It is clearly shown that the highest pressure occurs at the bottom centre of the tank and the distribution of pressure is symmetric about the central section (y axis). Figure 9 gives two snapshots of the pressure field in the tank just before impact and at the impact time t = 0.65 s. Therefore, we equally distribute 1500 mesh cells in the vertical direction and 200 mesh cells in the horizontal direction. To obtain an accurate prediction of the pressure peak, use of a fine mesh is recommended by other researchers (Braeunig et ah, 2009 Guilcher et ah, 2010). The impact pressure at this moment is of particular interest to ship structural engineers for this type of problem, since it is fundamentally a key issue for the safety of liquefied natural gas carriers. Under gravity, the water column will drop and impact upon the bottom of the tank at around t = 0.64 ~ 0.65 s. The initial pressure is p = 1 bar in the tank. A rectangular water column (p 2 = 1000 kg/m 3) with width 10 m and height 8 m is initially at rest in the closed tank and air (pi = 1 kg/m 3) fills the remainder of the tank. Figure 8 shows the setup for this problem. So the total pressure at a depth of 10.3 m is 2 atm-half from the water above and half from the air above.This benchmark test was proposed by the liquid sloshing community at the ISOPE 2010 conference to exam numerical codes’ capability of handling multidimensional water impact problems involving air entrapment (Dias and Brosset, 2010).
This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. Since water is nearly incompressible, we can neglect any change in its density over this depth. Just 10.3 m of water creates the same pressure as 120 km of air (height of uppermost layers of atmosphere).
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