|MadSci Network: Physics|
What a great question. First, let's understand a thing or two about solar panels. Solar panels adsorb heat from the sun; the amount of energy they adsorb is essentially fixed. Therefore, for a certain sized solar panel, your will only adsorb a specific number of calories, watts, or BTUs. The solar heat flux is roughly 1 kW/m2 (depending upon elevation, how clear the air is, orientation of the solar collectors, etc.), so each square meter of solar collector will produce roughly 1 kW of heat for a portion of the day. Whether your water runs fast or slow, hot or cold, there is still only the 1kW of heat per square meter available. The essence of your question is how to get as much of that 1kW of heat into your pool as possible. And that is entirely determined by heat loss. Heat loss is a function of temperature differences. A pipe carrying hot water will lose more heat to the surrounding air than a pipe carrying cooler water. A solar collector operating at high temperature will lose more heat to the surroundings than one that is cooler. Convective heat transfer is the movement of heat between a surface and a surrounding fluid (like air). The formula is Q=hA(T2-T1), where Q is the amount of heat that is lost, h is the convective heat transfer coefficient (typically 2 BTU/ft2hr, A is the area for heat transfer, and T2 is the temperature of the hot surface, and T1 is the temperature of air surrounding the surface. Anything you can do to reduce h and A will be useful in reducing Q, the heat loss. However, it is clear that if T2 and T1 are close in temperature, you will lose less energy than if there is a large difference. You can insulate the pipes carrying your water, you can insulate the solar collectors to decrease heat loss from the back. There are a lot of things that you can do to decrease heat loss. The simple fact of the matter is that more heat energy will be transferred to your pool by using a higher, cooler flow than by using a slow, high temperature flow. Higher flows, of course, require more pumping energy. However, that extra pumping energy also ends up as heat in your pool. You can use the same heat loss equation to get a rough idea of how much energy you are losing from your pool as a function of convective losses. Even more significant, some of the water in your pool is evaporating, taking about 540 BTUs of heat energy per pound that evaporates. So, perhaps one of the most effective things you can do to increase the temperature of the pool is to cover it when you are not using it to decrease evaporative losses. Good luck with keeping your pool a bit warmer.
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