wrote:

... the faster the water moves the more efficient the transfer process is.
... The trick is optimizing between pumping cost and heat rise.

OK. If it's 60 F outdoors and the pool is 70 F and 250 Btu/h-ft^2 of sun
is hitting 80 ft^2 of unglazed pool heaters 2' above the water surface,
how much water should we pump through the heaters to maximize the COP?

It looks like the answer is zero, with zero heat gain for the pool :-)

If 250x80 = 20K Btu/h = poolgain + airloss, and poolgain = 500gpm(Tf-70)
and airloss = (T-60)80ft^2x2Btu/h-F-ft^2, with final and average heater
temps Tf and T = (70+Tf)/2, Tf = (24K+35Kgpm)/(80+500gpm), which makes
COP = 89.5K/(gpm+6.25gpm^2), with a min COP = 0 at infinite gpm.

20 HEAD=2'pool heater head (feet)
30 FOR GPM=1 TO 5'heater flow
40 HP=HEAD*8.33*GPM/60/550'pump horsepower
50 PE=746*HP*3.412'pump power (Btu/h)
60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)
70 PS=60*8.33*GPM*(TF-70)'pool solar gain (Btu/h)
80 COP=PS/PE'coefficient of performance
90 PRINT GPM,TF,PS,COP
100 NEXT GPM

1 101.7241 15855.72 12338.92
2 87.03704 17030.23 6626.459
3 81.64557 17461.37 4529.477
4 78.84616 17685.23 3440.661
5 77.13178 17822.32 2773.866

Why pump more than 2 gpm? Going to 4 increases the pool heat gain by 4%
while halving the COP (to a thousand times more than an AC COP of 3 :-)

Nick