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Head Area Calc

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madgoose

Structural
Dec 15, 2005
2
I'm designing a paint calculator for our engineering dept. I'm having trouble calculating the area of the heads. any advice or help?
 
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Calculate the blank diameter of dished head. This is the diameter of circular plate before any forming of dished head takes place. Once you know the blank diameter, you can easily calculate the surface area.

By the way paint surface area does not have to be very accurate.

Also wastage factor for paint will be different for external and external surfaces. Wastage factor will be more for external surface, because of wind.
 
what types of heads are you trying to calculate their surface areas?
 
we do all types of heads. 2-1, dished, etc..
I'd like to get this as close as possible so any waste is not blamed on the programmer. (me.)
I thought about getting the square foot by the weight. This is ASME tanks so they know the empty weight.
 
These will get you very close.

(straight flange area not included where applicable, as you can treat that like a cylinder)

D=outside diameter in FEET

Surface Area in square FEET

2:1 Ellip.
1.09 x D*D

ASME F&D
0.918 x D*D

Hemi
1.5708 x D*D

Flat
0.7854 x D*D


Brian
 
madgoose,
I support the comment from PEHasan. A paint manufacture advised me to use wastage factor (loss) approx 30%.
 
Since all the heads including knuckles(normally considered arcs) have design parameters which you can easily find in the Perry chemical engineering handbook or from texts on pressure vessels, you can use the formula As=2*pie*r*s where As=surface area,r= perpendicular distance from the axis of revolution to the center of gravity,s=length of curve.You can then sum up the As's for the knuckle and head shape to get your info.
If you had Mechanics in college, it should not be hard to fing the center of gravities of arcs, ellipses and spherical shapes. It will take a little work even if you have a background in college engineering courses.
The suggestion above to get the surface area from a blank disk will put you in the ball park, however, it does not take into account the stretching of the metal during the forming process.
 
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