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jproj (Chemical)
17 Jul 02 10:32
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
Flareman (Petroleum)
17 Jul 02 13:06
I don't think that there's any easy way to do this without some slog because the heads are a compound of more than one geometry.  However, the general dimensions are sufficiently well defined in (ASME-VIII for example) to allow you to write a formula based on integrating the generalized linear geometry though the centerline.  Do that once and put it on a spreadsheet then you'll have all of them.
Even as I write, it seems like a nice little intellectual exercise.  Any of you guys out there got time on your hands?  Someone must already have done this!!
Helpful Member!(5)  mgp (Mechanical)
17 Jul 02 18:43
Did you try these:

Volume of 2:1 ellipsoidal head :
(pi/24)*D^3

Weight of 2:1 ellipsoidal head :
(pi/24)*(OD^3-ID^3)*Density

Volume of hemisphere:
(pi/12)*D^3

Weight of hemispherical head:
(pi/12)*(OD^3-ID^3)*Density

For a torispherical head It's somewhat more tricky as the geometry depends on the thickness and the formula is not straight forward.

regards
Mogens
Helpful Member!(2)  BabuGopakumar (Mechanical)
18 Jul 02 8:23
Hi,

I differ on Mr. Mogens views.

If the following equation is used, the weights that will be calculated will be almost half of the actual weight.(I tell this from my own experience as the actual weight was much more than the weight which I calculated using the formula which Mogens had suggested. So I tried to analyse the reason)
Weight of 2:1 ellipsoidal head :
(pi/24)*(OD^3-ID^3)*Density.
This can be explained by taking an example :

Consider the ID as 1000mm, say thickness of 10mm
OD becomes 1020mm.for ellipse; head height is D/4.
i.e,for a 2:1 ellipse profile, for 1000mm dia, the head height is 250mm & for 1020mm dia, the head height is 255mm.

Now in actual case, the outer head height is 260mm (250 + 10). This is beacuse, the inner side is an exact ellipse, and the outer side is not an exact ellipse)That means almost half of the thickness is not accounted when you calculate using the formula above.

If anybody needs better clarification , I can send them a sketch where this is better explained.

The best option is to find out the blank dia of the head(refer Brownell & Young) and then calculate the weight using the formula pi/4*(blankdia)^2*density*thickness.

Regards,

Babu
Email:babugopa@chemproindia.com
jproj (Chemical)
18 Jul 02 10:51
Well, after trying unsucessfully to calculate the weight of a 2:1 Elliptical head, I though I might try the Hemispherical head since the equation is simple.  I found something interesting that I can't explain...

The equation for the volume of a hemispherical head (shell portion) is (4*pi/6)*(ro^3-ri^3).  When comparing the calculated weight to the values in the book, I found that the values in the book were all heigher by an average factor of 1.234 (Book weight = my weight*1.234).  Can anyone explain this??

jproj
Flareman (Petroleum)
18 Jul 02 13:08
Jproj
It's probably because the formulae involving some power of D or R only consider the curved section of the head (and even then ignore the knuckle).  The weights you get from the book will probably include the flange and this is of variable length from one supplier to another.
jproj (Chemical)
18 Jul 02 13:29
Flareman :

I would consider that for a 2:1 of F&D head, but a hemispherical head is half of a sphere.  The Pressure Vessel Handbook (which I'm compairing my calculations to) says that in their calculations they use no straight flange for a hemispherical head and that they use 0.28333 lb/in3 for the density of steel.  That leaves the equation I listed in my last post and the mystery factor.

I believe Mogens has listed all the correct equations.  Babu is also correct regarding the head depth issue for a 2:1 elliptical head.  I'm begining to think the weights in the book are off.   I guess I shouldn't believe everything I read in books, but this is a fairly reliable text (this is probably the first time I've run across something I questioned...)

As an example, for an 10 ft. ID hemispherical head (1 in. thick) I have calculated a weight of 6,516 lb.  The Pressure Vessel Handbook lists a weight of 8,039 lb.

Any insight??

jproj
Flareman (Petroleum)
18 Jul 02 15:05
I have the 2nd ed of TPVH and I can't find a table of weights at all.  What page is it on?

You're probably correct to develop a healthy scepticism of published information!!
jproj (Chemical)
18 Jul 02 15:31
In the Tenth edition it's located on page 374.  It's listed in the Index under "Weights... shells and heads".  It may not be in the 2nd ed. if you can't find it... the table is pretty large (15 pages).

jproj
Flareman (Petroleum)
18 Jul 02 16:18
10th ????? I clearly need to buy a new book !!!
I didn't think I'd had it that long!!
Flareman (Petroleum)
18 Jul 02 16:22
Oh! A reality check.  I'm looking at Pressure Vessel DESIGN Handbook by Bednar (not by Megysey).  Sorry about that.  I'll try to read things a little better in future.
gunnarhole (Mechanical)
19 Jul 02 12:49
I have a formula that I was given by a head supplier a long time ago that gives the weight of a 2:1 elliptical head as:

Weight, lbf = 0.223*t*((OD-2*t)*1.22+SF+t)^2

where

OD = outside diameter, inches
t = head thickness, inches
SF = straight flange, inches

I pulled out my copy of Megysesy this morning and I see that an 120" OD elliptical head with a wall thickness of 1" weighs 5175 lbf (SF = 2"). In his table Megyesy is not specific as to whether the weights given refer to OD or ID heads. I checked a few values (8 diameters x 3 t's)in his table and the equation above is consistently ~4 to 6% low if I assume that the weights given are for ID heads.

For example, if we use 122" as the OD in the equation above the equation returns a weight of 4977 lbf, about 4% lower than the tabulated value.

Regards,

Gunnar
MJCronin (Mechanical)
19 Jul 02 13:27
Hey Everyone...

Just a thought about what may have been overlooked.....

Typically, when a pressure vessel head is sold, it includes a straight flange to ease the welding onto the shell. Is this included in any of the formulas or tabular values discussed above ???

Secondly, I am confused.....unless you are working for a fabricator or a head vendor, why do you need the weight of just the pressure vessel head anyway ??? Arn't you interested in the total weight of an empty vessel ??

I have used/developed an approximate method that gives the weight of the entire vessel, based on the volume and length of the straight shell.....the question of who owns the "straight flange" is moot when a calculation of the entire vessel weight is made.

Just my thoughts.....

MJC
jproj (Chemical)
19 Jul 02 14:32
MJCronin:

We are a small pressure vessel design company.  When we send out our bids, our customers usually want estimated tank weights.  Since we don't talk to fabricators unless we have the job, we must estimate the weights ourselves.... two heads plus one shell and assorted nozzles = tank weight.  Shell weights are easily calculated, but as shown by this thread, head weights are not so easily calculated.  I have head weights listed in a book, but I was mainly looking to define the weight using an equation.

Now, I can understand why it may be difficult to come up with an equation for an ASME F&D head, but a hemispherical head is half of a sphere... there should be no mystery about the equations to calculate the volume of a thin shelled half sphere (no straight flange).  With a little manipulation, the volume of 2:1 elliptical head is nearly as easy to calculate (with a 2 inch straight flange).

I am almost positive the equations I am using to calculate the 2:1 and hemi head weight are correct, but the weights don't match up with the pressure vessel handbook.  Not to say that the book is the end all / be all, but I don't have another reference with which to compare my results.

Best regards,

jproj
TD2K (Chemical)
19 Jul 02 15:04
jproj, I checked the weight in your last post for a 10' ID hemispherical head and got essentially the same value as you did (maybe we are both wrong?  ) and it's a pretty straightforward calc.  

This isn't my area of speciality but it's sure interesting to watch.
Helpful Member!  Guest (Visitor)
19 Jul 02 19:54
Jproj/TD2K,

The weight of a 10' I.D. hemispherical head, 1" thk.
is 8039 lbs. (Pressure Vessel Handbook by Megyesey).
The calculation is based on the material needed to
make the head (equivalent blank diameter=121 X 3.1416
X 0.5=190.067 in., Wt=0.7854 X 190.067^2 X 1 X 0.28333
=8039 lbs.). The weights given in the tables are correct.
GWK (Mechanical)
20 Jul 02 12:41
As an employee for an owner/user I sometimes need to confirm the accuracy of a vessel weight for development of equipment lift plans. This is not only for new construction, but also for maintenance where the equipment must be removed from the unit and sent out for repair. Weights of equipment shown on drawings can be significantly in error. I just checked one that was +80% low. I wonder if that empty vessel weight was penciled in rather than calculated. GWK
mgp (Mechanical)
23 Jul 02 5:22
JProj & Babu

You're both right about my formula's being incorrect, because the "outer" ellipsoid is not a 2:1 ellipsoid.
(It will be correct only with Zero thickness)

So here I try again, hopefully better off this time:

Volume of (any) ellipsoid

Vol=4/3*pi*a*b*c, where a,b,c is the radius of each major axis

A head is half an ellipsoid, i.e. Volume of ellipsoidal head:

Vol=2/3*pi*a*b*c


For an semi-ellipsoidal head, the inside of the head is 2:1, i.e

a=D/2, b=D/2, c=D/4, i.e.

Vol_inside = 2/3*pi*(D/2)^2*(D/4)
           = 1/24*pi*D^3

For the outside of a (inside) semi-ellipsoidal head:

a=D/2+t, b=D/2+t, c=D/4+t i.e.

Vol_outside = 2/3*pi*(D/2+t)^2*(D/4+t)


Then weight of shell is

Weight=2/3*pi*((D/2+t)^2*(D/4+t)-(D/2)^2*(D/4))*density

Assuming no variations in head thickness after rolling, this formula should give the correct value for all thicknesses.

regards
Mogens
mgp (Mechanical)
23 Jul 02 7:13
This missing formula for torispherical head volume kept bugging me, so I had a go at it. Here goes:

When:

D equals (inside) Diameter of Cylinder
R equals (inside) Radius of spherical part (dome)
r equals (inside) (secondary) radius of torus (knuckle radius)

Calculate

K=R/D (for ASME F/D head K=1.00)
L=r/D (for ASME F/D head L=0.06)

angle=ACOS((1/2-L)/(K-L)) (angle in radians)

angle is angle of torus part i.e. the angle where Knuckle radius = r) For ASME F/D head this angle is 1.084 radians (=62.09 degrees)

Then calculate inside volume by:

Vol=PI()/3*D^3*(2*K^3*(1-SIN(angle))-(K-L)*(1/2-L)^2
*SIN(angle)+L^2*((1/2-L)*3*angle+2*L*SIN(angle)))

(Remember angle must be in radians)

This formula should be valid for all shapes of torispherical heads, i.e. with different combinations of K and L.
I checked an ASME F/D head by drafting 3D CAD and measuring the volume but I'll be happy to share the background calculation if required.

Examples:

ASME F/D  (K=1, L=0.06)
inside Volume = 0.08100*D^3

DIN 28011 Kloepper form (K=1, L=0.10)
inside Volume = 0.09897*D^3

DIN 28013 Korbogen form(K=0.8 L=0.155)
inside Volume = 0.13116*D^3

Note that DIN 28013 really is a standard for semi ellipsoidal heads, but in reality ellipsoidal ends are rolled as torispherical, with a shape that comes close to an ellipsoid.

To calculate the shell weights, the outside volume must also be calculated. For this the short formulas cannot be used but new values of K,L and angle must be calculated and inserted:

K=(R+t)/(D+t)
L=(r+t)/(D+t)

It should be easy however if the formulas are entered in a spreadsheet.

Important note:
The short cylindrical ("Flange") part as mentioned by MJC is NOT included in this formula. Remember when calculating shell volume or weight to use tan-to-tan line, not just length of shell.

Hope this is useful, it took a while to get there

Regards
Mogens

BabuGopakumar (Mechanical)
25 Jul 02 0:26
Hi,

I Think Mogens has accuratelu summed up the volume calculations.
I am using almost the same thumb rules for volume calculations.
I am listing the volume calculation formulas for heads which I am listing out.

1. 6% Torispherical - 0.0846 D^3.
2. 10% Torispherical - 0.1 D^3.
3. 2:1 Ellipsoidal - 0.1313 D^3.( I have seen another refernce saying it as 0.1309D^3.

These values are almost the same as what Mogens has indicated. These have been taken from Perry's chemical engineers handbook, Ludwig etc.
For calculating the dish weights I agree with Mogens suggestion that it is better to do it in a spreadsheet.

Regards
Babu
Chiller (Mechanical)
25 Jul 02 8:23
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 (Mechanical)
5 Aug 02 10:11
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 (Mechanical)
5 Aug 02 17:15
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 (Mechanical)
6 Aug 02 14:38
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 (Industrial)
5 Feb 03 20:50
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 (Mechanical)
10 Feb 03 23:44
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)



    
Helpful Member!(2)  drdjones (Chemical)
23 May 03 17:02
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.

www.degussa.com
mechnasir (Mechanical)
24 May 03 6:30
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 (Chemical)
8 Jul 03 17:06
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.

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