Again, heat loss p/hr through an 8 hr. period from 5000 gal H2O 4

[Dutch500]

I want to use a vinyl, above ground pool, to produce fresh water marinelife. From seed to harvest is 90 days. During this 90 days the tempeture range of the water is critical. It must not fall below 65 degrees F. Any lower, the crop will die. Using a solar powered heat exchanger/boiler, I can heat the water during the daylight hours. At night, no sun, no heat. Gas or electric heaters are not an option. The pool is 18 ft. in diameter, giving a surface area of approx. 255 sq. ft. The sides are 3 ft. in ht. and have approx. 170 sq. ft. in area.(not a major concern, as they can be insulated) Capacity in the pool is 5,000 gals. The daytime water temp. stands at 81 deg. F. The night air temp. is 30 deg. F. Because of water flow for oxygenation and filtration, a cover coming in contact with the surface is not an option. At best, a cover will be five feet from the surface. What I need to know, very specificly, is this, given the ambient air temp. and surface area of exposure, the quantity of water total, and the starting temp. of 81 deg. F., what will the tempeture change of the water be per hr.? What is the formula to find, rate of heat lost per hour from this 5000 gals. of water in a deg./hr. format.

 

[prex]

Convective heat transfer from the upper surface of the pond may be estimated as follows (sorry for using those new fashioned SI units, you will have to convert!):
A=23.6 m²
deltat=28 °C
h=10 W/m²°C (at best)
power loss Q=hA deltat=6600 W
Considering an 8 hour overnight period, time=8*3600 sec, the temperature change t_loss of the water mass M=19000 kg with specific heat C=4186 W/kg°C will be:
t_loss=Q time/M/C= 2.4 °C.
Evaporative cooling from the open surface largely varies with air humidity: just to give a figure, you can account at least two times the convective heat transfer.
The conductive heat loss through the soil (uninsulated bottom) may be estimated to be the same order of magnitude as the convective heat loss though the top surface.
With those figures (quite optimistic, as the presence of wind will highly increase both the convective and evaporative heat losses) your are already beyond your limit of 9 °C loss, and cloudy weather is not accounted for!
To get closer to your goal you should definitely consider insulating also the bottom of the pool, painting in black the inner walls to collect solar heat, covering it with a transparent semi-insulating cover (preferably flat, to limit excessive drainage of droplets) and doubling the cover with a dark one during the night.
As I understand that your installation is at risk of a full loss if temperature gets excessively low, I would also consider the use of a heat accumulator (another pool or tank, but totally isulated). prex
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[gunsmith]

I agree to prex. The heat transfer coefficient for free convection to air is

alpha=1.52*deltat^0.25

The result for our example is 3.5 W/(m^2*K) for free convection to air. Multiplied with the surface and the temperature difference we get 2317 W. That means the energy loss is 8342 kJ per hour. So the heat transfer of the free convection is only the 1/3 of the result of prex. His result is more realistic than the simply calculation of free convection.

Andreas

 

[butelja]

Do not forget radiant cooling to the sky during the night. It can be a significant factor in addition to the convective and evaporative cooling. Is the pool indoors where it is sheltered from the wind? If not, a windy day will greatly increase evaporative cooling. If the pool could be located in a greenhouse type structure, you should be able to drastically lower evaporative cooling, as well as radiant cooling to the night sky.

 

[IRstuff]

my blackbody calculator came up with 0.0459W/cm^2 as the radiant power, which for your emitting area, works out to about 10.8 kW, with emissivity of 1 and temp of 300K.

While the values calculated show that the amount of heat lost will not alter the long term temperature of the pool, the surface layer temperature will most like get well below 65F, so there might be some die-off right at the surface, particularly if the air temp is already at 30F.

You might want to think about enclosing the pool in a hot house arrangement, which would minimize the evaporative loss as well as the radiant and convective loss. With double-pane glass, you could easily keep the air temp above 70F even at night.

TTFN

 

[gunsmith]

Lets take a greybody number of 3.5 W/(m^2*K^4) for practical use. The result is 6703 W. Knowing that W is nothing else than J/s we have an energy loss of 24131 kJ per hour and a loss of temperature of 2.42 K for a 8-hour-period during night for the 19000 kg of water only caused by heat radiation.

@IRstuff

Your blackbody theory has a result of 3.91 K temperature loss during night for the pool. I don´t know what is closer to reality. Someone with pool-experience should answer this question (not forgetting about the additional 2.4 K of prex´s calculation). I think that leads us to Your hot house arrangement.

Andreas

 

[Dutch500]

To Prex, Gunsmith, Butelja, & IRstuff,

Great responses from all of you, many thanks. Just FYI,
here is a more in depth description of the set-up. I have modified the design of a standard natural circulation boiler. Instrad of the upper drum being horizontal it is now vertical. The heating tubes extend from the center as spokes from the hub of a wheel, the outer end of the tube is joined to a vertical tube going down into the water. The vertical tube then joins a larger diameter tube going back to the bottom of the main drum. The whole apparatus is then put into the pool. The smaller tubes up top gathering solar heat, the larger tubes at the base acting as a heat exchanger. Because of the similarities of the boiler to an open parasol, it doubles as a greenhouse structure when covered with heavy plastic.
As for heat loss from the bottom, the vegetation grown in the pool is a by-product, an extremely prolific by- product. The pool will actually sit atop a 1 ft. tall box. Plant matter culled from the pool is placed in the box to decay into compost, this will provide some actual heat, how much I don't yet know. Beyond that, it will insulate the bottom of the pool with some where near a factor equivilant to an R-13.
The ultimate goal of my pilot project is to produce in a very small contained area, the same quantity, and much better quality, than a conventional method using three-hundred times more space and water. I also need to show that I can provide a continuous 12 month harvesting cycle, geared to the same amount of demand, so that price and amount of harvest are controlled by me and not the buyers that give 1/4 of what the public ends up paying. So far everything is on track. This winter and the tempeture cycle of the pool will tell me if it can be done.
Again, thankyou for the help, I am always open to suggestions and different perspectives, feel free to pass any futher info or oversights you see that I may not. Onward we go.

P.S. Conversions not ever a problem when the desired results are attained. terrific response.

 

[PacificSteve]

Being a former pool guy, and current solar heating guy, I would say that off the cuff, the wind will have the greatest effect on you heat loss, therefore some side shielding could be cheap and effective.

You may also wish to consider some external thermal storage in insulated tanks (multiple cheap solar storage units). Some $150 solar system controls could decide when to run a small pump to put hot water into the tanks during the day, and grudgingly give it back to the pool to maintain some "safety level" such as 70 degF. What thinks you?

Steve