FEA for Water Tank: Weight & Pressure Analysis

In summary, you calculate pressure using the weight of the water and the pressure due to the water, and the atmospheric pressure is not needed.
  • #1
Thomas Plank
6
0
Hi,

I'm trying to do some stress analysis on a tank filled with water, I have split the walls into sections and applied the appropriate pressures. However, I'm confused as to whether I apply the weight of the water+pressure at that height difference to the bottom of the plate; or will the pressure account for the weight already? But this doesn't seem likely as the formula didn't include mass.

Thank you.
 

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  • #2
You use the pressure only.
 
  • #3
Thomas Plank said:
Hi,

I'm trying to do some stress analysis on a tank filled with water, I have split the walls into sections and applied the appropriate pressures. However, I'm confused as to whether I apply the weight of the water+pressure at that height difference to the bottom of the plate; or will the pressure account for the weight already? But this doesn't seem likely as the formula didn't include mass.

Thank you.

Assume a tank with an area of the bottom 1 m2 and a height of 1 m filled with water. The weight of the contained water then would be 1 m3 of water multiplied with its density ##\rho##, hence 1000 kg corresponding to 9810 N.

Now let's calculate the pressure due to the water: ##p=\rho g h = 1000 \cdot 9.81 \cdot 1 = 9810\;N/m^2##. The force on an area of 1 m2 then is 9810 N, which is exactly the force due to the weight of the water. To find a total pressure at the bottom of course one has to add the atmospheric pressure at the water surface.
 
  • #4
stockzahn said:
Assume a tank with an area of the bottom 1 m2 and a height of 1 m filled with water. The weight of the contained water then would be 1 m3 of water multiplied with its density ##\rho##, hence 1000 kg corresponding to 9810 N.

Now let's calculate the pressure due to the water: ##p=\rho g h = 1000 \cdot 9.81 \cdot 1 = 9810\;N/m^2##. The force on an area of 1 m2 then is 9810 N, which is exactly the force due to the weight of the water. To find a total pressure at the bottom of course one has to add the atmospheric pressure at the water surface.
The atmospheric pressure doesn't have to be included if you're working with gauge pressures. Otherwise, you need to include the air pressure contribution from the outside of the tank as well.
 

1. What is FEA and how is it used for water tank analysis?

FEA stands for Finite Element Analysis, which is a computerized method used to solve complex engineering problems. For water tank analysis, FEA can be used to simulate the weight and pressure exerted on the tank by the water and determine its structural integrity.

2. How does FEA calculate the weight and pressure on a water tank?

FEA uses mathematical algorithms to divide a complex structure, such as a water tank, into smaller, simpler elements. These elements are then analyzed individually to determine the stresses and strains on the structure, which can be used to calculate the weight and pressure on the tank.

3. What factors influence the accuracy of FEA analysis for a water tank?

The accuracy of FEA analysis for a water tank depends on several factors, including the accuracy of the material properties and boundary conditions used in the simulation, the quality of the meshing (division of elements), and the complexity of the tank's geometry.

4. What are the benefits of using FEA for water tank analysis?

FEA can provide a more comprehensive and detailed analysis of a water tank's structural behavior compared to traditional methods. It can also save time and resources by allowing for virtual testing and optimization before physical construction.

5. Are there any limitations to using FEA for water tank analysis?

FEA is a powerful tool, but it also has its limitations. The accuracy of the results depends on the assumptions and simplifications made in the model, and it may not be suitable for highly nonlinear or dynamic problems. Additionally, FEA requires specialized software and expertise to use effectively.

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