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Analysis of Current Heat Exchanger with Increased Flow Rates

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Fattony11

Chemical
Jan 10, 2007
8
US
I am currently analyzing a shell and tube heat exchanger at our facility and I am having difficulties with my heat transfer fundamentals. I have all the data I need to determine my actual q, U, LMTD, etc. and I have done that.
If I increase the shell and tube side flow rates by proportional amounts (i.e. double both of them), what will be affected? Are any of the values mentioned above constant?
Any help would be appreciated.
 
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I'm having difficulties with something similar... i'm not changing the flow rates, but i'm pre-cooling one side, keeping the flow rates constant.

Try this:

* Take the initial Q for one side. If you increase the flow rate, the temperature should decrease => Q = mass flow x Cp x (Tout-Tin)
* Do the same for the other side. You get a new pair of exit temperatures (if the entrance temperatures are fixed).
* With the new temperatures, you get a new LMTD and a new correction factor F, if it applies to your design.
* You can keep constant UA (i'm not completely sure), because it depends mainly of the properties of the fluids.
* Now you have a new LMTD, F, and the same UA, you can find a new Q...
* With the new Q, start again and it will converge after 5 or 6 iterations.

It is similar to how i'm trying to solve my problem, but i haven't had any professional answer that confirms my method, so, if you know a better way to solve it, tell me please!

Best regards,

JuanMC
 
The U usually increases by the 0.7 power of the mass flowrate; correcting it for fouling and tube conductivity will slightly modify this .

The approximation that the heat transfer is proportional to :
Q=UA * LMTD is fairly good for the cases of (a) condensing on one side or boiling on one side or more than 4 passes on one side of the HX. If there is no boiling, no condensation, and less than 4 passes in the HX, then you would need to compute a configuration factor, usually using the e-NTU method. This procedure will output a HX effectiveness factor, and that factor will also vary by the effect of a changing U ( as previously stated, convective U varies by the 0.7 power of mass flowrate)
 
In the GPSA data book section 9 they replicate a table from several sources including Kern. For r2 = (f) * r1. Where r is film resistane (hr * ft^2 * Deg F/ BTU).

(f) tube side =( mass rate 1/ mass rate2)^.6
(f) shell side =( mass rate 1/ mass rate2)^.6


(f) shell side laminar =( mass rate 1/ mass rate2)^.6

there are other factors for viscosity, density, heat capacity, ect


 

The new friction drops and HE design pressures may be the limiting factors for such an exercise. A visit to thread391-104196 may help to enlighten other cost factors...
 
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