# Floor-air convection

1. Apr 21, 2014

### meeotch1

I'm trying to do a rough calculation to find the "operating temperature" of a concrete floor. I've found a paper that discusses calculations for rate of heat loss from the slab to the earth: here. (It's a Word file - don't freak out.) Which gave me a fairly reasonable figure of ~2400Btu/hr (~700W) for heat loss. Also, I've got the equation for convective heat transfer as: q = hc*A*dT. here

However, I'm not sure I can just assume that the heat loss to the earth will be exactly balanced by the heat transfer from the air. Or maybe I can, since the linked paper seems to give the heat loss figures for an equilibrium situation between the slab & the air? If I do that, and take 10.45 as the convective transfer coefficient (hc), A ~= 1000sf (~93m2), I get something < 1 degree for dT. That doesn't seem right.

Basically, I'm trying to predict the temperature of the floor, given adequate heating of the interior space by forced air. (With no air stratification, due to a gigantic fan.) It doesn't seem intuitive that a bare concrete floor of something like 8" thick would be almost exactly the same temperature as the air. Have I missed something?

EDIT: it occurs to me that the convection equation above may not apply anyway, since the action in this case it reversed... rather than floor heating air, hot air rising - it's air heats floor, cold air stays where it is, unless moved by some external force. But I'm told that this giant fan (8' diameter!) can reduce stratification to 1deg from ceiling to floor.

Last edited: Apr 21, 2014
2. Apr 21, 2014

### 256bits

The heat transfer coeficient has units of W/(square meter -degree C ) or Btu/(ft^2 - hour - degree F).

From your equation q = hc*A*dT, 1 watt will flow with an hc=1, through an area of 1 square meter with a temperature difference of 1 degree C ( between the surface temperature and the air temperature). Increasing the temperature difference, such as with hotter air will result in more heat flow, as will increasing the hc by blowing air more quickly over the surface.

Since you surface area is about 100 square meters, each square meter has to transfer 7 watts of heat on average.

You picked an hc at random of 10.45 from the site you visited no doubt equation (2), which does not seem to correlate as you have found out.

Thermal conductivity k of concrete is about 1 W/(m-K).

From this site,
http://en.wikipedia.org/wiki/Heat_transfer_coefficient
equation 2.3, For a hot surface facing down or a cold surface facing up,
you see that we use the Raleigh number,http://en.wikipedia.org/wiki/Rayleigh_Number, when it lies within a certain range.
Or the Nusselt number, or some of the other numbers listed at the bottom of the page.
so it can get quite involved.

Just picking out a Raleigh number of 1 million, which is just a GUESS, I get an hc of around 8. And doing the calculation on q = hc*A*dT, I get a dT of 1 C, which doesn't seem to make all that much of a difference either.
Perhaps the k for concrete is incorrect.
But if you want more finess, you will have to plug through the equations to determine Ra, or Nu, to try it out on your own.

Last edited: Apr 21, 2014
3. Apr 21, 2014

### meeotch1

Thanks for the response. And yes, I see I did impose my Imperial bias on the area constant. (I vaguely remember a big push for the conversion to metric in the U.S. somewhere around my 6th grade year, but apparently the country as a whole gave up after that. As evidenced by the fact that the linked paper is in Imperial, which is probably what threw me to begin with.)

20C does seem crazy high, though. That's like the difference between room temperature and 0C, and is actually more than the total "available" dT between room and earth. I see that the range for RaL in the wikipedia article would lead to a potential range of h of an order of magnitude or more. Looks like I'll have to try and refine that value.

Or maybe I can retrofit by laying low-temp heating elements down in the winter & covering them with a rug.

4. Apr 21, 2014

### 256bits

I re-edited after seeing an error, from 20C to 1 C. I guess you started posting and did not see the edit which I should have noted for you.

Do not take the calculation at full value since it is after all just a quick back of the napkin and in no way completely accurate. By just taking a meter squared element, rather than the whole floor, this has its own error of ignoring adjacent elements. And a basement being a closed box would be different than a flat horizontal plate surrounded by air.

Last edited: Apr 21, 2014
5. Apr 21, 2014

### meeotch1

Looking at the RaL range given in wikipedia, it looks like at the very low end of RaL, the difference in dT would be a factor of 1.35 (from your calculated figure). So yeah, very low sounding.

On the other hand, googling for people talking about their concrete floors in winter brings nothing but tales of woe, freezing feet, and aching joints.

I guess there's nothing to it but to live through one winter, and if I'm miserable, build a floating floor somehow without moving all my belongings out into the street.

fwiw, there is no basement - so it is in fact a giant flat plate, with air on one side.

6. Apr 22, 2014

### 256bits

Try this: Put a into the freezer a steel rod, a wooden dowel, and some cork.
After a time take them out. Which one of them feels colder to your hand? A good part of the cold floor feeling is the conductivity and thermal mass of the floor material.

7. Apr 22, 2014

### meeotch1

Argh - good point. Concrete does tend to feel cool, even in the summer. Though if it does actually come close to equalizing to room temp at the boundary as the napkin numbers suggest, maybe it'll be a matter of "cool" as opposed to "unbearable".

Someone was telling me that a normal 6" slab (mine is at least a couple inches thicker... though I don't generally feel the need to flaunt it) takes five days to warm up 1deg. The person in question wasn't a physicist, but that's a helluva mass.

Actually... getting out my own napkin: Area:93m2 * depth:.2m * density:2400 kg/m3 = mass:44640kg. Cp = 880 J/(kg*K) = 0.83 BTU / (kg*K). Assuming the 2400BTU/hr figure is correct for heat loss to the earth, that's 2400 / 0.83 / 44640 * 24 hours = 1.55C temp change / day. Faster, but not fast.

And it occurs to me: the linked paper above does take into account the interior (air) temp. (You can see the thermal circuit model, with a resistor endpoint in the middle of the air volume.) So maybe that makes the 2400BTU/hr a figure for heat lost from the *air* to the ground, through the slab. In which case, couldn't I ignore convection altogether, and use Rate = k * A * dT / depth as an approximation? That gives: ~700W = 1.5W/(m*C) * 93m2 / 0.2m * dT. Or dT = 1C.

Everything keeps coming up ~1C...

8. Apr 22, 2014

### 256bits

That really surprises me. As you, I always was under the impression that concrete slabs surface were colder than that.

After poking around, it seems that a good h to use is around 8 or 10 W/m^2K.

http://stuff.mit.edu/afs/athena/dept/cron/project/concrete-sustainability-hub/Literature%20Review/Building%20Energy/Thermal%20Mass/A%20simple%20method%20for%20estimating%20transient%20heat%20transfer%20in%20slab-on-ground%20floors%202004.pdf [Broken]

These guys with MIT use an h of 8.48 W/m^2-K as seen in Figure 1, but I did not see a justification for the value, for their analysis of heat loss trhough a slab.

http://vbn.aau.dk/files/13726805/Pa...ic_Heat_Storage_Capacity_of_Building_Elements

And from this site they show the dynamic temperature variation of a slab during a 24 hr period, taking a thermal mass and heat storage description for the slab and earth for interior comfort. From figure 1, the temperature dT between air and slab is maximum 2 degrees C, using an h = 10 W/m^2-K.
Increasing the h, with directed air flow will give the room more comfort at the expense of increased heat loss, but due to thermal mass effects, one might recuperate some of that back during parts of the day when the room temperature falls below the slab temperature.

This has been all very interesting.

Last edited by a moderator: May 6, 2017
9. Apr 22, 2014

### 256bits

I think I will now retract the above statement from post 2.

By the way, I used to like using BTU's for heat flow, since it seemed to be more "user friendly", for heating and cooling, but then that is progress. Just an opinion and not everyone will be in agreement. Once immersed in metric it all falls in place in the end.