Splitting and merging supply lines

[ari]

Hi, first off, I'd like to thank this forum for all the plumbing advice I've read as a guest. I haven't been able to find an answer to this particular question so I finally joined.

I am making a heat exchanger for my shower trap arm. I have a 2" section of copper drain pipe that I want to coil the incoming water supply line around, before the supply feeds into the 3/4" inlet of my on-demand water heater. I already tried and failed at coiling 3/4" soft copper tubing around the 2" pipe. I see in the commercially available ones, the supply line is split into multiple smaller lines which has the advantage of more surface area for contact in addition to the smaller diameter tubing being more workable. I'm considering this approach but am wondering if there are any implications for downstream pressure or restricted flow when splitting a line into multiple smaller lines and then merging them back together before feeding into the water heater. According to the table linked below, splitting into two 1/2" lines would maintain the same cross-sectional area as the single 3/4" line. The commercially available ones split into four lines but according to the table linked below, there is no diameter of pipe that will quadruple to match 3/4" cross sectional area. Is it better to have the cumulative cross-sections of the coiled lines exceed that of the 3/4" pipe so the coiled volume acts as a sort of reservoir?

I'm leaning towards two 1/2" lines because the tubing and fittings are easy to find at it would only require one tee on each end. Lastly, can anyone comment on whether it is reasonable to expect to coil 1/2" soft copper around a 2" diameter (wish I had asked about 3/4" before trying it).

*cross-sectional areas were taken from this chart: Copper Tubing Size Chart ASTM B-88 | Engineers Edge | www.engineersedge.com

thanks, ari

 

[wwhitney]

No offense, but I think unless you're a machinist and have specialized tools, it won't be cheaper to recreate a commercial DWHR than to buy one. With that aside, some possibly useful background:

The ability of a pipe to carry water flow is characterized by the frictional pressure loss per unit length for a given flow. That behavior follows the empirically derived Hazen-Williams formula:

https://en.wikipedia.org/wiki/Hazen–Williams_equation#U.S._customary_units_(Imperial)

If flow Q and inside pipe diameter d appeared in that formula with exponents in the ratio 1 : 2 (e.g. Q/d^2, or Q^1.5 / d^3, etc), then one pipe would be equivalent to two parallel pipes each of half the area. But the ratio is 1.852 : 4.8704 = 1 : 2.63. That means a pipe half the area can only carry 40% of the flow for the same pressure loss per unit length. Of course, in your application, for a fixed area of contact, splitting the pipe into two parallel lengths will also approximately half the length of each pipe.

The point of all this is that matching the cross sectional area of your 3/4" end connections isn't the proper criterion for sizing your parallel pipes. More to follow.

Cheers, Wayne

 

[Sponsor]

 

[wwhitney]

Rather, I would suggest something more like the following: pick a trial number of parallel pipes and inner diameter, then figure out what the equivalent length of 3/4" type L (0.785" ID) copper pipe is, and then decide if that's OK.

E.g. let's try 3 runs of 3/8" OD (0.311" ID) copper tubing. Wrapping 3/8" O.D. tube around a 2-1/8" OD pipe I'll estimate as working on a 2-1/2" OD cylinder. Per foot of contact region along the pipe, that's 2.5 \pi inch-feet of area. As the tube is 3/8" OD, that's 2.5 \pi / (3/8) = 21' of total tubing length, or 7' each run.

So consider Q = 1 gpm through 1' of 3/4" copper type L. Ignoring the constants that are going to cancel out in the end, the Hazen Williams formula gives us 1/(0.785)^4.8704 for the 3/4" copper. In comparison, we have 1/3 gpm through 7' of 0.311" ID copper, so the constant-less H-W formula is 7 * (1/3)^1.852 / (0.311)^4.8704. The ratio is 83, so this design would be equivalent to 83' feet of 3/4" copper type L, per foot of DWHR length.

That doesn't sound very good. For comparison, try (7) runs of the same tube, now each only 3' (not really practical, nor am I confident the approximation I used for length calculation would work). The constant-less H-W formula now gives 3 * (1/7)^1.852 / (0.311)^4.8704, or a ratio of 7.4. That's more reasonable.

The upshot of this is 3/8" OD copper tubing is too small; you can try the computations again with (2) runs of 5/8" OD copper tubing and see what that tells you.

Cheers, Wayne

 

[ari]

Wow thank you Wayne! This is more than I could have hoped for. Thanks so much for explaining so well and working through an example of an intimidating-looking equation. I got a ratio of 13 for two 7.85' runs of 0.545 ID pipe which I think will be okay considering the DWHR is only a couple feet long, my system has very few fixtures, short runs and high incoming pressure.

I know it would probably be more sensible to buy one but it's become an obsession and I feel oddly committed to getting this to work.

 

[Reach4]

Comment: it is better to feed the output side of the drain line with cold and draw pre-warmed water from the warmer end.

 

[ari]

Comment: it is better to feed the output side of the drain line with cold and draw pre-warmed water from the warmer end.

yep! that's how we've got it oriented.

 

[Mr tee]

I expect that wrapping 5/8 OD (1/2 nom.) neatly around 2 copper will range from difficult to impossible. 1/2 OD would probably also be something of a stinker.