tmengineer
Chemical
- Dec 4, 2013
- 23
Hello,
I'm using the standard method outlined in BS 6399-2:1997 to calculate the wind loading on circular vessels and am having some problems determining an appropriate net pressure coefficient.
I work for a manufacturing company that produces atmospheric tanks with a diameters in the range of 3 - 5 m and height in the range of 4 - 12 m. They are typically vented to the atmosphere through a nozzle with a diameter of approx. 8" and the shells are made from 5mm SS304 or SS316.
I assumed that the external pressure coefficient (Cpe) was +0.6 as this is recommended in "Wind loading" by N. J. Cook and that the internal pressure coefficient (Cpi) was 0. Assuming that the net pressure coefficient was 0 models the scenario where the wind does not affect the contents of the tank - which will largely be true except for the effects from the vent - as the shell is impermeable by wind.
This gave a net pressure coefficient (Cp) of +0.6, as Cp = Cpe - Cpi.
For the particular tank I did my calculation for I found a maximum force of 15.5kN acting on the side of the tank during gusts, however, I received an email from another engineering firm that said they had calculated a maximum force of 28.5kN for the same tank using EN 1991-1-4:2005 and I suspected that my coefficient assumption was incorrect.
I then looked through the standard and saw in table 20 that a value of Cp = 1.2 can be taken for circular sections with diameters approx. 200mm (mainly pipes and chimneys) and that this value can be used conservatively for larger diameter obstructions. Using this value would bring my answer to 31kN - much closer to the other firms answer. However, due to the ratio of H/D of 2.07 for this tank a reduction factor of 0.63 should be used according to figure 25 - giving an answer of 19.53kN.
Does this method sound appropriate? The answer it produces is significantly below the answer given by the other engineering firm.
If you are interested in checking the calcs:
Diameter = 3.814m
Height = 7.901m
Mass = 6500 kg
qs = dynamic pressure = 1067 pa^3
Ca = size affect factor = 0.95
Area = 24.7 m^2 Note! Area =/= H x D because there is a cone section at the bottom half of the tank and so the area is less. I have ignored the effect of the cone except for its affect on the area at the end - otherwise the vessel has been treated as a cylinder as an assumption.
Any feedback you have would be greatly appreciated!
I'm using the standard method outlined in BS 6399-2:1997 to calculate the wind loading on circular vessels and am having some problems determining an appropriate net pressure coefficient.
I work for a manufacturing company that produces atmospheric tanks with a diameters in the range of 3 - 5 m and height in the range of 4 - 12 m. They are typically vented to the atmosphere through a nozzle with a diameter of approx. 8" and the shells are made from 5mm SS304 or SS316.
I assumed that the external pressure coefficient (Cpe) was +0.6 as this is recommended in "Wind loading" by N. J. Cook and that the internal pressure coefficient (Cpi) was 0. Assuming that the net pressure coefficient was 0 models the scenario where the wind does not affect the contents of the tank - which will largely be true except for the effects from the vent - as the shell is impermeable by wind.
This gave a net pressure coefficient (Cp) of +0.6, as Cp = Cpe - Cpi.
For the particular tank I did my calculation for I found a maximum force of 15.5kN acting on the side of the tank during gusts, however, I received an email from another engineering firm that said they had calculated a maximum force of 28.5kN for the same tank using EN 1991-1-4:2005 and I suspected that my coefficient assumption was incorrect.
I then looked through the standard and saw in table 20 that a value of Cp = 1.2 can be taken for circular sections with diameters approx. 200mm (mainly pipes and chimneys) and that this value can be used conservatively for larger diameter obstructions. Using this value would bring my answer to 31kN - much closer to the other firms answer. However, due to the ratio of H/D of 2.07 for this tank a reduction factor of 0.63 should be used according to figure 25 - giving an answer of 19.53kN.
Does this method sound appropriate? The answer it produces is significantly below the answer given by the other engineering firm.
If you are interested in checking the calcs:
Diameter = 3.814m
Height = 7.901m
Mass = 6500 kg
qs = dynamic pressure = 1067 pa^3
Ca = size affect factor = 0.95
Area = 24.7 m^2 Note! Area =/= H x D because there is a cone section at the bottom half of the tank and so the area is less. I have ignored the effect of the cone except for its affect on the area at the end - otherwise the vessel has been treated as a cylinder as an assumption.
Any feedback you have would be greatly appreciated!
