2 replies to this topic

[CH3CH2OH]

Hello,

 

I am currently specifying a 10,000 gal pressure vessel that will need to be equipped with heating abilities.

 

I am looking to have a half-pipe jacket installed on the vessel that covers the bottom head, and the bottom half of the shell.  I know this will equate to a certain heat transfer area.

 

I would like to have a decent understanding of potential design of the jacket instead of blindly submitting it to a vendor.  I'm trying to figure out how to design this, but I've run into a couple of sticking points -- can anyone help?

 

In short, I just haven't found a good guide for doing this jacket design.

 

I would believe my steps are:

• Calculate total area of heat transfer (assuming 2" or 3" half pipe)
• Using Q = UAΔT_LM, calculate Q,max (ΔT_LM is calculated, U is assumed)
• Use Q,max to calculate steam flow rate (based on 90 psig steam available)
• Assume three or four zones, determine flow per zone based on max velocity of 25 m/s
• Complete pressure drop calculations on each zone
• Use steam flow rate to determine batch heating time

I've tried doing some of these steps, I'll share my results as needed, however, I'd like to make sure I'm on the right track.

 

Also, I'm very confused by:

• How do I calculate the total heat transfer area for the jacket?
• How do I calculate the cross-sectional area of the jacket?
  • I've seen some resources that require hydraulic diameter, some that require actual diameter, but no single, simple equation

 

 

[latexman]

We have dozens of these vessels.  For that size of vessel, 3" half-pipe is more reasonable.  For a long lasting, reliable product, you can't just bid any old vessel fabricator.  The half-pipe to shell weld requires a good, full penetration weld.  Forget, the lowest cost shops out there that will say they can do this.  Only consider shops that are high quality and have many years of experience doing this particular task.  I would guess three sections.  One on bottom head and two on the bottom half of shell.  The metal of the half-pipe where it is full penetration welded to the shell acts like an extended heat transfer surface, i.e. a fin.  The literature I've seen says handle the area between half-pipes as 70% effective compared to the area within the half-pipe.  But, you can't have too wide a gap.  What's your gap on 3"?  If you don't know, talk to an experienced shop.  They'll tell you what they can do.  Yes, pressure drop of non-circular conduit is not well understood by many.  What I do is verbatim from Crane TP410.  Use the hydraulic diameter just for the Reynolds Number to get the friction factor.  Then, use an equivalent diameter to the actual flow area, d = (4A/pi)^1/2 , for the other terms that use a diameter that generates a velocity.  It's kind of mixed up, right?  But that's what it says.  A lot of people use the hydraulic diameter for every diameter term.  That's wrong.  Petroleum Engineers have a different method they apply to annular flow.  I don't know if it applies or can be applied to other shapes of flow.  Maybe you can Google that.

Edited by latexman, 10 November 2017 - 10:35 AM.

[Kachisa]

A reference from one project I did : 

- 3" for half pipe

- Equivalent hydraulic diameter is not suitable for  pressure drop calculation 

Edited by Kachisa, 12 November 2017 - 07:00 AM.