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

 

[Flareman]

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!!

 

[mgp]

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

 

[BabuGopakumar]

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]

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]

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]

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]

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]

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]

10th ????? I clearly need to buy a new book !!!
I didn't think I'd had it that long!!

 

[Flareman]

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]

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&quot. 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]

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]

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]

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.

 

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]

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]

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]

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&quot 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]

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