

Thanks.

In a normal Procedure under Basic API 650:

- use 5.9.7.2 a. as "transposing width procedure" based on the top shell thickness. Note that this procedure is based on the fact a shell course of width W1 and thickness t1 has the same p_cr as a shell course of width W2 and thickness t2 when (t1^2.5)/W1=(t2^2.5)/W2;

- now you have a buckling equivalent (having thickness=top shell thickness) of the real shell;

- calculate H1=The maximum height of the unstiffened transposed shell:

a. API says H1=9.47*t*sqrt[(t/D)^3]*(190/V)^2*1.72/ Actual_total_pressure[kPa]
IMO, formula is not conservative and must be corrected.

b. Also IMO, H1=9.47*t*sqrt[(t/D)^3]*1.72/Actual_total_pressure[kPa]
in your case would be:
H1=9.47*t*sqrt[(t/D)^3]*1.72/[1.48*(V/190)^2+1] because EP=10mbar=1kPa (but note that is only my opinion, not API)

- consider stiffeners as the distance on transposed shell, between top shell- first stiffener and between stiffeners, is equal to H1; that means you got the number of stiffeners;

- calculate back the maximum distance between top shell- first stiffener and between stiffeners, on actual shell, using the inverse transformed shell formula;

- the actual distances between top shell- first stiffener and between stiffeners can be equal or less than the calculated distances.

OR

use Appendix V of API 650.

If your question is what you can do within TANK, I cannot answer.
Just keep in mind that TANK cannot correct API 650, must follow it!

Best regards.