Stop or Lower Heating T ° during absence?

[Christophe]

Uh completely ... It was didi who is right ... I am also disheveled by Dirk!

Here's a similar topic: https://www.econologie.com/forums/baisse-de- ... 12240.html

Everything is summarized in this image:



ps: not understand this ??

We lower from 20 to 17: we will win 90 (for 3 °)

We lower from 20 to 16: we will not win 120, but maybe 115

We lower from 20 to 15: we're not going to win 150, but maybe 130

[Remundo]

it is a "wet finger put in the wind" approximation of Did to explain that the reduction in consumption is not linear with the drop in the interior temperature

[Christophe]

Yes, that's what I thought I understood ... but in what unit is he speaking?

The heating consumption is linear with the delta of temp int - ext as we have just said ... after there are different yields which come in the "real figures" ...

[dirk pitt]

almost everything has been said.
I will just add that as the heat loss is proportional to the difference in temperature inside / outside, the lower the inside temperature, the lower the economy.
so don't worry too much about putting the house at 10 ° instead of 6 °. the difference should be minimal. much lower in any case than between 20 and 16 for example.

I withdraw what I said, it is false of course. I got carried away.
the consumption is proportional to the difference in temp therefore:
int: 20, ext: 5, consumption = X * 15
int: 16, ext: 5, consumption = X * 11
consumption gain = X * 4
int: 12, ext: 5, consumption = X * 7
gain in consumption compared to int = 16: X * 4 also

[Christophe]

Yes with the returns close it is that!

The "X" is not constant in reality: a boiler (well rather a heating system) which will run more often (delta> 15 °) will have a higher overall efficiency than in the off-season (delta <5 °)

Then there are also other thermal inputs: lighting, appliances, cooking, "inhabitants" ... Not necessarily negligible depending on the configuration of the house (the more it is insulated and compact, the more these free inputs are important in proportion ... and therefore minus the X is constant ...)

[Did67]

Yes, that's what I thought I understood ... but in what unit is he speaking?

The heating consumption is linear with the delta of temp int - ext as we have just said ... after there are different yields which come in the "real figures" ...

1) No units. I do like the kings of marketing, I balance figures that mean nothing !!!

2) No, take back the pretty curves that you put and which illustrate well what I was trying to write.

I spoke well for the same comfort setpoint, and of course the same external temperature.

You imagine that we lower even more than what is drawn: the temperature, at the beginning, does not drop any faster. It is the thermal inertia of the house. There is therefore no additional saving. Even without heating, the boiler off, the house "loses" its calories. At the same pace. Whether you program 17 ° or 8 °.

It is only towards the end, just before the rise, that the internal temperature curve would be lower. Instead of pulling asymptotically towards the 17 ° setpoint, it would continue to decrease more (not quite linearly however, since the interior-exterior delta is reduced!). And so there will be some additional savings!

But as it is lower, if you want to be back to 20 ° at the same time (8 hours), you will have to start again a little earlier too.

So in a daily rhythm, lowering more, at a given time, does not bring much more gain, for these two reasons:

- it does not descend faster / no further

- you have to go back earlier ...

My "anything" numbers just wanted to illustrate that.
If you lower it from 20 to 19 °, it will yield "x".
If you lower from 20 to 18 °, it will not be 2X but a little less.

And so on.

And I am not talking about efficiency, lower setpoint, etc ... Nor contributions "other ... I was in a stupid thermal reasoning (but taking into account the inertia of the building)

3) If we wanted savings to be proportional to lowering, we would have to imagine a building without any inertia :

- at the time of lowering, we "fall" instantly to the "lowered" set temperature (17 ° or 8 ° for example).

- and at the time of the passage in comfort, we would go up instantly.

There, indeed, and there only, the savings, represented by the hollow surface, would be proportional to the reduction.

But it's like dreaming of a car without weight, therefore without inertia, which would instantly accelerate with a moped engine!

4) For long absences, it is obvious that lowering more, to the limit of frost-free, brings additional savings. I wrote it too.

[Did67]

I withdraw what I said, it is false of course. I got carried away.
the consumption is proportional to the difference in temp therefore:
int: 20, ext: 5, consumption = X * 15
int: 16, ext: 5, consumption = X * 11
consumption gain = X * 4
int: 12, ext: 5, consumption = X * 7
gain in consumption compared to int = 16: X * 4 also

OK. I was afraid ! And asked myself two seconds if I had become an idiot. And then not finding the flaw in my reasoning ...

But be aware that your internal temperature is the temperature actually measured. Not the "programmed" one.

You can programmer 16 ° to 20 h.

But this temperature may be reached at midnight or at 2 a.m.

It all depends on the house, the insulation (ITI or ITE), etc.

[Gaston]

But be aware that your internal temperature is the temperature actually measured. Not the "programmed" one.

You can programmer 16 ° to 20 h.

But this temperature may be reached at midnight or at 2 a.m.

It all depends on the house, the insulation (ITI or ITE), etc.

Suddenly, the estimate of consumption is even more complicated over periods of a few hours since:

• As long as the indoor temperature is higher than the set point, consumption is zero.
• When the indoor temperature reaches the set temperature, the consumption is (on average) equal to the losses, therefore proportional to the temperature difference with the outside.
• When the setpoint is reset to a higher value, the consumption is greater than the losses (which allows the temperature to rise) and approximately equal to the maximum heating power.

[Did67]

No, it's easier.

Do not point yourself to the consumption of the boiler at all times.

From a thermal point of view:

- you have a "stock of calories" in your maion (it is its "thermal inertia")

- the boiler maintains this stock, if necessary by switching on / off or by modulating its power, thanks to regulation.

At the start of reduction, during the "descent" of the internal temperature curve, the boiler stops, but home continues to lose calories (since it is hotter than the outside). Less than if you maintain your comfort temperature.

In the middle / end of lowering, often, the boiler maintains a stable lowered temperature (eg 17 °): there is then a horizontal plate (not shown on the curves set by Christophe)

During the ascent (switching to comfort mode), the boiler compensates for the losses + replenish the calorie supply (she "pumps", more house losses, exactly what was destocked during the descent phase.

So if you run a cycle over 24 hours, the drop in stock at the start of the descent (stopping the boiler) is compensated by the "restocking" during the ascent (on-work of the boiler) (or vice versa).

Within 24 hours, the boiler will have produced exactly .... the losses of the house.. Destock / restock is canceled.


The losses of the house are figured by the surface between the curve of the actual internal temp and the curve of the actual external temperature.

The saving allowed by the lowering is represented by the area between the horizontal "comfort temp" and the curve of the real internal temp.

If you measure the surfaces (take graph paper) you can even "calculate" the% savings.

Without even knowing how much it is! (we lack the proportionality factor 1 cm² of surface = so and so many kWh!).

Just two recording thermometers and the mass is said!

Without worrying about the boiler.

Si more you measure the consumption of the boiler over these 24 hours, you will have proportionality: 1 cm² = so and so many liters of fuel, or so and so many kg of pellets or so and so many m3 of gas.

And there you can calculate the gain in kWh or euros.

Fastoche!