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.