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Whoa.! Copper Heat Exchanger Calc

Tinkerer
Tinkerer Member Posts: 6
Folks, (a die hard do-it-yourself'r here) I've spent several weeks (or months?) installing a solar thermal system for our house.



Somewhat sheepishly, its only after the fact that I'm trying to do a basic calculation of btu gain.



I have a home-made copper heat exchanger. It is four 60' coils of 1/2" soft L copper in parallel with 1" supply and return lines running into the tank.



So I figure the system GPM is split four ways within the heat exchanger. Can I thus divide system GPM by 4, and then look up the ft head loss on a chart for 1/2 copper for that number??? Now, do I multiply the result by 4 because there are four loops the water is pushing through? This is where I'm confused.



Same goes for the solar panels. Do I divide the system GPM by eight, look that head loss up on their pressure loss tables, and then multiply the result by eight?





System specifics:

We have (8) 4x10 AE-40 panels, 150' of 1" coper (round trip) from the panels to a 500 gal storage tank. Inside I build a coper heat exchanger with (4) 1/2" soft L coper coils in parallel. A Wilo S 21 pumps Dowfrost through the system. I'm curious how many GPM I'm pumping at the three speed settings.



I have the pump curve for my circulator, but calculating the head loss across the heat exchangers has me stumped. It seems really difficult to know how many GPM I am moving at a given speed setting.



Any help is much appreciated. Yeah, I know I should have done these calculations before I built everything, but that would have taken the fun out of it!

Comments

  • Kevin_in_Denver_2
    Kevin_in_Denver_2 Member Posts: 588
    adding resistors in parallel

    Here's a link that explains how to add it up:



    http://en.wikipedia.org/wiki/Resistors_in_parallel#Parallel



    But you have a special case where all the resistors are equal.



    Therefore, the head loss of the entire heat exchanger is equal to the head loss through just one of the loops.

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  • Tinkerer
    Tinkerer Member Posts: 6
    Ohms Law!

    Nuts! I've read that, but I'm confused at what flow to compute the head. Surely not the total system GPM? Do I devide the total system GPM / # coils of same length and look up the head loss for 1/2" pipe on a chart? And then your saying that number is the answer for the heat exchanger, I don't multiply or devide the result by # of coils?







    Not sure why ohms law would apply to fluid dynamics, amps and voltage don't seem to have obvious analogies with fluid. Ok, current could be flow, but.....
  • Kevin_in_Denver_2
    Kevin_in_Denver_2 Member Posts: 588
    edited May 2012
    Right

    The pressure drop across one coil equals the total drop across the whole thing.



    The flow through one coil is 1/4 the total flow.



    Period

    There was an error rendering this rich post.

  • Tinkerer
    Tinkerer Member Posts: 6
    of course!

    Okay, of course. The flow will equalize at the same pressure between all the four coils, i.e., the resistance of any one of them as long as they are all the same. Thanks!
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