Exchange area between parallel pipes

[ABerlinMML]

Hello,


What is the exchange area (from q=U * Ar * LMTD) between two identical parallel thru-holes drilled in a block? Is it pi*D*L*2 ?

Thanks and sorry for the basic question,

 

[ABerlinMML]

Hello, I realized this may sound like homework but it's not. I'm actually making this as a mock heat exchanger for a pair of servo motors. I was hoping to calculate if it will be sufficient to drill a couple of parallel NPT tapped channels in an aluminum block. I'm assuming for now this area is the projection between them.

 

[IRstuff]

The heat flow will spread out from the first tube, and then concentrate onto the second tube. You might want to do a search for "heat spread" for some examples that you might be able to modify

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[racookpe1978]

How much flow in each fluid would be optimum?
How much delta T between the two fluids?
How much are you trying to "improve things" between what you are using now?
How much money do you want to spend?
How "important" is the heat exchange, and which fluid (hot fluid or cold fluid) is more important to optimize? (Are you trying to heat the cold fluid by a little bit, cool the hot one by a lot, or "get as much as possible" out of the hot fluid "as cheap as possible"?

Those are a lot of questions, let me see if I can make their answers more obvious by using a couple of examples. We don't know enough to give you real answers, and you obviously don't want to pay for a real shell-and-tube heat exchanger.

Drill two holes 1 inch dia in the Al block, run both in parallel.
Drill two holes 1 inch dia, run the two counter-flow.
Drill 4x 1/2 holes for the coolant, and 4x 1/2 holes for the hot fluid.
In a 12 x 12 x 12 AL block, drill 4x 1 inch dia hot fluid holes, and 32 1/2 dia coolant fluid holes around each hot fluid hole in a octogon. Run fluids parallel or counter-flow. (Headers are tricker.)
In a 12 x 12 x 12 Al block, run 5x 1 inch dia hot fluid holes through East-West, alternating levels with 5x coolant fluid holes 1 inch dia running North-South. Put the headers across the ends to collect the outlets and dispurse the inlets.

 

[racookpe1978]

I think what I'm also trying to say is "What you have described will. But not very well. You can do a LOT of things to make it a better heat exchanger, but which one will make it optimum for cooling the hot fluid is going to depend on stuff I don't know.

 

[ABerlinMML]

Thanks for the replies.

Actually it's not a practical question, that's why I omitted all the application details. What I'm after is the theoretical 'area' from the overall heat transfer equation q=U * Ar * LMTD. If it was a length of tube in shell, then I know that area is just the surface area between the fluids, ie pi*D_tube*L. However with the drilled thru-holes in my block, they are not concentric. So what is that area in this case? Is it the rectangular projection between them?

 

[marty007]

I don't have the time to work through a detailed answer to your question, but one place to look is at a buried pipe analogy.

The textbook that I used in uni is called: Heat Transfer (9th ed) by J.P. Holman. This text includes a section titled "3-4 | The Conduction Shape Factor". As I remember from back in uni we had applied it to buried pipe systems, but you could probably use a similar approach.

This section looks at conduction between various shapes and their environment, but also has some examples of heat transfer between two shapes. In short, it defines the heat transfer term as q=k*S*delta(T)_overall. Next it gives a table of many different shapes, one of which is conduction between two isothermal cylinders of length L buried in infinite medium. For this system it defines S=(formula that depends on the diameter of each pipe and the distance between the pipes and inv-cosh trig [too complex to try and type in a forum]).

This type of approach might be a good place to start, see if your texts have any similar sections.

Cheers,

 

[IRstuff]

This article may be of interest:

http://www.scirp.org/journal/PaperInformation.aspx?paperID=43494#.VH3rS9LF_FI

It discusses using 45º as the spreading angle, i.e., expand the projected area along a 45º slope until halfway between the pipes, the collapse it back at a 45º angle until you reach the other pipe.

TTFN
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