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Preferential Flow in Aerial Cooler Configuration

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MatthewD

Chemical
Nov 17, 2004
35
Ok I have never been asked this question before so I need some help.

Currently I have 8 aerial coolers in a first in last out configuration (also called "U" configuration I believe) and people i.e my boss is concerned that the first cooler will have more flow due to its smaller pressure drop. He wants me to quantify "more flow". Ideally he wants a number like x kg/h into cooler 1, y kg/h into cooler 2 but Im not sure how to give him what he wants. Thanks to the goodness of 3-D models I know all my lengths and elevations (The gas is clean and single phase)

I had one idea of treating it like a electrc ciruit and using pressure loss like a resistance. Any suggestions are welcomed.
 
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With respect to your circuit analogy, pressure drop is like voltage rather than resistance.

As the first step to determining whether there will be a significant difference in flow between the coolers, you need to check to see how much pressure drop is expected in the common inlet and outlet headers using the total flow. If the common headers are large enough then each cooler and it's associated piping will have the same overall pressure drop. Using your circuit analogy, an equal pressure drop does not mean that the flows are the same unless the resistances are also the same for each cooler branch. Checking the headers first will greatly simplify your calculations, especially if you are doing them by hand.

best wishes, sshep

 
I agree with your post, just 1 point of clarification. If pressure drop is like voltage and flow is like current is "resistance" the same variable as described in Crane Flow of Fluids Technical Paper No. 410 (HL=K*v^2/2g
 
Your boss is right that you will have more flow through the first cooler. You will find a discussion about flow through coils(1800 bent pipes in Crane TP410, I can't recollect the page no.). Otherwise, the cooler manufacturer will be able to give you flowrates w.r.to pressure drops across the coolers.

Ideally, you should employ reversed return piping in this case(first in first out).

If you can provide balancing valves for each cooler, you can check and correct the flowrate by means of a manometer and valve performance curve.

Regards,


 
MatthewD,

I would be cautious about extending the circuit analogy to actual calculations of resistences in parallel, etc. For example in a circuit V=IR, in piping dP=KQ^2.

On the otherhand your calculation is simple if pressure drop in the common headers can be neglected as per the initial check I suggested above. Once you have calculated your K values for each branch (from Crane if that is your method), you merely need to trial and error dP values until the sum of the flows through all branches equals the total flow.

If you have to account for the common headers then it is not so simple because you will have 14 other pipe segment resistances to consider, i.e. 2*(8coolers-1). This is where some pipe network software comes in handy. You can still use a similar approach as per above by guessing an overall dP, but now you have to back calculate up the piping (using the total flow) to get a dP and flow through each branch- easy enough with a spreadsheet. You system is converged when you find a dP such that the flow through the last branch (as calculated by your flow equation) also satisfies the total flow constraint.

With respect to balancing the flows, the shortcut way is to try and adjust any individual exchanger butterfly valves that are avialable so as to get all the outlet temperatures the same.

best wishes, sshep
 
Thanks for your help unfortunatly it looks like I cannot ignore the pressure drop in the header. At least if I have to do it the hard way next time it should be easier.
 
Balancing the pressure drop in the header will be the kinetic energy of the gas. Thus the velocity head of the gas going to the last cooler will be effective in inducing flow through that branch. Less of the velocity head is available to induce flow through the upstream coolers.
Doug
 
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