Cool Down Test

Goal: To try and identify the discrepancy between actual cool down rate of the pond and the rate calculated based on R value.


It seems like the most likely possibility is that the larger than anticipated heat loss has something to do with the lid  and absorber configuration. 

To verify this, I opened the top cover glazing, and installed a 2 inch sheet of polyiso right over the top of the absorber.  To eliminate any air infiltration, it is weighted down, and then sealed around the edges with duct tape.



The extra layer of insulation over the existing lid/absorber.  Rocks to keep pressed against the edge lip of absorber, and duct tape just ot make sure.



This should cut the loss per sqft through the lid to at least as low as the loss through the walls and bottom.


If this reduces the loss to near the calculated value, then we can be pretty sure the excess loss is due something going on with the lid/absorber, water vapor, ...??



This is the overnight cool down -- black line is pond temperature, green is ambient.


The inside area of the pond box with the new sheet in place is 82.3 sqft.


The published R value for 2 inches of Atlas R-board is R12.1 (US)


Some temperature values extracted from the plot above:


Time T pond (F) T ambient (F)
6pm 07/08/07 138.5 80.6
9pm 137.5 71.8
12 midnight 136.4 64.9
3am 135.3 60.6
6am 134.1 60.9
9am 133.2 72.2
12 noon 132.5 77.7



Actual Loss rate:

Using the time span from 12 midnight to 6am


Q loss  rate = (136.4F - 134.1F)(190 gallons)(8.3 lb/gal)/ (6 hrs) = 605 BTU/hr 


The average ambient temperature for this time span is about 61F, and the average pond temperature is 135.2F.


The effective R value is then:


Q = A dT/R or R = A dT/Q


R = (82.3 sqft) (135.2F - 61F) / 605 BTU/hr = R 10.1


This is at least 20% worse than what would be expected, but its a lot closer than the factor of 1.8 that I was getting before.  So, it seems safe to conclude that most of the excess heat loss problem is in the lid, absorber, drains, pump outlet, lower glazing area?



If you assume that R10 could be obtained with some improvements (below) to the lid configuration, then the total heat loss for a 16 hour cool season night with a tank temp of 120F and an ambient of 40F would be:


Q typical night = (82.3 sqft)(120F - 40F)/R10 = 658 BTU/hr or 10500 BTU for 16 hour overnight.

Seems like more than you really want, given that the gain for a good day is not likely to be more than 26K BTU, and the demand might be near 15K BTU.


    - Insulate better?

    - Add more collector?

    - Reduce tank size and surface area to decrease losses?


It seems like some increase in collector area would be a good idea -- it is just barely adequate at best.

I am wondering why we need a tank as large as 190 gallons -- this is 4 to 6 days supply of hot water for two people.  Wouldn't two or three days be enough?

Making the tank smaller would reduce losses and costs, and allow higher insulation levels without running the cost up too much?  Maybe the big base, small tank configuration?


This still leaves the problem of figuring out why the loss is so much greater without the extra lid on the top.


Things that might be done to reduce losses:


    - Terminate the drains under the pond water surface  per Nick

    - Put floating insulating caps on the tops of the drain pipes per Alan

    - Figure out a way to get the absorber pond to drain fully

    - Add a layer of vegetable oil to top of pond as suggested by Jeroen


    - Others ideas?



On another subject, I have noticed white deposits on the EPDM absorber surface.  These look to me like mineral deposits from the water?  They are pretty significant considering that the collector has only been in operation for a few days.  Easy to rub off with a finger.

Use demineralized water?


I've also been wondering if (given the high stagnation temps on the absorber surface) if a metal absorber could be worked out that would not add significantly to cost and complexity?


Gary 7/9/2007