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After looking through the STSS website, it supports all the math that I have done. There is a certain amount of variables such as what you have listed above, but outside factors I would think only fudge the theoretical #s by less than +-5%.
It all starts with the flow in GPM that you want to move, that determines the pipe diameter, or how many of a certain diameter you want to use. The key is using a diameter or multiple coils to meet your flow requirements, and maintain the optimal heat transfer of copper which is 2-4ft/sec. which translates to 3-6gpm for 3/4 NOM copper. Nom means the inside diameter is 3/4". (refridgeration copper is OD)
So for domestic coils you want to support the usually at least 2.5-3gpm. Where 1/2" copper would support this, but it would be on the higher end of the velocity, or head pressure on the hot side.
So all of STSS's coils support that copper will transfer 60 BTU per hour per every sqft of copper, per degree of degree of delta T of water change.
So using one of there coils part no. 425 120ft of 3/4 NOM type L soft copper
(My #s will be in bold)
Temperature rise from 47 Deg. F. to 105 Deg F
delta T=58F. @ 2 GPM.
2GPM X 60 gal /min X 8.3 X delta t 58=57,768BTU/hr pretty close to below
Average tank temperature 120 Deg. F.
The way to calculate delta T use the difference between the tank temp to average DHW temp you want. In this case 120-average DHW temp of 76=44F delta T
Transfer rate 57,976 BTU’s per hour.
57,976 BTUs/60BTU/hr divided by 44F= 21.96 sqft of copper. 3/4" is .185 sqft per linear ft divide them= 118.4ft of copper pretty darn close if I say so myself
So let me see if can simplify the equation to length of ft. This is going to be for 3/4 NOM copper.
Example: 2.5gpm of DHW from 50-110F with a 140deg tank
1) design GPM x 8.3lbs per gal x delta T you are trying to achieve in DHW = BTU/min you are trying to heat.
2.5gpm x 8.3 x 60F= 1245 BTU/min
2) find average DHW temp and subtract from tank temp= design delta T
average of 50 to 110 is 80F. 140F-80F= Design Delta T of 60F
3) take design BTU/min and divide by design delta T = sq ft of copper in the coil.
1245 BTU/min divided by 60= 20.75sqft
4) .185sqft/linear ft of 3/4" copper.
20.75 sqft divided by .185 = 112.2linear ft of copper
5) from there I would add 5-10% for the wife kids, and Fluffy the dog
so that gives me 117-123ft of 3/4" copper
6) now what you can also count on is if the water sits in the coil for a long period of time its going to equalize with the tank and you will have a volume of that will be 140F that needs to be mixed down to 110F. I will spare you the math, but it equates to
roughly 2 minutes of "free heat" at 2.5gpm before you have to rely on the heat transfer of the copper to heat to instantaneously heat your hot water.
I hope this helps someone along the way.......or I really like to hear myself talk?[/quote]
I'll agree with the end conclusion of 120ft. The math appears to works in this particular case , but I think it may be overly simplified for other cases. You're averaging the low and the high temperature of the incoming water (50) and the desired outgoing temp (110) to get an average deltaT over the pipe (50+ 110)/2 = 80. Using this average dT may work fine when when the temperature difference between the tank and the desired DHW temp is fairly large, like it is in this example, but if you're trying to figure out a system where you want the outcoming DHW to be fairly close to the tank temp, I think the error in averaging the delta T becomes significant.
For instance, for your last example above, if you had a desired DHW temp of 120F and wanted to be able to operate your tank down close to 120, (for your solar heat in the summer, your weekly summer fire, or the cold tank morning Dave referes to), then you'll get a calculated length of 225ft. In acutality you would probably need a lot more because the heat flow rate becomes asymptotic approaching zero as the coil temperature approaches the tank temperature. The error of averaging the temperature becomes even more obvious if you want a DHW temp of 120 and you want to run the tank down to 115F. Common sense says it aint never gonna happen, but the calcualtion says...262ft.
Please don't read me wrong. I'm not trying to be righteous or argumentative. I'm just searching for the answer. I had some closer desired approach temperatures on my project. I ended up guestimating and everything worked out fine. But it still irks me that I don't know how to calculate it
!! And we want to make sure that Fluffy is happy.
JR