Posted by *nicksanspam* on March 12, 2008, 8:17 pm

*>The concept of a thermal time constant cannot be applied exactly...*

Sure it can :-)

*>For example, suppose I have a large thermal mass (call it a "water tank")*

*>inside the house, and suppose this mass has its own insulation between it and*

*>the the rest of the house... *

Model it separately. Its standby heat loss helps warm the house on an average

day, and it can keep the house warm for a few cloudy days with a distribution

system.

Nick

Posted by *Robert Scott* on March 12, 2008, 6:43 pm

On 12 Mar 2008 15:17:01 -0500, nicksanspam@ece.villanova.edu wrote:

*>>For example, suppose I have a large thermal mass (call it a "water tank")*

*>>inside the house, and suppose this mass has its own insulation between it and*

*>>the the rest of the house... *

*>Model it separately. Its standby heat loss helps warm the house on an average*

*>day, and it can keep the house warm for a few cloudy days with a distribution*

*>system.*

Yes, you can model it separately, in which case you get a separate time-constant

for those components of the house. In my example I considered a fairly simple

case of only two thermal masses - the water tank and everything else. In

practice, all the masses in the house have potentially their own separate time

constant, which can be modelled separately. All I was trying to show is that

there is no one time constant that can be said to stand for the whole house.

The temperature decay graph for a given spot in the house will not conform to

Temp0 * e ^ (-t/TC)

for any value of TC.

Robert Scott

Ypsilanti, Michigan

Posted by *nicksanspam* on March 12, 2008, 9:29 pm

*>>>For example, suppose I have a large thermal mass (call it a "water tank")*

*>>>inside the house, and suppose this mass has its own insulation between it*

*>>>and the the rest of the house... *

*>>*

*>>Model it separately. Its standby heat loss helps warm the house on an average*

*>>day, and it can keep the house warm for a few cloudy days with a distribution*

*>>system.*

*>Yes, you can model it separately, in which case you get a separate time-*

*>constant for those components of the house. In my example I considered*

*>a fairly simple case of only two thermal masses - the water tank and*

*>everything else.*

Sounds good to me. I figure the house cools like this overnight:

*> Temp0 * e ^ (-t/TC)*

to a constant outdoor temp, and I figure the tank cools from an average

tank temp to an average house temp, which can be the same for a few

cloudy days until the tank runs out of useful energy. This seems to

agree well with more calculus-intensive solutions and simulations.

And it works a lot better than architectural rules of thumb.

Nick

Posted by *daestrom* on March 13, 2008, 12:24 am

*> The concept of a thermal time constant cannot be applied exactly. It *

*> assumes*

*> that the heat gain/loss follows a single-pole exponential model. There *

*> are*

*> several things that can interfere with that model.*

*> For example, suppose I have a large thermal mass (call it a "water tank")*

*> inside the house, and suppose this mass has its own insulation between it *

*> and*

*> the the rest of the house. Start from a steady state where everything is *

*> at one*

*> fixed temperature. Then suddenly have the outside temperature change. If *

*> the*

*> single-pole model was accurate, the temperature of the air inside the *

*> house*

*> would follow an inverse exponential decay formula with a certain time *

*> constant.*

*> But what actually happens is this: At first the air temperature inside *

*> the*

*> house drops according to a fairly short time constant, determined by the *

*> thermal*

*> mass in immediate contact with the air and the insulation in the walls and*

*> windows. If you were to estimate the thermal time constant by the shape *

*> of that*

*> temperature decay graph, you would arrive at one figure. But if you wait *

*> a*

*> while, the "water tank" thermal mass begins to have more and more *

*> influence. The*

*> shape of the inside air temperature decay graph would begin to look like a*

*> longer time constant.*

*> The model of a single-pole exponential decay is accurate only if all the *

*> thermal*

*> mass in the house has approximately the same R-value between it and the *

*> outside.*

Quite right. If you plot the room temperature versus time on semi-log

graph, this shows up as a curved line. Whereas a 'simple time constant'

should be a straight line with slope -1/TC.

Quite often this can be accomodated by using two or three different time

constants, for small, medium and large temperature swings. For the small,

short duration changes, basically ignore the '900 lb gorilla' (I mean 'water

tank') and use the slope from the very first portion of the line. For part

of a day or hourly changes, this would probably be okay.

For long term calculations like a couple of days without sun or some such,

then you can't really ignore the 'water tank' and the math takes on another

degree of complexity. But at least the watertank and room are coupled in

both directions :-)

daestrom

Posted by *daestrom* on March 13, 2008, 12:18 am

*> Just wonder how to estimate the thermal time constant of a room.*

*> I'd like to get an idea of how long a room/house takes to cool off or *

*> heat up. I think knowing that will be helpful in passive heating, cooling *

*> techniques. At least in getting an idea of what the return on a different *

*> strategy would be.*

*> I imagine the "time constant" will be depend on the density of the *

*> weight of the walls (plaster lath or wood for me) and the mass of the *

*> contents. Somehow that couples with the rate of losses through the *

*> insulation and air infiltration. I don't think most furniture will have *

*> much effect.*

*> I suspect this has already been collated somewhere, I just can't find *

*> it.*

Rather than calculate it, it can be measured.

You simply pick a nice cold/calm/cloudy day and turn off the heat and plot

the temperature every 10 minutes or so for as long as your wife can stand it

:-). Knowing the inside/outside temperatures, you can get a pretty nice

plot of temperature versus time. When the temperature has reached 63% of

the way from starting point towards outside temperature, you've reached one

time-constant.

For example, turn off furnace when house is 70F and outdoor temperature is

40F (30F lower). 63% of the way would be 19F so when the house is down 19F

from 70F (51F), you've reached one time-constant.

If you take your temperature readings and plot on semi-log graph, ideally

you would get a straight line. Any curvature towards the horizontal with

time is an indication of either an unidentified heat source that will change

the 'final' temperature (such as yourself, or a dog, or a phantom electric

load), or some second-order effects due to a low thermal diffusivity between

a large capacitance and the measurement space.

If you then simply figure out the heat input over a period of several hours

needed to keep the house at a fixed temperature and divide by the difference

between indoor and outdoor temperature. With that, you can calculate the

'R' for the house. And knowing the time-constant and 'R' you can find 'C'

(thermal capacitance) if you still feel you need that parameter.

About as accurate as adding/estimating God knows how many slabs, walls,

fixtures IMHO.

If you have a long time-constant, you may have to estimate it from a smaller

temperature drop. For example, I found that shutting off my furnace at 2130

hours on a nice calm winter night with the house at 70F and the outdoors at

18F, the temperature is still 63F at 0530 hours the next morning. So with a

52F difference and only a 7F drop in 8 hours...

(7/52) = 1 - exp(-8/TC)

TC = -8 / ln(1-7/52) = 55.3 hours

{This ignores the heat input from the small wall-warts and refrigerator.

But with everyone snug in their beds upstairs, the downstair temperature

seems like a reasonable measure of temperature drop.}

Of course, the shorter your time span, or the smaller your temperature

change, the less accurate the results. And if the outside air temperature

changes wildly in the night, well, all bets are off then. How extreme you

get in the details will determine how accurate a result you get. You could

even stand outside in the snow and read the thermometer through the window

if you want to rule out body heat as interference, but *I* didn't go that

far :-)

daestrom

>The concept of a thermal time constant cannot be applied exactly...