existing HEX evaluation

[a1b2qs]

What is the best approach in evaluating an existing heat exchanger for a New Service??

 

[toabhijeet]

Go for a simulation mode of HTRI.

 

[sshep]

If the exchanger does not have a phase change (other than simple steam condensing), an affinity law approach can be used such as given in the GPSA manual. This requires the original datasheet information for pressure drop, heat transfer coefficient, etc.

If there are phase changes on the process side, then a model based rating is advised.

best wishes,
sshep

 

[a1b2qs]

So once the exchanger is rated the new service transfer rate has to be smaller, larger or same as the old service rate as shown on the TEMA??

 

[MJCronin]

Try the NTU or The Effectiveness method.

http://en.wikipedia.org/wiki/NTU_method

http://www.cheresources.com/hteffzz.shtml

or call these guys:

http://reratellc.com/

-MJC

 

[sshep]

The affinity law approach involves calculating the U required to meet the required duty for an exchanger of known geometry (area, and layout). The affinity calculations then compare the estimated U with the required U, and the estimated pressure drop vs the allowable. If U_est>U_req and dP_est<dP_allowed, then the exchanger should be adequate for the required service.

The affinity relationships include: viscosity, thermal conductivity, heat capacity, mass flowrate, density, and various geometry changes that could be considered.

best wishes,
sshep

 

[a1b2qs]

Thanks SShep for your help,

GPSA refers to the U calculated based on the affinity law the Available U, and the U based on the Q=UA(LMTD) the Required U. So Available U should be less than the Required U, Correct?? Wording kind of misleading...TEMA calls the Available U, Dirty U.

Thanks

 

[speco]

a1b2qs,

If you run a full check rating of the existing exchanger you should get a new U value and a new LMTD value. Your new heat load (Q) may be higher or lower than the original.

So the product of the new U, LMTD, and existing A (surface) needs to be equal to or greater than your new Q.

Of course, that assumes that there are no pressure drop or other problems on either side of the exchanger.

Regards,

Speco

 

[sshep]

Hey a1b2qs,

The affinity laws work on resistances. As U is calculated from resistances (1/sum(r)), it can be confusing, however: it only makes sense that U available should be equal to or greater than U required in order to be adequate.

I think that the GPSA example has a mistake in their text on this point, and I have written them for clarification.

With respect to clean vs service U, this is a different issue altogether. Fouling factor is a totally user defined input to the overal heat transfer coefficient (U), it cannot be calculated, rather it is estimated from experience.

If I hear back from GPSA, then I will forward their response. With luck, they may post a clarification for us.

best wishes always,
sshep

 

[rmw]

The fouling factor, what ever value you end up estimating, based on where it appears in the equation, just adds area to the minimum required area necessary to obtain the overall U so that when the Hx becomes fouled, there will still be adequate area to perform the duty. In some cases that can be beneficial, and others (P&F's) that can be detrimental.

If you have a service that you don't realistically expect to foul, (clean both sides) then the use of a generous fouling factor will produce an oversized Hx which may overperform.

rmw