Calculation of Head Weights 10

[jproj]

I am trying to find equations for the weights of 2:1 Elliptical, Hemispherical and ASME F&D heads (without much luck I might add). To find the weight of a cylindrical shell I found the inside and outside (total) volume and subtracted to find the shell volume. I am not having much luck using this method on vessel heads. I think it has to do with my equations for the volume based on the outside diameter... I have found and verified equations based on inside diameter.

I'm currently using a table from the Pressure Vessel Handbook (Eugene F. Megyesy) to lookup head weights based on diameter and thickness. Sometimes, there are cases where the weights are not listed in the book (say a tank with a diameter of > 12 ft). In this case, I have to extrapolate to find an approximate number (time consuming).

Does anyone know how to calculate the weight of these types of heads? Are there any references that address this topic?

Any help is greatly appreciated!!

jproj

 

[Chiller]

Just a thought - Considering that we have so much technology and drafting tools that can draw in 3D. Wouldn't it be easier to draw it. It takes only a couple of secs?

I personally draw it in Autocad and use the mass properties to find out its weight. Not only that, it also lists C of G's etc.

 

[DavidCR]

As a geometry fan, If I don´t trust on a formula, I´d use the "Pappus-Guldinus" theorem. I reckon, that combined with Autocad, would be the fastest and more general method to solve this. You can find this method in most Statics text books (such as Beer&Johnston).

 

[mgp]

To get the torispherical head formula, I used the "Guldin's theorem" which is the same as the "Pappa-Guldinus":

"If a plane figure is rotated about an axis in its plane then the volume of the solid body formed is equal to the product of the area with the distance travelled by the centre of gravity."

Regards
Mogens

 

[abeltio]

Regarding the calculations... when comparing the calculated weight with a tabulated weight listing a thickness for the formed head... please note that the listed thickness is the minimum thickness after forming...i.e. the thickness at the zone with the smallest radius of curvature...the so-called forming allowance can be as high as 25 or 30% depending on the material, thus the difference between the calculated and tabulated weights (this is of course exclusive of the skirt weight as properly noted before).
Hth.
a.

 

[Jeffro]

Area of a Hemi head = pi x Diameter squared Divided by 2

Area of 2:1 S.E head = 1.084 x Diameter Squared

Therefore Weight = Area x Thickness x Mass per cubic inch/foot/meter/centimeter or what ever you use.

 

[kyong]

In head weight calculation, it should be clear that what the thickness exactly mean. I noticed that many of the confusion or mistakes came from being mixed up which thickness is used, plate material thickness(normally called nominal thickness) or minimum thickness after forming.

Plate thickness is quite even and acurate even though it likey has a little bit of over tolerance.(I take 5% for this over tolerance if it's new plate.) On the other hand, the thickness after forming is not as straightforward as plate. Over surface of the entire head, thickness varies. Variance depends on manufacturing process(typically hot or cold), radius, etc. Some area has thinner thickness than plate material while other area is thicker than that.

Head is made of a circular plate of which the diameter is bigger than that of the head. It means thickness of edge becomes thicker. This is the reason why 1"(Minimum) 10' hemi head weighs more than 6516 lbs. To get minimum thickness of 1", the plate should be 1" or thicker. However, I don't think that the actual thickness will be the same as the plate of 121" x 3.14 / 2 dia x 1" thk or 8039 lbs. I would rather like to hope that manufacturers can use smaller plate in diameter with stretching radially during forming.

Not only to reduce material cost, but also to get better quality, manufacturers should know how to avoid too much thicker edge. And there is inevitable thinning out during forming(especiall when hot formed)

 

[drdjones]

I recently published a tank volume article giving rigorous equations for fluid volume in tanks with common type heads. A 2:1 ellipsoidal head would be a typical example. To find the weight of the head, calculate fluid volume using inside and outside dimensions, subtract, mult. by material density, and correct for flanges, construction variations (hard part), and other modifications. The article was in Nov 2002 Chemical Processing Magazine or can be downloaded in MS Word format at:

http;//unicon.netian.com/tank_vol_e.html

Use the vertical tank formulas for head weight calculations since they are simpler for full fluid volumes, which is what you would use for head weight calculations.

http://www.degussa.com

 

[mechnasir]

i also use the same method as jeffro mentioned in his reply. This method gives better (more value) result than weight calculated directly by using volume difference and multiply by density.

i have recently had a practical experience for subject thread. i calculted only volume of head by using both methods ie,

Method 1:
volume of 2:1 elliptical head= 1.084 x Diameter Squared x thickness

Method 2:

volume of 2:1 elliptical head = (pi/24)*(OD^3-ID^3)

and also

volume of blank : 3.14/4 * diameter squqre * thickness

i got difference values from method 1 and method 2

but value from method 1 is closer to volume of blank.

i have a point on it that it should be equal to volume of blank if we apply mass conservation theory on it.

Nasir

 

[drdjones]

Further comment on my May 23, 2003 posting on this subject (see above):

The method I suggested will not give correct theoretical head weights, only approximate, because two ellipsoids which are exactly x units apart along the x, y, & z-axes will not be exactly x units apart at every point on one of the ellipsoids. Another way of saying this is that both the inner and outer surfaces of an "ellipsoidal" head cannot both be ellipsoids if the thickness of the construction material is perfectly uniform. I had not really thought about this subtlety until I read mgp's first thread of July 23, 2002 above.

For torispherical heads, the method I suggested for torispherical heads will be rigorous. Two torispherical surfaces can be exactly the same distance apart at any point on one of the surfaces since spherical radii form the surfaces. So to find the rigorous theoretical weight of material in a torispherical head, define the parameters of the inner and outer surfaces such that the dish and knuckle radii differ by the head thickness, calculate full head volume of both surfaces and take the diference. This will be the rigorous theoretical volume of the head material of construction, except for flanges and such.

See reference in my posting above for the ways to calculate head volumes.