# Heat Transfer Coefs of My House

1. Jul 18, 2011

### alhaggerty

I’ve measured the temperature in my house over time - after turning off the air conditioner. I have a time series of internal temperatures for a fixed outdoor temperature. Is it possible to calculate heat transfer coefficients of my house using this data? Is it possible to relate these heat transfer coefficients to “R values” used in residential construction? I would like to make a meaningful comparison of home insulation properties using this data.

-Al

2. Jul 18, 2011

### AlephZero

You can only do that if you assume that all the heat transfer is by convection. If the sun was shining there would be a significant amount of heat input from radiation. Unless the house is airtight there willl also be heat transfer caused by air moving in and out of the house.

Even if you did calculate a heat transfer coefficient, the value would depend on the wind speed and direction, so a single number would not be much practical use.

3. Jul 18, 2011

### Drakkith

Staff Emeritus
Any way to get a decent estimate?

4. Jul 18, 2011

### turbo

Very tough problem. I helped a neighbor build a large addition on his home last fall, so that his daughter and 2 grand-daughters would have their own rooms. Recently, I have been helping him decide on how to insulate the place. We have been evaluating spray-in foam insulation (dense closed-cell, and lighter open-cell) and there are lots of variables even with that one choice. We built the whole addition out of full-dimension rough-sawn lumber with horizontally-boarded walls. I think we have settled on open-cell foam for the addition, though his house is insulated with that dense closed-cell foam.

Even if the place is well-insulated and all buttoned-up, can we calculate heat-loss/gain from solar input (dark enameled roofing material), thermal conductivity of siding, infiltration and solar gain from windows, etc? This stuff is an art, not a science, due to all the variables.

5. Jul 18, 2011

### gsal

how about simply calculating some kind of "all-encompassing" thermal resistance?

You know:

T2 - T1 = Q x R

T2 is the temperature outside,
T1 is the temperature inside,
Q is the heat going into the house
R is the thermal resistance between the inside and the ousdie...

this R would be all-encompassing in that it takes into account all walls, ceiling, windows, doors and the occasional opening of doors, etc.

If you take T2 to be some kind of average outside temperature and T1 the thermostat temperature...that means that the air inside the house is always at the same temperature...and all the heat coming in is being taken away by the air conditioning...then, all you have to do it see how much energy the air conditioning uses and depending on its efficiency you can estimate how much heat went into the house...isolate R from that equation and determine it...

6. Jul 18, 2011

### alhaggerty

Thanks for the replies!

This overall thermal resistance idea is what I had in mind. I'm looking for a rough approximation of energy efficiency that can be used to compare different homes.

It occurred to me that I could measure the duty cycle of the air conditioner times its output capacity over a period of time to estimate heat rate (Q). Then a simple overall thermal resistance could be calculated knowing indoor and outdoor temperature.

I was also wondering how the time series temperatures could be used? I read in a different posting on Physicsforums.com that a time series could be used to calculate the free convection coefficient of a tent. In this example, it seemed heat was calculated from the initial conditions before the temperature was allowed to relax. Could this technique be used here?

Thanks again,

-Al

7. Jul 18, 2011

### Staff: Mentor

Nope: You can't even calculate how much heat your house is losing, since you don't know the specific heat! (unless you calculate/add it up for every item in your house individually!)

8. Jul 18, 2011

### Staff: Mentor

How do you find Q?

9. Jul 18, 2011

### Staff: Mentor

That's much better, though as said that method doesn't differentiate between modes of heat transfer. You may want to try it after the sun has gone down on a hot night.

10. Jul 18, 2011

### turbo

Even so, the thermal mass of the house (I live in a log house) can corrupt that data since really massive walls cool slowly. I would expect very different behavior compared to a sided conventional-construction framed-wall house with siding , and a more common southern design with framed-wall construction with brick veneer (rare here, up north). In some municipalities here, new construction with very shiny metallic metal roofing is forbidden, but it's OK if you use enameled metal roofing. I'd hate to be in the AC business up here and be asked to engineer some tightly-constrained efficient system with all those variables. The best I'd hope to do is to spec out an over-designed system that wouldn't cost the owners too much by running at less-than-capacity.

11. Jul 18, 2011

### Staff: Mentor

Actually, that's precisely the problem that this method solves. You're not waiting for the house to cool, you're holding it at a steady state and measuring the energy required to keep it at that steady state. So the specific heat is irrelevant.
That's why we use software to do it for us.

12. Jul 18, 2011

### gsal

I guess it was not too clear, but in my previous post, Q the heat into the house, was meant to be equal to the heat absorbed by the A/C and equivalent to the energy used by the A/C (minus efficiency) ...giving that the thermostat is set always at the same temperature...

In any case, the problem could be discretized and perform some kind of integration over time....say every 12 hours...

12 hours during the day
12 hours during the night
12 hours during the next day
12 hours during the next night and so no

you could calculate average temperatures outside the house
and assume constant temperature inside
place a watt-o-meter on the A/C and know how much energy is consuming to keep the inside air at the same temperature and so this is the same as the heat going into the house since the air is being mantained at the same temp

For example, over two days and two nights, an average R could be calculated as

R = (1/4)x[ (Tday1 - Tin)/Qday1 + (Tnight1 - Tin)/Qnight1 +
(Tday2 - Tin)/Qday2 + (Tnight2 - Tin)/Qnight2 ]

or you could take only the days or whatever...you can play with what you want to take into account.

This, of course, is on a per-house basis since every house is different...then again, my neighborhood has 90 houses and just about every 5th is the same model...so, it might be useful, as mentioned by somebody else, to have the study for various types of constructions...

13. Jul 19, 2011

### Staff: Mentor

I should have realized that (or gone back and reread after getting to the end of the thread!).....yes, that makes sense.

14. Jul 19, 2011

### AlephZero

Obviosuly it depends on the construction of the house (and UK and US conventions are rather different there) but in a brick construction most of the thermal capacity will be in the brickwork not in the air inside, and solar heating will create a significant temperature gradient through the thickness of the bricks.

I can certainly "feel" the effect of that in my own house. The front of the house faces west and is in full sun from about noon fill sunset. It is noticeable that on sunny days the inside temperature keeps rising for a couple of hours AFTER sunset, while the walls dump the heat into the air inside. (The UK climate is temperate enough that hardly anybody uses aircon for cooling.)

As Russ said, making a realistic computer model of this, and validating it against measurements, is a lot of work - which is why most people just use the empirical results in building codes and reguilations instead.

15. Jul 19, 2011

### refusenik

Don't forget to add all lighting, fridge, electronics and 100W per each human occupant.