2 replies to this topic

Dear All,

I am currently working on a small task; of calculating how much water is required to cover the Agitator blades X, Y & Z tanks.

I know from the mechanical drawings and specs that the tanks I am working on have semi-ellipsoid Heads, I have been doing some research and found the following formula;

V=pi*ID3*f*(2-r-r3)/24
where
ID=diameter of circulare base
f=ratio of 2*(third axis) to ID (0.5 in your case)
r=ratio of unfilled height to total height of head or cap or semi-ellipsoid.
With r=1 the head is unfilled and you get V=0
With r=0 the head is full and you get the formula of my previous post.

I just wanted to confirm if the above formula is correct for the task in hand, or if any one as ever seen this formula before as it is new to me.

My calculations are with in reason

Agitator speed and vortex will be taken in to consideration.


Many Tanks,


James Silton (Rohm and Haas UK LTD)


Also does any one know the fromulas for; hemispherical and torispherical heads???

[Art Montemayor]

James:

I’ve answered this – and similar questions – before on our Forums and supplied a volume workbook that you can download in order to calculate total and partial volumes of vessels and different shaped heads.

But, I guess you haven’t done a search on the Search Engine and I’m just as lazy as the next person after having done this so many times. But being that it involves a good old boy from Birmingham, Alabama (joke) I’ll furnish the workbook one more time for effect.

In this workbook you will find all the equations and answers involving vessel volumes that I’ve been able to compile. I hope the workbook does the job for you – and others.

Tanks – but no tanks.

 Vessel_Volume.zip   772.91KB   1623 downloads

Thank you, ChE Jedi

Sorry for being lazy, I should have conducted a more in-depth search.....

The work book is excellent and I forwarded it on to many people, Thank you once again and keep up the good work,


James