OK, first thing to realize, is that regardless of it's type, figuring out the performance of a heat exchanger is a hairy and nasty problem... There are a bunch of variables, and most of them keep changing... The biggest of these is the Delta-T, or temperature difference between the fluid on one side, and the fluid on the other... Not only is this going to change over the course of a burn as the tank heats up, it will even change from one part of the HX to another at any given moment... Really the best that can be done is a series of approximations to try and make an educated guess as to the actual performance.
Cribbing a lot of info from the Engineering toolbox links above, we start out with a "still water" figure for a water to water HX, with copper as a separating medium, of 60 - 80(Btu / ft2 / hr / *F) or in metric 340 - 455(W / m2 / *K) In essence this says that you will see 60-80 BTU's per hour transferring across each square foot of HX area for each degree of difference in the two temperatures... I'll split the difference and use 70 (Btu / ft2 / hr / *F)
A nominal 1/2" tube will have a surface area of about 18.8in2 per foot of length, or 7.65' of tubing per square foot of surface area... Call this "one HX unit"
Since the overall performance gets worse as the tank heats up, and you get less and less difference between the boiler temp and the tank temp, we need to figure for something close to the worst case performance to get reasonable results, but going for the absolute worst case would need a near infinitely large HX as the temperature difference approaches zero, so we should probably solve for 3 cases just to see what difference it makes, say 5*F, 10*F and 20*F.
Per NoFo and other posters, we could assume that our boiler output is about 75% of rated power, but we also know that our peak tank heat should be happening towards the end of the burn as the fire is entering the coaling stage and the boiler is starting to cool slightly. Thus I'm going to assume a 150K BTU/Hr class boiler, with an actual heat to transfer of 100K BTU/Hr. (it makes the math a little easier!)
The formula above, assumes "still water" - how much difference does it make that this water is moving? I don't think it makes much, as if one takes a small enough time slice, the water is essentially still... Since the conditions are constantly changing, the number we are going to get is only going to apply for an instant in any case. Pumping water through will have the effect of increasing the temperature difference as there will be less time for the pumped water to transfer it's heat, obviously the faster the flow, the greater the difference...
So to figure for each temperature range, we get 70 Btu's of transfer per unit of HX, per degree of temp difference, so
CAP / (dT) (BTU) = HXU x (Sz) = L
where:
CAP = total amount of heat to transfer (BTU / hr)
dT = Difference in temperature (*F)
BTU = BTU transfer rate (BTU / hr / ft2 / *F)
HXU = 1 square foot of HX area
Sz = Length per HXU
L = Total length of HX tubing required
100,000 / (dT ) (70) = (HX units) x 7.65 = feet of 1/2" tube needed
100,000 / 5 x 70 = 285.7 x 7.65 = 2,186 ft at 5*F difference
100,000 / 10 x 70 = 142.9 x 7.65 = 1093 ft at 10*F difference
100,000 / 20 x 70 = 71.4 x 7.65 = 546 ft at 20*F difference
It seems fairly obvious from this, that we will probably be best off working with a 20*F difference in order to keep the HX an affordable size. If we assume a 190*F maximum boiler output temp, that would in theory give us a 170* max tank temp, but since as stated above, this is an "average" temp, the tank top is likely to be higher. Also, while the BTU/hr capacity goes down as the temp differential decreases, it will still transfer SOME heat as long as the boiler is hotter than the tank, so presumably one could approach the boiler output temperature although more slowly...
This equation can also be worked in the other direction - assuming the 546' HX from above, what sort of heat can we expect to transfer with a cold tank? Assume the same 190* boiler output temp, with protection to limit the boiler input to 140*. That gives a 50* dT. Reworking the equation, we get
(HXU) (dT) (70) = CAP
71.4 x 50 x 70 = 249,900 BTU / Hr - or considerably more than the nominal boiler output...
Gooserider