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Hi Team
I have few issues with the new Hydrostatic load definition in Nastran 2020.
1. I don't understand how I define the Fluid depth direction, what does it mean? (Reference image 1 and 2)
Should I do it in the direction of the fluid or the other way round?
Well I tried both ways and it seems that it gives me a correct result when the direction vector is (0,1,0) i.e, opposite to the direction of the fluid.
2. What is Surface pressure. Is it that in an open tank the surface pressure will be equivalent to the atmospheric pressure? (Reference image 2)
Thanks for the help in advance.
Regards,
Tarun
Solved! Go to Solution.
Hi @Anonymous
This looks like the tank tutorial model, so I assume you are trying to simulate a tank filled with water. Therefore, the pressure increases as you move in the -Y direction. Your image 2 shows the correct input: the fluid depth direction is [0, -1, 0], and the arrow are getting longer (higher pressure) toward the bottom of the tank.
You are correct that the Surface Pressure is the pressure on the free surface of the fluid. For atmospheric pressure, the value would be 0. For a pressurized vessel, the surface pressure would be the relative pressure of the air above the fluid (or the pressure of the fluid at the top of the tank). For a submarine in which the "point on the fluid surface" was set to the top of the submarine, the surface pressure would be (water density)*(depth of submarine). Whatever value you enter, the pressure at some point = surface pressure+hydrostatic pressure.
Hello I'm new in Nastran, I would like to ask regarding this hydrostatic load. I designing formwork with concrete pressure of 48kPa and concrete density is 24 kN/m3.
what i did I split face below and apply 48kPa. I want variable load of 0 on top and 48kpa to 2m height. my question is what box should i fill in the surface pressure or the fluid desity?
thank you in advance
Hi @felixjr11 Welcome to Inventor Nastran.
Hi @felixjr11
The density of the fluid is required. Otherwise, how would the software know if you are simulating molten steel or gaseous helium?
The pressure at the free surface needs to be entered when it is not 0.
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