RichardCHK
Industrial
- Nov 3, 2011
- 1
a. The velocity pressure is:
p = 0.00256Kz Kzt Kd V 2 I G = 1.48 kPa (31 lbf/ft2)
where
Kz = velocity pressure exposure coefficient = 1.04 for exposure C at a height of 40 ft,
Kzt = 1.0 for all structures except those on isolated hills or escarpments,
Kd = directionality factor = 0.95 for round tanks,
V = 3-second gust design wind speed = 190 km/h (120 mph) at 10 m (33 ft) above ground (see 5.2.1[j]),
I = importance factor = 1.0 for Category II structures,
G = gust factor = 0.85 for exposure C.
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).
b. The wind pressure is uniform over the theoretical buckling mode of the tank shell, which eliminates the need for a shape factor for the wind
loading.
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.
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 H1 shall be decreased by the ratio of 1.72 kPa (36 lbf/ft2) to the modified total pressure.
According to API 650 para 5.9.7.1, "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)."
Considering paragraph "d." above, there are factor a-c where factor "a" is a default with
1) for open-top tank - 1.48kpa +0.24kpa = 1.72 kpa.
2) for close-top tank = 1.72kpa maximum.
Again considering paragraph "d.", when other factor specify by Purchaser are great than 1.72kpa(either open/closed roof tank), then another ratio to be add on Main formula below,
H1 = 9.47t x [(t/D)^3}^0.5 x (190/V)^2
The question is do we have to add 0.24kPa into our so called "other factor(wind pressure)" calculation in order to make comparison to the default 1.72kPa? Or 0.24kPa is just needed to add during default calculation for open-top tanks only??
Or anybody know what is the "other factors" referring to at this point?
p = 0.00256Kz Kzt Kd V 2 I G = 1.48 kPa (31 lbf/ft2)
where
Kz = velocity pressure exposure coefficient = 1.04 for exposure C at a height of 40 ft,
Kzt = 1.0 for all structures except those on isolated hills or escarpments,
Kd = directionality factor = 0.95 for round tanks,
V = 3-second gust design wind speed = 190 km/h (120 mph) at 10 m (33 ft) above ground (see 5.2.1[j]),
I = importance factor = 1.0 for Category II structures,
G = gust factor = 0.85 for exposure C.
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).
b. The wind pressure is uniform over the theoretical buckling mode of the tank shell, which eliminates the need for a shape factor for the wind
loading.
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.
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 H1 shall be decreased by the ratio of 1.72 kPa (36 lbf/ft2) to the modified total pressure.
According to API 650 para 5.9.7.1, "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)."
Considering paragraph "d." above, there are factor a-c where factor "a" is a default with
1) for open-top tank - 1.48kpa +0.24kpa = 1.72 kpa.
2) for close-top tank = 1.72kpa maximum.
Again considering paragraph "d.", when other factor specify by Purchaser are great than 1.72kpa(either open/closed roof tank), then another ratio to be add on Main formula below,
H1 = 9.47t x [(t/D)^3}^0.5 x (190/V)^2
The question is do we have to add 0.24kPa into our so called "other factor(wind pressure)" calculation in order to make comparison to the default 1.72kPa? Or 0.24kPa is just needed to add during default calculation for open-top tanks only??
Or anybody know what is the "other factors" referring to at this point?