Calculation of inertia and lower heating requirements

[crashback]

Hi everyone I have a problem and I would like to be informed:

I am doing a study on the insulation of a house to show the differences that there is between a house well and badly insulated.

I made my calculation of thermal losses and I am now calculating the inertia because that which interests me of course is the money saving that can be done.
I was told that a better inertia could allow me to lower the temperature of my boiler so I earn money.

Here is the method I used to calculate the inertia:
- Inertia = thickness² / thermal diffusivity
- Thermal diffusivity = thermal conductivity / density x thermal capacity

The walls of the house before insulation work: brick and plasterboard
The walls of the house after insulation work: wooden cladding, polystyrene, brick and plaster

After calculation I find an inertia before works: 71h
after work: 216h

1ere question: Are these values ​​plausible?
2e question: If yes to the 1ère, what is it worth?
3th question: How with values, know how much I can earn on my heating bill?

Thank you in advance for your answers and for your

[dirk pitt]

inertia will not in itself "save" energy and therefore money on the bill.
the energy supplied must be equal to the energy that escapes (losses).
the inertia should be seen as a buffer tank that smooths the temperature variations providing additional comfort.
at the margin, this allows in theory a smaller dimensioning of the heat production device which can be satisfied with having a maximum power close to the average power required since the rapid variations in temperature are erased by the inertia "buffer" thermal.
but this "down-sizing" is marginal, I think economically.

[Did67]

... except the house has a "bioclimatic" tendency, in short, with significant openings on the south / west side. There, in mid-season, it can be important to have the capacity to passively store this solar energy and to benefit from it in the evening and until the morning (without having to restart the heating).

and, to a lesser extent, except that a smoothed interior temperature gives a better feeling of comfort, without being tempted to juggle with the thermostat ... Basically, it accelerates like speeding at the same time as accelerating / ceaselessly stop.

But it is true that the base is insulation to prevent losses.

[Did67]

The walls of the house before insulation work: brick and plasterboard
The walls of the house after insulation work: wooden cladding, polystyrene, brick and plaster

After calculation I find an inertia before works: 71h
after work: 216h
:

I am not at all a specialist, but there is intuitively something wrong!

Basically, the "before" and "after" inertia will be the same, since thermal inertia (i.e. the ability to "store calories") is related to the masses inside the body. insulated envelope: "massive" walls, thick slabs, etc ... This is what gives thermal inertia. However, the insulation (external if I understood correctly) will not change anything in terms of inertia! It will reduce wastage.

An insulating panel will only store a tiny amount of calories because it is ultra-light!

I think that your calculations are rather "phase shifts", that is to say the time that takes a thermal wave to cross the partition.

I'll say it again, I'm boozy (I'm just trying to be "logical").

[Capt_Maloche]

ALOA

In reality it is a simplified calculation that must be executed
The inertia of the building is used to determine a heating restart time from a maximum heating power and / or direct solar energy capture capacity (glazing)

You estimate the mass of each material constituting your interior (before insulation), you put them in a table and you note the CP (Massive heat of each material) http://www.techno-science.net/?onglet=g ... ition=3324

To simplify, drop the coefficient of thermal conductivity,

Mass Plaster Kg ---------------- CP = J / Kg. ° C ------------ TOTAL
Floor slab + tile Kg ------ CP = J / Kg. ° C ------------ TOTAL
Furniture
etc ...

This gives you a Nbr of J / ° C
the amount of energy you need to provide to increase the T ° of your interior by one degree Celsius (excluding losses)

1Watt = 1J / s
depending on your super heating power, you can determine a recovery time to reach your T ° of comfort after a period of reduced heating (absence)

you can do the same thing with your losses, to know how long you will lose 1 ° C

[swift2540]

Hello,
+ 1 Did67!
(short version without formula)
For the winter, the important thing is the lambda or the R. The bigger the R is (or lambda small), the more it is insulating. So the less you lose calories, the more you gain in $$$

On the other hand, for the summer, the important thing is phase shift. So keep the heat out. The best is a "church wall", so thick and heavy that in the evening the heat has not passed and the wall cools down at night.
In comparison, after 30 'in full sun, you no longer fit in a caravan. Why? Because there is no mass, simply.

As explained by Did67, the interest of the inertia is in inter-season (or full sun in winter) to be able to store solar gains without overheating, and to take advantage of these contributions not to have to reignite the heating. Hence the interest of the itite on the iti.

But in your case you do not change anything inside, so nothing to the inertia of the building. The only way to gain inertia = heavy slab on the ground or replacing the plasterboard with a double hard wall.

So in summary, you gain heating in winter (and it's already very good) but nothing in summer comfort. It's up to you to know what is necessary / interesting / too expensive / too restrictive.
@+

[Did67]

... but nothing in summer comfort.
@+

Uh, a little shortened anyway.

Insulation will also “slow down” the return of calories in summer, and therefore the heating of the house.

For the same inertia (same capacity to "swallow" calories), and the same outside temperature, the house will receive fewer calories, so it will heat up less (than the same house without insulation).

Of course it's even better with a lot of inertia ... because this reduced amount of calories (reduced not the insulation) will cause even less rise in temperature.

All this to consider also with the phase shift, because if the heat wave of the afternoon arrives inside the evening / beginning of night and if the situation / noise allows it, one opens the windows to ventilate ... and voila.

So take your calculators: a very well insulated house (prevents calories from going out in winter, coming in summer) with great inertia ("smooth" the need for power in winter and limit the "temperature rise" l 'summer) and a sufficient phase shift (allows you to refresh by opening the evening), and voila!

[swift2540]

A little shortened but hardly!
I can not find the internet link anymore, but I read that
for winter comfort (insulating), 10cm of pse = 175cm of concrete
for summer comfort (phase shift), 10cm of concrete = 175cm of pse!
(ladle is an order of magnitude)
The only thing that slows down is the mass. The pse being light, it brakes almost no.
If you replace the pse with ldb or cork, then it will brake a little better.
But not huge compared to a wall ...

[swift2540]

Here's what to get an idea about the insulation

[dedeleco]

Here is the method I used to calculate the inertia:
- Inertia = thickness² / thermal diffusivity
- Thermal diffusivity = thermal conductivity / density x thermal capacity

This, with the square, concerns the time taken by the heat to get through the insulating wall inwards but not the time necessary to cool or heat the house which is especially the ratio of the heat capacity of the house (the tilings with their cement and floors are hard to heat) divided by the heating or cooling power by respectively the boiler or the thermal conduction of the walls.
You seem to have replaced the brick with thick polystyrene and so the heat is 9 days, which seems high, because the defects (cladding and others) will shortcircuiter.
But the heat capacity of brick, plaster and slabs with tiling will increase this cooling time in winter in principle.
With your calculations, if you increase the time constant (which you call inertia) to 6month, by multiplying your thickness of insulation by 4 to 5, to multiply it by the square either 4x4 = 16 or 5x5 = 25 times more and so go from 9days to 144 or 225 days is more than 6mois, you will not need to heat the house, recovering in winter the heat of summer !!
Easy with the house under 3 meters of land, and heating by sun in summer which is kept until the hiver!!!

Finally the table shown by swift2540 is misleading because for the winter it uses a standard that tolerates more heat losses than the condition in summer of almost suppressing the conduction of heat from the sun inwards on a day by taking an insulator with high internal heat capacity !!
By putting in more on the inside against the bad insulation in summer a two layer of gypsum plasterboard with strong heat capacity you will need in summer only from the thinness of 5, 10 to 20 cm of insulation !!
I am categorical because I own a secondary house with 10cm of polystyrene and plasterboard that does not undergo any overheating in summer.
This table is mainly to sell these panels and therefore misleading on the correct physics.