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Let's say I have 2 exchangers that are completely the same, only difference being that one of them has half the number of tubes. Also let's assume the coolant is the same and has the same inlet temperature. I want the fluid to cool from 100 degrees to 70 degrees, and this works in the first exchanger. I wonder if reducing the flowrate of the fluid can make it possible for the second exchanger (with less tubes) to to the same job (to cool the fluid from 100 to 70 degrees).

Heat transfer rate is equal to:

$$Q_1=U \cdot \Delta T \cdot A$$

By reducing the number of tubes, area gets smaller and so does the heat transfer rate.

$$Q_2=U \cdot \Delta T \cdot \frac{1}{2}A= \frac{1}{2} Q_1$$

So if I want this exchanger with a halved heat transfer rate to do the same job (in terms of outlet temperatures), I assume I would need to halve the flowrate as well? Can someone tell me if I am correct here and maybe show it mathematically?

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  • $\begingroup$ Check out Heat Transfer by Siminson. $\endgroup$
    – Solar Mike
    Aug 3 '21 at 17:04
  • $\begingroup$ Can't find it... did you mean Engineering heat transfer by Simonson? $\endgroup$ Aug 3 '21 at 17:26
  • $\begingroup$ So did you read it? Did it answer your question? Should do... $\endgroup$
    – Solar Mike
    Aug 3 '21 at 17:46
  • $\begingroup$ Didn't answer it $\endgroup$ Aug 3 '21 at 17:50
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    $\begingroup$ You are on the right track with your train of thought. If I may make a suggestion, I would set up an example spreadsheet to test your theory. Program in the generic heat transfer balances you have in the last link you shared. A fantastic book on heat exchangers is Design of Thermal Mechanical Systems by William Janna. Also, keep in mind the following: in your original post you make the assumption that the heat transfer coefficient stays the same when you half the number of tubes. This may not be a valid assumption since the Nusselt number is dependent on the Reynolds number. Hope this helps! $\endgroup$
    – mechcad
    Aug 3 '21 at 19:32
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I assume you are referring to a exchanger like the following:

enter image description here

If :

  • the external dimensions of the heat exchanger remain the same.
  • the tubing diameter remains the same.
  • the number of tubes changes (halves) $n_2 = \frac{n_1}{2}$

In that case like you mentioned the exchange surface will halve, and that will reduce the heat exchange. As a result because the heat exchange will be less, the out temperature will be closer to the input temperatures.

My understanding of your question is,

if I remove half the tubes can I get the same temperature difference?

Since you have only half the transmitted heat energy, the seemingly obvious solution would be to reduce by half the flow rate of the cooled liquid (the one coming in at 100$^oC$). Although, if you reduce to half the flow rate then the average speed on the tubes will halve, and things should be ok (i.e. half the heat transferred, would reduce the temperature of half the flow.).

However, the problem is that the following, by changing the number of pipes, you are changing the internal configuration. That will affect the flow in the shell.

Because of those changes, then the Reynolds number might change.

The Reynolds number affects the Nusselt number which control the convection. Depending on the case , $Nu\propto Re^n$ where n is a number ranging from 0.5 to 1.

However that means that the heat transfer coefficient U changes (if for example the speed becomes too low then the flow becomes laminar and the heat exchange is significantly affected).

So the bottom line, is that this is not a simple problem to answer, and you need more details on the internal configuration of the heat exchanger to make an estimation.

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  • $\begingroup$ The important parts are correct but if the number of tubes and the flowrate are both halved, the volume flow rate in each tube and thus the flow conditions will remain the same.. $\endgroup$
    – mart
    Aug 4 '21 at 14:32
  • $\begingroup$ @mart you were right, the way I had written it did not make any sense. Hopefully, my edit makes it a bit more sense. Please let me know if you find something odd still. $\endgroup$
    – NMech
    Aug 4 '21 at 18:46

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