Discussion on the volume of liquid contained in a standard elliptical head at different liquid levels
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Today I met a client who requested to store XX liters of liquid in a standard elliptical head with dimensions of XX. So what is the liquid level height? I think this problem can be solved using integration, but after some time has passed, I’ve forgotten how to carry out the integration. I’m asking everyone here: how should one calculate it in such a situation? I’ve given a specific example here; the diagram is below. Feel free to discuss it.
Please take a look at this paper; it contains mathematical formulas:
Method 1: Use AutoCAD to draw the shape – create a region – rotate it around an axis to form a solid – then measure its volume or mass (the density is fixed at 1 and cannot be changed). Method 2: Use 3D modeling to measure the mass or volume
The method on the 2nd floor is too simple and practical; awesome!
In 3D, create a solid head, cut it, and then measure it. Or there are so many Excel sheets online for calculating the volume of elliptical heads; just look them up and download one
The calculation methods should be included in the quick reference manual for petrochemical equipment
a is the length of the major axis, and b is the length of the minor axis. The equation is as follows; solve it by yourself: 4b^2*pi*(x-x^3/(3*b^2)-2b/3)=V
The above equation applies to a standard elliptical head, with the liquid level not exceeding the tangent of the head. If it is not a standard elliptical head, replace 4b^2 in the first term with a^2. If the liquid level exceeds the tangent line of the head, it’s much simpler; the volume of an elliptical head (excluding the straight edges) is: V=2*a^2*b*pi/3
Sorry, the formula on floor 8 is incorrect; the symbol in the parentheses for the last term, 2b/3, should be + instead of something else-