UHX Fix Tubesheet 1

[laibme]

I am trying to go through the calculations for the fixed tubesheet using UHX. I got stuck when I was looking for the Kj value (the axial rigidity of expansion bellow).

Can someone tell me how to calculate the value or where it comes from?

I've tried looking at the examples in the UHX sections and they just seem to get the value from mid air.

Thanks.

laibme

 

[GenB]

Do you have an exp bellows in your design?
if not you do not need it and the value may be 1
I will study it further and advise.
genb

 

[laibme]

I am actually not trying to run an actual calculation, but to understand the calculations, as my job will require me to use it later. So I want to understand this for all cases.

I understand if there is no bellow the J value automatically goes to 1, as per pg 285 in UHX section 13.3. Which would mean that the Kj value is equal to Ks, since the equation for the J value is 1/(1+(Ks/Kj)).

I thought about using the Ks equation and substituting it for Kj, but it doesn't make since to me as the bellows do not act the same way as the shell, even if I substitue the values of ts, Ds, and L. Unless I am mistaken, please let me know.

Thank you.

laibme

 

[Swig]

laibme says:
"I understand if there is no bellow the J value automatically goes to 1, as per pg 285 in UHX section 13.3. Which would mean that the Kj value is equal to Ks, since the equation for the J value is 1/(1+(Ks/Kj)). "

If you don't have a bellow, then the Kj is equal to infinity (and not equal to Ks). You can consider 'no bellow' like 'one with an enormous axial rigidity'.

Ks/Kj = Ks/infinity = 0, so J equals 1/(1+0)=1.

swig

 

[PVGuy]

laibme,

The Kj value comes from the design of the bellows and is simply plugged into the design of the exchanger per UHX. If you are doing thin walled bellows and you are calculating them in house you can use appendix 26 in SC VIII-I which will give you this value. If you do thick walled bellows, then you need to go outside SC VIII-I to calculate the bellows and Kj; but you need to meet the stress criteria of App. 5 in SC VIII-I.

 

[orion952]

I am also trying to determine Kj for a flanged & flued expansion joint. From a copy of Design of Process Equipment (third edition), chapter 6 walks you thru a design. Since I only needed a value for Kj to input into my software I only calculated Fj from this book which is the force required to move the expansion joint 1 inch. For a shell od of 36", exp joint od of 46", wall thickness of 3/8", shell length of 264", exp jt radius's of 1.5,a delta value of 0.0824" and H = 5". for SA-516 Gr 70 material I get an Fj value of 22197 lb/in. Would someone like to check me and comment on the logic of this approach.

 

[PVGuy]

Orion 952,

The method you have sited is a generally accepted method, the book though has some errors though. The axial rigidity is not dependent on the delta value like Fj is. If you divide the Fj value by the value of Y (terms defined in your reference) then the result is the axial rigidity, it is not uncommon for this value to be near or in excess of 1,000,000 lb/in.