On the insulation of horizontal storage tanks
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Regarding the insulation of horizontal storage tanks: for example, if the cylindrical portion has a diameter of 2400 mm and a length of 5000 mm, while each of the two end heads has a length of 500 mm, how should the surface area be calculated? Points will be awarded for correct answers
The barrel section is very simple, so I won’t go into it. The surface area of an elliptical head can be calculated as follows: In CAD, draw the elliptical head to scale. Select the corresponding line, and use the LIST command to determine the length of that curve. Treat this length as the diameter of a developed circle; the area of this circle is then the surface area of the elliptical head.
It’s a math problem; the main difficulty lies in calculating the surface area of the head. The head can be spherical, elliptical, or oblate spheroidal in shape – all one needs to do is look up the relevant formula
The surface area of a horizontal tank is given by S = πr. Note: r: diameter of the horizontal tank, in meters; H: length of the horizontal tank (including the ends), in meters; h: length of the cylindrical portion of the tank, in meters; C: constant; for standard elliptical ends, this value is 0.760346
It seems that for DN greater than 1000 mm, it can be calculated as a flat wall. . It’s easy to calculate, and the error is relatively small. :)
Hello, is this zip file corrupted? I downloaded it but can’t open it. Could you please re-upload it? Thank you
There’s nothing wrong with the formula. However, based on the data provided by the OP, it can be seen that the head of this horizontal tank is likely not a standard one; therefore, the formula isn’t applicable