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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!

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## Comments

588MemberHere'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.

6MemberNuts! 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.....

588MemberThe 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

6MemberOkay, 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!