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The U.S. department of energy suggests electric radiant floor heat can be cost effective if the floor is preheated during off peak electricity hours if installed within an appropriately massive material such as concrete.

Is there a way I can predict the heat loss of the floor over the course of a winter day? In particular, I'm interested in installing it under ceramic tile installed on an uninsulated wooden subfloor with a basement underneath (i.e. air.)

The article states:

Because of the relatively high cost of electricity, electric radiant floors are usually only cost-effective if they include a significant thermal mass such as a thick concrete floor and your electric utility company offers time-of-use rates.

Time-of-use rates allow you to "charge" the concrete floor with heat during off-peak hours (approximately 9 p.m. to 6 a.m.). If the floor's thermal mass is large enough, the heat stored in it will keep the house comfortable for eight to ten hours without any further electrical input, particularly when daytime temperatures are significantly warmer than nighttime temperatures. This saves a considerable number of energy dollars compared to heating at peak electric rates during the day.

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The concept of the storage heater is very common in Britain: they usually include a phase change medium which means that a relatively small mass of material can contain a lot of heat without becoming very hot (the latent heat of fusion provides a "thermal cushion" where the medium can give off a lot of heat at a constant temperature, keeping the rate of heating constant).

But if you are stuck with concrete, you can just compute the heat capacity of the floor: multiply it by the maximum acceptable temperature, and you have your answer. As an example, taking a 5 x 8 meter room with a 20 cm thick concrete floor, you are talking about 5x8x0.2=8 cubic meters of concrete. With a typical density of 2400 kg/m^3 and heat capacity of 0.9 J/g/C, you can store about 8x2400x0.9=1.7E7 J/C. This means that a 4 kW heater could run for 1.7E7/4000=4320 seconds to heat the floor by 1 degree - a 12 hour overnight stint will get you up by 12x3600/4320=10 degrees Celsius.

So yes - you can store a reasonable amount of heat in a concrete floor over night. The numbers you need obviously depend on the dimensions of your room, but I hope the above gets you oriented.

If you put the heating element external to the floor, and use water as a medium to do the actual heating, then at some point in the future you could use other sources of heat (geothermal, solar) to heat the same water and get all the benefits of the warm flooring at a lower cost - and without having to rip out the old infrastructure. Worth considering.

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  • $\begingroup$ Thanks, this is a great start! In my case it's cost prohibitive to use a fluid heating system as well as concrete. I'm limited to electric with ceramic tile. I assume I can just plug in the density and thickness of ceramic tile and do the same math. $\endgroup$ Commented Oct 15, 2014 at 2:55
  • $\begingroup$ So the remaining unanswered part of the question is how do I compute the heat loss throughout the day once the heating elements are disabled? $\endgroup$ Commented Oct 15, 2014 at 2:56
  • $\begingroup$ Yes - that should get you some of the way. The key concept is "heat capacity". $\endgroup$
    – Floris
    Commented Oct 15, 2014 at 2:56
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    $\begingroup$ Heat loss depends on so many factors. But in essence, if you need 2 kW heater to heat for 24 hours a day, then 4kW overnight (12 hours) will in time achieve the same thing: if the heat is not lost to the air fast enough, the floor will be warmer at the end of the day, and this keeps going until it loses heat at the same rate that it gains. $\endgroup$
    – Floris
    Commented Oct 15, 2014 at 2:58
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    $\begingroup$ Having said that - the uninsulated connection to the space below means that space will also be heated - so there will be some additional losses (from that space to the outside world). This will make your system quite a bit less efficient. Your best bet would be to heat the space for a while, then monitor the temperature when you turn heating off. The rate at which temperature falls is a great measure of insulation. $\endgroup$
    – Floris
    Commented Oct 15, 2014 at 2:59

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